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Unformatted text preview: Problem 16.1 The beam in Figure 164 has length L = 4 m and the magnitude of the distributed load is w = 12 kN / m . The beam has a solid circular cross section with 80mm radius and is made of material with modulus of elasticity E = 200 GPa . Determine the beam’s deﬂection and slope at x = 1 m . Solution: The moment of inertia of the cross section is: I π 4 R 4 = π 4 (0 . 08 m) 4 = 3 . 22 × 10 − 5 m 4 The beam’s deﬂection (from page 626) is: v = w x 24 EI ( L 3 − 2 Lx 2 + x 3 ) The slope of the beam at any point is: v = w 24 EI ( L 3 − 6 Lx 2 + 4 x 3 ) At the point x = 1 m , the deﬂection of the beam is: v 1 m = ( 12 × 10 3 N / m ) (1 m) 24 200 × 10 9 N / m 2 (3 . 22 × 10 − 5 m 4 ) h (4 m) 3 − 2(4 m)(1 m) 2 + (1 m) 3 i ANS: v 1 m = 0 . 00443 m The slope of the beam at x = 1 m is: v 1 m = ( 12 × 10 3 N / m ) 24(200 × 10 9 N / m 2 )(3 . 22 × 10 − 5 m 4 ) h (4 m) 3 − 6 (4 m) (1 m) 2 + 4 (1 m) 3 i ANS: v 1 m = 0 . 00342 rad Problem 16.2 In Problem 16.1 what is the beam’s maximum deﬂection? What is the slope of the beam where the maximum deﬂection occurs? Solution: The beam’s deﬂection (from page 626) is: v = w x 24 EI ( L 3 − 2 Lx 2 + x 3 ) The slope of the beam at any point is: v = w 24 EI ( L 3 − 6 Lx 2 + 4 x 3 ) From the symmetrical loading and support of the beam, we see that maximum deﬂection occurs at x = 2 m . v 2 m = ( 12 × 10 3 N / m ) (2 m) 24 200 × 10 9 N / m 2 (3 . 22 × 10 − 5 m 4 ) h (4 m) 3 − 2 (4 m) (2 m) 2 + (2 m) 3 i ANS: v 2 m = 0 . 00622 m v 2 m = ( 12 × 10 3 N / m ) 24 200 × 10 9 N / m 2 (3 . 22 × 10 − 5 m 4 ) h (4 m) 3 − 6 (4 m) (2 m) 2 + 4 (2 m) 3 i ANS: v 2 m = 0 Problem 16.3 The beam in Figure 166a has length L = 72 in , and the magnitude of the distributed load is w = 14 lb / in . The beam’s moment of inertia I = 2 . 4 in 4 and it is made of material with a modu lus of elasticity E = 30 × 10 6 lb / sqin . Determine the deﬂection at x = 36 in , (a) by using Eq. (166); (b) by using Eq. (167). Solution: Using Equation (166) at x = 36 in : v 36 = w L 48 EI 9 Lx 2 − 4 x 3 = (14 lb / in) (72 in) 48(30 × 10 6 lb / in 2 )(2 . 4 in 4 ) h 9 (72 in) (36 in) 2 − 4 (36 in) 3 i ANS: v 36 = 0 . 1905 in Now using Equation (167) at x = 36 in : v 36 = w 384 EI 16 x 4 − 64 Lx 3 + 96 L 2 x 2 − 8 L 3 x + L 4 v 36 = (14 lb / in) 384(30 × 10 6 lb / in 2 )(2 . 4 in 4 ) h 16 (36 in) 4 − 64 (72 in) (36 in) 3 + 96 (72 in) 2 (36 in) 2 − 8 (72 in) 3 (36 in) + (72 in) 4 i ANS: v 36 = 0 . 1905 in Problem 16.4 In Problem 16.3 what is the beam’s maximum deﬂection? What is the slope of the beam where the maximum deﬂection occurs? Solution: It is apparent from the illustration that maximum deﬂection occurs at x = 72 in , Using the first derivative of Equation (167) at x = 72 in , to determine the slope at that point: v = w 384 EI 64 x 3 − 192 Lx 2 + 192 L 2 x − 8 L 3 v 72 = (14 lb / in) 384 30 × 10 6 lb / in 2 ( 2 . 4 in 4 ) h 64 (72 in) 3 − 192 (72 in) (72 in)...
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This note was uploaded on 04/03/2008 for the course MAE 101 taught by Professor Orient during the Winter '08 term at UCLA.
 Winter '08
 ORIENT
 Statics

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