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

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

Displacement

INTRODUCTION: STRUCTURAL ANALYSIS

Deflection Diagrams and the Elastic Curve

∆ = 0θ = 0

fixed support

∆ = 0

roller or rockersupport

inflection point

∆ = 0

pined supportθ

Moment diagram

inflection point

� Fixed-connected joint

fixed-connected joint

� Pined-connected jointpined-connected

Moment diagram

inflection point

Moment diagram

A B C D

inflection point

∆´B

fBB x RB

1∆´B + fBB RB = ∆B = 0

Method of Consistent DeformationsBeam 1 DOF

Ax = 0

∆´1

∆´2

Compatibility Equations.

∆´1 + f11R1 + f12R2 = ∆1 = 0

f12f22

Beam 2 DOF

∆´2 + f21R1 + f22R2 = ∆2 = 0

∆´1 ∆´2

f11 f21

+f11 f12

f12 f22

wP1 P2

f32f22

Compatibility Equations.

θ´1 + f11M1 + f12R2 + f13R3 = θ1 = 0

f33f23f13

∆´2 θ´1∆´3

Beam 3 DOF

∆´2 + f21M1 + f22R2 + f23R3 = ∆2 = 0

∆´3 + f31M1 + f32R2 + f33R3 = ∆3 = 0

∆´2

∆´3

f11 f12 f13

f21 f22 f23

f31 f32 f33

f21f11

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

∆´ n

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

1=i 2=j

Mj = mj

f12 = fij

f11 = fii

f22 = fjj

Mi = mi

Maxwell�s Theorem of reciprocal displacements

f21 = fji

∫=EIdxMmf jiij

∫=EIdxmm ji

∫=EIdxMmf ijji

∫=EIdxmm ij

jiij ff =

Example 1

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

6 m 6 m

=∆´B

6 m 6 m

SOLUTION

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

� Principle of superposition

∆´B

x RBfBB

Conjugate beam

∆´B

Real beam

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

Real beam

12 kN�m

∆´B = M´B = -9000/EI ,

fBB = M´´B = 576/EI,

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

0)576(9000: =+−↑+ BREIEI

∆´B

300 kN�m

x RB = 15.63 kN=

15.63 kN34.37 kN34.37 kN�m

1 kN1 kN

12 kN�m

� Substitute ∆´B and fBB in Eq. (1)

93.6/EI

-113/EI

6 m 6 m

Use conjugate beam in obtaining the displacement

113 kN�m

34.4 kN15.6 kN

EIEIEIM C

776)6(223)2(281' −=−=

↓−==∆ ,776'EI

Real Beam

Conjugate Beam

M (kN�m) x (m)

3.28 m 6 m 12 m

223/(EI)

4 m2 m

V´C 223/(EI)

281/(EI)

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

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

∆´B

3EI 2EI

2 m 2 m

Conjugate Beam

3EI 2EI

2 m 2 m

Real Beam

40 kN�m

� Use conjugate beam in obtaining ∆´B

26.66/EI

177.7/EI

V (kN) 10 10

x (m) M (kN�m)

40/3EI = 13.33/EI∆´B = M´B = 177.7/EI

22Conjugate Beam

3EI 2EI

2 m 2 m

Real Beam1 kN

4/(2EI)=2EI

� Use conjugate beam for fBB

8 kN�m

x (m)V (kN)-

x (m) M (kN�m)

83EI= 2.67

EI4/(3EI)=1.33EI

60.44/EI

12/EIfBB = M´B = 60.44/EI

1 kNx RB

∆´B

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

3EI 2EI

2 m 2 m 2.94 kN7.06 kN

16.48 kN�m

-2.94 -2.94

+- x (m)V (kN)

-16.48

+ x (m) M (kN�m)

2.33 m

1.67 m

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

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

� 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

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

αBBαAB

αBAαAA

θ´Bθ´Aθ´A

1 kN�m

5 kN/m

1 kN�m

A B4 m4 m

5 kN/m

SOLUTION

αAA αBA

BM×αAB

Compatibility equations:

θA θB

)1('0 −−−++== BABAAAAA MfMfθθ

)2('0 −−−++== BBBABABB MfMfθθ

� Use formula provided in obtaining θ´A, 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

EIEIEILM o

AA667.2

3===α

EIEIEILM o

BA333.1

6===α

EIEIEILM o

BB667.2

3===α

EIEIEIwL

B67.46

384)8)(5(7

EIEIEIwL

128)8)(5(3

4 m4 mθ´Bθ´Aθ´A

5 kN/m

EIEIEILM o

AB333.1

6===α

αBBαAB

Real Beam1 kN�m

αBBαAB

Real Beam1 kN�m

αAA αBAαAA

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

1 kN�m

5 kN/m

1 kN�m

A B4 m4 m

5 kN/m

θA θB

EIAA667.2

EIV BBA

333.1' ==α

EIB67.46' =θ

EIA60' =θ

EIAB333.1

=αEIBB667.2

Solve simultaneous equations,

MA = -18.33 kN�m, +

MB = -8.335 kN�m, +

Compatibility equation

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

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

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

RB = 3.753 kN,

Ra = 16.25 kN,

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

ΣFy = 0:+3.753

020 =−+ BA RR

A B4 m4 m

5 kN/m

18.33 kN�m 8.36 kN�m

3.75 kN16.25 kN

Vdiagram

Mdiagram

DeflectedCurve

3.25 m

-18.33

� Quantitative shear and bending diagram and qualitative deflected curve

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.

4 m4 m

2 kN/m

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

−−−=++∆−−−=++∆

CCCBBCC

CCBBBBB

RfRfRfRf

4 m4 m

2 kN/m

4 m4 m BA

2 kN/m

B'∆C'∆

4 m4 m B

BBf ' CBf '

1 kNAC4 m4 m B

CR×BCf ' CCf '

2 kN/m

� Solve equation

4 m4 m

2 kN/m

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

)2(067.17033.5333.149−−=++− CB R

↓−=

,29.0,71.3

kNRkNR

� Diagram

4 m4 m

2 kN/m

3.71 kN 0.29 kNkNAy 58.471.329.08 =−+=

mkNM A

•=−+=

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

V (kN)4.58

-3.422.29 m

M (kN�m)

x (m)Deflected shape

Point of inflection

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.

2 m 2 m 4 m

SOLUTION

∆´B

∆B = 5 mm

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

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

∆´B

Real beam

� Use conjugate beam method in obtaining ∆´B

2 m 2 m 4 m12 kN

M´diagram

Conjugatebeam

2 m 4 m

0)4(40)34(32'' =−+−

EIEIM B

V´´B

M´´B

+ ΣMB = 0:

↓−==∆ ,33.117'''EI

∆´B

Real beam

1 kN4 m 4 m

� Use conjugate beam method in obtaining fBB

0.5 kN 0.5 kN

m´diagram

-2Conjugate

−EI4

− EI4

v´´B

m´´B

− EI4

0)4(4)34(4'' =+−−

EIEIm B

+ ΣMB = 0:

↑== ,67.10''EI

mf BBB

,67.1033.1170; BREIEI

+−=↑+ RB = 11.0 kN,

xRB = 11.0 kN

=16 kN

11.0 kN= 6.5 kN = 1.5 kNRCRA

no settlement

↓−=∆ ,33.117'EIB

↑= ,67.10EI

12 kN 4 kN∆´B 1 kN

0.5 kN 0.5 kN

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

5.37 kN= 9.31 kN = 1.32 kNRCRA

with 5 mm settlement

↓−=∆ ,33.117'EIB

↑= ,67.10EI

12 kN 4 kN∆´B 1 kN+

0.5 kN 0.5 kN

� Quantitative shear and bending diagram and qualitative deflected curve

2 m 2 m 4 m11 kN 6.5 kN 1.5 kN

Vdiagram

Mdiagram

DeflectedCurve

∆B = 5 mmA

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

Vdiagram

-6.69-1.32

Mdiagram

18.625.24+

DeflectedCurve ∆B = 5 mm

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

2 kN/m

Compatibility equation:

∆´2 + f22R2 = ∆2

∆´2

4 m 6 m

w 2 kN/m1 2 3

↓−=−=

+××−−=

+−−=∆

,2.6)200)(200(

)1041024(24

)4)(2(

LLxxEI

wxprovidednsformulatioUse

↑+==

−−×

××−=

−−=

,48.0)200)(200(

)4610(106

xbLLEIPbxf

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)

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)

6.5 kN4.75 kN 8.75 kN

Basic Beams: Single span

- wL/2

APPENDIX

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

3845 4

� Find ∆∆∆∆1 by Castigliano�s

L/2 L/2

1 2 3P

+22PwL

+x1 V1

+ ΣΜ = 0;

MxPwLwxM

xPwLwxM

=++−=

=+−+

∫∫ ∂∂

∂=∆=∆

1max1 )()(

∫ ∂∂

1 )(2L

MxPwLwxM =++−=

1 ))22

(2 dxxPwLwxxEI

3845 4

L /2 L /2

P/2 P/2

EILfPf

11 =→==δ

Single span

L = 2l

l = L/2 l = L/2

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

max =∆

Double span

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

1.1wl1.1wl0.4wl 0.4wl

∆max = wl4

145 EI

V0.4wl

- 0.4wl- 0.6wl

-0.5wl

0.08 wl2 0.08 wl2

0.025 wl2

-0.1 wl2 -0.1 wl2

- -++M

Triple span

! Comparison Between Indeterminate andDeterminate

! Influence line for Statically IndeterminateBeams

! Qualitative Influence Lines for Frames

INFLUENCE LINES FOR STATICALLY INDETERMINATE BEAMS

AB CED

RA AB CED

Indeterminate Determinate

Comparison between Indeterminate and Determinate

AB CED

B CEDRA

AB CED

ME AB CED

AB CED

VD AB CED

Indeterminate Determinate

∆´1 = f1j

11 2 3j4

Redundant R1 applied

×R1×R1

Influence Lines for Reaction

011111 =∆=+ Rff j

11 ffR j−=

11 2 3j4

Redundant R2 applied

fjjf2j

Compatibility equation.

0' 22222 =∆=+∆ Rf

022222 =∆=+ Rff j

22 ffR j−=

1 2 3j4

Influence Lines for Shear

)1( jE ff

)1( jE fMα

1 2 3j4

Influence Lines for Bending Moment

R3 )(33

R2 )(22

1 2 3j4

� Influence line of Reaction

Using Equilibrium Condition for Shear and Bending Moment

11 =ff

22 =ff

33 =ff

1 2 3j4

V4 = R1

� Unit load to the right of 4

� Influence line of Shear

V4 = R1

V4 = R1 - 1

R2 R3R1

� Unit load to the left of 4

V4 = R1 - 1

01;0 41 =−−=Σ↑+ VRFy

0;0 41 =−=Σ↑+ VRFy

1 2 3j4

� Influence line of Bending moment

M4 = - l + x + l R1

M4 - 1 (l-x) - l R1 = 0 + Σ M4 = 0:

� Unit load to the left of 4

M4 = l R1

� Unit load to right of

R2 R3R1

1 1M4 = - l + x + l R1

M4 - l R1 = 0 + Σ M4 = 0:

M4 = lR1

Influence Line of VI

Maximum positive shear Maximum negative shear

Qualitative Influence Lines for Frames

Influence Line of MI

Maximum positive moment Maximum negative moment

15 m15 mAy GyDy

Influence Line for MOF

15 m15 m

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

BBf BBf BBf 1=BBf

� Influence line of RB

A BC D

2 m 2 m 2 m

ABf CBf DBfBBf

Real BeamA BC D

2 m 2 m 2 m

Conjugate Beam

� Find fxB by conjugate beam

6 kN�m

=−+EI

x 18726

32xV´x

fxB / fBB

A BC D

2 m 2 m 2 m

72/EI37.33/EI10.67/EI

EIEIxMf xxB

−+==

EI33.37

EI67.10

/72 = 0.518

/72 = 0.148

10.670

Influence line of RB

Influence line of RA

A BC D

2 m 2 m 2 m

Conjugate Beam

A BC D

2 m 2 m 2 m

Real Beam

� Find fxA by conjugate beam

6 kN�m

=−EIx

fxA / fAA

EIxMf xxA 6

A BC D

2 m 2 m 2 m

1 kNfAA fxA

72/EI 61.33 /EI34.67 /EI

EI67.34

EI33.61

61.3372

/72 = 1.0/72=0.852

/72 = 0.4821=

−==−+

=Σ↑+

A BC D

2 m 2 m 2 m

0.148.518

0.4820.852

1.01 kN

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

RA = 1- RB

0.1480.518

A BC D

2 m 2 m 2 mRA

0.8520.482

-0.148

Using equilibrium conditions for the influence line of VC

VC = 1 - RB

� Unit load to the left of C

=+−+

=Σ↑+

VC = - RB

=Σ↑+

RxMRxM

6606)6(1

++−==+−−−

A BC D

2 m 2 m 2 mRA

0.1480.518

-1.112 -0.892

Using equilibrium conditions for the influence line of MA

0.1480.518

A BC D

2 m 2 m 2 m

Using equilibrium conditions for the influence line of MC

0.592 RB

MC = 4RB

;0=+−

MC = -4 + x + 4RB

04)4(1;0

=+−−−=Σ

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

A B CD E F

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

A B CD E F

4 m 6 m2 m

Real beam

0.51.5

18.67/EI

Conjugate beam

EI33.5

EI67.10

EI67.10 EI

Mf AAA64' ==

=−EI

1 33.512

EIEIEIx 67.18646

32 −+=

V´ x1

M´ x1

EI33.5

Conjugate beam

4 m 8 m

EI33.5

EI67.18

M´ x2

V´ x2

EI67.18

CtoBforEI

xMf xxA ,33.512

1 −==

BtoAforEIEI

xMf xxA ,6467.186

22 +−==

fxA / fAA

-0.1562

-0.2188

0.4375

fAA fxA

A B CD E FG

4 m 6 m2 m

EI10−

EI16−

EI14−

-0.219 -0.25 -0.156

A B CD E FG

4 m 6 m2 m

Using equilibrium conditions for the influence line of RB

RB RCRA

1 -0.219 -0.25 -0.156

10.438

0.4850.8751.07810.59

0.5939

-0.1562

-0.2188

0.4375

0128;0

=−+−=Σ+

A B CD E FG

4 m 6 m2 m

1 -0.219 -0.25 -0.156

10.438

RB RCRA

-0.1562

-0.2188

0.4375

0.6719

0.1406

-0.0312

10.672

0.3750.141

-0.0312

Using equilibrium conditions for the influence line of RC

08)8(14;0

=−−−=Σ+

A B CD E FG

4 m 6 m2 m

� Check ΣFy = 0

RB RCRA

1 -0.219 -0.25 -0.156

10.438

0.490.8751.081

10.672

0.3750.141

-0.0312

=Σ↑+

0.6719

0.1406

-0.0312

-0.1562

-0.2188

0.4375

0.5939

1 -0.219 -0.25 -0.156

10.438

VG = RA

� Unit load to the right of G

-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

� Unit load to the left of G

01;0 =−−=Σ↑+ GAy VRF

10.672

0.3750.141

-0.0312

0.6250.328

0.0312

-0.141-0.375

A B CD E FG

4 m 6 m2 m

Using equilibrium conditions for the influence line of VE

� Unit load to the left of E

VE = - RC

0;0 =+=Σ↑+ ECy VRF

VE = 1 - RC

� Unit load to the right of E

01;0 =+−=Σ↑+ CEy RVF

1 -0.219 -0.25 -0.156

10.438

-0.438 -0.5 -0.312

A B CD E FG

4 m 6 m2 m

Using equilibrium conditions for the influence line of MG

MG = -2 + x + 2RA

� Unit load to the left of G

02)2(1;0

=−−+=Σ

MG = 2RA

� Unit load to the right of G

=−=Σ

10.672

0.3750.141

-0.0312

0.6881

4 m 6 m2 m

A B CD E FG

Using equilibrium conditions for the influence line of ME

-0.125

� Unit load to the left of E

ME = 4RC

+ ΣME = 0;

ME = - 4 + x+ 4RC

� Unit load to the right of E

04)4(1;0

=+−−−=Σ+

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

2EI 3EI

2 m 2 m

fAAfEA

/fAA /fAA /fAA

fAAfEA

2EI 3EI

2 m 2 m

42Conjugate Beam

2EI 3EI

2 m 2 m1Real Beam

8 kN�m

x (m)V (kN)1 1

x (m) M (kN�m)

fAA = M´A = 60.44/EI

60.44/EI

1.33EIEI67.2

EI67.2

60.44/EI

Conjugate Beam12/EI

1.33EI

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

RA = fxA/fAA

0.614 0.294 0.081

� Quantitative influence line of RA

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

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

-1.352-1.088 -1.648

0.614 0.294 0.081

4 m 2 m 2 m

Using equilibrium conditions for the influence line of VB

0.614 0.294 0.081

-1-0.386

-0.706-0.919

0 VB = RA -1

VB = RA - 1

4 m 2 m 2 m

Using equilibrium conditions for the influence line of VC and MC

0.614 0.294 0.081

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

VC = RAVC = RA - 1

-0.386 -0.706

0.3240.4561.176

0.294 0.081 0

4 m 2 m 2 m

0.614 0.294 0.081

10(0.081)=0.81 kN

MA (kN�m)+

-13.53

V (kN)

0.81 0.81

The quantitative shear and bending moment diagram and qualitative deflected curve

