INTRODUCTION: STRUCTURAL ANALYSISmtp.itd.co.th/ITD-CP/data/StructuralAnalysis2.pdf · Beam 1 DOF A...

1

! Deflected Shape of Structures! Method of Consistent Deformations! Maxwell�s Theorem of Reciprocal

Displacement

INTRODUCTION: STRUCTURAL ANALYSIS

2

-M

P

+M

-M

Deflection Diagrams and the Elastic Curve

∆ = 0θ = 0

fixed support

∆ = 0

roller or rockersupport

P

θ

inflection point

3

-M

P

∆ = 0

pined supportθ

4

Moment diagram

P

inflection point

inflection point

� Fixed-connected joint

fixed-connected joint

5

P

� Pined-connected jointpined-connected

joint

Moment diagram

6

P

inflection point

Moment diagram

7

-M

+Mx

M

P2

P1

A B C D

inflection point

8

x-M

+M

P1

P2

inflection point

9

+

AB

C

AB

C

P

=

AB

C

P

∆´B

fBB x RB

1∆´B + fBB RB = ∆B = 0

Method of Consistent DeformationsBeam 1 DOF

RBAy

Ax = 0

MA

10

w

w

=+

+

=∆1

∆2

∆´1

∆´2

R1

R2

×R2

Compatibility Equations.

∆´1 + f11R1 + f12R2 = ∆1 = 0

×R1

f12f22

0

0

Beam 2 DOF

∆´2 + f21R1 + f22R2 = ∆2 = 0

1

1

14

3

52

∆´1 ∆´2

f11 f21

+f11 f12

f12 f22

11

wP1 P2

=

wP1 P2

++

+1

1

f32f22

1

Compatibility Equations.

θ´1 + f11M1 + f12R2 + f13R3 = θ1 = 0

×R2

×R3

=

θ1

∆2

∆3

f31

f33f23f13

24

61

53

∆´2 θ´1∆´3

Beam 3 DOF

∆´2 + f21M1 + f22R2 + f23R3 = ∆2 = 0

∆´3 + f31M1 + f32R2 + f33R3 = ∆3 = 0

×M1

θ´1

∆´2

∆´3

+

f11 f12 f13

f21 f22 f23

f31 f32 f33

M1

R2

R3

0

0

0

f12

f21f11

12

If ∆1 = ∆2=��= ∆n = 0 ;

or [∆´] + [f][R] = 0

[R] = - [f]-1[∆´]

Compatibility Equation for n span.

∆´1 + f11R1 + f12R2 + f13R3 = ∆ 1∆´2 + f21R1 + f22R2 + f23R3 = ∆2

∆´n + fn1R1 + fn2R2 + fnnRn = ∆n

∆ ´1

∆´2 +

f11 f12 R1

R2 =

0

0

∆´ n

f13

f21 f22 f23

f n 1 f n 2 fn n R n 0

[fij] = Flexibility matrix dependent on

Equilibrium Equations

[Q] = [K][D] + [Qf]

Stiffness matrix

Fixed-end force matrix

Solve for displacement [D];

[D] = [K]-1[Q] - [Qf]

GJEAEI1,1,1

ijij f

k 1=

13

A B

A B

EI

1=i 2=j

Mj = mj

f12 = fij

f11 = fii

f22 = fjj

Mi = mi

Maxwell�s Theorem of reciprocal displacements

1

1

f21 = fji

∫=EIdxMmf jiij

∫=EIdxmm ji

∫=EIdxMmf ijji

∫=EIdxmm ij

jiij ff =

14

Example 1

Determine the reaction at all supports and the displacement at C.

AB

C

50 kN

6 m 6 m

15

50 kN

=∆´B

+

fBB

1 kN

x RB

AB

C

50 kN

6 m 6 m

SOLUTION

RB

MA

RA

0' =+∆ BBBB Rf -----(1)Compatibility equation :

� Principle of superposition

∆´B

x RBfBB

16

Conjugate beam

Conjugate beam

50 kN

∆´B

Real beam

50 kN

300 kN�m

6 m 6 m

/EI300 900/EI

6 + (2/3)6 = 10 m9000/EI

� Use conjugate beam in obtaining ∆∆∆∆´B and fBB

900/EI

12 /EI72/EI

(2/3)12 = 8 m576/EI

fBB

1 kN

Real beam

1 kN

12 kN�m

72/EI

∆´B = M´B = -9000/EI ,

fBB = M´´B = 576/EI,

BA

17

RB = +15.63 kN, (same direction as 1 kN)

0)576(9000: =+−↑+ BREIEI

50 kN

∆´B

50 kN

300 kN�m

x RB = 15.63 kN=

AB

C

50 kN

15.63 kN34.37 kN34.37 kN�m

fBB

1 kN1 kN

12 kN�m

+

� Substitute ∆´B and fBB in Eq. (1)

18

93.6/EI

-113/EI

6 m 6 m

AB

C

50 kN

Use conjugate beam in obtaining the displacement

113 kN�m

34.4 kN15.6 kN

EIEIEIM C

776)6(223)2(281' −=−=

↓−==∆ ,776'EI

M CC

Real Beam

Conjugate Beam

M (kN�m) x (m)

3.28 m 6 m 12 m

93.6

-113

223/(EI)

4 m2 m

M´C

V´C 223/(EI)

281/(EI)

∆C

19

10 kN

AC

B

4 m

3EI 2EI

2 m 2 m

Example 2

Determine the reaction at all supports and the displacement at C. Take E = 200GPa and I = 5(106) mm4

20

+

10 kN

=

1 kN

Compatibility equation: ∆´B + fBBRB = ∆B = 0

x RB

∆´B

fBB

10 kN

AC

B

4 m

3EI 2EI

2 m 2 m

21

Conjugate Beam

10 kN

AC

B

4 m

3EI 2EI

2 m 2 m

Real Beam

10 kN

40 kN�m

� Use conjugate beam in obtaining ∆´B

26.66/EI

177.7/EI

x (m)

V (kN) 10 10

+

x (m) M (kN�m)

40

-

40/3EI = 13.33/EI∆´B = M´B = 177.7/EI

22Conjugate Beam

AC

B

4 m

3EI 2EI

2 m 2 m

Real Beam1 kN

4/(2EI)=2EI

� Use conjugate beam for fBB

1 kN

8 kN�m

x (m)V (kN)-

-1 -1

x (m) M (kN�m)

84

+

83EI= 2.67

EI4/(3EI)=1.33EI

60.44/EI

12/EIfBB = M´B = 60.44/EI

23

+

10 kN

=

1 kNx RB

∆´B

fBB

10 kN

AC

B

4 m

3EI 2EI

2 m 2 m

Compatibility. equation: ∆´B + fBBRB = ∆B = 0

RB = +2.941 kN, (same direction as 1 kN)

0)44.60(7.177: =+−

−↑+ BREIEI

24

10 kN

AC

B

4 m

3EI 2EI

2 m 2 m 2.94 kN7.06 kN

16.48 kN�m

7.06

-2.94 -2.94

+- x (m)V (kN)

∆C

-16.48

11.76

-

+ x (m) M (kN�m)

2.33 m

1.67 m

� The quantitative shear and moment diagram and the qualitative deflected curve

25

10 kN

AC

B

4 m

3EI 2EI

2 m 2 m

↓−

=−= ,85.18)222.3(413.6)555.0(263.3EIEIEI

2.941 kN7.059 kN

16.48 kN�m

∆C

� Use the conjugate beam for find ∆∆∆∆C

Real beam

↓−=×

−==∆ ,85.18

)5200(85.18' mmM CC

16.483EI

11.763EI

11.762EI

2.335 m

1.665 m

Conjugate beam

6.413EI

3.263EI

(1.665)/3=0.555 m

1.665+(2/3)(2.335) = 3.222 m

M´C

26

Example 3

Draw the quantitative Shear and moment diagram and the qualitative deflectedcurve for the beam shown below.EI is constant. Neglect the effects of axial load.

A B4 m4 m

5 kN/m

27

αBBαAB

αBAαAA

θ´Bθ´Aθ´A

1 kN�m

5 kN/m

1 kN�m

A B4 m4 m

5 kN/m

SOLUTION


θ´B

=+

αAA αBA

AM×

αBB

BM×αAB

+

Compatibility equations:

θA θB

AM×

BM×

AM×

BM×

)1('0 −−−++== BABAAAAA MfMfθθ

)2('0 −−−++== BBBABABB MfMfθθ

28

� Use formula provided in obtaining θÁ, θ´B, αAA, αBA, αBB, αAB

Note: Maxwell�s theorem of reciprocal displacement, αAB = αBA

1 kN�m

αAA αBA8 m

1 kN�m

αBBαAB

8 m

EIEIEILM o

AA667.2

3)8(1

3===α

EIEIEILM o

BA333.1

6)8(1

6===α

EIEIEILM o

BB667.2

3)8(1

3===α

EIEIEIwL

B67.46

384)8)(5(7

3847'

33

===θ

EIEIEIwL

A60

128)8)(5(3

1283'

33

===θ

4 m4 mθ´BθÁθÁ

5 kN/m

θ´B

EIEIEILM o

AB333.1

6)8(1

6===α

29

αBBαAB

Real Beam1 kN�m

αBBαAB

8 m

αBA

Real Beam1 kN�m

αAA αBAαAA

8 m

Conjugate Beam

4/EI1/EI

(1/8)(1/8)

2.67/EI1.33/EI

(1/8) (1/8)

Conjugate Beam

1/EI4/EI

1.33/EI2.67/EI

EIV AAA

667.2' −==α

EIV BBB

667.2' ==α

EIV AAB

333.1' −==α

EIV BBA

333.1' ==α

� Use conjugate beam for αAA, αBA, αBB, αAB

30

1 kN�m

5 kN/m

1 kN�m

A B4 m4 m

5 kN/m

=+

+

θA θB

BM×

EIAA667.2

=α

AM×

EIV BBA

333.1' ==α

EIB67.46' =θ

EIA60' =θ

EIAB333.1

=αEIBB667.2

=α

Solve simultaneous equations,

MA = -18.33 kN�m, +

MB = -8.335 kN�m, +

Compatibility equation

+ 0)333.1()667.2(60=++ BA M

EIM

EIEI

+ 0)667.2()333.1(67.46=++ BA M

EIM

EIEI

31

MA = -18.33 kN�m,

MB = -8.335 kN�m,

A B4 m4 m

5 kN/m

18.33 kN�m 8.335 kN�m

RBRA

RB = 3.753 kN,

Ra = 16.25 kN,

+ ΣMA = 0: 0355.8)8()2(2033.18 =−+− BR

ΣFy = 0:+3.753

020 =−+ BA RR

32

A B4 m4 m

5 kN/m

18.33 kN�m 8.36 kN�m

3.75 kN16.25 kN

Vdiagram

Mdiagram

DeflectedCurve

-3.75

16.25

3.25 m

-18.33

6.67

-8.36

8.08

� Quantitative shear and bending diagram and qualitative deflected curve

33

Example 4

Determine the reactions at the supports for the beam shown and draw thequantitative shear and moment diagram and the qualitative deflected curve.EI is constant.

A

C

4 m4 m

2 kN/m

B

34

1 kN

)2(0''')1(0'''

−−−=++∆−−−=++∆

CCCBBCC

CCBBBBB

RfRfRfRf



A

C

4 m4 m

2 kN/m

B

4 m4 m BA

C

2 kN/m

B'∆C'∆

A

C

4 m4 m B

BBf ' CBf '

1 kNAC4 m4 m B

BR×

CR×BCf ' CCf '

35A

C

A

C

B

A

C

2 kN/m

1 kN

� Solve equation

A

C

4 m4 m

2 kN/m

B

1 kN

EIB64' −=∆ EIC

33.149' −=∆

EIf BB

33.21' = EIf CB

33.53' =

EIf BC

33.53' = EIf CC

67.170' =

)1(033.5333.2164−−=++− CB R

EIR

EIEI


)2(067.17033.5333.149−−=++− CB R

EIR

EIEI

↓−=

↑=

,29.0,71.3

kNRkNR

C

B

BR×

CR×

36

� Diagram

A

C

4 m4 m

2 kN/m

B

3.71 kN 0.29 kNkNAy 58.471.329.08 =−+=

mkNM A

•=−+=

48.3)4(71.3)8(29.0)2(8

x (m)

V (kN)4.58

0.29

-3.422.29 m

x (m)

M (kN�m)

-3.48

1.76

-1.16

x (m)Deflected shape

Point of inflection

37

Example 5

Draw the quantitative Shear and moment diagram and the qualitative deflectedcurve for the beam shown below.(a) The support at B does not settle(b) The support at B settles 5 mm.Take E = 200 GPa, I = 60(106) mm4.

16 kN

2 m 2 m 4 m

AB C

38

16 kN

2 m 2 m 4 m

AB C

16 kN

SOLUTION


∆´B

1 kN

fBB

BR×

∆´B

fBB

BR×=

+

∆B = 5 mm

Compatibility equation : BBBBB Rf+∆==∆ '0 -----(1) : no settlement+

BBBBB Rfm +∆=−=∆ '005.0 -----(2) : with settlement+

39

16 kN

∆´B

Real beam

A B C

� Use conjugate beam method in obtaining ∆´B

2 m 2 m 4 m12 kN

M´diagram

16

Conjugatebeam

EI24

EI24

EI72

34

32

2 m 4 m

24

EI40

EI56

4 kN

0)4(40)34(32'' =−+−

EIEIM B

EI16

EI32

EI40

V´´B

M´´B

34 4

32

+ ΣMB = 0:

↓−==∆ ,33.117'''EI

M BB

∆´B

Real beam

40

Real beam

1 kN4 m 4 m

AB

CfBB

� Use conjugate beam method in obtaining fBB

0.5 kN 0.5 kN

m´diagram

-2Conjugate

beam

EI2

−EI4

−EI4

− EI4

EI4

v´´B

m´´B

EI4

− EI4

34 4

32

0)4(4)34(4'' =+−−

EIEIm B

+ ΣMB = 0:

↑== ,67.10''EI

mf BBB

fBB

41


,67.1033.1170; BREIEI

+−=↑+ RB = 11.0 kN,

xRB = 11.0 kN

=16 kN

AB C

11.0 kN= 6.5 kN = 1.5 kNRCRA

no settlement

↓−=∆ ,33.117'EIB

↑= ,67.10EI

fBB

16 kN

12 kN 4 kN∆´B 1 kN

+

0.5 kN 0.5 kN

fBB

42


BREIEI

m 67.1033.117005.0; +−=−↑+

BREIm 67.1033.117)005.0( +−=−

RB = 5.37 kN,

BR67.1033.117)60200)(005.0( +−=×−

xRB = 5.37

=16 kN

AB C

5.37 kN= 9.31 kN = 1.32 kNRCRA

with 5 mm settlement

↓−=∆ ,33.117'EIB

↑= ,67.10EI

fBB

16 kN

12 kN 4 kN∆´B 1 kN+

0.5 kN 0.5 kN

fBB

43

� Quantitative shear and bending diagram and qualitative deflected curve

16 kN

AB C

2 m 2 m 4 m11 kN 6.5 kN 1.5 kN

Vdiagram

6.5

-9.5

1.5

Mdiagram

13

-6

+-

DeflectedCurve

16 kN

∆B = 5 mmA

B C

2 m 2 m 4 m5.37 kN 9.31 kN 1.32 kN

Vdiagram

9.31

-6.69-1.32

Mdiagram

18.625.24+

DeflectedCurve ∆B = 5 mm

44

Example 6

Calculate supports reactions and draw the bending moment diagrams for (a)D2 = 0 and (b) D2 = 2 mm.

4 m 6 m

w 2 kN/m1 2 3

45

+

2 kN/m

=

Compatibility equation:

∆´2 + f22R2 = ∆2

∆´2

1

f22

×R2

4 m 6 m

w 2 kN/m1 2 3

↓−=−=

+××−−=

+−−=∆

,2.6)200)(200(

248

)1041024(24

)4)(2(

)2(24

'

:

323

3232

mm

EI

LLxxEI

wxprovidednsformulatioUse

↑+==

−−×

××−=

−−=

,48.0)200)(200(

2.19

)4610(106

461

)(6

222

22222

mm

EI

xbLLEIPbxf

46

4 m 6 m

w 2 kN/m1 2 3

12.92 kN4.83 kN2.25 kN

Compatibility.equation:

-6.2 + 0.48R2 = -2R2 = 8.75 kN

4 m 6 m

w 2 kN/m1 2 3

For ∆2 = 2 mmFor ∆2 = 0


-6.2 + 0.48R2 = 0 R2 = 12.92 kN

x (m) V (kN)

2.25

-5.75

7.17

-4.83

+--

1.125 m

2.415 m

x (m) V (kN)

2.375 m

3.25 m

-++

-

4.75 5.5

-3.25-6.5

x (m) M (kN�m)

1.27

-7

5.85

+-

+

x (m) M (kN�m)

5.64

3

10.56

++

6.5 kN4.75 kN 8.75 kN

47

L

w

Basic Beams: Single span

wL/2

- wL/2

V

wL2/8

M

APPENDIX

wL/2wL/2L/3L/3

1

EIwL

3845 4

48

� Find ∆∆∆∆1 by Castigliano�s

w

L

L/2 L/2

x1 x2

1 2 3P

22PwL

+22PwL

+

22PwL

+x2

w

V2

M2

22PwL

+x1 V1

M1w

+ ΣΜ = 0;

21

21

1

1

21

1

)22

(2

0)22

(2

MxPwLwxM

xPwLwxM

=++−=

=+−+

49

0

∫∫ ∂∂

+∂

∂=∆=∆

2/

0

22

22/

0

11

1max1 )()(

LL

EIdxM

PM

EIdxM

PM

∫ ∂∂

=2/

0

11

1 )(2L

EIdxM

PM

21

21

1 )22

(2

MxPwLwxM =++−=

11

21

2/

0

1 ))22

(2

()2

(2 dxxPwLwxxEI

L

++−

= ∫

EIwL

3845 4

=

50

L

L /2 L /2

P

1

P/2 P/2

V

P/2

- P/2

+

-

PL/4

M

EILfPf

EIPL

4848

3

1111

3

11 =→==δ

Single span

51

L = 2l

l = L/2 l = L/2

1w

1.25 wl0.375 wl 0.375 wl

0.375 wl

-0.375 wl-0.625 wl

0.625 wl

+-

+-

0.070 wl2 0.070 wl2

-0.125 wl2

+ +

-

EIwl

185

4

max =∆

EIwl

185

4

max =∆

Double span

52

l = L/3 l = L/3 l = L/3

1 w 2

1.1wl1.1wl0.4wl 0.4wl

∆max = wl4

145 EI

V0.4wl

- 0.4wl- 0.6wl

0.5wl

-0.5wl

0.6wl

+

-

+-

+

-

0.08 wl2 0.08 wl2

0.025 wl2

-0.1 wl2 -0.1 wl2

+

- -++M

Triple span

1

! Comparison Between Indeterminate andDeterminate

! Influence line for Statically IndeterminateBeams

! Qualitative Influence Lines for Frames

INFLUENCE LINES FOR STATICALLY INDETERMINATE BEAMS

2

AB CED

RA AB CED

RA

Indeterminate Determinate

Comparison between Indeterminate and Determinate

11

3

AB CED

RAA

B CEDRA

AB CED

ME AB CED

ME

AB CED

VD AB CED

VD

1 1

1

Indeterminate Determinate

11

1

4

f1j

fjj

∆´1 = f1j

+

11 2 3j4

Redundant R1 applied

1

1

=

f11

fj1

×R1×R1

f11

Influence Lines for Reaction


011111 =∆=+ Rff j

)1(11

11 ffR j−=

)(11

11 f

fR j=

5

11 2 3j4

1

=+

×R2

Redundant R2 applied

fjjf2j

1

fj2

f22

Compatibility equation.

0' 22222 =∆=+∆ Rf

022222 =∆=+ Rff j

)1(22

22 ffR j−=

)(22

22 f

fR j=

6

1 2 3j4

fj4

1

1

Influence Lines for Shear

444

)1( jE ff

V =

f44

1

1

7

444

)1( jE fMα

=

1 2 3j4

Influence Lines for Bending Moment

1 1

α44

fj4

1 1

8

R3 )(33

33 f

fR j=

R2 )(22

22 f

fR j=

R1)(

11

11 f

fR j=

1 2 3j4

1

1

� Influence line of Reaction

1

Using Equilibrium Condition for Shear and Bending Moment

111

11 =ff

11

41

ff

11

1

ff j

122

22 =ff

22

2

ff j

22

42

ff

133

33 =ff

33

3

ff j

33

43

ff

9

1 2 3j4

4V4

V4 = R1

R1V4

M41

� Unit load to the right of 4

� Influence line of Shear

V4 = R1

V4 = R1 - 1

1

R2 R3R1

1

1

1

R1

1x

V4

M41

� Unit load to the left of 4

V4 = R1 - 1

01;0 41 =−−=Σ↑+ VRFy

1R1

0;0 41 =−=Σ↑+ VRFy

10

1 2 3j4

� Influence line of Bending moment

M4 = - l + x + l R1

M4 - 1 (l-x) - l R1 = 0 + Σ M4 = 0:

R1

1x

V4

M4l1

� Unit load to the left of 4

M4 = l R1

R1V4

M4

� Unit load to right of

l

4

1

1

R2 R3R1

1

l1 l2

l

M4

4

1R1

1 1M4 = - l + x + l R1

M4 - l R1 = 0 + Σ M4 = 0:

M4 = lR1

11

Influence Line of VI

Maximum positive shear Maximum negative shear

Qualitative Influence Lines for Frames

I

1

1

12

Influence Line of MI

Maximum positive moment Maximum negative moment

I

11

13

A D G

15 m15 mAy GyDy

MA

Dy

1.0

Ay

1.0

Gy

1.0

Influence Line for MOF

14

MH

MA

1

1

A H G

15 m15 m

D

15

RA

1.0

RB

1.0

MB

1

MG

1

A E

G

B C D

16

VG1

VF1

VH1

A E

G

B C D

17

Example 1

Draw the influence line for- the vertical reaction at A and B- shear at C- bending moment at A and C

EI is constant . Plot numerical values every 2 m.

A BC D

2 m 2 m 2 m

18

BBf BBf BBf 1=BBf

� Influence line of RB

1

A BC D

2 m 2 m 2 m

ABf CBf DBfBBf

19

Real BeamA BC D

2 m 2 m 2 m

Conjugate Beam

� Find fxB by conjugate beam

11

6 kN�m

x

=−+EI

xEIEI

x 18726

3

EI6

EI18

EI18

EI72

3x

32xV´x

M´x

x

EI72

EI18

EIx

2

2

EIx

20

x (m)

0

2

4

6

Point

B

D

C

A

x

fxB / fBB

1

0.518

0.148

0

A BC D

2 m 2 m 2 m

1

fBB

fxB

0

72/EI37.33/EI10.67/EI

EIx

EIEIxMf xxB

18726

'3

−+==

EI72

EI33.37

EI67.10

0

/72 = 0.518

/72 = 0.148

1

37.33

10.670

Influence line of RB

17272

==BB

BB

ff

fxB

21

1 kN

Influence line of RA

A BC D

2 m 2 m 2 m

AA

CA

ff

AA

DA

ff

1=AA

AA

ff

22

Conjugate Beam

xV´x

M´x

EIB y

18' =

EIx

2

2

EIx

3x

32x

A BC D

2 m 2 m 2 m

Real Beam

1 kN

� Find fxA by conjugate beam

1 kN

6 kN�m

EIM A

72' =

x

EIB y

18' =

=−EIx

EIx

618 3

EI18

EI6

23

Point

B

D

C

A

fxA / fAA

0

0.482

0.852

1.0

EIx

EIxMf xxA 6

18'3

−==

x (m)

0

2

4

6

x

A BC D

2 m 2 m 2 m

1 kNfAA fxA

72/EI 61.33 /EI34.67 /EI

fxA

0

EI67.34

EI33.61

EI72

34.67

61.3372

1 kN


/72 = 1.0/72=0.852

/72 = 0.4821=

AA

AA

ff

24

AB

BA

y

RRRRF

−==−+

=Σ↑+

101

;0

A BC D

2 m 2 m 2 m

1x

RA

MA

RB

0.148.518

1RB

1RB

0.4820.852

1.01 kN

RA

Alternate Method: Use equilibrium conditions for the influence line of RA

RA = 1- RB

25

VC

1

RB

0.1480.518

1

A BC D

2 m 2 m 2 mRA

MA

RB

0.8520.482

-0.148

1 x

1

Using equilibrium conditions for the influence line of VC

VC = 1 - RB

� Unit load to the left of C

RB

1 x

VC

MC

01

;0

=+−+

=Σ↑+

BC

y

RVF

RBVC

MC

VC = - RB

0;0

=++

=Σ↑+

CB

y

VRF


26

BA

BA

A

RxMRxM

M

6606)6(1

;0

++−==+−−−

=Σ

A BC D

2 m 2 m 2 mRA

MA

RB

1 x

1

MA

1

RB

0.1480.518

1

-1.112 -0.892

Using equilibrium conditions for the influence line of MA

+

27

MC

C

1

RB

0.1480.518

1

A BC D

2 m 2 m 2 m

1 x

RA

MA

RB

0.074

1

Using equilibrium conditions for the influence line of MC

0.592 RB

1 x

VC

MC


RBVC

MC

4 m

MC = 4RB

+04

;0=+−

=Σ

BC

C

RMM


4 m

MC = -4 + x + 4RB

04)4(1;0

=+−−−=Σ

BC

C

RxMM+

28

Example 2

Draw the influence line and plot numerical values every 2 m for- the vertical reaction at supports A, B and C- Shear at G and E- Bending moment at G and E

EI is constant.

A B CD E F

2@2=4 m 4@2 = 8 m

G

29


1

A B CD E F

2@2=4 m 4@2 = 8 m

G

1=AA

AA

ff

AA

DA

ff

AA

EA

ff

AA

FA

ff

30

4 /EI

1

A B CD E F

4 m 6 m2 m


Real beam

0.51.5

0

18.67/EI

64/EI

4/EI

4/EI

Conjugate beam

EI33.5

EI67.10

0

EI67.10 EI

Mf AAA64' ==

31

x1x2

=−EI

xEI

x 13

1 33.512

EIEIEIx 67.18646

32 −+=

V´ x1

M´ x1

x1EIx

21

EI33.5

EIx

4

21

32 1x

31x

Conjugate beam

4/EI

4 m 8 m

64/EI

EI33.5

EI67.18

M´ x2

V´ x2

x2

EIx2

EIx2

22

32x

32 2x

EI67.18

EI64

32

CtoBforEI

xEI

xMf xxA ,33.512

'3

1 −==

EI28

EI64

BtoAforEIEI

xEI

xMf xxA ,6467.186

' 23

22 +−==

fxA

0

0

EI16

−

EI10

−

EI14

−

x (m)

0

2

4

6

Point

C

F

E

D

B

A

8

12

G 10

fxA / fAA

0

-0.1562

-0.25

-0.2188

0

1

0.4375

x1x2

1

fAA fxA

A B CD E FG

4 m 6 m2 m

EI10−

EI16−

EI14−

EI28

EI64

1


-0.219 -0.25 -0.156

0.438

1=AA

AA

ff

33

A B CD E FG

4 m 6 m2 m

Using equilibrium conditions for the influence line of RB

RB RCRA

x1

RA

1 -0.219 -0.25 -0.156

10.438

RB

1

0.4850.8751.07810.59

RB

0

0.485

0.875

1.078

1

0.5939

0

x (m)

0

2

4

6

Point

C

F

E

D

B

A

8

12

G 10

RA

0

-0.1562

-0.25

-0.2188

0

0.4375

1

AB

AB

C

RxR

RxRM

812

8

0128;0

−=

=−+−=Σ+

34

A B CD E FG

4 m 6 m2 m

RA

1 -0.219 -0.25 -0.156

10.438

RB RCRA

x1

x (m)

0

2

4

6

Point

C

F

E

D

B

A

8

12

G 10

RA

0

-0.1562

-0.25

-0.2188

0

0.4375

1

RC

1

0.6719

0.375

0.1406

0

-0.0312

0 RC

1

10.672

0.3750.141

-0.0312

Using equilibrium conditions for the influence line of RC

18

5.0

08)8(14;0

+−=

=−−−=Σ+

xRR

RxRM

AC

CA

B

35

A B CD E FG

4 m 6 m2 m

� Check ΣFy = 0

RB RCRA

x1

RA

1 -0.219 -0.25 -0.156

10.438

RB

1

0.490.8751.081

0.59

RC1

10.672

0.3750.141

-0.0312

1

;0

=++

=Σ↑+

CBA

y

RRRF

RC

1

0.6719

0.375

0.1406

0

-0.0312

0

Point

C

F

E

D

B

A

G

RA

0

-0.1562

-0.25

-0.2188

0

0.4375

1

RB

0

0.485

0.875

1.078

1

0.5939

0

ΣR

1

1

1

1

1

1

1

36

VG

RA

1 -0.219 -0.25 -0.156

10.438

1x

VG = RA

RA

AVG

MG

� Unit load to the right of G

0.438

-0.219 -0.25 -0.156-0.5621

A B CD E FG

4 m 6 m2 m

Using equilibrium conditions for the influence line of VG

VG = RA - 1

RA

1x

AVG

MG

� Unit load to the left of G

01;0 =−−=Σ↑+ GAy VRF

37

VE

RC1

10.672

0.3750.141

-0.0312

1x

0.6250.328

0.0312

-0.141-0.375

1

A B CD E FG

4 m 6 m2 m

Using equilibrium conditions for the influence line of VE

RCVE

ME

� Unit load to the left of E

VE = - RC

0;0 =+=Σ↑+ ECy VRF

VE = 1 - RC

� Unit load to the right of E

RC

1 x

VE

ME

01;0 =+−=Σ↑+ CEy RVF

38

MG

RA

1 -0.219 -0.25 -0.156

10.438

1x

-0.438 -0.5 -0.312

1

A B CD E FG

4 m 6 m2 m

Using equilibrium conditions for the influence line of MG

0.876

MG = -2 + x + 2RA

� Unit load to the left of G

RA

1x

AVG

MG2 m

02)2(1;0

=−−+=Σ

AG

G

RxMM+

MG = 2RA

� Unit load to the right of G

RA

AVG

MG2 m

02;0

=−=Σ

AG

G

RMM+

39

RC1

10.672

0.3750.141

-0.0312

ME

1x

1.5

0.6881

4 m 6 m2 m

A B CD E FG

Using equilibrium conditions for the influence line of ME

-0.125

0.564

RCVE

ME

4 m

� Unit load to the left of E

ME = 4RC

+ ΣME = 0;

ME = - 4 + x+ 4RC

� Unit load to the right of E

RC

1 x

VE

ME4 m

04)4(1;0

=+−−−=Σ+

CE

E

RxMM

40

Example 3

For the beam shown(a) Draw quantitative influence lines for the reaction at supports A and B, andbending moment at B.(b) Determine all the reactions at supports, and also draw its quantitative shear,bending moment diagrams, and qualitative deflected curve for

- Only 10 kN downward at 6 m from A- Both 10 kN downward at 6 m from A and 20 kN downward at 4 m from A

10 kN

AC

B

4 m

2EI 3EI

2 m 2 m

20 kN

411

fCA

fAAfEA

fDA

1

/fAA /fAA /fAA

/fAA

fCA

fAAfEA

fDA

AC

B

4 m

2EI 3EI

2 m 2 m


42Conjugate Beam

AC

B

4 m

2EI 3EI

2 m 2 m1Real Beam


1 kN

8 kN�m

x (m)V (kN)1 1

+

x (m) M (kN�m)

84

+

2EI

fAA = M´A = 60.44/EI

60.44/EI

12/EI

1.33EIEI67.2

43

EI67.2

2EI

60.44/EI

Conjugate Beam12/EI

1.33EI

37.11/EI 17.77/EI 4.88/EI60.44/EI

0

RA = fxA/fAA

0.614 0.294 0.081

1

0

fxA

A C B

� Quantitative influence line of RA

44

A B

4 m 2 m 2 m

Using equilibrium conditions for the influence line of RB and MB

RB = 1 - RA

10.386

0.706 0.919

0

MB = 8RA - (8-x)(1)0

-1.352-1.088 -1.648

0

RA RB

MB

RA

0.614 0.294 0.081

1

0

1x

45

A B

4 m 2 m 2 m

Using equilibrium conditions for the influence line of VB

RA

0.614 0.294 0.081

1

0

-1-0.386

-0.706-0.919

0 VB = RA -1

RA

1x

VB = RA - 1

46

A B

4 m 2 m 2 m

C

Using equilibrium conditions for the influence line of VC and MC

RA

RA

0.614 0.294 0.081

1

0

1x

RB

MB

MC = 4RAMC = 4RA - (4-x)(1)

MC

VC = RAVC = RA - 1

VC1

-0.386 -0.706

0.3240.4561.176

0.294 0.081 0

47

A B

4 m 2 m 2 m

10 kN

RA

0.614 0.294 0.081

1

0

10(0.081)=0.81 kN

MA (kN�m)+

-

-13.53

4.86

V (kN)

-9.19

0.81 0.81

-

The quantitative shear and bending moment diagram and qualitative deflected curve

RB=9.19 kN

MB= 13.53 kN�m

48

A B

4 m 2 m 2 m

10 kNThe quantitative shear and bending moment diagram and qualitative deflected curve

20 kN

20(.294) +1(0.081)= 6.69 kN

V (kN)

-23.31

6.69 6.69

--13.31 -23.31

MA (kN�m)+

-

-46.48

26.760.14

RA

0.614 0.294 0.081

1

0

RB=23.31 kN

MB=46.48 kN�m

49

Example 1

Draw the influence line for - the vertical reaction at B

A BC D

2 m 2 m 2 m

APPENDIX�Muller-Breslau for the influence line of reaction, shear and moment�Influence lines for MDOF beams

50

1 kN


A BC D

2 m 2 m 2 m

1=AA

AA

ff AA

CA

ff

AA

DA

ff

51

Conjugate Beam

xV´x

M´x

EIB y

18' =

EIx

2

2

EIx

3x

32x

A BC D

2 m 2 m 2 m

Real Beam

1 kN


1 kN

6 kN�m

EIM A

72' =

x

EIB y

18' =

6 /EIEI18

=−EIx

EIx

618 3

52

Point

B

D

C

A

fxA / fAA

0

0.482

0.852

1.0

x (m)

0

2

4

6

x

A BC D

2 m 2 m 2 m

1 kNfAA fxA

72/EI 61.33 /EI34.67 /EI

EIx

EIxMf xxA 6

18'3

−==

fxA

0

EI67.34

EI33.61

EI72

34.67

61.3372

1 kN


/72 = 1.0/72=0.852

/72 = 0.4821=

AA

AA

ff

53

� Influence line of RB

1

A BC D

2 m 2 m 2 m

1=BB

BB

ff

BB

AB

ff

BB

CB

ff

BB

DB

ff

54

Real BeamA BC D

2 m 2 m 2 m

Conjugate Beam

� Find fxB by conjugate beam

11

6 kN�m

x

=−+EI

xEIEI

x 18726

3

EI6

EI18

EI18

EI72

3x

32xV´x

M´x

x

EI72

EI18

EIx

2

2

EIx

55

x (m)

0

2

4

6

Point

B

D

C

A

x

fxB / fBB

1

0.518

0.148

0

A BC D

2 m 2 m 2 m

1

fBB

fxB

0

72/EI37.33/EI10.67/EI

/72 = 0.518

/72 = 0.148/72 = 1

1

37.33

10.670

fBB

Influence line of RB

fBB= 1

72

EIx

EIEIxMf xxB

18726

'3

−+==

fxB

0

EI72

EI33.37

EI67.10

56

Example 2

For the beam shown(a) Draw the influence line for the shear at D for the beam(b) Draw the influence line for the bending moment at D for the beamEI is constant.Plot numerical values every 2 m.

A B CD E

2 m 2 m 2 m 2 m

57

A B CD ED

2 m 2 m 2 m 2 m

The influence line for the shear at D

1 kN

1 kN

DVD

1 kN

1 kNDD

ED

ff

1=DD

DD

ff

58

A B CD E

2 m 2 m 2 m 2 m

2 kN�m

1 kN

1 kN2 k

1 kN

1 kN2 kN�m

1 kN2 kN1 kN

� Using conjugate beam for find fxD

1 kN

1 kN

59

A B CD E

2 m 2 m 2 m 2 m

1 kN1 kN

Real beam

V( kN)

x (m)1

-1

M (kN �m) x (m)

4

Conjugate beam

4/EI

M´D

1 kN

1 kN2kN

60

2 m 2 m 2 m 2 m

M´D

Conjugate beam

4/EI

AB

CD E

4/EI

0

8/EI

m38

� Determine M´D at D

EI316

EI38

128/3EI

4/EI

8/EI

m38

EI316

EI340

61

=−=− )2)(3

8()32)(2(4

EIEIEI

EIEIEIEI 352)2)(

340(

3128)

32)(2( =−+=

EIEIEI 376)2)(

340()

32)(2( −=−=

128/3EI

2 m 2 m 2 m 2 m

Conjugate beam

4/EI

AB

CD E

EI38

EI340

V´DL

M´DL

2/EI2/EI

32

EI340

EI340 V´DR

M´DR

2/EI

32

128/3EI

2/EI2/EI

V´E

M´E

32

EI38

62

128/3EI = M´D = fDD

V ´x (m) θ

fxD = M ´ x (m) ∆

EI376

−

EI340

−EI334

−

EI32

EI316

−

EI38

EI352

EI4

−

Influence line of VD = fxD/fDD

76

52

4

0.406 =128

/128 = -0.594 /(128/3) = -0.094

Conjugate beam

EI38128/3EI

2 m 2 m 2 m 2 m

4/EI

AB

CD E

EI340

63

A B CD E

2 m 2 m 2 m 2 m

The influence line for the bending moment at D

1 kN �m1 kN �m

αDD

MD

DD

EDfα

DD

DDfα

64

1 kN�m

0.5 kN1 kN�m

0.5 kN1 kN0.5 kN

2 m 2 m 2 m 2 m

0.5 kN 0.5 kN

0.5 kN1 k

A B CD E

1 kN �m1 kN �m

� Using conjugate beam for find fxD

65

2 m 2 m 2 m 2 m

Real beamA B CD E

0.5 kN1 kN0.5 kN

1 kN �m1 kN �m

V (kN)

x (m)0.5

1

M (kN �m) x (m)

2

2/EI

Conjugate beam

66

2/EI

4/EI

m38

EI38

2/EI

0

4/EI

m38

2 m 2 m 2 m 2 m

2/EI

Conjugate beam

EI34

EI38

EI332

EI34

67

EIEIEI 326)2)(4()

32)(1( =+=

=−=− )2)(3

4()32)(1(2

EIEIEI

2 m 2 m 2 m 2 m

2/EI

Conjugate beam

EI332

EI34

EI4

M´D

V´D

1/EI1/EI

32

EI4

1/EI1/EI

V´E

M´E

32

EI34

68/(32/3) = -0.188-2Influence line of MD

813.03226

=

xD

xDfα

αDD = 32/3EI

2 m 2 m 2 m 2 m

2/EI

Conjugate beam

43EI

4EI

323EI

x (m)V ´ θEI4

EI38

−

EI31

EI34

EI317

−

EI5

fxD = M ´x (m) ∆

-2/EI

θD = 0.469 + 0.531 = 1 rad

θDL = 5/(32/3) = 0.469 rad.θDR = -17/32 = -0.531 rad.

26/3EI

69

Example 3

Draw the influence line for the reactions at supports for the beam shown in thefigure below. EI is constant.

