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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 7P
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 7P, Determine the approximate axial forces, shears, and moments for all the members of the frames shown

FIG. P12.7, P12.15

To determine

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

Explanation of Solution

Given information:

The axial force acting at point D (HD) is 50 k.

The vertical distance of the member AD, BE, and CF (L) is 20 ft.

The horizontal distance of the members AB and BC (l1) is 20 ft.

The horizontal distance of the members DE and EF (l2) is 20 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 aa as in Figure (2).

Colum 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=0S2SS+HD=0

Substitute 50 k for HD.

4S+50=04S=50S=504S=12.5k()

The shear force at the lower end of the columns (SADandSCF) is 12.5k().

Determine the shear force at the interior column.

SBE=2S

Substitute –12.5 k for S.

SBE=2×(12.5)=25k()

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

Column moments:

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

Determine the moment at the column member AD.

MAD=S×L2

Substitute 12.5 k for S and 20 ft for L.

MAD=12.5×202MAD=125k-ft(Counterclockwise)

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=125k-ft(counterclockwise).

Determine the moment at the column member BE.

MBE=2×MAD

Substitute –125 k-ft for MAD.

MBE=2×125=250k-ft(Counterclockwise)

The moment at the internal hinge of the horizontal member DE and EF has the equal maginitude and opposite direction of the moment at the vertical column member AD.

MDE=MEF=125k-ft(Clockwise).

Draw the free body diagram of frame with the column moments for frame portion ADE as in Figure (4).

Girder axial forces, moments, and shears:

Consider girder DE.

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

FX=0QDE+HDS=0

Substitute 50 k for HD and 12.5 k for S.

QDE+5012.5=0QDE=37.5kQDE=37.5k()

Determine the vertical shear at the point D.

Take moment about hinge at the horizontal member DE.

MH=0MDE+VDE×l12=0

Substitute 125k-ft for MDE and 20 ft for l1

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