Explanation Given information: The steel beam is supported on two springs, each having a stiffness of k = 8 MN/m . Young’s modulus or modulus of elasticity is E s t = 200 GPa = 200 × 10 9 N/m . Moment of inertia of the beam is I = 5 ( 10 6 ) mm 4 = 5 × 10 − 6 m 4 . 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 force in the spring at point B by taking moment about point A . ∑ M A = 0 ( F s p ) B ( 2 ) − ( 2 × 3 × 3 2 ) + ( 2 × 1 × 1 2 ) = 0 (1) Along the vertical direction: Determine the force in the spring at point A by resolving the vertical component of forces. ∑ F y = 0 ( F s p ) A + ( F s p ) B − 2 × 4 = 0 (2) Show the calculation of forces as follows: Conclusion: Solve Equation (1). ( F s p ) B ( 2 ) − 9 + 1 = 0 ( F s p ) B = 8 2 ( F s p ) B = 4 kN Substitute 4 kN for ( F s p ) B in Equation (2). ( F s p ) A + 4 − 8 = 0 ( F s p ) A = 4 kN Strain energy in the spring: Determine the strain energy in the spring using the Equation: ( U i ) s p = 1 2 k δ s p 2 (3) Here, k is stiffness and δ s p is spring deformation. Determine the spring deformation using the relation. δ s p = F s p k (4) Show the calculation of strain energy as follows: Conclusion: Substitute 4,000 N for F s p and ( 8 × 10 6 ) N/m for k in Equation (4). δ s p = 4 , 000 8 × 10 6 = 0.0005 m Substitute ( 8 × 10 6 ) N/m for k and 0.0005 m for δ s p in Equation (3). ( U i ) s p = 1 2 × 8 × 10 6 × ( 0.0005 ) 2 = 1 J Thus, the strain energy in each of the spring is ( U i ) s p = 1 J _ . 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 (5) Here, M is the moment in the beam and I is the moment of inertia of the beam. Region 0 ≤ x < 1 m from left end: Consider a section at a distance x 1 from support A . Due to symmetry of the beam section, determine the strain energy up-to half of the beam and multiply the value by twice. 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 = 0 M + 2 × x 1 × x 1 2 = 0 (6) Show the calculation of values as follows: Conclusion: Solve Equation (6). M = − x 1 2 Substitute 1 m for L , − x 1 2 for M , 200 × 10 9 N/m 2 for G , and 5 × 10 − 6 m 4 for I in Equation (5), and integrate. ( U 1 ) b = 2 × ∫ 0 1 m ( − x 1 2 ) 2 d x 2 × 200 × 10 9 × 5 × 10 − 6 = 10 6 200 × 10 9 × 5 × 10 − 6 ∫ 0 1 ( x 1 4 ) d x = 10 6 200 × 10 9 × 5 × 10 − 6 [ x 1 5 5 ] 0 1 = 10 6 200 × 10 9 × 5 × 10 − 6 [ ( 1 ) 5 5 − 0 ] 0 1 = 0

10th Edition

ISBN: 9780134319650

Author: Russell C. Hibbeler

Publisher: PEARSON

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Strain energy

The term strain indicates the variation in an object's length, area, and volume when the object is subjected to an external force. The external force may be of normal, shear, moment, torque, etc. The strain in an object depends on the types of external f…

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Chapter 14.2, Problem 14.18P

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The strain energy in each of the springs and the bending strain energy in the beam.

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Chapter 14.2, Problem 14.17P

Chapter 14.2, Problem 14.19P

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The strain energy in each of the springs and the bending strain energy in the beam.

Solution Summary: The author shows the free body diagram of the entire beam as in Figure 1. The steel beam is supported on two springs, each having a stiffness of k=8

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The strain energy in each of the springs and the bending strain energy in the beam.

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Chapter 14 Solutions