Explanation Given information: The diameter of the shaft is d = 100 mm . Assumptions: Young’s modulus of the 2014-T6 aluminum shaft is E = 73.1 × 10 6 kN/m 2 . Explanation: Virtual-work equation for the entire beam is: 1 ⋅ Δ = ∫ 0 L m M E I d x (1) Here, 1 is external virtual unit load acting on the beam in the direction of Δ , Δ is displacement caused by the real loads on the beam, m is internal virtual moment in the beam, expressed as a function of x and caused by the external virtual unit load, M is internal moment in the beam, expressed as a function of x and caused by the real loads, 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 ): Determine the reactions: Entire shaft: Show the free-body diagram of the entire shaft as in Figure 1. Moment about point A : Determine the vertical reaction at point B by taking moment about point A . ∑ M A = 0 B y ( 3 ) − 8 ( 2 ) − 8 ( 1 ) = 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 − 8 = 0 (3) Show the calculation of values as follows: Conclusion: Solve Equation (2). B y ( 3 ) − 16 − 8 = 0 B y = 8 kN Substitute 8 kN for B y in Equation (3). A y + 8 − 8 − 8 = 0 A y = 8 kN Region 0 ≤ x < 1 m : Show the free body diagram of the shaft for the region 0 ≤ x < 1 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 − A y ( x 1 ) = 0 (4) Substitute 8 kN for A y Equation (4). M 1 − 8 ( x 1 ) = 0 M 1 = 8 x 1 Region 1 m < x < 1.5 m : Show the free body diagram of the shaft for the region 1 m < x < 1.5 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 − A y ( 1 + x 2 ) + 8 ( x 2 ) = 0 (5) Substitute8 kN for A y in Equation (5). M 2 − 8 ( 1 + x 2 ) + 8 ( x 2 ) = 0 M 2 − 8 − 8 x 2 + 8 x 2 = 0 M 2 = 8 kN-m Internal virtual moment in the beam ( m ): Apply a unit moment at point A and calculate the moments. Determine the reactions: Entire shaft: Show the free-body diagram of the entire shaft as in Figure 4. Moment about point A : Determine the vertical reaction at point B by taking moment about point A . ∑ M A = 0 − 1 ( 1.5 ) + B y ( 3 ) = 0 (6) 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 − 1 = 0 (7) Show the calculation of values as follows: Conclusion: Solve Equation (6). 3 B y = 1.5 B y = 0.5 Substitute 0.5 for B y in Equation (7). A + 0.5 − 1 = 0 A y = 0.5 Region 0 ≤ x < 1 m : Show the free body diagram of the shaft for the region 0 ≤ x < 1 m as in Figure 5. Moment about the section: Determine the moment at the section by taking moment about the section. ∑ M x 1 = 0 m 1 − A y ( x 1 ) = 0 (8) Substitute 0.5for A y in Equation (8). m 1 − 0

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 14.7, Problem 14.104P

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The slope at A of the 2014-T6 aluminum shaft.

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Chapter 14.7, Problem 14.103P

Chapter 14.7, Problem 14.105P

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The slope at A of the 2014-T6 aluminum shaft.

Solution Summary: The author shows the free-body diagram of the 2014-T6 aluminum shaft as in Figure 1. Determine the vertical reaction at point A by resolving vertical component of forces.

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The slope at A of the 2014-T6 aluminum shaft.

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