Explanation Given information: The Young’s modulus ( E ) is 1,500 ksi. The moment of inertia ( I ) is 20 , 000 in . 4 . Calculation: Consider flexural rigidity EI of the beam is constant. Show the free body diagram of the beam as in Figure (1). Refer Figure (1), Consider upward force is positive and downward force is negative. Consider clockwise moment is negative and counterclockwise moment is positive. Determine the support reaction at D using the Equation of equilibrium; ∑ M A = 0 R D × 15 − ( 30 × 10 ) − ( 1.5 × 6 × 6 2 ) = 0 R D = 327 15 R D = 21.8 kips Determine the reaction at support A ; ∑ V = 0 R A + R D − ( 1.5 × 6 ) − 30 = 0 R A = 39 − 21.8 R A = 17.2 kips Show the reactions of the beam as in Figure (2). Refer Figure (2), Determine the bending moment at A ; M A = 21.8 × 15 − ( 30 × 10 ) − ( 1.5 × 6 × 6 2 ) = 327 − 300 − 27 = 0 Determine the bending moment at B ; M B = 21.8 × 9 − ( 30 × 4 ) = 76.2 kips-ft Determine the bending moment at C ; M C = 21.8 × 5 = 109 kips-ft Determine the bending moment at D ; M D = 17.2 × 15 − ( 1.5 × 6 × ( 6 2 + 9 ) ) − ( 30 × 5 ) = 258 − 108 − 150 = 0 Show the M / E I diagram for the given beam as in Figure (3). Show the elastic curve diagram as in Figure (4). Refer Figure (3), Determine the deflection between D and A using the relation; Δ D A = Moment of the area of the M / E I diagram between D and A about A = [ ( 1 2 b 1 h 1 ) ( 1 3 × b 1 ) + ( 2 3 × w l 2 8 × l × ( 15 2 + 3 4 × l ) ) + ( b 3 × h 3 × ( b 3 2 + 5 ) ) + ( 1 2 b 4 h 4 ) ( 5 + 1 3 × b 4 ) + ( 1 2 b 5 h 5 ) ( 2 3 × b 5 ) ] Here, b is the width and h is the height of respective rectangle and triangle, w is the uniformly distributed load, and l is the length of uniformly distributed load acting. Substitute 6 ft for b 1 , 76.2 E I for h 1 , 1.5 kips/ft for w , 6 ft for l , 4 ft for b 3 , 76.2 E I for h 3 , 4 ft for b 4 , ( 109 E I − 76.2 E I ) for h 4 , 5 ft for h 4 , and 109 E I for h 5 . Δ D A = [ ( 1 2 × 6 × 76.2 E I ) ( 9 + 1 3 × 6 ) + ( 2 3 × 1.5 × 6 2 8 × 6 × ( 15 2 + 3 4 × 6 ) ) + ( 4 × 76.2 E I × ( 4 2 + 5 ) ) + ( 1 2 × 4 × ( 109 E I − 76.2 E I ) ) ( 5 + 1 3 × 4 ) + ( 1 2 × 5 × 109 E I ) ( 2 3 × 5 ) ] = 2 , 514.6 + 324 + 2 , 133.6 + 415.5 + 908 E I = 6 , 296 kips-ft 3 E I Determine the slope at A using the relation; θ A = Δ D A L Substitute 6 , 296 kips-ft 3 E I for Δ D A and 15 ft for L . θ A = 6 , 296 kips-ft 3 E I 15 = 419.73 kips-ft 2 E I The maximum deflection occurs at point M from the support B at a distance of x M

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ISBN: 9780100469105

Author: KASSIMALI

Publisher: YUZU

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Beams

In structural engineering, a beam is a component made of a variety of materials (including steel alloys, and wood) that can resist loads applied laterally to the beam axis. Members, elements, rafters, shafts, and purlins are all terms used to describe be…

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Chapter 6, Problem 28P

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Find the maximum deflection Δmax of the given beam using the moment-area method.

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Find the maximum deflection Δ max of the given beam using the moment-area method.

Solution Summary: The author calculates the maximum deflection of the beam using the moment-area method. The Young's modulus is 1,500 ksi.

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Find the maximum deflection Δmax of the given beam using the moment-area method.

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Chapter 6 Solutions