Explanation Assumption: The self-weight of the cable is ignored. The cable is perfectly flexible and inextensible. The tensile force acting in the cable is always tangent to the cable at points along its length. Being inextensible, the cable has a constant length both before and after the load is applied. A cable or a segment of the cable can be treated as a rigid body. Method of joints is used to determine the force in the cable. Consider the state of member as tension where the force is pulling the member and as compression where the force is pushing the member. Consider the force indicating right side as positive and left side as negative in the horizontal components of forces. Consider the force indicating upward is taken as positive and downward as negative in the vertical components of forces. Consider clockwise moment as negative and anticlockwise moment as positive wherever applicable. Apply the Equation of equilibrium wherever applicable. Given information: Take the vertical external load P 2 = 600 N and sag at point B is y B = 3 m . Segment BCDE : Show the free body diagram of segment BCDE as in Figure (1). Using Figure (1), Determine the value of θ : tan θ = Opposite side Adjacent side (I) Moment about the point E : Determine the magnitude of external force P 1 by taking moment about point E . ∑ M E = 0 P 2 ( 3 ) + P 1 ( 9 ) + P 2 ( 15 ) − F A B cos θ 1 ( 4 ) − F A B sin θ 1 ( 15 ) = 0 (II) Conclusion: Substitute 3 m for the opposite side and 3 m for adjacent side in Equation (I). tan θ 1 = 3 3 θ 1 = 45 ° Substitute 500 N for P 2 , and 45 ° for θ 1 Equation (II). 600 ( 3 ) + P 1 ( 9 ) + 600 ( 15 ) − F A B cos 45 ° ( 4 ) − F A B sin 45 ° ( 15 ) = 0 1 , 800 + 9 P 1 + 9 , 000 − 2.828 F A B − 10.61 F A B = 0 9 P 1 − 13.438 F A B + 10 , 800 = 0 (III) Joint B : Show the free-body diagram of the joint B as in Figure (2). Using Figure (2), Along the vertical direction: Determine the force in the cable segments AB and BC by resolving the vertical component of forces. ∑ F y = 0 F A B sin θ 1 − P 2 − F B C sin θ 2 = 0 (IV) Along the horizontal direction: Determine the force in the cable segments AB and BC by resolving the horizontal component of forces. ∑ F x = 0 − F A B cos θ 1 + F B C cos θ 2 = 0 (V) Conclusion: Substitute 1 m for the opposite side and 6 m for the adjacent side in Equation (I). tan θ 2 = 1 6 θ 2 = 9.46 ° Substitute 45 ° for θ 1 , 600 N for P 2 , and 9.46 ° for θ 2 in Equation (IV). F A B sin 45 ° − 600 − F B C sin 9.46 ° = 0 0.707 F A B − 0.164 F B C = 600 (VI) Substitute 45 ° for θ 1 and 9.46 ° for θ 2 in Equation (V). − F A B cos 45 ° + F B C cos 9.46 ° = 0 − 0.707 F A B + 0.986 F B C = 0 (VII) Add Equation (VI) and Equation (VII). 0.707 F A B − 0.164 F B C − 0.707 F A B + 0.986 F B C = 600 + 0 0.822 F B C = 600 F B C = 729.93 N Substitute 729.93 N for F B C in Equation (VII). − 0.707 F A B + 0.986 ( 729.93 ) = 0 0.707 F A B = 719.712 F A B = 1 , 017.98 N Substitute 1,017.98 N for F A B in Equation (III). 9 P 1 − 13.438 ( 1 , 017.98 ) + 10 , 800 = 0 9 P 1 = 2 , 879

15th Edition

ISBN: 9780137514724

Author: HIBBELER

Publisher: PEARSON

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Types of loading

In mechanical terminology, the term load indicates that external force/weight, either constant or variable, acts on a mechanical member/structure. The weight of an object of a specific material is also an external load, and the external force that works …

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Chapter 7, Problem 95P

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The magnitude of P1 and sag yD of point D.

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The magnitude of P 1 and sag y D of point D .

Solution Summary: The author shows the free body diagram of segment BCDE as in Figure (1).

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