Explanation Given information: The width of the rectangular cross section is b . The height of the rectangular cross section is h . Assumption: Consider the force indicating right side as positive and left side as negative in horizontal components of forces. Consider the force indicating upward is taken as positive and downward as negative in vertical components of forces. Consider clockwise moment as negative and anti-clock wise moment as positive wherever applicable. Apply the Equation of equilibrium wherever applicable. The beam is symmetrical in cross section and external loads. The maximum bending moment will occur at the mid-span of the beam. Calculation: Show the free-body diagram of the entire beam as in Figure 1. Determine the vertical reaction at point B by taking moment about point A . ∑ M A = 0 B y ( L ) − 1 2 × w 0 × L 3 × ( 2 L 3 + 1 3 L 3 ) − w 0 × L 3 × ( L 3 + 1 2 L 3 ) − 1 2 × w 0 × L 3 × 2 3 L 3 = 0 B y ( L ) − w 0 L 6 × ( 7 L 9 ) − w 0 L 3 × ( L 2 ) − w 0 L 2 27 = 0 B y − 7 w 0 L 54 − w 0 L 6 − w 0 L 27 = 0 B y − 7 w 0 L 54 − 9 w 0 L 54 − 2 w 0 L 54 = 0 B y = w 0 L 3 Determine the vertical reaction at point A by resolving the vertical component of forces. ∑ F y = 0 A y + B y − 1 2 × w 0 × L 3 − w 0 × L 3 − 1 2 × w 0 × L 3 = 0 A y + w 0 L 3 − w 0 L 6 − w 0 L 3 − w 0 L 6 = 0 A y = w 0 L 3 Determine the horizontal direction at point B by resolving the horizontal component of forces. ∑ F x = 0 B x = 0 Show the calculation of shear force values as follows: Shear force at x = 0 : V | x = 0 ( Just left ) = 0 V | x = 0 ( Just right ) = A y = w 0 L 3 Shear force at x = L / 3 : V | x = L / 3 = w 0 L 3 − 1 2 × w 0 × L 3 = w 0 L 6 Shear force at x = L / 2 : V | x = L / 2 = w 0 L 6 − w 0 × L 6 = 0 Shear force at x = 2 L / 3 : V | x = 2 L / 3 = w 0 L 6 − w 0 × L 3 = − w 0 L 6 Shear force at x = L : V | x = L ( just left ) = − w 0 L 6 − 1 2 × w 0 × L 3 = − w 0 L 3 V | x = L ( just right ) = − w 0 L 3 + B y = − w 0 L 3 + w 0 L 3 = 0 Show the calculated shear force values as in Table 1. Distance, x Shear force, V 0 (Just left) 0 0 (Just right) w 0 L 3 L /3 w 0 L 6 L /2 0 2 L /3 − w 0 L 6 L (Just left) − w 0 L 3 L (Just right) 0 Show the shear force diagram as in Figure 2.

Statics and Mechanics of Materials (5th Edition)

5th Edition

ISBN: 9780134382593

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 11.4, Problem 82P

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Find the absolute maximum stress (σmax) in the beam.

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Chapter 11.4, Problem 81P

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Chapter 11 Solutions

Statics and Mechanics of Materials (5th Edition)

Ch. 11.2 - In each case, the beam is subjected to the...Ch. 11.2 - In each ease, express the shear and moment...Ch. 11.2 - In each ease, express the shear and moment...Ch. 11.2 - In each ease, express the shear and moment...Ch. 11.2 - In each ease, express the shear and moment...Ch. 11.2 - Prob. 5FP Ch. 11.2 - Prob. 6FP Ch. 11.2 - In each ease, draw the shear and moment diagrams...Ch. 11.2 - Prob. 8FP Ch. 11.2 - Prob. 1P

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Find the absolute maximum stress ( σ max ) in the beam.

Solution Summary: The author shows the free-body diagram of the entire beam as in Figure 1. The beam is symmetrical in cross section and external loads.

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Find the absolute maximum stress (σmax) in the beam.

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Chapter 11 Solutions