A DB C

5 m 5 m 5 m 5 m5 m 5 m

GE F

70

Influence line for RD

A DB C GE F

5 m 5 m 5 m 5 m5 m 5 m

1

fBDfCD fDD fED fFD

fXD

1

fXD/fDD = Influence line for RD

DD

BD

ff 1=

DD

DD

ff

DD

CD

ff

DD

ED

ff

DD

FD

ff

71Conjugate beam

15 + (2/3)(15)

EI5.112

EI15

� Use the consistency deformation method

1

+

x RG

- Use conjugate beam for find ∆´G and fGG

∆´G + fGGRG = 0 ------(1)

1

Real beam

15 m 15 m

A G

1

15

1

Real beam

30 mA G

1

30

1

A G3@5 =15 m 3@5 =15 m

fGG

∆´G

1

=

112.5/EI

EIM CC

5.2812'' ==∆

Conjugate beam

20 m

EI15 EI

450

EIMf GGG

9000''' ==

EI450

72

x RG = -0.3125 kN

090005.2812=+ GR

EIEI

↓−= ,3125.0 kNRG

Substitute ∆´G and fGG in (1) :

1

A G5.625

0.6875 0.3125

11

15

=

1

+

1

30

73

8.182 m

6.818 m

EI16.35

EI98.15

EI01.23

)3

(2

3125.0 22

2 xx−

22 '13.28xMx

EI=+

)2

(6875.0625.5 12

11 xEI

xx −=

)3

2(2

6875. 12

1 xEI

x+

1

fBDfCD fDD fED fFD

A G3@5 =15 m 3@5 =15 m

Real beam5.625

0.6875 0.3125

� Use the conjugate beam for find fXD

28.13EIx1

x2

x1 = 5 m -----> fBD = M´1= 56/EI

x1 = 10 m -----> fCD = M´1= 166.7/EI

x1 = 15 m -----> fDD = M´1= 246.1/EI

x2 = 5 m -----> fFD = M´2= 134.1/EI

x2 = 10 m -----> fED = M´2= 229.1/EI

x2 = 15 m -----> fDD = M´2= 246.1/EI

A G Conjugate beamEI625.5

EI688.4−

Ax1

(5.625-0.6875x1)/EI

V´1

M´1

EI625.5 EI

xx 211 6875.0625.5 −EI

x2

6875.0 21

G

x2

0.3125x2

V´2M´2

EI13.28

EIx

23125.0 2

2

74

� Influence Line for RD

Influence Line for RD

0.228 0.677 1.0 0.931 0.545

1

fXD

EI56

EI7.166

EI1.246

EI2.229

EI1.134

1

fXD/fDD

1.24656

1.2467.166

1.2461.246

1.2462.229

1.2461.134

75

Influence line for RG

A DB C GE F

5 m 5 m 5 m 5 m5 m 5 m

fXG

1fBG fCG

fGGfEGfFG

fXG/fGG

1GG

BG

ff

GG

CG

ff

GG

EG

ff

GG

FG

ff 1=

GG

GG

ff

76Conjugate beam

20 m

EI450

EI30

� Use consistency deformations

1

=

∆´D + fDDRD = 0 ------(2)

- Use conjugate beam for find ∆´D and fDD

1

+1

Real beam

30 mA G

1

30

1

Real beam

15 m 15 m

A G

1

15

X RD

fXG

13@5 =15 m 3@5 =15 m

∆´D

fDD

Conjugate beam15 + (2/3)(15)

EI15

EI5.112

450/EI

EI9000

77

EIEIEIEIMD

5.2812)15(4509000)3

15(5.112'' =−+==∆EIEI

MfDD1125)15

32(5.112'' =×==

↓=−==+ ,5.25.2,011255.2812 kNkNRREIEI DDSubstitute ∆´D and fDD in (2) :

x RD = -2.5 kN

=

1

11

30+

11

15

1.5

7.5

2.5

15 mV´

M´EI

5.112

EI5.1

450/EI

EI9000

M´´

V´´

EI15 EI

5.112

78

EIEIfBG

5.62)532(75.18

−=×−==

� Use the conjugate beam for find fXG

Real beam

1fBG fCG

fGGfEGfFG

3@5 =15 m 3@5 =15 m1.5

7.5

2.5

fGG = M´G = 1968.56/EI

168.75/EI

EIEIfCG

06.125)67.6(75.18−=−==

A G Conjugate beam

EI5.7−

EI15

10 m

15 + (10/3) = 18.33 m

25 + (2/3)(5) = 28.33 m

EI5.112

EI75

EI75.18

A5 m

V´1

M´1

EI5.7−

EI75.18

A6.67 m

V´2

M´2

EI75.18

EI5.7−

EI75.18−

79

=−+ 33

23 75.16856.1968)

3(

2x

EIEIxx

x = 5 m -----> fFG = M´= 1145.64/EI

x = 10 m -----> fEG = M´ = 447.73/EI

Influence line for RG-0.064-0.032

0.227 0.582 1.0

M´ Gx

fGG = M´G = 1968.56/EI

168.75/EI

x

V´2

2

2x

1

fXG

EI5.62−

EI125−

EI73.447

EI64.1145

EI56.1968

56.19685.62−

56.1968125−

56.196873.447

56.196864.1145

56.196856.1968

1

fXG/fGG

80

Using equilibrium condition for the influence line for Ay

A DB C GE F

5 m 5 m 5 m 5 m5 m 5 m

1x

MA

Ay RD RG

Unit load

1 1

Influence Line for RD

0.228 0.678 1.0 0.929 0.542

Influence line for RG-0.064-0.032

0.227 0.582 1.0

0.386Influence line for Ay

0.804

-0.156 -0.124

1.0

CDAy RRRF −−==Σ↑+ 1:0

81

Using equilibrium condition for the influence line for MA

A DB C GE F

5 m 5 m 5 m 5 m5 m 5 m

1x

MA

Ay RD RG

x 15

x 30

RD

0.228 0.678 1.0 0.929 0.542

1x5 10 15 20 25 30

RG-0.064-0.032

0.227 0.582 1.0

Influence line for MA

2.541.75

-0.745 -0.59

CDA RRxM 30151:0 −−=Σ+

1

! General Case! Stiffness Coefficients! Stiffness Coefficients Derivation! Fixed-End Moments! Pin-Supported End Span! Typical Problems! Analysis of Beams! Analysis of Frames: No Sidesway! Analysis of Frames: Sidesway

DISPLACEMENT METHOD OF ANALYSIS: SLOPE DEFLECTION EQUATIONS

2

Slope � Deflection Equations

settlement = ∆j

Pi j kw Cj

Mij MjiwP

θj

θi

ψ

i j

3

Degrees of Freedom

L

θΑ

A B

M

1 DOF: θΑ

PθΑ

θΒΒΒΒ

A BC 2 DOF: θΑ , θΒΒΒΒ

4

L

A B

1

Stiffness

kBAkAA

LEIkAA

4=

LEIkBA

2=

5

L

A B1

kBBkAB

LEIkBB

4=

LEIkAB

2=

6

Fixed-End ForcesFixed-End Moments: Loads

P

L/2 L/2

L

w

L

8PL

8PL

2P

2P

12

2wL12

2wL

2wL

2wL

7

General Case

settlement = ∆j

Pi j kw Cj

Mij MjiwP

θj

θi

ψ

i j

8

wP

settlement = ∆j

(MFij)∆ (MF

ji)∆

(MFij)Load (MF

ji)Load

+

Mij Mji

θi

θj

+

i jwPMij

Mji

settlement = ∆jθj

θi

ψ

=+ ji LEI

LEI θθ 24

ji LEI

LEI θθ 42

+=

,)()()2()4( LoadijF

ijF

jiij MMLEI

LEIM +++= ∆θθ Loadji

Fji

Fjiji MM

LEI

LEIM )()()4()2( +++= ∆θθ

L

9

Mji Mjk

Pi j kw Cj

Mji Mjk

Cj

j

Equilibrium Equations

0:0 =+−−=Σ+ jjkjij CMMM

10

+

1

1

i jMij Mji

θi

θj

LEIkii

4=

LEIk ji

2=

LEIkij

2=

LEIk jj

4=

iθ×

jθ×

Stiffness Coefficients

L

11

[ ]

=

jjji

ijii

kkkk

k

Stiffness Matrix

)()2()4( ijF

jiij MLEI

LEIM ++= θθ

)()4()2( jiF

jiji MLEI

LEIM ++= θθ

+

=

F

ji

Fij

j

iI

ji

ij

MM

LEILEILEILEI

MM

θθ

)/4()/2()/2()/4(

Matrix Formulation

12

[D] = [K]-1([Q] - [FEM])

Displacementmatrix

Stiffness matrix

Force matrixwP(MF

ij)Load (MFji)Load

+

+

i jwPMij

Mji

θj

θi

ψ ∆j

Mij Mji

θi

θj

(MFij)∆ (MF

ji)∆

Fixed-end momentmatrix

][]][[][ FEMKM += θ

]][[])[]([ θKFEMM =−

][][][][ 1 FEMMK −= −θ

L

13

L

Real beam

Conjugate beam

Stiffness Coefficients Derivation: Fixed-End Support

MjMi

L/3

θi

LMM ji +

EIM j

EIMi

θι

EILM j

2

EILMi

2

)1(2

0)3

2)(2

()3

)(2

(:0'

−−−=

=+−=Σ+

ji

jii

MM

LEI

LMLEI

LMM

)2(0)2

()2

(:0 −−−=+−=Σ↑+EI

LMEI

LMF jiiy θ ij

ii

LEIM

LEIM

andFrom

θ

θ

)2(

)4(

);2()1(

=

=

LMM ji +

i j

14

Conjugate beam

L

Real beam

Stiffness Coefficients Derivation: Pinned-End Support

Mi θi

θj

LM i

LMi

EIM i

EILMi

2

32L

θi θj

0)3

2)(2

(:0' =−=Σ+ LLEI

LMM ii

j θ

i j

0)2

()3

(:0 =+−=Σ↑+ jii

y EILM

EILMF θ

LEIM

EILM

ii

i3)

3(1 =→==θ

)6

(EI

LM ij

−=θ)

3(

EILM i

i =θ

15

Fixed end moment : Point Load

Real beam

8,0

162

22:0

2 PLMEI

PLEI

MLEI

MLFy ==+−−=Σ↑+

P

M

M EIM

Conjugate beamA

EIM

B

L

P

A B

EIM

EIML2

EIM

EIML2

EIPL

16

2

EIPL4 EI

PL16

2

16

L

P

841616PLPLPLPL

=+−

+−

8PL

8PL

2P

2PP/2

P/2

-PL/8 -PL/8

PL/8

-PL/16-PL/8

-

PL/4+

-PL/16-PL/8-

17

Uniform load

L

w

A B

w

M

M

Real beam Conjugate beam

A

EIM

EIM

B

12,0

242

22:0

23 wLMEI

wLEI

MLEI

MLFy ==+−−=Σ↑+

EIwL8

2

EIwL

24

3

EIwL

24

3

EIM

EIML2

EIM

EIML2

18

Settlements

M

M

L

∆

MjMi = Mj

LMM ji +L

MM ji +

Real beam

2

6LEIM ∆

=

Conjugate beam

EIM

A B

EIM

∆

,0)3

2)(2

()3

)(2

(:0 =+−∆−=Σ+L

EIMLL

EIMLM B

EIM

EIML2

EIMEI

ML2

∆

19

wP

A B

A B+θA θB

(FEM)AB(FEM)BA

Pin-Supported End Span: Simple Case

BA LEI

LEI θθ 24

+ BA LEI

LEI θθ 42

+

)1()()/2()/4(0 −−−++== ABBAAB FEMLEILEIM θθ

)2()()/4()/2(0 −−−++== BABABA FEMLEILEIM θθ

BABABBA FEMFEMLEIM )()(2)/6(2:)1()2(2 −+=− θ

2)()()/3( BA

BABBAFEMFEMLEIM −+= θ

wP

AB

L

20

AB

wP

A B

A BθA θB

(MF AB)load

(MF BA)load

Pin-Supported End Span: With End Couple and Settlement

BA LEI

LEI θθ 24

+ BA LEI

LEI θθ 42

+

L

(MF AB)∆

(MF BA) ∆

)1()()(24−−−+++== ∆

FABload

FABBAAAB MM

LEI

LEIMM θθ

)2()()(42−−−+++= ∆

FBAload

FBABABA MM

LEI

LEIM θθ

2)(

21])(

21)[(3:

2)1()2(2lim AF

BAloadFABload

FBABBAA

MMMMLEIMbyinateE ++−+=

−∆θθ

MA

wP

AB

∆

21

Fixed-End MomentsFixed-End Moments: Loads

P

L/2 L/2

P

L/2 L/2

8PL

8PL

163)]

8()[

21(

8PLPLPL

=−−+

12

2wL12

2wL

8)]

12()[

21(

12

222 wLwLwL=−−+

22

Typical Problem

0

0

0

0

A C

B

P1P2

L1 L2

wCB

P

8PL

8PL w12

2wL

12

2wL

8024 11

11

LPLEI

LEIM BAAB +++= θθ

8042 11

11

LPLEI

LEIM BABA −++= θθ

128024 2

222

22

wLLPLEI

LEIM CBBC ++++= θθ

128042 2

222

22

wLLPLEI

LEIM CBCB −

−+++= θθ

L L

23

MBA MBC

A C

B

P1P2

L1 L2

wCB

B

CB

8042 11

11

LPLEI


128024 2

222

22

wLLPLEI


BBCBABB forSolveMMCM θ→=−−=Σ+ 0:0

24

A C

B

P1P2

L1 L2

wCB

Substitute θB in MAB, MBA, MBC, MCB

MABMBA

MBC

MCB

0

0

0

0

8024 11

11

LPLEI

LEIM BAAB +++= θθ

8042 11

11

LPLEI


128024 2

222

22

wLLPLEI


128042 2

222

22

wLLPLEI

LEIM CBCB −

−+++= θθ

25

A BP1

MAB

MBA

L1

A CB

P1P2

L1 L2

wCB

ByR CyAy ByL

Ay Cy

MABMBA

MBC

MCB

By = ByL + ByR

B CP2

MBCMCB

L2

26

Example of Beams

27

10 kN 6 kN/m

A C

B 6 m4 m4 m

Example 1

Draw the quantitative shear , bending moment diagrams and qualitativedeflected curve for the beam shown. EI is constant.

28

PPL8 wwL2

30FEM

PL8

wL2

20

MBA MBC

B

Substitute θB in the moment equations:

MBC = 8.8 kN�m

MCB = -10 kN�m

MAB = 10.6 kN�m,

MBA = - 8.8 kN�m,

[M] = [K][Q] + [FEM]

10 kN 6 kN/m

A C

B 6 m4 m4 m

0

0

0

0

8)8)(10(

82

84

++= BAABEIEIM θθ

8)8)(10(

84

82

−+= BABAEIEIM θθ

30)6)(6(

62

64 2

++= CBBCEIEIM θθ

20)6)(6(

64

62 2

−+= CBCBEIEIM θθ

0:0 =−−=Σ+ BCBAB MMM

EI

EIEI

B

B

4.2

030

)6)(6(10)6

48

4(2

=

=+−+

θ

θ

29

10 kN 6 kN/m

A C

B 6 m4 m4 m

MBC = 8.8 kN�m

MCB= -10 kN�m

MAB = 10.6 kN�m,

MBA = - 8.8 kN�m,

= 5.23 kN = 4.78 kN = 5.8 kN = 12.2 kN

10 kN�m8.8 kN�m

10.6 kN�m8.8 kN�m

10 kN

10.6 kN�m

8.8 kN�m

A B

ByLAy

2 m

6 kN/m8.8 kN�m

10 kN�mB

CyByR

18 kN

30

10 kN 6 kN/m

A C

B 6 m4 m4 m

10 kN�m10.6 kN�m

5.23 kN 12.2 kN

4.78 + 5.8 = 10.58 kN

V (kN)x (m)

5.23

- 4.78

5.8

-12.2

+-

+

-

M (kN�m) x (m)

-10.6

10.3

-8.8 -10- -

+-

EIB4.2

=θ

Deflected shape x (m)

31

10 kN 6 kN/m

A C

B 6 m4 m4 m

Example 2


32

PPL8 wwL2

30FEM

PL8

wL2

20

[M] = [K][Q] + [FEM]

10 kN 6 kN/m

A C

B 6 m4 m4 m

)1(8

)8)(10(8

28

4−−−++= BAAB

EIEIM θθ

)2(8

)8)(10(8

48

2−−−−+= BABA

EIEIM θθ

)3(30

)6)(6(6

26

4 2

−−−++= CBBCEIEIM θθ

)4(20

)6)(6(6

46

2 2

−−−−+= CBCBEIEIM θθ

10

10

0

0

0

308

62:)1()2(2 −=− BBAEIM θ

)5(158

3−−−−= BBA

EIM θ

33

MBA MBC

B

)5(158

3−−−−= BBA

EIM θ

)3(30

)6)(6(6

4 2

−−−+= BBCEIM θ

0:0 =−−=Σ+ BCBAB MMM

EI

EIEI

B

B

488.7

)6(030

)6)(6(15)6

48

3(2

=

−−−=+−+

θ

θ

EI

EIEIinSubstitute

A

BAB

74.23

108

28

40:)1(

−=

−+=

θ

θθθ

Substitute θA and θB in (5), (3) and (4):

MBC = 12.19 kN�m

MCB = - 8.30 kN�m

MBA = - 12.19 kN�m)4(

20)6)(6(

62 2

−−−−= BCBEIM θ

34

10 kN 6 kN/m

A C

B 6 m4 m4 m

MBA = - 12.19 kN�m, MBC = 12.19 kN�m, MCB = - 8.30 kN�m

= 3.48 kN = 6.52 kN = 6.65 kN = 11.35 kN

12.19 kN�m

12.19 kN�m8.30 kN�m

10 kN

12.19 kN�m

A B

ByLAy

2 m

6 kN/m12.19 kN�m

8.30 kN�mC

CyByR

18 kN

B

35

Deflected shape x (m)EIB49.7

=θ

10 kN 6 kN/m

A C

B 6 m4 m4 m

V (kN)x (m)

3.48

- 6.52

6.65

-11.35

M (kN�m) x (m)

14

-12.2-8.3

11.35 kN3.48 kN

6.52 + 6.65 = 13.17 kN

EIA74.23−

=θ

36

10 kN 4 kN/m

A C

B6 m4 m4 m

2EI 3EI

Example 3


37

10 kN 4 kN/m

A C

B6 m4 m4 m

2EI 3EI(4)(62)/12 (4)(62)/12 (10)(8)/8(10)(8)/8

15

12

)1(8

)8)(10(8

)2(28

)2(4−−−++= BAAB

EIEIM θθ

)2(8

)8)(10(8

)2(48

)2(2−−−−+= BABA

EIEIM θθ

0 10

10

)2(8

)8)(10)(2/3(8

)2(3:2

)1()2(2 aEIM BBA −−−−=− θ

)3(12

)6)(4(6

)3(4 2


38

10 kN 4 kN/m

A C

B6 m4 m4 m

2EI 3EI(4)(62)/12 (4)(62)/12(3/2)(10)(8)/8

EIEIMM

B

BBCBA

/091.13151275.2:0

==+−==−−

θθ

15

12

)2(8

)8)(10)(2/3(8

)2(3 aEIM BBA −−−−= θ

)3(12

)6)(4(6

)3(4 2


mkNEIM

mkNEI

EIM

mkNEI

EIM

BCB

BC

BA

•−=−=

•=−=

•−=−=

91.10126

)3(2

18.1412)091.1(6

)3(4

18.1415)091.1(8

)2(3

θ

39

10 kN 4 kN/m

A C

B6 m4 m4 m

2EI 3EI

MBA = - 14.18 kN�m, MBC = 14.18 kN�m, MCB = -10.91 kN�m

14.18

140.18 kN�m

10 kN

A B

ByLAy = 6.73 kN= 3.23 kN

14.18 kN�m

10.91 kN�m

4 kN/m

C

CyByR

24 kN

14.18 10.91

= 11.46 kN= 12.55 kN

40

10 kN 4 kN/m

A C

B6 m4 m4 m

2EI 3EI

11.46 kN3.23 kN

10.91 kN�m

V (kN)x (m)

3.23

-6.73

12.55

-11.46

+-

+-

2.86

M (kN�m) x (m)

12.91

-14.18

5.53

-10.91

+

-+

-

6.77 + 12.55 = 19.32 kN

θB = 1.091/EIDeflected shape x (m)

41

10 kN 4 kN/m

A C

B6 m4 m4 m

2EI 3EI

Example 4


12 kN�m

42

10 kN 4 kN/m

A C

B6 m4 m4 m

2EI 3EI

12 kN�m

wL2/12 = 12 wL2/12 = 121.5PL/8 = 15

MBA

MBCB

12 kN�mMBA

MBC

EIB273.3

−=θ

EIA21.7

−=θ

)1(158

)2(3−−−−= BBA

EIM θ

)2(126

)3(4−−−+= BBC

EIM θ

)3(126

)3(2−−−−= BCB

EIM θ

8)8)(10(

8)3(2

8)2(4

++= BAABEIEIM θθ

0 -3.273/EI

012:int =−−− BCBA MMBJo

012)122()1575.0( =−+−−− BEIEI θ

mkNEI

EIM BA •−=−−= 45.1715)273.3(75.0

mkNEI

EIM BC •=+−= 45.512)273.3(2

mkNEI

EIM CB •−=−−= 27.1512)273.3(

43

10 kN

A B

4 kN/m

C

24 kN

17.45 kN�m5.45 kN�m

15.27 kN�m

2.82 kN 13.64 kN10.36 kN7.18 kN

12 kN�m10 kN 4 kN/m

A C

B13.64 kN2.82 kN

15.27 kN�m

17.54 kN

mkNEI

EIM BA •−=−−= 45.1715)273.3(75.0

mkNEI

EIM BC •=+−= 45.512)273.3(2

mkNEI

EIM CB •−=−−= 27.1512)273.3(

44

10 kN 4 kN/m

A C

B6 m4 m4 m

2EI 3EI

12 kN�m

13.64 kN2.82 kN

15.27 kN�m

17.54 kN

V (kN)x (m)3.41 m+

-+

-

2.82

-7.18

10.36

-13.64

M (kN�m) x (m)+

- -+

11.28

-17.45

-5.45-15.27

7.98


EIB273.3

=θEIA

21.7−=θ

EIB273.3

−=θ

EIA21.7

−=θ

45

Example 5

Draw the quantitative shear, bending moment diagrams, and qualitativedeflected curve for the beam shown. Support B settles 10 mm, and EI isconstant. Take E = 200 GPa, I = 200x106 mm4.

12 kN�m 10 kN 6 kN/m

A CB

6 m4 m4 m

2EI 3EI10 mm

46


A CB

6 m4 m4 m

2EI 3EI10 mm

[FEM]∆

∆

A

B

2

6LEI∆

2

6LEI∆

∆

BC

2

6LEI∆

2

6LEI∆

P w

[FEM]load

8PL

8PL

30

2wL30

2wL

)1(8

)8)(10(8

)01.0)(2(68

)2(28

)2(42 −−−+++=

EIEIEIM BAAB θθ

)2(8

)8)(10(8

)01.0)(2(68

)2(48

)2(22 −−−−++=

EIEIEIM BABA θθ

)3(30

)6)(6(6

)01.0)(3(66

)3(26

)3(4 2

2 −−−+−+=EIEIEIM CBBC θθ

)4(30

)6)(6(6

)01.0)(3(66

)3(46

)3(2 2

2 −−−−−+=EIEIEIM CBCB θθ

-12

0

0

47


A CB

6 m4 m4 m

2EI 3EI10 mm

Substitute EI = (200x106 kPa)(200x10-6 m4) = 200x200 kN� m2 :

)1(8

)8)(10(8

)01.0)(2(68

)2(28

)2(42 −−−+++=

EIEIEIM BAAB θθ

)2(8

)8)(10(8

)01.0)(2(68

)2(48

)2(22 −−−−++=

EIEIEIM BABA θθ

)1(10758

)2(28

)2(4−−−+++= BAAB

EIEIM θθ

)2(10758

)2(48

)2(2−−−−++= BABA

EIEIM θθ

)2(2/12)2/10(10)2/75(758

)2(3:2

)1()2(2 aEIM BBA −−−−−−−+=− θ

16.5

48

+ ΣMB = 0: - MBA - MBC = 0 (3/4 + 2)EIθB + 16.5 - 192.8 = 0

θB = 64.109/ EI

Substitute θB in (1): θA = -129.06/EI

Substitute θA and θB in (5), (3), (4):

MBC = -64.58 kN�m

MCB = -146.69 kN�m

MBA = 64.58 kN�m,

MBA MBC

B


A CB

6 m4 m4 m

2EI 3EI10 mm

MBC = (4/6)(3EI)θB - 192.8

MBA = (3/4)(2EI)θB + 16.5

49


A CB

6 m4 m4 m

64.58 kN�m 64.58 kN�m

= -1.57 kN= 11.57 kN

146.69 kN�m

= 47.21 kN= -29.21 kN

10 kN

A B

ByLAy

12 kN�m64.58 kN�m

2 m

6 kN/m

C

CyByR

18 kN

B

64.58 kN�m

146.69 kN�m

MBC = -64.58 kN�m

MCB = -146.69 kN�m

MBA = 64.58 kN�m,

50


A C

B6 m4 m4 m

2EI 3EI

V (kN)x (m)

11.57 1.57

-29.21-47.21

-+

M (kN�m) x (m)

1258.29 64.58

-146.69

+

-

47.21 kN

146.69 kN�m

11.57 kN

1.57 + 29.21 = 30.78 kN

10 mmθA = -129.06/EI

θB = 64.109/ EIθA = -129.06/EI


θB = 64.109/ EI

51

Example 6

For the beam shown, support A settles 10 mm downward, use the slope-deflectionmethod to(a)Determine all the slopes at supports(b)Determine all the reactions at supports(c)Draw its quantitative shear, bending moment diagrams, and qualitativedeflected shape. (3 points)Take E= 200 GPa, I = 50(106) mm4.

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

10 mm

52

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

)1(3

)2(4−−−= CCB

EIM θ

)2(1005.43

)5.1(23

)5.1(4−−−+++= ACCA

EIEIM θθ

)2(2

122

1002

)5.4(33

)5.1(3:2

)2()2(2 aEIM CCA −−−+++=− θ

)3(1005.43

)5.1(43

)5.1(2−−−+−+= ACAC

EIEIM θθ12

0.01 m

C

AmkN •=

××

1003

)01.0)(502005.1(62

mkN •100MF

∆

10 mm

6 kN/m

A C

5.412

)3(6 2

= 4.5MF

w

53

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

)1(3

)2(4−−−= CCB

EIM θ

)2(2

122

1002

)5.4(33

)5.1(3 aEIM CCA −−−+++= θ

10 mm

MCBMCA

C

0=+ CACB MM

02

122

1002

)5.4(33

)5.48(=+++

+C

EI θ

radEIC 0015.006.15

−=−

=θ

)3(1005.43

)5.1(4)06.15(3

)5.1(212 −−−+−+−

= AEI

EIEI θSubstitute θC in eq.(3)

radEIA 0034.022.34

−=−

=θ

� Equilibrium equation:

54

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

10 mm

radEIC 0015.006.15

−=−

=θ radEIA 0034.022.34

−=−

=θ

mkNEI

EIEIM CBC •−=−

== 08.20)06.15(3

)2(23

)2(2 θ

mkNEI

EIEIM CCB •−=−

== 16.40)06.15(3

)2(43

)2(4 θ

kN08.203

08.2016.40=

+kN08.20

B C40.16 kN�m20.08 kN�m

6 kN/m

AC

12 kN�m

40.16 kN�m

18 kN

8.39 kN26.39 kN

55

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

10 mm

6 kN/m

AC

12 kN�m

40.16 kN�m8.39 kN26.39 kN

B C40.16 kN�m20.08 kN�m

20.08 kN20.08 kN

V (kN)

x (m)

26.398.39

-20.08

+-

M (kN�m)

x (m)20.08

-40.16

12

radC 0015.0−=θ

radA 0034.0−=θ

Deflected shapex (m)

radC 0015.0=θ

radA 0034.0=θ

56

Example 7

For the beam shown, support A settles 10 mm downward, use theslope-deflection method to(a)Determine all the slopes at supports(b)Determine all the reactions at supports(c)Draw its quantitative shear, bending moment diagrams, and qualitativedeflected shape.Take E= 200 GPa, I = 50(106) mm4.

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

10 mm

57

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

10 mm

5.412

)3(6 2

=

6 kN/m

AC4.5

mkN •=

××

1003

)01.0)(502005.1(620.01 m

C

A

100

34 CEI∆

∆C

CB

34

3)2(6

2CC EIEI ∆

=∆

)2(3

43

)2(4−−−∆−= CCCB

EIEIM θ

)3(1005.43

)5.1(23

)5.1(4−−++∆++= CACCA EIEIEIM θθ

)4(1005.43

)5.1(43

)5.1(2−−+−∆++= CACAC EIEIEIM θθ

12

)1(3

43

)2(2−−−∆−= CCBC

EIEIM θ

)3(2

122

1002

)5.4(323

)5.1(3:2

)4()3(2 aEIEIM CCCA −−−+++∆+=− θ

CC EIEI

∆=∆

23)5.1(6∆C

C

A

CEI∆

58

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

10 mm

� Equilibrium equation:

)3

()( CBBCCBy

MMC +−=

6 kN/m

AC

12 kN�m

MCA

18 kN

AyB C

MCBMBC

By3

393

)5.1(1812)( +=

++= CACA

CAyMMC

C

MCB MCA

(Cy)CA(Cy)CB

*)1(0:0 −−−=+=Σ CACBC MMM

*)2(0)()(:0 −−−=+=Σ CAyCByy CCC

)5(15.628333.0167.4*)1( −−−−=∆− CC EIEIinSubstitute θ

)6(75.101167.35.2*)2( −−−−=∆+− CC EIEIinSubstitute θ

mmEIradEIandFrom CC 227.5/27.5200255.0/51.25)6()5( −=−=∆−=−=θ

59

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

10 mm

� Solve equation

radEIC 00255.051.25

−=−

=θ

mmEIC 227.527.52

−=−

=∆

Substitute θC and ∆C in (4)

radEIA 000286.086.2

−=−

=θ

Substitute θC and ∆C in (1), (2) and (3a)

mkNM BC •= 68.35

mkNMCB •= 67.1

mkNMCA •−= 67.1

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

35.68 kN�m

kN

Ay

55.56

68.3512)5.4(18

=

−−=

kNBy

45.12

55.518

=

−=

60

10 mm

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

35.68 kN�m

12.45 kN 5.55 kN

Deflected shape

x (m)

radC 00255.0−=θ

mmC 227.5−=∆

radA 000286.0−=θ

radC 00255.0=θradA 000286.0=θ

mmC 227.5=∆

M (kN�m)

x (m)1214.57

1.67

-35.68

-+

V (kN)

x (m)0.925 m12.45

-5.55

+

61

Example of Frame: No Sidesway

62

Example 6

For the frame shown, use the slope-deflection method to (a) Determine the end moments of each member and reactions at supports(b) Draw the quantitative bending moment diagram, and also draw the qualitative deflected shape of the entire frame.

10 kN

C

12 kN/m

A

B

6 m

40 kN

3 m

3 m

1 m

2EI

3EI

63

10 kN

C

12 kN/m

A

B

6 m

40 kN

3 m

3 m

1 m

2EI

3EIPL/8 = 30

PL/8 = 30

� Equilibrium equations

10

MBC

MBA

)3(18366

)2(3−−−++= BBC

EIM θ

*)1(010 −−−=−− BCBA MM

05430310 =−+− BEIθ

EIEIB667.4

)3(14 −

=−

=θ)1(30

6)3(2

−−−+= BABEIM θ

mkNMmkNM

mkNM

BC

BA

AB

•=•−=

•=

33.4933.39

33.25

36/2 = 18

(wL2/12 ) =3636

� Slope-Deflection Equations

)2(306

)3(4−−−−= BBA

EIM θ

Substitute (2) and (3) in (1*)

)3()1(667.4 toinEI

Substitute B−

=θ

64

10 kN

C

12 kN/m

A

B

6 m

40 kN

3 m

3 m

1 m

2EI

3EI39.33

25.33

49.33

A

B

40 kN

39.33

25.33

C

12 kN/m

B49.33

17.67 kN

27.78 kN

Bending moment diagram

-25.33

27.7

-39.3

Deflected curve

-49.33

20.58

10

θB = -4.667/EIθB

θB

MAB = 25.33 kN�m

MBA = -39.33 kN�m

MBC = 49.33 kN�m

65

A

B

C

D

E

25 kN

5 m

5 kN/m

60(106) mm4

240(106) mm4 180(106)

120(106) mm4

3 m 4 m3 m

Example 7

Draw the quantitative shear, bending moment diagrams and qualitativedeflected curve for the frame shown. E = 200 GPa.

66

A

B

C

D

E

25 kN

5 m

5 kN/m

60(106) mm4

240(106) mm4 180(106)

120(106) mm4

3 m 4 m3 m0

BAABEIEIM θθ

5)2(2

5)2(4

+=0

BABAEIEIM θθ

5)2(4

5)2(2

+=

75.186

)4(26

)4(4++= CBBC

EIEIM θθ

75.186

)4(46

)4(2−+= CBCB

EIEIM θθ

CCDEIM θ5

)(3=

104

)3(3+= CCE

EIM θ

0=+ BCBA MM

)1(75.18)68()

616

58( −−−−=++ CB EIEI θθ

0=++ CECDCB MMM

)2(75.8)49

53

616()

68( −−−=+++ CB EIEI θθ

EIEIandFrom CB

86.229.5:)2()1( =−

= θθ

PL/8 = 18.75 18.75

6.667 (wL2/12 ) = 6.667+ 3.333

67

MAB = −4.23 kN�m

MBA = −8.46 kN�m

MBC = 8.46 kN�m

MCB = −18.18 kN�m

MCD = 1.72 kN�m

MCE = 16.44 kN�m

Substitute θB = -1.11/EI, θc = -20.59/EI below

0

0BAAB

EIEIM θθ5

)2(25

)2(4+=

BABAEIEIM θθ

5)2(4

5)2(2

+=

75.186

)4(26

)4(4++= CBBC

EIEIM θθ

75.186

)4(46

)4(2−+= CBCB

EIEIM θθ

CCDEIM θ5

)(3=

104

)3(3+= CCE

EIM θ

68

MAB = -4.23 kN�m, MBA = -8.46 kN�m, MBC = 8.46 kN�m, MCB = -18.18 kN�m,MCD = 1.72 kN�m, MCE = 16.44 kN�m

A

B

5 m

C E

20 kN

A

B

5 m

4.23 kN�m

2.54 kN

(8.46 + 4.23)/5 = 2.54 kN

8.46 kN�m

C

25 kN

B 3 m 3 m2.54 kN 2.54 kN

8.46 kN�m(25(3)+8.46-18.18)/6 = 10.88 kN

14.12 kN18.18 kN�m

16.44 kN�m

(20(2)+16.44)/4= 14.11 kN

5.89 kN

0.34 kN

28.23 kN

14.12+14.11=28.23 kN1.72 kN�m

(1.72)/5 = 0.34 kN

10.88 kN

10.88 kN

2.54-0.34=2.2 kN

2.2 kN

69

Shear diagram

-2.54

-2.54

10.88

-14.12

14.11

-5.89

0.34

+

-

-

+-

+

2.82 m

1.18 m

Deflected curve

1.18 m

0.78 m2.33 m1.29 m

1.67m

1.29 m

0.78 m

Moment diagram

1.18 m

3.46

24.18

1.72

-18.18 -16.44

4.23

-8.46-8.46

2.33 m

+

-

+

- -

+

θB = −5.29/EI

θC = 2.86/EI1.67m

70

Example of Frames: Sidesway

71

A

B C

D

3m

10 kN

3 m

1 m

Example 8

Determine the moments at each joint of the frame and draw the quantitativebending moment diagrams and qualitative deflected curve . The joints at A andD are fixed and joint C is assumed pin-connected. EI is constant for each member

72

A

B C

D

3m

10 kN

3 m

1 m

MABMDC

Ax Dx

Ay Dy

� Boundary Conditions

θA = θD = 0

� Equilibrium Conditions

� Unknowns

θB and ∆

- Entire Frame

*)2(010:0 −−−=−−=Σ→+

xxx DAF

*)1(0:0 −−−=+=Σ BCBAB MMM

� Overview

B- Joint B

MBCMBA

73

)3(3

)(3−−−= BBC

EIM θ

)4(1875.0375.021375.0 −−−∆=∆−∆= EIEIEIM DC

)1(4

64

)1)(3(104

)(222

2

−−−∆

++=EIEIM BAB θ

5.625 0.375EI∆

A

B C

D3m

10 kN

3 m

1 m

(5.625)load

(1.875)load

(0.375EI∆)∆

(0.375EI∆)∆

(0.375EI∆)∆

(0.375EI∆)∆

:0=Σ+ CM

MDC

D

C

4 m

Dx

MAB

MBA

A

B

4 m

Ax

10 kN

∆ ∆

(1/2)(0.375EI∆)∆

:0=Σ+ BM

)5(563.11875.0375.0 −−−+∆+= EIEIA Bx θ4

)( BAABx

MMA +=

)6(0468.04

−−−∆== EIMD DCx

)2(4

64

)1)(3(104

)(422

2

−−−∆

+−=EIEIM BBA θ

5.625 0.375EI∆

� Slope-Deflection Equations

74

� Solve equation

*)2(010 −−−=−− xx DA*)1(0 −−−=+ BCBA MM

)3(3

)(3−−−= BBC

EIM θ

)4(1875.0 −−−∆= EIM DC

)1(375.0625.54

)(2−−−∆++= EIEIM BAB θ

)5(563.11875.0375.0 −−−+∆+= EIEIA Bx θ

)6(0468.0 −−−∆= EIDx

)2(375.0625.54

)(4−−−∆+−= EIEIM BBA θ

Equilibrium Conditions:

Slope-Deflection Equations:

Horizontal reaction at supports:


2EI θB + 0.375EI ∆ = 5.625 ----(7)


)8(437.8235.0375.0 −−−−=∆−− EIEI Bθ

From (7) and (8) can solve;

EIEIB8.446.5

=∆−

=θ

)6()1(8.446.5 toinEI

andEI

Substitute B =∆−

=θ

MAB = 15.88 kN�mMBA = 5.6 kN�mMBC = -5.6 kN�mMDC = 8.42 kN�mAx = 7.9 kNDx = 2.1 kN

75

A

BC

D

Deflected curve

A

BC

D


MAB = 15.88 kN�m, MBA = 5.6 kN�m, MBC = -5.6 kN�m, MDC = 8.42 kN�m, Ax = 7.9 kN, Dx = 2.1 kN,

A

B C

D

3m

10 kN

3 m

1 m

15.88

5.6

8.42

7.9 kN 2.1 kN

15.88

5.6

8.42

∆ = 44.8/EI ∆ = 44.8/EI

θB = -5.6/EI5.6

7.8

76

B C

DA

3m4 m

10 kN4 m

2 m

pin

2 EI

2.5 EI EI

Example 9

From the frame shown use the slope-deflection method to:(a) Determine the end moments of each member and reactions at supports(b) Draw the quantitative bending moment diagram, and also draw thequalitative deflected shape of the entire frame.

