Explanation Given information: The yield strength ( σ y ) of the beam is 240 MPa . Calculation: Consider the width and depth of the beam are b = 60 mm and d = 90 mm . Calculate the distance ( c ) between the neutral axis and the outer fiber as follows: c = d 2 (1) Substitute 90 mm for d in Equation (1). c = 90 2 = 45 mm × 1 m 1 , 000 mm = 0.045 m Calculate the moment of inertia of cross-section of the beam as follows: I = 1 12 b d 3 (2) Substitute 60 mm for b and 90 mm for d in Equation (2). I = 1 12 × 60 × 90 3 = 3 , 645 , 000 mm 4 × 1 m 4 10 12 mm 4 = 3.645 × 10 − 6 m 4 Consider the value of y is 45 mm . Consider the width and depth of the shaded area are b 1 = 60 mm and d 1 = 45 mm . Calculate the shaded area ( A 1 ) of the cross-section of the beam as follows: A 1 = b 1 d 1 (1) Substitute 60 mm for b 1 and 45 mm for d 1 in Equation (1). A 1 = 60 × 45 = 2 , 700 mm 2 × 1 m 2 10 6 mm 2 = 0.0027 m 2 Calculate the force ( R ) using the relation: R = σ Y A 1 (2) Substitute 240 MPa for σ Y and 0.0027 m 2 for A 1 in Equation (2). R = 240 MPa × ( 10 6 N m 2 1 MPa ) × 0.0027 m 2 = 648 × 10 3 N The value of d is 45 mm

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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 4.6, Problem 87P

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Find the residual stress at y=45 mm.

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Chapter 4.6, Problem 86P

Chapter 4.6, Problem 88P

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Find the residual stress at y = 45 mm .

Solution Summary: The author calculates the moment of inertia of cross-section of the beam as follows: I=112bd3.

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Find the residual stress at y=45 mm.

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