Explanation Given information: The allowable bending stress is σ allow = 150 MPa The allowable shear stress is τ allow = 85 MPa The outer radius of the cross section is 80 mm. The load P is 10 kN. Calculation: Determine the reactions: Entire beam: Show the free body diagram of beam as shown in Figure 1. Let, A y is the reaction at point A , B y is the reaction at B Moment about point A : Determine the vertical reaction at point B by taking moment about point A . ∑ M A = 0 B y × 3 − ( 10 × 2 ) − ( 10 × 1 ) = 0 3 B y − 20 − 10 = 0 3 B y − 30 = 0 3 B y = 30 B y = 10 kN Along the vertical direction: Determine the vertical reaction at point A by resolving the vertical component of forces. ∑ F y = 0 A y + B y = 20 (1) Substitute 10 kN for B y in Equation (1). A y + B y = 20 A y + 10 = 20 A y = 20 − 10 A y = 10 kN Sketch the calculated values as shown in Figure 2. Show the calculated values as shown in Figure 2. Calculated shear force as follows: Shear force at 0 m from left end. S F @ 0 m = 10 Shear force at 1 m from left end. S F @ 1 m = 10 − 10 = 0 Shear force at 2 m from left end. S F @ 2 m = 10 − 10 − 10 = 10 − 20 = − 10 kN Shear force at 3 m from left end. S F @ 3 m = 10 − 10 − 10 + 10 = 20 − 20 = 0 Show the shear force as shown in Figure 3. Calculated bending moment as follows: Bending moment at 0 m from left end. B M @ 0 m = 0 Bending moment at point 1 m from left end. B M @ 1 m = 10 × 1 = 10 kN ⋅ m Bending moment at point 2 m from left end. B M @ 2 m = 10 × 2 − ( 10 × 1 ) = 20 − 10 = 10 kN ⋅ m Bending moment at point 3 m from left end. B M @ 3 m = 10 × 3 − ( 10 × 2 ) − ( 10 × 1 ) = 30 − 20 − 10 = 10 − 10 = 0 Show the bending as shown in Figure 4. Refer to Figure 4: The maximum bending moment is 10 kN ⋅ m . Sketch the cross section of the shaft as shown in Figure 5. Bending stress : Calculate the moment of inertia ( I ) using the flexure formula. I = π 4 ( r o 4 − r i 4 ) (2) Here, r o is the outer radius of shaft and r i is the inner radius of shaft. Substitute 80 mm for r o in Equation (2). I = π 4 ( 80 4 mm ( 1 m 1 , 000 mm ) − r i 4 ) = π 4 ( 0.08 4 − r i 4 ) Find the inner radius r i as follows using flexure formula. σ allow = M max c I (3) Here, M max is the maximum bending moment, I is the moment of inertia, and c is the centroid distance of cross section. Substitute 150 MPa for σ allow , 10 kN ⋅ m for M max , 0.04 m for t , and π 4 ( 0.08 4 − r i 4 ) for I in Equation (3)

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Ch. 11.2 - Determine the minimum dimension a to the nearest...Ch. 11.2 - of the rod to safely support the load. The rod is...Ch. 11.2 - The wood has an allowable normal stress of allow =...Ch. 11.2 - of the beam's cross section to safely support the...Ch. 11.2 - Determine the minimum dimension b to the nearest...Ch. 11.2 - The beam is made of steel having an allowable...Ch. 11.2 - Determine its dimensions if it is to be...Ch. 11.2 - Determine the minimum width of the beam to the...Ch. 11.2 - if P=10 kip.Ch. 11.2 - If the allowable bending stress is allow = 22 ksi...

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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