RB=9.19 kN

MB= 13.53 kN�m

4 m 2 m 2 m

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

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

0.614 0.294 0.081

RB=23.31 kN

MB=46.48 kN�m

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

A BC D

2 m 2 m 2 m

Conjugate Beam

A BC D

2 m 2 m 2 m

Real Beam

6 kN�m

6 /EIEI18

=−EIx

fxA / fAA

A BC D

2 m 2 m 2 m

1 kNfAA fxA

72/EI 61.33 /EI34.67 /EI

EIxMf xxA 6

EI67.34

EI33.61

61.3372

/72 = 1.0/72=0.852

/72 = 0.4821=

� Influence line of RB

A BC D

2 m 2 m 2 m

Real BeamA BC D

2 m 2 m 2 m

Conjugate Beam

� Find fxB by conjugate beam

6 kN�m

=−+EI

x 18726

32xV´x

fxB / fBB

A BC D

2 m 2 m 2 m

72/EI37.33/EI10.67/EI

/72 = 0.518

/72 = 0.148/72 = 1

10.670

Influence line of RB

fBB= 1

EIEIxMf xxB

−+==

EI33.37

EI67.10

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

A B CD ED

2 m 2 m 2 m 2 m

The influence line for the shear at D

1 kNDD

A B CD E

2 m 2 m 2 m 2 m

2 kN�m

1 kN2 k

1 kN2 kN�m

1 kN2 kN1 kN

� Using conjugate beam for find fxD

A B CD E

2 m 2 m 2 m 2 m

1 kN1 kN

Real beam

V( kN)

x (m)1

M (kN �m) x (m)

Conjugate beam

1 kN2kN

2 m 2 m 2 m 2 m

Conjugate beam

� Determine M´D at D

128/3EI

=−=− )2)(3

8()32)(2(4

EIEIEI

EIEIEIEI 352)2)(

32)(2( =−+=

EIEIEI 376)2)(

32)(2( −=−=

128/3EI

2 m 2 m 2 m 2 m

Conjugate beam

2/EI2/EI

EI340 V´DR

128/3EI

2/EI2/EI

128/3EI = M´D = fDD

V ´x (m) θ

fxD = M ´ x (m) ∆

−EI334

Influence line of VD = fxD/fDD

0.406 =128

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

Conjugate beam

EI38128/3EI

2 m 2 m 2 m 2 m

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

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

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

M (kN �m) x (m)

Conjugate beam

2 m 2 m 2 m 2 m

Conjugate beam

EIEIEI 326)2)(4()

32)(1( =+=

=−=− )2)(3

4()32)(1(2

EIEIEI

2 m 2 m 2 m 2 m

Conjugate beam

1/EI1/EI

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

813.03226

αDD = 32/3EI

2 m 2 m 2 m 2 m

Conjugate beam

x (m)V ´ θEI4

fxD = M ´x (m) ∆

θD = 0.469 + 0.531 = 1 rad

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

26/3EI

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

Influence line for RD

A DB C GE F

5 m 5 m 5 m 5 m5 m 5 m

fBDfCD fDD fED fFD

fXD/fDD = Influence line for RD

71Conjugate beam

15 + (2/3)(15)

EI5.112

� Use the consistency deformation method

- Use conjugate beam for find ∆´G and fGG

∆´G + fGGRG = 0 ------(1)

Real beam

15 m 15 m

Real beam

30 mA G

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

∆´G

112.5/EI

EIM CC

5.2812'' ==∆

Conjugate beam

EI15 EI

EIMf GGG

9000''' ==

x RG = -0.3125 kN

090005.2812=+ GR

↓−= ,3125.0 kNRG

Substitute ∆´G and fGG in (1) :

A G5.625

0.6875 0.3125

8.182 m

6.818 m

EI16.35

EI98.15

EI01.23

3125.0 22

2 xx−

22 '13.28xMx

(6875.0625.5 12

11 xEI

xx −=

6875. 12

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

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−

(5.625-0.6875x1)/EI

EI625.5 EI

xx 211 6875.0625.5 −EI

6875.0 21

0.3125x2

V´2M´2

EI13.28

23125.0 2

� Influence Line for RD

Influence Line for RD

0.228 0.677 1.0 0.931 0.545

EI7.166

EI1.246

EI2.229

EI1.134

fXD/fDD

1.24656

1.2467.166

1.2461.246

1.2462.229

1.2461.134

Influence line for RG

A DB C GE F

5 m 5 m 5 m 5 m5 m 5 m

1fBG fCG

fGGfEGfFG

fXG/fGG

76Conjugate beam

� Use consistency deformations

∆´D + fDDRD = 0 ------(2)

- Use conjugate beam for find ∆´D and fDD

Real beam

30 mA G

Real beam

15 m 15 m

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

∆´D

Conjugate beam15 + (2/3)(15)

EI5.112

450/EI

EI9000

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

15 mV´

450/EI

EI9000

EI15 EI

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

fGG = M´G = 1968.56/EI

168.75/EI

EIEIfCG

06.125)67.6(75.18−=−==

A G Conjugate beam

EI5.7−

15 + (10/3) = 18.33 m

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

EI5.112

EI75.18

EI5.7−

EI75.18

A6.67 m

EI75.18

EI5.7−

EI75.18−

=−+ 33

23 75.16856.1968)

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

EI5.62−

EI125−

EI73.447

EI64.1145

EI56.1968

56.19685.62−

56.1968125−

56.196873.447

56.196864.1145

56.196856.1968

fXG/fGG

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

Ay RD RG

Unit load

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

-0.156 -0.124

CDAy RRRF −−==Σ↑+ 1:0

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

Ay RD RG

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+

! 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

Slope � Deflection Equations

settlement = ∆j

Pi j kw Cj

Mij MjiwP

Degrees of Freedom

1 DOF: θΑ

θΒΒΒΒ

A BC 2 DOF: θΑ , θΒΒΒΒ

Stiffness

kBAkAA

LEIkAA

LEIkBA

kBBkAB

LEIkBB

LEIkAB

Fixed-End ForcesFixed-End Moments: Loads

L/2 L/2

General Case

settlement = ∆j

Pi j kw Cj

Mij MjiwP

settlement = ∆j

(MFij)∆ (MF

ji)∆

(MFij)Load (MF

ji)Load

Mij Mji

i jwPMij

settlement = ∆jθj

=+ ji LEI

LEI θθ 24

ji LEI

LEI θθ 42

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

jiij MMLEI

LEIM +++= ∆θθ Loadji

Fjiji MM

LEIM )()()4()2( +++= ∆θθ

Mji Mjk

Pi j kw Cj

Mji Mjk

Equilibrium Equations

0:0 =+−−=Σ+ jjkjij CMMM

i jMij Mji

LEIkii

LEIk ji

LEIkij

LEIk jj

Stiffness Coefficients

Stiffness Matrix

)()2()4( ijF

jiij MLEI

LEIM ++= θθ

)()4()2( jiF

jiji MLEI

LEIM ++= θθ

LEILEILEILEI

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

Matrix Formulation

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

Displacementmatrix

Stiffness matrix

Force matrixwP(MF

ij)Load (MFji)Load

i jwPMij

ψ ∆j

Mij Mji

(MFij)∆ (MF

ji)∆

Fixed-end momentmatrix

][]][[][ FEMKM += θ

]][[])[]([ θKFEMM =−

][][][][ 1 FEMMK −= −θ

Real beam

Conjugate beam

Stiffness Coefficients Derivation: Fixed-End Support

LMM ji +

EILM j

−−−=

=+−=Σ+

)2(0)2

(:0 −−−=+−=Σ↑+EI

LMF jiiy θ ij

andFrom

);2()1(

LMM ji +

Conjugate beam

Real beam

Stiffness Coefficients Derivation: Pinned-End Support

Mi θi

θi θj

(:0' =−=Σ+ LLEI

LMM ii

(:0 =+−=Σ↑+ jii

y EILM

EILMF θ

3(1 =→==θ

−=θ)

EILM i

Fixed end moment : Point Load

Real beam

2 PLMEI

MLFy ==+−−=Σ↑+

Conjugate beamA

EIPL4 EI

841616PLPLPLPL

-PL/8 -PL/8

-PL/16-PL/8

-PL/16-PL/8-

Uniform load

Real beam Conjugate beam

23 wLMEI

MLFy ==+−−=Σ↑+

Settlements

MjMi = Mj

LMM ji +L

MM ji +

Real beam

6LEIM ∆

Conjugate beam

(:0 =+−∆−=Σ+L

EIMLM 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

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

(MF AB)∆

(MF BA) ∆

)1()()(24−−−+++== ∆

FABload

FABBAAAB MM

LEIMM θθ

)2()()(42−−−+++= ∆

FBAload

FBABABA MM

LEIM θθ

21)[(3:

2)1()2(2lim AF

BAloadFABload

FBABBAA

MMMMLEIMbyinateE ++−+=

−∆θθ

Fixed-End MomentsFixed-End Moments: Loads

L/2 L/2

8PLPLPL

=−−+

222 wLwLwL=−−+

Typical Problem

8PL w12

8024 11

LEIM BAAB +++= θθ

8042 11

LEIM BABA −++= θθ

128024 2

wLLPLEI

LEIM CBBC ++++= θθ

128042 2

wLLPLEI

LEIM CBCB −

−+++= θθ

MBA MBC

8042 11

128024 2

wLLPLEI

BBCBABB forSolveMMCM θ→=−−=Σ+ 0:0

Substitute θB in MAB, MBA, MBC, MCB

MABMBA

8024 11

LEIM BAAB +++= θθ

8042 11

128024 2

wLLPLEI

128042 2

wLLPLEI

LEIM CBCB −

−+++= θθ

ByR CyAy ByL

MABMBA

By = ByL + ByR

MBCMCB

Example of Beams

10 kN 6 kN/m

B 6 m4 m4 m

Example 1

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

PPL8 wwL2

MBA MBC

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

B 6 m4 m4 m

8)8)(10(

++= BAABEIEIM θθ

8)8)(10(

−+= BABAEIEIM θθ

30)6)(6(

++= CBBCEIEIM θθ

20)6)(6(

−+= CBCBEIEIM θθ

0:0 =−−=Σ+ BCBAB MMM

)6)(6(10)6

=+−+

10 kN 6 kN/m

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

10.6 kN�m

8.8 kN�m

6 kN/m8.8 kN�m

10 kN�mB

10 kN 6 kN/m

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)

- 4.78

M (kN�m) x (m)

-8.8 -10- -

EIB4.2

Deflected shape x (m)

10 kN 6 kN/m

B 6 m4 m4 m

Example 2

PPL8 wwL2

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

10 kN 6 kN/m

B 6 m4 m4 m

)8)(10(8

4−−−++= BAAB

EIEIM θθ

)8)(10(8

2−−−−+= BABA

EIEIM θθ

)6)(6(6

−−−++= CBBCEIEIM θθ

)6)(6(6

−−−−+= CBCBEIEIM θθ

62:)1()2(2 −=− BBAEIM θ

)5(158

3−−−−= BBA

EIM θ

MBA MBC

)5(158

3−−−−= BBA

EIM θ

)6)(6(6

−−−+= BBCEIM θ

0:0 =−−=Σ+ BCBAB MMM

)6(030

)6)(6(15)6

−−−=+−+

EIEIinSubstitute

40:)1(

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(

−−−−= BCBEIM θ

10 kN 6 kN/m

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

12.19 kN�m

6 kN/m12.19 kN�m

8.30 kN�mC

Deflected shape x (m)EIB49.7

10 kN 6 kN/m

B 6 m4 m4 m

V (kN)x (m)

- 6.52

-11.35

M (kN�m) x (m)

-12.2-8.3

11.35 kN3.48 kN

6.52 + 6.65 = 13.17 kN

EIA74.23−

10 kN 4 kN/m

B6 m4 m4 m

2EI 3EI

Example 3

10 kN 4 kN/m

B6 m4 m4 m

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

)8)(10(8

)2(4−−−++= BAAB

EIEIM θθ

)8)(10(8

)2(2−−−−+= BABA

EIEIM θθ

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

)2(3:2

)1()2(2 aEIM BBA −−−−=− θ

)6)(4(6

)3(4 2

10 kN 4 kN/m

B6 m4 m4 m

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

EIEIMM

/091.13151275.2:0

==+−==−−

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

)2(3 aEIM BBA −−−−= θ

)6)(4(6

)3(4 2

mkNEIM

•−=−=

•=−=

•−=−=

91.10126

18.1412)091.1(6

18.1415)091.1(8

10 kN 4 kN/m

B6 m4 m4 m

2EI 3EI

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

140.18 kN�m

ByLAy = 6.73 kN= 3.23 kN

14.18 kN�m

10.91 kN�m

4 kN/m

14.18 10.91

= 11.46 kN= 12.55 kN

10 kN 4 kN/m

B6 m4 m4 m

2EI 3EI

11.46 kN3.23 kN

10.91 kN�m

V (kN)x (m)

-11.46

M (kN�m) x (m)

-14.18

-10.91

6.77 + 12.55 = 19.32 kN

θB = 1.091/EIDeflected shape x (m)

10 kN 4 kN/m

B6 m4 m4 m

2EI 3EI

Example 4

12 kN�m

10 kN 4 kN/m

B6 m4 m4 m

2EI 3EI

12 kN�m

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

12 kN�mMBA

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(

++= BAABEIEIM θθ

0 -3.273/EI

012:int =−−− BCBA MMBJo

012)122()1575.0( =−+−−− BEIEI θ

EIM BA •−=−−= 45.1715)273.3(75.0

EIM BC •=+−= 45.512)273.3(2

EIM CB •−=−−= 27.1512)273.3(

4 kN/m

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

B13.64 kN2.82 kN

15.27 kN�m

17.54 kN

EIM BA •−=−−= 45.1715)273.3(75.0

EIM BC •=+−= 45.512)273.3(2

EIM CB •−=−−= 27.1512)273.3(

10 kN 4 kN/m

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+

-13.64

M (kN�m) x (m)+

-17.45

-5.45-15.27

EIB273.3

=θEIA

21.7−=θ

EIB273.3

−=θ

EIA21.7

−=θ

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

6 m4 m4 m

2EI 3EI10 mm

6 m4 m4 m

2EI 3EI10 mm

[FEM]∆

6LEI∆

[FEM]load

)8)(10(8

)01.0)(2(68

)2(42 −−−+++=

EIEIEIM BAAB θθ

)8)(10(8

)01.0)(2(68

)2(22 −−−−++=

EIEIEIM BABA θθ

)6)(6(6

)01.0)(3(66

)3(4 2

2 −−−+−+=EIEIEIM CBBC θθ

)6)(6(6

)01.0)(3(66

)3(2 2

2 −−−−−+=EIEIEIM CBCB θθ

6 m4 m4 m

2EI 3EI10 mm

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

)8)(10(8

)01.0)(2(68

)2(42 −−−+++=

EIEIEIM BAAB θθ

)8)(10(8

)01.0)(2(68

)2(22 −−−−++=

EIEIEIM BABA θθ

)1(10758

)2(4−−−+++= BAAB

EIEIM θθ

)2(10758

)2(2−−−−++= BABA

EIEIM θθ

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

)2(3:2

)1()2(2 aEIM BBA −−−−−−−+=− θ

+ 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

6 m4 m4 m

2EI 3EI10 mm

MBC = (4/6)(3EI)θB - 192.8

MBA = (3/4)(2EI)θB + 16.5

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

12 kN�m64.58 kN�m

6 kN/m

64.58 kN�m

146.69 kN�m

MBC = -64.58 kN�m

MCB = -146.69 kN�m

MBA = 64.58 kN�m,

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

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

3 m 3 m2EI 1.5EI

12 kN�m

6 kN/m

3 m 3 m2EI 1.5EI

12 kN�m

)2(4−−−= CCB

EIM θ

)2(1005.43

)5.1(23

)5.1(4−−−+++= ACCA

EIEIM θθ

)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

AmkN •=

)01.0)(502005.1(62

mkN •100MF

6 kN/m

)3(6 2

= 4.5MF

6 kN/m

3 m 3 m2EI 1.5EI

12 kN�m

)2(4−−−= CCB

EIM θ

)5.4(33

)5.1(3 aEIM CCA −−−+++= θ

MCBMCA

0=+ CACB MM

)5.4(33

)5.48(=+++

radEIC 0015.006.15

−=−

)3(1005.43

)5.1(4)06.15(3

)5.1(212 −−−+−+−

EIEI θSubstitute θC in eq.(3)

radEIA 0034.022.34

−=−

� Equilibrium equation:

6 kN/m

3 m 3 m2EI 1.5EI

12 kN�m

radEIC 0015.006.15

−=−

=θ radEIA 0034.022.34

−=−

EIEIM CBC •−=−

== 08.20)06.15(3

)2(2 θ

EIEIM CCB •−=−

== 16.40)06.15(3

)2(4 θ

kN08.203

08.2016.40=

+kN08.20

B C40.16 kN�m20.08 kN�m

6 kN/m

12 kN�m

40.16 kN�m

8.39 kN26.39 kN

6 kN/m

3 m 3 m2EI 1.5EI

12 kN�m

6 kN/m

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)

26.398.39

-20.08

M (kN�m)

x (m)20.08

-40.16

radC 0015.0−=θ

radA 0034.0−=θ

Deflected shapex (m)

radC 0015.0=θ

radA 0034.0=θ

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

3 m 3 m2EI 1.5EI

12 kN�m

6 kN/m

3 m 3 m2EI 1.5EI

12 kN�m

)3(6 2

6 kN/m

mkN •=

)01.0)(502005.1(620.01 m

34 CEI∆

2CC EIEI ∆

)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

)2(2−−−∆−= CCBC

EIEIM θ

)5.4(323

)5.1(3:2

)4()3(2 aEIEIM CCCA −−−+++∆+=− θ

CC EIEI

∆=∆

23)5.1(6∆C

CEI∆

6 kN/m

3 m 3 m2EI 1.5EI

12 kN�m

� Equilibrium equation:

()( CBBCCBy

MMC +−=

6 kN/m

12 kN�m

MCBMBC

)5.1(1812)( +=

++= CACA

CAyMMC

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

6 kN/m

3 m 3 m2EI 1.5EI

12 kN�m

� Solve equation

radEIC 00255.051.25

−=−

mmEIC 227.527.52

−=−

Substitute θC and ∆C in (4)

radEIA 000286.086.2

−=−

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

mkNM BC •= 68.35

mkNMCB •= 67.1

mkNMCA •−= 67.1

6 kN/m

3 m 3 m2EI 1.5EI

12 kN�m

35.68 kN�m

68.3512)5.4(18

−−=

55.518

6 kN/m

3 m 3 m2EI 1.5EI

12 kN�m

35.68 kN�m

12.45 kN 5.55 kN

Deflected shape

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

-35.68

V (kN)

x (m)0.925 m12.45

Example of Frame: No Sidesway

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.

