Explanation Given information: The A992 steel beam has a moment of inertia of I = 125 ( 10 6 ) mm 4 Method adopted: Castigliano’s theorem. Assumptions: Young’s modulus of the A992 steel beam is E = 200 × 10 6 kN/m 2 . Explanation: Castigliano’s theorem to determine the slope for the beam is as follows: θ = ∫ 0 L M ( ∂ M ∂ M ′ ) d x E I (1) Here, θ is the slope of the tangent θ at a point on the elastic curve is to be determined, M is internal moment in the beam, expressed as a function of x and caused by both the force P and the actual loads on the beam, M ′ is the external moment, E is modulus of elasticity of a member, and I is moment of inertia of the cross-sectional area about the neutral axis. Internal moment in the beam ( M ): Apply external moment M ′ at the point where the slope is to be determined. Determine the reactions: Entire beam: Show the free-body diagram of the entire beam as in Figure 1. Moment about point B : Determine the vertical reaction at point C by taking moment about point B . ∑ M B = 0 C y ( 6 ) + M ′ − 18 = 0 (2) Along the vertical direction: Determine the vertical reaction at point B by resolving the vertical component of forces. ∑ F y = 0 B y + C y = 0 (3) Along the horizontal direction: Determine the horizontal reaction at point C by resolving the horizontal component of forces. ∑ F x = 0 C x = 0 Show the calculation of values as follows: Conclusion: Solve Equation (2). C y ( 6 ) = 18 − M ′ C y = 3 − M ′ 6 Substitute ( 3 − M ′ 6 ) for C y in Equation (3). B y + 3 − M ′ 6 = 0 B y = M ′ 6 − 3 Region 0 ≤ x < 4 m from left end: Show the free body diagram of the beam for the region 0 ≤ x < 4 m as in Figure 2. Moment about the section: Determine the moment at the section by taking moment about the section. ∑ M x 1 = 0 M 1 + M ′ = 0 (4) SolveEquation (4). M 1 = − M ′ Region 4 m < x < 10 m from left end: Show the free body diagram of the beam for the region 4 m < x < 10 m as in Figure 3. Moment about the section: Determine the moment at the section by taking moment about the section. ∑ M x 2 = 0 M 2 + M ′ − B y ( x 2 ) = 0 (5) Substitute ( M ′ 6 − 3 ) for B y in Equation (5). M 2 + M ′ − ( M ′ 6 − 3 ) ( x 2 ) = 0 M 2 + M ′ − M ′ x 2 6 + 3 x 2 = 0 M 2 = − M ′ + M ′ x 2 6 − 3 x 2 Region 10 m < x ≤ 14 m : Show the free body diagram of the beam for the region 10 m < x ≤ 14 m as in Figure 4. Moment about the section: Determine the moment at the section by taking moment about the section. ∑ M x 3 = 0 − M 3 − 18 = 0 (6) Solve Equation (6). M 3 = − 18 kN-m Calculation: Tabulate the calculated values as in Table 1. Limit (m) M ∂ M ∂ M ′ M ( M ′ = 18 kN-m ) 0 ≤ x < 4 M 1 = − M ′ ∂ M 1 ∂ M ′ = − 1 M 1 = − 18 kN-m 4 < x < 10 M 2 = − M ′ + M ′ x 2 6 − 3 x 2 ∂ M 2 ∂ M ′ = − 1 + x 2 6 M 2 = − 18 kN-m 10 < x ≤ 14 M 3 = − 18 kN-m ∂ M 3 ∂ M ′ = 0 M 3 = − 18 kN-m Rearrange the Equation (1)

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Slope and Deflection

The slope in a deflected beam is described as an angle in radians made by the tangent of the section with the original axis of the beam. The deflection at any point on the axis of the beam is defined as the lateral displacement of the point before and af…

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Chapter 14.10, Problem 14.139P

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The slope at A.

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Chapter 14.10, Problem 14.138P

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The slope at A .

Solution Summary: The author explains Castigliano's theorem to determine the slope for the A992 steel beam.

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