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Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931

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Chapter
Section
BuyFindarrow_forward

Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931
Chapter 12, Problem 21P
Textbook Problem
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Determine the approximate axial forces, shears, and moments for all the members of the frames shown in Figs. P12.6 through P12.13 by using the cantilever method.

Chapter 12, Problem 21P, Determine the approximate axial forces, shears, and moments for all the members of the frames shown

FIG. P12.13, P12.21

To determine

Find the approximate axial forces, shears, and moments for the all members of the frames using cantilever method.

Explanation of Solution

Given information:

The axial force acting at point M (HM) is 7.5 k.

The axial force acting at point I (HI) is 15 k.

The axial force acting at point E (HE) is 15 k.

The vertical distance of the member JM and KL (L1) is 16 ft.

The vertical distance of the member EI, FJ, GK, and HL (L2) is 16 ft.

The vertical distance of the member AE, BF, CG, and DH (L3) is 16 ft.

The horizontal distance of the members AB, EF, and IJ (l1) is 30 ft.

The horizontal distance of the members BC, FG, and JK (l2) is 20 ft.

The horizontal distance of the members CD, GH, and KL (l3) is 30 ft.

Take the counterclockwise moment is positive and clockwise moment is negative.

The axial force in horizontal direction, towards right is positive and towards left side is negative.

The axial force in vertical direction, towards upward is positive and towards downward is negative.

Calculation:

Insert the internal hinges at the midpoints of all the members of the given frame to obtain the simplified frame for approximate analysis.

Draw the simplified frame as in Figure (1).

For the calculation of column axial forces of story of the frame, pass an imaginary section aa through the internal hinges at the midheights of columns JM and KN, pass an imaginary section bb through the internal hinges at the midheights of columns EI, FJ, GK, and HL, and pass an imaginary section cc through the internal hinges at the midheights of columns AE, BF, CG, and DH.

Draw the free body diagram of the frame portion with the passed imaginary lines as in Figure (2).

Column axial forces:

Above section aa:

Draw the free body diagram of the frame portion above the section aa as in Figure (3).

Determine the location of the centroid from support A using the relation.

x¯=AxA=Ax1+Ax2+Ax3+Ax44A        (1)

Here, A is the area of the column sections, x1 is the initial distance, x2 is the distance for the support B from support A, x3 is the distance for the support C from support A, and x4 is the distance for the support D from support A.

Substitute 0 ft for x1, 30 ft for x2, 50 ft for x3, and 80 ft for x4 in Equation (1).

x¯=A(0)+A(30)+A(50)+A(80)4A=40ft

The location of the centroid from the column member JM is 10 ft.

The given lateral load is acting on the frame to the right, therefore the axial force in column JM located to the left of the centroid, must be tensile whereas the axial force in column KN placed to the right of the centroid, must be compressive.

Determine the axial force in the column members JM and KN using equilibrium conditions.

Take moment about point O.

MO=0Q3×l2HM×L12=0

Substitute 20 ft for l2, 20 k for HM, and 16 ft for L1.

Q3×207.5×162=020Q3=60Q3=6020Q3=3k

The axial force in the column members JM and KN is QJM=3k() and QKN=3k().

Draw the free body diagram of the frame portion above the section aa with the axial forces in the column members as in Figure (4).

Draw the free body diagram of the frame portion above the section bb as in Figure (5).

Consider the axial forces in the columns are to be linearly proportional to their distances from centroid.

The given lateral load is acting on the frame to the right, therefore the axial force in column EI and FJ located to the left of the centroid, must be tensile whereas the axial force in column GK and HL placed to the right of the centroid, must be compressive.

Refer Figure (5).

Determine the relationship in column axial force between the member EI and FJ using the relation.

QFJ=x¯l1x¯QEI

Substitute 40 ft for x¯ and 30 ft for l1.

QFJ=403040QEI=1040QEI=QEI4

Determine the relationship in column axial force between the member EI and GK using the relation.

QGK=(l1+l2)x¯x¯QEI

Substitute 30 ft for l1, 20 ft for l2, and 40 ft for x¯.

QGK=(30+20)4040QEI=1040QEI=QEI4

Determine the axial force in the column members EI, FJ, GK, and HL using equilibrium conditions.

Take moment about point P.

MP=0QHL×(l1+l2+l3)+QGK×(l1+l2)QFJ×l1HM×(L1+L22)HI×L22=0

Substitute QEI for QHL, 30 ft for l1, 20 ft for l2, 30 ft for l3, QEI4 for QGK, QEI4 for QFJ, 7.5 k for HM, 16 ft for L1, 15 k for HI, and 16 ft for L2.

QEI×(30+20+30)+14QEI×(20+30)14QEI×307.5×(16+162)15×162=080QEI+504QEI304QEI=300QEI=30085QEI=3.53k

Determine the axial force in the column members FJ.

