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 ) + 18 + 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: Solve Equation (2). C y ( 6 ) = − M ′ C y = − M ′ 6 Substitute ( − M ′ 6 ) for C y in Equation (3). B y − M ′ 6 = 0 B y = M ′ 6 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 + 18 = 0 (4) SolveEquation (4). M 1 = − 18 kN-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 + 18 − B y ( x 2 ) + M ′ = 0 (5) Substitute M ′ 6 for B y in Equation (5). M 2 + 18 − M ′ 6 ( x 2 ) + M ′ = 0 M 2 = M ′ x 2 6 − 18 − M ′ 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 ′ = 0 ) 0 ≤ x < 4 M 1 = − 18 kN-m ∂ M 1 ∂ M ′ = 0 M 1 = − 18 kN-m 4 < x < 10 M 2 = M ′ x 2 6 − 18 − M ′ ∂ 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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Chapter 14.10, Problem 14.140P

To determine

The slope at B.

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Chapter 14.10, Problem 14.139P

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The slope at B .

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

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The slope at B.

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