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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 6P
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 6P, Determine the approximate axial forces, shears, and moments for all the members of the frames shown

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 E (HE) is 10 k.

The axial force acting at point C (HC) is 15 k.

The horizontal distance of the member AB, CD, and EF (L) is 20 ft.

The vertical distance of the members AC and BD (l1) is 12 ft.

The vertical distance of the members CE and DF (l2) is 12 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:

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

FX=0SS+HE+HC=0

Substitute 10 k for HE and 15 k for HC.

2S+15+10=02S=25S=252S=12.5k()

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

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

Determine the column shears at the upper end CD using equilibrium equations.

FX=0S1S1+HE=0

Substitute 10 k for HE.

2S1+10=02S1=10S1=102S1=5k()

The shear force at the end of the column members CE and DF (SCEandSDF) is 5k().

Draw the free body diagram of frame with the column shear value at the member CD as in Figure (4).

Column moments:

Determine the moment at the column member CE and DF.

MCE=S1×l12

Substitute 5 k for S1 and 12 ft for l1.

MCE=5×122MCE=30k-ft(Counterclockwise)

The moment at the column CE is equal to moment at the column DF. Therefore, the moment at the column member CE and DF is MCE=MEC=MDF=MFD=30k-ft(counterclockwise).

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

MEF=30k-ft(Clockwise).

Determine the moment at the column member AC and BD.

Take moment about the hinge in the member AC.

Mh=0MACS×l22=0

Substitute 12.5 k for S and 12 ft for l2.

MAC12.5×122=0MAC=75k-ft(Counterclockwise)

The moment at the column AC is equal to moment at the column BD. Therefore, the moment at the column member AC and BD is MAC=MBD=75k-ft(counterclockwise).

Determine the moment at the member CD using the relation.

MCD=MCE+MAC

Substitute 30k-ft for MCE and 75k-ft for MAC.

MCD=30+75=105k-ft(Clockwise)

The moment at the internal hinge of the horizontal member CD is MCD=105k-ft(Clockwise).

Draw the free body diagram of frame with the column moments as in Figure (5).

Girder axial forces, moments, and shears:

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

FX=0QEF+HES1=0

Substitute 10 k for HE and 5 k for S1.

QEF+105=0QEF=5kQEF=5k()

Determine the vertical shear at the point E.

Take moment about hinge at the horizontal member EF.

MH=0MEF+VEF×L2=0

Substitute 30k-ft for MEF and 20 ft for L.

30+VEF×202=010VEF=30VEF=3010VEF=3k

VEF=3k()

Determine the vertical shear at point F using the relation

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