Structural Analysis
Structural Analysis
6th Edition
ISBN: 9781337630931
Author: KASSIMALI, Aslam.
Publisher: Cengage,
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Chapter 16, Problem 31P
To determine

Find the member end moments and reactions for the frames.

Expert Solution & Answer
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Answer to Problem 31P

The reaction at point A (Ax), (Ay) and B (Bx), (By) are 13.5k_, 10.1k_, 13.5k_, and 10.1k_ respectively.

The end moment at the member (MAC,BD), (MCA,DB) (MCD,DC), (MCE,DF), (MEC,FD), and (MEF,FE) are 119kft_, 83.5kft_, 106.9kft_, 23.3kft_, 44.2kft_, and 44.2kft_ respectively.

Explanation of Solution

Fixed end moment:

Formula to calculate the relative stiffness for fixed support IL and for roller support (34)(IL).

Formula to calculate the fixed moment for point load with equal length are PL8.

Formula to calculate the fixed moment for point load with unequal length are Pab2L2 and Pa2bL2.

Formula to calculate the fixed moment for UDL is WL212.

Formula to calculate the fixed moment for UVL are WL230 and WL220.

Formula to calculate the fixed moment for deflection is 6EIΔL2

Calculation:

Consider the elastic modulus E of the frame is constant.

Show the free body diagram of the entire frame as in Figure 1.

Structural Analysis, Chapter 16, Problem 31P , additional homework tip  1

Refer Figure 1,

Calculate the relative stiffness KCA, KDB, KCE, and KDF for member CA, DB, CE and DF of the frame as below:

KCA=KDB=KCE=KDF=I15

Calculate the relative stiffness KCD and KEF for member CD and EF of the frame as below:

KCD=KEF=2I30=I15

Calculate the distribution factor DFCA for member CA of the frame.

DFCA=KCAKCA+KCD+KCE

Substitute I15 for KCA, I15 for KCD, and I15 for KCE.

DFCA=I15I15+I15+I15=13

Calculate the distribution factor DFCD for member CD of the frame.

DFCD=KCDKCA+KCD+KCE

Substitute I15 for KCA, I15 for KCD, and I15 for KCE.

DFCD=I15I15+I15+I15=13

Calculate the distribution factor DFCE for member CE of the frame.

DFCE=KCEKCA+KCD+KCE

Substitute I15 for KCA, I15 for KCD, and I15 for KCE.

DFCE=I15I15+I15+I15=13

Check for sum of distribution factor as below:

DFCA+DFCD+DFCE=1

Substitute 13 for DFCA, DFCD , and DFCE.

13+13+13=1

Hence, OK.

Calculate the distribution factor DFEC for member EC of the frame.

DFEC=KECKEC+KEF

Substitute I15 for KEC and I15 for KEF.

DFEC=I15I15+I15=12

Calculate the distribution factor DFEF for member EF of the frame.

DFEF=KEFKEC+KEF

Substitute I15 for KEC and I15 for KEF.

DFEF=I15I15+I15=12

Check for sum of distribution factor as below:

DFEC+DFEF=1

Substitute 12 for DFEC and 12 for DFEF.

12+12=1

Hence, OK.

Calculate the distribution factor DFFE for member EC of the frame.

DFFE=KFEKFD+KFE

Substitute I15 for KFD and I15 for KFE.

DFFE=I15I15+I15=12

Calculate the distribution factor DFFD for member FD of the frame.

DFFD=KFDKFD+KFE

Substitute I15 for KFD and I15 for KFE.

DFFD=I15I15+I15=12

Check for sum of distribution factor as below:

DFFE+DFFD=1

Substitute 12 for DFFE and 12 for DFFD.

12+12=1

Hence, OK.

Calculate the distribution factor DFDF for member of the frame.

DFDF=KDFKDF+KDC+KDB

Substitute I15 for KDF, I15 for KDC, and I15 for KDB.

DFDF=I15I15+I15+I15=13

Calculate the distribution factor DFDC for member DC of the frame.

DFDC=KDCKDF+KDC+KDB

Substitute I15 for KDF, I15 for KDC, and I15 for KDB.

DFDC=I15I15+I15+I15=13

Calculate the distribution factor DFDB for member DB of the frame.

DFDB=KDBKDF+KDC+KDB

Substitute I15 for KDF, I15 for KDC, and I15 for KDB.

DFDB=I15I15+I15+I15=13

Check for sum of distribution factor as below:

DFDF+DFDC+DFDB=1

Substitute 13 for DFDF, 13 for DFDC, and 13 for DFDB.

13+13+13=1

Hence, OK.

Show the translation Δ1 of the frame as in figure 2.

Structural Analysis, Chapter 16, Problem 31P , additional homework tip  2

Write the expression to calculate the Fixed-end moment of the member.

