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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 20P
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 20P, Determine the approximate axial forces, shears, and moments for all the members of the frames shown

FIG. P12.12, P12.20

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 I (HI) is 20 kN.

The axial force acting at point E (HE) is 40 kN.

The vertical distance of the member AE, BF, CG, and DH (L2) is 6 m.

The vertical distance of the member EI, FJ, GK, and HL (L1) is 4 m.

The horizontal distance of the members AB, EF, and IJ (l1) is 8 m.

The horizontal distance of the members BC, FG, and JK (l2) is 6 m.

The horizontal distance of the members CD, GH, and KL (l3) is 8 m.

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 EI, FJ, GK, and HL, and pass an imaginary section bb 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).

Refer Figure 3.

Determine the location of the centroid using the relation.

x¯=AxA=Ax1+Ax2+Ax3+Ax44A        (1)

Substitute 0 m for x1, 8 m for x2, 14 m for x3, and 22 m for x4 in Equation (1).

x¯=A(0)+A(8)+A(14)+A(22)4A=11m

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 HL and GK placed to the right of the centroid, must be compressive.

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

Apply similar triangle rule.

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

QFJ=x¯l1x¯QEI

Substitute 11 m for x¯ and 8 m for l1.

QFJ=11811QEI=311QEI

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

QGK=(l1+l2)x¯x¯QEI

Substitute 8 m for l1, 6 m for l2, and 11 m for x¯.

QGK=(8+6)1111QEI=311QEI

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

Take moment about point M.

MM=0QHL×(l1+l2+l3)+QGK×(l1+l2)QFJ×l1HI×L12=0

Substitute QEI for QHL, 8 m for l1, 6 m for l2, 8 m for l3, 311QEI for QGK, 311QEI for QFJ, 20 kN for HI, and 4 m for L1.

QEI×(8+6+8)+311QEI×(8+6)311QEI×820×42=0242QEI+42QEI24QEI=440QEI=440260QEI=1.69kN

Determine the axial force in the column members FJ.

QFJ=311QEI

Substitute 1.69 kN for QEI.

QFJ=311×1.69=0.46kN()

Determine the axial force in the column members GK.

QGK=311QEI

Substitute 1.69 kN for QEI.

QGK=311×1.69=0.46kN()

Determine the axial force in the column members HL.

QHL=QEI

Substitute 1.69 kN for QEI.

QHL=1.69kN()

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).

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 11 m for x¯ and 8 m for l1.

QBF=11811QAE=311QAE

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

QCG=(l1+l2)x¯x¯QAE

Substitute 8 m for l1, 6 m for l2, and 11 m for x¯.

QCG=(8+6)1111QAE=311QAE

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

Take moment about point N.

MN=0QDH×(l1+l2+l3)+QCG×(l1+l2)QBF×l1HI×(L1+L22)HE×L22=0

Substitute QAE for QDH, 8 m for l1, 6 m for l2, 8 m for l3, 311QAE for QBF, 311QAE for QCG, 20 kN for HI, 4 m for L1, 40 kN for HE, and 6 m for L2.

QAE×(8+6+8)+311QAE×(8+6)311QAE×820×(4+62)40×62=0242QAE+42QAE24QAE=2,860QAE=2,860260QAE=11kN

Determine the axial force in the column members BF.

QBF=311QAE

Substitute 11 kN for QAE.

QBF=311×11=3kN()

Determine the axial force in the column members CG.

QCG=311QAE

Substitute 11 kN for QAE.

QCG=311×11=3kN()

Determine the axial force in the column members DH.

QDH=QAE

Substitute 11 kN for QAE.

QDH=11kN()

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).

Girder shear and moments:

Consider girder IJ.

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

FY=0SIJ+QIE=0

Substitute 1.69 kN for QIE.

SIJ+1.69kN=0SIJ=1.69kNSIJ=1.69kN()

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

FY=0SIJ+SJI=0

Substitute 1.69 kN for SIJ.