12 kN/m

3EIPL/8 = 30

PL/8 = 30

� Equilibrium equations

)3(18366

)2(3−−−++= BBC

EIM θ

*)1(010 −−−=−− BCBA MM

05430310 =−+− BEIθ

EIEIB667.4

)3(14 −

=θ)1(30

−−−+= BABEIM θ

mkNMmkNM

•=•−=

33.4933.39

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−

12 kN/m

3EI39.33

12 kN/m

B49.33

17.67 kN

27.78 kN

Bending moment diagram

-25.33

Deflected curve

-49.33

θB = -4.667/EIθB

MAB = 25.33 kN�m

MBA = -39.33 kN�m

MBC = 49.33 kN�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.

5 kN/m

60(106) mm4

240(106) mm4 180(106)

120(106) mm4

3 m 4 m3 m0

BAABEIEIM θθ

BABAEIEIM θθ

75.186

)4(4++= CBBC

EIEIM θθ

75.186

)4(2−+= CBCB

EIEIM θθ

CCDEIM θ5

)3(3+= CCE

EIM θ

0=+ BCBA MM

)1(75.18)68()

58( −−−−=++ CB EIEI θθ

0=++ CECDCB MMM

)2(75.8)49

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

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

EIEIM θθ5

)2(4+=

BABAEIEIM θθ

75.186

)4(4++= CBBC

EIEIM θθ

75.186

)4(2−+= CBCB

EIEIM θθ

CCDEIM θ5

)3(3+= CCE

EIM θ

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

4.23 kN�m

2.54 kN

(8.46 + 4.23)/5 = 2.54 kN

8.46 kN�m

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

2.54-0.34=2.2 kN

2.2 kN

Shear diagram

-14.12

2.82 m

1.18 m

Deflected curve

1.18 m

0.78 m2.33 m1.29 m

1.29 m

0.78 m

Moment diagram

1.18 m

-18.18 -16.44

-8.46-8.46

2.33 m

θB = −5.29/EI

θC = 2.86/EI1.67m

Example of Frames: Sidesway

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

MABMDC

� 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

)(3−−−= BBC

EIM θ

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

)1)(3(104

−−−∆

++=EIEIM BAB θ

5.625 0.375EI∆

(5.625)load

(1.875)load

(0.375EI∆)∆

:0=Σ+ CM

∆ ∆

(1/2)(0.375EI∆)∆

:0=Σ+ BM

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

)( BAABx

MMA +=

)6(0468.04

−−−∆== EIMD DCx

)1)(3(104

−−−∆

+−=EIEIM BBA θ

5.625 0.375EI∆

� Slope-Deflection Equations

� Solve equation

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

)(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

=∆−

)6()1(8.446.5 toinEI

Substitute B =∆−

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

Deflected curve

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,

7.9 kN 2.1 kN

∆ = 44.8/EI ∆ = 44.8/EI

θB = -5.6/EI5.6

10 kN4 m

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.

10 kN4 m

2.5EI EI

MDCMAB

∆ 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

DA 3m4 m

10 kN4 m

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

= ∆ / cos 36.87° = 1.25 ∆

∆∆BC

∆ CD 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

∆BC= 0.75 ∆

0.375EI∆

6EI∆/(4) 2 = 0.375EI∆

∆´ B´

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

C´∆CD= 1.25 ∆

(1/2) 0.75EI∆

(1/2) 0.5625EI∆

PL/8 = 5

(6)(2.5EI)(1.25∆)/(5)2 = 0.75EI∆

0.75EI∆

� Slope-Deflection Equation

10 kN4 m

pin2 EI

2.5 EI EI

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

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

B CMBC

MBC/4MDC

� Horizontal reactions

� 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

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

)(22 −−−∆

++=EIEIM BAB θ

)(42 −−−∆

+−=EIEIM BBA θ

)75.0)(2(34

)2(32 −−−

∆−=

EIEIM BBC θ

)25.1)(5.2(32 −−−

EIM DC

)20(−−−

−+= ABBA

)6(443

−−−−

)7(050938.05.2 −−−=−∆+ EIEI Bθ

)8(05334.00938.0 −−−=−∆+ EIEI Bθ

Deflected shape

θB=1.45/EI

Bending-moment diagram

DA 11.19

∆ ∆

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

10 kN4 m

pin2 EI

2.5 EI EI11.19 kN�m

8.27 kN

0.478 kN1.73

0.478 kN

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.

20 kN/mpin

[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

[FEM]load

wL2/12 = 15

∆ ∆

[FEM]∆∆∆∆

6(2EI∆)/(3) 2= 1.333EI∆

6(2EI∆)/(3) 2= 1.333EI∆ 6(4EI∆)/(4) 2

= 1.5EI∆

1.5EI∆

BBCEIM θ3

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

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

BEIθ3= ----------(3)

∆= EI75.0 ----------(4)

∆+++= EIEIEIM BAAB 333.1153

)2(4 θθ0

∆+−+= EIEIEIM BABA 333.1153

)2(2 θθ0

∆+= EIEIM DDC 75.04

)4(3 θ0

(1/2)(1.5EI∆)

� Slope-Deflection Equation

Ax + ΣMC = 0:

:0=Σ+ BM

)6(188.04

−−−∆== EIMD DCx

)5(30889.0333.13

)5.1(60

−−−+∆+=

� Horizontal reactions

� Solve equation

*)1(0 −−−=+ BCBA MM

Equilibrium Conditions

Equation of moment

Horizontal reaction at support

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

EIEIB67.3451.5

=∆−

)6()1(67.3451.5 toinEI

Substitute B =∆−

)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

DMoment diagram

Deflected shape

∆ ∆

4m20 k

N/m 16.52 kN�m

53.87 kN�m

53.48 kN

6.52 kN26 kN�m

5.55 kN

mkNM AB •= 87.53mkNM BA •= 52.16

mkNM BC •−= 52.16mkNM DC •= 0.26

Ax = 53.48 kNDx = 6.52 kN

B16.52

! 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

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

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

COF = 0.5

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

B EI D

L1 /2 L1/2 L2 L3

LEIK 4

=LEIkCC

EIkDC2

COF = 0.5

Far-end member pinned or roller end support:

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

COF = 00

1LEIK 3

= LEIkAA

L1 /2 L1/2 L2 L3

K(CD) = 4EI/L3

Joint Stiffness Factor (K)

L1 /2 L1/2 L2 L3

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

Kjoint = KT = ΣΣΣΣKmember

01DF DFBC DFCB DFCDDFBA)

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

B C DA

Distribution Factor (DF)

L1 /2 L1/2 L2 L3

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

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

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

CB(DFBC) CB( DFBC)

CO=0.5CO=0

Distribution of Couple at Node

A CB D

L1 /2 L1/2 L2 L3

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

B C DA01DF DFBC DFCA DFCDDFBA

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

A CB3EI2EI

4 m 4 m 8 m 8 m

50 kN�m

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(DFBC) MF( DFBC)

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

B DL1 /2 L1/2 L2 L3

B C DA01DF 0.6 0.5 0.50.4

L1= L2 = 8 m, L3 = 10 m

wL22/12=161.5PL1/8=30 wL3

2/12=25

Moment Distribution for Beams

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.

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

13.17 kN 6.83 kN

11.33 kN�m25.34 kN�m

10.25 kN13.75 kN

0.5 0 0.5 0CO

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

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

3 kN/m

4 m 4 m 8 m

V (kN)x (m)

-13.17

-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

5 kN/m

B3EI2EI

4 m 4 m 8 m 8 m

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.

5 kN/m

4 m 4 m 8 m 8 m

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

5 kN/m

4 m 4 m 8 m 8 m

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)

36.22 kN�m43.25 kN�m

13.78 kN�m

CyLByR

40 kN43.25 kN�m

5 kN/m

4 m 4 m 8 m 8 m

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

5 kN/m

B2EI 3EID

50 kN�m

4 m 4 m 8 m 8 m

M(kN�m) x (m)

-36.22

13.7830.32

-43.25Deflected shape

V (kN) x (m)0.47

-14.59-27.13

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

3 m 3 m 9 m 3 m

50 kN�m

40 kN 40 kN

10 kN/m2EI

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.

3 m 3 m 9 m 3 m

50 kN�m

10 kN/m2EI

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 kN�m 40 kN

3 m 3 m 9 m 3 m

50 kN�m

40 kN 40 kN

10 kN/m2EI

= 27.75 kN = 12.25 kN

120 1204945.5

= 37.11 kN = 52.89 kN

= 40 kNByLAy

45.5 kN�m

A B1 kN�m

CyLByR

49 kN�m

90 kN120 kN�m

120 kN�m

50 kN�m

40 kN 40 kN

10 kN/m2EI

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)

-12.25

37.11 40

-52.893.71 m

M(kN�m) x (m)

-45.5 -49

19.8437.75

Deflected shape

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.

3EI4 m 4 m 8 m10 mm 161630

0.50.5

12 kN�m15 kN�m

[FEM]∆∆

B ∆BC

)2(6LEI ∆

375.92

)2(6)2(622 =

∆−

∆LEI

125.28)3(62 =

∆LEI 125.28)3(6

2 =∆

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

12.90 25.85Dist. 12.92CO

-8.72 -26.20-12 23.72ΣΣΣΣ

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

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

Note : Using the slope deflection

)2(75.18208

)2(2−−−+−+= BABA

EIEIM θθ

)2(2/12375.9308

)2(3:2

)1()2( aEIM BBA −−−−+−=− θ

)1(75.18208

)2(4−−−−++= BAAB

EIEIM θθ

(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

15 kN�m

EIM BCB

•−=

−−=

125.28168

)3(2 θ

)3(125.28168

)3(4−−−−+= BBC

EIM θ

)4(125.28168

)3(2−−−−−= BCB

EIM θ

4 m 4 m 8 m

A CB2EI

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

23.725 8.725 26.205

12 kN�m8.725 kN�m 26.205 kN�m

23.725 kN�m

Deflectedshape x (m)

A CB2EI

4 m 4 m 8 m

26.2057.41 kN 12.59+11.69 = 24.28 kN 12.31 kN

V (kN)x (m)

-12.59

-12.31

+-3.9 m

M (kN�m) x (m) M (kN�m) x (m)

-8.725

-26.205-23.725

10 mm θB = 23.9/EI

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

3 m 3 m2EI 1.5EI

12 kN�m

6 kN/m

C3 m 3 m2EI 1.5EI

12 kN�m

0.01 mC

A mkN •=

)01.0)(502005.1(62

100-(100/2) = 50

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

-20.08CO-22.59Dist. -40.16

[FEM]∆ 50[FEM]load 6.75

kN08.203

08.2016.40=

+kN08.20

B C40.16 kN�m20.08 kN�m

6 kN/m

12 kN�m

40.16 kN�m

8.39 kN26.39 kN

6 kN/m

C3 m 3 m2EI 1.5EI

12 kN�m

-10.16 12-20.08 40.16ΣΜΣΜΣΜΣΜ

6 kN/m

3 m 3 m2EI 1.5EI

12 kN�m

6 kN/m

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

Deflected shape

V (kN)

26.398.39

-20.08

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

3 m 3 m2EI 1.5EI

12 kN�m

6 kN/m

3 m 3 m2EI 1.5EI

12 kN�m

6 kN/m

12 kN�m

10 mmR

� Overview

*)1(0' −−−=+ CRR

� Artificial joint applied 6 kN/m

C3 m 3 m2EI 1.5EI

12 kN�m

0.01 mC

A mkN •=

)01.0)(502005.1(62

100-(100/2) = 50

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

-20.08CO-22.59Dist. -40.16

[FEM]∆ 50[FEM]load 6.75

kN08.203

08.2016.40=

+kN08.20

B C40.16 kN�m20.08 kN�m

6 kN/m

12 kN�m

40.16 kN�m

8.39 kN26.39 kN

6 kN/m

C3 m 3 m2EI 1.5EI

12 kN�m

-40.16 12-20.08 40.16ΣΜΣΜΣΜΣΜ

40.16 kN�m40.16 kN�m

26.39 kN20.08R

039.2608.20:0 =+−−=Σ↑+ RFy

kNR 47.46=

3 m 3 m2EI 1.5EI 0.50.5

� Artificial joint removed

=∆→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↑+

� Solve equation

*)1(67.66'47.46 inkNRandkNRSubstitute ==

6970.0067.6647.46

−==+

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

12 kN�m

R = 46.47 kN

20.08 kN�m

20.08 kN10 mm

8.39 kN

R´ = 66.67 kN

∆80 kN�m

46.67 kN 20 kN

6 kN/m

3 m 3 m2EI 1.5EI

12 kN�m

35.68 kN�m

12.45 kN 5.55 kN

Deflected shape

M (kN�m)

x (m)1214.57

-35.68

V (kN)

x (m)0.925 m12.45

A CB DL´ L L´

real beam

Symmetric Beam

� Symmetric Beam and Loading

+ ΣMC´ = 0: 0)2

)(()(' =+−LL

EIMLVB

EIMLVB 2' == θ

θLEIM 2

The stiffness factor for the center span is, therefore,

LEIK 2

conjugate beam

+ ΣMC´ = 0: 0)3

)((21)(' =+−

LLEIMLVB

EIMLVB 6' == θ

θLEIM 6

The stiffness factor for the center span is, therefore,

� Symmetric Beam with Antisymmetric Loading

L´ L L´

real beam

conjugate beam

V´BB´

)((21 L

LEIK 6

Example 5a

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

C6 m 4 m

15 kN/m

C6 m 4 m

15 kN/m

EILEIK AB ==

LEIK BC ==

,1)()(

)( ==AB

ABAB K

KDF ,692.0

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

308.0)6/2()4/3(

)6/2()()()(

[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

3 m 3 mDC

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

A B4 m

30 kN83

3 m 3 mDC

30 kN8336.07 kN�m 36.07 kN�m

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

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

Deflectedshape

V(kN) x (m)

-29.02

Example 5b

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

15 kN/m

4 m 3 m 4 m3 m

15 kN/mFixed End Moment

wL2/15 = 16 11wL2/192 = 30.938

5wL2/192 = 14.063

15 kN/m

wL2/15 = 165wL2/192 = 14.063

11wL2/192 = 30.938

15 kN/m

15 kN/m16.87516

16.875 16

A B4 m

30 kN83

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 ===

,1)( =ABDF ,429.0175.0

75.0)( =+

=BADF 571.0175.0

1)( =+

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

A B4 m

30 kN83

16.375 kN�m 16.375 kN�m

24.09 kN5.91 kN 27.96 kN27.96 kN

24.09 kN5.91 kN

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

Moment Distribution Frames: No Sidesway

48 kN 8 kN/m40 kN�m

5 m5 m

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.

40 kN�m

48 kN 8 kN/m

5 m5 m

2.5EI 2.5EI45 25

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

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

0.50.5

0.25 0.25DF 0.51 0 1

A B D C

BA BCMember BDAB DB CB

5 5Dist. 10

-10 -10Joint load -20

-10CO-45FEM 25

20 00 -5ΣΣΣΣ -50 -10

40 kN�m

2.5EI 2.5EI2010

16 kN24 kN

50 kN�m

48 kNA B 0

14 kN 34 kN

40 kNB C 3.75

40 kN�m

50243458 kN

14 kN 16 kN

Member BD DB CBBC20 0

AB0 -5ΣΣΣΣ

BA-50 -10

3.75 kN

3.75 kN10 kN�m

Moment diagram Deflected shape

5 m5 m

2.5EI 2.5EI

14 kN 16 kN

Moment Distribution for Frames: Sidesway

Single Frames

P∆∆

Artificial joint applied(no sidesway)

Artificial joint removed(sidesway)

0 = R + C1R´

0 =R2 + C1R2´ + C2R2´´

0 =R1 + C1R1´ + C2R2´´

R1´x C1

∆´∆´R2´

R1´x C1

R2´´

R1´´

R2´´

R1´´

∆´´ ∆´´

Multistory Frames

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.

artificial joint removed( sidesway)

artificial joint applied(no sidesway)

1 m 4 m

� Overview

R + C1R´ = 0 ---------(1)

� Artificial joint applied (no sidesway)

Ra = 4 mb = 1 m

Pb2a/L2 = 2.56

Pa2b/L2 = 10.24

Fixed end moment:

Equilibrium condition :

ΣFx = 0:+ Ax + Dx + R = 0

1 m 4 m

2.5610.24

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

1.32-2.88 2.72-5.78

ΣFx = 0:+ 1.73 - 0.81 + R = 0

R = - 0.92 kN

� Artificial joint removed ( sidesway)

Fixed end moment:

∆ ∆

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

∆ ∆

100 kN�m

100 kN�m0.5

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

60 kN�m

80 kN�m

100 kN�m

ΣFx = 0:+

-28 - 28 + R´ = 0

R´ = 56 kN

Equilibrium condition:

-0.92 + C1(56) = 0

R + C1R´ = 0

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

C1 =0.9256

1.73 kN 0.81

5.785.78

1 m 4 m

3.714.79

1.27 kN1.27

Bending moment diagram (kN�m)

1 m 4 m

3.714.79

1.27 kN 1.27 kN

2.99 kN13.01 kN∆ ∆

Deflected shape

4.794.79

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.

