Explanation Given information: The beam is given in the Figure. The maximum deflection Δ max ≤ L 360 . The value of E is 200 GPa. Apply the sign conventions for calculating reactions, forces and moments using the three equations of equilibrium as shown below. For summation of forces along x -direction is equal to zero ( ∑ F x = 0 ) , consider the forces acting towards right side as positive ( → + ) and the forces acting towards left side as negative ( ← − ) . For summation of forces along y -direction is equal to zero ( ∑ F y = 0 ) , consider the upward force as positive ( ↑ + ) and the downward force as negative ( ↓ − ) . For summation of moment about a point is equal to zero ( ∑ M at a point = 0 ) , consider the clockwise moment as negative and the counter clockwise moment as positive. Calculation: Consider the real system. Draw a diagram showing all the given real loads acting on it. Let the bending moment due to real load be M . Sketch the real system of the beam as shown in Figure 1. Consider the virtual system. Remove all the real loads and apply unit load where the point to find the deflection. Let the bending moment due to virtual load be M v . Sketch the virtual system of the beam with unit load at point C as shown in Figure 2. Find the equations for M and M v for the 2 segments of the beam as shown in Table 1. Segment x -coordinate M ( kN ⋅ m ) M v ( kN ⋅ m ) Origin Limits (m) AB A 0 − 6 − 400 − 50 x − 1 x BC A 6 − 12 − 400 − 50 x − 100 ( x − 6 ) − 1 x Find the maximum deflection that occurs at A using the virtual work expression: 1 ( Δ max ) = ∫ 0 L M v M E I d x (1) Rearrange Equation (1) for the limits 0 − 6 and 6 − 12 as follows. Δ max = 1 E I [ ∫ 0 6 ( M v M ) d x + ∫ 6 12 ( M v M ) d x ] Substitute − 1 x for M v , − 400 − 50 x for M for the limits 0 − 6 , − 1 x for M v , and − 400 − 50 x − 100 ( x − 6 ) for M for the limits 6 − 12