1.69+SJI=0SJI=1.69kN()

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

MIJ=SIJ×l12

Substitute 1.69 kN for SIJ and 8 m for l1.

MIJ=1.69×82=6.76kNm(Clockwise)

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

Take moment about point I.

MG=0MIJ+SJI×l1+MJI=0

Substitute 6.76kNm for MIJ, 1.69 kN for SJI, and 8 m for l1.

6.76+1.69×8+MJI=0MJI=6.76kNmMJI=6.76kNm(Clockwise)

Consider girder JK.

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

FY=0SJI+SJK+QJF=0

Substitute 1.69 kN for SJI and 0.46 kN for QJF.

1.69+SJK+0.46=0SJK=2.15kNSJK=2.15kN()

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

FY=0SJK+SKJ=0

Substitute 2.15 kN for SJK.

2.15+SJK=0SJK=2.15kN()

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

MJK=SJK×l22

Substitute 2.15 kN for SJK and 6 m for l2.

MJK=2.15×62=6.45kNm=6.45kNm(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 6.45kNm for MJK, 2.15 kN for SKJ, and 6 m for l2.

6.45+2.15×6+MKJ=0MKJ=6.45kNmMKJ=6.45kNm(Clockwise)

Consider girder KL.

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

FY=0SKJ+SKLQKG=0

Substitute 2.15 kN for SKJ and 0.46 kN for QKG.

2.15+SKL0.46=0SKL=1.69kNSKL=1.69kN()

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

FY=0SKL+SLK=0

Substitute 1.69 kN for SKL.

1.69+SKL=0SKL=1.69kN()

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

MKL=SKL×l32

Substitute 1.69 kN for SKL and 8 m for l3.

MKL=1.69×82=6.76kNm=6.76kNm(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 6.76kNm for MJK, 1.69 kN for SLK, and 8 m for l3.

6.76+1.69×8+MLK=0MLK=6.76kNmMLK=6.76kNm(Clockwise)

Consider girder EF.

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

FY=0SEFQEI+QEA=0

Substitute 1.69 kN for QEI and 11 kN for QEA.

SEF1.69+11=0SEF=9.31kNSEF=9.31kN()

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

FY=0SEF+SFE=0

Substitute 9.31 kN for SEF.

9.31+SFE=0SFE=9.31kN()

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

MEF=SEF×l12

Substitute 9.31 kN for SEF and 8 m for l1.

MEF=9.31×82=37.24kNm=37.24kNm(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 37.24kNm for MEF, 9.31 kN for SFE, and 8 m for l1.

37.24+9.31×8+MFE=0MFE=37.24kNmMFE=37.24kNm(Clockwise)

Consider girder FG.

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

FY=0SFE+SFGQFJ+QFB=0

Substitute 9.31 kN for SFE, 0.46 kN for QFJ, and 3 kN for QFB.

9.31+SFG0.46+3=0SFG=11.85kNSFG=11.85kN()

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

FY=0SFG+SGF=0

Substitute 11.85 kN for SFG.

11.85+SGF=0SGF=11.85kN()

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

MFG=SFG×l22

Substitute 11.85 kN for SFG and 6 m for l2.

MFG=11.85×62=35.55kNm=35.55kNm(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 35.55kNm for MFG, 11.85 kN for SGF, and 6 m for l2.

35.55+11.85×6+MGF=0MGF=35.55kNmMGF=35.55kNm(Clockwise)

Consider girder GH.

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

FY=0SGH+SGF+QGKQGC=0

Substitute 11.85 kN for SGF, 0.46 kN for QGK, and 3 kN for QGC.

SGH+11.85+0.463=0SGH=9.31kNSGH=9.31kN()

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

FY=0SGH+SHG=0

Substitute 9.31 kN for SGH.

9.31+SHG=0SHG=9.31kN()

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

MGH=SGH×l32

Substitute 9.31 kN for SGH and 8 m for l3.

MGH=9.31×82=37.24kNm=37.24kNm(Clockwise)

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

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