Sample Solution from
Mechanics of Materials (10th Edition)
10th Edition
ISBN: 9780134319650
Chapter 1
Problem 1.1RP
Try another sample solutionarrow_forward
To determine

To determine: The resultant internal loadings acting on cross sections located at D and E.

Answer

Answer:

The resultant internal loadings at cross section at D are ND=2.16kip,VD=0_, and MD=2.16kipft_.

The resultant internal loadings at cross section at E are NE=4.32kip,VE=0.54kip_, and ME=2.16kipft_.

Explanation of Solution

Explanation:

Given information:

The beam AB is pin supported at A and supported by a cable BC.

The weight of the beam AB is 120lb/ft.

The weight of the column FC is 180lb/ft.

Calculation:

Find the loading at the center of the beam AB (PAB):

PAB=Weight of beamAB×LengthofbeamAB

Substitute 120lb/ft for the weight of beam AB and 12 ft for the length of beam AB.

PAB=120×12=1,440lb

Convert the unit from lb to kip.

PAB=1,440lb×1kip1,000lb=1.44kip

Sketch the Free Body Diagram of the beam AB shown in Figure 1.

images

Refer to Figure 1.

Find the angle of cable BC to the horizontal (θ):

sinθ=112+32sinθ=13.1623θ=sin1(0.3162)θ=18.43°

Find the tension in cable BC as shown below.

Take moment about A is Equal to zero.

MA=0(FBCsin18.43°×12)+1.44×6=03.794FBC+8.64=03.794FBC=8.64

FBC=2.277kip

Find the support reaction at A as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

Fx=0Ax2.277cos18.43°=0Ax2.16=0Ax=2.16kip

Summation of forces along vertical direction is Equal to zero.

Fy=0Ay1.44+2.277sin18.43°=0Ay0.72=0Ay=0.72kip

Find the loading at the center of the beam AD (PAD):

PAD=Weight of beamAD×LengthofbeamAD

Substitute 120lb/ft for the weight of beam AD and 6 ft for the length of beam AD.

PAD=120×6=720lb

Convert the unit from lb to kip.

PAD=720lb×1kip1,000lb=0.72kip

Sketch the Free Body Diagram of the section for point D as shown in Figure 2.

images

Refer to Figure 2.

Find the internal loadings as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

Fx=0ND+2.16=0ND=2.16kip

Summation of forces along vertical direction is Equal to zero.

Fy=0VD+0.720.72=0VD=0

Take moment about D is Equal to zero.

MD=0MD0.72×3=0MD2.16=0MD=2.16kipft

Hence, the resultant internal loadings at cross section at D are ND=2.16kip,VD=0_, and MD=2.16kipft_.

Find the loading at the center of the column FC (PFC):

PFC=Weight of columnFC×LengthofcolumnFC

Substitute 180lb/ft for the weight of column FC and 16 ft for the length of column FC.

PFC=180×16=2,880lb

Convert the unit from lb to kip.

PFC=2,880lb×1kip1,000lb=2.88kip

Sketch the Free Body Diagram of the beam FC shown in Figure 3.

images

Refer to Figure 3.

Find the angle of cable CG to the horizontal.

sinθ=342+32sinθ=35θ=sin1(0.6)θ=36.87°

Find the tension in cable CG as shown below.

Summation of forces along horizontal direction is Equal to zero.

Fx=0FCGcos36.87°2.277sin18.43°=00.8FCG0.72=00.8FCG=0.72

FCG=0.9kip

Find the loading at the center of the column FE (PFE):

PFE=Weight of columnFE×LengthofcolumnFE

Substitute 180lb/ft for the weight of column FE and 4 ft for the length of column FC.

PFE=180×4=720lb

Convert the unit from lb to kip.

PFE=720lb×1kip1,000lb=0.72kip

Sketch the Free Body Diagram of the section for point E as shown in Figure 4.

images

Refer to Figure 4.

Find the internal loadings as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

Fx=0VE0.54=0VE=0.54kip

Summation of forces along vertical direction is Equal to zero.

Fy=0NE+0.725.04=0NE4.32=0NE=4.32kip

Take moment about E is Equal to zero.

ME=0ME+0.54×4=0ME+2.16=0ME=2.16kipft

Therefore, the resultant internal loadings at cross section at E are NE=4.32kip,VE=0.54kip_, and ME=2.16kipft_.

Not sold yet?Try another sample solutionarrow_forward