   # 5.29 through 5.51 Draw the shear and bending moment diagrams and the qualitative deflected shape for the beam shown.

#### Solutions

Chapter
Section
Chapter 5, Problem 50P
Textbook Problem
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## 5.29 through 5.51 Draw the shear and bending moment diagrams and the qualitative deflected shape for the beam shown. To determine

Plot the shear and bending moment diagram and the qualitative deflected shape.

### Explanation of Solution

Sign conversion:

Apply the sign convention for calculating the equations of equilibrium as below.

• For the horizontal forces equilibrium condition, take the force acting towards right side as positive (+) and the force acting towards left side as negative ().
• For the vertical forces equilibrium condition, take the upward force as positive (+) and downward force as negative ().
• For moment equilibrium condition, take the clockwise moment as negative and counter clockwise moment as positive.

Apply the following sign convention for calculating the shear and bending moments.

• When the portion of the beam considered is left of the section, then the external force acting to the left are considered as positive.
• When the portion of the beam considered is right of the section, then the external force acting to the right are considered as positive.
• When the portion of the beam considered is left of the section, then the external force acting upward are considered as positive.
• When the portion of the beam considered is right of the section, then the external force acting downward are considered as positive.
• When the portion of the beam considered is left of the section, then the clockwise moments are considered as positive.
• When the portion of the beam considered is right of the section, then the counterclockwise moments are considered as positive.

Calculation:

Consider the section CDE:

Show the free-body diagram of the section CDE as in Figure 1.

Find the vertical reaction at point E by taking moment about point C.

+MCCDE=0Ey(10)120(5)=010Ey600=0Ey=60kN

Find the vertical reaction at point C by resolving the vertical equilibrium.

+FyCDE=0Cy120+Ey=0Cy120+60=0Cy=60kN

Show the free-body diagram of the entire beam as in Figure 2.

Consider the section ABC.

+MCABC=0Ay(13)By(5)+15(13)(132)=013Ay5By+1267.5=013Ay+5By=1267.5 (1)

Resolve the vertical component of forces in the section ABC.

+FyABC=0Ay+By15(13)60=0Ay+By=2555Ay+5By=1275 (2)

Subtract Equation (2) from Equation (1);

13Ay+5By5Ay5By=1267.512758Ay=7.5Ay=0.94kN=0.94kN

Substitute –0.94 kN for Ay in Equation (1)

13(0.94)+5By=1267.512.22+5By=1267.55By=1279.72By=255.94kN

Consider the entire beam.

Find the vertical reaction at point F by taking moment about the point G.

+MG=00.94(36)+15(13)(132+23)255.94(28)+120(18)+10(13)(132)Fy(8)=033.84+5752.57166.32+2160+8458Fy=0Fy=203.13kN

Find the horizontal reaction at point A by resolving the horizontal equilibrium.

+Fx=0Ax=0Ax=0

Find the vertical reaction at point G by resolving the vertical equilibrium

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