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CE 201 - STATICSDr. Mustafa Y. Al-MandilDepartment of

Civil Engineering

LECTURE

18 134

EQUILIBRIUM OF RIGHD BODY

It is necessary and sufficient for the equilibrium of rigid body that the resultant

of all forces acting on the body is zero and that the resultant of all moments

taken about any point inside or outside the body is zero.

0~

F~ 0

~M~

0

00

z

y

x

M

MM

Ox

FR

MR

MC

MF

O

MC

&

0

00

z

y

x

F

FF ...

...

...

...

...

...

1F~

iF~

nF~

iF~

nF~

1F~

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CE 201 - STATICS

Support for Rigid Bodies Subjected to Two-Dimensional Forces System.

Table 5 - 1

Type of Connection Reaction Numbers of Unknowns

One unknown. The reaction is the tension force which

acts away from the member in the direction of the cable

One unknown. The reaction is a force which acts

along the axis of the link.


perpendicular to the surface at the point of contact.


perpendicular to the slot.


perpendicular to the surface at the point of contact.Rocker

Roller or pin in confined smooth slot

Roller

Weightless Link

Cable

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CE 201 - STATICS

Member fixed

connected tocollar on smooth rod

Support for Rigid Bodies Subjected to Two-Dimensional Forces System.

Table 5 - 1



perpendicular to the rod.

Two unknown. The reaction are two components offorce, or the magnitude and its direction of the resultantforce. Note that and are not necessarily equal.(usually not, unless the rod shown is a link as in (2)).

Two unknowns. The reactions are the couple moment

and the force which acts perpendicular to the rod.

Three unknowns. The reaction are the couple and the

two forces components, or the couple and the magnitude

and direction of the resultant force.



Fixed support

Smooth pin or hinge

Member pin connected to collar on smooth rod

Smooth contacting surface

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Civil Engineering

LECTURE

18 336

FREE BODY DIAGRAM

EQUILIBRIUM IN 2-DIMENIONS

FByFBxFB B

AD

CFc

OMo

Fy

Fx

FOR EQUILIBRIUM

OM

OF

OF

z

y

x

..(1)

..(2)

..(3)

RAx

RAyFcy RDy

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Civil Engineering

LECTURE

19 137

EXAMPLE

Determine the support reaction in the beam (ABCD).Consider beam weight equals 50N.

A

C4

3

5100N

B

3m 2m 5m

DX

A

B

C D

RAx

RAy

60N80N

50N RDy

SOLUTION

First we draw FBD for beam (ABCD)

+

(1)

(2)

Solve (1) & (2)

RDy = 49N ( ) RAy = 81N( )

+

1007R3R-

07R2503R

0M

130RR

0R5080R

0F

)(06R060R

0F

DyAy

DyAy

B

DyAy

DyAy

y

N

Ax

Ax

x

a

+

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Civil Engineering

LECTURE

19 238

A

B

C D

RAx

RAy

60N 80N

50N RDy

ALTERNATE SOLUTION I

SOLUTION

N

Ay

Ay

D

N

Dy

Dy

A

NAxAx

x

8110

560250R

010R780550

0M

)(4910

240250R

038055010R

0M

)(60R060R

0F +

+

+

CONCLUSION

0M0

M

0F

D

A

x

One force equationand two moment

equations on a linenot to the forcedirection.

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Civil Engineering

LECTURE

19 339

A B C D

RAx

RAy

100N

50N RDy

+4m

E

43

5

3m 2m 5m

ALTERNATE SOLUTION II

CONCLUSION

0M0

M

0M

E

D

A

Three momentequations taken

at three points noton a straight line.

)(60

4

240

4

250490R

04R5501049

0M

)(81R

0M

)(49R

0M

N

Ax

Ax

E

N

Ay

D

N

Dy

A

+

+

+

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CE 201 - STATICS

Given:

kA = kB = 15kN/m

Required:

c = ?

40 9.8 = 392N

A B

C

40kg

392N

1m 3m

RA=KAA RB=KBB

C

0.1045m

0.0784m

m

C

C

m

A

AA

AAB

y

m

B

BB

A

0.6532

3

0.1045-

1

0.10450.0784

0.078415000

1176

k

R

1176R0392RR

0F

15680.104515000

4392

043921k

0M

N

N

BR

+

+

A

B

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CE 201 - STATICS LECTURE20 2

41

TWO FORCE MEMBERS:

OR

THREE FORCE MEMBERS:

AF~

AF~

AF~

AF~

AF~

BF~

BF~

BF~

BA

F~

F~

A

BB

A

B

A

BF~

AF~

BF~

AF~

CF~

CF~

0F

0F

y

x

O

BA

C

CF~

AF~

BF~

Dr. Mustafa Y. Al-MandilDepartment of

Civil Engineering

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CE 201 - STATICS Dr. Mustafa Y. Al-MandilDepartment ofCivil Engineering

DISTRIBUTED LOADING

A) Uniform Distributed Loading:

B) Linear Distributed Loading:

C) Trapezoidal Distributed Loading:

