structanal- truss structures

8/13/2019 StructAnal- Truss Structures

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Structural Analysis of

Pin Jointed (Truss)Structures

Method of Joints and Sections

A

B C D E FG

HIJKL

M N O P

5.34 kN 6.67 kN 7.11 kN12.9 kN

0


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Method of Joints

If a truss is in equilibrium, then each of its

joints must also be in equilibrium

The method of joints consists of satisfying the

equilibrium conditions for the forces exerted

on the pin at each joint of the truss


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Truss members are all straight two-force

members lying in the same plane

The force system acting at each pin is coplanar

and concurrent (intersecting) Rotational or moment equilibrium is automatically

satisfied at the joint, only need to satisfy Fx = 0,

Fy = 0


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Establish the sense of the unknown forces

Always assume the unknown member forces

acting on the joints free-body diagram to be in

tension (pulling on the pin)

Assume what is believed to be the correct sense

of an unknown member force

In both cases a negative value indicates that thesense chosen must be reversed


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Procedure for Analysis

Draw the free-body diagram of a joint having

at least one known force and at most twounknown forces (may need to first determine

external reactions at the truss supports)


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Orient the x and y axes such that the forces

can be easily resolved into their x and y

components

Apply Fx = 0 and Fy = 0 and solve for

the unknown member forces and verify their

correct sense

Continue to analyze each of the other joints,choosing ones having at most two unknowns

and at least one known force


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Members in compression push on the joint

and members in tension pull on the joint

Mechanics of Materials and building codes

are used to size the members once the

forces are known


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2 m

2 m

500 N

45oA

B

C

The Method of Joints

Cy = 500 N

FBA

FBC

500 NB

45o

Joint B

x

ySFx= 0:

+500 -FBCsin45

o= 0

FBC= 707.11 N (C)

SFy= 0:+-FBA +FBCcos45

o = 0

FBA= 500 N (T)

Ax = 500 N

Ay = 500 N


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2 m

2 m

500 N

45oA

B

C

Cy = 500 N

Ax = 500 N

Ay = 500 N

Joint A

500 N

500 N

500 N

FAC

SFx= 0:+

500 -FAC= 0

FAC= 500 N (T)


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Zero Force Members

Truss analysis using the method of joints is

greatly simplified if one is able to determine

those members which support no loading

(zero-force members)

These zero-force members are used to

increase stability of the truss during

construction and to provide support if theapplied loading is changed


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If only two members form a truss joint and no

external load or support reaction is applied to

the joint, the members must be zero-force

members.

If three members form a truss for which two

of the members are collinear, the third

member is a zero-force member provided noexternal force or support reaction is applied.


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Zero-force members Frequently the analysis can be simplified by identifying

members that carry no load two typical cases are found

When only two members form a non-collinear joint and

there is no external force or reaction at that joint, then

both members must be zero-forceP

A BC

D

E

TCB

TCD

C

If either TCBor TCD

0, then C cannot be

in equilibrium, since

there is no restoring

force towards theright.

Hence both BC and

CD are zero-load

members here.


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When three members form a truss joint for

which two members are collinear and the

third is at an angle to these, then this thirdmember must be zero-force

in the absence of an external force or reaction

from a supportP

A BC

D

E

TBCTAB

TBD

B

Here, joint B has onlyone force in the

vertical direction.

Hence, this force

must be zero or B

would move (providedthere are no external

loads/reactions)

Also TAB= TBC


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While zero-force members can be removed in this

configuration, care should be taken

any change in the loading can lead to the member carrying aload

the stability of the truss can be degraded by removing the

zero-force member

P

A B C

D

E

You may think that we can

remove AD and BD to make atriangle

This satisfies the statics

requirements

However, this leaves a long

CE member to carry acompressive load. This long

member is highly susceptible

to failure by buckling.


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P

A

B C

DEDx

DyEy

C

FCD

FCB SFx= 0: FCB= 0+

SFy= 0: FCD= 0+

FAE

FAB

Aq

SFy= 0: FABsinq= 0, FAB = 0+SFx= 0: FAE+ 0= 0, FAE = 0

+


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Example 3-4

Using the method of joints, indicate all the members of the truss shown in the

figure below that have zero force.

