Chapter 4. F Fl F F. Solving for k.

Transcription

1 Chapter 4 4- For a torsion bar, k T = T/ = Fl/, and so = Fl/k T. For a cantilever, k l = F/, = F/k l. For the assembl, k = F/, or, = F/k = l + Thus F Fl F k kt kl Solving for k kk l T k ns. l kl l kt kt kl 4- 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 assembl, k = F/, or, = F/k = l + l + L. Thus F Fl F F k kt kl kl Solving for k kkk L l T k ns. l kk l Ll ktkl ktkl kt kl kl 4- (a) For a torsion bar, k =T/ =GJ/l. Two springs in parallel, with J =d 4 i /, and d = d = d, 4 4 JG JG d d k G x lx x lx 4 Gd x l x ns. () Deflection equation, Tx T l x JG JG T l x results in T () x From statics, T + T = T = 5. Substitute Eq. () Chapter 4 - Rev B, Page /8

2 l x x T T 5 T 5 x l ns. () Substitute into Eq. () resulting in l x T 5 l ns. (4) 4 (b) From Eq. (), k lbf in/rad 5 5 ns. 5 From Eq. (4), T 5 75 lbf in ns. 5 From Eq. (), T 5 75 lbf in ns. T 5 i From either section,. psi. kpsi d.5 ns. i 4-4 Deflection to be the same as Prob. 4- where T = 75 lbfin, l = l / = 5 in, and d =.5 in = = T 4 T T T 4 4 () dg dg.5 d d G Or, T T 5 d () 4 d () 4 T T T T Equal stress, d d d d (4) Divide Eq. (4) b the first two equations of Eq.() results in T T d d d.5 d (5) 4T 4T 4 4 d d Statics, T + T = 5 () Substitute in Eqs. () and (), with Eq. (5) gives 4 d 4 d Solving for d and substituting it back into Eq. (5) gives d =.88 8 in, d =.58 in ns. Chapter 4 - Rev B, Page /8

3 From Eqs. () and (), T = 5( )(.88 8) 4 = 4 lbfin T = ( )(.58 ) 4 = 57 lbfin ns. ns. Deflection of T is Tl rad JG 4 / T 5 8. lbf in ns..5 8 Spring constant is k The stress in d is T 4 d psi 9.7 kpsi ns. The stress in d is T 57 d psi 9.7 kpsi ns. 4-5 (a) Let the radii of the straight sections be r = d / and r = d /. Let the angle of the taper be where tan = (r r )/. Thus, the radius in the taper as a function of x is r = r + x tan, and the area is = (r + x tan ). The deflection of the tapered portion is l F F dx F dx E E E r x tan r xtan tan F F E rtan tanr ltan Etan r r F r r F ltan Fl Etan rr Etan rr rr E 4Fl ns. dd E (b) For section, l Fl 4Fl 4()() (.5 )()( ) E d E 4.4( ) in. ns For the tapered section, 4 Fl 4 () 4.( ) in ns. dd E (.5)(.75)()( ) For section, l Chapter 4 - Rev B, Page /8

4 Fl 4Fl 4()() 4.5( ) in ns. E d E (.75 )()( ) 4- (a) Let the radii of the straight sections be r = d / and r = d /. Let the angle of the taper be where tan = (r r )/. Thus, the radius in the taper as a function of x is r = r + x tan, and the polar second area moment is J = ( /) (r + x tan ) 4. The angular deflection of the tapered portion is l l T T dx T dx GJ G G r x r x 4 tan tan tan T T G r tan tanr ltan Gtan r r T r r T l r r Tl r rr r Gtan r r Gr r r r G r r Tl d dd d ns. G d d (b) The deflections, in degrees, are For section, Tl 8 Tl 8 (5)() 8.44 deg ns. (.5 ).5( ) 4 4 GJ d G For the tapered section, For section, Tl( d d d d ) 8 Gd d (5)().5 (.5)(.75) deg ns..5( )(.5 )(.75 ) Tl 8 Tl 8 (5)() 8.48 deg ns. 4 4 GJ dg (.75 ).5( ) 4-7 The area and the elastic modulus remain constant, however the force changes with respect to x. From Table -5 the unit weight of steel is =.8 lbf/in, and the elastic modulus is E = Mpsi. Starting from the top of the cable (i.e. x =, at the top). F = ()(lx) l Chapter 4 - Rev B, Page 4/8