77

B

C

DA

3m4 m

10 kN4 m

2 m

2EI

2.5EI EI

Ax Dx

Ay Dy

MDCMAB

B´

∆

∆

∆ CD C´∆BC

� Overview


θA = θD = 0


� Unknowns

θB and ∆

- Entire Frame

*)2(010:0 −−−=−−=Σ→+

xxx DAF

*)1(0:0 −−−=+=Σ BCBAB MMM

B- Joint B

MBCMBA

78

B

C

DA 3m4 m

10 kN4 m

2 m

2EI

2.5EI EI36.87° = ∆ tan 36.87° = 0.75 ∆

= ∆ / cos 36.87° = 1.25 ∆

B´

∆

∆∆BC

∆ CD C´

∆

∆BC

∆CD

C

C´

36.87°

)1(5375.04

)(2−−−+∆+= EIEIM BAB θ

)2(5375.04

)(4−−−−∆+= EIEIM BBA θ

)3(2813.04

)2(3−−−∆−= EIEIM BBC θ

)4(375.0 −−−∆= EIM DC

C

C´

BB´

∆BC= 0.75 ∆

0.375EI∆

6EI∆/(4) 2 = 0.375EI∆

B

A

∆´ B´

(6)(2EI)(0.75∆)/(4) 2 = 0.5625EI∆0.5625EI∆

D

C

C´∆CD= 1.25 ∆

(1/2) 0.75EI∆

(1/2) 0.5625EI∆

PL/8 = 5

5

(6)(2.5EI)(1.25∆)/(5)2 = 0.75EI∆

0.75EI∆

� Slope-Deflection Equation

79

B

C

DA

3m4 m

10 kN4 m

2 m

pin2 EI

2.5 EI EI

Dx= (MDC-(3/4)MBC)/4 ---(6)

MBC

4

Ax = (MBA+ MAB-20)/4 -----(5)

B CMBC

C

D

MBC/4MDC

MBC

4

A

B

MBA

10 kN

MAB

� Horizontal reactions

80

� Solve equations

*)2(010 −−−=−− xx DA*)1(0 −−−=+ BCBA MM

Equilibrium Conditions:

Slope-Deflection Equation:

Horizontal reactions at supports:



From (7) and (8) can solve;

EIEIB56.1445.1 −

=∆=θ

)6()1(56.1445.1 toinEI

andEI

Substitute B−

=∆=θ

MAB = 15.88 kN�mMBA = 5.6 kN�mMBC = -5.6 kN�mMDC = 8.42 kN�mAx = 7.9 kNDx = 2.1 kN

)1(4

654

)(22 −−−∆

++=EIEIM BAB θ

)2(4

654

)(42 −−−∆

+−=EIEIM BBA θ

)3(4

)75.0)(2(34

)2(32 −−−

∆−=

EIEIM BBC θ

)4(5

)25.1)(5.2(32 −−−

∆=

EIM DC

)5(4

)20(−−−

−+= ABBA

xMMA

)6(443

−−−−

=BCDC

x

MMD

)7(050938.05.2 −−−=−∆+ EIEI Bθ

)8(05334.00938.0 −−−=−∆+ EIEI Bθ

81

BC

DA

Deflected shape

θB=1.45/EI

θB=1.45/EI

Bending-moment diagram

BC

DA 11.19

1.91

5.35

5.46

1.91

∆ ∆

MAB = 11.19 kN�m MBA = 1.91 kN�m MBC = -1.91 kN�m MDC = 5.46 kN�m

Ax = 8.28 kN�m Dx = 1.72 kN�m

B

C

DA

3m4 m

10 kN4 m

2 m

pin2 EI

2.5 EI EI11.19 kN�m

8.27 kN

5.46

1.91

1.91

0.478 kN1.73

0.478 kN

82

Example 10

From the frame shown use the moment distribution method to:(a) Determine all the reactions at supports, and also(b) Draw its quantitative shear and bending moment diagrams, and qualitative deflected curve.

A

BC

D

3 m

3m

4m

20 kN/mpin

2EI

3EI

4EI

83

A

BC

D

3 m

3m

4m

20 k

N/m

2EI

3EI

4EI

[FEM]load

� Overview

∆ ∆


θA = θD = 0


� Unknowns

θB and ∆

- Entire Frame

*)2(060:0 −−−=−−=Σ→+

xxx DAF

*)1(0:0 −−−=+=Σ BCBAB MMM

B- Joint B

MBCMBA

84

A

BC

D

3 m

3m

4m

20 k

N/m

2EI

3EI

4EI

[FEM]load

wL2/12 = 15

wL2/12 = 15

A

B C

D

∆ ∆

[FEM]∆∆∆∆

6(2EI∆)/(3) 2= 1.333EI∆

6(2EI∆)/(3) 2= 1.333EI∆ 6(4EI∆)/(4) 2

= 1.5EI∆

1.5EI∆

BBCEIM θ3

)3(3=

∆++= EIEI B 333.115333.1 θ ----------(1)

∆+−= EIEI B 333.115667.2 θ ----------(2)

BEIθ3= ----------(3)

∆= EI75.0 ----------(4)

∆+++= EIEIEIM BAAB 333.1153

)2(23

)2(4 θθ0

∆+−+= EIEIEIM BABA 333.1153

)2(43

)2(2 θθ0

∆+= EIEIM DDC 75.04

)4(3 θ0

(1/2)(1.5EI∆)

� Slope-Deflection Equation

85

C

Dx

MDC

4 m

D

MAB

A

B

60 kN

MBA

1.5 m

1.5 m

Ax + ΣMC = 0:

:0=Σ+ BM

)6(188.04

−−−∆== EIMD DCx

)5(30889.0333.13

)5.1(60

−−−+∆+=

++=

EIEIA

MMA

Bx

ABBAx

θ

� Horizontal reactions

86

� Solve equation

*)1(0 −−−=+ BCBA MM

Equilibrium Conditions

Equation of moment

Horizontal reaction at support



From (7) and (8), solve equations;

EIEIB67.3451.5

=∆−

=θ

)6()1(67.3451.5 toinEI

andEI

Substitute B =∆−

=θ

)7(15333.1667.5 −−−=∆+ EIEI Bθ

)8(30077.1333.1 −−−−=∆−− EIEI Bθ

*)2(060 −−−=−− xx DA

)1(333.115333.1 −−−∆++= EIEIM BAB θ

)2(333.115667.2 −−−∆+−= EIEIM BBA θ

)3(3 −−−= BBC EIM θ

)4(75.0 −−−∆= EIM DC

)6(188.0 −−−∆= EIDx

)5(30889.0333.1 −−−+∆+= EIEIA Bx θ

mkNM AB •= 87.53mkNM BA •= 52.16

mkNM BC •−= 52.16mkNM DC •= 0.26

Ax = 53.48 kNDx = 6.52 kN

87

A

C

DMoment diagram

D

A

B C

Deflected shape

∆ ∆

A

BC

D

3 m

3m

4m20 k

N/m 16.52 kN�m

53.87 kN�m

53.48 kN

6.52 kN26 kN�m

5.55 kN

5.55 kN

26.

16.52

mkNM AB •= 87.53mkNM BA •= 52.16

mkNM BC •−= 52.16mkNM DC •= 0.26

Ax = 53.48 kNDx = 6.52 kN

53.87

B16.52

1

! Member Stiffness Factor (K)! Distribution Factor (DF)! Carry-Over Factor! Distribution of Couple at Node! Moment Distribution for Beams

! General Beams! Symmetric Beams

! Moment Distribution for Frames: No Sidesway! Moment Distribution for Frames: Sidesway

DISPLACEMENT MEDTHOD OF ANALYSIS: MOMENT DISTRIBUTION

2

Internal members and far-end member fixed at end support:

C D

K(BC) = 4EI/L2, K(CD) = 4EI/L3

COF = 0.5

1

Member Stiffness Factor (K) & Carry-Over Factor (COF)

P

A C

w

B EI D

CB

L1 /2 L1/2 L2 L3

LEIK 4

=LEIkCC

4= L

EIkDC2

=

COF = 0.5

B C

3

Far-end member pinned or roller end support:

C D

K(AB) = 3EI/L1, K(BC) = 4EI/L2, K(CD) = 4EI/L3

COF = 00

1LEIK 3

= LEIkAA

3=

P

A C

w

BEI

D

CB

L1 /2 L1/2 L2 L3

4

K(CD) = 4EI/L3

Joint Stiffness Factor (K)

P

A C

w

BEI

D

CB

L1 /2 L1/2 L2 L3

K(AB) = 3EI/L1 K(BC) = 4EI/L2,

Kjoint = KT = ΣΣΣΣKmember

5

01DF DFBC DFCB DFCDDFBA)

Notes:- far-end pined (DF = 1)- far-end fixed (DF = 0)

B C DA

Distribution Factor (DF)

P

A C

w

BEI

D

CB

L1 /2 L1/2 L2 L3

KKDF

Σ=

KAB/(K(AB) + K(BC) ) KBC/(K(AB) + K(BC) )

K(BC)/(K(BC) + K(CD) )

K(CD)/(K(BC) + K(CD) )

6

CB(DFBC) CB( DFBC)

CB(DFBC) CB( DFBC)

CO=0.5CO=0

Distribution of Couple at Node

CB

B

A CB D

CB

L1 /2 L1/2 L2 L3

(EI)3(EI)2(EI)1

B C DA01DF DFBC DFCA DFCDDFBA

7

50(.333)50(.333)

50(.667) 50(.667) CO=0.5CO=0

B C DA

01DF 0.667 0.5 0.50.333

L1= L2 = L3

50 kN�m

B

16.67

A CB3EI2EI

4 m 4 m 8 m 8 m

3EID

50 kN�m

8

P

A C

w

B DL1 /2 L1/2 L2 L3

B C DA01DF DFBC DFCB DFCDDFBA

L1= L2 = 8 m, L3 = 10 m

Distribution of Fixed-End Moments

MF

B

MF

MF(DFBC) MF( DFBC)

MF(DFBC) MF( DFBC)

MF

0

MF

0.5 0

B

(EI)13(EI)2(EI)1

9

P

A C

w

B DL1 /2 L1/2 L2 L3

B

145.6

8.4

5.6

8.4

4.2

B C DA01DF 0.6 0.5 0.50.4

L1= L2 = 8 m, L3 = 10 m

1630

0

30 16

14

0.5 0

wL22/12=161.5PL1/8=30 wL3

2/12=25

EI

B

10

Moment Distribution for Beams

11

20 kN

A CB3EI2EI

4 m 4 m 8 m

3 kN/m

Example 1

The support B of the beam shown (E = 200 GPa, I = 50x106 mm4 ).Use the moment distribution method to:(a) Determine all the reactions at supports, and also(b) Draw its quantitative shear and bending moment diagrams,and qualitative deflected shape.

12

20 kN

A CB3EI2EI

3 kN/m

4 m 4 m 8 m

K1 = 3(2EI)/8 K2 = 4(3EI)/8

161630

DF 0.3331 0.667 0

K1/(K1+ K2) K2/(K1+ K2)

[FEM]load 16 -16-30

4.662 9.338Dist. 4.669CO

-25.34 -11.3325.34ΣΣΣΣ

20 kNA B B C

24 kN

13.17 kN 6.83 kN

11.33 kN�m25.34 kN�m

10.25 kN13.75 kN

0.5 0 0.5 0CO

13

20 kN

A CB3EI2EI

3 kN/m

4 m 4 m 8 m

161620+10

+ ΣMB = 0: -MBA - MBC = 0

(0.75 + 1.5)EIθB - 30 + 16 = 0

θB = 6.22/EI

MBC = 25.33 kN�m

MBA = -25.33 kN�m,

MBA MBC

B

Note : Using the Slope Deflection

)1(308

)2(3−−−−= BBA

EIM θ

)2(168

)3(4−−−+= BBC

EIM θ

mkNEIM BCB •−=−= 33.11168

)3(2 θ

14

20 kN

A C

B

3 kN/m

4 m 4 m 8 m

V (kN)x (m)

6.83

-13.17

13.75

-10.25

+-

+-4.58 m

M (kN�m) x (m) M (kN�m) x (m)

-11.33

-25.33

27.326.13+

-+ -


11.336.83 kN 13.17 + 13.75 = 26.92 kN 10.25 kN

15

10 kN

A C

5 kN/m

B3EI2EI

4 m 4 m 8 m 8 m

3EID

50 kN�m

Example 2

From the beam shown use the moment distribution method to:(a) Determine all the reactions at supports, and also(b) Draw its quantitative shear and bending moment diagrams, and qualitative deflected shape.

16

10 kN

A C

5 kN/m

B2EI

4 m 4 m 8 m 8 m

3EID

50 kN�m

26.67 26.6715 400.5 0.50.5

Joint couple -16.65 -33.35

K1 = 3(2EI)/8 K2 = 4(3EI)/8 K3 = 3(3EI)/8

0.571DF 0.3331 0.4280.667 1

K1/(K1+ K2) K2/(K1+ K2) K2/(K2+ K3) K3/(K2+ K3)

50(.333)

50(.667)

50(.333)

50(.667) CO=0.5CO=050 kN�m

B

17

10 kN

A C

5 kN/m

B2EI

4 m 4 m 8 m 8 m

3EID

50 kN�m

26.67 26.6715 400.5 0.50.5

-26.667FEM 40-15 26.667 1.905 1.437Dist. -3.885 -7.782

2.218 1.673Dist. -0.317 -0.636

0.181 0.137Dist. -0.369 -0.740

Joint couple -16.65 -33.35CO -16.675

-3.891CO 0.953

-0.318 1.109CO

K1 = 3(2EI)/8 K2 = 4(3EI)/8 K3 = 3(3EI)/8

-43.28 43.25 -36.22 -13.78ΣΣΣΣ

0.571DF 0.3331 0.4290.667 1

K1/(K1+ K2) K2/(K1+ K2) K2/(K2+ K3) K3/(K2+ K3)

18

10 kN

A B

ByLAy

36.22 kN�m43.25 kN�m

C D

DyCyR

40 kN

13.78 kN�m

B C

CyLByR

40 kN43.25 kN�m

10 kN

A C

5 kN/m

B2EI

4 m 4 m 8 m 8 m

3EID

50 kN�m

13.78 43.2536.22 43.25

= 25.41 kN = 14.59 kN= 0.47 kN = 9.53 kN

= 12.87 kN = 27.13 kN

19

10 kN

AC

5 kN/m

B2EI 3EID

50 kN�m

4 m 4 m 8 m 8 m

M(kN�m) x (m)

21.29

1.88

-36.22

13.7830.32

-43.25Deflected shape

x (m)

V (kN) x (m)0.47

-9.53

12.87

-14.59-27.13

25.41

2.57 m

2.92 m

9.53+12.87=22.4 kN0.47 kN 14.59 kN

27.13+25.41=52.54 kN

20

3 m 3 m 9 m 3 m

50 kN�m

B

A

40 kN 40 kN

C

D

10 kN/m2EI

EI

Example 3

From the beam shown use the moment distribution method to:(a) Determine all the reactions at supports, and also(b) Draw its quantitative shear and bending moment diagrams, and qualitative deflected shape.

21

3 m 3 m 9 m 3 m

50 kN�m

B

A

40 kN

C

D

10 kN/m2EI

EI

DF 0.800 0.20 1

K1/(K1+ K2) K2/(K1+ K2)

30 30 101.25

K1 = 4(2EI)/6 K2 = 3(EI)/9

0.5 0.5

Dist. -9 -2.25

Joint couple 40 10

Dist.FEM 30 101.25-30CO 20 -60

-4.5CO

1 45.5 49 -120ΣΣΣΣ

-120

120 kN�m 40 kN

22

3 m 3 m 9 m 3 m

50 kN�m

B

A

40 kN 40 kN

C

D

10 kN/m2EI

EI

= 27.75 kN = 12.25 kN

1

120 1204945.5

= 37.11 kN = 52.89 kN

= 40 kNByLAy

45.5 kN�m

40 kN

A B1 kN�m

CyLByR

49 kN�m

B C

90 kN120 kN�m

120 kN�m

C D

CyR

40 kN

23

50 kN�m

B

A

40 kN 40 kN

C

D

10 kN/m2EI

EI

3 m 3 m 9 m 3 m12.25+37.11 = 49.36 kN

27.75 kN52.89+40 = 92.89 kN

V (kN) x (m)

27.75

-12.25

37.11 40

-52.893.71 m

+ +

--

+

M(kN�m) x (m)

-45.5 -49

-120

19.8437.75

--

+-

+ 1

Deflected shape

x (m)

24

20 kN

A CB3EI2EI

4 m 4 m 8 m

12 kN�m 3 kN/m15 kN�m

Example 4

The support B of the beam shown (E = 200 GPa, I = 50x106 mm4 ) settles 10 mm.Use the moment distribution method to:(a) Determine all the reactions at supports, and also(b) Draw its quantitative shear and bending moment diagrams, and qualitative deflected shape.

25

20 kN

A CB

2EI


3EI4 m 4 m 8 m10 mm 161630

0.50.5

12 kN�m15 kN�m

[FEM]∆∆

A

B ∆BC

2

)2(6LEI ∆

375.92

)2(6)2(622 =

∆−

∆LEI

LEI

125.28)3(62 =

∆LEI 125.28)3(6

2 =∆

LEI

DF 0.3331 0.667 0

K1/(K1+ K2) K2/(K1+ K2)

Joint couple -12 5 10

5[FEM]load 16 -16-30[FEM]∆ -28.125 -28.125 9.375

-6CO

12.90 25.85Dist. 12.92CO

-8.72 -26.20-12 23.72ΣΣΣΣ

K1 = 3(2EI)/8 K2 = 4(3EI)/8

26

20 kN

A CB

2EI


3EI

4 m 4 m 8 m10 mm16(16)load(20+10)load

Note : Using the slope deflection

)2(75.18208

)2(48

)2(2−−−+−+= BABA

EIEIM θθ

)2(2/12375.9308

)2(3:2

)1()2( aEIM BBA −−−−+−=− θ

)1(75.18208

)2(28

)2(4−−−−++= BAAB

EIEIM θθ

-12

(9.375 )∆

28.125

(28.125)∆

18.75-9.375

+ ΣMB = 0: - MBA - MBC + 15 = 0

(0.75 + 1.5)EIθB - 38.75 - 15 = 0

θB = 23.9/EI

MBC = 23.72 kN�m

MBA = -8.7 kN�m,

MBA MBC

B

15 kN�m

mkN

EIM BCB

•−=

−−=

2.26

125.28168

)3(2 θ

)3(125.28168

)3(4−−−−+= BBC

EIM θ

)4(125.28168

)3(2−−−−−= BCB

EIM θ

27

4 m 4 m 8 m

20 kN

A CB2EI


3EI

= 11.69 kN = 12.31 kN= 7.41 kN = 12.59 kN

23.725 8.725 26.205

20 kN

A B

ByLAy

12 kN�m8.725 kN�m 26.205 kN�m

23.725 kN�m

B C

CyByR

24 kN

28

Deflectedshape x (m)

20 kN

A CB2EI


3EI

4 m 4 m 8 m

26.2057.41 kN 12.59+11.69 = 24.28 kN 12.31 kN

V (kN)x (m)

7.41

-12.59

11.69

-12.31

+-

+-3.9 m

M (kN�m) x (m) M (kN�m) x (m)

12

-0.93

41.64

-8.725

-26.205-23.725

+

--

10 mm θB = 23.9/EI

29

Example 5

For the beam shown, support A settles 10 mm downward, use the momentdistribution method to(a)Determine all the reactions at supports(b)Draw its quantitative shear, bending moment diagrams, and qualitativedeflected shape.Take E= 200 GPa, I = 50(106) mm4.

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

10 mm

30

6 kN/m

B A

C3 m 3 m2EI 1.5EI

12 kN�m

10 mm

0.01 mC

A mkN •=

××

1003

)01.0)(502005.1(62

MF∆

100-(100/2) = 50

4.5

4.5+(4.5/2) = 6.75

DF 0.640 0.36 1

K1/(K1+ K2) K2/(K1+ K2)

-40.16 12-20.08 40.16ΣΣΣΣ

K1 = 4(2EI)/3 K2 = 3(1.5EI)/3

0.50.5

Joint couple 12

6CO

-20.08CO-22.59Dist. -40.16

[FEM]∆ 50[FEM]load 6.75

31

kN08.203

08.2016.40=

+kN08.20

B C40.16 kN�m20.08 kN�m

6 kN/m

AC

12 kN�m

40.16 kN�m

18 kN

8.39 kN26.39 kN

6 kN/m

B A

C3 m 3 m2EI 1.5EI

12 kN�m

-10.16 12-20.08 40.16ΣΜΣΜΣΜΣΜ

32

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

10 mm

6 kN/m

AC

12 kN�m

40.16 kN�m8.39 kN26.39 kN

B C40.16 kN�m20.08 kN�m

20.08 kN20.08 kN

M (kN�m)

x (m)20.08

-40.16

12

Deflected shape

x (m)

V (kN)

x (m)

26.398.39

-20.08

+-

33

Example 6

For the beam shown, support A settles 10 mm downward, use the momentdistribution method to(a)Determine all the reactions at supports(b)Draw its quantitative shear, bending moment diagrams, and qualitativedeflected shape.Take E= 200 GPa, I = 50(106) mm4.

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

10 mm

34

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

10 mm

6 kN/m

B A

C

12 kN�m

10 mmR

� Overview

B A

R´

∆

C×

*)1(0' −−−=+ CRR

35

� Artificial joint applied 6 kN/m

B A

C3 m 3 m2EI 1.5EI

12 kN�m

10 mm

0.01 mC

A mkN •=

××

1003

)01.0)(502005.1(62

MF∆

100-(100/2) = 50

4.5

4.5+(4.5/2) = 6.75

DF 0.640 0.36 1

K1/(K1+ K2) K2/(K1+ K2)

-40.16 12-20.08 40.16ΣΣΣΣ

K1 = 4(2EI)/3 K2 = 3(1.5EI)/3

0.50.5

Joint couple 12

6CO

-20.08CO-22.59Dist. -40.16

[FEM]∆ 50[FEM]load 6.75

36

kN08.203

08.2016.40=

+kN08.20

B C40.16 kN�m20.08 kN�m

6 kN/m

AC

12 kN�m

40.16 kN�m

18 kN

8.39 kN26.39 kN

6 kN/m

B A

C3 m 3 m2EI 1.5EI

12 kN�m

-40.16 12-20.08 40.16ΣΜΣΜΣΜΣΜ

C

40.16 kN�m40.16 kN�m

26.39 kN20.08R

039.2608.20:0 =+−−=Σ↑+ RFy

kNR 47.46=

37

B A C

3 m 3 m2EI 1.5EI 0.50.5

� Artificial joint removed

R´

∆

∆C

B

EI75

=∆→1003

)2(62 =

∆CEI

100 753

)75)(5.1(62 =EI

EI∆C

A75-(75/2)= 37.5

DF 0.640 0.36 1-100 -100[FEM]∆ +37.5

22.5Dist. 4020CO

-60-80 60ΣΣΣΣ

B C60 kN�m80 kN�m

AC60 kN�m

46.67 kN46.67 kN 20 kN20 kN C 20 kN46.67R´

kNRFy 67.66':0 ==Σ↑+

38

� Solve equation

*)1(67.66'47.46 inkNRandkNRSubstitute ==

6970.0067.6647.46

−==+

CC

6 kN/mB A C2EI 1.5EI

12 kN�m

35.68 kN�m

12.45 kN 5.55 kN

6970.0−=×C

6 kN/mB

A

C

12 kN�m

R = 46.47 kN

20.08 kN�m

20.08 kN10 mm

8.39 kN

BA

R´ = 66.67 kN

∆80 kN�m

46.67 kN 20 kN

39

10 mm

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

35.68 kN�m

12.45 kN 5.55 kN

Deflected shape

x (m)

M (kN�m)

x (m)1214.57

1.67

-35.68

-+

V (kN)

x (m)0.925 m12.45

-5.55

+

40

P P

A CB DL´ L L´

real beam

Symmetric Beam

� Symmetric Beam and Loading

θ θ

+ ΣMC´ = 0: 0)2

)(()(' =+−LL

EIMLVB

EIMLVB 2' == θ

θLEIM 2

=

The stiffness factor for the center span is, therefore,

LEIK 2

=

L2

L2

C´

V´C

B´

V´B

MEI

LMEI

conjugate beam

41

+ ΣMC´ = 0: 0)3

2)(2

)((21)(' =+−

LLEIMLVB

EIMLVB 6' == θ

θLEIM 6

=

The stiffness factor for the center span is, therefore,

� Symmetric Beam with Antisymmetric Loading

P

P

ACB

D

L´ L L´

real beam

θ

θ

conjugate beam

V´C

V´BB´

C´

MEI

MEI

)2

)((21 L

EIM

)2

)((21 L

EIM

L32

LEIK 6

=

42

Example 5a

Determine all the reactions at supports for the beam below. EI is constant.

A

B4 m

D

C6 m 4 m

15 kN/m

43

A

B4 m

D

C6 m 4 m

15 kN/m

,4

33)(

EILEIK AB ==

622

)(EI

LEIK BC ==

,1)()(

)( ==AB

ABAB K

KDF ,692.0

)6/2()4/3()4/3()(

)()(

)( =+

=+

=EIEI

EIKK

KDF

BCAB

ABBA

308.0)6/2()4/3(

)6/2()()()(

)( =+

=+

=EIEI

EIKK

KDF

BCAB

BCBC

[FEM]load 0 -16 +45

-20.07 -8.93Dist.

wL2/12 = 45wL2/15 = 16

DF 0.692 0.3081.0

A B4 m

30 kN83

B C

90 kN

3 m 3 mDC

4 m

30 kN8336.07 kN�m 36.07 kN�m

0.98 kN 29.02 kN

-36.07 +36.07ΣΜ

45 kN 45 kN 29.02 kN 0.98 kN

wL2/12 = 45wL2/15 = 16 wL2/15 = 16

44

A B4 m

30 kN83

B C

90 kN

3 m 3 mDC

4 m

30 kN8336.07 kN�m 36.07 kN�m

29.02 kN 45 kN 45 kN 29.02 kN 0.98 kN0.98 kN

A

B4 m

D

C6 m 4 m

15 kN/m

74.02 kN 74.02 kN 0.98 kN0.98 kN

M (kN�m) x (m)

+

--

-36.07 -36.07

31.42

Deflectedshape

V(kN) x (m)

-29.02

45

-45

29.02

-0.98

0.98

45

Example 5b

Determine all the reactions at supports for the beam below. EI is constant.

A

B

DC

15 kN/m

15 kN/m

4 m 3 m 4 m3 m

46

A

B

DC

15 kN/mFixed End Moment

wL2/15 = 16 11wL2/192 = 30.938

5wL2/192 = 14.063

A

B

DC

15 kN/m

wL2/15 = 165wL2/192 = 14.063

11wL2/192 = 30.938

A

B

DC

15 kN/m

15 kN/m16.87516

16.875 16

47

A B4 m

30 kN83

B C

45 kN

45 kN

DC4 m

30 kN

83

A

B

DC

15 kN/m

15 kN/m

4 m 3 m 4 m3 m

[FEM]load 0 -16 16.875

-0.375 -0.50Dist.

DF 0.429 0.5711.0

,75.04

33)( EIEI

LEIK AB === EIEI

LEIK BC ===

666

)(

,1)( =ABDF ,429.0175.0

75.0)( =+

=BADF 571.0175.0

1)( =+

=BCDF

16.87516

16.875 16

-16.37516.375ΣΜ

16.375 kN�m 16.375 kN�m

24.09 kN5.91 kN 27.96 kN27.96 kN 24.09 kN

5.91 kN

16.87516

48

A B4 m

30 kN83

B C

45 kN

45 kN

16.375 kN�m 16.375 kN�m

24.09 kN5.91 kN 27.96 kN27.96 kN

DC4 m

30 kN

83

24.09 kN5.91 kN

A

B

DC

15 kN/m

15 kN/m52.05 kN 52.05 kN 5.91 kN5.91 kN

V(kN) x (m)

5.91 5.91

-24.09 -24.09

27.96 27.96

-16.375

M (kN�m) x (m)

16.375

Deflected shape

49

Moment Distribution Frames: No Sidesway

50

AB C

D

4 m

48 kN 8 kN/m40 kN�m

5 m5 m

3EI

2.5EI 2.5EI

Example 6

From the frame shown use the moment distribution method to:(a) Determine all the reactions at supports(b) Draw its quantitative shear and bending moment diagrams,and qualitative deflected shape.

51

40 kN�m

AB C

D

4 m

48 kN 8 kN/m

5 m5 m

3EI

2.5EI 2.5EI45 25

KAB = KBC = 3(2.5EI)/5 = 1.5 EI

KBD = 4(3EI)/4 = 3EI

0.50.5

0.5

0.25 0.25DF 0.51 0 1

A B D C

BA BCMember BDAB DB CB

5 5Dist. 10

5CO

-10 -10Joint load -20

-10CO-45FEM 25

20 00 -5ΣΣΣΣ -50 -10

40 kN�m

52

AB C

D


3EI

2.5EI 2.5EI2010

50

16 kN24 kN

50 kN�m

48 kNA B 0

14 kN 34 kN

3.75

5820

40 kNB C 3.75

3.75

3.75

40 kN�m

20

10

50243458 kN

14 kN 16 kN

Member BD DB CBBC20 0

AB0 -5ΣΣΣΣ

BA-50 -10

5

3.75 kN

3.75 kN10 kN�m

5

58

58

53

Moment diagram Deflected shape

AB C

D

4 m


5 m5 m

3EI

2.5EI 2.5EI

58 kN

14 kN 16 kN

1635

5020

10

5

5

50

20

10

54

Moment Distribution for Frames: Sidesway

55

Single Frames

A

B C

D

P∆∆

A

BC

D

P

Artificial joint applied(no sidesway)

R´

A

BC

D

Artificial joint removed(sidesway)

0 = R + C1R´

x C1

RR´

x C1

R

56

0 =R2 + C1R2´ + C2R2´´

0 =R1 + C1R1´ + C2R2´´

P2

P1

P3

P4

∆2

∆1

P2

P1

P3

P4

R2

R1

R2

R1

R1´x C1

∆´∆´R2´

R1´x C1

R2´

x C1

R2´´

x C2

R1´´

R2´´

x C2

R1´´

x C2

∆´´ ∆´´

Multistory Frames

57

A

B C

D

5 m

16 kN

1 m 4 m

Example 7

From the frame shown use the moment distribution method to:(a) Determine all the reactions at supports, and also(b) Draw its quantitative shear and bending moment diagrams, and

qualitative deflected shape.EI is constant.

58

A

BC

D

+

artificial joint removed( sidesway)

A

BC

D

16 kN

=

artificial joint applied(no sidesway)

x C1

A

B C

D

5 m

16 kN

1 m 4 m

5 m

R´R

� Overview

R + C1R´ = 0 ---------(1)

59

� Artificial joint applied (no sidesway)

A

BC

D

16 kN

Ra = 4 mb = 1 m

Pb2a/L2 = 2.56

Pa2b/L2 = 10.24

Fixed end moment:

Equilibrium condition :

ΣFx = 0:+ Ax + Dx + R = 0

60

5 m

1 m 4 m

5 m

A

BC

D

16 kN

R

2.5610.24

0.5

0.5

0.5

0.50 0.50DF 0.500 0.50 0FEM -2.5610.24Dist. 1.28 -5.12-5.12 1.28CO -2.56-2.56 0.64 0.64Dist. 1.28 -0.32-0.32 1.28

CO -0.16-0.16 0.64 0.64Dist. 0.08 -0.32-0.32 0.08CO -0.16-0.16 0.04 0.04

Dist. 0.08 -0.02-0.02 0.08

5.78 1.32-2.88 2.72ΣΣΣΣ -5.78 -2.72

A B C D

2.5610.24

Ax = 1.73 kNDx = 0.81 kN

5.78 kN�m

2.88 kN�m

2.72 kN�m

1.32 kN�m

A

B

5 m

D

C

5 m

1.32-2.88 2.72-5.78

ΣFx = 0:+ 1.73 - 0.81 + R = 0

R = - 0.92 kN


61

� Artificial joint removed ( sidesway)

Fixed end moment:

5 m

5m

5 m

A

B C

D

R´

∆ ∆

100 kN�m

100 kN�m

100 kN�m

100 kN�m

Since both B and C happen to be displaced the same amount ∆, and AB and DChave the same E, I, and L so we will assume fixed-end moment to be 100 kN�m.


ΣFx = 0:+ Ax + Dx + R´ = 0

62

5 m

5m

5 m

A

B C

D

R´

∆ ∆

100 kN�m

100 kN�m

100 kN�m

100 kN�m0.5

0.5

0.5

0.50 0.50DF 0.500 0.50 0

A B C D

FEM 100100 100 100Dist. -50-50-50 -50CO -25.0-25.0 -25.0 -25.0Dist. 12.512.5 12.5 12.5

CO 6.56.5 6.5 6.5Dist. -3.125-3.125-3.125 -3.125CO -1.56-1.56 -1.56 -1.56

Dist. 0.780.78 0.78 0.78CO 0.390.39 0.39 0.39

Dist. -0.195-0.195-0.195 -0.195

-60 8080 60ΣΣΣΣ 60 -60

Ax = 28 kNDx = 28 kN

60 kN�m

80 kN�m

A

B

5 m

60 kN�m

80 kN�m

D

C

5 m

100 kN�m

100 kN�m

100 kN�m

100 kN�m

ΣFx = 0:+

-28 - 28 + R´ = 0

R´ = 56 kN

Equilibrium condition:

63

-0.92 + C1(56) = 0

R + C1R´ = 0

Substitute R = -0.92 and R´= 56 in (1) :

C1 =0.9256

A

B C

D

16 kN

R

1.73 kN 0.81

2.72

5.785.78

2.88

2.72

1.32

x C1

A

BC

D

R´

+

28 28

6060

60

80

60

80

A

BC

D

5 m

16 kN

1 m 4 m

5 m=

3.714.79

4.79

1.57

3.71

2.63

1.27 kN1.27

64

Bending moment diagram (kN�m)

A

BC

D

5 m

16 kN

1 m 4 m

5 m

3.714.79

4.79

1.57

3.71

2.63

1.27 kN 1.27 kN

8.22

3.71

2.99 kN13.01 kN∆ ∆

Deflected shape

1.57

4.794.79

3.71

2.63

65

Example 8

From the frame shown use the moment distribution method to:(a) Determine all the reactions at supports, and also(b) Draw its quantitative shear and bending moment diagrams, and qualitative deflected shape.

A

BC

D

3 m

3m

4m

20 kN/mpin

2EI

3EI

4EI

66

A

BC

D

3 m , 2EI

3m, 3EI

4m , 4EI

20 kN/m

=

A

B C

D


� Overview

R + C1R´ = 0 ----------(1)

x C1+

A

B C

D

artificial joint removed(sidesway)

R R´

67

R

A

B C

D

3 m , 2EI

3m , 3EI

4m, 4EI

20 kN/m

0.471 0.529DF 1.000 1.00 0FEM 15.00 -15.00Dist. 7.9357.065CO 3.533

0.5 0.5

0.5

15

15

KCD = 3(4EI)/4 = 3EI

KBA = 4(2EI)/3 = 2.667EI

KBC = 3(3EI)/3 = 3EI

A B C D

Ax = 33.53 kN

7.9418.53ΣΣΣΣ -7.94

7.94 kN�m

18.53 kN�m

A

B

3 m60


Dx = 0

0

D

C

4 m

ΣFx = 0: +

60 - 33.53 - 0 + R = 0

R = - 26.47 kN

68

� Artificial joint removed ( sidesway)

� Fixed end moment

R´

A

B C

D

3 m, 2EI

3m, 3EI

4m, 4EI

6(2EI∆)/(3) 2

6(2EI∆)/(3) 2

3(4EI∆)/(4) 2

∆ ∆

Assign a value of (FEM)AB = (FEM)BA = 100 kN�m

1003

)2(62 =

∆EI

∆AB = 75/EI

100 kN�m

100 kN�m

R´

A

B C

D

3 m, 2EI

3m, 3EI

4m, 4EI

100 kN�m

100 kN�m 3(4EI)(75/EI)/(4) 2 = 56.25 kN�m

∆ ∆

6(4EI)∆/(4) 2

69

R´

A

B C

D

3 m

3m

4m

0.471 0.529DF 1.000 1.00 0

CO -28.55

0.50.5

0.5

A B C D

Ax = 43.12kN

ΣFx = 0:+

-43.12 - 14.06 + R´ = 0

R´ = 57.18 kN

100

100

56.25

FEM 100 56.25 100Dist. -52.9-47.1 0

14.06 kN

∆ ∆

-52.976.45ΣΣΣΣ 52.9 56.25

56.25 kN�m

D

C

4 m

52.9. kN�m

76.45 kN�m

A

B

3 m

70

=

A

BC

D

3 m

3m

4m

20 kN/m

16.55 kN�m

53.92 kN�m

53.49 kN

6.51 kN

26.04 kN�m

R + C1R´ = 0

-26.47 + C1(57.18) = 0

C1 =26.4757.18

x C1+

R´

A

B C

D

52.9

52.9 kN�m

76.45 kN�m

43.12 kN

14.06

56.25

R

A

B C

D

33.53 kN

7.94 kN�m7.94 kN�m

18.53 kN�m

0

0

Substitute R = -26.37 and R´= 57.18 in (1) :

5.52 kN

5.52 kN

71

A

C

DMoment diagram

D

A

B C

Deflected shape

∆ ∆

A

BC

D

3 m

3m

4m

20 kN/m

16.55 kN�m

53.92 kN�m

53.49 kN

6.51 kN

26.04 kN�m

5.52 kN

5.52 kN 26.04

53.92

16.55

B16.55

72

BC

DA

3m4 m

10 kN

4 m

Example 8

From the frame shown use the moment distribution method to:(a) Determine all the reactions at supports, and also(b) Draw its quantitative shear and bending moment diagrams, and qualitative deflected shape.

EI is constant.

73

B C

DA

10 kN

=


x C1

R´

BC

DA

3m4 m

10 kN

4 m

R

+

B C

DA

artificial joint removed(sidesway)

R + C1R´ = 0 ---------(1)

� Overview

74


B C

DA

10 kNR

00

ΣFx = 0:+

10 + R = 0

R = - 10 kN


75

� Artificial joint removed (sidesway)

� Fixed end moment

R´B

C

DA

3m4 m

4 m= ∆ tan 36.87° = 0.75 ∆= 0.75(266.667/EI) = 200/EI

∆CD = ∆ / cos 36.87° = 1.25 ∆ = 1.25(266.667/EI) = 333.334/EI

∆AB = ∆

∆BC

∆CD

C

C´

36.87°

B´

∆

∆

C´∆ CD

36.87°

∆BC

R´B

C

DA

3m4 m

4 m

6EI∆AB/(4) 2

6EI∆AB/(4) 2

6EI∆BC/(4) 2

3EI∆CD/(5) 2

6EI∆CD/(5) 2

Assign a value of (FEM)AB = (FEM)BA = 100 kN�m : 1004

62 =∆ ABEI

∆AB = 266.667/EI

100 kN�m 100 kN�m

76

R´B

C

DA

100 kN�m

100 kN�m

6EI∆BC/(4) 2 = 6(200)/42 = 75 kN�m

3EI∆CD/(5) 2 = 3(333.334)/52 = 40 kN�m

∆BC= 200/EI, ∆CD = 333.334/EI


ΣFx = 0:+ Ax + Dx + R´ = 0

77

R´B

C

DA

3m4 m

4 m

0.50 0.50DF 0.6250 0.375 1

KBA = 4EI/4 = EI, KBC = 4EI/4 = EI, KCD = 3EI/5 = 0.6EI

100

100

40

7575

0.5

0.5

0.5

A B C D

FEM 100 100 40 -75 -75Dist. -12.5-12.5 13.125 21.875

CO -2.735 1.953 -2.735Dist. -0.977-0.977 1.026 1.709

-81.0591.02ΣΣΣΣ 81.05 -56.48 56.48

B C81.05

56.48

43.02 kN

34.38 kN34.38 kN

C

D

34.38 kN

34.38 kN

56.48

39.91 kNΣFx = 0:+

-43.02 - 39.91 + R´ = 0

R´ = 82.93 kN

A

B

91.02

81.0534.38 kN

34.38 kN

Dist. -5.469-5.469 2.344 3.906CO -6.25 10.938 -6.25

78

BC

DA

3m4 m

10 kN

4 m

5.19 kN

9.77

6.81

10.98

4.15 kN

9.77 6.81

4.15 kN

4.81 kN=

B C

DA

10 kN

00

00

0

R

x C1= 10/82.93+

B C

A 43.02 kN

81.05

56.48

91.02

34.38 kN

81.05 56.48

D

34.38 kN

39.91 kN

R´

C1 = 10/82.93

-10 + C1(82.93) = 0Substitute R = -10 kN and R´= 82.93 kN in (1) :

R + C1R´ = 0 ---------(1)

79

BC

DA

Bending moment diagram(kN�m)

BC

DA

3m4 m

10 kN

4 m

5.19 kN

9.77

6.81

10.98

4.15 kN

9.77 6.81

4.81 kN4.15 kN

BC

DA

Deflected shape

10.98

6.81

9.77

9.77

80

BC

DA

4 m3 m

2m2 m 3 m

40 kN

20 kN

4EI

3EI

4EI

Example 9

From the frame shown use the moment distribution method to:(a) Determine all the reactions at supports, and also(b) Draw its quantitative shear and bending moment diagrams,and

qualitative deflected shape.EI is constant.

