   # 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. P12.9, P12.17

#### Solutions

Chapter
Section
Chapter 12, Problem 17P
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 cantilever method.FIG. P12.9, P12.17 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 I (HI) is 20 k.

The axial force acting at point F (HF) is 40 k.

The vertical distance of the member AD, BE, and CF (L1) is 16 ft.

The vertical distance of the member DG, EH, and FI (L2) is 16 ft.

The horizontal distance of the members AB, DE, and GH (l1) is 24 ft.

The horizontal distance of the members BC, EF, and HI (l2) is 24 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 story of the frame, pass an imaginary section aa through the internal hinges at the midheights of columns AD, BE, and CF.

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+Ax2+Ax33A (1)

Substitute 0 ft for x1, 24 ft for x2, and 48 ft for x3 in Equation (1).

x¯=A(0)+A(24)+A(48)3A=24ft

The given lateral load is acting on the frame to the left, therefore the axial force in column DG located to the left of the centroid, must be compressive, whereas the axial force in column FI placed to the right of the centroid, must be tensile.

Determine the axial force in the column members DG and FI using equilibrium conditions.

MJ=0Q2×(l1+l2)+HI×L12QEH×l1=0

Substitute 24 ft for l1, 24 ft for l2, 20 k for HI, 16 ft for L1, and 0 k for QEH.

Q2×(24+24)+20×1620=048Q2=160Q2=16048Q2=3.33k

The axial force at the column member DG and FI is QDG=3.33k() and QFI=3.33k().

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).

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+Ax2+Ax33A (2)

Substitute 0 ft for x1, 24 ft for x2, and 48 ft for x3 in Equation (2).

x¯=A(0)+A(24)+A(48)3A=24ft

The given lateral load is acting on the frame to the left, therefore the axial force in column AD located to the left of the centroid, must be compressive, whereas the axial force in column CF placed to the right of the centroid, must be tensile.

Determine the axial force in the column members AD and CF using equilibrium conditions.

MK=0Q1×(l1+l2)+HI×(L1+L22)+HF×L12QBE×l1=0

Substitute 24 ft for l1, 24 ft for l2, 20 k for HI, 16 ft for L1, 16 ft for L2, 40 k for HF, and 0 k for QBE.

Q1×(24+24)+20×(16+162)+40×1620=048Q1=800Q2=80048Q2=16.67k

The axial force at the column member AD and CF is QAD=16.67k() and QCF=16.67k().

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 GH.

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

FY=0SGHQGD=0

Substitute 3.33 k for QGD.

SGH3.33=0SGH=3.33k()

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

FY=0SGH+SHG=0

Substitute 3.33 k for SGH.

3.33+SHG=0SHG=3.33kSHG=3.33k()

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

MGH=SGH×l12

Substitute 3.33 k for SGH and 24 ft for l1.

MGH=3.33×24240k-ft(Counterclockwise)

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

MG=0MGHSHG×l1+MHG=0

Substitute 40k-ft for MGH, 3.33 k for SHG, and 24 ft for l1.

403.33×24+MHG=0MHG=40k-ft(Counterclockwise)

Consider girder HI.

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

FY=0SHG+SHI=0

Substitute 3.33 k for SHG.

3.33+SHI=0SHI=3.33k()

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

FY=0SHI+SIH=0

Substitute 3.33 k for SHI.

3.33+SIH=0SIH=3.33kSIH=3.33k()

The sheare at the joint I is SIH=3.33k().

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

MHI=SHI×l22

Substitute 3.33 k for SHI and 24 ft for l2.

MHI=3.33×242=40k-ft(Counterclockwise)

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

MH=0MHISIH×l2+MIH=0

Substitute 40k-ft for MHI, 3.33 k for SIH, and 24 ft for l2.

403.33×24+MIH=0MIH=40k-ft(Counterclockwise)

Consider girder DE.

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

FY=0SDE+QDGQDA=0

Substitute 3.33 k for QDG and 16.67 k for QDA.

SDE+3.3316.67=0SDE=13.33k()

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

FY=0SDE+SED=0

Substitute 13.33 k for SDE.

13.33+SED=0SED=13.33kSED=13.33k()

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

MDE=SDE×l12

Substitute 13.33 k for SDE and 24 ft for l1.

MDE=13.33×242160k-ft(Counterclockwise)

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

MD=0MDESED×l1+MED=0

Substitute 160k-ft for MDE, 13.33 k for SED, and 24 ft for l1.

16013.33×24+MED=0MED=160k-ft(Counterclockwise)

Consider girder EF.

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

FY=0SED+SEF=0

Substitute 13.33 k for SED.

13.33+SEF=0SEF=13.33k()

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

FY=0SEF+SFE=0

Substitute 13.33 k for SEF.

13.33+SFE=0SFE=13.33kSFE=13.33k()

The sheare at the joint F is SF=13.33k().

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

MEF=SEF×l22

Substitute 13.33 k for SEF and 24 ft for l2.

MEF=13.33×242=160k-ft(Counterclockwise)

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

ME=0MEFSFE×l2+MFE=0

Substitute 160k-ft for MEF, 13.33 k for SFE, and 24 ft for l2.

16013.33×24+MFE=0MFE=160k-ft(Counterclockwise)

Column moments and shears:

Column moment for member DG, EH, and FI:

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

Apply the moment equilibrium of joints at G.

M=0MGD+MGH=0

Substitute 40k-ft for MGH.

MGD+40=0MGD=40k-ftMGD=40k-ft(Clockwise)

The moment at the column member GD is MGD=MDG=40k-ft(Clockwise).

Determine the moment at the column member EH.

Apply the moment equilibrium of joints at H.

M=0MHG+MHI+MHE=0

Substitute 40k-ft for MHG and 40k-ft for MHI

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