Explanation Assumption: Apply the sign convention for calculating the equations of equilibrium as below. For the horizontal forces equilibrium condition, take the force acting towards right side as positive ( → + ) and the force acting towards left side as negative ( ← − ) . For the vertical forces equilibrium condition, take the upward force as positive ( ↑ + ) and downward force as negative ( ↓ − ) . For moment equilibrium condition, take the clockwise moment as negative and counter clockwise moment as positive. Calculation: Consider the vertical reaction at A and I are denoted by A y and I y Calculate the end reactions at A and I as follows: Taking moment about A , ∑ M A = 0 − 2 ( 2.4 ) − 2 ( 4.8 ) − 2 ( 7.2 ) − 1 ( 9.6 ) + A y ( 9.6 ) = 0 − 4.8 − 9.6 − 14.4 − 9.6 + 9.6 A y = 0 − 38.4 + 9.6 A y = 0 9.6 A y = 38.4 A y = 4 kN Consider the equilibrium of forces along horizontal direction. ∑ F x = 0 A x = 0 Pass a section through member CD , DE, and DF . Sketch the free body diagram of roof truss as shown in Figure 1. Refer to Figure 1. Calculate the distance BJ and DE . B J = 2.16 9.6 = 9 40 D E = − 1 2.4 = − 5 12 Find the a value as shown in below: a = 0.46 slope BJ Substitute 9 40 for BJ

Vector Mechanics for Engineers: Statics, 11th Edition

11th Edition

ISBN: 9780077687304

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek

Publisher: McGraw-Hill Education

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Chapter 6.2, Problem 6.55P

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Find the force in member CE,DE, and DF.

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Chapter 6.2, Problem 6.54P

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6/86 The connecting rod AB of a certain internal-combustion engine weighs 1.2 lb with mass center at G and has a radius of gyration about G of 1.12 in. The piston and piston pin A together weigh 1.80 lb. The engine is running at a constant speed of 3000 rev/min, so that the angular velocity of the crank is 3000(2)/60 = 100л rad/sec. Neglect the weights of the components and the force exerted by the gas in the cylinder compared with the dynamic forces generated and calculate the magnitude of the force on the piston pin A for the crank angle 0 = 90°. (Suggestion: Use the alternative moment relation, Eq. 6/3, with B as the moment center.) Answer A = 347 lb 3" 1.3" B 1.7" PROBLEM 6/86

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Chapter 6 Solutions

Vector Mechanics for Engineers: Statics, 11th Edition

Ch. 6.1 - 6.1 through 6.8 Using the method of joints,...Ch. 6.1 - Prob. 6.2P Ch. 6.1 - Prob. 6.3P Ch. 6.1 - 6.1 through 6.8 Using the method of joints,...Ch. 6.1 - 6.1 through 6.8 Using the method of joints,...Ch. 6.1 - Using the method of joints, determine the force in...Ch. 6.1 - 6.1 through 6.8 Using the method of joints,...Ch. 6.1 - Prob. 6.8P Ch. 6.1 - 6.9 and 6.10 Determine the force in each member of...Ch. 6.1 - 6.9 and 6.10 Determine the force in each member of...

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