Explanation Given information: Set the angle of the moment from y - axis is θ = 30 ° . Explanation: Show the free-body diagram of the square cross section as in Figure 1. Determine the moment of inertia in y and z - direction of the square cross section using the formula. I z = I y = b d 3 12 Here, b is width of the square section and d is depth of the square section. Substitute 250 mm for b and 250 mm for d . I z = I y = 250 mm × 1 m 1,000 mm × ( 250 mm × 1 m 1,000 mm ) 3 12 = 0.325521 × 10 − 3 m 4 Show the free body diagram of the square cross section for resultant moments as in Figure 2. The resultant moment in y -direction is: M y = M cos θ = 850 × cos 30 ° = 736.12 N-m The resultant moment in z -direction is as follows: M z = M sin θ = 850 × sin 30 ° = 425 N-m Determine the resultant normal stress at any point on the cross section using the equation. σ = − M z y I z + M y z I y (1) Here, σ is the normal stress at the point. Tensile stress is positive and compressive stress is negative. Stress at point A : The distance in y -direction is y A = − 125 mm . The distance in z -direction is z A = − 125 mm . Thus, the y and z coordinates of point A is (–125 mm, –125 mm). Substitute 425 N-m for M z , –125 mm for y , 0.325521 × 10 − 3 m 4 for I z , 736.12 N-m for M y , –125 mm for z , and 0.325521 × 10 − 3 m 4 for I y in Equation (1). σ A = − 425 × ( − 125 mm × 1 m 1,000 mm ) 0.325521 × 10 − 3 + 736.12 × ( − 125 mm × 1 m 1,000 mm ) 0.325521 × 10 − 3 = 163 , 200 − 282 , 670 = − 119 , 470 N/m 2 × 1 kPa 10 3 N/m 2 = − 119.4 kPa ≈ 119 kPa Thus, the stress at point A is σ A = − 119 kPa _ . Stress at point B : The distance in y -direction is y B = − 125 mm . The distance in z -direction is z B = 125 mm . Thus, the y and z coordinates of point B is (–125 mm, 125 mm). Substitute 425 N-m for M z , –125 mm for y , 0.325521 × 10 − 3 m 4 for I z , 736.12 N-m for M y , 125 mm for z , and 0.325521 × 10 − 3 m 4 for I y in Equation (1). σ B = − 425 × ( − 125 mm × 1 m 1,000 mm ) 0

Statics and Mechanics of Materials (5th Edition)

5th Edition

ISBN: 9780134382593

Author: Russell C. Hibbeler

Publisher: PEARSON

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Any functioning component, such as machine parts, structural members, pressure vessels, and so on, are all under the influence of more than one force. When anybody is under the influence of more than one force such as shear forces, bending moments, axial…

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Chapter 11.5, Problem 97P

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Find the stress at each corner of the square cross section.

Sketch the stress distribution.

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Chapter 11.5, Problem 96P

Chapter 11.5, Problem 98P

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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 stress at each corner of the square cross section. Sketch the stress distribution.

Solution Summary: The author shows the free-body diagram of the square cross section for resultant moments as in Figure 1.