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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 13P
Textbook Problem
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12.6 through 12.13 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 portal method.

FIG. P12.13, P12.21

Chapter 12, Problem 13P, 12.6 through 12.13 Determine the approximate axial forces, shears, and moments for all the members

To determine

Find the approximate axial forces, shears, and moments for all members of the frame using portal 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 K (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.

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 shears in the columns of the frame, pass an imaginary section cc, bb, and aa through the columns just above the support level.

Draw the free body diagram of the frame portion above section cc as in Figure (2).

Column shears:

Consider the shear in the interior column BF and CG is twice as much as in the exterior columns AE and CG.

Determine the column shears at the lower end using equilibrium equations.

FX=0S12S12S1S1+HI+HE+HM=0

Substitute 15 k for HI, 15 k for HE, and 7.5 k for HM.

6S1+15+15+7.5=06S1=37.5S1=37.56S1=6.25k()

The shear force at the lower end of the columns (SAEandSCG) is 6.25k().

Determine the shear force at the interior columns.

S(BF,CG)=2S1

Substitute –6.25 k for S1.

S(BF,CG)=2×6.25=12.5k()

Draw the free body diagram of frame with the column shear value at the lower end column as in Figure (3).

Draw the free body diagram of the frame portion above section bb as in Figure (4).

Consider the shear in the interior column FJ and GK is twice as much as in the exterior columns EI and HL.

Determine the column shears at the second top lower end of the members EFGH using equilibrium equations.

FX=0S22S22S2S2+HI+HM=0

Substitute 15 k for HI and 7.5 k for HM.

6S2+15+7.5=06S2=22.5S2=22.56S2=3.75k()

The shear force at the end of the column members (SEIandSHL) is 3.75k().

Determine the shear force at the interior column members FJ and GK.

S(FJ,GK)=2S2

Substitute –3.75 k for S2.

S(FJ,GK)=2×(3.75)=7.5k()

Draw the free body diagram of frame with the column shear value at the end of the column member EFGH as in Figure (5).

Draw the free body diagram of the frame portion above section cc as in Figure (6).

Determine the column shears at the top lower end of the members JK using equilibrium equations.

FX=0S3S3+HM=0

Substitute 7.5 k for HM.

2S3+7.5=02S3=7.5S3=7.52S3=3.75k()

Draw the free body diagram of frame with the column shear value at the end of the column member JK as in Figure (7).

Column moments:

Consider thecolumn members JM and KN.

Determine the moment at the column member JM.

MJM=S3×L12

Substitute 3.75 k for S3 and 16 ft for L1.

MJM=3.75×162MJM=30k-ft(Counterclockwise)

The moment at the column JM is equal to moment at the column KN. Therefore, the moment at the column members JM and KN is MJM=MMJ=MKN=MNK=30k-ft(Counterclockwise).

The moment at the internal hinge of the horizontal member MN has the equal maginitude and opposite direction of the moment at the vertical column members JM and KN.

MMN=MNM=30k-ft(Clockwise)

Consider column members EI, FJ, GK, and HL.

Consider the anticlockwise moment is positive and clockwise moment is negative.

Determine the moment at the column member EI.

MEI=S2×L22

Substitute 3.75 k for S2 and 16 ft for L2.

MEI=3.75×162MEI=30k-ft(Counterclockwise)

The moment at the column EI is equal to moment at the column HL. Therefore, the moment at the column member EI and HL is MEI=MIE=MHL=MLH=30k-ft(Counterclockwise).

Determine the moment at the column members FJ and GK.

M(FJ,GK)=2×M(EI,HL)

Substitute 30k-ft for MEI,HL.

M(FJ,GK)=2×30=60k-ft(Counterclockwise)

The moment at the internal hinge of the horizontal member IJ and KL have the equal maginitude and opposite direction of the moment at the vertical column members EI and HL. The moment at the internal hinge of the horizontal member JK has the equal maginitude and opposite direction of the moment at the vertical column member FJ or GK.

MIJ=MKL=30k-ft(Clockwise) and MJK=60k-ft(Clockwise).

Consider lower end column members.

Consider the anticlockwise moment is positive and clockwise moment is negative.

Determine the moment at the column member AE.

MAE=S1×L32

Substitute 6.25 k for S1 and 16 ft for L3.

MAE=6.25×162MAE=50k-ft(Counterclockwise)

The moment at the column AE is equal to moment at the column DH. Therefore, the moment at the column member AE and DH is MAE=MEA=MDH=MHD=50k-ft(Counterclockwise).

Determine the moment at the column members BF and CG.

M(BF,CG)=2×M(AE,DH)

Substitute 50k-ft for M(AE,DH).

M(BF,CG)=2×50=100k-ft(Counterclockwise)

The moment at the internal hinge of the horizontal member EF, FG, and GH has the equal maginitude and opposite direction of the summazation of moment at the vertical column member (AE and EI) and (DH and HL).

Determine the moment at the internal hinge of the horizontal member EF.

MEF=MAE+MEI

Substitute 50k-ft for MAE and 30k-ft for MEI.

MEF=50+30=80k-ft(Clockwise)

Determine the moment at the internal hinge of the horizontal member GH.

MGH=MDH+MHL

Substitute 50k-ft for MDH and 30k-ft for MHL.

MGH=50+30=80k-ft(Clockwise)

Draw the free body diagram of frame with the column moments for frame portion JMNK as in Figure (8).

Girder axial forces, moments, and shears:

Consider girder MN.

Determine the girder end action at the upper left end joint M using the relation.

FX=0QM+HMS3=0

Substitute 7.5 k for HM and 3.75 k for S3.

QM+7.53.75=0QM=3.75kQM=3.75k()

The girder end action at the horizontal member MN is QMN=3.75k() and QNM=3.75k().

Determine the vertical shear at the point M.

Take moment about hinge at the horizontal member MN.

MHin=0MMN+VMN×l22=0

Substitute 30k-ft for MMN and 20 ft for l2.

30+VMN×202=010VMN=30VMN=3010VMN=3k()

Determine the vertical shear at point J using the relation.

FY=0VMN+VMN=0

Substitute 3 k for VMN.

3+VNM=0VNM=3k()

The vertical end action at column members JM and KN is QMJ=3k(), QJM=3k(), QNK=3k(), and QKN=3k().

Consider girder IJ.

Determine the girder end action at the upper left end joint I using the relation.

FX=0QI+HIS2=0

Substitute 15 k for HI and 3.75 k for S2.

QI+153.75=0QI=11.25kNQI=11.25()

The girder end action at the horizontal member IJ is QIJ=11.25k() and QJI=11.25k().

Determine the vertical shear at the point I.

Take moment about hinge at the horizontal member IJ.

MHin=0MIJ+VIJ×l12=0

Substitute 30k-ft for MIJ and 30 ft for l1.

30+VIJ×302=015VIJ=30VIJ=302VIJ=2k()

Determine the vertical shear at point J using the relation.

FY=0VIJ+VJI=0

Substitute 2 k for VIJ.

2+VJI=0VJI=2k()

The vertical end action at joint I and E is QIE=2k() and QEI=2k().

Consider girder JK.

Determine the girder end action at the joint J using the relation.

FX=0QJKQJI+SMJ=0

Substitute 11.25 k for QJI and 3.75 k for SMJ.

QJK=7.5k()

The girder end action at the horizontal member JK is QJK=7

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