20 kN/mpin

3 m , 2EI

3m, 3EI

4m , 4EI

20 kN/m

� Overview

R + C1R´ = 0 ----------(1)

artificial joint removed(sidesway)

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

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

Dx = 0

ΣFx = 0: +

60 - 33.53 - 0 + R = 0

R = - 26.47 kN

� Artificial joint removed ( sidesway)

� Fixed end moment

3 m, 2EI

3m, 3EI

4m, 4EI

6(2EI∆)/(3) 2

3(4EI∆)/(4) 2

∆ ∆

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

)2(62 =

∆AB = 75/EI

100 kN�m

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

0.471 0.529DF 1.000 1.00 0

CO -28.55

0.50.5

A B C D

Ax = 43.12kN

ΣFx = 0:+

-43.12 - 14.06 + R´ = 0

R´ = 57.18 kN

FEM 100 56.25 100Dist. -52.9-47.1 0

14.06 kN

∆ ∆

-52.976.45ΣΣΣΣ 52.9 56.25

56.25 kN�m

52.9. kN�m

76.45 kN�m

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

52.9 kN�m

76.45 kN�m

43.12 kN

33.53 kN

7.94 kN�m7.94 kN�m

18.53 kN�m

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

5.52 kN

DMoment diagram

Deflected shape

∆ ∆

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

B16.55

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.

artificial joint removed(sidesway)

R + C1R´ = 0 ---------(1)

� Overview

10 kNR

ΣFx = 0:+

10 + R = 0

R = - 10 kN

� Artificial joint removed (sidesway)

� Fixed end moment

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 = ∆

36.87°

C´∆ CD

36.87°

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

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

0.50 0.50DF 0.6250 0.375 1

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

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

43.02 kN

34.38 kN34.38 kN

34.38 kN

39.91 kNΣFx = 0:+

-43.02 - 39.91 + R´ = 0

R´ = 82.93 kN

81.0534.38 kN

34.38 kN

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

5.19 kN

4.15 kN

9.77 6.81

4.15 kN

4.81 kN=

x C1= 10/82.93+

A 43.02 kN

34.38 kN

81.05 56.48

34.38 kN

39.91 kN

C1 = 10/82.93

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

R + C1R´ = 0 ---------(1)

Bending moment diagram(kN�m)

5.19 kN

4.15 kN

9.77 6.81

4.81 kN4.15 kN

Deflected shape

4 m3 m

2m2 m 3 m

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.

4 m3 m

2m2 m 3 m

� Overview

R + C1R´ = 0 ----------(1)

2m2 m 3 m

artificial joint removed(no sidesway)

2m2 m 3 m

20 kN R15+(15/2)= 22.5 kN�m

PL/8 = 15

ΣFx = 0:+ Ax + Dx + R = 0

Fixed end moments:

2m2 m 3 m

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

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

40 kN13.5

15.5 kN

� 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

4EI 4EI

m 4.472 m

100 kN�m

33.69o

C´∆ CD

∆AB = 54.3/EI

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

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

∆CD = ∆/cos 26.57°= 50.4/EI

4 m3 m

2m2 m 3 m

23.8523.8568.34 kN

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

23.85 kN

30.24 kN�m

23.85 kN

B C71.55

-35.33 + C1(87.83) = 0

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

C1 = 35.33/87.83B C

20 kN35.33 kN

6.75 kN�m23.08 kN

24.5 kN

13.5 kN�m

7.75 kN

15.5 kN

90.59 kNB C

71.55 kN�m

85.78 kN�m

68.34 kN23.85 kN

19.485 kN

23.85 kN

30.24 kN�m

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

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

Deflected shape

12.1627.76

! 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-Dimension Trusses

Fundamentals of the Stiffness Method

� Node and Member Identification

� Global and Member Coordinates

� Degrees of Freedom

(x1, y1)

(x3, y3)

(x2, y2)

(x4, y4)

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

AE/LAE/L

x djAE/L x d´j

x diAE/LAE/L x d´i

Member Local Stiffness Matrix

x´y´

jii dL

AEq ''' −=

AE/LAE/Ld´ i = 1

d´ j = 1

jij dL

AEq ''' +−=

[q´] = [k´][d´] ----------(1)

x´y´

(xi,yi)

(xj,yj)x

Displacement and Force Transformation Matrices

22 )()(cos

ijijxx

−+−

−== θλ

22 )()(cos

ijijyy

−+−

−== θλ

Global

djyx´y´

� Displacement Transformation Matrices

yiyxixi ddd θθ coscos' +=

λλλλ00

yjyxjxj ddd θθ coscos' +=

yxTλλ

λλ00

λx λy

[d´] = [T][d] ----------(2)

Global

x´y´

� Force Transformation Matrices

xiix qq θcos'=

yiiy qq θcos'=

xjjx qq θcos'=

yjjy qq θcos'=

[q] = [T]T[q´] ----------(3)

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

00λyλx

λyλx00-11

AEL[ k ] =

[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

λλλλ00

λλλλ

Member Forces

x´y´

−−

−−=

λλλλλλλλ

q´j = AEL −λx −λy λx λy qj´

x´y´

Member Forces

qm = AEL −λx −λy λx λy qj´F+

x´y´

Member Forces

4 m 4 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.

Lixx ijij

�)(�)(� −+

−=λ

cosθx = λx cosθy = λy

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

(-4,3)

(4,-3)

4 m 4 m

5 m5 m

Member

λx λy

λyλx

λxλx

λyλy

λxλy

λyλx

λxλx

λyλy

λxλy

−λyλx

−λxλx

−λyλy

−λxλy

−λyλx

−λxλx

−λyλy

−λxλy

Vi Uj VjUi

[ k ]m = AEL

-4/5 = -0.8

4/5 = 0.8

-3/5 = -0.6

3/5 = 0.6

-3/5 = -0.6

1.08-0.48

1.92-0.48

-0.481

[K] = AE5

0.48-0.48

-0.48[ k ]1 =

2 3 41

[ k ]2 =

2 5 61

-0.480.48

[ k ]3 =

2 7 81

-0.480.48

Member

λx2 λx λy λy

0.64 0.48 0.36

0.64 -0.48 0.36

Global

Q1 = -50

Q2 = -80

-250.65/AE

-481.77/AE

4 m 4 m

5 m5 m

= -97.9 kN (C)

[q´F]1 = AE5 0.8 0.6 -0.8 -0.6 D2=

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)

50 kN 36.87o

97.9 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=

D8=D7=

-481.77/AE-250.65/AE

0.00.0

Member

#1#2#3

λx λy

[ ] [ ]F

xxyxmF q

LAEq ']'[ +

−−= λλλλ

50 kN 36.89o

97.9 kN

17.7 kN

Member

#1#2#3

-0.8-0.80.8

-0.60.6-0.6

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 :

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.

3 m+20o C

∆B = 2.5 mm

λyλx

λxλx

λyλy

λxλy

λyλx

λxλx

λyλy

λxλy

−λyλx

−λxλx

−λyλy

−λxλy

−λyλx

−λxλx

−λyλy

−λxλy

∆B = 2.5 mm

3 m+20o C

Vi Uj VjUi

[ k ]m = AEL

(-4,-3)

(-4,0)

(0,-3)

Member

λx λy

-4/4 = -1

-4/5 = -0.8

-3/5 = -0.6

-3/3 = -1

Lixx ijij

�)(�)(� −+

−=λ

0[k]1 = 8x103

2 3 41

[k]3 = 8x103

2 7 81

-0.333

Member

λx2/L λx λy/L λy

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

[k]2 = 8x103

2 5 61

(-4,-3)

(-4,0)

(0,-3)

[K] = 8x103

∆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

(-4,-3)

(-4,0)

(0,-3)

∆B = 2.5 mm

+20o C

-1.536

1.536-1.152

-1.536

-1.152+1

-2.5x10-3

6-0.096

-0.128

-0.072

-0.096+ 8x103

= 8x103

0.096 d1

d5 = 0d2

d6 = -2.5x10-3

= 8x103

2 5 61

-0.096

-0.128

-0.072

-0.096-0.096

-0.128

-0.072

-0.096

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

-1.536

-1.152+1

2= 8x103

0.096 d1

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

Global:

3 m+20o C

∆B = 2.5 mm

(-4,-3)

(-4,0)

(0,-3)

Q1 = -4

Q2 = -8= 8x103

0.096 D1

-1.536

-1.152+1.92

-0.8514x10-3 m

-2.356x10-3 m=

Member

λx λy

Local12

[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

2+20oC

1.92 kN

- 0.8 -0.6

[ ] [ ]F

xxyxmF q

LAEq ']'[ +

−−= λλλλ

6.28 kN

1.70 kN

2.87 kN

Member

cosθx

cosθy

[q´]m

1.70 kN

6.28 kN

2.87(0.8) = 2.30 kN

2.87(0.6) = 1.72 kN

∆ AD = + 3 mm

4 m 4 m∆ = - 4 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.

∆ AD = + 3 mm

4 m 4 m

∆ = - 4 mm

(-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

Member

λx λy

-4/5 =-0.8

4/5 = 0.8

4/4 = 1

-3/5 = -0.6

-3/3 = -1

-3/5 = -0.6

Lixx ijij

�)(�)(� −+

−=λ

[k]1 = 8x103

2 3 41

-0.096

-0.128

-0.072

-0.096-0.096

-0.128

-0.072

-0.096

[k]2 = 8x103

2 5 61

-0.333

[k]3 = 8x103

2 7 81

-0.096

-0.0960.096

-0.128

-0.072

-0.128

-0.072

0.096-0.096

-0.096

Member

λx2/L λy

2/Lλxλy/L

0.128 0.0720.096

0 0.3330

0.128 0.072-0.096

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

3 m 5 m

4 m 4 m

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

3 m 5 m

4 m 4 m

Member

λyλx

λx2/L λy

2/Lλxλy/L

0[k]5= 8x103

6 7 85

0.25 00

Global Stiffness Matrix

[K] = 8x103

2 5 71

[k]4= 8x103

4 5 63

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

[k]1 = 8x103

2 3 41

-0.096

-0.128

-0.072

-0.096-0.096

-0.128

-0.072

-0.096

[k]2 = 8x103

2 5 61

-0.333

[k]3 = 8x103

2 7 81

-0.096

-0.0960.096

-0.128

-0.072

-0.128

-0.072

0.096-0.096

-0.096

-0.250

0[k]4= 8x103

4 5 63

-0.250

0[k]5= 8x103

6 7 85

∆AE/L = 10.67 kN

∆ AD = + 3 mm1

3.84 kN

2.88 kN

3.84 kN

2.88 kN

∆ AD = + 3 mm

4 m 4 m

∆ = - 4 mm

Global Fixed end forces

0.00.0

1∆AE/L = 4.8 kN

4.8 kN

∆AE/L = 10.67 kN

10.67 kN∆ = -4 m

2 Fixed End3.84 kN

2.88 kN-3.84-2.88 + 10.67 = 7.79

∆ AD = + 3 mm

4 m 4 m

∆ = -4 mm

Global:

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

Q1 = 4

Q5 = 0Q2 = -8

Q7 = 0

= 8x103

2 5 71

-0.250.500.00.0

0.378-0.250.096

-0.128

-0.1280.00.0

0.0960.0

0.4770.0 D1

0.07.79

6.4426x10-3 m

2.6144x10-3 m-5.1902x10-3 m

5.2288x10-3 m

6.4426x10-3 m

2.6144x10-3 m-5.1902x10-3 m

5.2288x10-3 m

-4.8+D2

= -1.54 kN (C)

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

Member

λiyλix

-0.8 -0.6

4.8 kN

4.8 kN∆ AD

= + 3 mm

10.67 kN

[ ] [ ]F

xxyxmF q

LAEq ']'[ +

−−= λλλλ

6.4426x10-3 m

2.6144x10-3 m-5.1902x10-3 m

5.2288x10-3 m

[q´F]4 = 8x103 -1.0 0.0 1.0 0.04

= 5.23 kN (T)

[q´F]5 = 8x103 -1.0 0.0 1.0 0.04

= 5.23 kN (T)

[q´F]3 = 8x103 -0.8 0.6 0.8 -0.65

= -6.54 kN (C)

Member λx λy

0.8 -0.61 0

[ ] [ ]F

xxyxmF q

LAEq ']'[ +

−−= λλλλ

Member

λy [q´]

4 kN8 kN

3.17 kN

6.54 kN

1.54 kN

5.23 kN5.23 kN

4 kN8 kN

0.92 kN

3.17 kN 3.92 kN

-6.545.23

Special Trusses (Inclined roller supports)

λix = cos θi

λiy = sin θi

λjx = cos θj

λjy = sin θj

[ q* ] = [ T ]T[ q´ ] 1

q´iq´j

00λiyλix

λjyλjx00 [ T ] = [[ T ]T]T =

Transformation Matrices

[ 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

00λiyλix

λjyλjx00-11

AEL[ k ]m =

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.

4 m45o

Member 1:

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

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

[q*] = [T*]T[q´] + [T*]T[q´F]

q´iq´j

-0.7070.707

q´i q´j

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

0-0.707

00.707

-11[k*]1 =

-0.7070.707

Member 1:

[k*]1 = AE 63*4*

-0.1250.125

0-0.1768

0.125-0.125

00.1768

0.1768-0.1768

q´i q´j

4 m45o

AETk T

Member 2:q´i

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

90o+45o

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

0-0.707

-11[k*]2 =

-0.707-0.707

[k*]2 = AE

0.16670.1667-0.2357

-0.2357-0.23570.3333

4 m45o

AETk T

36.87o

Member 3:

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

-11[k]3 =

0.60.8

0.60.800

j36.87o

[k]3 = AE

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

4 m45o

[k*]1 = AE 63*4*

-0.1250.125

0-0.1768

0.125-0.125

00.1768

0.1768-0.1768

Global Stiffness:

[k*]2 = AE

0.16670.1667-0.2357

-0.2357-0.23570.3333

[k]3 = AE

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

00.0960.128

-0.23570.40530.096

0.2917-0.2357

Global :

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

Q1 = 30

Q3*= 0

Q2 = 0D1

D2 = AE 23*

00.0960.128

-0.23570.40530.096

0.2917-0.2357

D2 =352.5

-127.3-157.5AE

4 m45o

[q´F]1 = AE 00

-1 0 0.707 -0.7074

= -22.50 kN, (C)

[q´F]2 = AE 0 1 -0.707 -0.7073

= -22.50 kN, (C)

[q´F]3 = AE -0.8 -0.6 0.8 0.65

= 37.50 kN, (T)

D2 =352.5

-127.3-157.5AE

Member Forces :

Member

λiyλix λjx λjy

1 0 0.707 -0.707

0 -1 -0.707 -0.707

0.8 0.6 0.8 0.6

4 m45o

[ ] [ ]F

xxyxmF q

LAEq ']'[ +

−−= λλλλ

36.87o 45o45o

4 m45o

Member

Member Force (kN)

[q´]2[q´]1 [q´]3

-22.50 -22.50 37.50

22.50 kN

37.50 kN

22.50 kN

7.50 kN31.82 kN

Reactions :

4 m45o

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.

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

θij = -45o

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

[q*] = [T*]T[q´] + [T*]T[q´F]

q´iq´j

-0.7070.707

q´i q´j

4 m 4 m

6 3*4*

78Member 1:

4 m 4 m

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

0-0.707

00.707

-11[k*]1 =

-0.7070.707

[k*]1 = AE 63*4*

-0.1250.125

0-0.1768

0.125-0.125

00.1768

0.1768-0.1768

q´i q´j

6 3*4*

78Member 1:

AETk T

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

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

0-0.707

-11[k*]2 =

-0.707-0.707

[k*]2 = AE

0.16670.1667-0.2357

-0.2357-0.23570.3333

4 m 4 m

6 3*4*

78Member 2:

AETk T

90o+45o

36.87o

Member 3:

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

-11[k]3 =

0.60.8

0.60.800

j36.87o

[k]3 = AE

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

4 m 4 m

6 3*4*

AETk T

4 m 4 m

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

-11[k]1 =

q´i q´j

6 3*4*

78Member 4:

[k]4 = AE 278

0-0.25

AETk T

Member 5:

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

-0.14140

0.98990

j36.87o

4 m 4 m

6 3*4*

[k*]5 =00

-0.14140.9899

0.60.800

[k*]5 = AE

4* 83*

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

θ j = 36.87o ;

λ jx = cos (36.87o ) = 0.8,

λ jy = sin

(36.87o ) = 0.6

AETk T

550 -0.2357-0.2357

[k*]1 = AE 63*4*

-0.1250.125

0-0.1768

0.125-0.125

00.1768

0.1768-0.1768

[k*]3 = AE

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

4 m 4 m

Global Stiffness:1

6 3*4*

[k*]2 = AE

0.16670.1667-0.2357

-0.2357-0.23570.3333

[k]4 = AE 278

0-0.25

[k*]5 = AE

4* 83*

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

[K] = AE 23*

0.0960.378

0.40530.096

0.4877

Global :30 kN

4 m 4 m

6 3*4*

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

Q1 = 30

Q3*= 0

Q2 = 0D1

D2 =86.612

-13.791-28.535AE

= AE 23*

00.0960.378

-0.23570.40530.096

0.4877-0.2357

4 m 4 m

6 3*4*

D2 =86.612

-13.791-28.535AE

[q´F]1 = AE 00

-1 0 0.707 -0.7074

= -2.44 kN, (C)

[q´F]2 = AE 0 1 -0.707 -0.7073

= -6.26 kN, (C)

[q´F]3 = AE -0.8 -0.6 0.8 0.65

= 10.43 kN, (T)

Member Forces :

Member

λiyλix λjx λjy

1 0 0.707 -0.707

0 -1 -0.707 -0.707

0.8 0.6 0.8 0.6

[ ] [ ]F

xxyxmF q

LAEq ']'[ +

−−= λλλλ

Member

λiyλix λjx λjy

4 m 4 m

6 3*4*

D2 =86.612

-13.791-28.535AE

[q´F]4 = AE -1 0 1 04

= -21.65 kN, (C)

[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

xxyxmF q

LAEq ']'[ +

−−= λλλλ

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

6 3*4*

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

4 m36.87o

λjyλjx

λjxλjx

−λiyλjx

−λixλjx

λjyλjy

λjxλjy

−λiyλjy

−λixλjy

λiyλix

λixλix

−λjyλix

−λjxλix

λiyλiy

λixλiy

−λjyλiy

−λjxλiy

Vi Uj VjUi

[ k *]m = AEL

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.