81

BC

DA

4 m3 m

2m2 m 3 m

40 kN

20 kN

4EI

3EI

4EI

=x C1

R´

� Overview

R + C1R´ = 0 ----------(1)

CB

DA

2m2 m 3 m

40 kN

20 kN


R

artificial joint removed(no sidesway)

+

B C

DA

2m2 m 3 m

82


B C

DA

2m2 m 3 m

40 kN

20 kN R15+(15/2)= 22.5 kN�m

PL/8 = 15


ΣFx = 0:+ Ax + Dx + R = 0

Fixed end moments:

83

B C

DA

2m2 m 3 m

40 kN

20 kNR

0.60 0.40DF 1.000 1.00 022.5 15

KBA = 4(4EI)/3.6 = 4.444EI, KBC = 3(3EI)/3 = 3EI,

0.5 0.5

0.5

A B C D

Dist. -9.0-13.5CO -6.75

FEM 22.5

13.5-6.75ΣΣΣΣ -13.5

15.5 kN24.5 kN23.08 kN 7.75 kN

ΣFx = 0:+

23.08 + 20 -7.75 + R´ = 0

R´ = - 35.33 kN

24.5 kN

24.5 kN13.5 kN�m

6.75 kN�m

B

A

B C

40 kN13.5

C

D

15.5 kN

0

15.5 kN

84

B C

DA

R´

� Artificial joint removed (sidesway)

Fixed end moments:

3(3EI)∆BC/(3) 2

6(4EI)∆CD/(4.47) 2

3(4EI)∆CD/(4.472) 2

6(3EI)∆BC/(3) 2

6(4EI)∆AB/(3.61) 2

6(4EI)∆AB/(3.606) 2

Assign a value of (FEM)AB = (FEM)BA = 100 kN�m : 10061.3

)4(62 =∆ ABEI , ∆AB = 54.18/EI

B C

DA

3 m

4EI 4EI

R´

3.606

m 4.472 m

3EI

100 kN�m

100 kN�m

85

B C

DA

R´

33.69

o

26.5

7°

33.69o

C´∆ CD

C

B´

∆AB = 54.3/EI

B

EIEIEICBBC /59.52/05.30/54.22'' =+==∆

3(3EI)∆BC/(3) 2 = 3(3EI)(52.59/EI) /(3) 2 = 52.59 kN�m

3(4EI)∆CD/(4.472) 2 = 3(4EI)(50.4/EI)/(4.472) 2 = 30.24 kN�m

100 kN�m

100 kN�m B C

DA

R´

B´

BB´CC´

C´∆ = ∆ABcos 33.69° = 45.08/EI

26.57°

∆ tan 33.69 = 30.05/EI

∆ tan 26.57 = 22.54/EI∆ tan 26.57 = 22.54/EI

∆ tan 33.69 = 30.05/EI

C´∆ CD

C

∆CD = ∆/cos 26.57°= 50.4/EI

86

B C

DA

4 m3 m

2m2 m 3 m

4EI

3EI

4EI

R´

23.8523.8568.34 kN

0.5 0.5

0.5

0.60 0.40DF 1.000 1.00 0

A B C D

Dist. -18.96 -28.45FEM 100 100 30.24-52.59

-71.5585.78ΣΣΣΣ 71.55 30.24

19.49 kN

ΣFx = 0:+

-68.34 - 19.49 + R´ = 0

R´ = 87.83 kN

CO -14.223

23.85 kN

71.55 kN�mB

85.78kN�m

A

23.85 kN

C

D

23.85 kN

30.24 kN�m

23.85 kN

B C71.55

100

100

52.59

30.24

87

-35.33 + C1(87.83) = 0

x C1

Substitute R = -35.33 and R´= 87.83 in (1) :

C1 = 35.33/87.83B C

DA

40 kN

20 kN35.33 kN

6.75 kN�m23.08 kN

24.5 kN

13.5 kN�m

7.75 kN

0

15.5 kN

+

90.59 kNB C

D

A

71.55 kN�m

85.78 kN�m

68.34 kN23.85 kN

19.485 kN

23.85 kN

30.24 kN�m

=B

C

DA

40 kN20 kN

15.28 kN�m

27.76 kN�m4.41 kN

14.91 kN15.59 kN

25.09 kN

12.16 kN�m

88


BC

DA

BC

DA

40 kN20 kN

15.28 kN�m

27.76 kN�m

4.41 kN

14.91 kN15.59 kN

25.09 kN

12.16 kN�m

37.65

Deflected shape

B

DA

C

12.1627.76

15.28

1

! Fundamentals of the Stiffness Method! Member Local Stiffness Matrix! Displacement and Force Transformation Matrices! Member Global Stiffness Matrix! Application of the Stiffness Method for Truss

Analysis! Trusses Having Inclined Supports, Thermal Changes

and Fabrication Errors! Space-Truss Analysis

TRUSSES ANALYSIS

2

2-Dimension Trusses

3

Fundamentals of the Stiffness Method

� Node and Member Identification

� Global and Member Coordinates

� Degrees of Freedom

12

3 4

2

1

3

(x1, y1)

(x3, y3)

(x2, y2)

(x4, y4)

1

23

4

56

78

34

56

78

x

y

�Known degrees of freedom D3, D4, D5, D6, D7 and D8� Unknown degrees of freedom D1 and D2

4

AE/LAE/L

x djAE/L x d´j

AE/L

x diAE/LAE/L x dí

Member Local Stiffness Matrix

xý´

i

j

qí

q´j

AE/L

AE/L

jii dL

AEdL

AEq ''' −=

AE/LAE/Ld´ i = 1

d´ j = 1

x di

x dj

jij dL

AEdL

AEq ''' +−=

−

−=

j

i

j

i

dd

LAE

qq

''

1111

''

−

−=

1111

]'[L

AEk

[q´] = [k´][d´] ----------(1)

q´j

qí

xý´

xý´

5

x´y´

m

i

j

(xi,yi)

(xj,yj)x

y

Displacement and Force Transformation Matrices

θy

θx

22 )()(cos

ijij

ijijxx

yyxx

xxL

xx

−+−

−=

−== θλ

22 )()(cos

ijij

ijijyy

yyxx

yyL

yy

−+−

−=

−== θλ

6

x

y

Global

m

i

jdjx

djy

dix

diy

djx

djyxý´

m

i

j

Local

dí

d´j

dix

diy

d´j

dí

� Displacement Transformation Matrices

yiyxixi ddd θθ coscos' +=

=

jy

jx

iy

ix

yx

yx

j

i

dddd

dd

λλλλ00

00''

θy

θx

yjyxjxj ddd θθ coscos' +=

=

yx

yxTλλ

λλ00

00][

λx λy

[d´] = [T][d] ----------(2)

7

x

y

θy

θx

m

i

j

x

y

Global

x´y´

m

i

j

Local

q´i

q´j

� Force Transformation Matrices

θy

θx

xiix qq θcos'=

yiiy qq θcos'=

xjjx qq θcos'=

yjjy qq θcos'=

=

j

i

y

x

y

x

jy

jx

iy

ix

qq

qqqq

''

00

00

λλ

λλ

=

y

x

y

x

TT

where

λλ

λλ

00

00

][

λx

λy

[q] = [T]T[q´] ----------(3)

qjx

qjy

qix

qiy

8

Member Global Stiffness Matrix

[ q ] = [ T ]T ([k´][d´] + [q´F] ) = [ T ]T [ k´ ][T][d] + [ T ]T [q´F] = [k][d] + [qF]

[ k ] [ k ] = [ T ]T[ k´ ][T]

[qF] = [ T ]T [q´F]

[q] = [T]T[q´] ----------(3)

Substitute ( [q´] = [k´][d´] + [q´F] ) into Eq. 3, yields the result,

λyλx

λxλx

−λyλx

−λxλx

λyλy

λxλy

−λyλy

−λxλy

λyλx

λxλx

−λyλx

−λxλx

λyλy

λxλy

−λyλy

−λxλy

V U VU

[ k ] = AEL

VUV

U

00λyλx

λyλx00-11

1-1

AEL[ k ] =

00

λy

λx

λy

λx

00

9

[Qa] = [K][D] + [QF]

[Qk] = [K11][Du] + [K12][Dk] + [QF]

Reaction Boundary Condition

Unknown DisplacementJoint Load

Equilibrium Equation:

Partitioned Form:

Application of the Stiffness Method for Truss Analysis

[Du] = (([Qk] - [QF]) - [K12][Dk])[Ku] -1

Qk

Qu

Du

Dk

=K12

K22

K11

K21

+QF

k

QFu

10

+

−

−=

jF

iF

j

i

j

i

qq

dd

LAE

qq

''

''

1111

''

jy

jx

iy

ix

yx

yx

dddd

λλλλ00

00

+

−

−=

jF

iF

jy

jx

iy

ix

yx

yx

j

i

qq

DDDD

LAE

qq

''

0000

1111

''

λλλλ

Member Forces

x

y

θy

θx

x´y´

m

i

j

q´i

q´j

11

+

−−

−−=

jF

iF

jy

jx

iy

ix

yxyx

yxyx

j

i

qq

DDDD

LAE

qq

''

''

λλλλλλλλ

Dyi

Dxi

Dyj

Dxj

q´j = AEL −λx −λy λx λy qj´

F+

x

y

θy

θx

x´y´

m

i

j

Member Forces

q´i

q´j

12

Member Forces

Dyi

Dxi

Dyj

Dxj

qm = AEL −λx −λy λx λy qj´F+

x

y

θy

θx

x´y´

m

i

j

Member Forces

qm

13

3 m

3 m

4 m 4 m

80 kN

50 kN

5 m

5 m5 m

Example 1

For the truss shown, use the stiffness method to:(a) Determine the deflections of the loaded joint.(b) Determine the end forces of each member and reactions at supports.Assume EA to be the same for each member.

14

Ljyy

Lixx ijij

ij

�)(�)(� −+

−=λ

cosθx = λx cosθy = λy

1

2

3

434

56

78

1

2

(0,0)(-4,-3)

(-4,3)

(4,-3)

31

23 m

3 m

4 m 4 m

80 kN

50 kN

5 m

5 m5 m

Member

#1

#2

λx λy

#3

λyλx

λxλx

λyλy

λxλy

λyλx

λxλx

λyλy

λxλy

−λyλx

−λxλx

−λyλy

−λxλy

−λyλx

−λxλx

−λyλy

−λxλy

Vi Uj VjUi

[ k ]m = AEL

Vi

Uj

Vj

Ui

31

2

-4/5 = -0.8

-4/5 = -0.8

4/5 = 0.8

-3/5 = -0.6

3/5 = 0.6

-3/5 = -0.6

15

-0.48

1.08-0.48

1.92-0.48

0.64

0.36

-0.481

2

1 2

[K] = AE5

1

2

3

434

56

78

1

2

31

2

0.48

0.64

0.36

0.48-0.48

-0.64

-0.36

-0.48

-0.48

-0.64

-0.36

-0.48[ k ]1 =

2 3 41

AE5

23

4

1

[ k ]2 =

2 5 61

AE5

25

6

1

-0.48

0.64

0.36

-0.480.48

-0.64

-0.36

0.48

0.48

-0.64

-0.36

0.48

[ k ]3 =

2 7 81

AE5

27

8

1

-0.48

0.64

0.36

-0.480.48

-0.64

-0.36

0.48

0.48

-0.64

-0.36

0.48

31

2

34

56

78

Member

#1

#2

#3

λx

-0.8

- 0.8

0.8

λy

-0.6

0.6

-0.6

λx2 λx λy λy

2

0.64 0.48 0.36

0.64 -0.48 0.36

0.64 -0.48 0.36

0.48

0.64

0.36

0.48

-0.48

0.64

0.36

-0.48

16

34

56

78

1

2

31

2

Global

Q1 = -50

Q2 = -80

D1

D2

D1

D2=

-250.65/AE

-481.77/AE

3 m

3 m

4 m 4 m

80 kN

50 kN

5 m

5 m5 m

1

2

1

-0.48

2

-0.48

1.08

1.92

= AE5

80 kN

50 kN

0

0+1

2

17

1

2

3

434

56

78

1

2

31

2

Local

= -97.9 kN (C)

[q´F]1 = AE5 0.8 0.6 -0.8 -0.6 D2=

D1=

D4=D3=

-481.77/AE-250.65/AE

0.00.0

= +17.7 kN (T)

D2=D1=

D6=D5=

-481.77/AE-250.65/AE

0.00.0

= -17.7 kN (C)

80 kN

50 kN 36.87o

97.9 kN

17.7 kN

17.7 kN

#1 -0.8 -0.6#2 -0.8 0.6#3 0.8 -0.6

[q´F]2 = AE5 0.8 -0.6 -0.8 0.6

[q´F]3 = AE5 -0.8 +0.6 +0.8 -0.6 D2=

D1=

D8=D7=

-481.77/AE-250.65/AE

0.00.0

Member

#1#2#3

λx λy

[ ] [ ]F

yj

xj

yi

xi

xxyxmF q

DDDD

LAEq ']'[ +

−−= λλλλ

18

80 kN

50 kN 36.89o

97.9 kN

17.7 kN

17.7 kN

80 kN

50 kN

Member

#1#2#3

λx

-0.8-0.80.8

λy

-0.60.6-0.6

1

2

3

434

56

78

1

2

31

2

97.9(0.8)=78.32 kN

97.9(0.6)=58.74 kN

17.7(0.8)=14.16 kN

17.7(0.6)=10.62 kN

17.7(0.8)=14.16 kN

17.7(0.6)=10.62 kN

ΣFx ´ = 0:+ 17.7 + 17.7 +50cos 36.89 - 97.9cos73.78 - 80cos53.11 = 0, O.K

Check :

19

Example 2

For the truss shown, use the stiffness method to:(b) Determine the end forces of each member and reactions at supports.(a) Determine the deflections of the loaded joint.The support B settles downward 2.5 mm. Temperature in member BDincrease 20 oC. Take α = 12x10-6 /oC, AE = 8(103) kN.

A

BC

D

8 kN

4 kN

4 m

3 m+20o C

∆B = 2.5 mm

20

λyλx

λxλx

λyλy

λxλy

λyλx

λxλx

λyλy

λxλy

−λyλx

−λxλx

−λyλy

−λxλy

−λyλx

−λxλx

−λyλy

−λxλy

∆B = 2.5 mm

A

BC

D

8 kN

4 kN

4 m

3 m+20o C

Vi Uj VjUi

[ k ]m = AEL

Vi

Uj

Vj

Ui

12

3 4

(0,0)

(-4,-3)

(-4,0)

(0,-3)

2

1

3

1

23

4

56

78

Member

#1

#2

λx λy

#3

-4/4 = -1

-4/5 = -0.8

0

0

-3/5 = -0.6

-3/3 = -1

Ljyy

Lixx ijij

ij

�)(�)(� −+

−=λ


21

0.096

0.128

0.072

0.096

0

0

0.333

0

0

0.25

0

0

0

0.25

0

00

-0.25

0

0

0

-0.25

0

0[k]1 = 8x103

2 3 41

234

1

[k]3 = 8x103

2 7 81

278

1

0

0

0.333

00

0

-0.333

0

0

0

-0.333

0

Member

#1

#2

#3

λx

-1

- 0.8

0

λy

0

-0.6

-1

λx2/L λx λy/L λy

2/L

0.25 0 0

0.128 0.096 0.072

0 0 0.333

0.096

0.128

0.072

0.096

-0.096

-0.128

-0.072

-0.096-0.096

-0.128

-0.072

-0.096

[k]2 = 8x103

2 5 61

256

1

12

3 4

(0,0)

(-4,-3)

(-4,0)

(0,-3)

2

1

3

1

23

4

56

78

1

23

4

56

78

[K] = 8x103

21

21

0.096

0.378

0.405

0.096

22

∆B = 2.5 mm

1.536 kN

1.152 kN

2+20oC1.536 kN

1.152 kN

2+20oC

1.92 kΝ =α(∆T1)AE = (12x10-6)(20)(8x103)

1.92 kN

1.536 kN

1.152 kN

1.536 kN

1.152 kN

12

34

(0,0)

(-4,-3)

(-4,0)

(0,-3)

2

1

3

1

23

4

56

78

∆B = 2.5 mm

+20o C

-1.536

1.536-1.152

1.152

+

1

2

6

5

q1

q2

q1

q5

q2

q6

-1.536

-1.152+1

2

0

-2.5x10-3

5

6-0.096

-0.128

-0.072

-0.096+ 8x103

5 6

= 8x103

21

21

0.096

0.128

0.072

0.096 d1

d2

d1

d5 = 0d2

d6 = -2.5x10-3

0.096

0.128

0.072

0.096

= 8x103

2 5 61

256

1

0.096

0.128

0.072

0.096

-0.096

-0.128

-0.072

-0.096-0.096

-0.128

-0.072

-0.096

[q] = [k]m[d] + [qF]Member 2:

q1

q2

-1.536

-1.152+1

2= 8x103

21

0.096

0.128

0.072

0.096 d1

d2

1.92

1.44+

23

[Q] = [K][D] + [QF]

Global:

A

BC

D

8 kN

4 kN

4 m

3 m+20o C

∆B = 2.5 mm

12

3 4

(0,0)

(-4,-3)

(-4,0)

(0,-3)

2

1

3

1

23

4

56

78

Q1 = -4

Q2 = -8= 8x103

21

21

0.096

0.378

0.405

0.096 D1

D2

-1.536

-1.152+1.92

1.44+

D1

D2

-0.8514x10-3 m

-2.356x10-3 m=

24

Member

#1

#2

#3

λx λy

Local12

3 4

2

1

3

1

23

4

56

78

[q´F]1 = 8x103 1.0 0.0 -1.0 0.04

= -1.70 kN (C)

D2=D1=

D4=D3=

0.00.0

-0.8514x10-3

-2.356x10-3

-1.92+[q´F]2 = 8x103 0.8 0.6 -0.8 -0.65

= -2.87 kN (C)

D2=D1=

D6=D5=

-0.00250.0

-0.8514x10-3

-2.356x10-3

= -6.28 kN (C)

D2=D1=

D8=D7=

0.00.0

-0.8514x10-3

-2.356x10-3[q´F]3 = 8x103 0.0 1.0 0.0 -1.0

3

2+20oC

1.92 kN

1.92 kN

-1 0

- 0.8 -0.6

0 -1

[ ] [ ]F

yj

xj

yi

xi

xxyxmF q

DDDD

LAEq ']'[ +

−−= λλλλ

25

4 kN

8 kN

6.28 kN

1.70 kN

2.87 kN

12

3 4

2

1

3

1

23

4

56

78

Member

#1

#2

#3

cosθx

-1

- 0.8

0

cosθy

0

-0.6

-1

[q´]m

-1.70

-2.87

-6.28

4 kN

8 kN

2

1

3

1.70 kN

6.28 kN

2.87(0.8) = 2.30 kN

2.87(0.6) = 1.72 kN

26

∆ AD = + 3 mm

AB

C

3 m

D

8 kN

4 kN

4 m 4 m∆ = - 4 m

m

Example 3

For the truss shown, use the stiffness method to:(a) Determine the end forces of each member and reactions at supports.(b) Determine the displacement of the loaded joint.Take AE = 8(103) kN.

27

∆ AD = + 3 mm

AB

C

D

8 kN

4 kN

3 m

4 m 4 m

∆ = - 4 mm

21

4

3

5

1

2

34

56

78

1

2 3 4

(0,0)

(-4,-3) (4,-3)(0,-3)

λyλx

λxλx

λyλy

λxλy

λyλx

λxλx

λyλy

λxλy

−λyλx

−λxλx

−λyλy

−λxλy

−λyλx

−λxλx

−λyλy

−λxλy

Vi Uj VjUi

[ k ]m = AEL

Vi

Uj

Vj

Ui

Member

#1

#2

λx λy

#3

#4

#5

-4/5 =-0.8

0

4/5 = 0.8

4/4 = 1

4/4 = 1

-3/5 = -0.6

-3/3 = -1

-3/5 = -0.6

0

0

Ljyy

Lixx ijij

ij

�)(�)(� −+

−=λ


28

[k]1 = 8x103

2 3 41

234

1

0.096

0.128

0.072

0.096

-0.096

-0.128

-0.072

-0.096-0.096

-0.128

-0.072

-0.096

0.096

0.128

0.072

0.096

[k]2 = 8x103

2 5 61

256

1

0

0

0.333

00

0

-0.333

0

0

0

0.333

0

0

0

-0.333

0

[k]3 = 8x103

2 7 81

278

1

-0.096

0.128

0.072

-0.0960.096

-0.128

-0.072

0.096

0.096

-0.128

-0.072

0.096-0.096

0.128

0.072

-0.096

Member

#1

#2

λy

#3

λx

-0.8

0

0.8

-0.6

-1

-0.6

λx2/L λy

2/Lλxλy/L

0.128 0.0720.096

0 0.3330

0.128 0.072-0.096

21

4

3

5

1

2

34

56

78

1

2 3 4

(0,0)

(-4,-3) (4,-3)(0,-3)

5 m

3 m 5 m

4 m 4 m

29

21

4

3

5

1

2

34

56

78

1

2 3 4

(0,0)

(-4,-3) (4,-3)(0,-3)

5 m

3 m 5 m

4 m 4 m

Member

#4

#5

λyλx

1

1

0

0

λx2/L λy

2/Lλxλy/L

0

0.25

0

00

-0.25

0

0

0

0.25

0

0

0

-0.25

0

0

0

0.25

0

00

-0.25

0

0

0

0.25

0

0

0

-0.25

0

0[k]5= 8x103

6 7 85

678

5

0.25 00

0.25 00

Global Stiffness Matrix

[K] = 8x103

2 5 71

257

1

34 6

8

[k]4= 8x103

4 5 63

456

3

30

Global Stiffness Matrix

0.2560.0 0.477

0.0 0.00.0

0.0 0.0-0.128 0.096

-0.1280.096

0.50-0.25

-0.250.378

[K] = 8x103

2 5 71

257

1

[k]1 = 8x103

2 3 41

234

1

0.096

0.128

0.072

0.096

-0.096

-0.128

-0.072

-0.096-0.096

-0.128

-0.072

-0.096

0.096

0.128

0.072

0.096

[k]2 = 8x103

2 5 61

256

1

0

0

0.333

00

0

-0.333

0

0

0

0.333

0

0

0

-0.333

0

[k]3 = 8x103

2 7 81

278

1

-0.096

0.128

0.072

-0.0960.096

-0.128

-0.072

0.096

0.096

-0.128

-0.072

0.096-0.096

0.128

0.072

-0.096

0

0.250

-0.25

0

00

0

0

0.25

0

-0.250

0

0

0[k]4= 8x103

4 5 63

456

3

0

0.250

-0.25

0

00

0

0

0.25

0

-0.250

0

0

0[k]5= 8x103

6 7 85

678

5

31

∆AE/L = 10.67 kN

∆ AD = + 3 mm1

3.84 kN

2.88 kN

3.84 kN

2.88 kN

∆ AD = + 3 mm

AB

C

D

8 kN

4 kN

3 m

4 m 4 m

∆ = - 4 mm

1

2

21

4

3

534

56

78

Global Fixed end forces

0.00.0

257

1∆AE/L = 4.8 kN

4.8 kN

∆AE/L = 10.67 kN

10.67 kN∆ = -4 m

m

2 Fixed End3.84 kN

2.88 kN-3.84-2.88 + 10.67 = 7.79

32

∆ AD = + 3 mm

AB

C

D

8 kN

4 kN

3 m

4 m 4 m

∆ = -4 mm

Global:

1

2

21

4

3

534

56

78

[Q] = [K][D] + [QF]

Q1 = 4

Q5 = 0Q2 = -8

Q7 = 0

= 8x103

2 5 71

257

1

-0.250.500.00.0

0.378-0.250.096

-0.128

-0.1280.00.0

0.256

0.0960.0

0.4770.0 D1

D5

D2

D7

-3.84

0.07.79

0.0

+

D1

D5

D2

D7

6.4426x10-3 m

2.6144x10-3 m-5.1902x10-3 m

5.2288x10-3 m

=

8 kN

4 kN

33

D1

D5

D2

D7

6.4426x10-3 m

2.6144x10-3 m-5.1902x10-3 m

5.2288x10-3 m

=

-4.8+D2

D1

00

= -1.54 kN (C)

1

2

21

4

3

534

56

78

Member forces

= -3.17 kN (C)

10.67+

[q´F]1 = 8x103 0.8 0.6 -0.8 -0.65

[q´F]2 = 8x103 0.0 1.0 0.0 -1.03

D2

D1

0D5

Member

#1

#2

λiyλix

-0.8 -0.6

0 -1

1

4.8 kN

4.8 kN∆ AD

= + 3 mm

10.67 kN

10.67 kN

2

[ ] [ ]F

yj

xj

yi

xi

xxyxmF q

DDDD

LAEq ']'[ +

−−= λλλλ

34

D1

D5

D2

D7

6.4426x10-3 m

2.6144x10-3 m-5.1902x10-3 m

5.2288x10-3 m

=

1

2

21

4

3

534

56

78

00

0D5

[q´F]4 = 8x103 -1.0 0.0 1.0 0.04

= 5.23 kN (T)

0D5

0D7

[q´F]5 = 8x103 -1.0 0.0 1.0 0.04

= 5.23 kN (T)

D2

D1

0D7

[q´F]3 = 8x103 -0.8 0.6 0.8 -0.65

= -6.54 kN (C)

Member λx λy

#3#4

#5

0.8 -0.61 0

1 0

[ ] [ ]F

yj

xj

yi

xi

xxyxmF q

DDDD

LAEq ']'[ +

−−= λλλλ

35

1

2

21

4

3

534

56

78

-0.8

0

0.81

1

-0.6

-1

-0.60

0

Member

#1

#2

λx

#3#4

#5

λy [q´]

4 kN8 kN

3.17 kN

3.17 kN

6.54 kN

6.54 kN

1.54 kN

1.54 kN

5.23 kN5.23 kN

4 kN8 kN

0.92 kN

4 kN

3.17 kN 3.92 kN

-1.54

-3.17

-6.545.23

5.23

36

Special Trusses (Inclined roller supports)

37

λix = cos θi

λiy = sin θi

λjx = cos θj

λjy = sin θj

[ q* ] = [ T ]T[ q´ ] 1

2

3*

4*

56

78

3

2

4

5

1

x *

y *

θi

x

y

θj

1

i

j

q´i

q´j

q*3

q1

q*4

q2

q´iq´j

[T]T

00λiyλix

λjyλjx00 [ T ] = [[ T ]T]T =

=00

λiy

λix

λjy

λjx

00

1

i

j1

2

3*

4*

Transformation Matrices

38

[ k ] = [ T ]T[ k´ ][T]

λjyλjx

λjxλjx

−λiyλjx

−λixλjx

λjyλjy

λjxλjy

−λiyλjy

−λixλjy

λiyλix

λixλix

−λjyλix

−λjxλix

λiyλiy

λixλiy

−λjyλiy

−λjxλiy

Vi Uj VjUi

[ k ]m = AEL

Vi

Uj

Vj

Ui

00λiyλix

λjyλjx00-11

1-1

AEL[ k ]m =

00

λiy

λix

λjy

λjx

00

39

Example 5

For the truss shown, use the stiffness method to:(b) Determine the end forces of each member and reactions at supports.(a) Determine the displacement of the loaded joint. AE is constant.

3 m

4 m

45o

30 kN

40

3 m

4 m45o

2

1

3

1

2

5

63*

4*

Member 1:

15

63*

4*

[q*]θi = 0,λix = cos 0 = 1,λiy = sin 0 = 0

θij = -45o = 135o,λix = cos (-45o) = 0.707,λiy = sin(- 45o) = -0.707

45o

[q*] = [T*]T[q´] + [T*]T[q´F]

q5

q3*

q6

q4*

q´iq´j

[T*]T

=0001

-0.7070.707

00

q´i q´j

i j1

41

15

63*

4*

[q*]θi = 0 ;λix = cos 0 = 1,λiy = sin 0 = 0


45o

0-0.707

00.707

00

10

AE4

1-1

-11[k*]1 =

0001

-0.7070.707

00

Member 1:

[k*]1 = AE 63*4*

53*

-0.1250.125

0-0.1768

4*

0.125-0.125

00.1768

5

0.1768-0.1768

00.25

6

0000

q´i q´j

i j1

3 m

4 m45o

2

1

3

1

2

5

63*

4*

[ ]TL

AETk T

−

−=

1111

][*][

42

Member 2:q´i

q´j

i

j

22

1

2

3*4*

θi = -90o = 270o,λix = cos(-90o) = 0,λiy = sin(-90o) = -1

90o+45o

=135o

90o

θj = -135o = 215o ,λix = cos (-135o) = -0.707,λiy = sin(- 135o) = -0.707

0-0.707

0-0.707

-10

00

AE3

1-1

-11[k*]2 =

00-10

-0.707-0.707

00

[k*]2 = AE

2 4*1

23*4*

1

0.16670.1667-0.2357

0

0.16670.1667-0.2357

0

0000

-0.2357-0.23570.3333

03*

3 m

4 m45o

2

1

3

1

2

5

63*

4*

[ ]TL

AETk T

−

−=

1111

][*][

43

[ ]TL

AETk T

−

−=

1111

][][

36.87o

Member 3:

θi = θj = 36.87o ;λix = λjx = cos (36.87o) = 0.8,λiy = λjy = sin(36.87o) = 0.6

00.6

00.8

0.60

0.80

AE5

1-1

-11[k]3 =

00

0.60.8

0.60.800

3

1

2

5

63

q´i

q´j

i

j36.87o

[k]3 = AE

6 25

612

5

0.0960.128-0.096-0.128

0.0720.096-0.072-0.096

-0.096-0.1280.0960.128

-0.072-0.0960.0720.096

1

3 m

4 m45o

2

1

3

1

2

5

63*

4*

44

3 m

4 m45o

2

1

3

1

2

5

63*

4*

[k*]1 = AE 63*4*

53*

-0.1250.125

0-0.1768

4*

0.125-0.125

00.1768

5

0.1768-0.1768

00.25

6

0000

5

6 4*

Global Stiffness:

[k*]2 = AE

2 4*1

23*4*

1

0.16670.1667-0.2357

0

0.16670.1667-0.2357

0

0000

-0.2357-0.23570.3333

03*

[k]3 = AE

6 25

612

5

0.0960.128-0.096-0.128

0.0720.096-0.072-0.096

-0.096-0.1280.0960.128

-0.072-0.0960.0720.096

1[K] = AE 2

3*

11

00.0960.128

2

-0.23570.40530.096

0.2917-0.2357

03*

45

Global :

[Q] = [K][D] + [QF]

Q1 = 30

Q3*= 0

Q2 = 0D1

D3*

D2 = AE 23*

11

00.0960.128

2

-0.23570.40530.096

0.2917-0.2357

03*

D1

D3*

D2 =352.5

-127.3-157.5AE

1

3 m

4 m45o

30 kN

3 m

4 m45o

2

1

3

1

2

5

63*

4*

5

6 4*

46

[q´F]1 = AE 00

0D3*

-1 0 0.707 -0.7074

= -22.50 kN, (C)

D2

D1

0D3*

[q´F]2 = AE 0 1 -0.707 -0.7073

= -22.50 kN, (C)

00

D2

D1

[q´F]3 = AE -0.8 -0.6 0.8 0.65

= 37.50 kN, (T)

D1

D3*

D2 =352.5

-127.3-157.5AE

1

Member Forces :

Member

#1

λiyλix λjx λjy

#2

#3

1 0 0.707 -0.707

0 -1 -0.707 -0.707

0.8 0.6 0.8 0.6

3 m

4 m45o

2

1

3

1

2

5

63*

4*

5

6 4*

[ ] [ ]F

yj

xj

yi

xi

xxyxmF q

DDDD

LAEq ']'[ +

−−= λλλλ

47

36.87o 45o45o

3 m

4 m45o

2

1

3

1

2

5

63*

4*

Member

Member Force (kN)

[q´]2[q´]1 [q´]3

-22.50 -22.50 37.50

22.50 kN

22.50 kN

37.50 kN

22.50 kN

7.50 kN31.82 kN

Reactions :

3 m

4 m45o

30 kN

48

Example 6

For the truss shown, use the stiffness method to:(b) Determine the end forces of each member and reactions at supports.(a) Determine the displacement of the loaded joint. AE is constant.

3 m

4 m

45o

30 kN

4 m

49

15

63*

4*

[q*]θi = 0o,λix = cos 0o = 1,λiy = sin 0o = 0

θij = -45o

λix = cos (-45o) = 0.707,λiy = sin(- 45o) = -0.707

45o

[q*] = [T*]T[q´] + [T*]T[q´F]

q5

q3*

q6

q4*

q´iq´j

[T*]T

=0001

-0.7070.707

00

q´i q´j

i j1

3 m

4 m 4 m

2

1

3

4

5

12

5

6 3*4*

78Member 1:

50

3 m

4 m 4 m

15

63*

4*

[q*]θi = 0o,λix = cos 0o = 1,λiy = sin 0o = 0


45o

0-0.707

00.707

00

10

AE4

1-1

-11[k*]1 =

0001

-0.7070.707

00

[k*]1 = AE 63*4*

53*

-0.1250.125

0-0.1768

4*

0.125-0.125

00.1768

5

0.1768-0.1768

00.25

6

0000

q´i q´j

i j1

2

1

3

4

5

12

5

6 3*4*

78Member 1:

[ ]TL

AETk T

−

−=

1111

][*][

51

q´i

q´j

i

j

22

1

2

3*4*

θi = -90o,λix = cos(-90o) = 0,λiy = sin(-90o) = -1

90o

θj = -135o = 215o,λix = cos (-135o) = -0.707,λiy = sin(- 135o) = -0.707

0-0.707

0-0.707

-10

00

AE3

1-1

-11[k*]2 =

00-10

-0.707-0.707

00

[k*]2 = AE

2 4*1

23*4*

1

0.16670.1667-0.2357

0

0.16670.1667-0.2357

0

0000

-0.2357-0.23570.3333

03*

3 m

4 m 4 m

2

1

3

4

5

12

5

6 3*4*

78Member 2:

[ ]TL

AETk T

−

−=

1111

][*][

90o+45o

=135o

52

36.87o

Member 3:

θi = θj = 36.87o ;λix = λjx = cos (36.87o) = 0.8,λiy = λjy = sin(36.87o) = 0.6

00.6

00.8

0.60

0.80

AE5

1-1

-11[k]3 =

00

0.60.8

0.60.800

3

1

2

5

63

q´i

q´j

i

j36.87o

[k]3 = AE

6 25

612

5

0.0960.128-0.096-0.128

0.0720.096-0.072-0.096

-0.096-0.1280.0960.128

-0.072-0.0960.0720.096

1

3 m

4 m 4 m

2

1

3

4

5

12

5

6 3*4*

78

[ ]TL

AETk T

−

−=

1111

][*][

53

3 m

4 m 4 m

θi = θij = 0o; λix = λjx = cos 0o = 1, λiy = λjy = sin 0o = 0

00

01

00

10

AE4

1-1

-11[k]1 =

0001

0100

q´i q´j

i j1

2

1

3

4

5

12

5

6 3*4*

78Member 4:

47

8

[q]1

2

[k]4 = AE 278

17

00.25

0-0.25

8

0000

1

0-0.25

00.25

2

0000

[ ]TL

AETk T

−

−=

1111

][*][

54

Member 5:

θi = - 8.13o;λix = cos (- 8.13o) = 0.9899,λiy = sin(- 8.13o) = -0.1414

00.6

00.8

-0.14140

0.98990

AE5

1-1

-11

5

q´i

q´j

i

j36.87o

3 m

4 m 4 m

2

1

3

4

5

12

5

6 3*4*

78

53*4*

7

8

[k*]5 =00

-0.14140.9899

0.60.800

[k*]5 = AE

4* 83*

4*78

3*

0.0960.128

0.02263-0.1584

0.0720.096

0.01697-0.1188

-0.1188-0.1584-0.0280.196

0.016970.022630.004-0.028

7

θ j = 36.87o ;

λ jx = cos (36.87o ) = 0.8,

λ jy = sin

(36.87o ) = 0.6

[ ]TL

AETk T

−

−=

1111

][*][

8.13o

550 -0.2357-0.2357

0

[k*]1 = AE 63*4*

53*

-0.1250.125

0-0.1768

4*

0.125-0.125

00.1768

5

0.1768-0.1768

00.25

6

0000

[k*]3 = AE

6 25

612

5

0.0960.128-0.096-0.128

0.0720.096-0.072-0.096

-0.096-0.1280.0960.128

-0.072-0.0960.0720.096

1

3 m

4 m 4 m

2

1

3

4

5

Global Stiffness:1

2

5

6 3*4*

78

5

6 4*

78

[k*]2 = AE

2 4*1

23*4*

1

0.16670.1667-0.2357

0

0.16670.1667-0.2357

0

0000

-0.2357-0.23570.3333

03*

[k]4 = AE 278

17

00.25

0-0.25

8

0000

1

0-0.25

00.25

2

0000

[k*]5 = AE

4* 83*

4*78

3*

0.0960.128

0.02263-0.1584

0.0720.096

0.01697-0.1188

-0.1188-0.1584-0.0280.196

0.016970.022630.004-0.028

7

[K] = AE 23*

11 2

0.0960.378

0.40530.096

3*

0.4877

56

Global :30 kN

4 m 4 m

3 m2

1

3

4

5

12

5

6 3*4*

78

5

6 4*

78

[Q] = [K][D] + [QF]

Q1 = 30

Q3*= 0

Q2 = 0D1

D3*

D2

D1

D3*

D2 =86.612

-13.791-28.535AE

1

= AE 23*

11

00.0960.378

2

-0.23570.40530.096

0.4877-0.2357

03*

57

3 m

4 m 4 m

2

1

3

4

5

12

5

6 3*4*

78

5

6 4*

78

D1

D3*

D2 =86.612

-13.791-28.535AE

1

[q´F]1 = AE 00

0D3*

-1 0 0.707 -0.7074

= -2.44 kN, (C)

D2

D1

0D3*

[q´F]2 = AE 0 1 -0.707 -0.7073

= -6.26 kN, (C)

00

D2

D1

[q´F]3 = AE -0.8 -0.6 0.8 0.65

= 10.43 kN, (T)

Member Forces :

Member

#1

λiyλix λjx λjy

#2

#3

1 0 0.707 -0.707

0 -1 -0.707 -0.707

0.8 0.6 0.8 0.6

[ ] [ ]F

yj

xj

yi

xi

xxyxmF q

DDDD

LAEq ']'[ +

−−= λλλλ

58

Member

#4

λiyλix λjx λjy

#5

3 m

4 m 4 m

2

1

3

4

5

12

5

6 3*4*

78

5

6 4*

78

D1

D3*

D2 =86.612

-13.791-28.535AE

1

D2

D1

00

[q´F]4 = AE -1 0 1 04

= -21.65 kN, (C)

0D3*

00

[q´F]5 = AE -0.9899 0.141 0.8 0.65

= 2.73 kN, (T)1 0 1 0

0.9899 -0.141 0.8 0.6

Member Forces :

[ ] [ ]F

yj

xj

yi

xi

xxyxmF q

DDDD

LAEq ']'[ +

−−= λλλλ

59

36.87o 45o45o

81.87o

36.87o

Member

Member Force (kN)

[q´]1 [q´]2 [q´]3 [q´]4 [q´]5

-2.44 -6.26 10.43 -21.65 2.73

Reactions :30 kN

4 m 4 m

3 m2

1

3

4

5

12

5

6 3*4*

78

5

6 4*

78

6.26 kN10.43 kN

21.65 kN

2.73 kN

2.44 kN 2.44 kN5.90 kN

6.26 kN

19.47 kN

1.64 kN

6.54 kN

60

AB

C

3 m

D

8 kN

4 kN

4 m36.87o

4 m

λjyλjx

λjxλjx

−λiyλjx

−λixλjx

λjyλjy

λjxλjy

−λiyλjy

−λixλjy

λiyλix

λixλix

−λjyλix

−λjxλix

λiyλiy

λixλiy

−λjyλiy

−λjxλiy

Vi Uj VjUi

[ k *]m = AEL

Vi

Uj

Vj

Ui

Example 7

For the truss shown, use the stiffness method to:(b) Determine the end forces of each member and reactions at supports.(a) Determine the displacement of the loaded joint.Take AE = 8(103) kN.

