Explanation Given information: The rectangular cross section size is 20 × 40 mm . Calculation: Refer to Figure P8.51 in the textbook. Forces at H are as follows: The force in x direction, F x = − 3 kN . The force in y direction, F y = − 0.5 kN The force in z direction, F z = − 10 kN Moments at H are as follows: Find the moment about y axis as follows: M y = ( 0.150 × 10 kN ( 10 3 N 1 kN ) ) = 0.150 × 10 , 000 = 1 , 500 N ⋅ m Find the moment about z axis as follows: M z = − ( 0.150 × 500 ) = − 75 N ⋅ m M x = 0 Sketch the force couple system containing at point H as shown in Figure 1. Find the area of cross section ( A ) using the equation: A = b h (1) Here, b is the width of the cross section and h is the height of the cross section. Substitute 20 mm for h and 40 mm for b in Equation (1). A = 20 × 40 = 800 mm 2 ( 1 m 10 3 mm ) 2 = 800 × 10 − 6 m 2 Find the moment of inertia ( I z ) of section using the relation: I z = b h 3 12 (2) Substitute 20 mm for h and 40 mm for b in Equation (2). I z = 40 × 20 3 12 = 320 , 000 12 = 26.667 × 10 3 mm 4 ( 1 m 10 3 mm ) 4 = 26.667 × 10 − 9 m 4 Determine the normal stress at point H using the relation: σ H = P A − M z y I z (3) Here P is the centric force, A is the area of cross section, M z is the moment about z axis, y is the distance from y axis, and I z is the moment of inertia about z axis. Substitute − 3 kN for P , 26.667 × 10 − 9 m 4 for I z , − 75 N ⋅ m for M z , 10 mm for y , and 800 × 10 − 6 m 2 for A in Equation (3). σ H = − 3 kN ( 10 3 N 1 kN ) 800 × 10 − 6 m 2 + − 75 × 10 mm ( 1 m 10 3 mm ) 26.667 × 10 − 9 = − 3 , 000 800 × 10 − 6 − − 75 × 0.01 26.667 × 10 − 9 = − 3 , 750 , 000 + 28 , 124 , 648.44 = 24 , 374 , 648.44 Pa ( 1 MPa 10 6 Pa ) σ H = 24.375 MPa Determine the shearing stress at point H using the relation: τ H = 3 2 V z A (4) Here, V z is the shear about z axis. Substitute − 10 kN for V z and 800 × 10 − 6 m 2 for A in Equation (4). τ H = 3 2 10 kN ( 10 3 N 1 kN ) 800 × 10 − 6 = 18 , 750 , 000 Pa ( 1 MPa 10 6 Pa ) = 18.75 MPa Sketch represents the stresses as shown in Figure 2.

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ISBN: 9780100257061

Author: BEER

Publisher: YUZU

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Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial …

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Chapter 8.3, Problem 52P

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The principal stresses and the maximum shearing stress at point H.

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Chapter 8 Solutions

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Ch. 8.2 - A W10 = 39 rolled-steel beam supports a load P as...Ch. 8.2 - Solve Prob. 8.1, assuming that P = 22.5 kips and a...Ch. 8.2 - An overhanging W920 449 rolled-steel beam...Ch. 8.2 - Solve Prob. 8.3, assuming that P = 850 kN and a =...Ch. 8.2 - 8.5 and 8.6 (a) Knowing that all = 160 MPa and all...Ch. 8.2 - 8.5 and 8.6 (a) Knowing that all = 160 MPa and all...Ch. 8.2 - 8.7 and 8.8 (a) Knowing that all = 24 ksi and all...Ch. 8.2 - 8.7 and 8.8 (a) Knowing that all = 24 ksi and all...Ch. 8.2 - 8.9 through 8.14 Each of the following problems...Ch. 8.2 - 8.9 through 8.14 Each of the following problems...

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