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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 14P
Textbook Problem
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12.14 through 12.21 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.

FIG. P12.6, P12.14

Chapter 12, Problem 14P, 12.14 through 12.21 Determine the approximate axial forces, shears, and moments for all the members

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 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 CE and DF (l1) is 12 ft.

The vertical distance of the members AC and BD (l2) is 12 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 second story of the frame, pass an imaginary section aa through the internal hinges at the midheights of columns CE and DF.

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).

Determine the location of the centroid using the relation.

x=AxA=Ax1+Ax22A (1)

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

x=A(0)+A(20)2A=10ft

The given lateral loads are acting on the frame to the right, therefore the axial force in column CE located to the left of the centroid, must be tensile, whereas the axial force in column DF placed to the right of the centroid, must be compressive.

Determine the axial force in the column members CE and DF using equilibrium conditions.

Take moment about point G.

MG=0Q2×LHE×l12=0

Substitute 20 ft for L, 10 k for HE, and 12 ft for l1.

Q2×2010×122=020Q2=60Q2=6020Q2=3k

The axial force at the column member CE and DF is QCE=3k() and QDF=3k().

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).

Above section bb:

Draw the free body diagram of the frame portion above the section bb as in Figure (5).

Determine the location of the centroid using the relation.

x¯=AxA=Ax1+Ax22A (2)

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

x¯=A(0)+A(20)2A=10ft

The given lateral loads are acting on the frame to the right, therefore the axial force in column AC located to the left of the centroid, must be tensile, whereas the axial force in column BD placed to the right of the centroid, must be compressive.

Determine the axial force in the column members AC and BD using equilibrium conditions.

Take moment about point H.

MH=0Q1×LHC×l22HE×(l1+l22)=0

Substitute 20 ft for L, 15 k for HC, 12 ft for l2, 10 k for HE, and 12 ft for l1.

Q1×2015×12210×(12+122)=020Q1=270Q1=27020Q1=13.5k

The axial force at the column member AC and BD is QAC=13.5k() and QBD=13.5k().

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 EF.

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

FY=0SEFQCE=0

Substitute 3 k for QCE.

SEF3=0SEF=3k()

The shear at the girder EF is SEF=3k().

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

FY=0SEF+SFE=0

Substitute 3 k for SEF.

3+SFE=0SFE=3k()

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

MEF=SEF×L2

Substitute 3 k for SEF and 20 ft for L.

MEF=3×202=30k-ft=30k-ft(Clockwise)

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

Take moment about point E.

ME=0MEF+SFE×LMFE=0

Substitute 30k-ft for MEF, 3 k for SFE, and 20 ft for L.

30+3×20MFE=0MFE=30k-ftMFE=30k-ft(Clockwise)

Consider girder CD.

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

FY=0SCD+QCASEF=0

Substitute 13.5 k for QCA and 3 k for SEF.

FY=0SCD+13.53=0SCD=10.5kSCD=10.5k()

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

FY=0SCD+SDC=0

Substitute 10.5 k for SCD.

10.5+SDC=0SDC=10.5k()

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

MCD=SCD×L2

Substitute 10.5 k for SCD and 20 ft for L.

MCD=10.5×202=105k-ft=105k-ft(Clockwise)

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

Take moment about point C.

MC=0MCD+SDC×LMDC=0

Substitute 105k-ft for MCD, 10.5 k for SDC, and 20 ft for L.

105+10.5×20MDC=0MDC=105k-ftMDC=105k-ft(Clockwise)

Column moments and shears:

Column moment for member CE and AC:

Determine the moment at the column member CE using moment equilibrium of joints.

Apply the moment equilibrium of joints at E.

M=0MECMEF=0

Substitute 30k-ft for MEF.

MEC30=0MEC=30k-ft(Counterclockwise)

The moment at the column member CE is MCE=MEC=30k-ft(Counterclockwise)

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