5 l l l.8 5() ( l x) dx.9 in lx x () l Fdx w c o E E E E From the weight at the bottom of the cable, Wl 4Wl 4(5) 5() W E d E (.5 )( ) 5.9 in c W in ns. The percentage of total elongation due to the cable s own weight.9 ().% ns F = = R F R = F M = = M Fa M = Fa V B = F, M B =F (x a ), V BC = M BC = Section B: F x B Fx adx ax C EI EI () B = at x = C = F x F x x B ax dx a EI C () EI B = at x = C = Fx B x a ns. EI Section BC: BC dx C EI From Eq. (), at x = a (with C = ), F a aa ( ) Fa EI EI Fa BC EI Fa Fa BC dx x C4 EI EI () = C. Thus, Chapter 4 - Rev B, Page 5/8

6 F a a Fa From Eq. (), at x = a (with C = ), a. Thus, from Eq. () EI EI Fa Fa Fa ac4 C4 Substitute into Eq. () EI EI EI Fa Fa Fa EI EI EI x a x BC ns. The maximum deflection occurs at x= l, Fa max a l ns. EI 4-9 M C = = F (l /) R l R = F / F = = F / + R F R = F / Break at x l /: V B = R = F /, M B = R x = Fx / Break at l / x l : V BC = R F = R = F /, M BC = R x F ( x l / ) = F (l x) / Section B: Fx F x B dx C EI EI 4 From smmetr, B = at x = l / l F Fl C C. Thus, 4EI EI F x Fl F B 4x l () EI 4 EI EI F F 4x B 4x l dx l x C EI EI Chapter 4 - Rev B, Page /8

7 B = at x = C =, and, Fx () 48EI B 4x l BC is not given, because with smmetr, Eq. () can be used in this region. The maximum deflection occurs at x =l /, l F l Fl max 4 l ns. 48EI 48EI 4- From Table -, for each angle, I = 7 cm 4. Thus, I = (7) ( 4 ) = 4.4( ) mm 4 - From Table -9, use beam with F = 5 N, a = mm, and l = mm; and beam with w = N/mm and l = mm. 4 Fa l max ( a l) w EI 8EI 4 5() ()() () (7) (4.4) 8(7)( )(4.4)( ) 5.4 mm ns. M Fa( w l /) O = 5() [( )/] = 9.5( ) Nmm F rom Table -, from centroid to upper surface is = 9 mm. From centroid to bottom surface is = 9. = 7 mm. The maximum stress is compressive at the bottom of the beam at the wall. This stress is M 9.5( )( 7) max MPa ns. I 4.4( ) Chapter 4 - Rev B, Page 7/8

8 4- R R O C 4 (45) () 45 lbf (45) () 85 lbf M = 45() =.48( ) lbfin M =.48( ) +5(4) = 4.( ) lbfin M max 4. max 5 Z.8 in Z Z For deflections, use beams 5 and of Table -9 Fa [ l ( l/)] l l Fl xft a l EIl 48EI 45(7)() (4 ) ()( ) I(4) 48()( ) I 4 4 I. in I /. in Select two 5 in-.7 lbf/ft channels from Table -7, I = (7.49) = 4.98 in 4, Z =(.) =. in. midspan.4 in max 5.7 kpsi I (.5 ).485 in 4 From Table -9 b superposition of beams and 7, at x = a = 5 in, with b = 4 in and l = 9 in Fba a [ a b l ] (la a l ) EIl w 4EI t x = l / = 9.5 in 4(4) () (.485)9 (5 /)(5) 4() (.485) (9)(5 ) in. ns Chapter 4 - Rev B, Page 8/8

9 Fa [ l ( l /)] l ( /) l l a l l l l EIl 4EI 4(5)(9.5) ()( )(.485)(9) 4()( )(.485) w (5 /)(9.5) l ns (9)(9.5 ) in % difference () 5.% ns..978 ()( ).84 mm 4- I 4 From Table -9-, beam Fa C ( l a) EI Fax B l x EIl d B Fa ( l x ) dx EIl db t x =, dx Fal Fal EIl EI Fa l O a EI With both loads, Fa l Fa O ( l a) EI EI Fa 4( ) (l a) (5) ().7 mm ns. EI (7) (.84) t midspan, Fa( l / ) l Fal 4()(5 ) E l. mm ns. EIl 4 EI I (.5 ).59 in 4 Chapter 4 - Rev B, Page 9/8

