To determine The maximum bending stress acting on the beam and the orientation of the neutral axis. Explanation Given information: The beam is subjected to the internal moment of M = 1 , 200 kN-m . Calculation : Cross-section: Show the free-body diagram of the cross section as in Figure 1. Internal moment components: Sketch the distribution of moment in the y and z direction as in Figure 2. Resolve for the moment in y -direction. M y = 1 , 200 × sin 30 ° = 600 kN-m Resolve for the moment in z -direction. M z = − 1 , 200 × cos 30 ° = − 1 , 039.23 kN-m Centroid distance: Determine the centroid distance y ¯ using the relation. y ¯ = ∑ y ¯ A ∑ A = y 1 A 1 − y 2 A 2 A 1 − A 2 (1) Here, A 1 is cross sectional area of the large rectangular section, y 1 is the centroid distance of the large rectangular section, A 2 is cross sectional area of the smaller rectangular section, and y 2 is the centroid distance of the smaller rectangular section. Here, y 1 = 300 mm , A 1 = 300 mm × 600 mm y 2 = 375 mm , A 2 = 150 mm × 150 mm Substitute 300 mm for y 1 , 300 mm × 600 mm for A 1 , 375 mm for y 2 , and 150 mm × 150 mm for A 2 in Equation (1). y ¯ = 300 × 300 × 600 − 375 × 150 × 150 300 × 600 − 150 × 150 = 289.3 mm Moment of Inertia in y -axis: Determine the moment of inertia of the cross section using the equation: I y = d 1 b 1 3 12 − d 2 b 2 3 12 (2) Here, b 1 is width of the larger rectangular section, d 1 is depth of the larger rectangular section, b 2 is width of the smaller rectangular section, and d 2 is depth of the smaller rectangular section. Substitute 300 mm for b 1 , 600 mm for d 1 , 150 mm for b 2 , and 150 mm for d 2 , in Equation (2). I y = 600 × 300 3 12 − 150 × 150 3 12 = 1350 × 10 6 − 42.1875 × 10 6 = 1 , 307.8125 × 10 6 mm 4 × ( 1 m 1,000 mm ) 4 = 1 , 307.8125 × 10 − 6 m 4 Moment of Inertia in z -axis: Determine the moment of inertia of the cross section using the equation: I z = ∑ ( I ¯ + A d 2 ) = b 1 d 1 3 12 + A 1 ( y ¯ − y 1 ) 2 − b 2 d 2 3 12 + A 2 ( y ¯ − y 2 ) 2 (3) Substitute 300 mm for b 1 , 600 mm for d 1 , 300 mm × 600 mm for A 1 , 289.3 mm for y ¯ , 300 mm for y 1 , 150 mm for b 2 , 150 mm for d 2 , 150 mm × 150 mm for A 2 , and 375 mm for y 2 in Equation (3). I z = ( 300 × 600 3 12 + 300 × 600 ( 289.3 − 300 ) 2 − 150 × 150 3 12 − 150 × 150 ( 289.3 − 375 ) 2 ) = 5 , 400 × 10 6 + 20.6082 × 10 6 − 42.1875 × 10 6 − 165.251 × 10 6 = 5 , 213.17 × 10 6 mm 4 × ( 1 m 1,000 mm ) 4 = 5 , 213.17 × 10 − 6 m 4 Maximum bending stress at point A : Determine the maximum bending stress in the beam using the equation

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ISBN: 2810014920922

Author: HIBBELER

Publisher: PEARSON

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Chapter 6.5, Problem 6.111P

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The maximum bending stress acting on the beam and the orientation of the neutral axis.

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Chapter 6.5, Problem 6.110P

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