Explanation Given information: The yield strength ( σ y ) of the beam is 240 MPa . Calculation: Consider the outer width and depth of the beam are b = 60 mm and d = 90 mm . Consider the width and depth of hollow portion of the beam are b 1 = 30 mm and d 1 = 30 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 − 1 12 b 1 d 1 3 (2) Substitute 60 mm for b , 90 mm for d , 30 mm for b 1 , and 30 mm for d 1 in Equation (2). I = 1 12 × 60 × 90 3 − 1 12 × 30 × 30 3 = ( 3 , 645 , 000 − 67500 ) mm 4 × 1 m 4 10 12 mm 4 = 3.5775 × 10 − 6 m 4 Consider the value of y is 45 mm . Calculate the value of R 1 using the relation: R 1 = σ Y A 1 (1) Substitute 240 MPa for σ Y and ( 60 mm × 30 mm ) for A 1 in Equation (1). R 1 = 240 MPa × 1 N/mm 2 1 MPa × ( 60 mm × 30 mm ) = 240 × 1 , 800 = 432 , 000 N The distance y 1 between the neutral axis and the centroid of rectangle 1 is 30 mm . Calculate the value of R 2 using the relation: R 2 = σ Y A 2 (2) Substitute 240 MPa for σ Y and ( 30 mm × 15 mm ) for A 2 in Equation (2). R 2 = 240 MPa × 1 N/mm 2 1 MPa × ( 30 mm × 15 mm ) = 240 × 450 = 108 , 000 N The distance y 2 between the neutral axis and the centroid of rectangle 2 is y 2 = 1 2 ( 15 mm ) = 7

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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 88P

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Find the residual stress at y=45 mm.

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Chapter 4.6, Problem 87P

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Mechanics of Materials, 7th Edition

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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.

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Find the residual stress at y=45 mm.

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