10 From Table -5, E =.4 Mpsi From Table -9, beams and, b superposition Fl 4() () B Fa (.4) (.59) (.4) (.59) B ( a l) () (4)() EI EI B.94 in ns. 4-5 From Table -7, I = (.85) =.7 in 4 From Table -5, E =. Mpsi From Table -9, beams and, b superposition 4 4 Fl ( w w ) 5( ) 5 (5 /) ( ) c l.8 in ns. EI 8EI () (.7) 8() (.7) 4 4- I d 4 From Table -5, E 7( ) MPa From Table -9, beams 5 and 9, with F C = F = F, b superposition Fl B Fa B (4a l ) I FBl Fa(4a l ) 48EI 4EI 48E B I 55( ) 75 (5) 4(5 ) ( ) 48(7) 5.4 mm (5.4). mm. d I ns 4-7 From Table -9, beams 8 (region BC for this beam with a = ) and (with a = a), b superposition M Fax B x lx l x l x EIl EIl M x lx l xfaxl x n. EIl s d M F( x l) BC x lx l x ( x l) [( x l) a(x l dx EIl )] EI xl M l F( x l) ( x l) [( xl) a( x l) ] EI EI Chapter 4 - Rev B, Page /8

11 ( ) ( ) ( x l) M lf xl a xl ns. EI 4-8 Note to the instructor: Beams with discontinuous loading are better solved using singularit functions. This eliminates matching the slopes and displacements at the discontinuit as is done in this solution. a wa M C Rl w ala R la l ns. wa wa F lar w a R l l ns. V R wa B l a la x a l w wx = wx = l ns. a VBC R w l ns. w x M B VBdx lax a xc l M B atx C M wx B al a lx l ns. wa wa M BC VBCdx dx xc l l wa wa M BC at xl C MBC ( lx) l ns. M B wx w B dx al a lxdx alx a x lx C EI EI l EI l w EI l B Bdx alx a x lx C dx w EI l at x C B 4 alx a x lx Cx C4 4 BC B M a a EI EI w l EI w l at xa BC BC dx ( l x) dx lx x C5 w 4 wa wa ala a la C la a C5 C C EI l EI l 5 () Chapter 4 - Rev B, Page /8

12 a a BC BCdx lx x C5 dx lx x C5x EI w w C l EI l wal BC at x l C C5l wa BC lx x l C5( xl) EI l at xa B BC w 5 4 wa ala a la Ca la a l C5( a l) l l wa Ca la 4 l C5( al) () 4l wa Substituting () into () ieldsc5 a 4l. Substituting this back into () gives 4l wa C 4ala 4l. Thus, 4l w 4 4 B 4alx a x lx 4a lx a x 4a l x 4EIl wx ( ) B ax l a lx a l a ns. 4EIl w 4 4 BC alx ax ax4alxal ns. 4EIl This result is sufficient for BC. However, this can be shown to be equivalent to w 4 4 w 4 BC 4alx a x lx 4a l x 4 a lx a x ( x a) 4EIl 4EI w 4 BC B ( xa) ns. 4EI b expanding this or b solving the problem using singularit functions. 4-9 The beam can be broken up into a uniform load w downward from points to C and a uniform load upward from points to B. wx x B bx ( l b) lx b l b w ax ( la) lx a la 4EIl 4EIl wx bx ( l b) b l b ax ( l a) a l a ns. 4EIl w 4 BC bx ( lb) lx b xlb 4EIl alx a x lx 4a l x4 a lxa xl( xa) ns. Chapter 4 - Rev B, Page /8

