Explanation Given information: Moment of inertia of the beam is I = 53.8 in . 4 . Method adopted: Castigliano’s theorem. Assumptions: Young’s modulus of the A992 steel beam is E = 29 ( 10 3 ) ksi . Explanation: Castigliano’s theorem to determine the slope for the beam is as follows: Δ = ∫ 0 L M ( ∂ M ∂ P ) d x E I (1) Here, Δ is displacement of the point caused by the real loads acting on the beam, P is an external force of variable magnitude applied to the beam at the point and in the direction of Δ , 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, 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 an external force P at the point where the displacement 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 A : Determine the vertical reaction at point B by taking moment about point A . ∑ M A = 0 8 ( 5 ) − B y ( 10 ) − P ( 15 ) = 0 (2) Along the vertical direction: Determine the vertical reaction at point A by resolving the vertical component of forces. ∑ F y = 0 A y − B y − 8 − P = 0 (3) Along the horizontal direction: Determine the horizontal reaction at point B by resolving the horizontal component of forces. ∑ F x = 0 B x = 0 Show the calculation of values as follows: Solve Equation (2). 40 − 10 B y − 15 P = 0 10 B y = 40 − 15 P B y = 4 − 1.5 P Substitute ( 4 − 1.5 P ) for B y in Equation (3). A y − 4 + 1.5 P − 8 − P = 0 A y = 12 − 0.5 P Region 0 ≤ x < 5 ft : Show the free body diagram of the beam for the region 0 ≤ x < 5 ft 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 + 8 ( x 1 ) = 0 (4) SolveEquation (4). M 1 = − 8 x 1 Region 5 ft < x < 15 ft : Show the free body diagram of the beam for the region 5 ft < x < 15 ft 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 + 8 ( 5 + x 2 ) − A y ( x 2 ) = 0 (5) Substitute ( 12 − 0.5 P ) for A y in Equation (5). M 2 + 40 + 8 x 2 − ( 12 − 0.5 P ) ( x 2 ) = 0 M 2 + 40 + 8 x 2 − 12 x 2 + 0.5 P x 2 = 0 M 2 = 4 x 2 − 40 − 0.5 P x 2 Region 5 ft < x ≤ 20 ft : Show the free body diagram of the beam for the region 5 ft < x ≤ 20 ft 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 + 8 ( 15 + x 3 ) − A y ( x 3 + 10 ) + B y ( x 3 ) = 0 (6) Substitute ( 12 − 0.5 P ) for A y and ( 4 − 1.5 P ) for B y in Equation (6). M 3 + 8 ( 15 + x 3 ) − ( 12 − 0.5 P ) ( x 3 + 10 ) + ( 4 − 1.5 P ) ( x 3 ) = 0 M 3 + 120 + 8 x 3 − 12 x 3 − 120 + 0.5 P x 3 + 5 P + 4 x 3 − 1.5 P x 3 = 0 M 3 + 0.5 P x 3 + 5 P − 1.5 P x 3 = 0 M 3 = P x 3 − 5 P Calculation: Tabulate the calculated values as in Table 1

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Moment of Inertia

Inertia can be termed as a resistance offered by a body to change its state. A rigid body initially at rest will apply resistance when acted by an external force that tends to change its state of rest or velocity, or a body initially at motion will provi…

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Chapter 14.10, Problem 14.135P

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The displacement at point C of the beam made from A992 steel.

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Chapter 14.10, Problem 14.134P

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The displacement at point C of the beam made from A992 steel.

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

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