Explanation Given information: The W 8 × 28 rolled steel beam. Calculation: Write the properties of the rolled steel section W 8 × 28 as shown in table 1: Dimensions Unit(in.) d 8.06 in . b f 6.54 in . t f 0.465 in . t w 0.285 in . I x 98.0 in 4 I y 21.7 in 4 A 8.24 in 2 Here, d is depth of the section, b f is breadth of the flange, t f is thickness of the flange, I x is moment of inertia about x -axis, and I y is moment of inertia about y -axis. Sketch the force and couple moment as shown in Figure 1. Calculate the shear force and bending moment along x , y , and z at point a . Axial force of the section is P = 20 kips . Shear force of the section is as follows: V y = 5 kips V x = − 3.2 kips Here, V x is shear force in x direction and V y is shear force in y direction. Calculate the bending moment at the plane of point a . The bending moment along x axis, M x = − ( 20 × V y ) − ( d 2 × P ) (1) Substitute 8.06 in . for d , 20 kips for P , and 5 kips for V y in Equation (1). M x = − ( 20 × 5 ) − ( 8.06 2 × 20 ) = − 100 − 80.6 = − 180.6 kip ⋅ in . The bending moment along y axis, M y = − ( 20 + 4 ) × 3.2 = − 76.8 kip ⋅ in Find the distance x value at point a using the relation: x a = − b f 2 + 3 (2) Substitute 6.54 in . for b f in Equation (2). x a = − 6.54 2 + 3 = − 0.27 in . Find the distance y value at point a using the relation: y a = d 2 (3) Substitute 8.06 in . for d in Equation (3). y a = 8.06 2 = 4.03 in . Find the normal stress ( σ a ) using the relation: σ a = − P A + M x y a I x − M y x a I y (4) Here, P is centric force, A is the cross sectional area, M x , M y couple moment, I x is moment of inertia about x axis, I y is moment of inertia about y axis, y a is distance about y axis, and x a is distance about x axis. Substitute 20 kips for P , 8.24 in 2 for A , 21.7 in 4 for I y , 98.0 in 4 for I x , − 180.6 kip ⋅ in . for M x , − 76.8 kip ⋅ in for M y , − 0.27 in . for x a , and 4.03 in for y a in Equation (4). σ a = − 20 8.24 + − 180.6 × 4.03 98.0 − − 76.8 × − 0.27 21.7 = − 2.424 + 7.4267 − 0.95555 = 4.0439 ksi Sketch the shearing stress at point a due to V x as shown in figure 2. Determine the shearing stress at point a due to V x using the relation: Refer to Figure 2. Find the area ( A ) of section using the relation: A = b f t f (5) Substitute 3 in . for b f and 0.465 in . for t f in Equation (5). A = 3 × 0.465 = 1.395 in 2 Find the value of x ¯ using the relation: x ¯ = − b f 2 + 3 2 Substitute 6.54 in . for b f . x ¯ = − 6.54 2 + 3 2 = − 1.77 in . Find the first moment area ( Q ) using the relation: Q = − A x ¯ (6) Substitute 1.395 in 2 for A and − 1.77 in . for x ¯ in Equation (6). Q = − 1.395 × ( − 1.77 ) = 2.4692 in 3 Find the shear stress ( τ x z ) at point a using the relation: τ x z = V x Q I y t (7) Here, Q is the first moment area and t is thickness of section

7th Edition

ISBN: 9780073398235

Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek

Publisher: McGraw-Hill Education

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Chapter 8.3, Problem 57P

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The principal stresses and maximum shearing stress at point a.

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Chapter 8 Solutions

Mechanics of Materials, 7th Edition

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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The principal stresses and maximum shearing stress at point a .

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