13 w CD 4blx b x lx 4b l x 4 b lx b x l( x b) 4EIl w alx a x lx 4a l x 4 a lx a x l( x a) 4EIl w 4 4 ( xb) ( xa) B ns. 4EI 4- Note to the instructor: See the note in the solution for Problem 4-8. wa wa F RB w a RB la ns. l l For region BC, isolate right-hand element of length (l + a x) wa VB R, VBC wlax ns. l wa w M B Rx x, MBC lax ns. l wa EI B M Bdx x C 4l wa EIB x Cx C l wa B = at x = C = EIB x Cx l al B = at x = l C w wa wa l wa x wa x EIB x x l x B l x ns. l l EIl w EI BC MBCdx lax C w EI 4 BC l a x Cx C wa wa BC = at x = l Cl C4 C4 Cl 4 4 () al al a B = BC at x = l w w wa C C w la 4 wa Substitute C into Eq. () gives C4 a 4lla 4. Substitute back into BC 4 w 4 wa wa wa l BC l a x xl a EI la 4 4 w 4 4 lax 4 a lxlaa ns. 4EI Chapter 4 - Rev B, Page /8

14 4- Table -9, beam 7, wl () R R 5 lbf wx x B lx x l () x x 4EI 4.5 x x x.7778 d dx w 4EI Slope: B B lx 4x l l w B xl 4EI w 4EI w l BC B xl xl x.7778 x xl 4EI 4() (.5) t x = l, ll 4l l From Prob. 4-, wa 4 wa 4 R RB l a l () l () wax 4 x B l x x x EIl.5 8 lbf () 4 48 lbf x w 4 4 BC lax 4a lxlaa 4EI x 44 x x 89x9 Superposition, R lbf R lbf ns. B x x x x x ns. B BC.7778 x x 89x9 ns. The deflection equations can be simplified further. However, the are sufficient for plotting. Using a spreadsheet, x x Chapter 4 - Rev B, Page 4/8

15 x x (a) Useful relations F 48EI k l kl 8 I.58 in 48E 48() 4 From I = bh /, and b = h, then I = 5 h 4 /, or, I (.58) 5 5 h in h is close to / in and 9/ in, while b is close to 5.4 in. Changing the height drasticall changes the spring rate, so changing the base will make finding a close solution easier. Trial and error was applied to find the combination of values from Table -7 that ielded the closet desired spring rate. h (in) b (in) b/h k (lbf/in) / 5 8 / 5½ 78 / 5¾ / / Chapter 4 - Rev B, Page 5/8

16 h = ½ in, b = 5 ½ in should be selected because it results in a close spring rate and b/h is still reasonabl close to. (b) 4 I 5.5(.5) /.579 in Mc ( Fl / 4) c 4 I 4() (.579) F 58 lbf I I lc (.5) Fl (58).84 in ns. 48EI 48() (.579) 4- From the solutions to Prob. -8, T lbf and T 4 lbf 4 4 d (.5) 4 I. 98 in 4 4 From Table -9, beam, Fbx Fb x z ( x b l ) ( x b l ) EIl EIl ( 575)()() 4 () (.98)(4) 4()() () (.98)(4) () (.98)(4) xin 4. in. xin ns dz d Fbx Fb x ( x b l ) ( x b l ) dx xin dx EIl EIl Fb Fb ( x b l ) ( x b l ) EIl EIl (575)() 4 4() 4 () (.98)(4) 4.( ) rad ns. xin 4-4 From the solutions to Prob. -9, T 88 N and T 4 N I 4 4 d () 9.7 mm Chapter 4 - Rev B, Page /8

17 The load in between the supports supplies an angle to the overhanging end of the beam. That angle is found b taking the derivative of the deflection from that load. From Table -9, beams (subscript ) and (subscript ), a () BC C beam beam x a lx lx x a l d Fa l x Fa dx EIl EIl BC C Fa l a EIl Equation () is thus Fa Fa l a a ( la) EIl EI () 7( ) (7) (9.7) (5) (7) (9.7) xl xl mm ns. The slope at, relative to the z axis is z Fa d F( xl) ( l a ) ( x l) a( x l) EIl dx EI Fa F l a ( x l) a( x l) a( x l) EIl EI Fa F ( l a ) a la EIl EI () (7) (9.7) (5) 7 (7) (9.7) 5 ( ) (5)() x l a xla.4 rad ns. 4-5 From the solutions to Prob. -7, T 9. lbf and T 58.8 lbf 4 4 d () I.49 9 in 4 4 From Table -9, beam, 4 Chapter 4 - Rev B, Page 7/8