P

A B

C

DEFG

H


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SOLUTION

P

A B

C

DEFG

H

Joint D

SFy= 0:+ FDCsinq= 0, FDC= 0SFx= 0:

+ FDE+ 0 = 0, FDE= 0x

y

D

q

FDC

FDE

0

FEC

FEF

P

EFEF= 0SFx= 0:

+Joint E

Ax

Ax

Gx

0

00


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P

A B

C

DEFG

H

FHF

FHA FHB

x

y

H

Joint H

SFy= 0:+

FHB= 0

Ax

Ax

Gx

G

Gx FGF

FGAJoint G

SFy= 0:+ FGA= 0

00

0 0

0


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Method of Sections

Based on the principle that if a body is inequilibrium, then any part of the body is alsoin equilibrium

Procedure for analysis Section or cut the truss through the members

where the forces are to be determined

Before isolating the appropriate section, it may be

necessary to determine the trusss externalreactions (then 3 equs. of equilibrium can be usedto solve for unknown memberforces in thesection).


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Draw the free-body diagram of that part of the

sectioned truss that has the least number of

forces acting on it

Establish the sense of the unknown member

forces


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Apply 3 equations of equilibrium trying to avoid

equations that need to be solved simultaneously

Moments should be summed about a point that

lies at the intersection of the lines of action oftwo unknown forces

If two unknown forces are parallelsum forces

perpendicular to the direction of these

unknowns


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C

The Method of Sections

100 N

A

B C D

EFG

2 m

2 m 2 m 2 m

Ex

Dx

Dya

a

100 N

A

B

G

2 m

45

o

FGF

FGC

FBC

+ SMG= 0:100(2) -FBC(2) = 0

FBC= 100 N (T)

SFy= 0:+-100 +FGCsin45o= 0

FGC= 141.42 N (T)

+ SMC= 0:100(4) -FGF(2) = 0

FGF= 200 N (C)


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Example 3-6

Determine the force in members GFand GD of the truss shown in the figure

below. State whether the members are in tension or compression. The reactions at

the supports have been calculated.

6 kN 8 kN 2 kN

Ax= 0

Ay= 9 kN Ey= 7 kN

A

B C D

E

F

G

H

3 m 3 m 3 m 3 m

3 m 4.5 m


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2 kN Ey= 7 kN

D E

F

Section a-a

3 m

6 kN 8 kN 2 kN

Ax= 0

Ay= 9 kN Ey= 7 kN

A

B C D

E

F

G

H

3 m 3 m 3 m 3 m

3 m4.5 m

SOLUTION a

a

FFG

FDG

FDC

+ SMD= 0:FFGsin26.6

o(3.6) + 7(3) = 0,

FFG= -17.83 kN (C)

3 m

26.6o O

+ SMO= 0:- 7(3) + 2(6) +FDGsin56.3

o(6) = 0,

FDG= 1.80 kN (C)

26.6o

56.3o


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Example 3-7

Determine the force in membersBCand MC of the K-truss shown in the figure

below. State whether the members are in tension or compression. The reactions atthe supports have been calculated.

AB C D E F

G

HIJKL

M N O P

5.34 kN 6.67 kN 8 kN 7.11 kN12.9 kN

03 m

3 m

4.6 m 4.6 m 4.6 m 4.6 m 4.6 m 4.6 m

a


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A

B C D E FG

HIJKL

M N O P

5.34 kN 6.67 kN 8 kN 7.11 kN12.9 kN

03 m

3 m

4.6 m 4.6 m 4.6 m 4.6 m 4.6 m 4.6 m

SOLUTIONa

a

Section a-a

A

B

L

5.34 kN12.9 kN

4.6 m

6 m

FLK

FBM

FLM

FBC

+ SML= 0:F

BC(6) - 12.9(4.6) = 0,

FBC= 9.89 kN (T)


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A

B C D E F

G

HIJKL

M N O P

5.34 kN 6.67 kN 8 kN 7.11 kN12.9 kN

03 m

3 m

4.6 m 4.6 m 4.6 m 4.6 m 4.6 m 4.6 m

b

b

C D E FG

HIJK

N O P

6.67 kN 8 kN7.11 kN

3 m

3 m

4.6 m 4.6 m 4.6 m 4.6 m

FKL

FKMFCM

9.89 kN

+ SMK= 0: -FCMcos33.1o(6) - 9.89(6) - 8(4.6) + 7.11(18.4) = 0FCM= 6.90 kN (T)

33.1o


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