Fundamentals of Structural Analysis
Fundamentals of Structural Analysis
5th Edition
ISBN: 9780073398006
Author: Kenneth M. Leet Emeritus, Chia-Ming Uang, Joel Lanning
Publisher: McGraw-Hill Education
Question
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Chapter 9, Problem 27P
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 (A) of all the bars is 1000mm2.

Identify Zero Force Members:

In a two member forces, 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.

Calculation:

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

Fundamentals of Structural Analysis, Chapter 9, Problem 27P , additional homework tip  1

Refer Figure 1.

Find the length of the member AB and BC as follows:

Consider triangle ABD.

AB=(4+2)2+82=62+82=100=10m

By symmetry of truss,

AB=BC=10m

Find the length of the member AD and DC as follows:

Consider triangle ADC.

AD=22+82=22+82=68=8.246m

By symmetry of truss,

AD=DC=8.246m

Find the angle θ1 as follows:

tanθ1=28θ1=tan1(28)θ1=14.036°

Find the angle θ2 as follows:

tanθ2=68θ2=tan1(68)θ2=36.869°

Consider the horizontal reaction at C as the redundant.

Remove the redundant at C to get the released structure.

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

Fundamentals of Structural Analysis, Chapter 9, Problem 27P , additional homework tip  2

Refer Figure 2.

Find the reactions at A, D, and C as follows:

Apply Equation of Equilibrium,

Fx=0Ax1+30=0Ax1=30kN

MA=0Cy1×16(18×8)(30×6)=0Cy1=(32416)Cy1=20.25kN

Fy=0Ay1+Cy1=18Ay1=18Cy1Ay1=18(20.25)Ay1=2.25kN

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

Refer Figure 2.

Consider Joint C.

Apply Equation of Equilibrium,

Fx=0FCD1cosθ1FCB1cosθ2=0FCD1cos(14.036°)FCB1cos(36.869°)=0FCD1=FCB1[cos(36.869°)cos(14.036°)]FCD1=FCB1(0.80.97)FCD1=0.824FCB1

Fy=0FCD1sinθ1+FCB1sinθ2+Cy1=0(0.824FCB1)sin(14.036°)+FCB1sin(36.869°)+20.25=0(0.824FCB1)sin(14.036°)+FCB1sin(36.869°)+20.25=00.2FCB1+0.6FCB1=20.25

FCB1=20.250.4FCB1=50.625kNFCB1=50.625kN(C)

FCD1=0.824FCB1=0.824(50.625)=41.715kN(T)

Consider joint B.

Apply Equation of Equilibrium,

Fx=030FAB1cosθ2+FBC1cosθ2=030FAB1cos(36.869°)+FBC1cos(36.869°)=0300.8FAB1+0.8FBC1=0300.8FAB1+0.8(50.625)=0FAB1=10.50.8FAB1=13.125kN(C)

Consider joint D.

Apply Equation of Equilibrium,

Fx=0FAD1cosθ2+FDC1cosθ2=0FAD1cosθ2=FDC1cosθ2FAD1=FDC1FAD1=41.75kN(T)

Fy=0FDB118FAD1sinθ1FDC1sinθ1=0FDB11841.75sin(14.036°)41.75sin(14.036°)=0FDB11820.2513=0FDB1=38.25kN(T)

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

MembersFP forces
AB13.125kN(C)
BC50.625kN(C)
AD41.715kN(T)
DC41.715kN(T)
DB38.25kN(T)

Table 1

Show the released structure loaded with unit horizontal load at C as shown in Figure 2.

Fundamentals of Structural Analysis, Chapter 9, Problem 27P , additional homework tip  3

Refer Figure 3.

By symmetry of the truss,

Fx=0Ax+1=0Ax=1kN

Consider Joint D.

Apply Equation of Equilibrium,

Fx=0FCDcosθ1FCBcosθ2=1FCDcos(14.036°)FCBcos(36.869°)=1        (1)

Fy=0FCDsinθ1+FCBsinθ2+Cy=0FCDsin(14.036°)+FCBsin(36.869°)=0        (2)

Solve Equation (1) and (2).

FCD=1.546kN(T)FCB=0.625kN(C)

By symmetry of the truss,

FAD=FCD=1.546kN(T)FAB=FCB=0.625kN(C)

Consider Joint B.

Apply Equation of Equilibrium,

Fy=0FABsinθ2FBCcosθ2FBD=0FBD=FCDsin(36.869°)+FCBsin(36.869°)FBD=(0.625)sin(36.869°)+(0.625)sin(36.869°)FBD=0.75FBD=0.75kN(T)

Show the Q forces in the members of the truss due to the unit load at C as shown in Table 2.

MembersFQ forces
AB0.625kN(C)
BC0.625kN(C)
CD1.546kN(T)
AC1.546kN(T)
DB0.75kN(T)

Table 2

Find the deflection ΔCO of the released structure due to the applied load and the deflection δCC of the released structure due to the unit load as follows:

MembersFP forces(kN)FQ forces(kN)L(m)ΔCX=FQFPLAEδCC=FQ2LAE
AB13.125kN0.625kN1082.03AE3.90625AE
BC50.625kN0.625kN10316.4063AE3.90625AE
AD41.715kN1.546kN8.246531.796AE19.7089AE
DC41.715kN1.546kN8.246531.796AE19.7089AE
DB38.25kN0.75kN4114.75AE2.25AE

Table 3

Refer Table 3.

ΔCX=FQFPLAEΔCX=1576AE

δCC=FQ2LAE=49.48AE

Find the horizontal reaction at C as follows:

ΔC=ΔCX+δCCRC0=1576AE+49.48AERCRC=157649.48RC=31.89kN

Thus, the horizontal reaction at C is RC=31.89kN_.

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

MembersFP(kN)FQ(kN)RC()(kN)Fi=FP+FQRC (kN)
AB13.125kN0.625kN-31.896.80
BC50.625kN0.625kN-31.89-30.7
AD41.715kN1.546kN-31.89-7.5
DC41.715kN1.546kN-31.89-7.5
DB38.25kN0.75kN-31.8914.34

Table 4

Show the final support reaction as shown in Table 5.

Support Reactions
MembersPi(kN)FQ(kN) forcesRC(kN)()Reaction=Pi+FQRCkN
HA-30-1-31.891.89
RA-2.250-31.89-2.25
RC20.251-31.8920.25

Table 5

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

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