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

#### Solutions

Chapter
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Chapter 5, Problem 48P
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:

Show the free-body diagram of the entire beam as in Figure 1.

Consider the section DE.

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

+MDDE=0Ey(35)2(35)(352)=035Ey1225=0Ey=35k

Consider the section BCDE.

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

+MBBCDE=0Cy(15)2(65)(652)+35(65)=015Cy4225+2275=0Cy=130k

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

+Fx=0Ax=0Ax=0

Find the moment at point A by taking moment about point A.

+MA=0MA2(100)(1002)+130(50)+35(100)=0MA10,000+6,500+3,500=0MA=0

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

+Fy=0Ay2(100)+Cy+Ey=0Ay200+130+35=0Ay=35k

Show the reactions of the beam as in Figure 2.

Shear diagram:

Point A:

SA,L=0SA,R=35k

Point B:

SB=352(35)=35k

Point C:

SC,L=352(50)=65kSC,R=65+130=65k

Point D:

SD=652(15)=35k

Point E:

SE,L=352(35)=35kSE,R=35+35=0

Plot the shear force diagram of the beam as in Figure 3

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