Explanation Given information: The beam is given in the Figure. The maximum deflection is Δ max ≤ L 360 . The value of E is 29,000 ksi. 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. Let the bending moment due to real load be M . Sketch the real system of the beam as shown in Figure 1. Find the reactions at the supports: Summation of moments about A is equal to 0. ∑ M A = 0 20 B y − 3 ( 20 ) ( 20 2 ) = 0 B y = 30 k Summation of forces along y -direction is equal to 0. + ↑ ∑ F y = 0 A y + B y − 3 ( 20 ) = 0 A y + 30 − 3 ( 20 ) = 0 A y = 30 k ↑ 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 reactions at the supports: Summation of moments about A is equal to 0. ∑ M A = 0 20 B y − 1 ( 10 ) = 0 B y = 0.5 k Summation of forces along y -direction is equal to 0. + ↑ ∑ F y = 0 A y + B y − 1 = 0 A y + 0.5 − 1 = 0 A y = 0.5 k ↑ Find the equations for M and M v for the 2 segments of the beam as shown in Table 1. Segment x -coordinate M ( k-ft ) M v ( k-ft ) Origin Limits (ft) AM A 0 − 10 30 x − 3 x 2 2 0.5 x BM B 0 − 10 30 x − 3 x 2 2 0.5 x Find the maximum deflection that occurs at M 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 − 10 and 0 − 10 as follows