Explanation Given information: Magnitude of the uniformly varying load is 0 ≤ w ≤ 0.8 kip/ft . Magnitude of Force applied at the end is F = 1.5 kip . Length of the beam A B is l A B = 9 ft . Length of the beam B C is l B C = 9 ft . Width of the rectangular cross-section is b = 6 in . Height of the rectangular cross-section is h = 12 in . Modulus of elasticity of a wooden beam is E = 1.6 × 10 3 ksi . For the beam to be in equilibrium. ∑ F y = 0 R A − 1 2 × l A B × w max + R B − F = 0 R A + R B = 1 2 × l A B × w max + F R A + R B = 1 2 × 9 ft × 0.8 kip/ft + 1.5 kip R A + R B = 5.1 kip .................. 1 For the beam to be in equilibrium. ∑ M A = 0 1 2 × l A B × w max × 2 3 l A B − R B × l A B + F × l A C = 0 1 2 × 9 ft × 0.8 kip/ft × 2 3 × 9 ft + 1.5 kip × 18 ft = R B × 9 ft R B = 48.6 kip-ft 9 ft R B = 5.4 kip .......................... 2 From (1) and (2). R A + 5.4 kip = 5.1 kip R A = − 0.3 kip Bending moment Equation is given by M I = E R 1 R = d 2 y d x 2 M I = E d 2 y d x 2 d 2 y d x 2 = M E I Bending moment at section X-X is given by using the Macauley equation. M x = R A x − 0 − 1 2 × x − 0 × w max x − 0 l A B × 1 3 × x − 0 + R B x − l A B + 1 2 × l A B × w max × x − 2 3 × l A B = − 0.3 x − 0 kip − 1 6 × 0.8 kip x − 0 3 9 ft + 5.4 kip x − 9 ft − 1 2 × 9 ft × 0.8 kip × x − 2 3 × 9 ft = − 0.3 x − 0.0148 x 3 + 5.4 x − 9 − 3.6 × x − 6 kip-ft Moment of inertia of the wooden beam is given by I = b h 3 12 = 6 in × 12 3 in 3 12 = 864 in 4 Calculation: Bending moment Equation is given by d 2 y d x 2 = M E I = 1 E I − 0

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ISBN: 9780137605460

Author: Russell C. Hibbeler

Publisher: Pearson Education (US)

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Chapter 12.3, Problem 48P

To determine

The equation of elastic curve and deflection at C .

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Chapter 12.3, Problem 47P

Chapter 12.3, Problem 49P

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