6m 3m

6m 4m

5N/m 30N

45N

15N/m

10N/m

60N

30N

20N/m

6m 4m

3m

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CE 201 - STATICS

Given: Beam ABCRequired: Find Reactions

)(408040R

0401005

4R

0F

)(306030R

0301005

3R

0F

100R

4004R

010405R5

4

0M

kN

Ay

Ay

y

kN

Ax

Ax

x

kN

B

B

B

A

a3

2d

2

ad

2

d

2

a

0)2

d

2

a

(W)2

a

(dW

0Mo

+

+

(

+

50kN

34

5m

5m

3m

4m

5mA B

D

C

B CA 5m

34

RBRAy

RAx

40kN

30kN

RB

RD3m

4mTwo Force Member

RB= RD d

D/2 D/2

Wa

o

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CE 201 - STATICS

Q.


Civil Engineering

Determine the support reaction of beam ABCD ?3m4m

B

D

CA

2m

w = 15N/m3

4

50N

3m4m2m

RAx

RByRAy

A B C60N 40N

30N

60Nm

Free body Diagram

A. Draw F.B.D.Transfer the (50N) force from point (D) to point (C)

)(85

4

340R

340602801204R

06070404R260

0M

)(30R

030R0F

N

By

By

By

A

N

Ax

Ax

x

)(15

408560R

0408560R

0F

N

Ay

Ay

y

+

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CE 201 - STATICS

Support for Rigid Bodies Subjected to Three-Dimensional Forces System.

Table 5 - 2


One unknown. The reaction is a force which acts away

from the member in the direction of the cable.





Three unknowns. The reactions are three rectangular

force components.

Four unknowns. The reactions are two force and two

couple components which act perpendicular to the shaft.

Cable

smooth surface support

Roller on a smooth surface

Ball and socket

Single journal bearing

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CE 201 - STATICS

Support for Rigid Bodies Subjected to Three-Dimensional Forces System.

Table 5 - 2


Five unknowns. The reactions are three force and

two couple components.

Five unknowns. The reactions are three force andtwo couple components.

Five unknowns. The reactions are three force and

two couple components.

Six unknowns. The reactions are three force and

three couple components.

Single thrust bearing

Single smooth pin

Single hinge

Fixed support

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Civil Engineering

1m

1.5m

4m

D

A

C

B

y

x

z

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Civil Engineering

A

0.3m

0.2m

0.6m

0.5m

1m

0.5m

100 kg

C

B E

D

x

y

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Civil Engineeringz

x

y

A

C

B

3ft3ft

4ft

6ft

D

E

RAz

RAx

8ft

4ft

RAyT

CD

TCD

TCE

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Civil Engineering

TRUSSES

Chep. 6

Atruss is an assembly of prismaticmembers connected at their ends

by smooth pin.Forces can only be

applied at such connections (joints)

Simple truss:

Assumptions for Analysis & Design:

1- Loading and reactions are applied

at joints (nodes) only. Member weight is

distributed equally to the adjacent nodes.

2- Members are loaded at their ends only

(i.e. two-force members). So, they undergo

axial forces only (tension or compression). RxyRAy

RAx

B

A C E

D G

H

K

J

LECTURE

23 151

LECTURE

23 151

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CE 201 - STATICS LECTURE 232

52


Civil Engineering

Analyze the truss shown below:

)(40R0F

)(45R0F

)(45R

06203204R

0M

kN

Ayy

kNAxx

kN

Bx

Bx

A

F.B.D. For Whole Truss :

Take Joint (E) :

(C)15F0255

3F

0F

(T)25F020F54

0F

kN

EDED

x

kNECEC

y

+

+

4m

RAy

RAx

RBx

A C

B DE

3m 3m

20kN 20kN

FEC

FEDE

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CE 201 - STATICS LECTURE 23

Analyze the truss shown below:

252


Civil Engineering

Take Joint (C) :

(C)20F0F

(T)15F0F

kN

CDy

kN

CAx

Take Joint (A) :

0

0505

440

0

)(50

05

34515

0

AB

AD

y

kN

AD

AD

x

F

F

F

TF

F

F

+

FCA

FCD25kN

C

40kN

45kN 15kNA

FAD

FAB

CE 201 STATICSD M t f Y Al M dil

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53


Civil Engineering ZERO-FORCE MEMBERS

Members which carry no forces in a truss either because of particular

loading pattern or because they enhance geometric stability:.

EQUAL-FORCE MEMBERSF3

F1

F2

F2 F1

F1= F2

F3

= 0

F3

F1

F2F4

F1= F3

F2= F4

30kN

35kN25kN

30kN40kN

A

B

D I LP

T

G K

C E H J M N Q

ZERO-FORCE MEMBERS

BC, BEKM, KN

PQ, PN, LN

(C)40F

(T)25F

kN

IJ

kN

HG


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CE 201 - STATICS

By sectioning the truss into two or more parts (each part containingmore then one joint), then we can find forces in certain numbers directly.

LECTURE

24 254


Civil Engineering METHOD OF SECTIONS

Use Section A-A

Find FGH ?

