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chapter 14 answers

chapter 14 answers - Problem 14.1 In Example 7-1 determine...

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Problem 14.1 In Example 7-1, determine the internal forces and moment at C if the distance from A to C is L/ 2 . Free Body Diagram: Solution: We first determine the reactions at the ends of the beam: Σ M B = 0 = F L 4 A y ( L ) A y = F/ 4 Σ F y = 0 = F + A y + B y = F + F/ 4 + B y B y = 3 F/ 4 Σ F x = 0 = A x A x = 0 Using the FBD of the left-hand portion of the beam: Σ F x = 0 = C x ANS: C x = 0 Σ F y = 0 = A y V C = F/ 4 V C ANS: V C = F/ 4 Σ M A = 0 = F 4 L 2 + M C ANS: M C = FL/ 8
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Problem 14.2 Determine the internal forces and mo- ment at A , B , and C . STRATEGY – In this case you don’t need to determine the reactions at the built-in support. Cut the beam at the point where you want to determine the internal forces and moment and draw the free-body diagram of the part of the beam to the left of your cut. Remember that P , V , and M must be in their defined positive directions in your free-body diagrams. Solution: Cut the beam through point A and draw the FBD: Σ F x = 0 = P A ANS: P A = 0 Σ F y = 0 = 2 , 000 N V A ANS: V A = 2 , 000 N Σ M A = 0 = (2 , 000 N)(1 m) + M A ANS: M A = 2 , 000 N m Cut the beam through point B and draw the FBD: Σ F x = 0 = P B ANS: P B = 0 Σ F y = 0 = 2 , 000 N V B ANS: V B = 2 , 000 N Σ M B = 0 = (2 , 000 N)(2 m) + M B ANS: M B = 4 , 000 N m Cut the beam through point C and draw the FBD: Σ F x = 0 = P C ANS: P C = 0 Σ F y = 0 = 2 , 000 N V C ANS: V C = 2 , 000 N Σ M C = 0 = (2 , 000 N)(3 m) + M C ANS: M C = 6 , 000 N m
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Problem 14.3 Determine the internal forces and the moment at A , B , and C . Solution: The sum of forces exerted by a couple is zero in every direction, so the couple can be omitted from any force equations. Cut through the beam at point A and draw the FBD: Σ F x = 0 = P A ANS: P A = 0 Σ F y = 0 = V A ANS: V A = 0 Σ M A = 0 = 2 , 000 ft lb M A ANS: M A = 2 , 000 ft lb Cut through the beam at point B and draw the FBD: Σ F x = 0 = P B ANS: P B = 0 Σ F y = 0 = V B ANS: V B = 0 Σ M B = 0 = 2 , 000 ft lb M B ANS: M B = 2 , 000 ft lb Cut through the beam at point C and draw the FBD: Σ F x = 0 = P C ANS: P C = 0 F y = 0 = V C ANS: V C = 0 M C 0 = 2 , 000 ft lb M C ANS: M C = 2 , 000 ft lb
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Problem 14.4 Determine the internal forces and mo- ment at A . Solution: Cut the bar through point A and draw the FBD: Σ F x = 0 = (1 , 000 lb)( cos 30 ) P A ANS: P A = 866 lb Σ F y = 0 = (1 , 000 lb)( sin 30 ) V A ANS: V A = 500 lb Σ M A = 0 = (1 , 000 lb)( sin 30 )(6 ft) M A ANS: M A = 3 , 000 ft lb Free Body Diagram: Problem 14.5 Determine the internal forces and moment (a) at B ; (b) at C . Solution: Draw the FBD of the entire beam and determine the reactions at points A and D . Σ M A = 0 = (80 lb)(4 ft) + D y (12 ft) D y = 26 . 67 lb Σ F x = 0 = A x A x = 0 Σ F y = 80lb+ D y + A y = 80lb+26 . 67lb+ A y A y = 53 . 33lb (a) Cut the beam through point B and draw the FBD: Σ F y = 0 = 26 . 67 lb V B Σ F x = 0 = P B ANS: V B = 26 . 67 lb P B = 0 Σ M A = 0 : 80(4)+6( V B ) M B = 320+6(26 . 67) M B = 0 ANS: M B = 480 ft lb (b) Cut the beam through point C and draw the FBD: Σ F y = 0 = 26 . 67 lb V C Σ F x = 0 = P C ANS: V C = 26 . 67 lb P C = 0 Σ M D = 0 = V C (3 ft) M C = (26 . 67 lb)(3 ft) M C ANS: M C = 80 ft lb Free Body Diagram:
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Problem 14.6 Determine the internal forces and mo- ment at B (a) if x = 250 mm ;
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