   # Determine the axial forces, shears, and bending moments at points A and B of the structure shown. FIG.P5.5

#### Solutions

Chapter
Section
Chapter 5, Problem 5P
Textbook Problem
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## Determine the axial forces, shears, and bending moments at points A and B of the structure shown. FIG.P5.5

To determine

Find the axial force, shear, and bending moment at points A and B of the beam.

### 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 axial forces, 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.

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

+Fx=0100(12)+Cx+100(15)=070.71+Cx+44.72=0Cx=25.99kN

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

+MC=0100(12)(3)15060(5)90(10)Cy(15)+100(25)(22)=0212.1315030090015Cy+1967.74=0Cy=27.04kN

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

+Fy=0100(12)+Cy6090Dy+100(25)=070.71+Cy15027.04+89.44=0Cy=16.89kN

Pass the sections aa and bb at points A and B respectively.

Show the sections aa and bb as in Figure 2.

Consider section aa:

Consider the left side of the section aa for calculation of internal forces.

Show the free-body diagram of the left side of the section aa as in Figure 3.

Find the axial force at point A by resolving the horizontal equilibrium.

+Fx=0100(12)25

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