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

36.87o

73.74o

36.87o

q´iq´j

0.960.28

0.60.8

[k]1 = 8x103

4* 1 23*

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

36.87o

AETk T

36.87x

yMember 2:

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

[q*] = [T*]T[q´] + [T*]T[q´F]

36.87o

λjx = cos 0o = 1,λjy = sin 0o = 0

q´iq´j

[k] = [TT] AEL

[T]1-1

q´i q´j

0.60.8

i j2 5

[k]2 = 8x103

4* 5 63*

-0.15-0.2

0-0.20.120.16

0-0.150.090.12

270o 1

Member 3:

36.87o

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

Member 4:

[k]4 = 8x103

2 7 81

-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

0.3330

-0.3330

2 5 61

10.333

-0.3330

323.13o3

yMember 5:

36.87o

λx = cos 0o = 1,λy = sin 0o = 0

0-0.25

[k]5 = 8x103

6 7 85

660.17568

-0.2-0.20.5

-0.0336-0.0448

-0.0448 -0.03360.0

0.2560.4740.0

[k]1 = 8x103

4* 1 23*

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

-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

0-0.25

[k]5 = 8x103

6 7 85

[k]2 = 8x103

4* 5 63*

-0.15-0.2

0-0.20.120.16

0-0.150.090.12

[k]3 = 8x103

0.3330

-0.3330

2 5 61

10.333

-0.3330

[K] = 8x103

2 3* 51

36.87o

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

1.988x10-3 m

1.996x10-4 m-2.0824x10-3 m

7.984x10-5 m

Q1 = 4

Q3*= 0

Q2 = -8

Q5 = 0

0.17568-0.2

-0.20.5

-0.0336-0.0448

-0.0448 -0.03360.0

0.2560.4740.0

= 8x103

2 3* 51

36.87o

Member forces

1.988x10-3 m

1.996x10-4 m-2.0824x10-3 m

7.984x10-5 m

[q´F]1 = 8x103 -0.28 -0.96 0.8 0.65

= 0.46 kN, (T)

[q´F]2 = 8x103 -0.8 -0.6 1 04

= -0.16 kN, (C)

Member

λiyλix λjx λjy

0.28 0.96 0.8 0.6

0.8 0.6 1 0

[ ] [ ]F

xxyxmF q

LAEq ']'[ +

−−= λλλλ

1.988x10-3 m

1.996x10-4 m-2.0824x10-3 m

7.984x10-5 m

[q´F]3 = 8x103 D2

0 1 0 -13

= -5.55 kN

[q´F]4 = 8x103 -0.8 0.6 0.8 - 0.65

= -4.54 kN

[q´F]5 = 8x103 -1 0 1 04

= -0.16 kN

Member

λiyλix λjx λjy

0 -1 0 -1

0.8 -0.6 0.8 -0.6

1 0 1 0

[ ] [ ]F

xxyxmF q

LAEq ']'[ +

−−= λλλλ

36.87o

36.87o 36.87o

36.87o

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

Space-Truss Analysis

[ ] [ ]F

dddddd

EA '1111

λλλλλλ000

[q´] = [k´][d´] + [q´F]

= [k´][T][d] + [q´F]

Member Local Stiffness [k´]:

Member Global Stiffness [km]:

[km]= [T]T[k´] [T]

m LEAk

λλλλλλ

λλλ

000000

Global equilibrium matrix:

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

KII,II

Reaction Support Boundary Condition

Fixed End Forces

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

qjzqiy

[ ] [ ] [ ]F

zyxzyxmj q

dddddd

LEAq '' +

−−−= λλλλλλ

4 m4 m

3 m3 m

80 kN60 kN

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.

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

λx λy λz

-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

(0, 0, 10)

(-4, 3, 0)

(4, 3, 0)

(4, -3, 0)

(-4, -3, 0)4

4 m4 m

3 m3 m

80 kN60 kN

Member

λx λy λz

-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

+0.80-0.240

+0.320

+0.320-0.096

+0.128

-0.240+0.072

-0.096

[k11]1 =AEL

+0.80-0.240

-0.320

-0.320+0.096

+0.128

-0.240+0.072

+0.096

[k11]2 =AEL

+0.80+0.240

-0.320

-0.320-0.096

+0.128

+0.240+0.072

-0.096

[k11]3 =AEL

+0.80+0.240

+0.320

+0.320+0.096

+0.128

+0.240+0.072

+0.096

[k11]4 =AEL

3.20.0

0.00.0

0.00.288

[KI,I] =AEL

(0, 0, 10)

(-4, 3, 0)

(4, 3, 0)

(4, -3, 0)

(-4, -3, 0)4

4 m4 m

3 m3 m

80 kN60 kN

0.0-80

0.00.0

3.20.0

0.00.0

0.00.288

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

Global equilibrium matrix:

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

KII,II

Reaction Support Boundary Condition

Fixed End Forces

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

0.0-80

0.00.0

3.20.0

0.00.0

0.00.288

0.0 mm-15.53 mm

6.551 mm=

0.0-277.8

+117.2

Member forces:

Member

λx λy λz

-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 = AEL −λx −λy −λz λx λy λz

= +116.5 kN (T)

+0.3578 -0.2683 +0.8944[q´j]1 =AEL

-277.8L

4 m4 m

3 m3 m

80 kN60 kN

Member

λx λy λz

-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

-277.8L

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

-277.8L

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

-277.8L

4 m4 m

3 m3 m

80 kN60 kN

32.6 kN116.5 kN

116.5 kN

32.6 kN

Member

λx λy λz

-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

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

! Development: The Slope-DeflectionEquations

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

BEAM ANALYSIS USING THE STIFFNESSMETHOD

Slope � Deflection Equations

settlement = ∆j

Pi j kw Cj

Mij MjiwP

1 DOF: θΑ

θΒΒΒΒ

A BC 2 DOF: θΑ , θΒΒΒΒ

� Stiffness Definition

kBAkAA

LEIkAA

LEIkBA

kBBkAB

LEIkBB

LEIkAB

� Fixed-End ForcesFixed-End Forces: Loads

L/2 L/2

� General Case

settlement = ∆j

Pi j kw Cj

Mij MjiwP

settlement = ∆j

(MFij)∆ (MF

ji)∆

(MFij)Load (MF

ji)Load

Mij Mji

i jwPMij

settlement = ∆jθj

=+ ji LEI

LEI θθ 24

ji LEI

LEI θθ 42

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

jiij MMLEI

LEIM +++= ∆θθ Loadji

Fjiji MM

LEIM )()()4()2( +++= ∆θθ

Mji Mjk

Pi j kw Cj

Mji Mjk

� Equilibrium Equations

0:0 =+−−=Σ+ jjkjij CMMM

i jMij Mji

LEIkii

LEIk ji

LEIkij

LEIk jj

� Stiffness Coefficients

Stiffness Matrix

)()2()4( ijF

jiij MLEI

LEIM ++= θθ

)()4()2( jiF

jiji MLEI

LEIM ++= θθ

LEILEILEILEI

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

� Matrix Formulation

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

Displacementmatrix

Stiffness matrix

Force matrixwP(MF

ij)Load (MFji)Load

i jwPMij

ψ ∆j

Mij Mji

(MFij)∆ (MF

ji)∆

Fixed-end momentmatrix

][]][[][ FEMKM += θ

]][[])[]([ θKFEMM =−

][][][][ 1 FEMMK −= −θ

Real beam

Conjugate beam

� Stiffness Coefficients Derivation

LMM ji +

EILM j

−−−=

=+−=Σ+

)2(0)2

(:0 −−−=+−=Σ↑+EI

LMF jiiy θ ij

andFrom

);2()1(

LMM ji +

� Derivation of Fixed-End Moment

Real beam

2 PLMEI

MLFy ==+−−=Σ↑+

Conjugate beamA

EIPL4 EI

Point load

841616PLPLPLPL

-PL/8 -PL/8

-PL/16-PL/8

-PL/16-PL/8-

Uniform load

Real beam Conjugate beam

23 wLMEI

MLFy ==+−−=Σ↑+

Settlements

MjMi = Mj

LMM ji +L

MM ji +

Real beam

6LEIM ∆

Conjugate beam

(:0 =+−∆−=Σ+L

EIMLM B

� Typical Problem

8024 11

LEIM BAAB +++= θθ

8042 11

128024 2

wLLPLEI

128042 2

wLLPLEI

LEIM CBCB −

−+++= θθ

MBA MBC

8042 11

128024 2

wLLPLEI

BBCBABB forSolveMMCM θ→=−−=Σ+ 0:0

Substitute θB in MAB, MBA, MBC, MCB

MABMBA

8024 11

LEIM BAAB +++= θθ

8042 11

128024 2

wLLPLEI

128042 2

wLLPLEI

LEIM CBCB −

−+++= θθ

ByR CyAy ByL

MABMBA

By = ByL + ByR

MBCMCB

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

E, I, A, L

Beam-Member Stiffness Matrix

d4 = 1

AE/LAE/L AE/L AE/L

1 2 3 4 5 6

d1 = 1

k11 = = k44

- AE/LAE/L

-AE/L AE/L

k41 k14

d2 = 1

1 2 3 4 5 6

- AE/LAE/L

-AE/L AE/L

6EI/L2

d5 = 1

6EI/L2

12EI/L312EI/L3 12EI/L3

12EI/L3

6EI/L2

12EI/L3

- 6EI/L2

- 12EI/L3

6EI/L2

-12EI/L3

- 6EI/L2

12EI/L3

k22 = = k55

= k32 = k65

E, I, A, L6

1 2 3 4 5 6

- AE/LAE/L

-AE/L AE/L

6EI/L2

12EI/L3

- 6EI/L2

- 12EI/L3

6EI/L2

-12EI/L3

- 6EI/L2

12EI/L3

6EI/L2 6EI/L2

-6EI/L2

d6 = 1

d3 = 1 2EI/L 4EI/L

6EI/L26EI/L26EI/L2

4EI/L 2EI/L

6EI/L2

k23 = = k53

= k63 = k36

= k56 k26 =

E, I, A, L6

MFi MF

FFxjFF

FFyi FF

6EI/L2

12EI/L312EI/L3

x ∆i

12EI/L3

6EI/L2

12EI/L3

4EI/L 2EI/L

6EI/L26EI/L2

6EI/L2

4EI/L 2EI/L

6EI/L2

x δjAE/L

AE/L AE/L

6EI/L2

12EI/L312EI/L3

x ∆j

6EI/L2

12EI/L3

6EI/L26EI/L2

6EI/L2

� Member Equilibrium Equations

i jFxj

Fyi Fyj

MiE, I, A, L

x δiAE/L AE/L

AE/LAE/L

−−−−

−−−

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/

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()/(

−=−−−=

−=−=−+=

+++−++=

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

Fixed-end force matrix

End-force matrix

Stiffness matrix

Displacement matrix

6x6 Stiffness Matrix

−−−−

−−−

LAELAE

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

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

00/00/

θi ∆j θj∆i δjδi

−−−−

−−

LEILEILEILEILEILEILEILEILEILEILEILEILEILEILEILEI

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

θi ∆j θj∆i

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

θi θj

=× LEILEI

LEILEIk

/4/2/2/4

21 312

General Procedures: Application of the Stiffness Method for Beam Analysis

wP P2y

Global

Member

3121 2

Global

wP P2y

(q´F4 )2

(q´F3 )2

(q´F1)2

(q´F2)2

(q´F4)1

(q´F3)1

(q´F2)1

(q´F1 )1

1[FEF]

3121 2

Global

wP P2y

[q] = [T]T[q´]

3121 2

Global

[k] = [T]T[q´] [T]

1000010000100001

1´ 4´2´ 3´12341

[q] = [T]T[q´]

3121 2

Global

[k] = [T]T[q´] [T]

1000010000100001

1´ 4´2´ 3´34562

Stiffness Matrix:

21 312

[k]1 =

1 2 3 4

[k]2 =

3 4 5 6

Member 1

1 2 3 4 5 6

Member 2

Node 2

QQQQQQ

DDDDDD

QQQQQQ

2 234156

5 612 3 4

21 312

ReactionQu

Joint LoadQk

Du=Dunknown

Dk = DknownQF

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

[ ] [ ] [ ] [ ])(1 FAkAAu QQKD −+= −

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

DDDDDD

Global:

21 312

(q´F6 )2

(q´F5 )2

(qF4)2

(qF3)2

(qF4)1

(qF3)1

(qF2)1

(qF1 )1

1[FEF]

Node 2

Member 1

123456

1 2 3 4 5 6

Member 2

Node 2=

QQQQQQ

qqQqqQ

)()()()(

1 2 312

Node 2

Member 1:

21 312

DdDdDd

1 2 3 4

21 312

Member 2:

1 2 3 4

9 m 3 m

10 kN1 kN/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.

21Global

1Members

9 m 3 m

10 kN1 kN/m

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

1.5 m1.5 m

1 kN/m

9 m 1[FEF]

9 m 3 m

[k]1= EI4/9

2[k]2 = EI

[K] = EI

(4/9)+(4/3)

Stiffness Matrix: θi θj

=× LEILEI

LEILEIk

/4/2/2/4

10 kN1 kN/m

Equilibrium equations: MCB = 0MBA + MBC = 0

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

=2.423/EI

0.779/EI

9 m 1.5 m1.5 m6.756.75

+-3.75

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

MCB = 01

2/3 = EI

(4/9)+(4/3)

3.753.756.75

= -6.40

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

= 0.779/EIθB

+ -6.75

Substitute θB and θC in the member matrix,

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

1 kN/m

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

Substitute θB and θC in the member matrix,

2 = EI

= 2.423/EI

= 0.779/EI

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

1.5 m1.5 m

7.13 kN 2.87 kN

6.40 kN�m

PL/8 = 3.75PL/8 = 3.75

+-3.75

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

M (kN�m) x (m)

-6.92 -6.40

3.48 4.32

V (kN) x (m)

-4.44 -2.874.56 m

θB = +0.779/EIθC = +2.423/EI

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.

0.5625

-0.375

[K] = EI

Use 4x4 stiffness matrix,

0.375EI

-0.75EI2EI

0.75EI - 0.375EI

0.375EI

0.75EI

-0.75EI

0.75EI

-0.375EI

-0.75EI

1 2 3 4

0.375EI

-0.1875EI

-0.375EI

0.1875EI

-0.375EIEI

0.375EI - 0.1875EI

0.1875EI

0.375EI

-0.375EI

3 4 5 6

2EI EI

θi ∆j θj

−−−−

−−

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

20 kN9kN/m 40 kN�m

4 m 4 m

12 kN�m18 kN

12 kN�m

18 kN 18 kN

Global:

9.697/EI

-61.09/EI

Q4 = 40

Q3 = -203

4 + -12

40 kN�m

12 kN�m

2EI EI

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

0.5625

-0.375

-0.375= EI

Member 1:

9 kN/m

A B12 kN�m

12 kN�m

18 kN 18 kN

9 kN/m

53.21 kN�m

48.18 kN67.51 kN�m

12.18 kN

2EI 12 kN�m 12 kN�m

18 kN 18 kN

-12.18

12(2EI)/43

-0.75EI2EI

0.75EI - 0.375EI

0.375EI

0.75EI

-0.75EI

0.75EI

-0.375EI

-0.75EI

1 2 3 4

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

d3 = -61.09/EI

d4 = 9.697/EI

d2 = 0

d1 = 0

Member 2:

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

0.375EI

-0.1875EI

-0.375EI

0.1875EI

-0.375EIEI

0.375EI - 0.1875EI

0.1875EI

0.375EI

-0.375EI

3 4 5 6

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

V (kN) x (m)

-7.818

-7.818-

M (kN�m) x (m)

-18.08-

-67.51

9 kN/m

53.21 kN�m

48.18 kN67.51 kN�m

12.18 kN2

18.06 kN�m

7.82 kN

13.21 kN�m

7.82 kN

12.18+

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

A CBEI2EI

4 m 2 m

9kN/m 40 kN�m10 kN

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.

Global1 2

A CB EI2EI

4 m 2 m 2 m

10 kN5 kN�m

5 kN�m

212 kN�m

12 kN�m

9 kN/m

1[FEF]

θi ∆j θj

−−−−

−−

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

1 2 3 4 m 2 m 2 m

0.50.375 0.5 1

[K] = EI

0.5625

-0.375

12(2EI)/43

[k]1 =-0.75EI2EI

0.75EI - 0.375EI

0.375EI

0.75EI

-0.75EI

0.75EI

-0.375EI

-0.75EI

1 2 3 4

[k]2 =0.375EI

0.375EI

-0.1875EI

-0.375EI

0.1875EI

-0.375EIEI

0.375EI - 0.1875EI

0.1875EI

0.375EI

-0.375EI

3 4 5 6

MBA+MBC = 40

VBL+VBR = -20

MCB = 0θB

-12 + 5 = -7-5

= -7.667/EI

-116.593/EI

52.556/EI

0.5625

-0.375

0.50.375 0.5 1

1 2 3 10 kN5 kN�m

5 kN�m

212 kN�m

12 kN�m

9 kN/m

1[FEF]

40 kN�m

5 kN�m5 kN�m

12 kN�m

+18 + 5 = 23

0.375EI

= -0.75EI2EI

0.75EI - 0.375EI

0.375EI

0.75EI

-0.75EI

0.75EI

-0.375EI

-0.75EI

1 2 3 4

=91.78

-19.97

91.78 kN�m

55.97 kN

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

18=-116.593/EI

=-7.667/EI

12 kN�m

9 kN/m

19.97 kN

60.11 kN�m4 m

12 kN�m

18 kN18 kN

=-20.11

-0.0278

10.03=52.556/EI

=-116.593/EI

=-7.667/EIθB

10 kN5 kN�m

5 kN�m

10.03 kN

= 0.375EI

0.375EI

-0.1875EI

-0.375EI

0.1875EI

-0.375EIEI

0.375EI - 0.1875EI

0.1875EI

0.375EI

-0.375EI

3 4 5 6

0.0278 kN

20.11 kN�m

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

45 kN�m

5 kN�m

4 m 2 m

10.03 kN

91.78 kN�m

55.97 kN

V (kN) x (m)

-0.0278

19.97+

-10.03 -10.03-

91.78 kN�m

55.97 kN 19.97 kN60.11 kN�m4 m

110.03 kN0.0278 kN

20.11 kN�m2 m

10 kN2 m

91.78 kN�m

60.11 kN�m

20.11 kN�m

M (kN�m) x (m)

-91.78

+20.11

θB = -7.667/EI

θC = 52.556/EI

∆B = -116.593/EI

8 m 4 m 4 m

2EI EI

6 kN/m

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.