QFJ=14QEI

Substitute 3.53 k for QEI.

QFJ=14×3.53=0.882k()

Determine the axial force in the column members GK.

QGK=14QEI

Substitute 3.53 k for QEI.

QGK=14×3.53=0.882k()

Determine the axial force in the column members HL.

QHL=QEI

Substitute 3.53 k for QEI.

QHL=3.53k()

Draw the free body diagram of the frame portion above the section bb with the axial forces in the column members as in Figure (6).

Draw the free body diagram of the frame portion above the section cc as in Figure (7).

The given lateral load is acting on the frame to the right, therefore the axial force in column AE and BF located to the left of the centroid, must be tensile, whereas the axial force in column CG and DH placed to the right of the centroid, must be compressive.

Determine the relationship in column axial force between the member AE and BF using the relation.

QBF=x¯l1x¯QAE

Substitute 40 ft for x¯ and 30 ft for l1.

QBF=403040QAE=1040QAE=QAE4

Determine the relationship in column axial force between the member AE and CG using the relation.

QCG=(l1+l2)x¯x¯QAE

Substitute 30 ft for l1, 20 ft for l2, and 40 ft for x¯.

QCG=(30+20)4040QAE=1040QAE=QAE4

Determine the axial force in the column members AE, BF, CG, and DH using equilibrium conditions.

Take moment about point Q.

MQ=0{QDH×(l1+l2+l3)+QCG×(l1+l2)QBF×l1HM×(L1+L2+L32)HI×(L2+L32)HE×L32}=0

Substitute QAE for QDH, 30 ft for l1, 20 ft for l2, 30 ft for l3, QAE4 for QCG, QAE4 for QBF, 7.5 k for HM, 16 ft for L1, 16 ft for L2, 16 ft for L3, 15 k for HI, and 15 k for HE.

{QAE×(30+20+30)+14QAE×(20+30)14QAE×307.5×(16+16+162)15×(16+162)15×162}=080QAE+504QAE304QAE=780QAE=78085QAE=9.18k()

Determine the axial force in the column members BF.

QBF=14QAE

Substitute 9.18 k for QAE.

QBF=14×9.18=2.29k()

Determine the axial force in the column members CG.

QCG=14QAE

Substitute 9.18 k for QAE.

QCG=14×9.18=2.29k()

Determine the axial force in the column members DH.

QDH=QAE

Substitute 2.29 k for QAE.

QDH=2.29k()

Draw the free body diagram of the frame portion above the section bb with the axial forces in the column members as in Figure (8).

Girder shear and moments:

Consider girder MN.

Determine the shear at upper left end joint M using equilibrium equation.

FY=0SMNQJM=0

Substitute 3 k for QJM.

SMN+3=0SMN=3kSMN=3k()

Determine the shear at upper right end joint N using equilibrium equation.

FY=0SMN+SNM=0

Substitute 3 k for SNM.

3+SNM=0SNM=3k()

Determine the moment at left end of the girder MN using equilibrium equations.

MMN=SMN×l22

Substitute 3 k for SMN and 20 ft for l2.

MMN=3×202=30k-ft(Clockwise)

Determine the moment at right end of the girder NM using equilibrium equations.

Take moment about point M.

MM=0MMN+SNM×l2+MNM=0

Substitute 30k-ft for MMN, 3 k for SNM, and 20 ft for l2.

30+3×20+MNM=0MNM=30k-ftMNM=30k-ft(Clockwise)

Consider girder IJ.

Determine the shear at upper left end joint I using equilibrium equation.

FY=0SIJ+QIE=0

Substitute 3.53 k for QIE.

SIJ+3.53=0SIJ=3.53kSIJ=3.53k()

Determine the shear at upper right end joint J using equilibrium equation.

FY=0SIJ+SJI=0

Substitute 3.53 k for SIJ.

3.53+SJI=0SJI=3.53k()

Determine the moment at left end of the girder IJ using equilibrium equations.

MIJ=SIJ×l12

Substitute 3.53 k for SIJ and 30 ft for l1.

MIJ=3.53×30253k-ft(Clockwise)

Determine the moment at right end of the girder IJ using equilibrium equations.

Take moment about point I.

MI=0MIJ+SJI×l1+MJI=0

Substitute 53k-ft for MIJ, 3.53 k for SJI, and 30 ft for l1.

53+3.53×30+MJI=0MJI=53k-ftMJI=53k-ft(Clockwise)

Consider girder JK.

Determine the shear at left end joint J using equilibrium equation.

FY=0SJI+SJK+QJFQJM=0

Substitute 3.53 k for SJI, 0.882 k for QJF, and 3 k for QJM.

