Questions 16 through 24: Member AD in the assembly of Fig 4 is too short by...
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Questions 16 through 24: Member AD in the assembly of Fig 4 is too short by an amount d. It must be stretched downward to be pinned at point A to the rigid member ABC. Member BE has that same cross- sectional area A and Young's Modulus E. (Q16 to Q20: 2 points each, Q21 to Q24: 1 point each) 16) The downward force that must be applied onto point A of member AD in order to stretch it enough to pin it to rigid member ABC is dL a) EA EAL c) = d L b) EA EAd d) L y L IP L x A,E L+d rigid A B C L L Fig 4 (Questions 16-24) 17) The corresponding strain in AD as it is stretched to point A on the rigid member ABC is dL a) A L b) c) 1/14 d) / 18) After AD is pinned at A to rigid member ABC and the applied force is removed, a) member AD will be in tension and BE will be in compression b) member BE will be in tension and AD will be in compression c) both AD and BE will be in tension. d) both AD and BE will be in compression. 19) After AD is pinned at A to rigid member ABC and the applied force is removed, the magnitude F of the force supported by AD is related to the magnitude of the force P supported by BE as a) F = P b) F = = 2P c) P = 2F d) P = 3F 20) Assuming small displacements of the rigid member ABC, the displacement d at point A and dB at point B are approximately related by a) d = 2dB b) 2d = dB c) d = dB d) ddB = 2L 21) After AD is pinned at A to rigid member ABC and the applied force is removed, the magnitude F of the force supported by AD depends on the upward displacement d of the rigid member ABC by a) EAA L EA(d-dA) b) L c) dAL EA (d-dA)L d) e) EAL EA d A 22) After AD is pinned at A to rigid member ABC and the applied force is removed, point A of the rigid member ABC displacement upward by an amount a) d b) d c) d d) d 23) After AD is pinned at A to rigid member ABC and the applied force is removed, the residual stress in member AD Ed a) L b) 2Ed L Ed c) 5L 4Ed d) 5L 24) After AD is pinned at A to rigid member ABC and the applied force is removed, the residual stress in member BE a) -4Ed 5L 4Ed b) 5L -2Ed -5Ed c) d) 5L 2L Questions 25 through 30: A B P* FIG 5a Questions 25 to 30 P P" P* Y X P' Members AB and CD, made of different types of steels are fixed at ends A and C respectively, and attached to a rigid plate at ends B and D. The rigid plate can move only vertically as shown in Fig. 5a. The cross section areas of AB and CD are A1 and A2 respectively. Initial lengths of AB and CD are both equal to L. The elastic moduli of AB and CD are E1 and E2. Both steels are elastic - perfectly plastic. The yield stress of AB is Y1 and that of BC is y2. Assume y2 > Oy1 N 0 8' 8* 8" FIG 5b Questions 25 to 30 A load P* is applied to the rigid plate as shown, and then reduced slowly to zero. On the left is shown the combined load displacement diagram of the system. Q25 to Q30: 2 points each 25) The coordinates of point X are (8', P'). The value of 8' is a) y1L/E1 b) y1A1/E1 c) (y1+02)L/E1 d) (y1A1+ y2A2)L/E1 e) P'L/E1A1 26) The coordinates of point X are (8', P'). The value of P' is a) E1A1'/L b) y1A1+ y2A2 c) (Oy1+Oy2)A1 d) (y1+y2)( A1+A2) e) y1 (E1A1+ E2A2)/E1 27) The coordinates of point Y are (8", P"). The value of 8" is e) 28' a) y2L/E2 b) y2A2/E2+8' c) (y1+y2)L/E1 d) (y1A1+ y2A2)L/E1 28) The coordinates of point Y are (8", P"). The value of P" is a) 2P* b) y1A1+ y2A2 c) (y1+y2)(A1+A2) d) A11+A202 29) The unloading slope when decreasing the load from P* to zero is equal to a) (E1A1+E2A2)/L b) P*/8* c) (P"-P')/(8"-8') d) 0 e) Oy2(E1A1+ E2A2)/E1 e) (E1A2+E2A1)/L 30) The permanent set after reducing load from P* to 0 is equal to a) 8*/P' b) 8*-( S'P*/P') c) 8*-8' d) (8*+8')/2 e) 8*-(y1+y2)( A1+A2)/L Questions 16 through 24: Member AD in the assembly