Explanation Given information: Weight of the block is 75 kg. The overhang beam is made of 2014-T6 aluminum. Falling height of the block is h = 0.75 m . Speed of the block falling is v = 3 m/sec . Young’s modulus or modulus of elasticity is E = 73.1 GPa = 73.1 × 10 9 N/m 2 . Maximum elastic stress in the beam is σ Y = 414 MPa . Explanation: Determine the reactions: Entire beam: Show the free-body diagram of the entire beam as in Figure 1. Moment about the point A : Determine the vertical reaction at point B by taking moment about point A . ∑ M A = 0 B y ( 4 ) − P ( 6 ) = 0 (1) 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 − P = 0 (2) Show the calculation of reaction as follows: Conclusion: Solve Equation (1). 4 B y = 6 P B y = 1.5 P Substitute 1.5 P for B y in Equation (2). − A y + 1.5 P − P = 0 A y = 0.5 P Strain energy due to Bending: Determine the strain energy in the beam due to bending using the equation. ( U i ) b = ∫ 0 L M 2 d x 2 E I (3) Here, M is the moment in the beam, E is Young’s modulus, and I is the moment of inertia of the beam. Moment of Inertia: Determine the moment of inertia of the rectangular beam section using the formula. I = b d 3 12 (4) Here, b is width of beam and d is depth of beam. Region 0 ≤ x < 4 m from left end A : Consider a section at a distance x 1 from left support A . Show the free-body diagram of the section as in Figure 2. Moment about the section: Determine the moment at section by taking moment about the section. ∑ M x 1 = 0 M 1 + A y ( x 1 ) = 0 (5) Show the calculation of values as follows: Conclusion: Substitute 0.075 m for b and 0.15 m for d in Equation (4). I = 0.075 × 0.15 3 12 = 2.1094 × 10 − 5 m 4 Substitute 0.5 P for A y in Equation (5). M 1 + 0.5 P ( x 1 ) = 0 M 1 = − 0.5 P x 1 Substitute 4 m for L , − 0.5 P x 1 for M 1 , 73.1 × 10 9 N/m 2 for E , and 2.1094 × 10 − 5 m 4 for I in Equation (3), and integrate. ( U 1 ) b = ∫ 0 4 m ( − 0.5 P x 1 ) 2 d x 2 × 73.1 × 10 9 × 2.1094 × 10 − 5 = 0.25 P 2 2 × 73.1 × 10 9 × 2.1094 × 10 − 5 ∫ 0 4 ( x 1 2 ) d x = 0.25 P 2 2 × 73.1 × 10 9 × 2.1094 × 10 − 5 [ x 1 3 3 ] 0 4 = 0.25 P 2 2 × 73.1 × 10 9 × 2.1094 × 10 − 5 [ ( 4 ) 3 3 − 0 ] = 1.7294 × 10 − 6 P 2 Region 0 ≤ x < 2 m from right end C : Consider a section at a distance x 2 from right end. Show the free-body diagram of the section as in Figure 3. Moment about the section: Determine the moment at section by taking moment about the section. ∑ M x 2 = 0 − M 2 − P × x 2 = 0 (6) Show the calculation of values as follows: Conclusion: Solve Equation (6). M 2 = − P x 2 Substitute 2 m for L , − P x 1 for M 1 , 73.1 × 10 9 N/m 2 for E , and 2.1094 × 10 − 5 m 4 for I in Equation (3), and integrate. ( U 2 ) b = ∫ 0 2 m ( − P x 1 ) 2 d x 2 × 73.1 × 10 9 × 2.1094 × 10 − 5 = P 2 2 × 73.1 × 10 9 × 2.1094 × 10 − 5 ∫ 0 2 ( x 1 2 ) d x = P 2 2 × 73

Mechanics of Materials Plus Mastering Engineering with Pearson eText - Access Card Package (10th Edition)

10th Edition

ISBN: 9780134518121

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 14.4, Problem 14.66P

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The maximum bending stress developed in the beam.

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Mechanics of Materials Plus Mastering Engineering with Pearson eText - Access Card Package (10th Edition)

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The maximum bending stress developed in the beam.

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