[k]1 =4

2EI EI

3[K] = EI

3[k]2 = EI

Use 2x2 stiffness matrix:θi θj

=× LEILEI

LEILEIk

/4/2/2/4

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

=-61.27/EI rad

185.64/EI rad

2EI EI

6 kN/m

∆B = -10 mm 1 2

[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

6(2EI)∆/L2 = 75 kN�m

[FEM]load

32 kN�mwL2/12 = 32 kN�m

6 kN/m

75 kN�m

6(2EI)∆/L2 = 75 kN�m10 mm

1[FEM]∆∆∆∆

6 kN/m

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

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

40 kN�mPL/8 = 40 kN�m 2

37.5 kN�m

6(EI)D/L2 = 37.5 kN�m10 mm

[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

+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

Alternate method: Use 4x4 stiffness matrix

2EI EI

-0.375

12(2)/82

-0.375

1 2 3 41

-0.1875

0.1875

3 4 5 63

2-0.75

-0.1875-0.75

0.1875-0.75

-0.754

-0.18750.75

-0.3751.5

-0.375-1.5

-0.7512

0.5625-0.758

EI=[K]

123456

1 4 5 62 3

8 m 4 m 4 m

2EI EI

6 kN/m

∆B = -10 mm 1 2

40 kN40 kN�m

40 kN�m

232 kN�m

32 kN�m

6 kN/m

1[FEF]

-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

Q4 = 0

-0.75-0.75 D5 = -0.01)

8200200( ×

Q4 = 0

-0.375

12(2)/82

-0.375

1 2 3 41

= (200x200)8

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∆B = -10 mm 32 kN�m

32 kN�m

6 kN/m

1[FEF]load

d2 = 0

d1 = 0

d4 = -1.532x10-3

d3 = -0.01

6 kN/m

31.26 kN

76.37 kN�m

16.74 kN

18.27 kN�m

Member 2:

d4 = -1.532x10-3

d3 = -0.01

d6 = 4.641x10-3

d5 = 0

-0.1875

0.1875

3 4 5 63

= (200x200)8

18.27 kN�m

22.28 kN

0 kN�m

17.72 kN17.72 kN

18.27 kN�m

22.28 kN

-10 mm = ∆B

40 kN40 kN�m

40 kN�m

2[FEF]

8 m 4 m 4 m

2EI EI

6 kN/m

∆B = -10 mm31.26 kN

76.37 kN�m

16.74 + 22.28 kN 17.72 kN

V (kN)

x (m)+ +--

-16.74

-17.725.21 m

M (kN�m) x (m)

-76.37

-18.27

D6 = θC = 4.641x10-3 rad

d4 = θB = -1.532x10-3 rad

Deflected Curve

∆B = -10 mm

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.

8 m 4 m 4 m

2EI EI

0.6 kN/m

C∆C = -10 mm

20 kN�m

2EI EI

1 2-10 mm

0.75 2

0.7524

-0.375

12(2)/82

-0.375

1 2 3 41

-0.1875

0.1875

3 4 5 63

-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

200200( ×+

4 kN4 kN�m

4 kN�m

23.2 kN�m

2.4 kN

3.2 kN�m

2.4 kN

0.6 kN/m

1[FEF]

8 m 4 m 4 m2EI EI

0.6 kN/m

20 kN�m2EI EI

0.5625

0.75-0.75

-0.75122

D4 +9.375

37.537.5

2.4+2 = 4.4

-4.0-3.2+4 = 0.8+

-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

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

8.55 kN

43.19 kN�m

3.75 kN

6.03 kN�m

13.2 kN�m

2.4 kN

3.2 kN�m

2.4 kN

0.6 kN/m

1[FEF]load 8 m

-0.375

12(2)/82

-0.375

1 2 3 41

8200200×

Member 2:

d4 = -1.538x10-3

d3 = -9.433x10-3

d6 = 1.863x10-3

d5 = -0.01

-6.0 kN�m

3.75 kN

20.0 kN�m

0.25 kN0.25 kN

6.0 kN�m

3.75 kN

10 mm2

4 4 kN4 kN�m

4 kN�m

2[FEF]

-0.1875

0.1875

3 4 5 63

8200200×

20 kN�m

8 m 4 m 4 m2EI EI

0.6 kN/m

20 kN�m

Deflected Curve

0.6 kN/m

8.55 kN

43.19 kN�m

3.75 kN

6.03 kN�m

V (kN)x (m)

8.553.75

0.25 kN

6.0 kN�m

3.75 kN

20 kN�m

M (kN�m) x (m)

-43.19

D3 = -9.433 mm

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

D5 = -10 mm

4 m 2 m

9kN/mHinge

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

[K] = EI 1

4 m 2 m 2 m

[k]1 =EI

1[k]2 =

Use 2x2 stiffness matrix,θi θj

=× LEILEI

LEILEIk

/4/2/2/4

15 kN�mB C

15 kN�m30 kN

12 kN�m

12 kN�m 9kN/m

1[FEF]

= EI 1

0.0006 rad

-0.0015 rad

Global matrix:

30 kN9kN/m

12 kN�m 15 kN�m1

Member 1:

= 0.0006

12 kN�m

12 kN�m 9kN/m

1[FEF]A

18 kN�m22.5 kN 13.5 kN

9 kN/m

B C30 kN

15 kN�mB C

15 kN�m 30 kN

= -0.0015

-22.5=

Member 2:

22.5 kN�m

20.63 kN9.37 kN

4 m 2 m

9kN/mHinge

-20.63

V (kN) x (m)

22.59.37

M (kN�m) x (m)

θBL= 0.0006 rad θBR= -0.0015 rad

13.5 kN22.5 kN

22.5 kN�mB C

20.63 kN9.37 kN

A B18 kN�m

9 kN/m

9.37 kN13.5 kN

22.87 kN

A CBEI2EI

4 m 4 m

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.

A CB1 2

2EI EI 6

-0.75 2.0

0[K] = EI 1

0.5625

[k]1 =0.75EI

EI-0.75EI

0.375EI0.75EI

0.75EI-0.375EI

2EI0.75EI

EI-0.75EI

-0.75EI- 0.375EI

0.375EI-0.75EI

4 5 1 24

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

Q1 = -20

Q3 = 0.0

Q2 = 0.0

D2 =-0.02382 m

0.008933 rad

-0.008333 rad

12 kN�mA B

12 kN�m

18 kN18 kN1

A CB1 2

2EI EI

Global:

= EI 1

0.5625

12 kN�m

12 kN�mA B

12 kN�m

18 kN18 kN1

Member 1:

0.75EI

-0.75EI

0.375EI

0.75EI

-0.375EI

0.75EI

-0.75EI

- 0.375EI

0.375EI

-0.75EI

4 5 1 2

= -0.02382

= -0.00833

= 107.32

107.32 kN�m

8.83 kN44.83 kN

Member 2:

44.66 kN�mB C

11.16 kN11.16 kN

= -0.02382

= 0.008933

0.375EI

-0.1875EI

-0.375EI

0.1875EI

-0.375EIEI

0.375EI - 0.1875EI

0.1875EI

0.375EI

-0.375EI

1 3 6 71

-11.16

-44.66

A CB4 m 4 m

V (kN) x (m)

-11.16 -11.16

M (kN�m) x (m)

-107.32

-44.66-

107.32 kN�m

8.83 kN44.83 kN

144.66 kN�m

11.16 kN11.16 kN

θBL= -0.008333 rad θBR= 0.008933 rad

BEI2EI

4 m 2 m

9kN/mHinge

40 kN�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.

[K] = EI 1

4 m 2 m 2 m

[k]1 =EI

1[k]2 =

Use 2x2 stiffness matrix: θi θj

=× LEILEI

LEILEIk

/4/2/2/4

15 kN�mB C

15 kN�m30 kN

12 kN�m

12 kN�m 9kN/m

1[FEM]

Global matrix:

12 kN�m 15 kN�m

BEI2EI

9kN/mHinge

40 kN�m

Q1 = 40

Q2 = 0.0

0.0026 rad

-0.0015 rad

= EI 1

40 kN�m

12 kN�m

12 kN�m 9kN/m

1[FEF]A

40 kN�mA B38 kN�m

9 kN/m

1.5 kN37.5 kN

Member 1:

= 0.0026

15 kN�mB C

15 kN�m 30 kN

Member 2:

= -0.0015

-22.5=

22.5 kN�mB C

20.63 kN9.37 kN

9.37 kN

1.5 kN

4 m 2 m

9kN/mHinge

40 kN�m

-20.63

V (kN) x (m)

9.371.5+

M (kN�m) x (m)

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

20.63 kN9.37 kN

A CBEI2EI

4 m 4 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

A CB1 2

A CBEI2EI

4 m 4 m

Hinge40 kN�m

12 kN�m

9 kN/m

1[FEM]θi ∆j θj

−−−−

−−

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

1 22A CB

[K] = EI 1

0.5625

[k]1 = 0.75EI

EI-0.75EI

0.375EI0.75EI

0.75EI-0.375EI

2EI0.75EI

EI-0.75EI

-0.75EI- 0.375EI

0.375EI-0.75EI

4 5 1 24

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

2EI EI1

Q1 = -20

Q3 = 0.0

Q2 = 40

D2 =-0.01316 m

0.0049333 rad

-0.002333 rad

A CB1 2

2EI EI

40 kN�m

12 kN�m

9 kN/m

1[FEF]

= EI 1

0.5625

12 kN�m

18 kNGlobal:

0.75EI

-0.75EI

0.375EI

0.75EI

-0.375EI

0.75EI

-0.75EI

- 0.375EI

0.375EI

-0.75EI

4 5 1 2

= 87.37

-13.85

12 kN�m

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

= -0.01316

= -0.002333

12 kN�m

18 kN1

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

0.375EI

-0.1875EI

-0.375EI

0.1875EI

-0.375EIEI

0.375EI - 0.1875EI

0.1875EI

0.375EI

-0.375EI

1 3 6 71

-24.69

24.69 kN�m

6.18 kN6.18 kN

= -0.01316

= 0.004933

4 m 4 m

9kN/m 40 kN�m

24.69 kN�mB C

6.18 kN6.18 kN

87.37 kN�m 13.85 kN49.85 kN

40 kN�m

M (kN�m) x (m)

-87.37

--24.69

θBL= -0.002333 rad θBR = 0.004933 rad

V (kN) x (m)-6.18

-13.85+

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

� Curvature

Temperature Effects

Tm> TR

� Thermal Fixed-End Forces (FEF)

mTTT +

Tb > Tt

Rmaxial TTT −=∆ )(

tbbending TTT −=∆ )(

)()(dTyTy

∆=∆

)(tandT

dTT tb ∆

σaxialσaxial

Tb - Tt

Room temp = TR

- Axial

σaxialσaxial

axialF

A ∫= σ

dAE axial∫= ε

dATE axial∫ ∆= )(α

axialTEA )(∆= α

AETTF RmAF

axial )()( −= α

dATE axial ∫∆= )(α

Tb > Tt

Tm> TR

Rmaxial TTT −=∆ )(

- Bending

yFA ∫= σ

dATyE y∫ ∆= )(α

dAdTyyE∫

∆= )(α

∫∆

= dAydTE 2)(α

TTF ulA

Fbending )()( −

dAyE∫= ε

tbbending TTT −=∆ )(

)()(dTyTy

∆=∆

y β Tt

Afterdeformation

Before deformation

ds = dx y

ρ)()( θρ ddx =

)( θdy )(1dxdθ

� Elastic Curve: Bending

dxdTyyd )()( ∆

= αθ

dxdTd )()( ∆

= αθ

� Bending Temperature

Tl > Tu

mTTT +

∆T = Tb - Tt

== )(1)( αρ

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.

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

4 m 4 m

T2 = 40oC

T1 = 25oC

182 mm

mkNEIdTF F

bending •=××−

×=∆

= − 78.19)502002)(182.0

2540)(1012()2)(( 6α

[M1] = 3EIθ1 + (1.978x10-3)EI0

Element 1:

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

+ 1.978

-1.978(10-3) EI= 1

Element 2:

q1 + 0

0= 0.5

θ1 = -0.659x10-3 rad

BEI2EI

4 m 4 m

19.78 kN�m19.78 kN�m182 mm

Element 2:

q1= + 0

= -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:

M2 = q1

q2 + 19.78

-19.78

4.95 kN4.95 kN

26.37 kN�m 6.59 kN�m

6.59 kN�m 3.3 kN�m

2.47 kN2.47 kN

BEI2EI

4 m 4 m

19.78 kN�m19.78 kN�m182 mm

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)

V (kN) x (m)

-4.95-2.47

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

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

4 m 4 m

T2 = 40oC

T1 = 25oC1

Element 1:

mkNmLAE /)10(2

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

mmkNLEI

•=×××

)10(20)4(

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

mmkNLEI

•=×××

)10(10)4(

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

mmkNLEI )10(5.7

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

2 =×××

EI /)10(75.3)4(

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

3 =×××

Fixed-end forces due to temperatures

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

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

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

+12 oC

19.78 kN�m

432 kN432 kN

Element 1:

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

- 2x106

10x103

20x103

-2x106

20x103

10x103

-19.78

EI, AE2EI, 2AE

4 m 4 m

T2 = 40oC

T1 = 25oC

182 mm1

Element 2:

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

- 1x106

10x103

-1x106

10x103

7 8 91 2 3

EI, AE2EI, 2AE

4 m 4 m

T2 = 40oC

T1 = 25oC

182 mm1

D2 =0.000144 m

-719.3x10-6 rad

-0.0004795 m

Q1 = 0.0

Q3 = 0.0

Q2 = 0.0

0.0+= 1

30x103

Global:

EI, AE2EI, 2AE

4 m 4 m

T2 = 40oC

T1 = 25oC

182 mm1

= 144x10-6

= -479.5x10-6

= -719.3x10-6

Element 1:

- 2x106

10x103

20x103

-2x106

20x103

10x103

-19.78

144.0 kN

-23.38 kN�m

-3.60 kN

-144.0 kN

3.60 kN

9.00 kN�m

A B144 kN

3.60 kN

23.38 kN�m

144 kN

3.60 kN

9 kN�m

= 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

144 kNB C

3.60 kN

9 kN�m

144 kN

3.60 kN

5.39 kN�m

Element 2:

- 1x106

10x103

-1x106

10x103

7 8 91 2 3

4 m 4 m

T2 = 40oC

T1 = 25oC

Isolate axial part from the system

RA RA = RC+

Compatibility equation: dC/A = 0

RA = 144 kN

RC = -144 kN

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

)4( 6 =−×++ −

AER AA

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)

M (kN�m) x (m)

23.388.98

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

4 m 4 m

T2 = 40oC

T1 = 25oC

D2 = -0.0004795 mm

Skew Roller Support

- Force Transformation- Displacement Transformation- Stiffness Matrix

x´y´

� Displacement and Force Transformation Matrices

θyθx

λx λy

4´5´

Force Transformation

Lxx ij

−=λ

Lyy ij

−=λ

yx qqq θθ coscos 5'4'4 −=

xy qqq θθ coscos 5'4'5 −=

6'6 qq =

λλλλ

1000000000000000010000000000

qqqqqq

λλλλ

qqqqqq

[ ] [ ] [ ]'qTq T=

θyθx

d4 d 4 cos θ x

d 4 cos θ yθy

d 5 cos θ x

d 5 cos θ y

Displacement Transformation

6λx λy

yx θdθdd coscos' 544 +=

xy θdθdd' coscos 545 +−=

66' dd =

λλλλ

1000000000000000010000000000

dddddd

λλλλ

dddddd

[ ] [ ][ ]dTd ='

[ ] [ ] [ ]'qTq T=

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

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

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

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

[ ] [ ] [ ]FTF qTq '=

[ ] [ ] [ ]'qTq T=

[ ] [ ][ ]dTd ='

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

θjθi 4 *

5 * 6*

λix = cos θi

λiy = sin θi

λjx = cos θj

λjy = sin θj

[q*] = [T]T[q´]

Stiffness matrix

5´ 6´

2´3´

qqqqqq

1000000000000000010000000000

λλλλ

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

[ T ]T

qqqqqq

1000000000000000010000000000

λλλλ

1 4 5 62 3123456

−−−−

−−−

LAELAE

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

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

00/00/

1 2 3 4 5 6

[ k ] = [ T ]T[ k´ ][T] =

- λiy6EIL2

λix6EIL2

λjy6EIL2

- λjx6EIL2

Ui Vi Mi

- λiy6EIL2

λix6EIL2

λjy6EIL2

λjx6EIL2

- λixλiy)(12EIL3

λiy2 +

12EIL3

λix2 )(

λix6EIL2

λiyλjx -

12EIL3

λixλjy)-(

λixλjx +

12EIL3

λiyλjy)-(

λjy6EIL2

λjx2 +

12EIL3

λjy2 )(

λjy6EIL2

λjx6EIL2

- λjxλjy)(12EIL3

- λjx6EIL2

λixλjy -

12EIL3

λiyλjx)-(

- λixλiy)(12EIL3

- λiy6EIL2

λixλjx +

12EIL3

λiy λjy)-(

λix2 +

12EIL3

λiyλjy +

12EIL3

λix λjx )-(

λiyλjy +

12EIL3

λixλjx)-(

12EIL3

λjx2 )

- ( λixλjy- λiyλjx )12EIL3

λiyλjx -

12EIL3

λixλjy)-(

- λjxλjy)(12EIL3

AEL λjy

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.