FEMAC=FEMCA=FEMBD=FEMDB=6EIΔ1(15)2

Assume the Fixed-end moment of the members AC, CA, BD and DB as 100kft.

FEMAC=FEMCA=FEMBD=FEMDB=100kft

Write the expression to calculate the Fixed-end moment of the member.

FEMCE=FEMEC=FEMDF=FEMFD=6EIΔ1(15)2

Assume the Fixed-end moment of the members CE, EC, DF and FD as 100kft.

FEMCE=FEMEC=FEMDF=FEMFD=100kft

Show the calculation of MQ1 moments using moment distribution method as in Table 1.

Structural Analysis, Chapter 16, Problem 31P , additional homework tip  3

Show the free body diagram of the frame with unknown reaction Q21 as in Figure 3.

Structural Analysis, Chapter 16, Problem 31P , additional homework tip  4

Consider member EC:

Calculate the horizontal reaction at the joint C by taking moment about point E.

+ME=0Cx1(15)85.261.8=0Cx1(15)=147Cx1=14715Cx1=9.8k

Calculate the horizontal reaction at joint E by resolving the horizontal equilibrium.

+Fx=0Ex1+Cx1=0Ex1+9.8=0Ex1=9.8k

Consider member FD:

Calculate the horizontal reaction at the joint D by taking moment about point F.

+MF=0Dx1(15)85.261.8=0Dx1(15)=147Dx1=14715Dx1=9.8k

Calculate the horizontal reaction at joint F by resolving the horizontal equilibrium.

+Fx=0Fx1+Dx1=0Fx1+9.8=0Fx1=9.8k

Calculate the reaction Q21 using the relation:

Q21=Ex1+Fx1=9.8+9.8=19.6k

Show the free body diagram of the frame with unknown reaction Q11 as in Figure 4.

Structural Analysis, Chapter 16, Problem 31P , additional homework tip  5

Consider member AC:

Calculate the horizontal reaction at the joint A by taking moment about point C.

+MC=0Ax1(15)+97.1+94.2=0Ax1(15)=191.3Ax1=191.315Ax1=12.75k

Calculate the horizontal reaction at joint C by resolving the horizontal equilibrium.

+Fx=0Ax1+Cx11=0Cx1112.75=0Cx11=12.75k

Consider member BD:

Calculate the horizontal reaction at the joint B by taking moment about point D.

+MD=0Bx1(15)+94.2+97.1=0Bx1(15)=191.3Bx1=191.315Bx1=12.75k

Calculate the horizontal reaction at joint D by resolving the horizontal equilibrium.

+Fx=0Bx1+Dx11=012.75+Dx11=0Dx11=12.75k

Calculate the reaction Q11 using the relation:

Q11=Cx11+Dx11+Ex1+Fx1=12.75+12.75+9.8+9.8=45.1k

Show the translation Δ2 of the frame as in Figure 5.

Structural Analysis, Chapter 16, Problem 31P , additional homework tip  6

Write the expression to calculate the Fixed-end moment of the member.

FEMCE=FEMEC=FEMDF=FEMFD=6EIΔ1(15)2

Assume the Fixed-end moment of the members CE, EC, DF and FD as 100kft.

FEMCE=FEMEC=FEMDF=FEMFD=100kft

Show the calculation of MQ2 moments using moment distribution method as in Table 2.

Structural Analysis, Chapter 16, Problem 31P , additional homework tip  7

Show the free body diagram of the frame with unknown reaction Q22 as in Figure 6.

Structural Analysis, Chapter 16, Problem 31P , additional homework tip  8

Consider member EC:

Calculate the horizontal reaction at the joint C by taking moment about point E.

+ME=0Cx2(15)+58.9+53=0Cx2(15)=111.9Cx2=111.915Cx2=7.46k

Calculate the horizontal reaction at joint E by resolving the horizontal equilibrium.

+Fx=0Ex2+Cx2=0Ex27.46=0Ex2=7.46k

Consider member FD:

Calculate the horizontal reaction at the joint D by taking moment about point F.

+MF=0Dx2(15)+53+58.9=0Dx2(15)=111.9Dx2=111.915Dx2=7.46k

Calculate the horizontal reaction at joint F by resolving the horizontal equilibrium.

+Fx=0Fx2+Dx2=0Fx27.46=0Fx2=7.46k

Calculate the reaction Q21:

Q21=Ex2+Fx2=7.46+7.46=14.9k

Show the free body diagram of the frame with unknown reaction Q12 as in Figure 7.

Structural Analysis, Chapter 16, Problem 31P , additional homework tip  9

Consider member AC:

Calculate the horizontal reaction at the joint A by taking moment about point C.

+MC=0Ax2(15)11.823.5=0Ax2(15)=35.3Ax2=35.315Ax2=2.35k

Calculate the horizontal reaction at joint C by resolving the horizontal equilibrium.

