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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 19P
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 19P, Determine the approximate axial forces, shears, and moments for all the members of the frames shown

FIG. P12.11, P12.19

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 15 k.

The axial force acting at point F (HF) is 25 k.

The vertical distance of the member AD, BE, and CF (L2) is 12 ft.

The vertical distance of the member DG, EH, and FI (L1) is 12 ft.

The horizontal distance of the members AB, DE, and GH (l1) is 20 ft.

The horizontal distance of the members BC, EF, and HI (l2) is 15 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 DG, EH, and FI and pass an imaginary section bb through the internal hinges at the midheights of columns AD, BE, and CF.

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+Ax33A        (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, and x3 is the distance for the support C from support A.

Substitute 0 ft for x1, 20 ft for x2, and 35 ft for x3 in Equation (1).

x¯=A(0)+A(20)+A(35)3A=18.33ft

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

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

Refer Figure (3).

Apply similar triangle rule.

Determine the relationship in column axial force between the member DG and EH using the relation.

QEH=l1x¯x¯QDG

Substitute 20 ft for l1 and 15 ft for x¯.

QEH=2018.3318.33QDG=0.091QDG

Determine the relationship in column axial force between the member DG and FI using the relation.

QFI=(l1+l2)x¯x¯QDG

Substitute 20 ft for l1, 15 ft for l2, and 18.33 ft for x¯.

QFI=(20+15)18.3318.33QDG=0.909QDG

Determine the axial force in the column members DG and FI using equilibrium conditions.

Take moment about point J.

MJ=0Q2×(l1+l2)+HI×L12+QEH×l2=0

Substitute 20 ft for l1, 15 ft for l2, 15 k for HI, 12 ft for L1, and 0.091Q2 for QEH.

Q2×(20+15)+15×122+0.091Q2×15=035Q2+1.365Q2=90Q2=9033.635Q2=2.68k

The axial force at the column member is QDG=2.68k().

Determine the axial force in the column members EH.

QEH=0.091QDG

Substitute 2.68 k for QDG.

QEH=0.091×2.68=0.243k()

Determine the axial force in the column members FI.

QFI=0.909QDG

Substitute 2.68 k for QDG.

QFI=0.909×2.68=2.43()

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).

Refer Figure 5.

Determine the location of the centroid using the relation.

x¯=AxA=Ax1+Ax2+Ax33A        (2)

Substitute 0 ft for x1, 20 ft for x2, and 35 ft for x3 in Equation (2).

x¯=A(0)+A(20)+A(35)3A=18.33ft

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

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 AD and BE using the relation.

QBE=l1x¯x¯QAD

Substitute 20 ft for l1 and 18.33 ft for x¯.

QBE=2018.3318.33QAD=0.091QAD

Determine the relationship in column axial force between the member AD and CF using the relation.

QCF=(l1+l2)x¯x¯QAD

Substitute 20 ft for l1, 15 ft for l2, and 18.33 ft for x¯.

QCF=(20+15)18.3318.33QAD=0.909QAD

Determine the axial force in the column members AD and CF using equilibrium conditions.

Take moment about point K.

MK=0Q1×(l1+l2)+HI×(L1+L22)+HF×L12+QBE×l2=0

Substitute 20 ft for l1, 15 ft for l2, 15 k for HI, 12 ft for L1, 12 ft for L2, 25 k for HF, and 0.091Q1 for QBE.

Q1×(20+15)+15×(12+122)+25×122+0.091Q1×15=033.635Q1=420Q1=42033.635Q1=12.49k

The axial force at the column member is QAD=12.49k().

Determine the axial force in the column members BE.

QBE=0.091QAD

Substitute 12.49 k for QAD.

QBE=0.091×12.49=1.14k()

Determine the axial force in the column members FI.

QCF=0.909QAD

Substitute 12.49 k for QAD.

QCF=0.909×12.49=11.35k()

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 GH.

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

FY=0SGHQGD=0

Substitute 2.68 k for QGD.

SGH2.68=0SGH=2.68k()

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

FY=0SGH+SHG=0

Substitute 2.68 k for SGH.

2.68+SHG=0SHG=2.68kSHG=2.68k()

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

MGH=SGH×l12

Substitute 2.68 k for SGH and 20 ft for l1.

MGH=2.68×202=26.8k-ft(Counterclockwise)

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

Take moment about point G.

MG=0MGHSHG×l1+MHG=0

Substitute 26.8k-ft for MGH, 2.68 k for SHG, and 20 ft for l1.

26.82.68×20+MHG=0MHG=26.8k-ft(Counterclockwise)

Consider girder HI.

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

FY=0SHG+SHI+QHE=0

Substitute 2.68 k for SHG and 0.243 k for QHE.

2.68+SHI+0.243=0SHI=2.437k()

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

FY=0SHI+SIH=0

Substitute 2.437 k for SHI.

2.437+SIH=0SIH=2.437kSIH=2.437k()

The sheare at the joint I is SI=2.437k().

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

MHI=SHI×l22

Substitute 2.437 k for SHI and 15 ft for l2.

MHI=2.437×152=18.28k-ft(Counterclockwise)

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

Take moment about point H.

MH=0MHISIH×l2+MIH=0

Substitute 18.28k-ft for MHI, 2.43 k for SIH, and 15 ft for l2.

18.282.437×15+MIH=0MIH=18.28k-ft(Counterclockwise)

Consider girder DE.

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

FY=0SDE+QDGQDA=0

Substitute 2.68 k for QDG and 12.49 k for QDA.

SDE+2.6812.49=0SDE=9.81k()

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

FY=0SDE+SED=0

Substitute 9.81 k for SDE.

9.81+SED=0SED=9.81kSED=9.81k()

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

MDE=SDE×l12

Substitute 9.81 k for SDE and 20 ft for l1.

MDE=9.81×202=98.1k-ft=98.1k-ft(Counterclockwise)

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

Take moment about point D.

MD=0MDESED×l1+MED=0

Substitute 98.1k-ft for MDE, 9.81 k for SED, and 20 ft for l1.

98.19.81×20+MED=0MED=98.1k-ft(Counterclockwise)

Consider girder EF.

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

FY=0SED+SEF+QBEQEH=0

Substitute 9.81 k for SED, 1.14 k for QBE, and 0.243 k for QEH.

9.81+SEF+1.140.243=0SEF=8.91k()

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

FY=0SEF+SFE=0

Substitute 8.91 k for SEF.

8.91+SFE=0SFE=8.91kSFE=8.91k()

The sheare at the joint F is SF=8.91k().

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

MEF=SEF×l22

Substitute 8

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