Explanation Consider a section at a distance x from left end A of the section AC . Show the free-body diagram of the section AC as in Figure 1. Determine the moment at the section by taking moment about the section. ∑ M = 0 M − A y ( x ) = 0 M = A y x Write the second order differential equation as follows; d 2 y d x 2 = M ( x ) E I (1) Here, the moment at the corresponding section is M ( x ) , the modulus of elasticity of the material is E , and the moment of inertia of the section is I . Substitute A y x for M ( x ) in Equation (1). d 2 y d x 2 = A y x E I E I d 2 y d x 2 = A y x Integrate the equation with respect to x ; E I d y d x = A y x 2 2 + C 1 (2) Integrate the Equation (2) with respect to x . E I y = A y x 3 6 + C 1 x + C 2 (3) Show the free-body diagram of the section BC as in Figure 2. Determine the moment at the section by taking moment about the section. ∑ M = 0 M − A y ( x ) + [ 1 2 × 2 w 0 L × ( x − L 2 ) × ( x − L 2 ) × 1 3 × ( x − L 2 ) ] = 0 M − A y x + w 0 3 L × ( x − L 2 ) 3 = 0 M = A y x − w 0 3 L × ( x − L 2 ) 3 (4) Substitute ( A y x − w 0 3 L × ( x − L 2 ) 3 ) for M ( x ) in Equation (1). d 2 y d x 2 = A y x − w 0 3 L × ( x − L 2 ) 3 E I E I d 2 y d x 2 = A y x − w 0 3 L × ( x − L 2 ) 3 Integrate the equation with respect to x ; E I d y d x = A y x 2 2 − w 0 12 L × ( x − L 2 ) 4 + C 3 (5) Integrate the Equation (5) with respect to x . E I y = A y x 3 6 − w 0 60 L × ( x − L 2 ) 5 + C 3 x + C 4 (6) Boundary condition 1: At the point A ; x = 0 ; y = 0 . Substitute 0 for x and 0 for y in Equation (3). E I ( 0 ) = A y ( 0 ) 3 6 + C 1 ( 0 ) + C 2 0 = 0 + 0 + C 2 C 2 = 0 Boundary condition 2: At the point B ; x = L ; d y d x = 0 . Substitute L for x and 0 for d y d x in Equation (5). E I ( 0 ) = A y L 2 2 − w 0 12 L × ( L − L 2 ) 4 + C 3 0 = A y L 2 2 − w 0 L 3 192 + C 3 C 3 = − A y L 2 2 + w 0 L 3 192 Boundary condition 3: At the point C ; x = L 2 ; d y d x = d y d x . Equate Equation (2) and (5). A y x 2 2 + C 1 = A y x 2 2 − w 0 12 L × ( x − L 2 ) 4 + C 3 Substitute L 2 and x . A y ( L 2 ) 2 2 + C 1 = A y ( L 2 ) 2 2 − w 0 12 L × ( L 2 − L 2 ) 4 + C 3 C 1 = C 3 C 1 = − A y L 2 2 + w 0 L 3 192 Boundary condition 4: At the point C ; x = L 2 ; y = y . Equate Equation (3) and (6). A y x 3 6 + C 1 x + C 2 = A y x 3 6 − w 0 60 L × ( x − L 2 ) 5 + C 3 x + C 4 Substitute L 2 for x , 0 for C 2 , ( − A y L 2 2 + w 0 L 3 192 ) for C 1 , and ( − A y L 2 2 + w 0 L 3 192 ) for C 3

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ISBN: 8220102804487

Author: BEER

Publisher: YUZU

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Chapter 9.2, Problem 27P

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Find the reaction at the roller support of the beam.

Draw the bending moment diagram of the beam.

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Find the reaction at the roller support of the beam. Draw the bending moment diagram of the beam.

Solution Summary: The author analyzes the reaction at the roller support of the beam and the bending moment diagram of it.