+Fx=0Ax2+Cx22=0Cx22+2.35=0Cx22=2.35k

Consider member BD:

Calculate the horizontal reaction at the joint B by taking moment about point D.

+MD=0Bx2(15)23.511.8=0Bx2(15)=35.3Bx2=35.315Bx2=2.35k

Calculate the horizontal reaction at joint D by resolving the horizontal equilibrium.

+Fx=0Bx2+Dx22=02.35+Dx22=0Dx11=2.35k

Calculate the reaction Q11 using the relation:

Q11=Cx22+Dx22+Ex2+Fx2=2.35+2.35+7.46+7.46=19.6k

Write the equation by superimposing the horizontal forces at joints C,

45.1c119.6c2=18 (1).

Write the equation by superimposing the horizontal forces at joints E,

19.6c1+14.9c2=9 (2).

Calculate the value of c1 and c2 by solving equation (1) and (2).

c1=1.545c2=2.636

Calculate the actual member end moments of the member AC and BD:

MAC,BD=c1(MQ1)AC,BD+c2(MQ2)AC,BD

Substitute 97.1kft for (MQ1)AC,BD, 11.8kft for (MQ2)AC,BD,1.545 for c1 and 2.636 for c2.

MAC,BD=(1.545×97.1)+2.636×(11.8)=15031=119kft

Calculate the actual member end moments of the member CA and DB:

MCA,DB=c1(MQ1)CA,DB+c2(MQ2)CA,DB

Substitute 94.2kft for (MQ1)CA,DB, 23.5kft for (MQ2)CA,DB,1.545 for c1 and 2.636 for c2.

MCA,DB=(1.545×94.2)+2.636×(23.5)=145.562=83.5kft

Calculate the actual member end moments of the member CD and DC:

MCD,DC=c1(MQ1)CD,DC+c2(MQ2)CD,DC

Substitute 8.8kft for (MQ1)CD,DC, 35.4kft for (MQ2)CD,DC,1.545 for c1 and 2.636 for c2.

MCD,DC=(1.545×(8.8))+2.636×(35.4)=13.693.3=106.9kft

Calculate the actual member end moments of the member CE and DF:

MCE,DF=c1(MQ1)CE,DF+c2(MQ2)CE,DF

Substitute 85.2kft for (MQ1)CE,DF, 58.9kft for (MQ2)CE,DF,1.545 for c1 and 2.636 for c2.

MCE,DF=(1.545×(85.2))+2.636×(58.9)=132+155.3=23.3kft

Calculate the actual member end moments of the member EC and FD:

MEC,FD=c1(MQ1)EC,FD+c2(MQ2)EC,FD

Substitute 61.8kft for (MQ1)EC,FD, 53kft for (MQ2)EC,FD,1.545 for c1 and 2.636 for c2.

MEC,FD=(1.545×(61.8))+2.636×(53)=95.48+139.71=44.2kft

Calculate the actual member end moments of the member EC and FD:

MEF,FE=c1(MQ1)EF,FE+c2(MQ2)EF,FE

Substitute 61.7kft for (MQ1)EF,FE, 52.8kft for (MQ2)EF,FE,1.545 for c1 and 2.636 for c2.

MEF,FE=(1.545×(61.8))+2.636×(53)=95.48139.71=44.2kft

Show the section free body diagram of the member as in Figure 8.

Structural Analysis, Chapter 16, Problem 31P , additional homework tip  10

Consider member EF:

Calculate the vertical reaction at the joint E by taking moment about point F.

+MF=0Ey(30)44.244.2=0Ey(30)=88.4Ey=88.430Ey=2.95k

Calculate the vertical reaction at joint F by resolving the horizontal equilibrium.

+Fy=0Ey+Fy=02.95+Fy=0Fy=2.95k

Consider member CD:

Calculate the vertical reaction at the joint C by taking moment about point D.

+MD=0Cy(30)106.9106.9=0Cy(30)=213.8Cy=213.830Cy=7.13k

Calculate the vertical reaction at joint D by resolving the horizontal equilibrium.

+Fy=0Cy+Dy=07.13+Dy=0Dy=7.13k

Calculate the reaction at joint A using the relation:

Ay=Ey+Cy=2.95k+7.13k=10.1k

Calculate the reaction at joint B using the relation:

By=Fy+Dy=2.95k+7.13k=10.1k

Consider member AC:

Calculate the horizontal reaction at the joint A by taking moment about point C.

+MC=0Ax(15)+119+83.5=0Ax(15)=202.5Ax=202.515Ax=13.5k

Consider member BD:

Calculate the horizontal reaction at the joint B by taking moment about point D.

+MD=0Bx(15)+119+83.5=0Bx(15)=202.5Bx=202.515Bx=13.5k

Show the reactions of the frame as in Figure 9.

Structural Analysis, Chapter 16, Problem 31P , additional homework tip  11

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