Explanation Given Information: The concentrated live load ( P ) is 25 k. The uniformly distributed live load ( w L ) is 2 k/ft. The uniformly distributed dead load ( w D ) is 0.5 k/ft. Calculation: Apply a 1 k unit moving load at a distance of x from right end D . Sketch the free body diagram of beam as shown in Figure 1. Refer Figure 1. Find the equation of support reaction ( B y ) at B using equilibrium equation: Take moment about point D . Consider moment equilibrium at point D . Consider clockwise moment as positive and anticlockwise moment as negative. Sum of moment at point D is zero. Σ M D = 0 B y ( 20 ) − 1 ( x ) = 0 20 B y = x B y = x 20 (1) Find the equation of support reaction ( D y ) at D using equilibrium equation: Apply vertical equilibrium equation of forces. Consider upward force as positive ( ↑ + ) and downward force as negative ( ↓ − ) . B y + D y = 1 Substitute x 20 for B y . x 20 + D y = 1 D y = 1 − x 20 (2) Find the equation of shear at C . Find the equation of shear force at C of portion DC ( 0 ≤ x < 10 ft ) . Apply 1 k load at just right of C . Sketch the free body diagram of the section DC as shown in Figure 2. Refer Figure 2. Apply equilibrium equation of forces. Consider upward force as positive ( ↑ + ) and downward force as negative ( ↓ − ) . Σ F y = 0 S C + D y − 1 = 0 S C = − D y + 1 Substitute 1 − x 20 for D y . S C = − ( 1 − x 20 ) + 1 = − 1 + x 20 + 1 = x 20 Find the equation of shear force at C of portion CA ( 10 ft < x ≤ 25 ft ) . Apply 1 k load at just left of C . Sketch the free body diagram of the section CA as shown in Figure 3. Refer Figure 3. Apply equilibrium equation of forces. Consider upward force as positive ( ↑ + ) and downward force as negative ( ↓ − ) . Σ F y = 0 B y − 1 − S C = 0 S C = B y − 1 Substitute x 20 for B y . S C = x 20 − 1 Thus, the equations of the influence line for S C are, S C = x 20 0 ≤ x < 10 ft (3) S C = x 20 − 1 10 ft < x ≤ 25 ft (4) Find the value of influence line ordinate of shear force at various points of x using the Equations (3) and (4) and summarize the value as in Table 1. x ( ft ) from right end S C ( k/k ) 0 0 10 + 1 2 10 − − 1 2 20 0 25 1 4 Draw the influence lines for the shear force at point C using Table 1 as shown in Figure 4. Refer Figure 4. The maximum positive ILD ordinate at point C is 1 2 k/k . The maximum negative ILD ordinate at point C is − 1 2 k/k . Find the positive area ( A 1 ) of the influence line diagram of shear force at point C . A 1 = 1 2 ( L A B ) ( 1 4 ) + 1 2 ( L C D ) ( 1 2 ) Here, L A B is the length of beam between A to B and L C D is the length of beam between C to D . Substitute 5 ft for L A B and 10 ft for L C D . A 1 = 1 2 ( 5 ) ( 1 4 ) + 1 2 ( 10 ) ( 1 2 ) = 0.625 + 2.5 = 3.125 ft Find the negative area ( A 2 ) of the influence line diagram of shear force at point C . A 2 = 1 2 ( L B C ) ( − 1 2 ) Here, L B C is the length of beam between B to C . Substitute 10 ft for L B C . A 2 = 1 2 ( 10 ) ( − 1 2 ) = − 2.5 ft Find the maximum positive shear at point C using the equation. Maximum positive S C = P ( maximum positive ILD ordinate at C ) + w L ( A 1 ) + w D ( A 1 + A 2 ) Substitute 25 k for P , 1 2 k/k for maximum positive ILD ordinate at C , 2 k/ft for w L , 3.125 ft for A 1 , 0.5 k/ft for w D , and − 2.5 ft for A 2 . Maximum positive S C = 25 ( 1 2 ) + 2 ( 3.125 ) + 0.5 ( 3.125 − 2.5 ) = 12.5 + 6.25 + 0.3125 = 19.06 k Therefore, the maximum positive shear at point C is 19.06 k _ . Find the maximum negative shear at point C using the equation. Maximum negative S C = P ( maximum negative ILD ordinate at C ) + w L ( A 2 ) + w D ( A 1 + A 2 ) Substitute 25 k for P , − 1 2 k/k for maximum positive ILD ordinate at C , 2 k/ft for w L , 3