The beam AB is pin supported at A and supported by a cable BC. A separate cable CG is used to hold up the frame. If AB weighs 120 lb/ft and the column FC has a weight of 180 lb/ft, determine the resultant internal loadings acting on cross sections located at points D and E. Neglect the thickness of both the beam and column in the calculation.
Find the resultant internal loadings acting on cross sections located at D and E.
Answer to Problem 1RP
The resultant internal loadings at cross section at D are
The resultant internal loadings at cross section at E are
Explanation of Solution
Given information:
The beam AB is pin supported at A and supported by a cable BC.
The weight of the beam AB is
The weight of the column FC is
Calculation:
Find the loading at the center of the beam AB
Substitute
Convert the unit from lb to 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
Find the tension in cable BC as shown below.
Take moment about A is Equal to zero.
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.
Summation of forces along vertical direction is Equal to zero.
Find the loading at the center of the beam AD
Substitute
Convert the unit from lb to 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.
Summation of forces along vertical direction is Equal to zero.
Take moment about D is Equal to zero.
Hence, the resultant internal loadings at cross section at D are
Find the loading at the center of the column FC
Substitute
Convert the unit from lb to 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.
Find the tension in cable CG as shown below.
Summation of forces along horizontal direction is Equal to zero.
Find the loading at the center of the column FE
Substitute
Convert the unit from lb to 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.
Summation of forces along vertical direction is Equal to zero.
Take moment about E is Equal to zero.
Therefore, the resultant internal loadings at cross section at E are
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