Explanation The Young’s modulus ( E ) is 200 GPa. The moment of inertia ( I ) is 600 × 10 6 mm 4 . Calculation: Consider elastic modulus E of the beam is constant. Show the given beam as in Figure (1). Refer Figure (1), Consider upward is positive and downward is negative. Consider clockwise is negative and counterclowise is positive. Determine the support reaction at A using the Equation of equilibrium; ∑ M C = 0 R A × 24 − ( 80 × 12 ) = 0 R A = 960 24 R A = 40 kN Determine the reaction at support C ; ∑ V = 0 R A + R C − 80 = 0 R C = 80 − 40 R C = 40 kN Show the free body diagram of the given beam as in Figure (2). Refer Figure (2), Determine the bending moment at A ; M A = ( 40 × 24 ) − ( 80 × 12 ) = 0 Determine the bending moment at B ; M B = ( 40 × 12 ) = 480 kN ⋅ m The bending moment at B is 480 kN ⋅ m , the moment of inertia between A and B is I but the moment of inertia of portion B and C is 2 I . Therefore, the in M / E I diagram the moment can be labeled as 240 E I . Show the M / E I diagram for the given beam as in Figure (3). Show the M / E I diagram for the conjugate-beam as in Figure (4). Consider counterclockwise is positive and clockwise is negative. Take moment about C : Determine the support reaction at A using the Equation of equilibrium; R A ( 24 ) − 1 2 ( 480 E I ) ( 12 ) ( 12 + 1 3 × 12 ) − 1 2 ( 480 E I ) ( 12 ) ( 2 3 × 12 ) = 0 R A = 57 , 600 kN ⋅ m 3 24 E I R A = 2 , 400 kN ⋅ m 2 E I The maximum bending moment in the conjugate beam occur at point M , at a distance of x M from the support A . Shear force at point M is equal to zero because the value of shear force at a point is equal to zero where the bending moment is high. Determine the intensity of load at x M using the relation; w M x M = 480 kN ⋅ m 12 m w M = 40 x M Determine the shear force at point M at a distance of x using the relation; S M = − R A + ( 1 2 × b 1 × h 1 ) = 0 Here, b is the width and h is the height of the triangle