(C)155F

04F480650

0M

kN

GH

GH

J

80kN

50kN

JG

FGH

+

4m

50kN80kN

3m

3m

3m

3m

3

m

A A

F3

F2

F1

0

0Mo

F1 =


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55


Civil Engineering THEORY OF PLANE TRUSSES

M = Number of members

R = Number of Unknown ReactionsJ = Number of joints

if M + R > 2J Statically Indeterminate

if M + R = 2J Statically Determinate

if M + R < 2J Statically Unstable

M = 9

R = 3

J = 6

M + R = 2J

Determinate

M + R > 2J

Indeterminate

M = 10

R = 3

J = 6

M + R < 2J

UNSTABLE

M = 8R = 3

J = 6

B

C E

D

A G

CE 201 STATICSDr Mustafa Y Al Mandil

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CE 201 - STATICS

Frames are structural systemscomposed of members connected

and loaded at unrestricted

locations along their axis.

LECTURE

27 158


Civil Engineering

ANALYSIS & DESIGN

1- SOLVE FOR EXTERNAL REACTIONS

( partially or totally ).

2- DIS-ASSEMBLE & TREAT EACH COMPONENTAS RIGID BODY UNDER EQUILIBRIUM.

3- FORMULATE SETS OF SIMULTANEOUS

EQUATIONS & SOLVE FOR UNKNOWNS.

RAx

RAy RDy

A D

B

C

P2

P1

M

RAx RGx

RAy RGy

P1

P3P2

C

G

E

A

B

D

CE 201 STATICSDr Mustafa Y Al Mandil

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CE 201 - STATICS

Analyze Frame (ABCDE)

LECTURE

27 258


Civil Engineering

Take whole Frame Take Part (ABC) As F.B.D.

kN

Ay

Ay

y

kN

Ey

Ey

A

25R

0R4015

0F

1512

180

R

03406012R

0M

)(3.75

8

120150R

06R8R340

0M)(15R

04025R

0F

kN

Ax

AyAx

C

kN

Cy

Ey

y

Take Whole Frame Again

)(3.75R03.75R

0F

kN

Exex

x

RAy

A E

C

40kN

60kNm

REy

4m

4m

DB

3m3

m

3

m

3

m

RAx

CRCy

RAy

RCx

A

B

45kN

40kN

+

+

CE 201 STATICSDr Mustafa Y Al-Mandil

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56


Civil Engineering GENERAL EXAMPLE

(a) (b)

(c) (d)

(e)

(g) (h)

(f)

Howe Truss Pratt Truss

Warren Truss Warren Truss with verticals

K Truss Sub-divided Warren Truss

Sub-divided Pratt Trussor Baltimore Truss

Baltimore Truss with inclinedchord or Petit Truss

(a)

(d)

(c)

(b)

Howe Truss

Pratt Truss

Fan Truss

Fink Truss

Conventional Roof Trusses

Conventional Bridge Trusses

CE 201 STATICSDr Mustafa Y Al-Mandil

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57


Civil Engineering GENERAL EXAMPLE

CE 201 STATICS LECTUREDr Mustafa Y Al-Mandil

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59a

Dr. Mustafa Y. Al MandilDepartment of

Civil EngineeringEXAMPLE

AnalyzetheFra

me

Take whole Frame :

)(35R

0R80115

0F

)(1158

920R

015404808R

0M

kN

Ay

Ay

y

kN

Hy

Hy

A

+

)(40R0F

)(40R

0M

kN

eyy

kN

Cy

E

Take ( C.D.E. ) :

+

RAx RHx

RAy RHy

80kN

C

H

G

A

B

E40kN D

12m

3m

6m

9m

4m 4m

RCx C

RCy

ED80kN

REy

REx

CE 201 - STATICS LECTUREDr. Mustafa Y. Al-Mandil

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59b

Dr. Mustafa Y. Al MandilDepartment of

Civil EngineeringEXAMPLE

Take whole Frame :

kN

Exx

kNGB

GB

y

kNHx

Hx

J

120R0F

213.6F

040F8.54

3115

0F

80R

015R1640-16115

0M +

Analyze the Frame

RHx

H

G

6m

9m

115kN

FGB

JREx

40kN

16m E

3

8


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60a

usta a a dDepartment of

Civil Engineering

Analyze the Frame

Take whole Frame :

N

AyAy

y

N

Ax

Ax

x

N

Ex

Ex

A

100R0100R

0F

5461R

05461R

0F

61.54R0810013R

0M +

Take Pulley as F.B.D. :

100N

100N

100N

100N

100N

2m6m

H

E

D

C

B

2m

4m

5m

2m

REx

RAx

RAy

3m

G

A


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CE 201 STATICS LECTURE28 2

60b

Department ofCivil Engineering

Take GCH as FBD :

Analyze the Frame

G C H 100N

100NTGB

RCy

RCx

N

Cy

Cy

y

N

Cx

Cx

x

N

GB

GB

c

300R

0100R2505

40F

50R

0100R2505

3

0FT250T

061003T5

4

0M +

CE 201-A4 p2_l18_l28

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