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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 11P
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 portal method.

Chapter 12, Problem 11P, 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 all members of the frame using portal 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 (L1) is 12 ft.

The vertical distance of the member DG, EH, and FI (L2) 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.

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 aa through the columns just above the support level.

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

Column shears:

Consider the shear in the interior column BE is twice as much as in the exterior columns AD and CF.

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

FX=0S1+2S1+S1HFHI=0

Substitute 25 k for HF and 15 k for HI.

4S12515=04S1=40S1=404S1=10k()

The shear force at the lower end of the columns (SADandSCF) is 10k().

Determine the shear force at the interior column.

SBE=2S1

Substitute 10 k for S1.

SBE=2×(10)=20k()

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 aa as in Figure (4).

Consider the shear in the interior column EH is twice as much as in the exterior columns DG and FI.

Determine the column shears at the lower end of the member DEF using equilibrium equations.

FX=0S2+2S2+S2HI=0

Substitute 15 k for HI.

4S215=04S2=15S2=154S2=3.75k()

The shear force at the end of the columns (SDGandSFI) is 3.75k().

Determine the shear force at the interior column.

SEH=2S2

Substitute 3.75 k for S2.

SEH=2×(3.75)=7.5k()

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

Column moments:

Consider upper end column members.

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

Determine the moment at the column member DG.

MDG=S2×L12

Substitute 3.75 k for S2 and 12 ft for L1.

MDG=3.75×122=22.5k-ft(Clockwise)

The moment at the column DG is equal to moment at the column FI. Therefore, the moment at the column member DG and FI is MDG=MGD=MHI=MIH=22.5k-ft(Clockwise).

Determine the moment at the column member EH.

MEH=2×MDG

Substitute –22.5 k-ft for MDG.

MEH=2×22.5=45k-ft(Clockwise)

The moment at the internal hinge of the horizontal member GH and HI has the equal maginitude and opposite direction of the moment at the vertical column member DG and FI.

MGH=MHI=22.5k-ft(Counterclockwise).

Consider lower end column members.

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

Determine the moment at the column member AD.

MAD=S1×L22

Substitute 10 k for S1 and 12 ft for L2.

MAD=10×122=60k-ft(Clockwise)

The moment at the column AD is equal to moment at the column CF. Therefore, the moment at the column member AD and CF is MAD=MDA=MCF=MFC=60k-ft(Clockwise).

Determine the moment at the column member BE.

MBE=2×MAD

Substitute –60 k-ft for MAD.

MBE=2×60=120k-ft(Clockwise)

The moment at the internal hinge of the horizontal member DE and EF has the equal maginitude and opposite direction of the summazation of moment at the vertical column member (AD and DG) and(CF and FI).

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

MDE=MDG+MAD

Substitute 22.5 k-ft for MDG and 60 k-ft for MAD.

MDE=22.5+60=82.5k-ft(Counterclockwise)

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

MEF=MCF+MFI

Substitute 60 k-ft for MCF and 22.5 k-ft for MFI.

MEF=60+22.5=82.5k-ft(Counterclockwise)

Draw the free body diagram of frame with the column moments for frame portion FIH as in Figure (6).

Girder axial forces, moments, and shears:

Consider girder HI.

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

FX=0QHIHI+S2=0

Substitute 15 k for HI and 3.75 k for S2.

QHI15+3.75=0QHI=11.25k()

Determine the vertical shear at the point I using the relation.

Take moment about hinge at the horizontal member HI.

MHin=0MIHVIH×l22=0

Substitute 22.5k-ft for MIH and 15 ft for l2.

22.5VIH×152=07.5VIH=22.5VIH=22.57.5VIH=3k()

Determine the vertical shear at point H using the relation.

FY=0VIH+VHI=0

Substitute 3

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