Fundamentals Of Structural Analysis:
Fundamentals Of Structural Analysis:
5th Edition
ISBN: 9781260083330
Author: Leet, Kenneth
Publisher: MCGRAW-HILL HIGHER EDUCATION
Question
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Chapter 9, Problem 29P
To determine

Find the reactions and all bar forces for the truss.

Expert Solution & Answer
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Explanation of Solution

Given information:

The Young’s modulus E of the beam is 200GPa.

The area of all the bars is 1000mm2.

Calculation:

Show the free body diagram of the truss as shown in Figure 1.

Fundamentals Of Structural Analysis:, Chapter 9, Problem 29P , additional homework tip  1

Refer Figure 1.

Find the length of the inclined members of the truss as follows:

Consider the length of the inclined members AE, EB, BD, DC is l.

l=32+42=25=5m

By symmetry of truss,

AE=EB=BD=DC=5m

Find the angle θ as follows:

tanθ=43θ=tan1(43)θ=53.130°

Consider the horizontal reaction at D as the redundant.

Remove the redundant at D to get the released structure.

Show the released structure with applied load as shown in Figure 2.

Fundamentals Of Structural Analysis:, Chapter 9, Problem 29P , additional homework tip  2

Refer Figure 2.

Find the reactions at A and D as follows:

Apply Equation of Equilibrium,

Fx=0Ax1=0

MA=0(Dy1×9)(100×3)(60×6)(60×12)=0Dy1=(13809)Dy1=153.3kN

Fy=0Ay1+Dy1=60+60+100Ay1=220Dy1Ay1=220153.3Ay1=66.7kN

Find the forces FP in the bars of released structure due to applied load as follows:

Refer Figure 2.

Consider Joint A.

Apply Equation of Equilibrium,

Fy=0Ay1FAE1sinθ=066.7FAE1sin(53.130°)=066.70.8FAE1=0FAE1=83.3kN(T)

Fx=0Ax1+FAB1+FAE1cosθ=00+FAB1+FAE1cos(53.130°)=0FAB1+0.6(83.3)=0FAB1=50kNFAB1=50kN(C)

Consider joint C.

Apply Equation of Equilibrium,

Fy=0FCD1sinθ60=0FCD1sin(53.130°)60=00.8FCD160=0FCD1=75kN(C)

Fx=0FBC1FCD1cosθ=0FBC1=(75)cos(53.130°)FBC1=45kN(T)

Consider joint B.

Apply Equation of Equilibrium,

Fy=060FBE1sin(53.130°)FBD1sin(53.130°)=00.8FBE1+0.8FBD1=60        (1)

Fx=0FBA1FBE1cos(53.130°)+FBD1cos(53.130°)+FBC1=0(50)0.6FBE1+0.6FBD1+45=00.6FBE1+0.6FBD1=95        (2)

Solve Equation (1) and (2).

FBE1=41.7kN(T)FBD1=116.7kN(C)

Consider joint D.

Apply Equation of Equilibrium,

Fx=0FEA1cosθ+FED1+FEB1cosθ=0(83.3)cos(53.130°)+FED1+(41.7)cos(53.130°)=050+FED1+25=0FED1=25kN(T)

Show the P forces in the members of the truss due to the applied load at D as shown in Table 1.

MembersFP forces (kN)
AB50 (C)
BC45 (T)
CD75 (C)
DE25 (T)
EA83.3 (T)
BE41.7 (T)
BD116.7 (C)

Table 1

Show the released structure loaded with unit horizontal load at D as shown in Figure 3.

Fundamentals Of Structural Analysis:, Chapter 9, Problem 29P , additional homework tip  3

Refer Figure 3.

By symmetry of the truss,

Fx=0Ax1=0Ax=1kN

MA=0Dy×9(1×4)=0Dy=0.444kN

Fy=0Ay+Dy=0Ay+0.444kN=0Ay=0.444kN

Find the FQ forces in the members of the released structure due to unit load as follows:

Refer Figure 3.

Consider Joint A.

Apply Equation of Equilibrium,

Fy=0AyFAEsinθ=00.444FAEsin(53.130°)=00.4440.8FAE=0FAE=0.556kN(C)

Fx=0Ax+FAB+FAEcosθ=01+FAB+FAEcos(53.130°)=0FAB+0.6(0.556)=1FAB=0.664kNFAB=0.664kN(C)

Consider joint C.

In a two member forces BC and CD, if both the members are not parallel and the no external loads or reactions acts at that joint. Then, the force in both the members is zero.

The forces in the member BC and CD is FBC=FCD=0.

Consider joint B.

Apply Equation of Equilibrium,

Fy=0FBEsin(53.130°)FBDsin(53.130°)=00.8FBE+0.8FBD=0FBE=FBD        (3)

Fy=0FBAFBEcos(53.130°)+FBDcos(53.130°)+FBC=0(50)0.6FBE1+0.6FBD1+45=00.6FBE1+0.6FBD1=95        (4)

Solve Equation (1) and (2).

FBE1=41.7kN(T)FBD1=116.7kN(C)

Consider joint D.

Apply Equation of Equilibrium,

Fx=0FEAcosθ+FED+FEBcosθ=0(0.556)cos(53.130°)+FED+(0.556)cos(53.130°)=0(0.556)cos(53.130°)+FED+(0.556)cos(53.130°)=00.3336+FED+0.3336=0FED=0.667kN(C)

Show the Q forces in the members of the truss due to the unit load at D as shown in Table 1.

MembersFQ forces (kN)
AB0.667 (C)
BC0
CD0
DE0.667 (C)
EA0.556 (C)
BE0.556 (T)
BD0.556 (C)

Table 2

Find the deflection ΔDO of the released structure due to the applied load and the deflection δDD of the released structure due to the unit load as follows:

MembersFP forces(kN)FQ forces (kN)L(m)ΔDO=FQFPLAEδDD=FQ2LAE
AB-50-0.6676200.1AE2.669AE
BC450600
CD-750500
DE25-0.6676100.05AE2.669AE
EA83.3-0.5565231.574AE1.5456AE
BE41.70.5565115.926AE1.5456AE
BD-116.7-0.5565324.426AE1.5456AE

Table 3

Refer Table 3.

ΔDO=FQFPLAEΔDO=309AE

δDD=FQ2LAE=10AE

Find the horizontal reaction at D as follows:

ΔD=ΔDD+δDDRD0=309AE+10AERDRD=30910RD=30.9kN()

Thus, the horizontal reaction at D is RD=30.9kN()_.

Find final forces in the truss as shown in Table 4.

MembersFP(kN)FQ(kN)RD()(kN)Fi=FP+FQRD(kN)
AB-50-0.667-30.9-29.4
BC450-30.945
CD-750-30.9-75
DE25-0.667-30.945
EA83.3-0.556-30.9100.5
BE41.70.556-30.924.5
BD-116.7-0.556-30.9-99.5

Table 4

Show the final support reaction as shown in Table 5.

Support Reactions
MembersPi(kN)PQ(kN) forcesRD(kN)()Reaction=Pi+PQRDkN
HA01-30.9-30.9
RA66.7-0.444-30.980.4
RD153.30.444-30.9139.6

Table 5

Thus, the reactions and all bar forces for the truss is tabulate in table 4 and table 5.

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