40 kN40 kN�m

40 kN�m

[ q´F ][FEF]

1Global

4 m4 m 22.02o

20sin22.02=7.520cos22.02=18.54

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

2* 3 *

� Transformation matrix

Member 1: [ q ] = [ T ]T[q´]

i j1´

Global Local

λjy = cos 67.98o = 0.3749

qqqqqq

10000009271.03749.000003749.09271.0000000100000010000001

1´ 4´ 5´ 6´2´ 3´4561*

1000000000000000010000000000

λλλλ

[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

0.0000.000

-150.00.0000.000

0.0000.93753.7500.000

-0.93753.750

0.0003.75020.000.000

-3.75010.00

i j1´

� Local stiffness matrix

−−−−

−−−

LAELAE

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

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

00/00/

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

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

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

0.0000.000

-139.0-56.250.000

0.0000.93753.7500.351

-0.8693.750

0.0003.75020.001.406

-3.75010.00

2* 3 * 1

i j1´

Global Local

Global Equilibrium:

Q1 = 0.0

Q3 = 0.0

36.37x10-6 m

0.002 rad

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

= 1031*

2* 3 *

4 m4 m 22.02 o

40 kN40 kN�m

40 kN�m

[ q´F ]20sin22.02=7.5

20cos22.02=18.54

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

40.020

-7.5418.54-40.0

q2*q3*

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

0.0000.000

-139.0-56.250.000

0.0000.93753.7500.351

-0.8693.750

0.0003.75020.001.406

-3.75010.00

36.4x10-6

02x10-3

6027.5

013.48

13.48 kN

60 kN�m

27.5 kN

5.06 kN

4 m4 m 22.02 o

2* 3 *

40 kN40 kN�m

7.518.54

40 kN�m

4 m4 m 22.02 o

5.05 kN27.51 kN

60.05 kN�m

13.47 kN

Bending moment diagram (kN�m)

Deflected shape

-60.05

0.002 rad

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.

22.02 o8 m 4 m 4 m

2EI, 2AE EI, AE

6 kN/m

8 * 9 *

Global

2´3´

5´6´

•×=

10150)8(

)/10200)(006.0(

mmkNLEI

•×=

1020)8(

)0002.0)(/10200(44

mkNLEI

•×= 310102

mmkNLEI

1075.3)8(

)0002.0)(/10200(66

mmkNLEI

/109375.0)8(

)0002.0)(/10200(1212

[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

0.0000.000

-150.00.0000.000

0.0000.93753.7500.000

-0.93753.750

0.0003.75020.000.000

-3.75010.00

8 * 9 *

Global Local : Member 1

[ q´]

[q] = [q´] Thus, [k] = [k´]

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

q8*q9*

0.92710.3749

-0.37490.9271

1´ 4´ 5´ 6´2´ 3´

8 * 9 *

λjy = cos 67.98o = 0.3749

8 * 9* [ q* ]

2´3´

5´6´

[ q´ ]

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

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

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

0.0000.000

-139.0-56.250.000

0.0000.93753.7500.351

-0.8693.750

0.0003.75020.001.406

-3.47710.00

x´7 *

8 * 9 * [ q* ]

2´3´

5´6´

[ q´ ]

8 * 9 *

[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

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

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

0.0000.000

-139.0-56.250.000

0.0000.93753.7500.351

-0.8693.750

0.0003.75020.001.406

-3.47710.00

Global:

=-509.84x10-6 rad

18.15x10-6 m

0.00225 rad

58.73x10-6 mD3

-32 + 40 = 80

-40-7.5

32 kN�m

6 kN/m

40 kN6 kN/m

= 1030

101.406

1.406-139

1.40620

1 3 7* 9*

40 kN40 kN�m

7.518.54

40 kN�m

40 kN�m32 kN�m

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

024-32

d1=18.15x10-6

0d3=-509.84 x10-6

= 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

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

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

6 kN/m

q8*q9*

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

0.0000.000

-139.0-56.250.000

0.0000.93753.7500.351

-0.8693.750

0.0003.75020.001.406

-3.75010.00

-7.518.54-40

18.15x10-6

-509.84x10-6

58.73x10-6

00.00225

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

20sin22.02=7.520cos22.02=18.54

40 kN40 kN�m

40 kN�m

[ q´F ]

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

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

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)

-21.8-

V (kN)x (m)

+20.18

-27.82

14.51cos 22.02o

=13.45 kN

-52.39-

12.14 +

-13.45

! 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

Simple Frames

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

-6EI/L2

- 6EI/L2

12EI/L3

-6EI/L2

6EI/L2

-12EI/L3

x´y´

3´ 6´2´ 4´1´ 5´

4´5´6´

1´2´3´

6EI/L24EI/L

6EI/L2

12EI/L3

d 1́ = 1

d 2́ = 1

12EI/L3

d 3́ = 1

6EI/L2

6EI/L26EI/L2AE/L

6EI/L2

6EI/L2d 6́

6EI/L2

6EI/L22EI/L

6EI/L2

12EI/L3

4EI/L4EI/L12EI/L3

12EI/L3

6EI/L2

6EI/L2AE/L

6EI/L2d 4́ = 1

d 5́ = 1AE/L

x´y´

Displacement and Force Transformation Matrices

θyθx

q4 = q4´ cos θx - q5´ cos θy

q5 = q4´ cos θy + q5´ cos θx

q6 = q6´

4´5´

Force Transformation

Lxx ij

−=λ

Lyy ij

−=λ

λλλλ

1000000000000000010000000000

qqqqqq

λλλλ

qqqqqq

[ ] [ ] [ ]'qTq T=

[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]

[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] =

- λiy6EIL2

λix6EIL2

λjy6EIL2

- λjx6EIL2

Ui Vi Mi

- λiy6EIL2

λix6EIL2

λjy6EIL2

λjx6EIL2

- λixλiy)(12EIL3

λiy2 +

12EIL3

λix2 )(

λix6EIL2

λiyλjx -

12EIL3

λixλjy)-(

λixλjx +

12EIL3

λiyλjy)-(

λjy6EIL2

λjx2 +

12EIL3

λjy2 )(

λjy6EIL2

λjx6EIL2

- λjxλjy)(12EIL3

- λjx6EIL2

λixλjy -

12EIL3

λiyλjx)-(

- λixλiy)(12EIL3

- λiy6EIL2

λixλjx +

12EIL3

λiy λjy)-(

λix2 +

12EIL3

λiyλjy +

12EIL3

λix λjx )-(

λiyλjy +

12EIL3

λixλjx)-(

12EIL3

λjx2 )

λiyλjx -

12EIL3

λixλjy)-(

- λjxλjy)(12EIL3

AEL λjy

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

C kN/m666.667(6m)

)m10)(60mkN1012(20012

3 =××

kN/m200006m

)mkN10)(200m10(600 2

kN2000(6m)

)m10)(60mkN106(2006

2 =××

mkN80006m

)m10)(60mkN104(2004

•=××

mkN40006m

)m10)(60mkN102(2002

•=××

Global :

A B14´

5´ 6´

2´ 3´

Local :

1´ 3´

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

A B14´

5´ 6´

2´ 3´

Local :

[q] = [q´]

-> [k]1 = [k´]1

Stiffness Matrix: Member 1

Global:

4 6 5 1 2 34

-20000

666.667

-666.667

-20000

-666.667

666.667

[k]1 =

Local:

1´ 3´

[q]2 = [ T ]T[ q´]2

q4´q5´

Stiffness Matrix: Member 2

123789

4´0000

5´000100

6´000001

2´100000

3´001000

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

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

Global:

[k]2 = [ T ]T[ k´ ]2[ T ]

1´ 2´ 3´ 4´ 5´ 6´1´

-20000

666.667

-666.667

-20000

-666.667

666.667

[k´]2 =

1 2 3 7 8 9

666.667 20001

-666.667

20000 0

-20000

8000 -2000

666.667 0

20000 0

0 8000

[k]2 =

[k]14 6 5 1 2 3

-20000

0666.667

-666.667

-200000

020000

0-666.667

-20000

666.667

1 2 3 7 8 9666.667 20001

666.667

-666.667

20000 0

-20000

8000 -2000-2000

666.667 0

20000 0

0 8000

Global Stiffness Matrix:

-20000

20666.667

4 5 1 2 3

Global:

Q4 = 0Q5 = 0

Q1 = 5Q2 = 0

Q3 = 0

0.01316 m

0.01316 m9.199(10-4) rad

-9.355(10-5) m

-1.887(10-3) rad

Global:

4 5 1 2 3-20000 0 0

20666.667 00

20666.667 -2000

-2000 16000

0-2000

08000 0 -2000 4000

-200000

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

D4 = 0.01316

D5 = 9.199(10-4)

D6 = 0

D1 = 0.01316

D2 = -9.355(10-5)

D3 = -1.887(10-3)

-11.22

Member 1

1.87 kN 11.22 kN�m1.87 kN

4 6 5 1 2 3

-20000

666.667

-666.667

-20000

-666.667

666.667

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

1.87 kN 11.22 kN�m

5 kN1.87 kN

18.77 kN�m

Member 2

D1 = 0.01316

D3 = -1.887(10-3)

D2 = -9.355(10-5)

D7 = 0

D8 = 0

D9 = 0

1 2 3 7 8 9

666.667 20001

-666.667

20000 0

-20000

8000 -2000

666.667 0

20000 0

0 8000

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

1.87 kN 11.22 kN�m1.87 kN5 kN

1.87 kN

18.77 kN�m5 kN

1.87 kN

Shear diagram

-1.87-

-11.22

1.87 kN 11.22 kN�m

1.87 kN

18.77 kN�m

Deflected shape

Bending moment diagram +

-11.22

D3=-0.00189 rad

D1=13.16 mmD4=13.16 mm

0.01316 m

0.01316 m9.199(10-4) rad

-9.355(10-5) m

-1.887(10-3) rad

Global :

D5=0.00092 rad

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.

Global 2

[FEM] 9 kN�m

9 kN�m

3 kN/m

Members1

6 m 6 m

3 kN/m

6 m 6 m

3 kN/m

λx = cos θx = 6/7.5 = 0.8

Member 1:

λy = cos θy = 4.5/7.5 = 0.6

/160005.7

)/10200)(10600( 2626

mmkNLEI

/33.341)5.7(

)1060)(/10200(12123

mmkNLEI

1280)5.7(

)1060)(/10200(662

mmkNLEI

64005.7

)1060)(/10200(44 4626

mmkNLEI

32005.7

)1060)(/10200(22 4626

[ km ] = [ T ]T[ k´ ][T ] =

λx = cos θx

Member m:

θy Uj

λy = cos θy

- λiy6EIL2

λix6EIL2

λjy6EIL2

- λjx6EIL2

Ui Vi Mi

- λiy6EIL2

λix6EIL2

λjy6EIL2

λjx6EIL2

- λixλiy)(12EIL3

λiy2 +

12EIL3

λix2 )(

λix6EIL2

λiyλjx -

12EIL3

λixλjy)-(

λixλjx +

12EIL3

λiyλjy)-(

λjy6EIL2

λjx2 +

12EIL3

λjy2 )(

λjy6EIL2

λjx6EIL2

- λjxλjy)(12EIL3

- λjx6EIL2

λixλjy -

12EIL3

λiyλjx)-(

- λixλiy)(12EIL3

- λiy6EIL2

λixλjx +

12EIL3

λiy λjy)-(

λix2 +

12EIL3

λiyλjy +

12EIL3

λix λjx )-(

λiyλjy +

12EIL3

λixλjx)-(

12EIL3

λjx2 )

λiyλjx -

12EIL3

λixλjy)-(

- λjxλjy)(12EIL3

AEL λjy

λx = cos θx = 6/7.5 = 0.8

Member 1:

λy = cos θy = 4.5/7.5 = 0.6

6 m 6 m

3 kN/m

-10362.879

-7516.162

10362.879

7516.162

-7516.162

-5978.451

7516.162

5978.451

10362.879

7516.162

-10362.879

-7516.162

7516.162

5978.451

-7516.162

-5978.451

λx = cos 0o = 1.0, λy = cos 90o = 0

Member 2:

6 m 6 m

3 kN/m

7.5 mmkN

/200006

)/10200)(10600( 2626

mmkNLEI

/667.666)6(

)1060)(/10200(12123

mmkNLEI

2000)6(

)1060)(/10200(662

mmkNLEI

)1060)(/10200(44 4626

mmkNLEI

)1060)(/10200(22 4626

[k2] =0

-6EI/L2

- 6EI/L2

6EI/L2

-12EI/L3

6EI/L2

- 6EI/L2

- 12EI/L3

6EI/L2

-6EI/L2

0 - AE/L

12EI/L3

3 92 71 8

[k2] =0

- 2000

-666.667

-20000

- 2000

- 666.667

0 - 20000

666.667

3 92 71 8

-10362.879

-7516.162

10362.879

7516.162

-7516.162

-5978.451

7516.162

5978.451

10362.879

7516.162

-10362.879

-7516.162

7516.162

5978.451

-7516.162

-5978.451

[k2] =0

- 2000

-666.667

-20000

- 2000

- 666.667

0 - 20000

666.667

3 92 71 8

D2 =4.575(10-4) m

-5.278(10-4) rad

-1.794(10-3) m

Global:

6 m 6 m

3 kN/m

7.5 m9 kN

9 kN�m

30362.9

7516.16

6645.12

14400=

9 kN�m

λx = cos θx = 6/7.5 = 0.8

Member 1:

λy = cos θy = 4.5/7.5 = 0.6

D1 = 4.575(10-4)

D2 = -1.794(10-3)

D3 = -5.278(10-4)

5 1 2 34 6

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

Member 2:

D1 = 4.575(10-4)

D2 = -1.794(10-3)

D3 = -5.278(10-4)

2 7 8 91 3

-14.70

9 kN�m

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

3 kN/m

29.15 kN

6.75 kN

1.19 kN�m

9.15 kN11.25 kN

14.70 kN�m

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

Shear diagram (kN)

Deflected shape

3 kN/m

29.15 kN

6.75 kN

1.19 kN�m

9.15 kN11.25 kN

14.70 kN�m

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

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

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.

6 m 3 m

BC15 kN

20 kN�m

3 kN/m

[FEM] 13 kN/m 11.25 kN

wL/2 = 11.25 kN

wL2/12 = 14.06 kN�m

14.06 kN�m

Global

6 m 3 m

BC15 kN

20 kN�m

3 kN/m

7.5 kN�m

λx = cos θx = 6/7.5 = 0.8

Members1

θy = 53.13

θx =36.87

λy = cos θy = 4.5/7.5 = 0.6

6.75 kN

11.25(0.6) = 6.75 kN

11.25(0.8) = 9 kN

Global:

Q1 = 15

Q3 = 20

Q2 = 0

-14.06 + 7.5

9 + 5+1

30362.9

7516.16

6645.2

14400=

[FEM]13 kN/m 11.25 kN

11.25 kN

14.06 kN�m

14.06 kN�m7.5 kN�m

7.5 kN�m

6.75 kN

11.25(0.6) = 6.75 kN

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

D2 =1.751(10-3) m

2.049(10-3) rad

-4.388(10-3) m

Member 1:

D1 = 1.751(10-3)

D2 = -4.388(10-3)

D3 = 2.049(10-3)

-14.06

5 1 2 34 6

-20.01

13 kN/m 11.25 kN

11.25 kN

14.06 kN�m

6.75 kN

11.25(0.6) = 6.75 kN

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

14.06 kN�m

6.75 kN

11.25(0.6) = 6.75 kN

3 kN/m

7.5 m1

19.71 kN

7.07 kN

4.89 kN�m

15.43 kN

26.46 kN�m

-20.01

θx = 36.87o

θy = 53.13o 1

3 kN/m

20.01 kN

4.89 kN�m

6.51 kN

24.17 kN

26.46 kN�m

6.17 kN

Member 2:

D1 = 1.751(10-3)

D2 = -4.388(10-3)

D3 = 2.049(10-3)

2 7 8 91 3

-35.02

35.02 kN

6.17 kN

15.12 kN�m

35.02 kN3.83 kN

8.08 kN�m

7.5 kN�m

10 kN 2

7.5 kN�m

3 kN/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

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

Shear diagram (kN)

3 kN/m

7.5 m1

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

15.43D1

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

-15.12 -8.08

Special Frames

θjθi 4 *

5 * 6*

λix = cos θi

λiy = sin θi

λjx = cos θj

λjy = sin θj

[ q* ] = [ T ]T[ q´ ]

Stiffness matrix

5´ 6´

2´3´

[ T ]T

1000000000000000010000000000

qqqqqq

λλλλ

qqqqqq

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

1´ 2´ 3´ 4´ 5´ 6´

1000000000000000010000000000

λλλλ

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

[ k ] = [ T ]T[ k´ ][T] =

- λiy6EIL2

λix6EIL2

λjy6EIL2

- λjx6EIL2

Ui Vi Mi

- λiy6EIL2

λix6EIL2

λjy6EIL2

λjx6EIL2

- λixλiy)(12EIL3

λiy2 +

12EIL3

λix2 )(

λix6EIL2

λiyλjx -

12EIL3

λixλjy)-(

λixλjx +

12EIL3

λiyλjy)-(

λjy6EIL2

λjx2 +

12EIL3

λjy2 )(

λjy6EIL2

λjx6EIL2

- λjxλjy)(12EIL3

- λjx6EIL2

λixλjy -

12EIL3

λiyλjx)-(

- λixλiy)(12EIL3

- λiy6EIL2

λixλjx +

12EIL3

λiy λjy)-(

λix2 +

12EIL3

λiyλjy +

12EIL3

λix λjx )-(

λiyλjy +

12EIL3

λixλjx)-(

12EIL3

λjx2 )

λiyλjx -

12EIL3

λixλjy)-(

- λjxλjy)(12EIL3

AEL λjy

4 m 4 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.

