Chapter_4_Solutions

Chapter_4_Solutions - Chapter 4 For a torsion bar, k T = T/...

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Chapter 4 4-1 For a torsion bar, k T = T/ = Fl/ , and so = Fl/k T . For a cantilever, k l = F/ , = F/k l . For the assembly, k = F/y , or, y = F/k = l + Thus 2 Tl FF l F y kk k  Solving for k 2 2 1 . 1 lT kA l kl k n s ______________________________________________________________________________ 4-2 For a torsion bar, k T = T/ = Fl/ , and so = Fl/k T . For each cantilever, k l = F/ l , l = F/k l , and , L = F/k L . For the assembly, k = F/y , or, y = F/k = l + l + L . Thus 2 l y k k  L Solving for k 2 2 1 . 11 LlT lL TL TlL kkk l kkl k k k n s ______________________________________________________________________________ 4-3 (a ) For a torsion bar, k =T/ =GJ/l . Two springs in parallel, with J = d i 4 /32, and d 1 = d 1 = d , 44 12 1 2 4 32 .( 1 ) 32 JG JG d d kG x lx x lx Gd Ans xlx      Deflection equation,  2 1 2 1 results in (2) Tl x Tx JG JG T x From statics, T 1 + T 2 = T = 1500. Substitute Eq. (2) Chapter 4 - Rev B, Page 1/81
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22 2 1500 1500 . (3) lx x TT T A n s xl     Substitute into Eq. (2) resulting in 1 1500 . (4) TA n l s ( b ) From Eq. (1),   46 3 11 0.5 11.5 10 28.2 10 lbf in/rad . 32 5 10 5 kA  n s From Eq. (4), 1 10 5 1500 750 lbf in . 10 T Ans  From Eq. (3), 2 5 1500 750 lbf in . 10 T Ans  From either section,   3 3 3 16 1500 16 30.6 10 psi 30.6 kpsi . 0.5 i i T Ans d ______________________________________________________________________________ 4-4 Deflection to be the same as Prob. 4-3 where T 1 = 750 lbf in, l 1 = l / 2 = 5 in, and d 1 = 0.5 in 1 = 2 =     12 3 44 4 4 6 750 5 60 10 (1) 0.5 32 32 32 dd dG G  Or, 34 15 10 (2) Td 10 10 (3) Equal stress, 1 2 33 3 3 1 2 16 16 (4) T T d d    Divide Eq. (4) by the first two equations of Eq.(1) results in 21 1.5 (5) Statics, T 1 + T 2 = 1500 (6) Substitute in Eqs. (2) and (3), with Eq. (5) gives 4 3 15 10 10 10 1.5 1500 Solving for d 1 and substituting it back into Eq. (5) gives d 1 = 0.388 8 in, d 2 = 0.583 2 in Ans. Chapter 4 - Rev B, Page 2/81
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From Eqs. (2) and (3), T 1 = 15(10 3 )(0.388 8) 4 = 343 lbf in Ans. T 2 = 10(10 3 )(0.583 2) 4 = 1 157 lbf in Ans. Deflection of T is    11 1 46 1 343 4 0.053 18 rad / 32 0.388 8 11.5 10 Tl JG  Spring constant is 3 1 1500 28.2 10 lbf in . 0.053 18 T k Ans The stress in d 1 is   3 1 1 3 3 1 16 343 16 29.7 10 psi 29.7 kpsi . 0.388 8 T Ans d The stress in d 1 is   3 2 2 3 3 2 16 1157 16 29.7 10 psi 29.7 kpsi . 0.583 2 T Ans d ______________________________________________________________________________ 4-5 ( a ) Let the radii of the straight sections be r 1 = d 1 /2 and r 2 = d 2 /2. Let the angle of the taper be where tan = ( r 2 r 1 )/2. Thus, the radius in the taper as a function of x is r = r 1 + x tan , and the area is A = ( r 1 + x tan ) 2 . The deflection of the tapered portion is 2 1 00 1 0 1 21 12 1 tan tan tan 1 tan tan tan tan tan tan tan 4 . l ll FF d x F dx AE E E r x rx Er r l E r r rr l F l E E rrE Fl Ans ddE 2 1          ( b ) For section 1, 4 1 22 6 1 4 4(1000)(2) 3.40(10 ) in . (0.5 )(30)(10 ) Fl Fl Ans AE d E  For the tapered section, 4 6 4 4 1000(2) 2.26(10 ) in .
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This note was uploaded on 11/20/2011 for the course MCE 325 taught by Professor Ahm during the Spring '11 term at American Dubai.

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Chapter_4_Solutions - Chapter 4 For a torsion bar, k T = T/...

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