Explanation Given information: The mass of the crane is 500 kg. Calculation: Show the free body diagram of member as shown in Figure 1. Here C x and C y denotes horizontal and vertical reaction, the point of force of action F D E is 0.4m above the horizontal axis member CD . Refer Figure 1: Write the equilibrium moment about C . ∑ M C = 0 F D E sin 30 ° ( 6 ) + F D E cos 30 ° ( 0.4 ) − ( 500 × 9.81 × 2 ) = 0 3 F D E + 0.346 F D E − 9 , 810 = 0 3.346 F D E − 9 , 810 = 0 F D E = 2 , 931.50 Show the equilibrium of free body diagram of the boom right cut segment as shown in Figure 2. Here, N is normal force, M is moment, and V is vertical shear force acting at downward direction. Refer Figure 2: The equilibrium force along x direction. ∑ F x = 0 N − 2931.50 cos 30 ° = 0 N − 2 , 538.75 = 0 N = 2 , 538.75 N The equilibrium force along y direction. ∑ F y = 0 2931.50 sin 30 ° − V = 0 1 , 465.75 − V = 0 V = 1 , 465.75 N ∑ M o = 0 2931.50 sin 30 ° ( 2 ) + 2 , 931.50 cos 30 ° ( 0.4 ) − M = 0 2931.50 + 1 , 015.50 − M = 0 M = 3 , 947 N ⋅ m Show the I section as shown in Figure 3. Refer Figure 3: Consider the size of big rectangle as 150 mm × 300 mm . and the size of two small rectangles as 130 mm × 260 mm . The moment of inertia of I- section is the subtraction of moment of inertia of big rectangle of 150 mm × 300 mm . and two small rectangles of 130 mm × 260 mm Determine the cross sectional area ( A ) as follows: A = b 1 d 1 − 2 ( b 2 d 2 ) (1) Here b 1 d 1 area of big rectangle and b 2 d 2 area of small rectangle. Substitute 150 mm for b 1 , 300 mm for d 1 , 65 mm for b 2 , and 260 mm for d 2 in Equation (1). A = 150 × 300 − 2 ( 65 × 260 ) = 45 , 000 − 33 , 800 = 11 , 200 mm 2 ( 1 m 2 10 6 mm 2 ) = 0.0112 m 2 Determine the moment of inertia ( I ) as shown in below: I = b 1 d 1 3 12 − 2 ( b 2 d 2 3 ) 12 (2) Substitute 150 mm for b 1 , 300 mm for d 1 , 65 mm for b 2 , and 260 mm for d 2 in Equation (2). I = 150 × 300 3 12 − 2 ( 65 × 260 ) 12 = 337 , 500 , 000 − 190 , 406 , 666.7 = 147 , 093 , 333

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 8.2, Problem 8.40P

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The state of stress (σA) at point A for boom.

The shear stress (τA) at point A for boom.

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Chapter 8.2, Problem 8.39P

Chapter 8.2, Problem 8.41P

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what is shear stress and normal? how to tell them while calculating?

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The state of stress ( σ A ) at point A for boom. The shear stress ( τ A ) at point A for boom.

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The state of stress (σA) at point A for boom.

The shear stress (τA) at point A for boom.