3.53+SJK+0.8823=0SJK=1.41kSJK=1.41k()

Determine the shear at right end joint K using equilibrium equation.

FY=0SJK+SKJ=0

Substitute 1.41 k for SJK.

1.41+SJK=0SJK=1.41k()

Determine the moment at left end of the girder JK using equilibrium equations.

MJK=SJK×l22

Substitute 1.41 k for SJK and 20 ft for l2.

MJK=1.41×202=14.1k-ft=14.1k-ft(Clockwise)

Determine the moment at right end of the girder JK using equilibrium equations.

Take moment about point J.

MJ=0MJK+SKJ×l2+MKJ=0

Substitute 14.1k-ft for MJK, 1.41 k for SKJ, and 20 ft for l2.

14.1+1.41×20+MKJ=0MKJ=14.1k-ftMKJ=14.1k-ft(Clockwise)

Consider girder KL.

Determine the shear at left end joint K using equilibrium equation.

FY=0SKJ+SKLQKG+QKN=0

Substitute 1.41 k for SKJ, 0.882 k for QKG, and 3 k for QKN.

1.41+SKL0.882+3=0SKL=3.53kSKL=3.53k()

Determine the shear at right end joint L using equilibrium equation.

FY=0SKL+SLK=0

Substitute 3.53 k for SKL.

3.53+SKL=0SKL=3.53k()

Determine the moment at left end of the girder KL using equilibrium equations.

MKL=SKL×l32

Substitute 3.53 k for SKL and 30 ft for l3.

MKL=3.53×30253k-ft=53k-ft(Clockwise)

Determine the moment at right end of the girder JK using equilibrium equations.

Take moment about point K.

MK=0MKL+SLK×l3+MLK=0

Substitute 53k-ft for MJK, 3.53 k for SLK, and 30 ft for l3.

53+3.53×30+MLK=0MLK=53k-ftMLK=53k-ft(Clockwise)

Consider girder EF.

Determine the shear at left end joint E using equilibrium equation.

FY=0SEFQEI+QEA=0

Substitute 3.53 k for QEI and 9.18 k for QEA.

SEF3.53+9.18=0SEF=5.65kSEF=5.65k()

Determine the shear at right end joint F using equilibrium equation.

FY=0SEF+SFE=0

Substitute 5.65 k for SEF.

5.65+SFE=0SFE=5.65k()

Determine the moment at left end of the girder EF using equilibrium equations.

MEF=SEF×l12

Substitute 5.65 k for SEF and 30 ft for l1.

MEF=5.65×302=84.75k-ft=84.75k-ft(Clockwise)

Determine the moment at right end of the girder EF using equilibrium equations.

Take moment about point E.

ME=0MEF+SFE×l1+MFE=0

Substitute 84.75k-ft for MEF, 5.65 k for SFE, and 30 ft for l1.

84.75+5.65×30+MFE=0MFE=84.75k-ftMFE=84.75k-ft(Clockwise)

Consider girder FG.

Determine the shear at left end joint F using equilibrium equation.

FY=0SFE+SFGQFJ+QFB=0

Substitute 5.65 k for SFE, 0.882 k for QFJ, and 2.29 k for QFB.

5.65+SFG0.882+2.29=0SFG=7.06kSFG=7.06k()

Determine the shear at right end joint G using equilibrium equation.

FY=0SFG+SGF=0

Substitute 7.06 k for SFG.

7.06+SGF=0SGF=7.06k()

Determine the moment at left end of the girder FG using equilibrium equations.

MFG=SFG×l22

Substitute 7.06 k for SFG and 20 ft for l2.

MFG=7.06×202=70.6k-ft=70.6k-ft(Clockwise)

Determine the moment at right end of the girder FG using equilibrium equations.

Take moment about point F.

MF=0MFG+SGF×l2+MGF=0

Substitute 70.6k-ft for MFG, 7.06 k for SGF, and 20 ft for l2.

70.6+7.06×20+MGF=0MGF=70.6k-ftMGF=70.6k-ft(Clockwise)

Consider girder GH.

Determine the shear at left end joint G using equilibrium equation.

FY=0SGH+SGF+QGKQGC=0

Substitute 7.06 k for SGF, 0.882 k for QGK, and 2.29 k for QGC.

SGH+7.06+0.8822.29=0SGH=5.65kSGH=5.65k()

Determine the shear at right end joint H using equilibrium equation.

FY=0SGH+SHG=0

Substitute 5.65 k for SGH.

5.65+SHG=0SHG=5.65k()

Determine the moment at left end of the girder GH using equilibrium equations.

MGH=SGH×l32

Substitute 5.65 k for SGH and 30 ft for l3.

MGH=5.65×302=84.75k-ft=84.75k-ft(Clockwise)

Determine the moment at right end of the girder GH using equilibrium equations

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