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: 9780100257061

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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Chapter 9.2, Problem 26P

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Problem 2. (a) Draw the moment diagram for the beam. (b) Observing the direction of load and sign of the moment diagram, and knowledge of the boundary conditions, sketch the elastic curve, that is, the vertical deflection of the axial line thru the centroid of the beam section, (c) Using integration, determine the equation of the elastic curve for the beam using the x coordinate that is valid for 0 < x < L/2. (d) Specify the slope at A and the beam's maximum deflection. El is constant. For parts (a) and (b), model only one-half of the beam, from x=0 to x=L/2, and use the symmetry dv boundary condition, slope = 0, at the vertical line of symmetry at x=L/2. dx X NA B

Problem 3. (a) Draw the moment diagram for the beam. (b) Observing the direction of load and sign of the moment diagram, and knowledge of the boundary conditions, sketch the elastic curve, that is, the vertical deflection of the axial line thru the centroid of the beam section, (c) Determine the equation of the elastic curve for the beam using the x coordinate. (d) Specify the slope at A and the maximum deflection. El is constant. Mo L B

if the moments at B are 5wL^2/192, where does that formula come from? (Only Handwritten solution Need)

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EBK MECHANICS OF MATERIALS

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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.