of Fig 4 is too short by an amount d. It must be stretched downward to be pinned at point A to the rigid member ABC. Member BE has that same cross- sectional area A and Young's Modulus E. (Q16 to Q20: 2 points each, Q21 to Q24: 1 point each) 16) The downward force that must be applied onto point A of member AD in order to stretch it enough to pin it to rigid member ABC is dL a) EA EAL c) = d L b) EA EAd d) L y L IP L x A,E L+d rigid A B C L L Fig 4 (Questions 16-24) 17) The corresponding strain in AD as it is stretched to point A on the rigid member ABC is dL a) A L b) c) 1/14 d) / 18) After AD is pinned at A to rigid member ABC and the applied force is removed, a) member AD will be in tension and BE will be in compression b) member BE will be in tension and AD will be in compression c) both AD and BE will be in tension. d) both AD and BE will be in compression. 19) After AD is pinned at A to rigid member ABC and the applied force is removed, the magnitude F of the force supported by AD is related to the magnitude of the force P supported by BE as a) F = P b) F = = 2P c) P = 2F d) P = 3F 20) Assuming small displacements of the rigid member ABC, the displacement d at point A and dB at point B are approximately related by a) d = 2dB b) 2d = dB c) d = dB d) ddB = 2L 21) After AD is pinned at A to rigid member ABC and the applied force is removed, the magnitude F of the force supported by AD depends on the upward displacement d of the rigid member ABC by a) EAA L EA(d-dA) b) L c) dAL EA (d-dA)L d) e) EAL EA d A 22) After AD is pinned at A to rigid member ABC and the applied force is removed, point A of the rigid member ABC displacement upward by an amount a) d b) d c) d d) d 23) After AD is pinned at A to rigid member ABC and the applied force is removed, the residual stress in member AD Ed a) L b) 2Ed L Ed c) 5L 4Ed d) 5L 24) After AD is pinned at A to rigid member ABC and the applied force is removed, the residual stress in member BE a) -4Ed 5L 4Ed b) 5L -2Ed -5Ed c) d) 5L 2L Questions 25 through 30: A B P* FIG 5a Questions 25 to 30 P P" P* Y X P' Members AB and CD, made of different types of steels are fixed at ends A and C respectively, and attached to a rigid plate at ends B and D. The rigid plate can move only vertically as shown in Fig. 5a. The cross section areas of AB and CD are A1 and A2 respectively. Initial lengths of AB and CD are both equal to L. The elastic moduli of AB and CD are E1 and E2. Both steels are elastic - perfectly plastic. The yield stress of AB is Y1 and that of BC is y2. Assume y2 > Oy1 N 0 8' 8* 8" FIG 5b Questions 25 to 30 A load P* is applied to the rigid plate as shown, and then reduced slowly to zero. On the left is shown the combined load displacement diagram of the system. Q25 to Q30: 2 points each 25) The coordinates of point X are (8', P'). The value of 8' is a) y1L/E1 b) y1A1/E1 c) (y1+02)L/E1 d) (y1A1+ y2A2)L/E1 e) P'L/E1A1 26) The coordinates of point X are (8', P'). The value of P' is a) E1A1'/L b) y1A1+ y2A2 c) (Oy1+Oy2)A1 d) (y1+y2)( A1+A2) e) y1 (E1A1+ E2A2)/E1 27) The coordinates of point Y are (8", P"). The value of 8" is e) 28' a) y2L/E2 b) y2A2/E2+8' c) (y1+y2)L/E1 d) (y1A1+ y2A2)L/E1 28) The coordinates of point Y are (8", P"). The value of P" is a) 2P* b) y1A1+ y2A2 c) (y1+y2)(A1+A2) d) A11+A202 29) The unloading slope when decreasing the load from P* to zero is equal to a) (E1A1+E2A2)/L b) P*/8* c) (P"-P')/(8"-8') d) 0 e) Oy2(E1A1+ E2A2)/E1 e) (E1A2+E2A1)/L 30) The permanent set after reducing load from P* to 0 is equal to a) 8*/P' b) 8*-( S'P*/P') c) 8*-8' d) (8*+8')/2 e) 8*-(y1+y2)( A1+A2)/L
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Income Tax Fundamentals 2013
ISBN: 9781285586618
31st Edition
Authors: Gerald E. Whittenburg, Martha Altus Buller, Steven L Gill
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