Chapter 3Chapter 3

TrussesTrusses

Trusses in Building

Trusses Types for Building

Structural Analysis IDr. Mohammed Arafa

6 Chapter 2 Dr. Mohammed Arafa

Truss Bridge

Truss Bridge

Truss Bridge Types

Assumptions for Design

All members are connected at both ends by smooth

frictionless pins.

All loads are applied at joints (member weight is

negligible)

Centroids of all joint members coincide at the joint.

All members are straight.

All load conditions satisfy Hookeís law.

Classification of coplanar TrussSimple Truss

Compound Truss

Stability of Compound Truss

Complex Truss

Determinacy

For plane truss

If b+r = 2j statically determinate

If b+r > 2j statically indeterminate

Where

b = number of bars

r = number of external support reaction

j = number of joints

Stability

For plane truss

If b+r < 2j unstable

The truss can be statically determinate or indeterminate (b+r >=2j)

but unstable in the following cases:

External Stability: truss is externally unstable if all of its

reactions are concurrent or parallel

Internal stability:

Internally stable Internally unstable

Internally unstable

Classify each of the truss as stable , unstable,

statically determinate or indeterminate.

Stable & edeterminat Statically

22)11(22

22

11 3 19

j

rb

jrb

Stable & ateindetermin Statically

18)9(22

19

9 4 15

j

rb

jrb

Structural Analysis IDr. Mohammed Arafa

Stable & edeterminat Statically

12)6(22

12

6 3 9

j

rb

jrb

Unstable

16)8(22

15

8 3 12

j

rb

jrb

STEPS FOR ANALYSIS1. If the support reactions are not given, draw a FBD of the entire

truss and determine all the support reactions using the

equations of equilibrium.

2. Draw the free-body diagram of a joint with one or two unknowns.

Assume that all unknown member forces act in tension (pulling

the pin) unless you can determine by inspection that the forces

are compression loads.

3. Apply the scalar equations of equilibrium, FX = 0 and

FY = 0, to determine the unknown(s). If the answer is

positive, then the assumed direction (tension) is correct, otherwise

it is in the opposite direction (compression).

4. Repeat steps 2 and 3 at each joint in succession until all the

required forces are determined.

The Method of Joints

The Method of Joints

Structural Analysis IDr. Mohammed Arafa

Example 1

Example 2Solve the following truss

Example 2Solve the following truss

0;

4 sin 30 0 8 C

0;

8cos30 0 6.93 T

y

AG AG

x

AB AB

F

F F KN

F

F F KN

0;

sin 30 3cos30 0 3.0 C

0;

8 3sin 30 3cos30 0 5.0 C

y

GB GB

x

GF GF

F

F F kN

F

F F kN

0;

sin 60 3.0sin 30 0 1.73 T

0;

1.73cos 60 3.0cos30 6.93 0

4.46 T

y

BF BF

x

BC

BC

F

F F KN

F

F

F KN

ProblemDetermine the force in each member

Zero Force Member

1- If a joint has only two non-colinear members and there is no

external load or support reaction at that joint, then those two

members are zero-force members

Zero Force Member2- If three members form a truss joint for which two of the

members are collinear and there is no external load or reaction at that

joint, then the third non-collinear member is a zero force member.

Example 3Find Zero force member of the following truss

Method of Section

The Method of Section

1. Decide how you need to “cut” the truss. This is based on:

a) where you need to determine forces, and, b) where the total

number of unknowns does not exceed three (in general).

2. Decide which side of the cut truss will be easier to work with

(minimize the number of reactions you have to find).

3. If required, determine the necessary support reactions by

drawing the FBD of the entire truss and applying the E. of E.

The Method of SectionSTEPS FOR ANALYSIS

4. Draw the FBD of the selected part of the cut truss. We need to

indicate the unknown forces at the cut members. Initially we

assume all the members are in tension, as we did when using the

method of joints. Upon solving, if the answer is positive, the

member is in tension as per our assumption. If the answer is

negative, the member must be in compression. (Please note that you

can also assume forces to be either tension or compression by inspection as was

done in the previous example above.)

