Explanation Given information: The curved member is symmetric and is subjected to a moment of M = 600 lb-ft . Calculation: Sketch the free body diagram of the trapezoidal cross section passing through the point A and B as in Figure 1. Area of the trapezoidal section: Determine the area of the trapezoidal section using the formula. A = b 1 d 1 + 1 2 b 2 d 2 + 1 2 b 3 d 3 (1) Here, b 1 is width of the rectangular section 1, d 1 is the depth of the rectangular section, b 2 is width of the triangular section 2, d 2 is depth of the triangular section, b 3 is width of the triangular section 3, and d 3 is depth of the triangular section 3. Substitute 0.5 in. for b 1 , 2 in. for d 1 , 0.5 in. for b 2 , 2 in. for d 2 , 0.5 in. for b 3 , and 2 in. for d 3 in Equation (1). A = ( 0.5 × 2 ) + ( 1 2 × 0.5 × 2 ) + ( 1 2 × 0.5 × 2 ) = 1 + 0.5 + 0.5 = 2 in . 2 Properties of section 1: The cross sectional area of the section 1 is as follows: A 1 = b 1 d 1 = 0.5 × 2 = 1 in . 2 The centroidal distance is r 1 = 9 in . . Properties of section 2: The cross sectional area of the section 2 is as follows: A 2 = 1 2 b 2 d 2 = 1 2 × 0.5 × 2 = 0.5 in . 2 The centroidal distance is as follows: r 2 = 8 + ( 1 3 × 2 ) = 8.667 in . Properties of section 3: The cross sectional area of the section 3 is as follows: A 3 = 1 2 b 3 d 3 = 1 2 × 0.5 × 2 = 0.5 in . 2 The centroidal distance is as follows: r 3 = 8 + ( 1 3 × 2 ) = 8.667 in . Location of the Centroid: Determine the location of the centroid r ¯ using the relation: r ¯ = ∑ r ˜ A ∑ A = r 1 A 1 + r 2 A 2 + r 3 A 3 A 1 + A 2 + A 2 (2) Substitute 9 in. for r 1 , 1 in . 2 for A 1 , 8.667 in. for r 2 , 0.5 in . 2 for A 2 , 8.667 in. for r 3 , and 0.5 in . 2 for A 3 in Equation (2). r ¯ = 9 × 1 + 8.667 × 0.5 + 8.667 × 0.5 1 + 0.5 + 0.5 = 8.8333 in . Find the value of ∫ A d A r : Refer to Table 6.1, The value of ∫ A d A r for rectangular cross section is as follows: ∫ A d A r = b ln r 2 r 1 (3) Substitute 0.5 in. for b , 10 in. for r 2 , and 8 in. for r 1 in Equation (3). ∫ A d A r = b ln r 2 r 1 = 0.5 ln 10 8 = 0.111572 in . The value of ∫ A d A r for triangular section: ∫ A d A r = [ b r 2 ( r 2 − r 1 ) ( ln r 2 r 1 ) − b ] (4) Substitute 0

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 6.9, Problem 6.144P

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The bending stress in the member at points A and B.

To show: the stress acting on volume elements located at points A and B.

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Chapter 6.9, Problem 6.143P

Chapter 6.9, Problem 6.145P

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The bending stress in the member at points A and B . To show: the stress acting on volume elements located at points A and B .

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The bending stress in the member at points A and B. To show: the stress acting on volume elements located at points A and B.

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The bending stress in the member at points A and B.

To show: the stress acting on volume elements located at points A and B.