Explanation 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. + → ∑ F x = 0 − 100 ( 1 2 ) + C x + 100 ( 1 5 ) = 0 − 70.71 + C x + 44.72 = 0 C x = 25.99 kN → Find the vertical reaction at point D by taking moment about point C . + ↺ ∑ M C = 0 − 100 ( 1 2 ) ( 3 ) − 150 − 60 ( 5 ) − 90 ( 10 ) − C y ( 15 ) + 100 ( 2 5 ) ( 22 ) = 0 − 212.13 − 150 − 300 − 900 − 15 C y + 1967.74 = 0 C y = 27.04 kN ↓ Find the vertical reaction at point C by resolving the vertical equilibrium. + ↑ ∑ F y = 0 100 ( 1 2 ) + C y − 60 − 90 − D y + 100 ( 2 5 ) = 0 70.71 + C y − 150 − 27.04 + 89.44 = 0 C y = 16.89 kN ↑ 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. + ← ∑ F x = 0 100 ( 1 2 ) − 25