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. Given information: The inclined point load acting at the left end of the beam ( P 1 ) is 75 kN. The point load acting at the distance of 10 m from the left support ( P 2 ) is 90 kN. The inclined point load acting at the right end of the beam ( P 3 ) is 75 kN. Calculation: Draw the free body diagram of the entire beam as in Figure (1). Determine the angle for the left end point load of the beam. tan θ 1 = 3 4 θ 1 = tan − 1 ( 3 4 ) θ 1 = 36.86 ° Determine the angle for the right end point load of the beam. tan θ 2 = 4 3 θ 2 = tan − 1 ( 4 3 ) θ 2 = 53.13 ° Determine the horizontal force at the right support using equilibrium condition. + → ∑ F x = 0 D x + P 1 cos θ 1 − P 3 cos θ 2 = 0 Substitute 75 kN for P 1 , 36.86 ° for θ 1 , 75 kN for P 3 , and 53.13 ° for θ 2 . D x − 75 cos 36.86 ° + 75 cos 53.13 ° = 0 D x = 15 kN ( → ) Consider clockwise moment as positive and counter clockwise moment as negative. Determine the vertical force at the left support using equilibrium condition. Taking moment about point D . + ↺ ∑ M D = 0 { P 1 sin θ 1 × ( l 1 + l 2 + l 3 + l 4 ) − C y × ( l 2 + l 3 + l 4 ) + P 2 × l 4 + M L + M 1 − P 3 sin θ 2 × ( l 5 + l 6 ) } = 0 Substitute 75 kN for P 1 , 36.86 ° for θ 1 , 4 m for l 1 , 3 m for l 2 , 3 m for l 3 , 6 m for l 4 , 90 kN for P 2 , 80 kN ⋅ m for M L , 100 kN ⋅ m for M 1 , 75 kN for P 3 , 53.13 ° for θ 2 , 3 m for l 5 , and 3 m for l 6 . { 75 sin 36.86 ° × ( 4 + 3 + 3 + 6 ) − C y × ( 3 + 3 + 6 ) + 90 × 6 + 80 + 100 − 75 sin 53.13 ° × ( 2 + 2 ) } = 0 − 12 C y = − 1 , 200 C y = 1 , 200 12 C y = 100 kN ( ↑ ) Determine the vertical force at the right support using equilibrium condition