61

Member 1:

λix = cos 73.74o = 0.28,λiy = sin 73.74o = 0.96

[q*] = [T*]T[q´] + [T*]T[q´F]

λjx = cos 36.87o = 0.8,λjy = sin 36.87o = 0.6

AB

C

3 m

D

8 kN

4 kN

4 m

36.87o

4 m

x *

y *

73.74o

x

y

36.87o

1

i

j1

2

3*

4*1

i

j

q´i

q´j

q3*

q1

q4*

q2

q´iq´j

=00

0.960.28

0.60.8

00

[T*]T

3

2

4

5

1

1

2

56

78

3*

4*

62

[k]1 = 8x103

4* 1 23*

4*

12

3*

0.0960.128

-0.1536-0.0448

0.0720.096

-0.1152-0.0336

-0.0336-0.04480.053760.01568

-0.1152-0.15360.184320.05376

000.960.280.60.800

[k]1 =00

0.960.28

0.60.8

00

8x103

51

-1-11

AB

C

3 m

D

8 kN

4 kN

4 m

36.87o

4 m

3

2

4

5

1

1

2

56

78

3*

4*

[ ]TL

AETk T

−

−=

1111

][][

63

x *

y*

36.87x

yMember 2:

λix = cos 36.87o = 0.8,λiy = sin 36.87o = 0.6

[q*] = [T*]T[q´] + [T*]T[q´F]

AB

C

3 m

D

8 kN

4 kN

4 m

36.87o

4 m

λjx = cos 0o = 1,λjy = sin 0o = 0

q3*

q5

q4*

q6

q´iq´j

[T*]T

[k] = [TT] AEL

[T]1-1

-11

q´i q´j

i j2

=00

0.60.8

0100

i j2 5

6

3*

4*

[k]2 = 8x103

4* 5 63*

4*

56

3*

00.25

-0.15-0.2

0000

0-0.20.120.16

0-0.150.090.12

3

2

4

5

1

1

2

56

78

3*

4*

64

x

y

270o 1

2

784

Member 3:

AB

C

3 m

D

8 kN

4 kN

4 m

36.87o

4 m

λx = cos 323.13o = 0.8,λy = sin 323.13o = -0.6

Member 4:

[k]4 = 8x103

2 7 81

278

1

-0.0960.1280.096

-0.128

0.072-0.096-0.0720.096

0.096-0.128-0.0960.128

-0.0720.0960.072

-0.096

λx = cos 270o = 0, λy = sin 270o = -1

[k]3 = 8x103

0000

0.3330

-0.3330

00

00

2 5 61

256

10.333

0

-0.3330

x

y

323.13o3

1

2

56

3

2

4

5

1

1

2

56

78

3*

4*

65

x

yMember 5:

AB

C

3 m

D

8 kN

4 kN

4 m

36.87o

4 m

λx = cos 0o = 1,λy = sin 0o = 0

00.25

0-0.25

0000

00.25

0-0.25

00

00

[k]5 = 8x103

6 7 85

678

5

5 78

56

3

2

4

5

1

1

2

56

78

3*

4*

660.17568

-0.2-0.20.5

00

0 0

-0.0336-0.0448

-0.0448 -0.03360.0

0.2560.4740.0

[k]1 = 8x103

4* 1 23*

4*

12

3*

0.0960.128

-0.1536-0.0448

0.0720.096

-0.1152-0.0336

-0.0336-0.04480.053760.01568

-0.1152-0.15360.184320.05376

[k]4 = 8x103

2 7 81

278

1

-0.0960.1280.096

-0.128

0.072-0.096-0.0720.096

0.096-0.128-0.0960.128

-0.0720.0960.072

-0.096

00.25

0-0.25

0000

00.25

0-0.25

00

00

[k]5 = 8x103

6 7 85

678

5

[k]2 = 8x103

4* 5 63*

4*

56

3*

00.25

-0.15-0.2

0000

0-0.20.120.16

0-0.150.090.12

[k]3 = 8x103

0000

0.3330

-0.3330

00

00

2 5 61

256

10.333

0

-0.3330

[K] = 8x103

2 3* 51

23*

5

1

3

2

4

5

1

12

56

78

3*

4* 6

784*

67

3

2

4

5

1

2

3*

4*

56

78A

BC

3 m

D

8 kN

4 kN

4 m

36.87o

4 m

1

Global: [Q] = [K][D] + [QF]

D1

D3*

D2

D5

1.988x10-3 m

1.996x10-4 m-2.0824x10-3 m

7.984x10-5 m

=

Q1 = 4

Q3*= 0

Q2 = -8

Q5 = 0

D1

D3*

D2

D5

0.17568-0.2

-0.20.5

00

0 0

-0.0336-0.0448

-0.0448 -0.03360.0

0.2560.4740.0

= 8x103

2 3* 51

23*

5

1

8 kN

4 kN

68

AB

C

3 m

D

8 kN

4 kN

4 m

36.87o

4 m

3

2

4

5

1

2

3*

4*

56

78

1

Member forces

D1

D3*

D2

D5

1.988x10-3 m

1.996x10-4 m-2.0824x10-3 m

7.984x10-5 m

=

0D3*

D2

D1

[q´F]1 = 8x103 -0.28 -0.96 0.8 0.65

= 0.46 kN, (T)

0D3*

0D5

[q´F]2 = 8x103 -0.8 -0.6 1 04

= -0.16 kN, (C)

Member

#1

#2

λiyλix λjx λjy

0.28 0.96 0.8 0.6

0.8 0.6 1 0

[ ] [ ]F

yj

xj

yi

xi

xxyxmF q

DDDD

LAEq ']'[ +

−−= λλλλ

69

D1

D*3

D2

D5

1.988x10-3 m

1.996x10-4 m-2.0824x10-3 m

7.984x10-5 m

=

3

2

4

5

1

2

3*

4*

56

78

1

[q´F]3 = 8x103 D2

D1

0D5

0 1 0 -13

= -5.55 kN

D2

D1

00

[q´F]4 = 8x103 -0.8 0.6 0.8 - 0.65

= -4.54 kN

0D5

00

[q´F]5 = 8x103 -1 0 1 04

= -0.16 kN

Member

#3

λiyλix λjx λjy

#4

#5

0 -1 0 -1

0.8 -0.6 0.8 -0.6

1 0 1 0

[ ] [ ]F

yj

xj

yi

xi

xxyxmF q

DDDD

LAEq ']'[ +

−−= λλλλ

70

y*

x *

36.87o

36.87o 36.87o

AB

C

3 m

D

8 kN

4 kN

4 m

36.87o

4 m

3

2

4

5

1

2

3*

4*

56

78

1

0.46 -0.16 -5.55 -4.54 -0.16

0.46 kN

5.55 kN0.36 kN

4.54 kN

3.79 kN

2.72 kN

Member

Member Force (kN)

[q]2[q]1 [q]3 [q]4 [q]5

0.16 kN 0.16 kN

5.55 kN

71

Space-Truss Analysis

72

+

−

−=

Fj

Fi

j

i

j

i

qq

dd

LEA

qq

''

''

1111

''

[ ] [ ]F

jz

jy

jx

iz

iy

ix

q

dddddd

TL

EA '1111

+

−

−=

[ ]

=

zyx

zyxT

where

λλλλλλ000

000,

[q´] = [k´][d´] + [q´F]

= [k´][T][d] + [q´F]

Member Local Stiffness [k´]:

73

Member Global Stiffness [km]:

[km]= [T]T[k´] [T]

[ ]

−

−

=zyx

zyx

z

y

x

z

y

x

m LEAk

λλλλλλ

λλλ

λλλ

000000

1111

000

000

74

Global equilibrium matrix:

[Q] = [K][D] + [QF]

QI

QII

Du

Dk

KI,II

KII,II

KI,I

KII,I

Reaction Support Boundary Condition


QFI

QFII

= +

Fixed End Forces

75

q´j

q´i

i

j

m

i

j

m

=EAL

λxλx

λzλx

λyλx

λxλy

λyλy

λzλy

λxλz

λyλz

λzλz

λxλx

λzλx

λyλx

λxλy

λyλy

λzλy

λxλz

λyλz

λzλz

−λxλx

−λzλx

−λyλx

−λxλy

−λyλy

−λzλy

−λxλz

−λyλz

−λzλz

−λxλx

−λzλx

−λyλx

−λxλy

−λyλy

−λzλy

−λxλz

−λyλz

−λzλz

diy

dix

djx

diz

djy

djz

qiy

qix

qjx

qiz

qjy

qjz

qjy

qjx

qjzqiy

qix

qiz

+

−

−=

Fj

Fi

j

i

j

i

qq

dd

LEA

qq

''

''

1111

''

[ ] [ ] [ ]F

jz

jy

jx

iz

iy

ix

zyxzyxmj q

dddddd

LEAq '' +

−−−= λλλλλλ

76

x

z

y

4 m4 m

3 m3 m

10 m

80 kN60 kN

O

Example 6

For the truss shown, use the stiffness method to:(b) Determine the end forces of each member.(a) Determine the deflections of the loaded joint.Take E = 200 GPa, A = 1000 mm2.

77

λ1 = (-4/11.18)i + (3/11.18)j + (-10/11.18)k

= -0.3578 i + 0.2683 j - 0.8944 k

λ2 = (+4/11.18)i + (3/11.18)j + (-10/11.18)k= +0.3578 i + 0.2683 j - 0.8944 k

λ3 = (+4/11.18)i + (-3/11.18)j + (-10/11.18)k

= +0.3578 i - 0.2683 j - 0.8944 k

= -0.3578 i - 0.2683 j - 0.8944 k

λ4 = (-4/11.18)i + (-3/11.18)j + (-10/11.18)k

Member

#1

#2

#3

λx λy λz

#4

-0.3578 +0.2683 -0.8944

+0.3578 +0.2683 -0.8944

+0.3578 -0.2683 -0.8944

-0.3578 -0.2683 -0.8944

λm = λxi + λyj + λzk

12

34

12

3

(0, 0, 10)

(-4, 3, 0)

(4, 3, 0)

(4, -3, 0)

(-4, -3, 0)4

1

2

53

x

z

y

4 m4 m

3 m3 m

10 m

80 kN60 kN

O

78

Member

#1#2

#3

λx λy λz

#4

-0.3578 +0.2683 -0.8944+0.3578 +0.2683 -0.8944

+0.3578 -0.2683 -0.8944-0.3578 -0.2683 -0.8944

Member Stiffness Matrix [k]6x6

[k]m =[k12]3x3

[k22]3x3

[k11]3x3

[k21]3x3

2 31

23

1

+0.80-0.240

+0.320

+0.320-0.096

+0.128

-0.240+0.072

-0.096

[k11]1 =AEL

2 31

23

1

+0.80-0.240

-0.320

-0.320+0.096

+0.128

-0.240+0.072

+0.096

[k11]2 =AEL

2 31

23

1

+0.80+0.240

-0.320

-0.320-0.096

+0.128

+0.240+0.072

-0.096

[k11]3 =AEL

2 31

23

1

+0.80+0.240

+0.320

+0.320+0.096

+0.128

+0.240+0.072

+0.096

[k11]4 =AEL

2 31

23

1

3.20.0

0.0

0.00.0

0.512

0.00.288

0.0

[KI,I] =AEL

79

12

34

12

3

(0, 0, 10)

(-4, 3, 0)

(4, 3, 0)

(4, -3, 0)

(-4, -3, 0)4

1

2

53

x

z

y

4 m4 m

3 m3 m

10 m

80 kN60 kN

O

0.0-80

60

D3

D2

D1

0.00.0

0.0+

2 31

23

1

3.20.0

0.0

0.00.0

0.512

0.00.288

0.0

=AEL

[Q] = [K][D] + [QF]

80

Global equilibrium matrix:

[Q] = [K][D] + [QF]

QI

QII

Du

Dk

KI,II

KII,II

KI,I

KII,I

Reaction Support Boundary Condition


QFI

QFII

= +

Fixed End Forces

(AE/L) = (1x10-3)(200x106)/(11.18) = 17.89x103 kN

0.0-80

60

D3

D2

D1

0.00.0

0.0+

2 31

23

1

3.20.0

0.0

0.00.0

0.512

0.00.288

0.0

=AEL

0.0 mm-15.53 mm

6.551 mm=

D3

D2

D1

= LAE

0.0-277.8

+117.2

81

Member forces:

q´F+

Member

#1#2

#3

λx λy λz

#4

-0.3578 +0.2683 -0.8944+0.3578 +0.2683 -0.8944

+0.3578 -0.2683 -0.8944-0.3578 -0.2683 -0.8944

dyi

dxi

dxj

dzi

dyj

dzj

[q´j]m = AEL −λx −λy −λz λx λy λz

D3

D2

D1

[ 0 ]

= +116.5 kN (T)

+0.3578 -0.2683 +0.8944[q´j]1 =AEL

0.0

117.2

-277.8L

AE

x

z

y

4 m4 m

3 m3 m

10 m

80 kN60 kN

O

82

Member

#1#2

#3

λx λy λz

#4

-0.3578 +0.2683 -0.8944+0.3578 +0.2683 -0.8944

+0.3578 -0.2683 -0.8944-0.3578 -0.2683 -0.8944

= +32.61 kN (T)-0.3578 -0.2683 +0.8944[q´j]2 =AEL

0.0

117.2

-277.8L

AE

= -116.5 kN (T)-0.3578 +0.2683 +0.8944[q´j]3 =AEL

0.0

117.2

-277.8L

AE

= -32.61 kN (T)+0.3578 +0.2683 +0.8944[q´j]4 =AEL

0.0

117.2

-277.8L

AE

x

z

y

4 m4 m

3 m3 m

10 m

80 kN60 kN

O

83

5

32.6 kN116.5 kN

116.5 kN

32.6 kN

Member

#1#2

#3

λx λy λz

#4

-0.3578 +0.2683 -0.8944+0.3578 +0.2683 -0.8944

+0.3578 -0.2683 -0.8944-0.3578 -0.2683 -0.8944

[q´j]m

116.5 32.6

-32.6-116.5

12

34

80 kN60 kN

R5x = (-32.6)(-0.3578) = 11.66 kN

R5y = (-32.6)(-0.2683) = 8.75 kN

R5z = (-32.6)(-0.8944) = 29.16 kN

1

! Development: The Slope-DeflectionEquations

! Stiffness Matrix! General Procedures! Internal Hinges! Temperature Effects! Force & Displacement Transformation! Skew Roller Support

BEAM ANALYSIS USING THE STIFFNESSMETHOD

2

Slope � Deflection Equations

settlement = ∆j

Pi j kw Cj

Mij MjiwP

θj

θi

ψ

i j

3


L

θΑ

A B

M

1 DOF: θΑ

PθΑ

θΒΒΒΒ

A BC 2 DOF: θΑ , θΒΒΒΒ

4

L

A B

1

� Stiffness Definition

kBAkAA

LEIkAA

4=

LEIkBA

2=

5

L

A B1

kBBkAB

LEIkBB

4=

LEIkAB

2=

6

� Fixed-End ForcesFixed-End Forces: Loads

P

L/2 L/2

L

w

L

8PL

8PL

2P

2P

12

2wL12

2wL

2wL

2wL

7

� General Case

settlement = ∆j

Pi j kw Cj

Mij MjiwP

θj

θi

ψ

i j

8

wP

settlement = ∆j

(MFij)∆ (MF

ji)∆

(MFij)Load (MF

ji)Load

+

Mij Mji

θi

θj

+

i jwPMij

Mji

settlement = ∆jθj

θi

ψ

=+ ji LEI

LEI θθ 24

ji LEI

LEI θθ 42

+=

,)()()2()4( LoadijF

ijF

jiij MMLEI

LEIM +++= ∆θθ Loadji

Fji

Fjiji MM

LEI

LEIM )()()4()2( +++= ∆θθ

L

9

Mji Mjk

Pi j kw Cj

Mji Mjk

Cj

j

� Equilibrium Equations

0:0 =+−−=Σ+ jjkjij CMMM

10

+

1

1

i jMij Mji

θi

θj

LEIkii

4=

LEIk ji

2=

LEIkij

2=

LEIk jj

4=

iθ×

jθ×

� Stiffness Coefficients

L

11

[ ]

=

jjji

ijii

kkkk

k

Stiffness Matrix

)()2()4( ijF

jiij MLEI

LEIM ++= θθ

)()4()2( jiF

jiji MLEI

LEIM ++= θθ

+

=

F

ji

Fij

j

iI

ji

ij

MM

LEILEILEILEI

MM

θθ

)/4()/2()/2()/4(

� Matrix Formulation

12

[D] = [K]-1([Q] - [FEM])

Displacementmatrix

Stiffness matrix

Force matrixwP(MF

ij)Load (MFji)Load

+

+

i jwPMij

Mji

θj

θi

ψ ∆j

Mij Mji

θi

θj

(MFij)∆ (MF

ji)∆

Fixed-end momentmatrix

][]][[][ FEMKM += θ

]][[])[]([ θKFEMM =−

][][][][ 1 FEMMK −= −θ

L

13

L

Real beam

Conjugate beam

� Stiffness Coefficients Derivation

MjMi

L/3

θi

LMM ji +

EIM j

EIMi

θι

EILM j

2

EILMi

2

)1(2

0)3

2)(2

()3

)(2

(:0'

−−−=

=+−=Σ+

ji

jii

MM

LEI

LMLEI

LMM

)2(0)2

()2

(:0 −−−=+−=Σ↑+EI

LMEI

LMF jiiy θ ij

ii

LEIM

LEIM

andFrom

θ

θ

)2(

)4(

);2()1(

=

=

LMM ji +

i j

14

� Derivation of Fixed-End Moment

Real beam

8,0

162

22:0

2 PLMEI

PLEI

MLEI

MLFy ==+−−=Σ↑+

P

M

M EIM

Conjugate beamA

EIM

B

L

P

A B

EIM

EIML2

EIM

EIML2

EIPL

16

2

EIPL4 EI

PL16

2

Point load

15

L

P

841616PLPLPLPL

=+−

+−

8PL

8PL

2P

2PP/2

P/2

-PL/8 -PL/8

PL/8

-PL/16-PL/8

-

PL/4+

-PL/16-PL/8-

16

Uniform load

L

w

A B

w

M

M

Real beam Conjugate beam

A

EIM

EIM

B

12,0

242

22:0

23 wLMEI

wLEI

MLEI

MLFy ==+−−=Σ↑+

EIwL8

2

EIwL

24

3

EIwL

24

3

EIM

EIML2

EIM

EIML2

17

Settlements

M

M

L

∆

MjMi = Mj

LMM ji +L

MM ji +

Real beam

2

6LEIM ∆

=

Conjugate beam

EIM

A B

EIM

∆

,0)3

2)(2

()3

)(2

(:0 =+−∆−=Σ+L

EIMLL

EIMLM B

EIM

EIML2

EIMEI

ML2

∆

18

� Typical Problem

0

0

0

0

A C

B

P1P2

L1 L2

wCB

8024 11

11

LPLEI

LEIM BAAB +++= θθ

8042 11

11

LPLEI


128024 2

222

22

wLLPLEI


128042 2

222

22

wLLPLEI

LEIM CBCB −

−+++= θθ

P

8PL

8PL

L

w12

2wL

12

2wL

L

19

MBA MBC

A C

B

P1P2

L1 L2

wCB

B

CB

8042 11

11

LPLEI


128024 2

222

22

wLLPLEI


BBCBABB forSolveMMCM θ→=−−=Σ+ 0:0

20

A C

B

P1P2

L1 L2

wCB

Substitute θB in MAB, MBA, MBC, MCB

MABMBA

MBC

MCB

0

0

0

0

8024 11

11

LPLEI

LEIM BAAB +++= θθ

8042 11

11

LPLEI


128024 2

222

22

wLLPLEI


128042 2

222

22

wLLPLEI

LEIM CBCB −

−+++= θθ

21

A BP1

MAB

MBA

L1

A CB

P1P2

L1 L2

wCB

ByR CyAy ByL

Ay Cy

MABMBA

MBC

MCB

By = ByL + ByR

B CP2

MBCMCB

L2

22

2EI EI

Stiffness Matrix

� Node and Member Identification

� Global and Member Coordinates


�Known degrees of freedom D4, D5, D6, D7, D8 and D9� Unknown degrees of freedom D1, D2 and D3

1 21 2 31

2 3

7

8

9

4

5

6

1

2 3

7

8

9

4

5

6

23

i j

E, I, A, L

Beam-Member Stiffness Matrix

d4 = 1

AE/LAE/L AE/L AE/L

64

5

1

2

3

[k] =

1 2 3 4 5 6

3

2

1

4

6

5

d1 = 1

k11 = = k44

0

0

0

0

0

0

0

0

- AE/LAE/L

-AE/L AE/L

k41 k14

24

d2 = 1

[k] =

1 2 3 4 5 6

3

2

1

4

6

5

0

0

0

0

0

0

0

0

- AE/LAE/L

-AE/L AE/L

6EI/L2

6EI/L2

d5 = 1

6EI/L2

6EI/L2

12EI/L312EI/L3 12EI/L3

12EI/L3

6EI/L2

12EI/L3

- 6EI/L2

- 12EI/L3

0

0

6EI/L2

-12EI/L3

0

0

- 6EI/L2

12EI/L3

k22 = = k55

= k52

k62 =

= k32 = k65

= k25

= k35

i j

E, I, A, L6

4

5

1

2

3

25

[k] =

1 2 3 4 5 6

3

2

1

4

6

5

0

0

0

0

0

0

0

0

- AE/LAE/L

-AE/L AE/L

6EI/L2

12EI/L3

- 6EI/L2

- 12EI/L3

0

0

6EI/L2

-12EI/L3

0

0

- 6EI/L2

12EI/L3

4EI/L

6EI/L2 6EI/L2

2EI/L

0

0

2EI/L

-6EI/L2

0

0

-6EI/L2

4EI/L

d6 = 1

d3 = 1 2EI/L 4EI/L

6EI/L26EI/L26EI/L2

4EI/L 2EI/L

6EI/L2

k33 =

k23 = = k53

= k63 = k36

= k56 k26 =

= k66

i j

E, I, A, L6

4

5

1

2

3

26

FFxi

FFyi

MFi

FFxj

FFyj

MFj

+

MFi MF

j

FFxjFF

xi

FFyi FF

yj

1

6EI/L2

6EI/L2

12EI/L312EI/L3

+

x ∆i

12EI/L3

6EI/L2

6EI/L2

12EI/L3

1

+

4EI/L 2EI/L

6EI/L26EI/L2

x θi

6EI/L2

4EI/L 2EI/L

6EI/L2

x δjAE/L

1

AE/L

+

AE/L AE/L

6EI/L2

6EI/L2

1

12EI/L312EI/L3

x ∆j

6EI/L2

6EI/L2

12EI/L3

12EI/L3

1

+

2EI/L

6EI/L26EI/L2

4EI/L

x θj

6EI/L2

2EI/L

6EI/L2

4EI/L

Fxi

Fyi

MiFxj

Fyj

Mj

� Member Equilibrium Equations

Mj

i jFxj

Fxi

Fyi Fyj

=

MiE, I, A, L

x δiAE/L AE/L

1

AE/LAE/L

27

+

−−−−

−−−

−

=

Fj

Fyi

Fxj

Fi

Fyi

Fxi

j

j

j

i

i

i

j

yj

xj

i

yj

xi

MFFMFF

θ∆δθ∆δ

LEILEILEILEILEILEILEILEI

LAELAELEILEILEILEILEILEILEILEI

LAELAE

MFFMFF

/4/60/2/60/6/120/6/120

00/00//2/60/4/60/6/120/6/120

00/00/

22

2323

22

2323

Fjjjjiiij

Fyjjjjiiiyj

Fxijjjiiixj

Fijjjiiixi

Fyijjjiiiyi

Fxijjjiiixi

MθLEI∆LEIδθLEI∆LEIδMFθLEI∆δθLEI∆LEIδFFθ∆δLAEθ∆δLAEFMθLEI∆LEIδθLEI∆LEIδMFθLEI∆LEIδθLEI∆LEIδFFθ∆δLAEθ∆δLAEF

)/4()/6()0()/2()/6()0()/6()0()0()/6()/12()0(

)0()0()/()0()0()/()/2()/6()0()/4()/6()0()/6()/12()0()/6()/12()0(

)0()0()/()0()0()/(

22

223

22

2323

−=−−−=

−=−=−+=

+++−++=

[q] = [k][d] + [qF]

Fixed-end force matrix

End-force matrix

Stiffness matrix

Displacement matrix

28

6x6 Stiffness Matrix

[ ]

−−−−

−−−

−

=×



LAELAE

k

/4/60/2/60/6/120/6/120

00/00//2/60/4/60/6/120/6/120

00/00/

22

2323

22

2323

66

θi ∆j θj∆i δjδi

Mi

Vj

Mj

Vi

Nj

Ni


[ ]

−−−−

−−

=×

LEILEILEILEILEILEILEILEILEILEILEILEILEILEILEILEI

k

/4/6/2/6/6/12/6/12/2/6/4/6/6/12/6/12

22

2323

22

2323

44

θi ∆j θj∆i

Mi

Vj

Mj

Vi

29

Comment: - When use 4x4 stiffness matrix, specify settlement. - When use 2x2 stiffness matrix, fixed-end forces must be included.


θi θj

Mi

Mj[ ]

=× LEILEI

LEILEIk

/4/2/2/4

22

30

21 312

1

2

3

4

5

6

General Procedures: Application of the Stiffness Method for Beam Analysis

wP P2y

M2M1

Local

1´

2´

3´

4´

1

1´

2´

3´

4´

2

Global

31

Member

1

2

3

4

1

3

4

5

6

2

1

2

3

4

5

6

3121 2

Global

wP P2y

M2M1

Local

1´

2´

3´

4´

1

1´

2´

3´

4´

2

32

(q´F4 )2

(q´F3 )2

(q´F1)2

(q´F2)2

w

2

P

(q´F4)1

(q´F3)1

(q´F2)1

(q´F1 )1

1[FEF]

1

2

3

4

5

6

3121 2

Global

Local

1´

2´

3´

4´

1

1´

2´

3´

4´

2

wP P2y

M2M1

33

[q] = [T]T[q´]

1

2

3

4

5

6

3121 2

Global

Local

1´

2´

3´

4´

1

[k] = [T]T[q´] [T]

=

'4

'3

'2

'1

4

3

2

1

1000010000100001

qqqq

qqqq

1´ 4´2´ 3´12341

34

[q] = [T]T[q´]

1

2

3

4

5

6

3121 2

Global

Local

1´

2´

3´

4´

2

[k] = [T]T[q´] [T]

=

'4

'3

'2

'1

6

5

4

3

1000010000100001

qqqq

qqqq

1´ 4´2´ 3´34562

35

Stiffness Matrix:

21 312

1

2

3

4

5

6

[k]1 =

1

2

3

4

1 2 3 4

[k]2 =

3

4

5

6

3 4 5 6

Member 1

[K] =

1

23

4

5

6

1 2 3 4 5 6

Member 2

Node 2

1 2

36

+

=

F

F

F

F

F

F

QQQQQQ

DDDDDD

QQQQQQ

6

5

1

4

3

2

6

5

1

4

3

2

6

5

1

4

3

2 234156

5 612 3 4

000

M1z

-P2y

M3z

KBA

KAB

KBB

KAA

21 312

1

2

3

4

5

6

ReactionQu

Joint LoadQk

Du=Dunknown

Dk = DknownQF

B

QFA

[ ] [ ][ ] [ ]FAuAAk QDKQ +=

[ ] [ ] [ ] [ ])(1 FAkAAu QQKD −+= −

[ ] [ ][ ] [ ]FqdkqForceMember +=:

37

+

6

5

4

3

2

1

DDDDDD

Global:

21 312

1

2

3

4

5

6

1 2

2

3

4

51

6

(q´F6 )2

(q´F5 )2

(qF4)2

(qF3)2

w

2

P

(qF4)1

(qF3)1

(qF2)1

(qF1 )1

1[FEF]

=−=

=

24

23

12

MQPQ

MQ

y

4

3

2

DDD

+F

F

F

QQQ

4

3

2

2 3 4

Node 2

=

234

Member 1

123456

1 2 3 4 5 6

Member 2

Node 2=

6

5

4

3

2

1

QQQQQQ

+=+=

F

F

F4

F4

F

F3

F3

F

F

F

QQ

qqQqqQ

QQ

6

5

214

213

2

1

)()()()(

M1

-P2y

M2

0

00

38

1 2 312

1

2

3

4

5

6

4

3

2

DDD

2 3 4

Node 2

=

234

−

=−=

=

F

F

F

QQQ

MQPQ

MQ

4

3

2

24

2y3

12

39

Member 1:

21 312

1

2

3

4

5

6

P

qF4

qF3

qF2

qF1

1

[FEF]

4

3

2

1

qqqq

====

44

33

22

1 0

DdDdDd

d

+

F

F

F

F

qqqq

4

3

2

11234

1 2 3 4

= 1k

40

21 312

1

2

3

4

5

6

Member 2:

qF6

qF5

qF4

qF3

w

2

[FEM]

6

5

4

3

qqqq

==

==

00

6

5

44

33

dd

DdDd

+

F

F

F

F

qqqq

6

5

4

31234

1 2 3 4

= 1k

41

AC

B

9 m 3 m

10 kN1 kN/m

1.5 m

Example 1

For the beam shown, use the stiffness method to:(a) Determine the deflection and rotation at B.(b) Determine all the reactions at supports.(c) Draw the quantitative shear and bending moment diagrams.

42

21Global

123

1Members

23

2

12

AC

B

9 m 3 m

10 kN1 kN/m

1.5 m

wL2/12 = 6.75wL2/12 = 6.75 PL/8 = 3.75PL/8 = 3.75

10 kN

1.5 m1.5 m

2

1 kN/m

9 m 1[FEF]

43

4/3

2/3

2/3

21

123

9 m 3 m

[k]1= EI4/9

2/9

3

4/9

2/92

3

2[k]2 = EI

4/3

4/3

2/3

2/3

2 12

1

[K] = EI

2 12

1

(4/9)+(4/3)

21

123

Stiffness Matrix: θi θj

Mi

Mj[ ]

=× LEILEI

LEILEIk

/4/2/2/4

22

44

A CB

10 kN1 kN/m

123

Equilibrium equations: MCB = 0MBA + MBC = 0

Global Equilibrium: [Q] = [K][D] + [QF]

=2.423/EI

0.779/EI

θC

θB

9 m 1.5 m1.5 m6.756.75

+-3.75

-6.75 + 3.75 = -3MBA + MBC = 0

MCB = 01

2

θC

θB

4/3

2/3

2/3 = EI

2 12

1

(4/9)+(4/3)

3.753.756.75

45

= -6.40

6.92

Member 1 : [q]1 = [k]1[d]1 + [qF]1

MBA

MAB

2

3

= 0.779/EIθB

θA

0

+ -6.75

6.75

Substitute θB and θC in the member matrix,

1

23

wL2/12 = 6.75wL2/12 = 6.75

1 kN/m

9 m 1[FEF]

4.56 kN 4.44 kN

6.92 kN�m 6.40 kN�m

wL2/12 = 6.75wL2/12 = 6.75

= EI4/9

2/9

3

4/9

2/92

1 kN/m

19 m

46

Member 2 : [q]2 = [k]2[d]2 + [qF]2

Substitute θB and θC in the member matrix,

MCB

MBC

1

2 = EI

4/3

4/3

2/3

2/3

2 1

= 0

6.40

θC

θB

= 2.423/EI

= 0.779/EI

2

12

[FEF]PL/8 = 3.75PL/8 = 3.75

10 kN

1.5 m1.5 m

2

7.13 kN 2.87 kN

6.40 kN�m

10 kN

2

PL/8 = 3.75PL/8 = 3.75

+-3.75

3.75

47

A CB

10 kN1 kN/m

9 m 1.5 m1.5 m

6.92 kN�m

11.57 kN 2.87 kN4.56 kN

4.56 kN 4.44 kN 7.13 kN 2.87 kN

1 kN/m6.92 6.40

6.40

10 kN

M (kN�m) x (m)

-6.92 -6.40

3.48 4.32

V (kN) x (m)

4.56

-4.44 -2.874.56 m

7.13

θB = +0.779/EIθC = +2.423/EI

48

20 kN

A CBEI2EI

4 m 4 m

9kN/m 40 kN�m

Example 2

For the beam shown, use the stiffness method to:(a) Determine the deflection and rotation at B.(b) Determine all the reactions at supports.

49

1 2 3

0.5625

3

-0.375

-0.375

1

2

3

4

5

6

[K] = EI

3 4

3

4

Use 4x4 stiffness matrix,

[k]1

0.375EI

-0.75EI2EI

0.75EI - 0.375EI

0.375EI

0.75EI

EI

-0.75EI

2EI

0.75EI

0.75EI

-0.375EI

EI

-0.75EI

-0.75EI

1 2 3 4

1

2

34

[k]2

0.375EI

0.375EI

-0.1875EI

0.5EI

-0.375EI

-0.375EI

0.1875EI

-0.375EIEI

0.375EI - 0.1875EI

0.1875EI

0.375EI

0.5EI

-0.375EI

EI

3 4 5 6

3

4

5

6

2EI EI

1 2

1

2

5

6

θi ∆j θj

Mi

∆i

Vj

Mj

Vi

[ ]

−−−−

−−

=


k

/4/6/2/6/6/12/6/12/2/6/4/6/6/12/6/12

22

2323

22

2323

50

20 kN9kN/m 40 kN�m

4 m 4 m

12 kN�m18 kN

12 kN�m

18 kN 18 kN

D4

D3

Global:

D4

D3=

9.697/EI

-61.09/EI

Q4 = 40

Q3 = -203

4 + -12

18 3

4

20 kN

40 kN�m

12 kN�m

1 2

1

2

3

4

5

6

1 2 3

2EI EI

[Q] = [K][D] + [QF]

0.5625

3

-0.375

-0.375= EI

3 4

3

4

51

Member 1:

1

9 kN/m

A B12 kN�m

[qF]1

12 kN�m

18 kN 18 kN

1

9 kN/m

A B

[q]1

53.21 kN�m

48.18 kN67.51 kN�m

12.18 kN

1

1

2

3

4

1 2

2EI 12 kN�m 12 kN�m

18 kN 18 kN

q2

q1

q4L

q3L

67.51

48.18

53.21

-12.18

12

18

-12

18

12(2EI)/43

-0.75EI2EI

0.75EI - 0.375EI

0.375EI

0.75EI

EI

-0.75EI

2EI

0.75EI

0.75EI

-0.375EI

EI

-0.75EI

-0.75EI

1 2 3 4

1

2

34

[q]1 = [k]1[d]1 + [qF]1

d3 = -61.09/EI

d4 = 9.697/EI

d2 = 0

d1 = 0

52

2B C

[q]2

Member 2:

[q]2 = [k]2[d]2 + [qF]2

2

3

4

5

6

2 3

EI

q4R

q3R

q6

q5

0.375EI

0.375EI

-0.1875EI

0.5EI

-0.375EI

-0.375EI

0.1875EI

-0.375EIEI

0.375EI - 0.1875EI

0.1875EI

0.375EI

0.5EI

-0.375EI

EI

3 4 5 6

3

4

5

6

0

0

0

0

d3 =-61.09/EI

d4 = 9.697/EI

d6 = 0

d5 = 0

18.06 kN�m

7.82 kN

13.21 kN�m

7.82 kN

-13.21

-7.818

-18.06

7.818

53

V (kN) x (m)

-7.818

48.18

-7.818-

M (kN�m) x (m)

13.21

-18.08-

-67.51

-

1

9 kN/m

A B

[q]1

53.21 kN�m

48.18 kN67.51 kN�m

12.18 kN2

B C

[q]2

18.06 kN�m

7.82 kN

13.21 kN�m

7.82 kN

12.18+

53.21

+

20 kN

4 m 4 m

9kN/m 40 kN�m

7.818 kN

18.06 kN�m67.51 kN�m

48.18 kN

D3 = ∆B = -61.09/EI

D4 = θB = +9.697/EI

54

20 kN

A CBEI2EI

4 m 2 m

9kN/m 40 kN�m10 kN

2 m

Example 3

For the beam shown, use the stiffness method to:(a) Determine the deflection and rotation at B.(b) Determine all the reactions at supports.