18 Fbx ( 5)(4)(8) in. x b l ns EIl x8in () (.49 9)() Fbx ( 45.98)()(8) ( ) 8.48 in ns. z x b l EIl x8in () (.49 9)() The displacement magnitude is z in ns. d d Fbx Fb x b l ( a b l ) dx dx EIl EIl z xa () (.499)() xa ( 5)(4) rad. dz d Fbx Fb dx dx EIl EIl ( x b l ) a b l xa (45.98)() () (.499)() xa ns 8.5 rad ns rad ns. The slope magnitude is 4- From the solutions to Prob. -7, T 5 N and T 7.5 N 4 4 d () I mm 4 4 Fbx 45sin 45 (55)() x b l (7) (7 854)(85). mm ns. Fbx z Fbx z ( x b l ) ( x b l ) EIl EIl mm o ( ) EIl xmm o 45cos 45 (55)() (7) (7 854)(85) 87.5(5)() (7) (7 854)(85) 4 x mm ns. z..5.7 mm ns. The displacement magnitude is Chapter 4 - Rev B, Page 8/8

19 d d Fbx Fb x b l ( a b l ) dx dx EIl EIl z xa xa o 45sin 45 (55) (7) (7 854)(85) rad. dz d Fbx Fbx dx dx EIl EIl z x b l x b l xa Fb Fb EIl EIl a b l a b l z o 45cos 45 (55) (7) (7 854)(85) (7) (7 85 ns xa 87.5(5) rad ns. 4)(85) The slope magnitude is rad ns. 4-7 From the solutions to Prob. -7, FB 75 lbf 4 4 d (.5) 4 I.98 in 4 4 From Table -9, beams (subscript ) and (subscript ) Fbx F ax EIl EIl x b l l x xin o o cos (4)() 75sin (9)() 4 () (.9 8)() () (.9 8)().85 in ns. Fbx F ax z EIl o z z x b l l x EIl xin o sin (4)() 75 cos (9)() 4 () (.9 8)() () (.9 8)().9 in ns. The displacement magnitude is z in ns. Chapter 4 - Rev B, Page 9/8

20 d d Fbx F ax dx dx EIl EIl x b l l x z Fb xa xa a b l l a EIl EIl o cos (4) () (.9 8)() F a 4 o 75sin (9) 5 () (.9 8)() 8. rad ns. dz d Fbx F ax dx dx EIl EIl z z x b l l x xa Fb F a EIl EIl z a b l l a z xa o 4 o sin (4) 75 cos (9) () (.9 8)() () (.9 8)().5 rad ns. The slope magnitude is rad ns. 4-8 From the solutions to Prob. -7, F B =.8 ( ) N 4 4 d 5 I.8 mm 4 4 From Table -9, beam, Fbx F bx ( x b l ) ( x b l ) EIl EIl (7) (.8) (5) 4 4mm o sin (5)(4) (7) (.8) (5) o.8 sin 5 ()(4) mm ns. x Chapter 4 - Rev B, Page /8

21 Fbx z Fbx z z ( x b l ) ( x b l ) EIl EIl (7) (.8) (5) 4mm o cos (5)(4) o.8 cos 5 ()(4) mm ns. (7) (.8) (5) The displacement magnitude is x z mm ns. d d Fbx Fbx dx dx EIl EIl z z x b l x b l z Fb xa Fb a b l a b l EIl EIl (7) (.8) (5) o sin (5) o.8 sin 5 () 4 5 (7) (.8) (5).57 rad ns. z z x b l x b l xa o cos (5) o.8 cos 5 () xa dz d Fbx Fbx dx dx EIl EIl Fb z Fb z a b l a b l EIl EIl (7) (.8) (5) 4 5 (7) (.8) (5).489 rad ns. The slope magnitude is xa rad ns. 4-9 From the solutions to Prob. -8, T = lbf and T = 4 lbf, and Prob. 4-, I =.9 8 in 4. From Table -9, beam, Chapter 4 - Rev B, Page /8

