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 4 in 2 in 3 in AB 40 lb 20 lb C D E Problem 4-b The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect.

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Presentation on theme: " 4 in 2 in 3 in AB 40 lb 20 lb C D E Problem 4-b The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect."— Presentation transcript:

1  4 in 2 in 3 in AB 40 lb 20 lb C D E Problem 4-b The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine the reactions at C, D, and E when  = 30 o.

2 1. Draw a free-body diagram of the body. This diagram shows the body and all the forces acting on it.  4 in 2 in 3 in AB 40 lb 20 lb C D E Solving Problems on Your Own The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine the reactions at C, D, and E when  = 30 o. Problem 4-b

3 2. Write equilibrium equations and solve for the unknowns. For two-dimensional structure the three equations might be:  F x = 0  F y = 0  M O = 0 where O is an arbitrary point in the plane of the structure or  F x = 0  M A = 0  M B = 0 where point B is such that line AB is not parallel to the y axis or  M A = 0  M B = 0  M C = 0 where the points A, B, and C do not lie in a straight line.  4 in 2 in 3 in AB 40 lb 20 lb C D E Solving Problems on Your Own The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine the reactions at C, D, and E when  = 30 o. Problem 4-b

4  4 in 2 in 3 in AB 40 lb 20 lb C D E 4 in A 40 lb 20 lb C 2 in 3 in B D E E C D 30 o Draw a free-body diagram of the body. Problem 4-b Solution

5 4 in A C 40 lb20 lb C 2 in 3 in B D E E D 30 o Write equilibrium equations and solve for the unknowns.  F y = 0: E cos 30 o _ 20 _ 40 = 0 E = = lb E = 69.3 lb 60 o 60 lb cos 30 o Problem 4-b Solution

6 4 in A C 40 lb 20 lb C 2 in 3 in B D E E D 30 o  M D = 0: ( 20 lb)( 4 in) _ ( 40 lb)( 4 in) _ C ( 3 in) + E sin 30 o ( 3 in) = 0 _ 80 _ 3C ( 0.5 )( 3 ) = 0 C = lb C = 7.97 lb Problem 4-b Solution

7 4 in A C 40 lb 20 lb C 2 in 3 in B D E E D 30 o +  F x = 0: E sin 30 o + C _ D = 0 ( lb )( 0.5 ) lb _ D = 0 D = 42.6 lb


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