55

Global1 2

1

2

3

4

5

6

1 2 3

20 kN

A CB EI2EI


4 m 2 m 2 m

10 kN5 kN�m

5 kN

5 kN�m

5 kN

212 kN�m

18 kN

12 kN�m

18 kN

9 kN/m

1[FEF]

θi ∆j θj

Mi

∆i

Vj

Mj

Vi

[ ]

−−−−

−−

=×


k

/4/6/2/6/6/12/6/12/2/6/4/6/6/12/6/12

22

2323

22

2323

44

56

1 2

1

2

3

4

5

6

1 2 3 4 m 2 m 2 m

1

2

5

1 2

0.375

0.50.375 0.5 1

[K] = EI

3 4 6

3

46

0.5625

3

-0.375

-0.375

12(2EI)/43

[k]1 =-0.75EI2EI

0.75EI - 0.375EI

0.375EI

0.75EI

EI

-0.75EI

2EI/L

0.75EI

0.75EI

-0.375EI

EI

-0.75EI

-0.75EI

1 2 3 4

1

2

34

[k]2 =0.375EI

0.375EI

-0.1875EI

0.5EI

-0.375EI

-0.375EI

0.1875EI

-0.375EIEI

0.375EI - 0.1875EI

0.1875EI

0.375EI

0.5EI

-0.375EI

EI

3 4 5 6

3

4

5

6

57

MBA+MBC = 40

VBL+VBR = -20

MCB = 0θB

∆B

θC

-12 + 5 = -7-5

θB

∆B

θC

= -7.667/EI

-116.593/EI

52.556/EI


= EI

0.5625

3

-0.375

-0.375

0.375

0.50.375 0.5 1

3 4 6

3

46

20 kN

A CB


21

1

2

3

4

5

6

1 2 3 10 kN5 kN�m

5 kN

5 kN�m

5 kN

212 kN�m

18 kN

12 kN�m

18 kN

9 kN/m

1[FEF]

20 kN

40 kN�m

5 kN�m5 kN�m

5 kN

12 kN�m

18 kN

+18 + 5 = 23

58

0.375EI

= -0.75EI2EI

0.75EI - 0.375EI

0.375EI

0.75EI

EI

-0.75EI

2EI/L

0.75EI

0.75EI

-0.375EI

EI

-0.75EI

-0.75EI

1 2 3 4

1

2

34

=91.78

55.97

60.11

-19.97

91.78 kN�m

55.97 kN

Member 1: [q]1 = [k]1[d]1 + [qF]1

MAB

VA

MBA

VBL+

12

18

-12

18=-116.593/EI

=-7.667/EI

0

0

θB

∆B

1

1

2

3

4

1 2

12 kN�m

18 kN

12 kN�m

18 kN

9 kN/m

1

[FEF]

19.97 kN

60.11 kN�m4 m

9kN/m

1

12 kN�m

18 kN

12 kN�m

18 kN18 kN

59

VC

MBC

VBR

MCB

+ 5

5

5 -5

=-20.11

-0.0278

0

10.03=52.556/EI

=-116.593/EI

=-7.667/EIθB

∆B

θC

0

10 kN5 kN�m

5 kN

5 kN�m

5 kN

2

[FEM]

10.03 kN

2 m

10 kN

2 m2

= 0.375EI

0.375EI

-0.1875EI

0.5EI

-0.375EI

-0.375EI

0.1875EI

-0.375EIEI

0.375EI - 0.1875EI

0.1875EI

0.375EI

0.5EI

-0.375EI

EI

3 4 5 6

3

4

5

6

0.0278 kN

20.11 kN�m

Member 2: [q]1 = [k]1[d]1 + [qF]1

223

5

6

3

45 kN�m

5 kN

5 kN�m

5 kN

60

20 kN

4 m 2 m


2 m

10.03 kN

91.78 kN�m

55.97 kN

V (kN) x (m)

55.97

-0.0278

19.97+

-10.03 -10.03-

91.78 kN�m

55.97 kN 19.97 kN60.11 kN�m4 m

9kN/m

110.03 kN0.0278 kN

20.11 kN�m2 m

10 kN2 m

2

91.78 kN�m

60.11 kN�m

20.11 kN�m

M (kN�m) x (m)

-91.78

-

+20.11

60.11

θB = -7.667/EI

θC = 52.556/EI

∆B = -116.593/EI

61

40 kN

8 m 4 m 4 m

2EI EI

6 kN/m

AB

C∆B = -10 mm

Example 4

For the beam shown:(a) Use the stiffness method to determine all the reactions at supports.(b) Draw the quantitative free-body diagram of member.(c) Draw the quantitative shear diagram, bending moment diagramand qualitative deflected shape.Take I = 200(106) mm4 and E = 200 GPa and support B settlement 10 mm.

62

[k]1 =4

8

8

4

1 2

1

2

EI8

1 2

1 2 3

2EI EI

1

2

42

2 3

2

3[K] = EI

8

12

2

4

4

2

2 3

2

3[k]2 = EI

8

Use 2x2 stiffness matrix:θi θj

Mi

Mj[ ]

=× LEILEI

LEILEIk

/4/2/2/4

22

63

75 kN�m

6(2EI)∆/L2 = 75 kN�m10 mm

1 37.5 kN�m

6(EI)∆/L2 = 37.5 kN�m10 mm

2[FEM]∆∆∆∆

6(EI)∆/L2 = 37.5 kN�m

D2

D3

=-61.27/EI rad

185.64/EI rad

40 kN

2EI EI

6 kN/m

A

B C

∆B = -10 mm 1 2

1 2 3

[FEM]load

32 kN�mwL2/12 = 32 kN�m

6 kN/m

1 40 kN�mPL/8 = 40 kN�m 2

40 kN8 m 4 m 4 m

32 kN�m 40 kN�mPL/8 = 40 kN�m

75 kN�m 37.5 kN�m

+-32 + 40 + 75 -37.5 = 45.5Q2 = 0

Q3 = 0 -40 - 37.5 = -77.5

D2

D3

2

42

2 3

2

3

12

8EI

=


64

6(2EI)∆/L2 = 75 kN�m

[FEM]load

32 kN�mwL2/12 = 32 kN�m

6 kN/m

1

75 kN�m

6(2EI)∆/L2 = 75 kN�m10 mm

1[FEM]∆∆∆∆

6 kN/m

18 m

1

2

2EI

1

q1

q2

d1 = 0

d2 = -61.27/EI=

76.37 kN�m

-18.27 kN�m

Member 1: [q]1 = [k]1[d]1 + [qF]1

32 kN�mwL2/12 = 32 kN�m

75 kN�m

+32 + 75 = 107

-32 + 75 = 43

31.26 kN

76.37 kN�m

16.74 kN

18.27 kN�m

4

8

8

4

1 2

1

28EI

=

65

Member 2: [q]2 = [k]2[d]2 + [qF]2

2

2 3

40 kN�mPL/8 = 40 kN�m 2

40 kN

37.5 kN�m

6(EI)D/L2 = 37.5 kN�m10 mm

2

[FEM]load

[FEM]∆∆∆∆

40 kN�m

37.5 kN�m

6(EI)D/L2 = 37.5 kN�m

PL/8 = 40 kN�m

=18.27 kN�m

0 kN�m

q2

q3

+40 - 37.5 = 2.5

-40 - 37.5 = -77.5

d2 = -61.27/EI

d3 = 185.64/EI

17.72 kN

18.27 kN�m

22.28 kN

40 kN

2

2

4

4

2

2 3

2

38EI

=

66

Alternate method: Use 4x4 stiffness matrix

2EI EI

1 2

2

5

6

3

4

1

8EI

= 1.5

1.5

-0.375

12(2)/82

4

-1.5

8

1.5

-1.5

-1.5

-0.375

0.375

1.5

4

-1.5

8

1 2 3 41

2

3

4

[k]1

8EI

=0.75

0.75

-0.1875

12/82

2

-0.75

4

0.75

-0.75

-0.75

-0.1875

0.1875

0.75

2

-0.75

4

3 4 5 63

4

5

6

[k]2

2-0.75

-0.1875-0.75

0.1875-0.75

0.752

-0.754

-0.18750.75

1.5 4

1.54

0.375

-0.3751.5

1.58

-1.5

-0.375-1.5

00

00

00

00

-0.7512

0.5625-0.758

EI=[K]

123456

1 4 5 62 3

67


40 kN

8 m 4 m 4 m

2EI EI

6 kN/m

A

B C

∆B = -10 mm 1 2

2

5

6

3

4

1

40 kN40 kN�m

20 kN

40 kN�m

20 kN

232 kN�m

24 kN

32 kN�m

24 kN

6 kN/m

1[FEF]

53

2

1

D4

D6=

-61.27/EI = -1.532x10-3 rad

185.64/EI = 4.641x10-3 rad

+(200x200/8)(-0.75)(-0.01) = 37.5

(200x200/8)(0.75)(-0.01) = -37.5

D4

D6

Q4 = 0

Q6= 0

2

4

12

2+

8

-40

4 6

4

68EI

=

-0.75-0.75 D5 = -0.01)

8200200( ×

+ 46

5

D4

D6

Q4 = 0

Q6= 0

2

4

12

2

4 6

4

68EI

= +8

-40

68

+32

24

-32

24

q1

q2

q4

q3

1.5

1.5

-0.375

12(2)/82

4

-1.5

8

1.5

-1.5

-1.5

-0.375

0.375

1.5

4

-1.5

8

1 2 3 41

2

3

4

= (200x200)8

q1

q2

q4

q3=

76.37 kN�m

31.26 kN

-18.27 kN�m

16.74 kN

Member 1: [q]1 = [k]1[d]1 + [qF]1

1

2

3

4

1∆B = -10 mm 32 kN�m

24 kN

32 kN�m

24 kN

6 kN/m

1[FEF]load

d2 = 0

d1 = 0

d4 = -1.532x10-3

d3 = -0.01

6 kN/m

18 m

31.26 kN

76.37 kN�m

16.74 kN

18.27 kN�m

69

Member 2:

+40

20

-40

20

q3

q4

q6

q5

d4 = -1.532x10-3

d3 = -0.01

d6 = 4.641x10-3

d5 = 0

0.75

0.75

-0.1875

12/82

2

-0.75

4

0.75

-0.75

-0.75

-0.1875

0.1875

0.75

2

-0.75

4

3 4 5 63

4

5

6

= (200x200)8

q3

q4

q6

q5=

18.27 kN�m

22.28 kN

0 kN�m

17.72 kN17.72 kN

18.27 kN�m

22.28 kN

40 kN

2

2

5

6

3

4

-10 mm = ∆B

40 kN40 kN�m

20 kN

40 kN�m

20 kN

2[FEF]

70

40 kN

8 m 4 m 4 m

2EI EI

6 kN/m

A

B C

∆B = -10 mm31.26 kN

76.37 kN�m

16.74 + 22.28 kN 17.72 kN

V (kN)

x (m)+ +--

31.26

-16.74

22.28

-17.725.21 m

M (kN�m) x (m)

-76.37

5.06

-18.27

70.85

+

--

D6 = θC = 4.641x10-3 rad

d4 = θB = -1.532x10-3 rad

Deflected Curve

∆B = -10 mm

71

Example 5

For the beam shown:(a) Use the stiffness method to determine all the reactions at supports.(b) Draw the quantitative free-body diagram of member.(c) Draw the quantitative shear diagram, bending moment diagramand qualitative deflected shape.Take I = 200(106) mm4 and E = 200 GPa and support C settlement 10 mm.

4 kN

8 m 4 m 4 m

2EI EI

0.6 kN/m

AB

C∆C = -10 mm

20 kN�m

72

2EI EI

1 2-10 mm

2

5

6

3

4

1

2

51

Q3

Q6

Q4

0.75 2

0.7524

1.5

1.5

-0.375

12(2)/82

4

-1.5

8

1.5

-1.5

-1.5

-0.375

0.375

1.5

4

-1.5

8

1 2 3 41

2

3

4

[k]1

8EI

=

8EI

=0.75

0.75

-0.1875

12/82

2

-0.75

4

0.75

-0.75

-0.75

-0.1875

0.1875

0.75

2

-0.75

4

3 4 5 63

4

5

6

[k]2

D3

D6

D4

-0.1875

-0.75-0.75 D5 = -0.01

� Global: [Q] = [K][D] + [QF]

�Member stiffness matrix [k]4x4

0.5625-0.75

-0.7512

346

3 4 6

8EI

= )8

200200( ×+

346

5QF

3

QF6

QF4+

73

4 kN4 kN�m

2 kN

4 kN�m

2 kN

23.2 kN�m

2.4 kN

3.2 kN�m

2.4 kN

0.6 kN/m

1[FEF]

4 kN

8 m 4 m 4 m2EI EI

0.6 kN/m

A

BC

10 mm

20 kN�m2EI EI

1 2

6

3

42

51

346

0.5625

0.75-0.75

-0.75122

0.752

3 4 6

48EI

=

D3

D6

D4 +9.375

37.537.5

2.4+2 = 4.4

-4.0-3.2+4 = 0.8+

D3

D6

D4

-377.30/EI = -9.433x10-3 m

+74.50/EI = +1.863x10-3 rad-61.53/EI = -1.538x10-3 rad=


Q3 = 0

Q6 = 20Q4 = 0

74

+3.2

2.4

-3.2

2.4

q1

q2

q4

q3

q1

q2

q4

q3=

43.19 kN�m

8.55 kN

6.03 kN�m

-3.75 kN

Member 1: [q]1 = [k]1[d]1 + [qF]1

d2 = 0

d1 = 0

d4 = -1.538x10-3

d3 = -9.433x10-3

0.6 kN/m

18 m

8.55 kN

43.19 kN�m

3.75 kN

6.03 kN�m

1

2

3

4

13.2 kN�m

2.4 kN

3.2 kN�m

2.4 kN

0.6 kN/m

1[FEF]load 8 m

1.5

1.5

-0.375

12(2)/82

4

-1.5

8

1.5

-1.5

-1.5

-0.375

0.375

1.5

4

-1.5

8

1 2 3 41

2

3

4

8200200×

=

75

Member 2:

+4

2

-4

2

q3

q4

q6

q5

d4 = -1.538x10-3

d3 = -9.433x10-3

d6 = 1.863x10-3

d5 = -0.01

q3

q4

q6

q5=

-6.0 kN�m

3.75 kN

20.0 kN�m

0.25 kN0.25 kN

6.0 kN�m

3.75 kN

4 kN

2

10 mm2

5

6

3

4 4 kN4 kN�m

2 kN

4 kN�m

2 kN

2[FEF]

8 m

0.75

0.75

-0.1875

12/82

2

-0.75

4

0.75

-0.75

-0.75

-0.1875

0.1875

0.75

2

-0.75

4

3 4 5 63

4

5

6

8200200×

=

20 kN�m

76

4 kN

8 m 4 m 4 m2EI EI

0.6 kN/m

A

BC

10 mm

20 kN�m

Deflected Curve

0.6 kN/m

18 m

8.55 kN

43.19 kN�m

3.75 kN

6.03 kN�m

V (kN)x (m)

+

8.553.75

-0.25

0.25 kN

6.0 kN�m

3.75 kN

4 kN

2

20 kN�m

M (kN�m) x (m)

-43.19

621+

-

20

D3 = -9.433 mm

D6 = +1.863x10-3 radD4 = -1.538x10-3 rad

D5 = -10 mm

77

30 kN

A CB

EI2EI

4 m 2 m

9kN/mHinge

2 m

Example 6

For the beam shown, use the stiffness method to:(a) Determine all the reactions at supports.(b) Draw the quantitative shear and bending moment diagramsand qualitative deflected shape.E = 200 GPa, I = 50x10-6 m4.

Internal Hinges

78

[K] = EI 1

2

1 2

0.0

0.0

1.0

2.0

A C

B

2EI

4 m 2 m 2 m

EI1 2

43

1

2

[k]1 =EI

2EI

2EI

EI

3 1

3

1[k]2 =

0.5EI

1EI

1EI

0.5EI

2 4

2

4

3 4

Use 2x2 stiffness matrix,θi θj

Mi

Mj[ ]

=× LEILEI

LEILEIk

/4/2/2/4

22

79

0.0

0.0

D1

D2

15 kN�mB C

15 kN�m30 kN

2

12 kN�m

A B

12 kN�m 9kN/m

1[FEF]

= EI 1

2

1 2

0.0

0.0

1.0

2.0

D1

D2=

0.0006 rad

-0.0015 rad

-12

15+

Global matrix:

30 kN9kN/m

Hinge

A C

B1 2

43

12 kN�m 15 kN�m1

2

80

Member 1:

q3

q1

12

-12+=

EI

2EI

2EI

EI

3 1

3

1

18

0.0=

d3

d1

= 0.0

= 0.0006

12 kN�m

A B

12 kN�m 9kN/m

1[FEF]A

B1

31

18 kN�m22.5 kN 13.5 kN

A B

9 kN/m

1

81

B C30 kN

CB2

4

2

15 kN�mB C

15 kN�m 30 kN

2

q2

q4

= -0.0015

= 0.0

d2

d4

15

-15+=

0.5EI

1EI

1EI

0.5EI

2 4

2

4

0.0

-22.5=

Member 2:

22.5 kN�m

20.63 kN9.37 kN

82

30 kN

A CB

EI2EI

4 m 2 m

9kN/mHinge

2 m

-20.63

V (kN) x (m)

22.59.37

-13.5

+

2.5 m

M (kN�m) x (m)

-18

10.13

-22.5

18.75

+-

+--

θBL= 0.0006 rad θBR= -0.0015 rad

13.5 kN22.5 kN

22.5 kN�mB C

30 kN

20.63 kN9.37 kN

A B18 kN�m

9 kN/m

9.37 kN13.5 kN

22.87 kN

83

20 kN

A CBEI2EI

4 m 4 m

9kN/m

Hinge

Example 7

For the beam shown, use the stiffness method to:(a) Determine all the reactions at supports.(b) Draw the quantitative shear and bending moment diagramsand qualitative deflected shape.E = 200 GPa, I = 50x10-6 m4.

84

A CB1 2

2EI EI 6

7

4

5

4

5

6

7

0.375

0.375

1.0

-0.75

-0.75 2.0

0.0

0[K] = EI 1

2

3

1 2 3

0.5625

[k]1 =0.75EI

EI-0.75EI

2EI/L

0.375EI0.75EI

0.75EI-0.375EI

2EI0.75EI

EI-0.75EI

-0.75EI- 0.375EI

0.375EI-0.75EI

4 5 1 24

51

2

0.375EI

0.375EI-0.1875EI

0.5EI -0.375EI

-0.375EI

[k]2 =

0.1875EI

-0.375EIEI0.375EI - 0.1875EI

0.1875EI

0.375EI0.5EI

-0.375EI EI

1 3 6 71

36

7

1

2

3

85

Q1 = -20

Q3 = 0.0

Q2 = 0.0

D1

D3

D2

18

0.0

-12+

D1

D3

D2 =-0.02382 m

0.008933 rad

-0.008333 rad

[FEM]

12 kN�mA B

12 kN�m

18 kN18 kN1

9kN/m

20 kN

A CB

9kN/m

A CB1 2

6

7

4

5

2EI EI

Global:

= EI 1

2

3

1

0.5625

0.375

-0.75

2

-0.75

2.0

0.0

3

0.375

0

1.0

12 kN�m

18 kN

20 kN

1

2

3

86

A B1

2EI1

24

5

[FEF]

12 kN�mA B

12 kN�m

18 kN18 kN1

9kN/m

+12

18

-12

18

Member 1:

q4

q5

q1

q2

=

0.75EI

EI

-0.75EI

2EI/L

0.375EI

0.75EI

0.75EI

-0.375EI

2EI

0.75EI

EI

-0.75EI

-0.75EI

- 0.375EI

0.375EI

-0.75EI

4 5 1 2

4

5

1

2

= 0.0

= 0.0

= -0.02382

= -0.00833

d5

d4

d2

d1

= 107.32

44.83

0.0

-8.83

A B

107.32 kN�m

8.83 kN44.83 kN

9kN/m

1

87

Member 2:

CB2

EI1

3

6

7

44.66 kN�mB C

11.16 kN11.16 kN

2

+0

0

0

0

= -0.02382

= 0.008933

= 0.0

= 0.0

d3

d1

d7

d6

q1

q3

q7

q6

0.375EI

0.375EI

-0.1875EI

0.5EI

-0.375EI

-0.375EI

=

0.1875EI

-0.375EIEI

0.375EI - 0.1875EI

0.1875EI

0.375EI

0.5EI

-0.375EI

EI

1 3 6 71

3

6

7

= 0.0

-11.16

-44.66

11.16

88

20 kN

A CB4 m 4 m

9kN/m

V (kN) x (m)

44.83

8.83

-11.16 -11.16

+

-

M (kN�m) x (m)

-107.32

-44.66-

-

A B

107.32 kN�m

8.83 kN44.83 kN

9kN/m

144.66 kN�m

B C

11.16 kN11.16 kN

2

θBL= -0.008333 rad θBR= 0.008933 rad

89

30 kN

A C

BEI2EI

4 m 2 m

9kN/mHinge

40 kN�m

2 m

Example 8

For the beam shown, use the stiffness method to:(a) Determine all the reactions at supports.(b) Draw the quantitative shear and bending moment diagramsand qualitative deflected shape.40 kN�m at the end of member AB. E = 200 GPa, I = 50x10-6 m4.

90

[K] = EI 1

2

1 2

0.0

0.0

1.0

2.0

A C

B

2EI

4 m 2 m 2 m

EI1 2

43

[k]1 =EI

2EI

2EI

EI

3 1

3

1[k]2 =

0.5EI

1EI

1EI

0.5EI

2 4

2

4

3 4

1

2

Use 2x2 stiffness matrix: θi θj

Mi

Mj[ ]

=× LEILEI

LEILEIk

/4/2/2/4

22

91

15 kN�mB C

15 kN�m30 kN

2

12 kN�m

A B

12 kN�m 9kN/m

1[FEM]

Global matrix:

A C

B1 2

43

1

2

12 kN�m 15 kN�m

30 kN

A C

BEI2EI

9kN/mHinge

40 kN�m

Q1 = 40

Q2 = 0.0

D1

D2

-12

15+

D1

D2=

0.0026 rad

-0.0015 rad

= EI 1

2

1 2

0.0

0.0

1.0

2.0

40 kN�m

92

12 kN�m

A B

12 kN�m 9kN/m

1[FEF]A

B1

31

40 kN�mA B38 kN�m

9 kN/m

1.5 kN37.5 kN

Member 1:

38

40=

q3

q1

= 0.0

= 0.0026

d3

d1

12

-12+=

EI

2EI

2EI

EI

3 1

3

1

93

CB2

4

2

15 kN�mB C

15 kN�m 30 kN

2

Member 2:

q2

q4

= -0.0015

= 0.0

d2

d4

15

-15+=

0.5EI

1EI

1EI

0.5EI

2 4

2

4

0.0

-22.5=

22.5 kN�mB C

30 kN

20.63 kN9.37 kN

94

9.37 kN

1.5 kN

30 kN

A C

B

4 m 2 m

9kN/mHinge

40 kN�m

2 m

-20.63

V (kN) x (m)

37.5

9.371.5+

M (kN�m) x (m)

-38

40

-22.5

+-

18.75+

--

θBL= 0.0026 radθBR=- 0.0015 rad

7.87 kN40 kN�mA B

38 kN�m

9 kN/m

1.5 kN37.5 kN

22.5 kN�mB C

30 kN

20.63 kN9.37 kN

95

20 kN

A CBEI2EI

4 m 4 m

9kN/m

Hinge40 kN�m

Example 9

For the beam shown, use the stiffness method to:(a) Determine all the reactions at supports.(b) Draw the quantitative shear and bending moment diagramsand qualitative deflected shape.40 kN�m at the end of member AB. E = 200 GPa, I = 50x10-6 m4

96

A CB1 2

6

7

4

51

2

3

20 kN

A CBEI2EI

4 m 4 m

9kN/m

Hinge40 kN�m

12 kN�m

18 kN

12 kN�m

18 kN

9 kN/m

1[FEM]θi ∆j θj

Mi

∆i

Vj

Mj

Vi

[ ]

−−−−

−−

=×


k

/4/6/2/6/6/12/6/12/2/6/4/6/6/12/6/12

22

2323

22

2323

44

97

1 22A CB

6

7

4

5

[K] = EI 1

2

3

1 2 3

6

7

4

5

0.5625

0.0

0

0.375

0.375

1.0

-0.75

-0.75

2.0

[k]1 = 0.75EI

EI-0.75EI

2EI

0.375EI0.75EI

0.75EI-0.375EI

2EI0.75EI

EI-0.75EI

-0.75EI- 0.375EI

0.375EI-0.75EI

4 5 1 24

51

2

1

0.375EI

0.375EI-0.1875EI

0.5EI -0.375EI

-0.375EI

[k]2 =

0.1875EI-0.375EIEI

0.375EI - 0.1875EI

0.1875EI

0.375EI0.5EI

-0.375EI EI

1 3 6 71

36

7

2EI EI1

2

3

98

Q1 = -20

Q3 = 0.0

Q2 = 40

D1

D3

D2

18

0.0

-12+

D1

D3

D2 =-0.01316 m

0.0049333 rad

-0.002333 rad

A CB1 2

6

7

4

5

2EI EI

20 kN

EI

9kN/m

40 kN�m

20 kN

40 kN�m

12 kN�m

18 kN

12 kN�m

18 kN

9 kN/m

1[FEF]

= EI 1

2

3

1 2 3

0.5625

0.0

0

0.375

0.375

1.0

-0.75

-0.75

2.0

12 kN�m

18 kNGlobal:

1

2

3

99

1

2EI

[q]1

1

12

1

2EI4

5

+12

18

-12

18

q4

q5

q1

q2

=

0.75EI

EI

-0.75EI

2EI

0.375EI

0.75EI

0.75EI

-0.375EI

2EI

0.75EI

EI

-0.75EI

-0.75EI

- 0.375EI

0.375EI

-0.75EI

4 5 1 2

4

5

1

2

= 87.37

49.85

40

-13.85

12 kN�m

18 kN

12 kN�m

18 kN

9 kN/m

1[FEF]

Member 1: [q]1 = [k]1[d]1 + [qF]1

13.85 kN

40 kN�m87.37 kN�m

49.85 kN

= 0.0

= 0.0

= -0.01316

= -0.002333

d5

d4

d2

d1

12 kN�m

18 kN

12 kN�m

18 kN1

100

22 CB

1

3

EI 6

7

Member 2: [q]2 = [k]2[d]2 + [qF]2

+0

0

0

0

q1

q3

q7

q6

0.375EI

0.375EI

-0.1875EI

0.5EI

-0.375EI

-0.375EI

=

0.1875EI

-0.375EIEI

0.375EI - 0.1875EI

0.1875EI

0.375EI

0.5EI

-0.375EI

EI

1 3 6 71

3

6

7

= 0.0

-6.18

-24.69

6.18

24.69 kN�m

6.18 kN6.18 kN

= -0.01316

= 0.004933

= 0.0

= 0.0

d3

d1

d7

d6

22

[q]2

101

20 kN

A CB

EI2EI

4 m 4 m

9kN/m 40 kN�m

24.69 kN�mB C

6.18 kN6.18 kN

A B

87.37 kN�m 13.85 kN49.85 kN

40 kN�m

M (kN�m) x (m)

-

-87.37

+40

--24.69

θBL= -0.002333 rad θBR = 0.004933 rad

49.85

-6.18

V (kN) x (m)-6.18

-13.85+

102

� Fixed-End Forces (FEF)- Axial- Bending

� Curvature

Temperature Effects

103

Tm> TR

� Thermal Fixed-End Forces (FEF)

Tl

Tt

2bt

mTTT +

=

Tt

Tb > Tt

βF

AFF

BF

BFBMF

AM

Rmaxial TTT −=∆ )(

tbbending TTT −=∆ )(

)()(dTyTy

∆=∆

y β

)(tandT

dTT tb ∆

=−

=β

A

A B

σaxialσaxial

A B

σy

y

TR Tm

Tt

Tb

Tb - Tt

Room temp = TR

ct

cbd

TR

NA

104

- Axial

A B

σaxialσaxial

FAF F

BF

dAFA

axialF

A ∫= σ

dAE axial∫= ε

dATE axial∫ ∆= )(α

axialTEA )(∆= α

AETTF RmAF

axial )()( −= α

dATE axial ∫∆= )(α

Tt

Tb > Tt

Tm> TR

Rmaxial TTT −=∆ )(

TR Tm

105

- Bending

A B

σy

y

dAyMA

yFA ∫= σ

dATyE y∫ ∆= )(α

dAdTyyE∫

∆= )(α

∫∆

= dAydTE 2)(α

EId

TTF ulA

Fbending )()( −

= α

FBMF

AM

dAyE∫= ε

tbbending TTT −=∆ )(

)()(dTyTy

∆=∆

y β Tt

Tb

106

O´

ds´

dx

Afterdeformation

dθ

dθ

Before deformation

y dx

ds = dx y

ρ)()( θρ ddx =

)( θdy )(1dxdθ

ρ=

� Elastic Curve: Bending

107

dxdTyyd )()( ∆

= αθ

dxdTd )()( ∆

= αθ

Tb

Tt

yy

dT∆

� Bending Temperature

dx

Tt

Tb

Tl > Tu

y

Tl > Tu

Tu

O

dθ

2bt

mTTT +

= dθ

MM

Tb

cb

ct

Tt

∆T = Tb - Tt

dT∆

=β

EIM

dT

dxd

=∆

== )(1)( αρ

θ

108

A CB

EI2EI

4 m 4 m

T2 = 40oC

T1 = 25oC

182 mm

Example 10

For the beam shown, use the stiffness method to:(a) Determine all the reactions at supports.(b) Draw the quantitative shear and bending moment diagramsand qualitative deflected shape.Room temp = 32.5oC, α = 12x10-6 /oC, E = 200 GPa, I = 50x10-6 m4.

109

12 3

1 2

Mean temperature = (40+25)/2 = 32.5Room temp = 32.5oC

FEM+7.5 oC-7.5 oC

a(∆T /d)EI19.78 kN�m = 19.78 kN�m

A CB

EI2EI

4 m 4 m

T2 = 40oC

T1 = 25oC

182 mm

mkNEIdTF F

bending •=××−

×=∆

= − 78.19)502002)(182.0

2540)(1012()2)(( 6α

110

[M1] = 3EIθ1 + (1.978x10-3)EI0

Element 1:

M1

M2

q1

q2

[q] = [k][d] + [qF]

+ 1.978

-1.978(10-3) EI= 1

2

2

1

2 1

2

1EI

Element 2:

M3

M1

q3

q1 + 0

0= 0.5

1

1

0.5

1 3

1

3EI

θ1 = -0.659x10-3 rad

12 3

A C

BEI2EI

4 m 4 m

1 2

19.78 kN�m19.78 kN�m182 mm

111

Element 2:

M3

M1

q3

q1= + 0

0

0.5

1

1

0.5

1 3

1

3EI

= -0.659x10-3

= 0= 6.59 kN�m

-26.37 kN�m

= -0.659x10-3

= 0= -3.30 kN�m

-6.59 kN�m

Element 1:

M1

M2 = q1

q2 + 19.78

-19.78

1

2

2

1

2 1

2

1EI

4.95 kN4.95 kN

26.37 kN�m 6.59 kN�m

A B

6.59 kN�m 3.3 kN�m

B C

2.47 kN2.47 kN

12 3

A C

BEI2EI

4 m 4 m

1 2

19.78 kN�m19.78 kN�m182 mm

112

A CB

4 m 4 m

+7.5 oC

-7.5 oC 3.3 kN�m

26.37 kN�m

4.95 kN2.47 kN

2.47 kN

M (kN�m) x (m)

26.37

6.59

-3.30

+-

V (kN) x (m)

-4.95-2.47

- -

θB = -0.659x10-3 radDeflected curve x (m)

113

A CB

EI, AE2EI, 2AE

4 m 4 m

T2 = 40oC

T1 = 25oC

182 mm

Example 11

For the beam shown, use the stiffness method to:(a) Determine all the reactions at supports.(b) Draw the quantitative shear and bending moment diagramsand qualitative deflected shape.Room temp = 28 oC, a = 12x10-6 /oC, E = 200 GPa, I = 50x10-6 m4,A = 20(10-3) m2

114

A

CB

EI2EI

4 m 4 m

T2 = 40oC

T1 = 25oC1

2 3

7

8

9

1 24

5

6

Element 1:

mkNm

mkNmLAE /)10(2

)4()/10200)(1020(2)2( 6

2623

=××

=−

mkNm

mmkNLEI

•=×××

=−

)10(20)4(

)1050)(/10200(24)2(4 34626

mkNm

mmkNLEI

•=×××

=−

)10(10)4(

)1050)(/10200(22)2(2 34626

kNm

mmkNLEI )10(5.7

)4()1050)(/10200(26)2(6 3

2

4626

2 =×××

=−

mkNm

mmkNL

EI /)10(75.3)4(

)1050)(/10200(212)2(12 33

4626

3 =×××

=−

115

Fixed-end forces due to temperatures

Mean temperature(Tm) = (40+25)/2 = 32.5 oC ,TR = 28 oC

A CB

EI, AE2EI, 2AE

4 m 4 m

T2 = 40oC

T1 = 25oC

182 mm

mkNEIdTF F

bending •=××−

×=∆

= − 78.19)502002)(182.0

2540)(1012()2)(( 6α

19.78 kN�m19.78 kN�m

432 kN432 kNA B

+12 oC

-3 oC

kNmkNmAETF Faxial 432)/10200)(310202)(285.32)(1012()( 2626 =×−××−×=∆= −α

116

FEM

+12 oC

-3 oC

19.78 kN�m

A B

19.78 kN�m

432 kN432 kN

Element 1:

[q] = [k][d] + [qF]

0

0

0

d1

d2

d3

q4

q6

q5

q1

q2q3

=

4

5

6

1

2

3

4 1

- 2x106

0.00

0.00

2x106

0.00

0.00

2

0.00

-3750

-7500

0.00

3750

-7500

3

0.00

7500

10x103

0.00

-7500

20x103

2x106

0.00

0.00

-2x106

0.00

0.00

5

0.00

3750

7500

0.00

-3750

7500

6

0.00

7500

20x103

0.00

-7500

10x103

432

-19.78

0.00

-432

0.00

19.78

+

A CB

EI, AE2EI, 2AE

4 m 4 m

T2 = 40oC

T1 = 25oC

182 mm1

2 3

7

8

9

1 24

5

6

117

Element 2:

[q] = [k][d] + [qF]

d1

d3

d2

0

0

0

0

0

0

0

0

0

+

q1

q3

q2

q7

q8q9

- 1x106

0.00

0.00

1x106

0.00

0.00

0.00

-1875

-3750

0.00

1875

-3750

0.00

3750

5x103

0.00

-3750

10x103

1x106

0.00

0.00

-1x106

0.00

0.00

0.00

1875

3750

0.00

-1875

3750

0.00

3750

10x103

0.00

-3750

5x103

7 8 91 2 3

=

1

2

3

7

8

9

A CB

EI, AE2EI, 2AE

4 m 4 m

T2 = 40oC

T1 = 25oC

182 mm1

2 3

7

8

9

1 24

5

6

118

D1

D3

D2 =0.000144 m

-719.3x10-6 rad

-0.0004795 m

Q1 = 0.0

Q3 = 0.0

Q2 = 0.0

D1

D3

D2

-432

19.78

0.0+= 1

2

3

1

3x106

0.0

0.0

2

0.0

5625

-3750

3

0.0

-3750

30x103

Global:

A CB

EI, AE2EI, 2AE

4 m 4 m

T2 = 40oC

T1 = 25oC

182 mm1

2 3

7

8

9

1 24

5

6

119

= 144x10-6

= -479.5x10-6

= -719.3x10-6

Element 1:

0

0

0

d1

d2

d3

q4

q6

q5

q1

q2q3

=

4

5

6

1

2

3

4 1

- 2x106

0.00

0.00

2x106

0.00

0.00

2

0.00

-3750

-7500

0.00

3750

-7500

3

0.00

7500

10x103

0.00

-7500

20x103

2x106

0.00

0.00

-2x106

0.00

0.00

5

0.00

3750

7500

0.00

-3750

7500

6

0.00

7500

20x103

0.00

-7500

10x103

432

-19.78

0.00

-432

0.00

19.78

+

=

q4

q6

q5

q1

q2q3

144.0 kN

-23.38 kN�m

-3.60 kN

-144.0 kN

3.60 kN

9.00 kN�m

A B 1

2 3

4

5

6

A B144 kN

3.60 kN

23.38 kN�m

144 kN

3.60 kN

9 kN�m

120

= 144x10-6

= -479.5x10-6

= -719.3x10-6

144 kN

-9 kN�m

-3.6 kN

-144 kN

3.6 kN

-5.39 kN�m

=

q1

q3

q2

q7

q8q9

B C 7

8 9

1

2

3

144 kNB C

3.60 kN

9 kN�m

144 kN

3.60 kN

5.39 kN�m

Element 2:

d1

d3

d2

0

0

0

0

0

0

0

0

0

+

q1

q3

q2

q7

q8q9

=

- 1x106

0.00

0.00

1x106

0.00

0.00

0.00

-1875

-3750

0.00

1875

-3750

0.00

3750

5x103

0.00

-3750

10x103

1x106

0.00

0.00

-1x106

0.00

0.00

0.00

1875

37500

0.00

-1875

3750

0.00

3750

10x103

0.00

-3750

5x103

7 8 91 2 3

1

2

3

7

8

9

121

A CB

EI2EI

4 m 4 m

T2 = 40oC

T1 = 25oC

Isolate axial part from the system

RCRA

RA RA = RC+

Compatibility equation: dC/A = 0

RA = 144 kN

RC = -144 kN

0)4)(285.32(1012)4(2

)4( 6 =−×++ −

AER

AER AA

122

A B

144 kN

3.60 kN

23.38 kN�m

144 kN

3.60 kN

9 kN�m

144 kN

B C3.60 kN

9 kN�m

144 kN

3.60 kN

5.39 kN�m

V (kN) x (m)

-3.6

-

M (kN�m) x (m)

23.388.98

-5.39

+-

D3 = -719.3x10-6 radDeflected curve x (m)

A CB

EI2EI

4 m 4 m

T2 = 40oC

T1 = 25oC

D2 = -0.0004795 mm

123

Skew Roller Support

- Force Transformation- Displacement Transformation- Stiffness Matrix

124

m

i

j

m

i

j

x´y´

x

y

� Displacement and Force Transformation Matrices

12

3

45

6

θyθx

4´

5´

6´

1´

2´

3´

125

λx λy

i

jθy

θx

y´

x´

4´5´

6´

1´

2´

3´

θy

θx

m

i

j

x

y

12

3

45

6

Force Transformation

Lxx ij

x

−=λ

Lyy ij

y

−=λ

yx qqq θθ coscos 5'4'4 −=

xy qqq θθ coscos 5'4'5 −=

6'6 qq =

6

5

4

qqq

−=

10000

xy

yx

λλλλ

6'

5'

4'

qqq

−

−

=

6'

5'

4'

3'

2'

1'

6

5

4

3

2

1

1000000000000000010000000000

qqqqqq

λλλλ

λλλλ

qqqqqq

xy

yx

xy

yx

[ ] [ ] [ ]'qTq T=

126x

y

θyθx

x

y

θyθx

d4 d 4 cos θ x

d 4 cos θ yθy

θx

d5

θx

θy

d 5 cos θ x

d 5 cos θ y

x´

Displacement Transformation

j4´

5´

6´j4

5

6λx λy

yx θdθdd coscos' 544 +=

xy θdθdd' coscos 545 +−=

66' dd =

6'

5'

4'

ddd

−=

10000

xy

yx

λλλλ

6

5

4

ddd

−

−

=

6

5

4

3

2

1

6'

5'

4'

3'

2'

1'

1000000000000000010000000000

dddddd

λλλλ

λλλλ

dddddd

xy

yx

xy

yx

[ ] [ ][ ]dTd ='

127

[ ] [ ] [ ]'qTq T=

[ ] [ ][ ] [ ])'( FT qd'k'T +=

[ ] [ ][ ] [ ] [ ]FTT qTd'k'T '+=

[ ] [ ] [ ][ ][ ] [ ] [ ] [ ][ ] [ ]FFTT qdkqTdTk'Tq +=+= '

[ ] [ ] [ ][ ]TkTkTherefore T ', =

[ ] [ ] [ ]FTF qTq '=

[ ] [ ] [ ]'qTq T=

[ ] [ ][ ]dTd ='

[ ] [ ] [ ][ ]TkTk T '=

128

i j

θjθi 4 *

5 * 6*

1*

2*3 *

λix = cos θi

λiy = sin θi

λjx = cos θj

λjy = sin θj

[q*] = [T]T[q´]

1

Stiffness matrix

i j

4´

5´ 6´

1´

2´3´

*6

5*

*4

*3

2*

*1

qqqqqq

−

−

=

1000000000000000010000000000

jxjy

jyjx

ixiy

iyix

λλλλ

λλλλ

1 4 5 62 31*2*3*4*5*6*

[ T ]T

6'

5'

4'

3'

2'

1'

qqqqqq

129

[ ]

−

−

=

1000000000000000010000000000

jxjy

jyjx

ixiy

iyix

λλλλ

λλλλ

T

1 4 5 62 3123456

[ ]

−−−−

−−−

−

=



LAELAE

k

/4/60/2/60/6/120/6/120

00/00//2/60/4/60/6/120/6/120

00/00/

'

22

2323

22

2323

1

34

6

2

5

1 2 3 4 5 6

130

[ k ] = [ T ]T[ k´ ][T] =

Ui

Vi

Mi

Uj

Vj

Mj

Vj Mj

- λiy6EIL2

λix6EIL2

2EIL

λjy6EIL2

- λjx6EIL2

4EIL

Ui Vi Mi

- λiy6EIL2

λix6EIL2

4EIL

λjy6EIL2

λjx6EIL2

-

2EIL

Uj

AEL

- λixλiy)(12EIL3

AEL

λiy2 +

12EIL3

λix2 )(

λix6EIL2

λix6EIL2

AEL

λiyλjx -

12EIL3

λixλjy)-(

AEL

λixλjx +

12EIL3

λiyλjy)-(

λjy6EIL2

AEL

λjx2 +

12EIL3

λjy2 )(

λjy6EIL2

λjx6EIL2

-

AEL

- λjxλjy)(12EIL3

- λjx6EIL2

AEL

λixλjy -

12EIL3

λiyλjx)-(

AEL

- λixλiy)(12EIL3

- λiy6EIL2

- λiy6EIL2

AEL

λixλjx +

12EIL3

λiy λjy)-(

)(AEL

λix2 +

12EIL3

λiy2

AEL

λiyλjy +

12EIL3

λix λjx )-(

AEL

λiyλjy +

12EIL3

λixλjx)-(

12EIL3

λjx2 )

AEL

- ( λixλjy- λiyλjx )12EIL3

AEL

λiyλjx -

12EIL3

λixλjy)-(

AEL

- λjxλjy)(12EIL3

AEL λjy

2 + (

131

40 kN

4 m4 m 22.02 o

Example 12

For the beam shown:(a) Use the stiffness method to determine all the reactions at supports.(b) Draw the quantitative free-body diagram of member.(c) Draw the quantitative bending moment diagramsand qualitative deflected shape.Take I = 200(106) mm4, A = 6(103) mm2, and E = 200 GPa for all members.Include axial deformation in the stiffness matrix.