22 dz d Fbx Fbx dx dx EIl EIl x Fb z Fb 575() EIl EIl () (.9 8)(4) z z O x b l x b l C z b l b l 4 4() 4.48 rad ns. () (.9 8)(4) x z x a lx x a lx dz d F a l x F a l x z dx dx EIl EIl xl Fa Fa EIl EIl Fa Fa z l a l a EIl EIl z lx l x a lx l x a z z 575() 4 4(8) rad ns. () (.9 8)(4) () (.9 8)(4) 4- From the solutions to Prob. -9, T = 88 N and T = 4 N, and Prob. 4-4, I = 9.7 ( ) mm 4. From Table -9, beams and d d Fbx Fax O ( ) ( ) z x b l l x dx dx EIl EIl x x Fb Fa Fb Fal ( x b l ) ( l x ) ( b l ) EIl EIl x EIl EI (8) 7() (5) 8 5 (7) (9.7) (5) (7) (9.7). rad ns. d d Fa ( lx) Fa x ( ) ( ) x a lx l x C z dx dx EIl EIl xl Fa Fa Fa Fal (lx l x a) ( l x ) ( l a ) EIl EIl xl EIl EI () 7()(5) (5 ) (7) (9.7) (5) (7) (9.7).9 rad ns. 4- From the solutions to Prob. -7, T = 9.9 lbf and T = 58.8 lbf, and Prob. 4-5, I =.49 9 in 4. From Table -9, beam xl xl xl Chapter 4 - Rev B, Page /8

23 d d Fbx Fb x b l ( b l ) z dx dx EIl EIl x x 5(4) O () (.499)() 4.7 rad ns. dz d F b x F b dx dx EIl EIl x x 45.98() z z O x b l b l () (.499)().4 rad ns. The slope magnitude is rad ns. O d d F a ( lx) z dx dx EIl xl Fa F a lx l x a ( l a ) EIl EIl xl 5(8) C x a lx () (.49)() xl 8.5 rad. dz d F a ( lx) dx dx EIl z C x a lx xl F a F a EIl xl EIl 45.98() z lx l x a l a z () (.499)() ns xl ns.84 rad rad ns. The slope magnitude is C 4- From the solutions to Prob. -7, T =5 N and T =7.5 N, and Prob. 4-, I = mm 4. From Table -9, beam d d Fbx Fb x b l ( b l ) dx dx EIl EIl O z x x o 45sin 45 (55) (7) (7 854)(85) rad ns. Chapter 4 - Rev B, Page /8

24 dz d Fbx Fbx dx dx EIl EIl z z O x b l x b l x o Fb 45cos 45 (55) z Fb EIl EIl (7) (7 854)(85) z b l b l (5) (7) (7 854)(85) rad ns. The slope magnitude is O rad ns. d d F a ( lx) F a dx dx EIl EIl xl C x a lx lx l x a z o F 45sin 45 () a ( l a ) rad ns. EIl (7) (7 854)(85) dz d F a ( lx) F a ( lx) dx dx EIl EIl z z C x a lx x a lx xl o Fa 45cos 45 () z Fa EIl EIl (7) (7 854)(85) z l a l a (7) (7) (7 854)(85 ) xl rad ns. The slope magnitude is 5 C rad ns. 4- From the solutions to Prob. -7, F B = 75 lbf, and Prob. 4-7, I =.9 8 in 4. From Table -9, beams and d d Fbx F ax O x b z l l x dx dx EIl EIl x x Fb F a Fb F al x b l l x b l EIl EIl EIl EI x o o cos (4) 75sin 4 (9)().75 rad ns. () (.9 8)() () (.9 8) x xl xl Chapter 4 - Rev B, Page 4/8

25 dz d Fbx F ax dx dx EIl EIl z z O x b l l x x Fb z Fza Fb z Fzal x b l l x b l EIl EIl EIl EI o o sin (4) 75cos 4 (9)().4 rad ns. () (.9 8)() () (.9 8) The slope magnitude is O x x rad ns. d d F a ( lx) F a x dx dx EIl EIl xl C x a lx l x z Fa F a Fa F al lx l x a l x ( l a ) EIl EIl EIl EI o cos () () (.9 8)() xl () (.9 8) xl o 75sin (9)().9 rad ns. dz d Fa( lx) Fax dx dx EIl EIl z z C x a lx l x xl Fa z Fa z Fa z Fal z lx l x a l x l a EIl EIl EIl EI o sin () () (.9 8)() xl () (.9 8) The slope magnitude is xl o 75 cos (9)().9 rad ns rad ns. C 4-4 From the solutions to Prob. -7, F B =.8 kn, and Prob. 4-8, I =.8 ( ) mm 4. From Table -9, beam d d Fbx Fbx dx dx EIl EIl O x b l x b l z x x Fb EIl EIl (7) (.8) (5) o sin (5) Fb b l b l 5 5 o.8 sin 5 () (7) ( 5.5 rad ns..8) (5) Chapter 4 - Rev B, Page 5/8

