Explanation:

Given information:

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

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

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

Calculation:

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

PAB=Weight of beam AB×Length of beam AB

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

PAB=120×12=1,440 lb

Convert the unit from lb to kip.

PAB=1,440 lb×1 kip1,000 lb=1.44 kip

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

Refer to Figure 1.

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

sinθ=112+32sinθ=13.1623θ=sin−1(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=−2.277 kip

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=0Ax−2.277cos18.43°=0Ax−2.16=0Ax=2.16 kip

Summation of forces along vertical direction is Equal to zero.

∑Fy=0Ay−1.44+2.277sin18.43°=0Ay−0.72=0Ay=0.72 kip

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

PAD=Weight of beam AD×Length of beam AD

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

PAD=120×6=720 lb

Convert the unit from lb to kip.

PAD=720 lb×1 kip1,000 lb=0.72 kip

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

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=−2.16 kip

Summation of forces along vertical direction is Equal to zero.

∑Fy=0VD+0.72−0.72−=0VD=0

Take moment about D is Equal to zero.

∑MD=0MD−0.72×3=0MD−2.16=0MD=2.16 kip⋅ft

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

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

PFC=Weight of column FC×Length of column FC

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

PFC=180×16=2,880 lb

Convert the unit from lb to kip.

PFC=2,880 lb×1 kip1,000 lb=2.88 kip

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

Refer to Figure 3.

Find the angle of cable CG to the horizontal.

sinθ=342+32sinθ=35θ=sin−1(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°=00.8FCG−0.72=00.8FCG=0.72

FCG=0.9 kip

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

PFE=Weight of column FE×Length of column FE

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

PFE=180×4=720 lb

Convert the unit from lb to kip.

PFE=720 lb×1 kip1,000 lb=0.72 kip

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

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=0VE−0.54=0VE=0.54 kip

Summation of forces along vertical direction is Equal to zero.

∑Fy=0NE+0.72−5.04=0NE−4.32=0NE=4.32 kip

Take moment about E is Equal to zero.

∑ME=0−ME+0.54×4=0−ME+2.16=0ME=2.16 kip⋅ft

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