Explanation Given information: The modulus of elasticity ( E a ) of Aluminum is 105 GPa . The modulus of elasticity ( E c ) of Copper is 75 GPa . The moment ( M ) in the beam is 35 N ⋅ m . Calculation: Show the cross-section of the composite bar as shown in Figure 1. Refer Figure 1. Consider the Copper and Aluminum is represented by rectangle 2 and 1. The width and depth of the rectangle 1 are b 1 = 24 mm and d 1 = 9 mm . The width and depth of the rectangle 2 are b 2 = 24 mm and d 1 = 3 mm . Consider Aluminum as the reference material, then the value of n 1 = 1 . Calculate the ratio n 1 for Copper using the relation: n 1 = E c E a (1) Substitute 105 GPa for E c and 75 GPa for E a in Equation (1). n 1 = 105 GPa 75 GPa = 1.4 Calculate the distance ( Y ¯ 0 ) between the neutral axis and the bottom of the beam using the relation: Y ¯ 0 = ( n 1 A 1 y 1 + n 2 A 2 y 2 n 1 A 1 + n 2 A 2 ) (2) Substitute ( 24 mm × 9 mm ) for A 1 , ( 24 mm × 3 mm ) for A 2 , 1.4 for n 1 , 1 for n 2 , 7.5 mm for y 1 , 1.5 mm for y 2 in Equation (2). Y ¯ 0 = ( 1 × ( 24 × 9 ) × 7.5 + 1.4 × ( 24 × 3 ) × 1.5 1.4 × ( 24 × 9 ) + 1 × ( 24 × 3 ) ) = 1 , 620 + 151.8 216 + 100.8 = 1 , 771.2 316.8 = 5.5909 mm Consider the entire cross-section of the composite bar is transformed into Aluminum. Calculate the moment of inertia ( I 1 ) of the rectangle 1 using the relation: I 1 = n 1 12 b 1 d 1 3 + n 1 A 1 ( y 1 − Y ¯ 0 ) 2 (3) Substitute 24 mm for b 1 , 9 mm for d 1 , 1 for n 1 , 5.5909 mm for Y ¯ 0 , and 7

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Chapter 4.5, Problem 44P

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Find the radius of curvature produced by a moment of 35 N⋅m.

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Chapter 4.5, Problem 43P

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Find the radius of curvature produced by a moment of 35 N ⋅ m .

Solution Summary: The author shows the cross-section of the composite bar as shown in Figure 1. Consider the Copper and Aluminum is represented by rectangle 2 and 1.

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