Explanation Given information: The allowable bending stress for the beam is σ allow = 170 MPa . The allowable shear stress for the beam is τ allow = 95 MPa . Calculation: Show the cross section of the beam as shown in Figure 1. Refer to Figure 1. Consider that the width and depth of the rectangle 1 are b 1 = 150 mm and d 1 = 15 mm . Consider that the width and depth of the rectangle 2 are b 2 = 15 mm and d 2 = 150 mm . Calculate the areas of rectangles 1 and 2 as follows: A 1 = b 1 d 1 = 150 × 15 = 2250 mm 2 A 2 = b 2 d 2 = 15 × 150 = 2250 mm 2 The distances between the top fiber and the centroid of rectangles 1 and 2 are denoted as y 1 = 7.5 mm and y 2 = 90 mm . The distance between the top fiber and the neutral axis is denoted by y ¯ . Calculate the value of y ¯ using the given relation: y ¯ = y 1 A 1 + y 2 A 2 A 1 + A 2 (1) Substitute 2250 mm 2 for A 1 , 7.5 mm for y 1 , 2250 mm 2 for A 2 , 90 mm for y 2 in Equation (1). y ¯ = 7.5 × 2250 + 90 × 2250 2250 + 2250 = 219 , 375 4 , 500 = 48.75 mm Calculate the moment of inertia ( I ) as follows: I = b 1 d 1 3 12 + A 1 ( y ¯ − y 1 ) 2 + b 2 d 2 3 12 + A 2 ( y ¯ − y 2 ) 2 (2) Substitute 2250 mm 2 for A 1 , 7.5 mm for y 1 , 2250 mm 2 for A 2 , 90 mm for y 2 , 150 mm for b 1 , 15 mm for d 1 , 15 mm for b 2 , 150 mm for d 2 , and 48.75 mm for y ¯ in Equation (2). I = 150 × 15 3 12 + 2 , 250 ( 48.75 − 7.5 ) 2 + 15 × 150 3 12 + 2 , 250 ( 48.75 − 90 ) 2 = 0.04218 × 10 6 + 3.8285 × 10 6 + 4.2187 × 10 6 + 3.8285 × 10 6 = 11.917 × 10 6 mm 4 Calculate the value of the section modulus S as follows: S = I c (3) Here, I is the moment of inertia and c is the distance between the neutral axis and the outermost fiber. Calculate the value of c as follows: c = Overall Depth − y ¯ = 165 mm − 48.75 mm =116 .25 mm Substitute 116 .25 mm for c and 11.917 × 10 6 mm 4 for I in Equation (3). S = 11.917 × 10 6 mm 4 116 .25 mm = 0.1025 × 10 6 mm 3 Show the cross section of the beam as shown in Figure 2. Calculate the first moment of area ( Q max ) using the given relation: Q max = y ¯ ′ A ′ (4) Here, A ′ is the shaded area, and y ¯ ′ is the centroid of the shaded area. Refer to Figure 1. Substitute ( 165 − 48.75 ) 2 mm for y ¯ ′ and ( 165 − 48.75 ) × 15 mm 2 for A ′ in Equation (4). Q max = ( 165 − 48.75 ) 2 × ( 165 − 48.75 ) × 15 = 6757

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 11.2, Problem 11.27P

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Find the maximum value of load P that that is safely supported by the beam.

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Chapter 11.2, Problem 11.26P

Chapter 11.2, Problem 11.28P

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Chapter 11 Solutions

Mechanics of Materials

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Ch. 11.2 - The allowable bending stress is allow = 24 ksi and...Ch. 11.2 - The allowable bending stress is allow = 22 ksi and...Ch. 11.2 - 11–7. Draw the shear and moment diagrams for the...Ch. 11.2 - *11–8. The simply supported beam is made of timber...Ch. 11.2 - The beam has an allowable normal stress of allow =...Ch. 11.2 - The beam has an allowable normal stress of allow...Ch. 11.2 - 11–11. The timber beam is to be loaded as shown....Ch. 11.2 - If each beam is to be designed to carry 90 lb/ft...Ch. 11.2 - 11–13. Select the lightest steel wide-flange beam...Ch. 11.2 - 11–14. Select the lightest-weight steel...Ch. 11.2 - Prob. 11.15P Ch. 11.2 - The beam has an allowable normal stress of allow....Ch. 11.2 - Determine the maximum cable force P that can...Ch. 11.2 - to safely support the load. The wood has an...Ch. 11.2 - and the wood has an allowable normal stress of...Ch. 11.2 - Prob. 11.20P Ch. 11.2 - Prob. 11.21P Ch. 11.2 - Prob. 11.22P Ch. 11.2 - 11–23. The beam is constructed from three boards...Ch. 11.2 - Prob. 11.24P Ch. 11.2 - Prob. 11.25P Ch. 11.2 - Select the lightest-weight wide-flange beam from...Ch. 11.2 - Prob. 11.27P Ch. 11.2 - * 11–28. The joist AB used in housing construction...Ch. 11.2 - If the maximum bending stress is not to exceed...Ch. 11.2 - 11–30. The simply supported beam supports a load...Ch. 11.4 - Determine the variation in the width was a...Ch. 11.4 - Prob. 11.32P Ch. 11.4 - Prob. 11.33P Ch. 11.4 - The beam is made from a plate that has a constant...Ch. 11.4 - Determine the variation in the depth d of a...Ch. 11.4 - Determine the variation of the radius r of the...Ch. 11.4 - Prob. 11.37P Ch. 11.4 - Determine the variation in the width b as a...Ch. 11.4 - The tubular shaft has an inner diameter of 15 mm....Ch. 11.4 - Prob. 11.40P Ch. 11.4 - Prob. 11.41P Ch. 11.4 - Prob. 11.42P Ch. 11.4 - Prob. 11.43P Ch. 11.4 - Prob. 11.44P Ch. 11.4 - Prob. 11.45P Ch. 11.4 - Prob. 11.46P Ch. 11 - The cantilevered beam has a circular cross...Ch. 11 - Select the lightest-weight wide-flange overhanging...Ch. 11 - Prob. 11.49RP Ch. 11 - Determine the shaft's diameter to the nearest...Ch. 11 - Select the lightest-weight wide-flange beam from...Ch. 11 - The simply supported joist is used in the...Ch. 11 - The simply supported joist is used in the...Ch. 11 - by 4-in. pieces of wood braced as shown. If the...

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