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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Any functioning component, such as machine parts, structural members, pressure vessels, and so on, are all under the influence of more than one force. When anybody is under the influence of more than one force such as shear forces, bending moments, axial…

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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.