132

40 kN40 kN�m

20 kN

40 kN�m

20 kN

[ q´F ][FEF]

1Global

i j

1

i j

Local

40 kN

4 m4 m 22.02o

20sin22.02=7.520cos22.02=18.54

1´

2´

3´

4´

5´

6´

4

5

6

x*

x´1 *

2* 3 *

λix = cos 0o = 1, λiy = cos 90o = 0 λjx = cos 22.02o = 0.9271,

λjy = cos 67.98o = 0.3749

22.02 o

133

x*

x´1

4

5

6

1 *

2* 3 *

� Transformation matrix

Member 1: [ q ] = [ T ]T[q´]

1

i j1´

2´

3´

4´

5´

6´

Global Local


λjy = cos 67.98o = 0.3749

*3

2*

*1

6

5

4

qqqqqq

6'

5'

4'

3'

2'

1'

qqqqqq

−=

10000009271.03749.000003749.09271.0000000100000010000001

1´ 4´ 5´ 6´2´ 3´4561*

2*

3*

[ ]

−

−

=

1000000000000000010000000000

jxjy

jyjx

ixiy

iyix

TT

λλλλ

λλλλ

134

[k´]1 = 103

1´2´ 3´4´5´6´

1´ 4´ 5´ 6´2´ 3´-150.00.0000.000150.0

0.0000.000

0.000-0.9375-3.7500.000

0.9375-3.750

0.0003.75010.000.000

-3.75020.00

150.0

0.0000.000

-150.00.0000.000

0.0000.93753.7500.000

-0.93753.750

0.0003.75020.000.000

-3.75010.00

1

i j1´

2´

3´

4´

5´

6´

Local

� Local stiffness matrix

[ ]

−−−−

−−−

−

=×



LAELAE

k

/4/60/2/60/6/120/6/120

00/00//2/60/4/60/6/120/6/120

00/00/

22

2323

22

2323

66


Mi

Vj

Mj

Vi

Nj

Ni

135

Stiffness matrix [k´]:

[k´]1 = 103

1´2´ 3´4´5´6´

1´ 4´ 5´ 6´2´ 3´-150.00.0000.000150.0

0.0000.000

0.000-0.9375-3.7500.000

0.9375-3.750

0.0003.75010.000.000

-3.75020.00

150.0

0.0000.000

-150.00.0000.000

0.0000.93753.7500.000

-0.93753.750

0.0003.75020.000.000

-3.75010.00

Stiffness matrix [k*]: [ k* ] = [ T ]T[ k´ ][T]

[k*]1 = 103

4561*

2*

3*

4 1* 2* 3*5 6-139.00.3511.406129.0

1.40651.82

-56.25-0.869-3.75051.8221.90

-3.476

0.0003.75010.001.406

-3.47620.00

150.0

0.0000.000

-139.0-56.250.000

0.0000.93753.7500.351

-0.8693.750

0.0003.75020.001.406

-3.75010.00

x*

x´1

4

5

6

1 *

2* 3 * 1

i j1´

2´

3´

4´

5´

6´

Global Local

4

5

6

2*

136

Global Equilibrium:

Q1 = 0.0

Q3 = 0.0

D1*

D3*=

36.37x10-6 m

0.002 rad

[Q] = [K][D] + [QF]

= 1031*

3*

1*

129

1.406

3*

1.406

20.0

D1*

D3*

-7.5

-40+

x*

x´14

5

6

1 *

2* 3 *

4 m4 m 22.02 o

40 kN

x*

x´4

5

6

2*

40 kN40 kN�m

20 kN

40 kN�m

20 kN

[ q´F ]20sin22.02=7.5

20cos22.02=18.54

137

Member Force : [q] = [k*][D] + [qF]

0

40.020

-7.5418.54-40.0

+

q4

q6

q5

q1*

q2*q3*

= 103

4561*

2*

3*

4 1* 2* 3*5 6-139.00.3511.406129.0

1.40651.82

-56.25-0.869-3.75051.8221.90

-3.476

0.0003.75010.001.406

-3.47620.00

150.0

0.0000.000

-139.0-56.250.000

0.0000.93753.7500.351

-0.8693.750

0.0003.75020.001.406

-3.75010.00

0

00

36.4x10-6

02x10-3

-5.06

6027.5

013.48

0

=

40 kN

13.48 kN

60 kN�m

27.5 kN

5.06 kN

4 m4 m 22.02 o

40 kN

x*

x´1

4

5

6

1 *

2* 3 *

40 kN40 kN�m

7.518.54

40 kN�m

20 kN

4

5

6

2*

138

4 m4 m 22.02 o

40 kN

5.05 kN27.51 kN

60.05 kN�m

13.47 kN

40 kN

+

Bending moment diagram (kN�m)

Deflected shape

50.04

-

-60.05

0.002 rad

139

40 kN

22.02o

8 m 4 m 4 m2EI, 2AE EI, AE

6 kN/m

Example 13

For the beam shown:(a) Use the stiffness method to determine all the reactions at supports.(b) Draw the quantitative free-body diagram of member.(c) Draw the quantitative bending moment diagramsand qualitative deflected shape.Take I = 200(106) mm4, A = 6(103) mm2, and E = 200 GPa for all members.Include axial deformation in the stiffness matrix.

140

40 kN

22.02 o8 m 4 m 4 m

2EI, 2AE EI, AE

6 kN/m

1 21

2

3

4

5

6

7 *

8 * 9 *

Global

21´

2´3´

4´

5´6´

11´

2´

3´

4´

5´

6´

Local

mkNm

mkNmL

AE

•×=

×=

3

262

10150)8(

)/10200)(006.0(

mkNm

mmkNLEI

•×=

×=

3

426

1020)8(

)0002.0)(/10200(44

mkNLEI

•×= 310102

kNm

mmkNLEI

3

2

426

2

1075.3)8(

)0002.0)(/10200(66

×=

×=

mkNm

mmkNLEI

/109375.0)8(

)0002.0)(/10200(1212

3

3

426

2

×=

×=

141

[k]1 = [k´]1 = 2x103

456123

4 1 2 35 6-150.00.0000.000150.0

0.0000.000

0.000-0.9375-3.7500.000

0.9375-3.750

0.0003.75010.000.000

-3.75020.00

150.0

0.0000.000

-150.00.0000.000

0.0000.93753.7500.000

-0.93753.750

0.0003.75020.000.000

-3.75010.00

1 21

2

3

4

5

6

7 *

8 * 9 *

Global Local : Member 1

14

5

6

1

2

3

[ q ]

11´

2´

3´

4´

5´

6´

[ q´]

[q] = [q´] Thus, [k] = [k´]

142

Member 2: Use transformation matrix, [q*] = [T]T[q´]

q1´

q3´

q2´

q4´

q5´

q6´

q1

q3

q2

q7*

q8*q9*

0.92710.3749

-0.37490.9271

000

000

000

0 0

001

10

01

=

1237*

8*

9*

1´ 4´ 5´ 6´2´ 3´

0 0

001

000

000

000

1 21

2

3

4

5

6

7 *

8 * 9 *


λjy = cos 67.98o = 0.3749

21

2

3

x*

x´7*

8 * 9* [ q* ]

21´

2´3´

4´

5´6´

[ q´ ]

143

Stiffness matrix [k´]:

[k´]2 = 103

1´2´ 3´4´5´6´

1´ 4´ 5´ 6´2´ 3´-150.00.0000.000150.0

0.0000.000

0.000-0.9375-3.7500.000

0.9375-3.750

0.0003.75010.000.000

-3.75020.00

150.0

0.0000.000

-150.00.0000.000

0.0000.93753.7500.000

-0.93753.750

0.0003.75020.000.000

-3.75010.00

Stiffness matrix [k*]: [k*] = [T]T[k´][T]

[k*]2 = 103

1237*

8*

9*

1 7* 8* 9*2 3-139.00.3511.406129.0

1.40651.82

-56.25-0.869-3.47751.8221.90

-3.476

0.0003.75010.001.406

-3.47620.00

150.0

0.0000.000

-139.0-56.250.000

0.0000.93753.7500.351

-0.8693.750

0.0003.75020.001.406

-3.47710.00

21

23

x*

x´7 *

8 * 9 * [ q* ]

21´

2´3´

4´

5´6´

[ q´ ]

144

1 21

2

3

4

5

6

7 *

8 * 9 *

1

2

3

4

5

6

8 *

[k]1= 2x103

456123

4 1 2 35 6-150.00.0000.000150.0

0.0000.000

0.000-0.9375-3.7500.000

0.9375-3.750

0.0003.75010.000.000

-3.75020.00

150.0

0.0000.000

-150.00.0000.000

0.0000.93753.7500.000

-0.93753.750

0.0003.75020.000.000

-3.75010.00

[k*]2 = 103

1237*

8*

9*

1 7* 8* 9*2 3-139.00.3511.406129.0

1.40651.82

-56.25-0.869-3.47751.8221.90

-3.476

0.0003.75010.001.406

-3.47620.00

150.0

0.0000.000

-139.0-56.250.000

0.0000.93753.7500.351

-0.8693.750

0.0003.75020.001.406

-3.47710.00

145

Global:

=-509.84x10-6 rad

18.15x10-6 m

0.00225 rad

58.73x10-6 mD3

D1

D9*

D7*

00

00 +

-32 + 40 = 80

-40-7.5

D3

D1

D9*

D7*

32 kN�m

24 kN

32 kN�m

24 kN

6 kN/m

1

40 kN6 kN/m

1 21

2

3

4

5

6

7*

8 *

9*

1

2

3

4

5

6

8 *

= 1030

0-139

450

101.406

600

1.406

1.406-139

129

010

1.40620

1 3 7* 9*

137*

9*

40 kN40 kN�m

7.518.54

40 kN�m

20 kN

40 kN�m

7.5

40 kN�m32 kN�m

146

Member 1: [ q ] = [k ][d] + [qF]

0

3224

024-32

+

0

00

d1=18.15x10-6

0d3=-509.84 x10-6

q4

q6

q5

q1

q2q3

= 2x103

456123

4 1 2 35 6-150.00.0000.000150.0

0.0000.000

0.000-0.9375-3.7500.000

0.9375-3.750

0.0003.75010.000.000

-3.75020.00

150.0

0.0000.000

-150.00.0000.000

0.0000.93753.7500.000

-0.93753.750

0.0003.75020.000.000

-3.75010.00

-5.45 kN

21.80 kN�m20.18 kN

5.45 kN27.82 kN-52.39 kN�m

=

q4

q6

q5

q1

q2q3

6 kN/m

5.45 kN

20.18 kN

21.80 kN�m

5.45 kN

27.82 kN

52.39 kN�m

32 kN�m

24 kN

32 kN�m

24 kN

6 kN/m

114

5

6

1

2

3

147

q1

q3

q2

q7*

q8*q9*

21

3

x*

x´7*

9 *

= 103

1237*

8*

9*

1 7* 8* 9*2 3-139.00.3511.406129.0

1.40651.82

-56.25-0.869-3.75051.8221.90

-3.476

0.0003.75010.001.406

-3.47620.00

150.0

0.0000.000

-139.0-56.250.000

0.0000.93753.7500.351

-0.8693.750

0.0003.75020.001.406

-3.75010.00

0

4020

-7.518.54-40

+

18.15x10-6

-509.84x10-6

0

58.73x10-6

00.00225

q1

q3

q2

q7*

q8*q9*

-5.45 kN

52.39 kN�m26.55 kN

0 kN14.51 kN0 kN�m

= 5.45 kN

26.55 kN

52.39 kN�m

14.51 kN

40 kN

20sin22.02=7.520cos22.02=18.54

40 kN40 kN�m

20 kN

40 kN�m

20 kN

[ q´F ]

2

2 8 *

Member 2: [ q ] = [k ][d] + [qF]

148

5.45 kN26.55 kN

52.39 kN�m

14.51 kN

40 kN6 kN/m

5.45 kN

27.82 kN

52.39 kN�m

5.45 kN

20.18 kN

21.80 kN�m

40 kN

22.02o

8 m 4 m 4 m

6 kN/m

5.45 kN

20.18 kN

21.80 kN�m

14.51 kN54.37 kN

3.36 m

M (kN�m) x (m)

53.81

-21.8-

V (kN)x (m)

26.55

+20.18

+- -

-27.82

14.51cos 22.02o

=13.45 kN

-52.39-

12.14 +

-13.45

1

! Simple Frames! Frame-Member Stiffness Matrix! Displacement and Force Transformation Matrices! Frame-Member Global Stiffness Matrix

! Special Frames! Frame-Member Global Stiffness Matrix

FRAME ANALYSIS USING THE STIFFNESSMETHOD

2

Simple Frames

3

Frame-Member Stiffness Matrix

0 0 00 - AE/LAE/L

4EI/L - 6EI/L2 2EI/L6EI/L2 00

6EI/L2 - 12EI/L3 6EI/L212EI/L3 00

0

2EI/L

-6EI/L2

0

- 6EI/L2

12EI/L3

0

-6EI/L2

4EI/L

0

6EI/L2

-12EI/L3

AE/L

0

0

-AE/L

0

0

m

x´y´

i

j

3´

5´

6´

2´

4´

1´

3´ 6´2´ 4´1´ 5´

[k´]

4´5´6´

1´2´3´

6EI/L24EI/L

6EI/L24EI/L

AE/L

AE/L

6EI/L2

6EI/L2

12EI/L3

12EI/L3

2EI/L

d 1́ = 1

AE/L

AE/L

d 2́ = 1

12EI/L3

12EI/L3

d 3́ = 1

2EI/L

6EI/L2

6EI/L2

6EI/L26EI/L2AE/L

2EI/L

6EI/L2

6EI/L2d 6́

= 1

6EI/L2

6EI/L22EI/L

6EI/L2

12EI/L3

12EI/L3

4EI/L4EI/L12EI/L3

12EI/L3

6EI/L2

6EI/L2AE/L

6EI/L2d 4́ = 1

d 5́ = 1AE/L

AE/L

4

m

i

j

m

i

j

x´y´

x

y

Displacement and Force Transformation Matrices

12

3

45

6

θyθx

4´

5´

6´

1´

2´

3´

5

λx

q4 = q4´ cos θx - q5´ cos θy

q5 = q4´ cos θy + q5´ cos θx

q6 = q6´

λy

i

jθy

θx

y´

x´

4´5´

6´

1´

2´

3´

θy

θx

m

i

j

x

y

12

3

45

6

Force Transformation

Lxx ij

x

−=λ

Lyy ij

y

−=λ

−=

6'

5'

4'

6

5

4

10000

qqq

λλλλ

qqq

xy

yx

−

−

=

6'

5'

4'

3'

2'

1'

6

5

4

3

2

1

1000000000000000010000000000

qqqqqq

λλλλ

λλλλ

qqqqqq

xy

yx

xy

yx

[ ] [ ] [ ]'qTq T=

6

[q] = [T]T[q´]

= [T]T ( [k´][d´] + [q´F] )

= [T]T [k´][d´] + [T]T [q´F]

[q] = [T]T [k´][T][d] + [T]T [q´F] = [k][d] + [qF]

Therefore, [k] = [T]T [k´][T]

[qF] = [T]T [q´F]

[q] = [T]T[q´]

[d´] = [T][d]

[k] = [T]T [k´][T]

7

[q] = [T]T[q´]= [T]T ( [k´][d´] + [q´F] ) = [T]T[k´][d´] + [T]T[q´F] = [T]T [k´][T][d] + [T]T [q´F]

Frame Member Global Stiffness Matrix

[k] [qF][ k ] = [ T ]T[ k´ ][T] =

Ui

Vi

Mi

Uj

Vj

Mj

Vj Mj

- λiy6EIL2

λix6EIL2

2EIL

λjy6EIL2

- λjx6EIL2

4EIL

Ui Vi Mi

- λiy6EIL2

λix6EIL2

4EIL

λjy6EIL2

λjx6EIL2

-

2EIL

Uj

AEL

- λixλiy)(12EIL3

AEL

λiy2 +

12EIL3

λix2 )(

λix6EIL2

λix6EIL2

AEL

λiyλjx -

12EIL3

λixλjy)-(

AEL

λixλjx +

12EIL3

λiyλjy)-(

λjy6EIL2

AEL

λjx2 +

12EIL3

λjy2 )(

λjy6EIL2

λjx6EIL2

-

AEL

- λjxλjy)(12EIL3

- λjx6EIL2

AEL

λixλjy -

12EIL3

λiyλjx)-(

AEL

- λixλiy)(12EIL3

- λiy6EIL2

- λiy6EIL2

AEL

λixλjx +

12EIL3

λiy λjy)-(

)(AEL

λix2 +

12EIL3

λiy2

AEL

λiyλjy +

12EIL3

λix λjx )-(

AEL

λiyλjy +

12EIL3

λixλjx)-(

12EIL3

λjx2 )

AEL


AEL

λiyλjx -

12EIL3

λixλjy)-(

AEL

- λjxλjy)(12EIL3

AEL λjy

2 + (

8

5 kN

6 m

6 m

AB

C

Example 1

For the frame shown, use the stiffness method to:(a) Determine the deflection and rotation at B.(b) Determine all the reactions at supports.(c) Draw the quantitative shear and bending moment diagrams.E = 200 GPa, I = 60(106) mm4, A = 600 mm2

9

5 kN

6 m

6 m

AB

C kN/m666.667(6m)

)m10)(60mkN1012(20012

3

462

6

3 =××

=

−

LEI

kN/m200006m

)mkN10)(200m10(600 2

626

=××

=

−

LAE

kN2000(6m)

)m10)(60mkN106(2006

2

462

6

2 =××

=

−

LEI

mkN80006m

)m10)(60mkN104(2004

462

6

•=××

=

−

LEI

mkN40006m

)m10)(60mkN102(2002

462

6

•=××

=

−

LEI

Global :

AB

C

1

2

78 9

4

6 5

12 3

10

Global :

AB

C

1

2

78 9

4

6 5

12 3

A B14´

5´ 6´

1´

2´ 3´

Local :

5´

2´

1´ 3´

4´

6´

2

Using Transformation Matrix:

� Member Stiffness Matrix

[ ]

−−−−

−−−

−

=



LAEAE/L

/4/60/2/60/6/120/6/120

00/00//2/60/4/60/6/120/6/120

00/00

k'

22

2323

22

2323

Mi

Vj

Mj

Vi

Nj

Ni


11

A B14´

5´ 6´

1´

2´ 3´

Local :

[q] = [q´]

-> [k]1 = [k´]1

Stiffness Matrix: Member 1

Global:

AB

C

1

2

78 9

4

6 5

12 3

4 6 5 1 2 34

6

5

1

2

3

20000

0

0

-20000

0

0

0

666.667

2000

0

-666.667

2000

0

2000

8000

0

-2000

4000

-20000

0

0

20000

0

0

0

-666.667

-2000

0

666.667

-2000

0

2000

4000

0

-2000

8000

[k]1 =

12

Local:

5´

2´

1´ 3´

4´

6´

2

[q]2 = [ T ]T[ q´]2

q1´

q3´

q2´

q4´q5´

q6´

q1

q3

q2

q7

q8q9

[T]T

Stiffness Matrix: Member 2

=

123789

4´0000

0-1

5´000100

6´000001

1´0

0-1

000

2´100000

3´001000

90o

λjx = cos (-90o) = 0λjy = sin (-90o) = -1

λix = cos (-90o) = 0λiy = sin (-90o) = -1

Global:

AB

C

1

2

78 9

4

6 5

12 3

13

[k]2 = [ T ]T[ k´ ]2[ T ]

1´ 2´ 3´ 4´ 5´ 6´1´

2´

3´

4´

5´

6´

20000

0

0

-20000

0

0

0

666.667

2000

0

-666.667

2000

0

2000

8000

0

-2000

4000

-20000

0

0

20000

0

0

0

-666.667

-2000

0

666.667

-2000

0

2000

4000

0

-2000

8000

[k´]2 =

1 2 3 7 8 9

666.667 20001

2

3

7

8

9

2000

0

0

-666.667

-666.667

0

0

2000

2000

20000 0

0

0

0

-20000

-20000

0

0

8000 -2000

-2000

0

0

4000

4000

666.667 0

0

-2000

-2000

20000 0

0 8000

[k]2 =

14

[k]14 6 5 1 2 3

4

6

5

1

2

3

20000

0

0

-20000

0

0

0666.667

20000

-666.667

2000

02000

80000

-2000

4000

-200000

020000

0

0

0-666.667

-20000

666.667

-2000

02000

40000

-2000

8000

1 2 3 7 8 9666.667 20001

2

3

7

8

9

2000

00

666.667

-666.667

0

0

2000

2000

20000 0

0

0

0

-20000

-20000

0

0

8000 -2000-2000

0

0

4000

4000

666.667 0

0

-2000

2000

20000 0

0 8000

[k]2

Global Stiffness Matrix:

20000

0

-20000

0

0

0

8000

0

-2000

4000

0

-2000

0

4000

-20000

0

0

20666.667

-2000

2000

-2000

16000

20666.667

0

2000

[K]

4

5

1

2

3

4 5 1 2 3

Global:

AB

C

1

2

78 9

4

6 5

12 3

15

AB

C

Q4 = 0Q5 = 0

Q1 = 5Q2 = 0

Q3 = 0

D4

D5

D1

D2

D3

+

0 0

0 0

0

D4

D5

D1

D2

D3

=

0.01316 m

0.01316 m9.199(10-4) rad

-9.355(10-5) m

-1.887(10-3) rad

5 kN

6 m

6 m

AB

C

1

2

Global:

1

2

78 9

4

6 5

12 3

5 kN

=

4

5

1

2

3

4 5 1 2 3-20000 0 0

20666.667 00

2000

2000

20666.667 -2000

-2000 16000

0-2000

4000

08000 0 -2000 4000

-200000

20000

0

0

[Q] = [K][D] + [QF]

16

D4 = 0.01316

D5 = 9.199(10-4)

D6 = 0

D1 = 0.01316

D2 = -9.355(10-5)

D3 = -1.887(10-3)

15 kN

6 m

6 m

AB

C

2

q4

q5

q6

q1

q2

q3

0

0

-1.87

0

1.87

-11.22

Member 1

A B11

2 3

4

6 5

A B1

1.87 kN 11.22 kN�m1.87 kN

4 6 5 1 2 3

4

6

5

1

2

3

20000

0

0

-20000

0

0

0

666.667

2000

0

-666.667

2000

0

2000

8000

0

-2000

4000

-20000

0

0

20000

0

0

0

-666.667

-2000

0

666.667

-2000

0

2000

4000

0

-2000

8000

[q]1 = [k]1[d]1 + [qF]1

17

5 kN

1.87 kN 11.22 kN�m

5 kN1.87 kN

18.77 kN�m

Member 2

q1

q3

q2

q7

q8

q9

D1 = 0.01316

D3 = -1.887(10-3)

D2 = -9.355(10-5)

D7 = 0

D8 = 0

D9 = 0

5

11.22

-1.87

-5

1.87

18.77

1 2 3 7 8 9

666.667 20001

2

3

7

8

9

2000

0

0

-666.667

-666.667

0

0

2000

2000

20000 0

0

0

0

-20000

-20000

0

0

8000 -2000

-2000

0

0

4000

4000

666.667 0

0

-2000

-2000

20000 0

0 8000

[q]2 = [k]2[d]2 + [qF]2

15 kN

6 m

6 m

AB

C

2 1

2 3

78 9

2 2

18

AB

C


A B1

1.87 kN 11.22 kN�m1.87 kN5 kN

6 m

6 m

AB

C

1.87 kN

18.77 kN�m5 kN

1.87 kN

AB

C

Shear diagram

5

5

+

-1.87-

+

18.77

11.22

-11.22

-

5 kN

1.87 kN 11.22 kN�m

5 kN

1.87 kN

18.77 kN�m

2

19

Deflected shape

AB

C

AB

C

Bending moment diagram +

18.77

11.22

-11.22

-

D3=-0.00189 rad

D3=-0.00189 rad

D1=13.16 mmD4=13.16 mm

D4

D5

D1

D2

D3

=

0.01316 m

0.01316 m9.199(10-4) rad

-9.355(10-5) m

-1.887(10-3) rad

Global :

AB

C

1

2

78 9

4

6 5

12 3

D5=0.00092 rad

20

4.5 m

6 m 6 m

3 kN/m

Example 2

For the beam shown, use the stiffness method to:(a) Determine the deflection and rotation at B(b) Determine all the reactions at supports(c) Draw the quantitative shear and bending moment diagrams.E = 200 GPa, I = 60(106) mm4, A = 600 mm2 for each member.

A

B C

21

Global 2

1

1

2 3

4

56

7

8 9

2 7

8 9

1

2 3

[FEM] 9 kN�m

9 kN

9 kN�m

9 kN

3 kN/m

2

Members1

2 3

4

56 1

4.5 m

6 m 6 m

3 kN/m

22

4.5 m

6 m 6 m

3 kN/m

7.5 m

λx = cos θx = 6/7.5 = 0.8

Member 1:

θx

θy 1

2 3

4

56 1

λy = cos θy = 4.5/7.5 = 0.6

mkNm

mkNmL

AE

/160005.7

)/10200)(10600( 2626

=

××=

−

mkNm

mmkNLEI

/33.341)5.7(

)1060)(/10200(12123

4626

3

=

××=

−

kNm

mmkNLEI

1280)5.7(

)1060)(/10200(662

4626

2

=

××=

−

mkNm

mmkNLEI

•=

××=

−

64005.7

)1060)(/10200(44 4626

mkNm

mmkNLEI

•=

××=

−

32005.7

)1060)(/10200(22 4626

23

[ km ] = [ T ]T[ k´ ][T ] =

λx = cos θx

Member m:

θx

θy Uj

VjMj

Ui

Vi

Mim

λy = cos θy

Ui

Vi

Mi

Uj

Vj

Mj

Vj Mj

- λiy6EIL2

λix6EIL2

2EIL

λjy6EIL2

- λjx6EIL2

4EIL

Ui Vi Mi

- λiy6EIL2

λix6EIL2

4EIL

λjy6EIL2

λjx6EIL2

-

2EIL

Uj

AEL

- λixλiy)(12EIL3

AEL

λiy2 +

12EIL3

λix2 )(

λix6EIL2

λix6EIL2

AEL

λiyλjx -

12EIL3

λixλjy)-(

AEL

λixλjx +

12EIL3

λiyλjy)-(

λjy6EIL2

AEL

λjx2 +

12EIL3

λjy2 )(

λjy6EIL2

λjx6EIL2

-

AEL

- λjxλjy)(12EIL3

- λjx6EIL2

AEL

λixλjy -

12EIL3

λiyλjx)-(

AEL

- λixλiy)(12EIL3

- λiy6EIL2

- λiy6EIL2

AEL

λixλjx +

12EIL3

λiy λjy)-(

)(AEL

λix2 +

12EIL3

λiy2

AEL

λiyλjy +

12EIL3

λix λjx )-(

AEL

λiyλjy +

12EIL3

λixλjx)-(

12EIL3

λjx2 )

AEL


AEL

λiyλjx -

12EIL3

λixλjy)-(

AEL

- λjxλjy)(12EIL3

AEL λjy

2 + (

24

λx = cos θx = 6/7.5 = 0.8

Member 1:

θx

θy 1

2 3

4

56 1

λy = cos θy = 4.5/7.5 = 0.6

4.5 m

6 m 6 m

3 kN/m

7.5 m

=

4

5

6

1

2

3

4 1

-10362.879

-7516.162

768

10362.879

768

7516.162

2

-7516.162

-5978.451

-1024

7516.162

5978.451

-1024

3

-768

1024

3200

768

-1024

6400

10362.879

-768

7516.162

-10362.879

-7516.162

-768

5

7516.162

5978.451

1024

-7516.162

-5978.451

1024

6

-768

1024

6400

768

-1024

3200

[k1]

25

λx = cos 0o = 1.0, λy = cos 90o = 0

Member 2:

2 7

8 9

1

2 3

4.5 m

6 m 6 m

3 kN/m

7.5 mmkN

mmkNm

LAE

/200006

)/10200)(10600( 2626

=

××=

−

mkNm

mmkNLEI

/667.666)6(

)1060)(/10200(12123

4626

3

=

××=

−

kNm

mmkNLEI

2000)6(

)1060)(/10200(662

4626

2

=

××=

−

mkNm

mmkNLEI

•=

××=

−

80006

)1060)(/10200(44 4626

mkNm

mmkNLEI

•=

××=

−

40006

)1060)(/10200(22 4626

26

[k2] =0

2EI/L

-6EI/L2

- 6EI/L2

6EI/L2

0

6EI/L2

-12EI/L3

0

0

0

0

-AE/L

0

0

0

6EI/L2

0

- 6EI/L2

- 12EI/L3

0

0

6EI/L2

2EI/L

0

-6EI/L2

0 - AE/L

0

0

4EI/L

12EI/L3

4EI/L

12EI/L3

AE/L

AE/L

3

8

9

2

7

1

3 92 71 8

[k2] =0

4000

-2000

- 2000

2000

0

2000

-666.667

0

0

0

0

-20000

0

0

0

2000

0

- 2000

- 666.667

0

0

2000

4000

0

-2000

0 - 20000

0

0

8000

666.667

8000

666.667

20000

20000

3

8

9

2

7

1

3 92 71 8

2 7

8 9

1

2 3

27

=

4

5

6

1

2

3

4 1

-10362.879

-7516.162

768

10362.879

768

7516.162

2

-7516.162

-5978.451

-1024

7516.162

5978.451

-1024

3

-768

1024

3200

768

-1024

6400

10362.879

-768

7516.162

-10362.879

-7516.162

-768

5

7516.162

5978.451

1024

-7516.162

-5978.451

1024

6

-768

1024

6400

768

-1024

3200

[k1]

[k2] =0

4000

-2000

- 2000

2000

0

2000

-666.667

0

0

0

0

-20000

0

0

0

2000

0

- 2000

- 666.667

0

0

2000

4000

0

-2000

0 - 20000

0

0

8000

666.667

8000

666.667

20000

20000

3

8

9

2

7

1

3 92 71 8

28

D1

D3

D2 =4.575(10-4) m

-5.278(10-4) rad

-1.794(10-3) m

Global:

1

21

2 3

4

5 6

7

8 9

4.5 m

6 m 6 m

3 kN/m

7.5 m9 kN

9 kN�m

9 kN

0

0

0

Q1

Q3

Q2

D1

D3

D2

0

9

9+1

2

3

1

30362.9

768

7516.16

2

7516.16

6645.12

976

3

768

976

14400=

9 kN�m

9 kN

29

λx = cos θx = 6/7.5 = 0.8

Member 1:

θx

θy 1

2 3

4

56 1

λy = cos θy = 4.5/7.5 = 0.6

0

0

0

D1 = 4.575(10-4)

D2 = -1.794(10-3)

D3 = -5.278(10-4)

0

0

0

0

0

0

+ =

q4

q6

q5

q1

q2q3

4

5

6

1

2

3

5 1 2 34 6

k1

1

11.37 kN

11.37 kN

0.09 kN

1.19 kN�m

0.09 kN

0.50 kN�m 1

9.15 kN6.75 kN 1.19 kN�m

9.15 kN

6.75 kN

0.50 kN�m

9.15

0.50

6.75

-9.15

-6.75

-1.19

=

30

Member 2:

2 7

8 9

1

2 3

0

9

9

0

9

-9

+=

q1

q3

q2

q7

q8q9

D1 = 4.575(10-4)

D2 = -1.794(10-3)

D3 = -5.278(10-4)

0

0

0

1

2

3

7

8

9

2 7 8 91 3

k2

9.15

1.19

6.75

-9.15

11.25

-14.70

=

9 kN�m

9 kN

9 kN�m

9 kN

3 kN/m

2[FEM]

3 kN/m

29.15 kN

6.75 kN

1.19 kN�m

9.15 kN11.25 kN

14.70 kN�m

31

3 kN/m

29.15 kN

6.75 kN

1.19 kN�m

9.15 kN11.25 kN

14.70 kN�m

1

9.15 kN6.75 kN 1.19 kN�m

9.15 kN

6.75 kN

0.50 kN�m

3 kN/m

All Reactions

9.15 kN11.25 kN

14.70 kN�m

9.15 kN

6.75 kN

0.50 kN�m

32

-1.19

Shear diagram (kN)

Deflected shape

-0.09

-0.09

6.75

3 kN/m

29.15 kN

6.75 kN

1.19 kN�m

9.15 kN11.25 kN

14.70 kN�m

1

11.37 kN

11.37 kN

0.09 kN

1.19 kN�m

0.09 kN0.50 kN�m

D3 =-5.278(10-4) rad

D2 = -1.79 mm

D1 = 0.46 mm

D1

D3

D2 =4.575(10-4) m

-5.278(10-4) rad

-1.794(10-3) m -11.25

-

+

Bending-moment diagram (kN�m)

-14.70

0.5

33

Example 3

For the beam shown, use the stiffness method to:(a) Determine the deflection and rotation at B.(b) Determine all the reactions at supports.(c) Draw the quantitative shear and bending moment diagrams.E = 200 GPa, I = 60(106) mm4, A = 600 mm2 for each member.