1Local

4 m 4 m

7.416 m

22.02 o

20 kN200 kN�m

6 kN/m

8 922.02 o

Global

2´ 3´

5´ 6´

2´3´

5´ 6´

2* 3 *

AE /101508

)/10200)(006.0( 3262

mmkNLEI

•×=×

= 3426

)0002.0)(/10200(44

mkNLEI

•×= 310102

mmkNLEI 3

2 1075.3)8(

)0002.0)(/10200(66×=

mmkNLEI /109375.0

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

3 ×=×

−−−−

−−−

LAEAE/L

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

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

2´ 3´

5´ 6´

2´3´

5´ 6´

- 3750- 937.5

0375010000

0-3750

0 - 150000

- 3750

-937.5

-150000

[ k´]1 = [ k´]2 =20000

150000

θi =22.02o

Member 1:

2 * 3 *

Global1´

2´3´

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

[ T ]1T

0.9270.375

- 0.375 0.927

[ q* ] = [ T ]T[ q´ ]

1´ 4´ 5´ 6´2´ 3´

θi =22.02o1

2 * 3 *

Global1´

2´3´

5´ 6´Local

θj = 0o

[ k* ]1 = [ T ]1T[ k´ ]1[ T ]1

[ k* ]1 = 103

1*2*3*456

129.04651.811-1.406

-139.0580.351-1.406

51.811

21.892

3.476-56.240

-0.8693.476

-1.4063.47620.000.00-3.7510.00

-139.058-56.240

0.351-0.869-3.75

00.938-3.75

-1.4063.47610.00

0-3.75

θi =22.02o1

2 * 3 *

Globalθj = 0o

40 kN40 kN�m40 kN�m

20 kN20 kN

[ qF* ] = [ T ]T[ qF´]

[ qF*] = [ T ]1T

40 kN40 kN�m

40 kN�m

18.547.5

-7.5018.5440020-40

Member 2

[ T ]2T

0.927-0.375

0.3750.927

0.927-0.375

0.3750.927

[ q ] = [ T ]T[ q´ ]

1´ 4´ 5´ 6´2´ 3´

6 kN/m2

32 kN�m

[ q´F]

2´ 3´

5´ 6´

[ q´]

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

[ q ]7

8 922.02o

22.02o

2´ 3´

5´ 6´

[ q´ ]

[ k ]2 = [ T ]2T[ k´]2[ T ]2

[ q ]7

8 922.02o

22.02o

[ k ]2 = 103

129.046-51.811

1.406-129.046

51.8111.4056

-51.81121.8923.476

51.811-21.892

1.4063.476

20-1.406-3.476

-129.04651.811-1.406

129.046-51.811-1.406

51.811-21.892-3.476

-51.81121.892-3.476

1.4063.476

10-1.406-3.476

6 kN/m2

32 kN�m

[ q´F ]

[ qF* ] = [ T ]T[ qF´ ]

[ qF ] = [ T ]2T

8.99822.249328.99822.249-32

6 kN/m2

32 kN�m

[ qF ]

22.249 kN

8.998kN

22.249 kN

8.998 kN

[ q ]7

8 922.02o

22.02o

[ 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

50.351-0.869-3.75

00.938-3.75

6-1.4063.47610.00

0-3.75

Global Stiffness:

2* 3 *

[ k ]2 = 103

129.046-51.811

1.406-129.046

51.8111.4056

-51.81121.8923.476

51.811-21.892

1.4063.476

20-1.406-3.476

-129.04651.811-1.406

129.046-51.811-1.406

51.811-21.892-3.476

-51.81121.892-3.476

1.4063.476

10-1.406-3.476

Global: 40 kN 20 kN200 kN�m

6 kN/m 12

2 * 3 *

6 kN/m2

32 kN�m

[ qF ]

22.249 kN

8.998 kN

22.249 kN

8.998 kN

40 kN40 kN�m

40 kN�m

18.54 kN7.5

[ q*F ]

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

40 kN�m

40 kN�m 32 kN�m

20 kN 22.249 kN

8.998 kN

20 kN200 kN�m

= -200

129.046

-1.406

-139.058

-1.406

-139.058

279.046

-51.811

22.829

-0.274

6-1.406

-0.274

+ 8.998

20 +22.249

-40 + 32

Global: 40 kN 20 kN200 kN�m

6 kN/m 12

2 * 3 *

-0.0205 m

-0.0112 rad

-0.0191 m

-0.0476 m

-0.0024 rad

2 * 3 *

40 kN 40 kN�m

40 kN�m

18.547.5

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

D1*=-0.0205

D3*=-0.0112D2*= 0.0

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

022.63 kN

-8.49 kN19.02 kN7.87 kN�m

22.63 kN

8.49 kN

19.02 kN

7.87 kN�m

1*2*3*456

129.04651.811-1.406

-139.0580.351-1.406

51.811

21.892

3.476-56.240

-0.8693.476

-1.4063.47620.000.00-3.7510.00

-139.058-56.240

0.351-0.869-3.75

00.938-3.75

-1.4063.47610.00

0-3.75

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

3222.249

8.99822.249

[ q ] 7

8 96 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

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

129.046-51.811

1.406-129.046

51.8111.4056

-51.81121.8923.476

51.811-21.892

1.4063.476

20-1.406-3.476

-129.04651.811-1.406

129.046-51.811-1.406

51.811-21.892-3.476

-51.81121.892-3.476

1.4063.476

10-1.406-3.476

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

8.49 kN

20.98 kN

207.87

-248.04

Deflected shape

D3*=-0.0112 rad

D1*=- 20.5 mm

D6=-0.0024 rad

-0.0205 m

-0.0112 rad

-0.0191 m

-0.0476 m

-0.0024 radD4=-19.1 mmD5=-47.6 mm

80 kN�m20 kN/m

4 m 2 m

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.

Global

1´2´

4´5´

80 kN�m20 kN/m

4 m 2 m

2´ 3´

5´ 6´1

76.31o

80 kN�m20 kN/m

4 m 2 m

Member 1:

kN/m103000m4

)kN/m10)(200m10(60

mkN1080m4

)m10)(400kN/m104(2004

•×=

mkN1040m4

)m10)(400kN/m102(2002

•×=

kN1030m)(4

)m10)(400kN/m106(2006

kN/m1015m)(4

)m10)(400kN/m1012(20012

-26.67

+= 103

1 4 5 6

-30000

80 kN�m20 kN/m

4 m 2 m

26.67 kN�m26.67 kN�m

40 kN40 kN

20 kN/m

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

56.31o

76.31o

Member 2:

1´2´

4´5´

kN/m103324m3.61

)kN/m10)(200m10(60

mkN1088.64m3.61

)m10)(400kN/m104(2004

•×=

kN1036.83m)(3.61

)m10)(400kN/m106(2006

kN/m1020.41m)(3.61

)m10)(400kN/m1012(20012

mkN1044.32m3.61

)m10)(400kN/m102(2002

•×=

[ k´ ]2 = 103

1´2´3´4´5´6´

1´ 4´ 5´ 6´3324

-332400

20.4136.83

0-20.4136.83

36.8388.64

0-36.8344.32

-332400

0-20.41-36.83

020.41

-36.83

036.8344.32

0-36.8388.64

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

[ q* ] = [ T ]T[ q´ ]

q4´q5´q6´

q7*q8*q9*

4567*8*9*

1´ 4´000

0-0.97

5´000

0.970.24

6´000001

0-0.83

2´0.830.55

3´001000

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

56.31o

1´2´

4´5´

[ 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

[k]1 = 103

123456

1 4 5 60

-300000

0-1530

-300000

0-15-300

[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

Global:

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

-26.6740

-2.199x10-5 m

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

0.979x10-3 m6.161x10-4 rad

80 kN�m20 kN/m

4 m 2 m

C 26.67 kN�m

26.67 kN�m

40 kN40 kN

20 kN/m

80 kN�m50 kN

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

-26.67

+= 103

1 4 5 6

-30000

Member 1:

65.97 kN

24.59 kN�m

36.12 kN

-65.97 kN

43.88 kN-40.11 kN�m

40.11 kN�m24.59 kN�m

43.88 kN36.12 kN

20 kN/m

65.97 kN65.97 kN

q8*q9*

Member 2:

q8*q9*

15.97 kN

-39.89 kN�m

-43.88 kN

46.69 kN0 kN�m

4 7* 8* 9*5 6-455.21651.27

-35.73

210.67

-35.73

-769.1

1769.34-2678.93

-769.13128.82

30.5720.26

-35.73-8.84

1036.923

30.57-1508.14

-455.211769.34

-1508.142296.15

651.27-2678.93

30.5720.26

-35.73-8.84

15.97 kN43.88 kN 39.89 kN�m

46.69 kN

40.11 kN�m24.59 kN�m

43.88 kN36.12 kN

20 kN/m

65.97 kN65.97 kN

15.97 kN43.88 kN 39.89 kN�m

46.69 kN

-24.59-

-40.11

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

! 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

Trusses

a(b)R1

V = R1

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.

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

+ 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

V = 10 kN

FAE= FFFB= F

θ = 36.87o

F20 kN

θ6.67 kN

8.33FAF

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

F = FEC

F = FDB

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

10 kN 20 kN 20 kN 20 kN 10 kN

4 m 4 m 4 m 4 m

FBH = 0

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

ΣFy = 0:+ FJA = 40 kN (C)

FAI = 0

FJB V = 30 kN

ΣFy = 0:+ 40 - 10 - FJBcos 45o = 0

FJB = 42.43 kN (T)

FAI = 0

10 kN 20 kN

+ 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

FIC V = 10 kN

10 kN 20 kN

30 kN0

042.3 kNFBI

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

Vertical Loads on Building Frames

typical building frame

0.21L0.21Lpoint of zero moment

approximate case

assumed points of zero moment0.1L 0.1L

� Assumptions for Approximate Analysis

0.1L 0.1L0.8L

Simply supported(c)

column column

girder

Point ofzeromoment

Example 3

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

1 kN/m

4.8 kN

0.6 kN 2.4 kN

0.6(0.3) + 2.4(0.6) = 1.62 kN�m

0.6 kN2.4 kN

1.62 kN�m

0.1L=0.6 m 0.6 m

2.4 kN 2.4 kN

Portal Frames and Trusses

� Frames: Pin-Supported∆ ∆

assumedhinge

P/2 P/2

Ph/2Ph/2

� Frames : Fixed-SupportedP

Ph/2lP/2

∆ ∆

assumedhinges

P/2 P/2

Ph/2lPh/4 Ph/4

Ph/4Ph/4

� Frames : Partial Fixity

h/3h/3

assumedhinges

� Trusses

P/2 P/2h/2

assumedhinges

∆ ∆

Example 4

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

4 m 2 m2 m

40/2 = 20 kN = V

20 kN = V

I3.5 m

V = 20 kN

20 kN = VM M

4 m 2 m2 m

+ ΣMJ = 0:

N(8) - 40(5.5) = 0

N = 27.5 kN

+ ΣMA = 0:

M - 20(3.5) = 0

M = 70 kN�m

27.5 kN

FBDFBH

+ 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)

27.5 kN

20 kN = V

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

FDH = 38.9 kN (C)

D75 kN

38.9 kN

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

45o45o

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

FHE = 38.9 kN (T)

= inflection point

Lateral Loads on Building Frames: Portal Method

(b)V V V V

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

B D F G

I J K L

4 m 4 m 4 m

I J K L

V V2V 2V

Iy Jy Ky Ly

ΣFx = 0: 5 - 6V = 0

V = 0.833 kN

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

0.625 kN

0.833kN

0.625 kN

0.833 kN

L1.5 m

0.833 kN 0.625 kN 1.25 kN�m

J1.5 m

1.666 kN

1.666kN 2.50 kN�m

0.835 kN

K1.5 m

1.666 kN

1.666kN 2.5 kN�m

0.833 kN0.625 kN 1.25 kN�m

0.833 kN

2.501 kN

0.625kN

1.666 kN1.5 m

2 m2 mN O

0.833 kN

0.625 kN 0.835 kN

0.625 kN

1.666 kN

2 m 2 m

4.167 kN

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.

8 m 8 m

ΣFx = 0: 20 - 4V = 0 V = 5 kN+

V´ Ky

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

20 kN2.5 m

Oy = 3.125

20 kN2.5 m

G R4 m

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

15.625kN12.5 kN

Bx = 25 kNBx = 0 MB = 75 kN�m

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

3 m4 m4 m

22.5 kN

12.5 kN

H SR15 kN

3.125 kN

4 m4 m

3.125 kN

12.5 kN

M2.5 m

3 m 4 m

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.

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.

3)(6)(0~=

==∑∑

3 m 3 m

30 kNC

+ ΣMM = 0: -30(2) + 3Hy + 3Ky = 0

The unknowns can be related by proportional triangles, that is

yyyy KHor

kNKH yy 10==

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

kNLG yy 35==

30 kN2 m

C 3 m Ix = 15 kN

Iy = 10 kN

Hx = 15 kN

DI15 kN

Kx = 15 kN

2 m 3 m Jx = 7.5 kN

Jy = 25 kN

Gx = 22.5 kN35 kN

7.5 kN

Lx = 22.5 kN

35 kN22.5 kN

Ax = 22.5 kNAx = 35 kN MA = 45 kN�m

Fx = 22.5 kNFy = 35 kN MF = 45 kN�m

22.5 kN35 kN

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.

A B C D

L M N O

E F G H

4 m 8 m

5000mm2

4000 mm2

6000 mm2

A B C D

L M N O

E F G H

6 m 4 m 8 m

6000 mm2 6000 mm24000 mm25000mm2

x 48.86000400050006000

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

++++++

==∑∑

35 kN2 m

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

L M N O

I J K4 m

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

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

Ix= 114.702 kN

Iy= 15.536 kN

Lx= 5.262 kN

Ex= 64.44 kN

Px= 29.738 kN

Py= 3.508 kN

19.044 kN

5.262 kN3.508 kN

E19.044 kN

64.44 kN

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

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

Documents

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

i ¥ @q ¹ ? @l ¿ Ì º l m W

C A M D E N T O W N2

RETROCAVAL URETER: A REPORT OF TWO CASES URETERE RETROCAVE : A PROPOS DE DEUX CAS M. OULED YEHDIH, M. MAATOUK, M.A. JELLALI, W. MNARI, A. ZRIG, W. HARZALLAH,

M t j W # ;y

0νββ TPC : AXEL · 2020. 12. 30. · AXEL U Z + w A t m M o x ² t ô ¤ É ç ª Ç á µ p º p ^ d o M h i M o S | f w w U Z + A É t m M o x f j ° ` o M h i Z q ¾ M p K [1]

LES JOURS ENQUÊTE SUR UNE MORT SOLITAIRE w w w ......LES JOURS ENQUÊTE SUR UNE MORT SOLITAIRE w w w. parismatc M € L'EPREUVE DU DE DE JEN'ÅBÅNDONNERAI PAS LES CHRISTOPHE DOMINICI

1 J'.) 1 1MI1.''oe- M M-Jh -. * .Jj Ajg -* . tarchive.ical.ir/files/debate/ip/07/07-222.pdf · 1 a'!&.e. p J'.)Q)) V o'tv'o''.lw-f:bL:''- w - @ w - ... 1 1MI1.''oe- M M-Jh -. * .Jj

FT31 - SAGITRACE ind II...d’une tresse cuivre étamée pour mise à la terre, revêtue d’une MISE HORS GEL MAINTIEN EN TEMPERATURE SAPGS SAMT 15 W/m 25 W/m 165 m 125 m FT 31 ind.

payettecounty.infopayettecounty.info/news/1944/07/1944_07_27_04_pie.pdf · Wi K\ iM Ki-M W' Hi m KI w KI-.w v 4-4'H' 1 'H' i tnkfUte'ly fat Uiyvtte I tlifs y lti thtf wllt'jst'endth.Is"A\wh

{M « ~ Ì {w ¯ÔA¨tmMo

Magazine Municipal VRIL 2018 a c h i m . c o m w w w . … · 2018-05-07 · Ecole ouverte Cette année, nous avons ouvert notre école à tous les parents d’élèves. Ainsi, le

Arch ici&itd pp1

D p w m u k2

- Hanukkah Fun · Hanukkah Israel Latkes Light Maccabees Matityahu Menorah Nun Oil Shamash Shin Sufganiyot Temple . o M C C B E D B o s G o w M M s E p B D w s M D w M E C G L E o

&¾'g ± Û W i9 b) ;o0b ± Û Û/¡ ¦ » ¦ »6× KDVHJDZD W# ...Q b S u _ W i9 c M m A ^ b [ 6 : W>8 ± Û b3å Û"á \ "á b S x W 1 *º º v ¥>8 º v 1* P1ß / G'Å / )ª3H>83å

W>m UNIVERSITÉ DE m SHERBROOKE

“M— ;. . . . ', “35$: “W”; ’l 1:23 · mmm' mwamr mu “‘1'” MIJIHIHEEIEH___ mung "“ “I” I" twain-----w m' a I N “am 1 mm “M— mhé m MM.——"——' m—u—I-rn—.

Geonaute - FR W 700XC M Swip · 2019. 1. 15. · Les montres Swip Analog M, Swip Digital, A 300 M Swip, A 300+ M Swip, W 500 M Swip, W 500+ M Swip, W 700 XC M Swip sont uniquement

Mécanisme de leffet de serre GES anthropique -2 W/m 2 Aérosols anthropiques +1 W/m 2.

confinementpedagogie.ac-guadeloupe.fr/.../dossier_de_presentation_memwa_kaz… · M EM W A KA ZÈN É : P O U M O U N TO U JO U S O N JÉ Q U O I? "M e m w a kaz è né : p ou m oun

&¾'g ± Û W i9 b) ;o0b ± Û Û/¡ ¦ » ¦ »6× KDVHJDZD W# ...Q b S u _ W i9 c M m A ^ b [ 6 : W>8 ± Û b3å Û"á \ "á b S x W 1 º º v ¥>8 º v 1 P1ß / G'Å / )ª3H>83å