5. Apply the equations of equilibrium (EofE) to the selected cut

section of the truss to solve for the unknown member forces.

Please note that in most cases it is possible to write one equation

to solve for one unknown directly.

Example 4Solve the CF & GC members in the truss

C 4.3460)93.6(300)12(30sin

;0

IbFF

M

CFCF

E

0;

(12) 300(6.93) 346.4sin 30 12 0 346.4 T

A

GC GC

M

F F Ib

Determine the force in members GJ and CO of

the roof truss shown.

Example 5

Example 5

Example 5

Example 6Solve the GF & GD members in the truss

C 8.10)6(2)3(7)6(3.56sin0

C 83.70)3(7)3(6.26cos0

kNFFM

kNFFM

GFGDo

GFGFD

Example 7Solve the ED & EB members in the truss

Example 8Solve the BC & MC members in the K-truss

T 21750)20()15(29000 IbFFM BCBCL

TIbFF MBy 12000 BJoint At

TIbFFF MLMLy 29001200120029000

part cutting totalFor the

TIbF

CIbF

FFF

FFF

MC

MK

MKMCy

MKMCx

1532

1532

0120029000

00

MJoint At

13

2

13

2

13

3

13

3

Problem 2Solve All members

Problem 3Solve All members

Problem 4Solve members CH , CI

Example 1 Compound TrussSolve the truss

C 46.30)60sin4()2(4)4(50 kNFFM HGHGC

F

& FF

& FF

& FF

& FF

JC

BJBC

IBIJ

HJHI

AIAB

JJoint

BJoint

IJoint

HJoint

AJoint

Example 2 Compound Truss

Example 3 Compound TrussSolve the truss

F

& FF

EB

ABAE

BJoint

AJoint

AGAF

AE

& FF

F

AJoint

SolveAfter

Complex Truss

iii xsSS '

Force in members

?

0'

x

xsSS ECECEC

1

1

Complex Trusses

+=

X x

S´EC + x sEC = 0

x =S´EC

sEC

P

A

B

C

DF

Er + b = 2j,

3 9 2(6)

•Determinate

•Stable

F´ECP

A

B

C

DF

E

= FAD

SAD

sEC

A

B

C

DF

E

Si = S´i + x si

Member

AB

AC

AF

FE

BE

ED

FC

EC

'iS is ixsiS

iii xsSS '

?

0'

x

xsSS ECECEC

Example 1 Complex Truss

Example 2

P

b + r < 3j

b + r = 3j

Statically indeterminate -Check stability

• Determinacy and Stability

Space Truss

Unstable truss

b + r > 3j

Statically determinate -Check stability

•x, y, z, Force Components

Space Truss

222 zyxl

)(l

xFFx

)(l

yFFy

)(l

zFFz

222

zyx FFFF

x

y

z

y

z

x

y

z

x

z

x

y

x

y

z

Fy

Fz

Fz

Fx

Fx

Fz

Fy

short link

y

x

z

roller

z

x

yslotted roller

constrained

in a cylinder

y

z

x ball-and -socket

Support Reactions

Zero Force Member1- If all but one of the members connected to a joint lie on the

same plane, and provided no external load act on the joint, then

the member not lying in the plane of the other members must

subjected to zero force.

SFz = 0 , FD = 0

Zero Force Member2- If it has been determined that all but two of several members

connected at a joint support zero force, then the two remaining

members must also support zero force, provided they don’t lie a

long the same line and no external load act on the joint.

SFz = 0 , FB = 0

SFy = 0 , FD = 0

Example 1 Space Truss

Structural Analysis IDr. Mohammed Arafa

Example 2 Space Truss

Copyright © 2009 Pearson Prentice Hall Inc.