4.5 m

6 m 3 m

10 kN

A

BC15 kN

20 kN�m

3 kN/m

3 m

34

[FEM] 13 kN/m 11.25 kN

wL/2 = 11.25 kN

wL2/12 = 14.06 kN�m

14.06 kN�m

Global

21

2 7

8 9

1

2 3

4.5 m

6 m 3 m

10 kN

A

BC15 kN

20 kN�m

3 kN/m

3 m

7.5 kN�m

5 kN

7.5 kN�m

5 kN

2

10 kN

λx = cos θx = 6/7.5 = 0.8

1

Members1

2 3

4

56

1

2 3

4

56

7

8 9

θy = 53.13

θx =36.87

λy = cos θy = 4.5/7.5 = 0.6

6.75 kN

9 kN

11.25(0.6) = 6.75 kN

11.25(0.8) = 9 kN

35

Global:

1

21

2 3

4

5 6

7

8 9

Q1 = 15

Q3 = 20

Q2 = 0

D1

D3

D2

-6.75

-14.06 + 7.5

9 + 5+1

2

3

1

30362.9

768

7516.16

2

7516.16

6645.2

976

3

768

976

14400=

[FEM]13 kN/m 11.25 kN

11.25 kN

14.06 kN�m

14.06 kN�m7.5 kN�m

5 kN

7.5 kN�m

5 kN

2

10 kN

6.75 kN

9 kN

11.25(0.6) = 6.75 kN

9 kN

10 kN

A

BC15 kN

20 kN�m

3 kN/m

14.06 kN�m7.5 kN�m

6.75 kN

5 kN9 kN

15 kN20 kN�m

36

D1

D3

D2 =1.751(10-3) m

2.049(10-3) rad

-4.388(10-3) m

1

21

2 3

4

5 6

7

8 9

37

Member 1:

0

0

0

D1 = 1.751(10-3)

D2 = -4.388(10-3)

D3 = 2.049(10-3)

-6.75

14.06

9

-6.75

9

-14.06

+ =

q4

q6

q5

q1

q2q3

4

5

6

1

2

3

5 1 2 34 6

k1

6.51

26.46

24.17

-20.01

-6.17

4.89

=

[FEM]

13 kN/m 11.25 kN

11.25 kN

14.06 kN�m

14.06 kN�m

6.75 kN

9 kN

11.25(0.6) = 6.75 kN

9 kN

θx

θy 1

2 3

4

56 1

λx = cos 36.87o = 0.8, λy = cos 53.13o = 0.6

14.06 kN�m

14.06 kN�m

6.75 kN

9 kN

11.25(0.6) = 6.75 kN

9 kN

38

3 kN/m

7.5 m1

19.71 kN

19.71 kN

7.07 kN

4.89 kN�m

15.43 kN

26.46 kN�m

=

q4

q6

q5

q1

q2q3

6.51

26.46

24.17

-20.01

-6.17

4.89

θx = 36.87o

θy = 53.13o 1

2 3

4

56 1

3 kN/m

7.5 m

20.01 kN

4.89 kN�m

6.51 kN

24.17 kN

26.46 kN�m

6.17 kN

39

Member 2:

2 7

8 9

1

2 3

0

7.5

5

0

5

-7.5

+=

q1

q3

q2

q7

q8q9

D1 = 1.751(10-3)

D2 = -4.388(10-3)

D3 = 2.049(10-3)

0

0

0

1

2

3

7

8

9

2 7 8 91 3

k2

35.02

15.12

6.17

-35.02

3.83

-8.08

=

35.02 kN

6.17 kN

15.12 kN�m

35.02 kN3.83 kN

8.08 kN�m

[FEM]

7.5 kN�m

5 kN

7.5 kN�m

5 kN

2

10 kN 2

10 kN

7.5 kN�m

5 kN

7.5 kN�m

5 kN

40

3 kN/m

7.5 m

20.01 kN

4.89 kN�m

6.51 kN

24.17 kN

26.46 kN�m

6.17 kN

35.02 kN

6.17 kN

15.12 kN�m

35.02 kN3.83 kN

8.08 kN�m

2

10 kN

10 kN

A

BC15 kN

20 kN�m

3 kN/m

6 m 3 m3 m

6.51 kN

24.17 kN

26.46 kN�m

35.02 kN3.83 kN

8.08 kN�m

41

Shear diagram (kN)


3 kN/m

7.5 m1

19.71 kN

19.71 kN

7.07 kN

4.89 kN�m

15.43 kN

26.46 kN�m

35.02 kN

6.17 kN

15.12 kN�m

35.02 kN3.83 kN

8.08 kN�m

2

10 kN

15.43D1

D3

D2 =1.751(10-3) m

2.049(10-3) rad

-4.388(10-3) m

Deflected shape

D3 = 2.05(10-3) rad

D2 = -4.39 mm

D1 =1.75 mm

-26.46

4.89

-15.12 -8.08

-7.07

6.17

-3.83

42

Special Frames

43

i j

θjθi 4 *

5 * 6*

1*

2*3 *

λix = cos θi

λiy = sin θi

λjx = cos θj

λjy = sin θj

[ q* ] = [ T ]T[ q´ ]

1

Stiffness matrix

i j

4´

5´ 6´

1´

2´3´

[ T ]T

−

−

=

6'

5'

4'

3'

2'

1'

*6

5*

*4

*3

2*

*1

1000000000000000010000000000

qqqqqq

λλλλ

λλλλ

qqqqqq

jxjy

jyjx

ixiy

iyix1*2*3*4*5*6*

1´ 2´ 3´ 4´ 5´ 6´

44

[ ]

−

−

=

1000000000000000010000000000

jxjy

jyjx

ixiy

iyix

λλλλ

λλλλ

T

1´2´3´4´5´6´

1* 2* 3* 4* 5* 6*

� Member Stiffness Matrix

1´2´3´4´5´6´

1´ 2´ 3´ 4´ 5´ 6´

[ ]

−−−−

−−−

−

=



LAEAE/L

/4/60/2/60/6/120/6/120

00/00//2/60/4/60/6/120/6/120

00/00

k'

22

2323

22

2323

45

[ k ] = [ T ]T[ k´ ][T] =

Ui

Vi

Mi

Uj

Vj

Mj

Vj Mj

- λiy6EIL2

λix6EIL2

2EIL

λjy6EIL2

- λjx6EIL2

4EIL

Ui Vi Mi

- λiy6EIL2

λix6EIL2

4EIL

λjy6EIL2

λjx6EIL2

-

2EIL

Uj

AEL

- λixλiy)(12EIL3

AEL

λiy2 +

12EIL3

λix2 )(

λix6EIL2

λix6EIL2

AEL

λiyλjx -

12EIL3

λixλjy)-(

AEL

λixλjx +

12EIL3

λiyλjy)-(

λjy6EIL2

AEL

λjx2 +

12EIL3

λjy2 )(

λjy6EIL2

λjx6EIL2

-

AEL

- λjxλjy)(12EIL3

- λjx6EIL2

AEL

λixλjy -

12EIL3

λiyλjx)-(

AEL

- λixλiy)(12EIL3

- λiy6EIL2

- λiy6EIL2

AEL

λixλjx +

12EIL3

λiy λjy)-(

)(AEL

λix2 +

12EIL3

λiy2

AEL

λiyλjy +

12EIL3

λix λjx )-(

AEL

λiyλjy +

12EIL3

λixλjx)-(

12EIL3

λjx2 )

AEL


AEL

λiyλjx -

12EIL3

λixλjy)-(

AEL

- λjxλjy)(12EIL3

AEL λjy

2 + (

46

40 kN

4 m 4 m

3 m

7.416 m

22.02 o

20 kN200 kN�m

6 kN/m

Example 4

For the beam shown:(a) Use the stiffness method to determine all the reactions at supports.(b) Draw the quantitative free-body diagram of member.(c) Draw the quantitative bending moment diagrams and qualitativedeflected shape.Take I = 200(106) mm4 , A = 6(103) mm2, and E = 200 GPa for all members.Include axial deformation in the stiffness matrix.

47

1Local

12

40 kN

4 m 4 m

3 m

7.416 m

22.02 o

20 kN200 kN�m

6 kN/m

4

5 6

7

8 922.02 o

Global

2

1´

2´ 3´

4´

5´ 6´

1´

2´3´

4´

5´ 6´

1 *

2* 3 *

48

mkNm

mkNmL

AE /101508

)/10200)(006.0( 3262

×=×

=

mkNm

mmkNLEI

•×=×

= 3426

10208

)0002.0)(/10200(44

mkNLEI

•×= 310102

kNm

mmkNLEI 3

2

426

2 1075.3)8(

)0002.0)(/10200(66×=

×=

mkNm

mmkNLEI /109375.0

)8()0002.0)(/10200(1212 3

3

426

3 ×=×

=

[ ]

−−−−

−−−

−

=



LAEAE/L

/4/60/2/60/6/120/6/120

00/00//2/60/4/60/6/120/6/120

00/00

k'

22

2323

22

2323

Mi

Vj

Mj

Vi

Nj

Ni


49

2

1´

2´ 3´

4´

5´ 6´

1´

2´3´

1 4´

5´ 6´

Local

8 m

8 m

03750

0

- 3750- 937.5

0

0375010000

0-3750

0 - 150000

00

0

10000

-3750

- 3750

3750

0

3750

-937.5

0

0

00

-150000

0

0

[ k´]1 = [ k´]2 =20000

937.5

20000

937.5

150000

150000


Mi

Vj

Mj

Vi

Nj

Ni

50

θi =22.02o

Member 1:

1

1 *

2 * 3 *

4

5 6

Global1´

2´3´

1 4´

5´ 6´Local

θj = 0o

λjx = cos (0o) = 1,λjy = sin (0o) = 0

λix = cos (22.02o) = 0.927,λiy = sin (22.02o) = 0.375

40 kN40 kN�m40 kN�m

20 kN20 kN

1

[FEM]

q1´

q3´

q2´

q4´

q5´

q6´

q*1

q*3

q*2

q4

q5q6

[ T ]1T

00

0

00

0

00

0

000

000

000

1 0

1

01

00

0 0

0.9270.375

- 0.375 0.927

10 0

00

[ q* ] = [ T ]T[ q´ ]

=

1´ 4´ 5´ 6´2´ 3´

1*

2*

3*

4*

5*

6*

51

θi =22.02o1

1 *

2 * 3 *

4

5 6

Global1´

2´3´

1 4´

5´ 6´Local

θj = 0o

[ k* ]1 = [ T ]1T[ k´ ]1[ T ]1

[ k* ]1 = 103

1*2*3*456

1*

129.04651.811-1.406

-139.0580.351-1.406

2*

51.811

21.892

3.476-56.240

-0.8693.476

3*

-1.4063.47620.000.00-3.7510.00

4

-139.058-56.240

01500

0

5

0.351-0.869-3.75

00.938-3.75

6

-1.4063.47610.00

0-3.75

20

52

θi =22.02o1

1 *

2 * 3 *

4

5 6

Globalθj = 0o

40 kN40 kN�m40 kN�m

20 kN20 kN

1

[FEM]

[ qF* ] = [ T ]T[ qF´]

[ qF*] = [ T ]1T

0

40

20

0

20

-40

40 kN40 kN�m

20 kN

1

40 kN�m

18.547.5

-7.5018.5440020-40

=

1*

2*

3*

4

5

6

53

Member 2

q1´

q3´

q2´

q4´

q5´

q6´

q4

q6

q5

q7

q8q9

[ T ]2T

0.927-0.375

0.3750.927

1

00

0

00

0

00

0

000

000

000

1

0.927-0.375

0.3750.927

00

0 0

0 0

00

[ q ] = [ T ]T[ q´ ]

=

1´ 4´ 5´ 6´2´ 3´

456

78

9

6 kN/m2

24 kN

24 kN

32 kN�m

32 kN�m

[ q´F]

2

1´

2´ 3´

4´

5´ 6´

[ q´]

λix = λjx = cos (-22.02o) = 0.927,λiy = λjy = sin (-22.02o) = -0.375

2

45 6

[ q ]7

8 922.02o

22.02o

54

2

1´

2´ 3´

4´

5´ 6´

[ q´ ]

[ k ]2 = [ T ]2T[ k´]2[ T ]2

2

45 6

[ q ]7

8 922.02o

22.02o

[ k ]2 = 103

456

78

9

4

129.046-51.811

1.406-129.046

51.8111.4056

5

-51.81121.8923.476

51.811-21.892

3.476

6

1.4063.476

20-1.406-3.476

10

7

-129.04651.811-1.406

129.046-51.811-1.406

8

51.811-21.892-3.476

-51.81121.892-3.476

9

1.4063.476

10-1.406-3.476

20

55

6 kN/m2

24 kN

24 kN

32 kN�m

32 kN�m

[ q´F ]

[ qF* ] = [ T ]T[ qF´ ]

[ qF ] = [ T ]2T

0

32

24

0

24

-32

8.99822.249328.99822.249-32

=

4

5

6

7

8

9

6 kN/m2

32 kN�m

32 kN�m

[ qF ]

22.249 kN

8.998kN

22.249 kN

8.998 kN

2

45 6

[ q ]7

8 922.02o

22.02o

56

[ k* ]1 = 103

1*2*3*456

1*129.04651.811-1.406

-139.0580.351-1.406

2*51.811

21.892

3.476-56.240-0.8693.476

3*-1.4063.47620.000.00-3.7510.00

4-139.058-56.240

01500

0

50.351-0.869-3.75

00.938-3.75

6-1.4063.47610.00

0-3.75

20

Global Stiffness:

12

1 *

2* 3 *

4

5 6

7

8 9

2*

7

8 9

[ k ]2 = 103

456

78

9

4

129.046-51.811

1.406-129.046

51.8111.4056

5

-51.81121.8923.476

51.811-21.892

3.476

6

1.4063.476

20-1.406-3.476

10

7

-129.04651.811-1.406

129.046-51.811-1.406

8

51.811-21.892-3.476

-51.81121.892-3.476

9

1.4063.476

10-1.406-3.476

20

57

Global: 40 kN 20 kN200 kN�m

6 kN/m 12

1 *

2 * 3 *

4

5 6

7

8 9

6 kN/m2

32 kN�m

32 kN�m

[ qF ]

22.249 kN

8.998 kN

22.249 kN

8.998 kN

40 kN40 kN�m

20 kN

1

40 kN�m

18.54 kN7.5

[ q*F ]

2 *

7

8 9

[ Q ] = [ K ][ D ] + [ QF ]

40 kN�m

7.5

40 kN�m 32 kN�m

20 kN 22.249 kN

8.998 kN

20 kN200 kN�m

D1*

D4

D3*

D5

D6

= 0.0

= 0.0

= 0.0

= -20

= -200

Q1*

Q4

Q3*

Q5

Q6

= 103

1*

3*

4

5

6

1*

129.046

-1.406

-139.058

0.352

-1.406

3*

-1.406

20

0

-3.75

10

4

-139.058

0

279.046

-51.811

1.406

5

0.351

-3.75

-51.811

22.829

-0.274

6-1.406

10

1.406

-0.274

40

+ 8.998

-7.5

40

20 +22.249

-40 + 32

58

Global: 40 kN 20 kN200 kN�m

6 kN/m 12

1 *

2 * 3 *

4

5 6

7

8 9

2 *

7

8 9

D1*

D4

D3*

D5

D6

=

-0.0205 m

-0.0112 rad

-0.0191 m

-0.0476 m

-0.0024 rad

59

1

1 *

2 * 3 *

4

5 6

[ q ]

40 kN 40 kN�m

20 kN

1

40 kN�m

18.547.5

[ qF* ]Member 1: [ q*] = [ k*][ d*] + [ qF*]

q1*

q3*

q2*

q4

q5q6

D1*=-0.0205

D3*=-0.0112D2*= 0.0

D4= -0.0191D5= -0.0476D6=-0.0024

-7.50

40

18.54

0

20

-40

+

q1*

q3*

q2*

q4

q5

q6

0

022.63 kN

-8.49 kN19.02 kN7.87 kN�m

=

40 kN

22.63 kN

8.49 kN

19.02 kN

7.87 kN�m

= 103

1*2*3*456

1*

129.04651.811-1.406

-139.0580.351-1.406

2*

51.811

21.892

3.476-56.240

-0.8693.476

3*

-1.4063.47620.000.00-3.7510.00

4

-139.058-56.240

01500

0

5

0.351-0.869-3.75

00.938-3.75

6

-1.4063.47610.00

0-3.75

20

60

Member 2 : [ q ] = [ k ][ d ] + [ qF ]

q4

q6

q5

q7

q8q9

8.998

3222.249

8.99822.249

-32

+

2

4

5 6

[ q ] 7

8 96 kN/m

2

32 kN�m

32 kN�m

[ qF ]

22.249 kN

8.998 kN

22.249 kN

8.998 kN

D4= -0.0191

D6=-0.0024D5= -0.0476

D7 = 0D8 = 0D9 = 0

q4

q6

q5

q7

q8

q9

8.49 kN

-207.87 kN�m-39.02 kN

9.51 kN83.51 kN-248.04 kN�m

= 8.49 kN

39.02 kN

207.87 kN�m

9.51 kN83.51 kN

248.04 kN�m

6 kN/m

=103

456

78

9

4

129.046-51.811

1.406-129.046

51.8111.4056

5

-51.81121.8923.476

51.811-21.892

3.476

6

1.4063.476

20-1.406-3.476

10

7

-129.04651.811-1.406

129.046-51.811-1.406

8

51.811-21.892-3.476

-51.81121.892-3.476

9

1.4063.476

10-1.406-3.476

20

61

40 kN

22.63 kN

8.49 kN

19.02 kN

7.87 kN�m

8.49 kN

39.02 kN

207.87 kN�m

9.51 kN

83.51 kN

248.04 kN�m6 kN/m

22.5 kN81 kN

22.5 kN

33 kN

8.49 kN

20.98 kN


-

+

207.87

-248.04

83.56

+7.87

Deflected shape

D3*=-0.0112 rad

D1*=- 20.5 mm

D6=-0.0024 rad

D1*

D4

D3*

D5

D6

=

-0.0205 m

-0.0112 rad

-0.0191 m

-0.0476 m

-0.0024 radD4=-19.1 mmD5=-47.6 mm

62

3 m

80 kN�m20 kN/m

20o

50 kN

4 m 2 m

A B

C

Example 5

For the beam shown:(a) Use the stiffness method to determine all the reactions at supports.(b) Draw the quantitative free-body diagram of member.(c) Draw the quantitative bending moment diagramsand qualitative deflected shape.Take I = 400(106) mm4 , A = 60(103) mm2, and E = 200 GPa for all members.

63

Global

1

2 3

4

5 6

7*9*

8*

1

2 2

1´2´

3´

4´5´

6´

3 m

80 kN�m20 kN/m

20o

50 kN

4 m 2 m

A B

C

Local

1´

2´ 3´

4´

5´ 6´1

76.31o

64

3 m

80 kN�m20 kN/m

20o

50 kN

4 m 2 m

A B

C

1

2 3

4

5 61

Member 1:

kN/m103000m4

)kN/m10)(200m10(60

3

2623

×=

××=

−

LAE

mkN1080m4

)m10)(400kN/m104(2004

3

4626

•×=

××=

−

LEI

mkN1040m4

)m10)(400kN/m102(2002

3

4626

•×=

××=

−

LEI

kN1030m)(4

)m10)(400kN/m106(2006

3

2

4626

2

×=

××=

−

LEI

kN/m1015m)(4

)m10)(400kN/m1012(20012

3

3

4626

3

×=

××=

−

LEI

65

d1

d3

d2

d4

d5

d6

q1

q3

q2

q4

q5q6

0

26.67

40

040

-26.67

+= 103

123

45

6

1 4 5 6

030

40

0-30

80

3000

0

0

-30000

0

2

015

30

0-15

30

3

030

80

0-30

40

-30000

0

3000

0

0

0-15

-30

015

-30

3 m

80 kN�m20 kN/m

20o

50 kN

4 m 2 m

A B

C

26.67 kN�m26.67 kN�m

40 kN40 kN

1

20 kN/m

Member 1: [ q ] = [ k ][ d ] + [ qF ]

1

2 3

4

5 61

56.31o

76.31o

66

Member 2:

2

1´2´

3´

4´5´

6´

45 6

7*9*

8*

2

i

j

kN/m103324m3.61

)kN/m10)(200m10(60

3

2623

×=

××=

−

LAE

mkN1088.64m3.61

)m10)(400kN/m104(2004

3

4626

•×=

××=

−

LEI

kN1036.83m)(3.61

)m10)(400kN/m106(2006

3

2

4626

2

×=

××=

−

LEI

kN/m1020.41m)(3.61

)m10)(400kN/m1012(20012

3

3

4626

3

×=

××=

−

LEI

mkN1044.32m3.61

)m10)(400kN/m102(2002

3

4626

•×=

××=

−

LEI

[ k´ ]2 = 103

1´2´3´4´5´6´

1´ 4´ 5´ 6´3324

00

-332400

2´0

20.4136.83

0-20.4136.83

3´0

36.8388.64

0-36.8344.32

-332400

3324

00

0-20.41-36.83

020.41

-36.83

036.8344.32

0-36.8388.64

67

45 6

7*9*

8*

2

i

j 76.31oλjx = cos (-76.31o) = 0.24,λjy = sin (-76.31o) = -0.97

[ q* ] = [ T ]T[ q´ ]

q1´

q3´

q2´

q4´q5´q6´

q4

q6

q5

q7*q8*q9*

=

4567*8*9*

1´ 4´000

0.24

0-0.97

5´000

0.970.24

0

6´000001

0.55

0-0.83

000

2´0.830.55

0000

3´001000

λix = cos (-56.31o) = 0.55,λiy = sin (-56.31o) = -0.83

56.31o

2

i

j

1´2´

3´

4´5´

6´

[ k* ]2 = [ T ]T[ k´]2[ T ] = 103

4567*8*9*

4 7* 8* 9*5 6-452.884643.585-35.786205.452

-35.786-759.668

1787.474-2689.923

-8.717-759.6683139.053

-8.717

30.64620.43144.321

-35.786-8.71788.643

1036.923

30.646-1524.780

-452.8841787.474

30.646

-1524.7802307.582

20.431643.585

-2689.92320.431

30.64620.43188.643

-35.786-8.71744.321

68

[k]1 = 103

123456

1 4 5 60

30400

-3080

3000

00

-300000

20

1530

0-1530

30

30800

-3040

-300000

3000

00

0-15-300

15-30

1

2 3

4

5 6

7*9*

8*

1

2

[k*]2 = 103

4567*8*9*

4 7* 8* 9*5 6-452.884643.585-35.786205.452

-35.786-759.668

1787.474-2689.923

-8.717-759.6683139.053

-8.717

30.64620.43144.321

-35.786-8.71788.643

1036.923

30.646-1524.780

-452.8841787.474

30.646

-1524.7802307.582

20.431643.585

-2689.92320.431

30.64620.43188.643

-35.786-8.71744.321

69

Global:

= -50= 0= -80= 0= 0

Q4

Q6

Q5

Q7*

Q9*

D4

D6

D5

D7*

D9*

0

-26.6740

00

+

D4

D6

D5

D7*

D9*

-2.199x10-5 m

-2.840x10-4 rad-3.095x10-4 m

0.979x10-3 m6.161x10-4 rad

=

1

2 3

4

5 6

7*9*

8*

1

23 m

80 kN�m20 kN/m

20o

50 kN

4 m 2 m

A B

C 26.67 kN�m

26.67 kN�m

40 kN40 kN

1

20 kN/m

80 kN�m50 kN

= 103

45679

4 7* 9*5 6

205.452

-452.884643.585-35.786

-35.786

30.646-9.56944.321

-35.78688.643

-452.884

4036.923

30.646-1524.780

30.646

-1524.780232.582

-9.569643.58520.431

30.646-9.569

168.643-35.78644.321

70

q1

q3

q2

q4

q5q6

0

26.67

40

040

-26.67

+= 103

123

45

6

1 4 5 6

030

40

0-30

80

3000

0

0

-30000

0

2

015

30

0-15

30

3

030

80

0-30

40

-30000

0

3000

0

0

0-15

-30

015

-30

Member 1:

q1

q3

q2

q4

q5q6

65.97 kN

24.59 kN�m

36.12 kN

-65.97 kN

43.88 kN-40.11 kN�m

=

0

0

0

D4

D5

D6

40.11 kN�m24.59 kN�m

43.88 kN36.12 kN

1

20 kN/m

65.97 kN65.97 kN

1

2 3

4

5 61

71

q4

q6

q5

q7*

q8*q9*

Member 2:

q4

q6

q5

q7*

q8*q9*

15.97 kN

-39.89 kN�m

-43.88 kN

0 kN

46.69 kN0 kN�m

=

= 103

456

7*8*

9*

4 7* 8* 9*5 6-455.21651.27

-35.73

210.67

-35.73

-769.1

1769.34-2678.93

-8.84

-769.13128.82

-8.84

30.5720.26

44.32

-35.73-8.84

88.64

1036.923

30.57-1508.14

-455.211769.34

30.57

-1508.142296.15

20.26

651.27-2678.93

20.26

30.5720.26

88.64

-35.73-8.84

44.32

D4

D6

D5

D*7

0

D*9

15.97 kN43.88 kN 39.89 kN�m

46.69 kN

2

45 6

7*9*

8*

2

i

j

72


+

40.11 kN�m24.59 kN�m

43.88 kN36.12 kN

1

20 kN/m

65.97 kN65.97 kN

15.97 kN43.88 kN 39.89 kN�m

46.69 kN

2

-24.59-

-40.11

- +

39.89

D4

D6

D5

D7*D9*

-2.199x10-5 m

-2.840x10-4 rad-3.095x10-4 m

0.979x10-3 m6.161x10-4 rad

=

Deflected shape

D4=-2.2x10-5 m

D5=-3.1x10-4 m

D7*=0.979x10-3 mD6 = -2.84x10-4 rad

D9*=6.161x10-4 rad

1

2 3

4

5 6

7*9*

8*

1

2

1

! Trusses! Vertical Loads on Building Frames! Lateral Loads on Building Frames: Portal

Method! Lateral Loads on Building Frames: Cantilever

Method Problems

APPROXIMATE ANALYSIS OF STATICALLYINDETERMINATE STRUCTURES

2

P1 P2

(a)

Trusses

R1 R2

a

a(b)R1

a

a

V = R1

F1

F2

Fb

Fa

Method 1 : If the diagonals are intentionally designed to be long and slender,it is reasonable to assume that the panel shear is resisted entirely by the tension diagonal,whereas the compressive diagonal is assumed to be a zero-force member.

Method 2 : If the diagonal members are intended to be constructed from largerolled sections such as angles or channels, we will assume that the tension and compressiondiagonals each carry half the panel shear.

3

Example 1

Determine (approximately) the forces in the members of the truss shown inFigure. The diagonals are to be designed to support both tensile and compressiveforces, and therefore each is assumed to carry half the panel shear. The supportreactions have been computed.

10 kN 20 kN4 m 4 m

3 m

A BC

DEF

4

+ ΣMA = 0: FFE(3) - 8.33cos36.87o(3) = 0

FFE = 6.67 kN (C)

ΣFy = 0:+ 20 - 10 - 2Fsin(36.87o) = 0

F = 8.33 kN

FFB = 8.33 kN (T)

FAE = 8.33 kN (C)

+ ΣMF = 0: FAB(3) - 8.33cos36.87o(3) = 0

FAB = 6.67 kN (T)

ΣFy = 0:+ FAF - 10 - 8.33sin(36.87o) = 0

FAF = 15 kN (T)

10 kN 20 kN4 m 4 m

3 m

A BC

DEF

10 kN

20 kN

V = 10 kN

FFE

FAB

FAE= FFFB= F

θ = 36.87o

10 kN

3 m

A

F20 kN

θ

θ6.67 kN

8.33FAF

10 kN

θA

5

C

D

10 kN

3 mθ

θ

ΣFy = 0:+ 10 - 2Fsin(36.87o) = 0

F = 8.33 kN

FDB = 8.33 kN (T)

FEC = 8.33 kN (C)

+ ΣMC = 0: FED(3) - 8.33cos36.87o(3) = 0

FED = 6.67 kN (C)

+ ΣMD = 0: FBC(3) - 8.33cos36.87o(3) = 0

FBC = 6.67 kN (T)

θ = 36.87o

V = 10 kN

θ = 36.87oD

θ6.67 kN

8.33 kNFDC

ΣFy = 0:+ FDC - 8.33sin(36.87o) = 0

FDC = 5 kN (C)

ΣFy = 0:+

FEB = 2(8.33sin36.87o) = 10 kN (T)

Eθ θ

6.67 kN6.67 kN

8.33 kN 8.33 kN

θ = 36.87o

FEB

FED

FBC

F = FEC

F = FDB

6

Example 2

Determine (approximately) the forces in the members of the truss shown inFigure. The diagonals are slender and therefore will not support a compressiveforce. The support reactions have been computed.

40 kN 40 kN

4 m

A B C

GHJ

D E

FI

10 kN 20 kN 20 kN 20 kN 10 kN

4 m 4 m 4 m 4 m

7

FBH = 0

FIH

FBC

FIC

40 kN

4 m

A

J

10 kN

45o

V = 10 kN

FBH = 0

ΣFy = 0:+ 40 - 10 - 20 - FICcos 45o = 0

FIC = 14.14 kN (T)

+ ΣMA = 0: FJI(4) - 42.43sin 45o(4) = 0

FJI = 30 kN (C)

+ ΣMJ = 0: FAB(4) = 0

FAB = 0

0

0FJA

40 kN

θA

ΣFy = 0:+ FJA = 40 kN (C)

FAI = 0

FJI

FAB

FJB V = 30 kN

ΣFy = 0:+ 40 - 10 - FJBcos 45o = 0

FJB = 42.43 kN (T)

FAI = 0

40 kN

4 m

A B

J I

10 kN 20 kN

4 m

45o

8

+ ΣMB = 0:

FIH(4) - 14.14sin 45o(4) + 10(4) - 40(4) = 0

FJH = 40 kN (C)

FBC = 30 kN (T)

+ ΣMI = 0: FBC(4) - 40(4) + 10(4) = 0

FBH = 0

FIH

FBC

FIC V = 10 kN

40 kN

4 m

A B

J I

10 kN 20 kN

4 m

45oB

30 kN0

042.3 kNFBI

45o

45o45o

ΣFy = 0:+ FBI = 42.3 sin 45o = 30 kN (T)

ΣFy = 0:+

FBI = 2(14.1442.3 sin 45o) = 20 kN (C)

C30 kN30 kN

14.14 kN14.14 kNFCH

45o45o

9

Vertical Loads on Building Frames

typical building frame

10

w

A B

L(b)

0.21L0.21Lpoint of zero moment

approximate case

w

L

(d)

assumed points of zero moment0.1L 0.1L

model

w

(e)

� Assumptions for Approximate Analysis

0.1L 0.1L0.8L

w

A B

L

Simply supported(c)

column column

girder

w

A BL

(a)

Point ofzeromoment

Point ofzeromoment

11

Example 3

Determine (approximately) the moment at the joints E and C caused by membersEF and CD of the building bent in the figure.

B

D

F

A

C

E

6 m

1 kN/m

1 kN/m

12

1 kN/m

1 kN/m

4.8 m

4.8 kN

0.6 m

0.6 kN 2.4 kN

3 kN

0.6(0.3) + 2.4(0.6) = 1.62 kN�m

0.6 m

0.6 kN2.4 kN

3 kN

1.62 kN�m

0.1L=0.6 m 0.6 m

4.8 m

2.4 kN 2.4 kN

13

P

h

l(a)

P

h

l(b)

Portal Frames and Trusses

� Frames: Pin-Supported∆ ∆

assumedhinge

h

L/2

(c)

P

h

L/2

Ph/l

P/2

Ph/l

P/2

P/2 P/2

Ph/l

Ph/2

Ph/2Ph/2

Ph/2

(d)

Ph/l

14

h/2

� Frames : Fixed-SupportedP

h

l(a)

P/2

Ph/2lP/2

Ph/2l

P

h

l(b)

∆ ∆

assumedhinges

(c)

h/2

L/2P

h/2

L/2

P/2 P/2

Ph/2l

P/2

Ph/2l

P/2

Ph/2l

Ph/2l

Ph/2l

P/2 P/2

Ph/2lPh/4 Ph/4

Ph/4

Ph/4Ph/4

Ph/4

(d)

Ph/4Ph/4

15

� Frames : Partial Fixity

P

(b)

P

h

l(a)

h/3h/3

assumedhinges

θ θ

� Trusses

P

h

l(a)

P

l(b)

P/2 P/2h/2

assumedhinges

∆ ∆

16

Example 4

Determine by approximate methods the forces acting in the members of theWarren portal shown in the figure.

40 kN

7 m

8 m

2 m

4 m

A

B

C

4 m 2 m2 m

D E F

H G

I

17

40/2 = 20 kN = V

N N

20 kN = V

I3.5 m

A

V = 20 kN

NN

V = 20 kN

V = 20 kN

N N

20 kN = VM M

40 kN

3.5 m

2 m

B

C

4 m 2 m2 m

D E F

H G

J K

+ ΣMJ = 0:

N(8) - 40(5.5) = 0

N = 27.5 kN

+ ΣMA = 0:

M - 20(3.5) = 0

M = 70 kN�m

18

20 kN

27.5 kN

40 kN

3.5 m

2 m

B

C

2 m

D

J

FCD

FBDFBH

45o

+ ΣMB = 0: FCD(2) - 40(2) - 20(3.5) = 0

FCD = 75 kN (C)

+ ΣMD = 0: FBH(2) + 27.5(2) - 20(5.5) = 0

FBH = 27.5 kN (T)

ΣFy = 0:+ -27.5 + FBDcos 45o = 0

FBD = 38.9 kN (T)

+ ΣMG = 0: FEF(2) - 20(3.5) = 0

FEF = 35 kN (T)

+ ΣME = 0: FGH(2) + 27.5(2) - 20(5.5) = 0

FGH = 27.5 kN (C)

ΣFy = 0:+ 27.5 - FEGcos 45o = 0

FEG = 38.9 kN (C)

3.5 m

2 m

27.5 kN

20 kN = V

2 m

E F

G

K

FGH

FEG

FEF

45o

19

ΣFy = 0:+ FDHsin 45o - 38.9sin 45o = 0

FDH = 38.9 kN (C)

D75 kN

38.9 kN

FDE

FDH

x

y

45o45o

75 - 2(38.9 cos 45o) - FDE = 0

FDE = 20 kN (C)

ΣFx = 0:+

H 27.5 kN27.5 kN

38.9 kN FHE

x

y

45o45o

ΣFy = 0:+ FHEsin 45o - 38.9sin 45o = 0

FHE = 38.9 kN (T)

20

= inflection point

P

(a)

Lateral Loads on Building Frames: Portal Method

(b)V V V V

21

Example 5

Determine by approximate methods the forces acting in the members of theWarren portal shown in the figure.

4 m 4 m 4 m

5 kN

3 m

A

B

C

D

E

F G

H

22

5 kN

1.5 m

B D F G

I J K L

M N O

4 m 4 m 4 m

5 kN

3 m

A

B

C

D

E

F G

H

I J K L

M N O

V V2V 2V

Iy Jy Ky Ly

ΣFx = 0: 5 - 6V = 0

V = 0.833 kN

+

23

Ly 0.625 kN =

0.625kN

Ky = 0

4.167 kN

0.625 kN

Iy = 0.625kN

2.501kN

0.625kN

Jy = 0

1.5 mA

I

0.625 kN

0.833kN

0.625 kN

0.833 kN

H

L1.5 m

0.833 kN 0.625 kN 1.25 kN�m

C

J1.5 m

1.666 kN

1.666kN 2.50 kN�m

0.835 kN

E

K1.5 m

1.666 kN

1.666kN 2.5 kN�m

0.833 kN0.625 kN 1.25 kN�m

5 kN

1.5 m

B 2 m

0.833 kN

M

I

2.501 kN

0.625kN

F

1.666 kN1.5 m

2 m2 mN O

K

G

0.833 kN

1.5 m

2 mO

L

0.625 kN 0.835 kN

0.625 kN

D

1.666 kN

2 m 2 m

1.5 m

N

J

4.167 kN

24

Example 6

Determine (approximately) the reactions at the base of the columns of the frameshown in Fig. 7-14a. Use the portal method of analysis.

20 kN

30 kN

8 m 8 m

5 m

6 m

G H I

A B C

D E F

25

20 kN

30 kN

8 m 8 m

5 m

6 m

G H I

A B C

D E F

S

N

J K L

O P Q

R

M

26

Py

VOy

VPy

2V

ΣFx = 0: 20 - 4V = 0 V = 5 kN+

Ly

V´Jy

V´ Ky

2V´

ΣFx = 0: 20 + 30 - 4V´ = 0 V´ = 12.5 kN+

20 kN2.5 m

G I

20 kN

30 kN

5 m

3 m

G H I

D E F

27

Oy = 3.125

20 kN2.5 m

G R4 m

5 kN

Rx = 15 kN

Ry = 3.125 kN

Sx = 5 kN

Sy = 3.125 kN

Py = 0 kN

Mx = 22.5 kN

My = 12.5 kN

Jy = 15.625 kN

Nx = 7.5 kN

Ny = 12.5 kN

Ky = 0 kN

B

K3 m

25 kN

A

J3 m

15.625kN12.5 kN

Bx = 25 kNBx = 0 MB = 75 kN�m

Ax = 12.5 kNAx = 15.625 kN MA = 37.5 kN�m

25 kN

2.5 m

N

K

P

M

3 m4 m4 m

10 kN

22.5 kN

12.5 kN

10 kN

H SR15 kN

3.125 kN

2.5 m

4 m4 m

5 kN

3.125 kN

12.5 kN

30 kN

J

O

M2.5 m

3 m 4 m

28

Lateral Loads on Building Frames: Cantilever MethodP

beam(a)

building frame(b)

In summary, using the cantilever method, the following assumptions apply to a fixed-supported frame.

1. A hinge is place at the center of each girder, since this is assumed to be point of zero moment.

2. A hinge is placed at the center of each column, since this is assumed to be a point of zero moment.

3. The axial stress in a column is proportional to its distance from the centroid of the cross-sectional areas of the columns at a given floor level. Since stress equals force per area, then in the special case of the columns having equal cross-sectional areas, the force in a column is also proportional to its distance from the centroid of the column areas.

29

Example 7

Determine (approximately) the reactions at the base of the columns of the frameshown. The columns are assumed to have equal crossectional areas. Use thecantilever method of analysis.

30 kN

15 kN

6 m

CD

BE

A F

4 m

4 m

30

30 kN

15 kN

6 m

CD

BE

A F

4 m

4 m

x

6 m

3)(6)(0~=

++

==∑∑

AAAA

AAx

x

G

H

I

J

K

L

31

I

3 m 3 m

Hy

Hx

Ky

Kx

30 kNC

D2 m

M

+ ΣMM = 0: -30(2) + 3Hy + 3Ky = 0

The unknowns can be related by proportional triangles, that is

yyyy KHor

KH==

33

kNKH yy 10==

Gy

Gx

Ly

LxN

30 kN

15 kN

CD

BE

4 m H

I

J

K

3 m 3 m2 m

+ ΣMN = 0: -30(6) - 15(2) + 3Gy + 3Ly = 0

The unknowns can be related by proportional triangles, that is

yyyy LGor

GG==

33

kNLG yy 35==

32

10 kN

30 kN2 m

C 3 m Ix = 15 kN

Iy = 10 kN

Hx = 15 kN

10 kN

DI15 kN

10 kN

2 m

3 m

Kx = 15 kN

35 kN

15 kN

10 kN

15 kN

G

H

J2 m

2 m 3 m Jx = 7.5 kN

Jy = 25 kN

Gx = 22.5 kN35 kN

2 m

L

K

J

2 m

3 m

15 kN

7.5 kN

25 kN

10 kN

Lx = 22.5 kN

A

G2 m

35 kN22.5 kN

Ax = 22.5 kNAx = 35 kN MA = 45 kN�m

Fx = 22.5 kNFy = 35 kN MF = 45 kN�m

F

L2 m

22.5 kN35 kN

33

Example 8

Show how to determine (approximately) the reactions at the base of the columnsof the frame shown. The columns have the crossectional areas show. Use thecantilever of analysis.

35 kN

45 kN

6 m

6 m

4 m

P Q R

A B C D

L M N O

E F G H

I J K

4 m 8 m

5000mm2

5000mm2

4000 mm2

4000 mm2

6000 mm2

6000 mm2

6000 mm2

6000 mm2

34

35 kN

45 kN

6 m

4 m

P Q R

A B C D

L M N O

E F G H

I J K

6 m 4 m 8 m

6 m 4 m 8 m

6000 mm2 6000 mm24000 mm25000mm2

x

mAAx

x 48.86000400050006000

)18(6000)10(4000)6(5000)0(6000~=

++++++

==∑∑

35

35 kN2 m

P Q R

Ly

Lx My

Mx

Ny

Nx

Oy

Ox

1.52 m2.48 m8.48 m 9.52 m

+ ΣMNA = 0: -35(2) + Ly(8.48) + My(2.48) + Ny(1.52) + Oy(9.52) = 0 -----(1)

)2())10(6000

(48.848.2

)10(5000;

48.848.2;

48.848.2 66 −−−−−=== −−yy

LMLM LM

σσσσ

)3())10(6000

(48.852.1

)10(4000;

48.852.1;

48.852.1 66 −−−−−=== −−yy

LNLN LN

σσσσ

)4())10(6000

(48.852.9

)10(6000;

48.852.9;

48.852.9 66 −−−−−=== −−yy

LOLO LO

σσσσ

Since any column stress σ is proportional to its distance from the neutral axis

Solving Eqs. (1) - (4) yields Ly = 3.508 kN My = 0.855 kN Ny = 0.419 kN Oy = 3.938 kN

36

35 kN

45 kN

3 m

L M N O

I J K4 m

Ey

Ex Fy

Fx

Gy

Gx

Hy

Hx

1.52 m2.48 m8.48 m 9.52 m

+ ΣMNA = 0: -45(3) - 35(7) + Ey(8.48) + Fy(2.48) + Gy(1.52) + Hy(9.52) = 0 -----(5)

)6())10(6000

(48.848.2

)10(5000;

48.848.2;

48.848.2 66 −−−−−=== −−yy

EFEF EF

σσσσ

)7())10(6000

(48.852.1

)10(4000;

48.852.1;

48.852.1 66 −−−−−=== −−yy

EGEG EG

σσσσ

)8())10(6000

(48.852.9

)10(6000;

48.852.9;

48.852.9 66 −−−−−=== −−yy

EHEH EH

σσσσ

Since any column stress σ is proportional to its distance from the neutral axis ;

Solving Eqs. (1) - (4) yields Ey = 19.044 kN Fy = 4.641 kN Gy = 2.276 kN Hy = 21.38 kN

37

One can continue to analyze the other segments in sequence, i.e., PQM, then MJFI, then FB, and so on.

35 kN2 m

3.508 kN

3 m

Ix= 114.702 kN

Iy= 15.536 kN

Lx= 5.262 kN

Ex= 64.44 kN

Px= 29.738 kN

Py= 3.508 kN

45 kN

3 m

I2 m

19.044 kN

5.262 kN3.508 kN

3 m

3 m

E19.044 kN

64.44 kN

Ax = 64.44 kNAx = 19.044 kN MA = 193.32 kN�m

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