26 dz d Fbx Fbx dx dx EIl EIl x Fb z Fb z b l b l EIl EIl o cos (5) 5 5 z z O x b l x b l (7) (.8) (5) o.8 cos 5 () (7) x 5.47 rad ns. (.8) (5) rad ns. The slope magnitude is O d d F a ( lx) F a ( lx) dx dx EIl EIl xl C x a lx x a lx z F a F a (lx l x a ) lx l x a EIl EIl F a F a l a l a EIl EIl o.8 sin 5 (75) (7) (.8) (5) xl o sin (4) 5 4 (7) (.8) (5) rad ns. dz d F a ( lx) F a ( lx) dx dx EIl EIl z z C x a lx x a lx xl Fa Fa EIl EIl z lx l x a lx l x a z Fa Fa EIl EIl z l a l a z (7) (.8) (5) o.8 cos 5 (75) rad ns. xl o cos (4) 5 4 (7) (.8) (5) xl The slope magnitude is C...74 rad ns. 4-5 The required new slope in radians is new =.( /8) =.5 rad. In Prob. 4-9, I =.9 8 in 4, and it was found that the greater angle occurs at the bearing at O where ( O ) =.48 rad. Since is inversel proportional to I, xl Chapter 4 - Rev B, Page /8

27 4 new I new = old I old I new = d /4 = old I old / new new or, d new 4 old new I old /4 The absolute sign is used as the old slope ma be negative dnew in..5 ns 4- The required new slope in radians is new =.( /8) =.5 rad. In Prob. 4-, I = 9.7 ( ) mm 4, and it was found that the greater angle occurs at the bearing at C where ( C ) =.9 rad. See the solution to Prob. 4-5 for the development of the equation /4 d new 4 old new I old /4 4.9 dnew 9.7. mm..5 ns /4 4-7 The required new slope in radians is new =.( /8) =.5 rad. In Prob. 4-, I =.49 in 4, and the maximum slope is C =.4 rad. See the solution to Prob. 4-5 for the development of the equation /4 4 old dnew Iold new 4.4 dnew in..5 ns 4-8 The required new slope in radians is new =.( /8) =.5 rad. In Prob. 4-, I = mm 4, and the maximum slope is O =.75 rad. See the solution to Prob. 4-5 for the development of the equation /4 Chapter 4 - Rev B, Page 7/8

28 d new 4 old new I old / dnew mm..5 ns 4-9 The required new slope in radians is new =.( /8) =.5 rad. In Prob. 4-, I =.9 8 in 4, and the maximum slope =. rad. See the solution to Prob. 4-5 for the development of the equation /4 d new 4 old new I old /4 4. dnew in..5 ns 4-4 The required new slope in radians is new =.( /8) =.5 rad. In Prob. 4-4, I =.8 ( ) mm 4, and the maximum slope is C =.74 rad. See the solution to Prob. 4-5 for the development of the equation /4 d new 4 old new I old / dnew.8.9 mm..5 ns /4 4-4 I B = 4 /4 =.499 in 4, J B = I B =.988 in 4, I BC = (.5)(.5) / =.7 in 4, I CD = (/4) 4 /4 =.55 in 4. For Eq. (-4), p., b/c =.5/.5 = =.99. The deflection can be broken down into several parts. The vertical deflection of B due to force and moment acting on B ( ).. The vertical deflection due to the slope at B, B, due to the force and moment acting on B ( = CD B = B ). Chapter 4 - Rev B, Page 8/8

29 . The vertical deflection due to the rotation at B, B, due to the torsion acting at B ( = BC B = 5 B ). 4. The vertical deflection of C due to the force acting on C ( 4 ). 5. The rotation at C, C, due to the torsion acting at C ( = CD C = C ).. The vertical deflection of D due to the force acting on D ( 5 ).. From Table -9, beams and 4 with F = lbf and M B = () = 4 lbfin 4.47 in From Table -9, beams and 4 d Fx MBx Fx MBx B xl x l dx EI EI EI EI xl xl l Fl M B rad EI.499 = (.47) =.85 in. The torsion at B is T B = 5() = lbfin. From Eq. (4-5) TL.54 rad B JG B = 5(.54) =.57 in 4. For bending of BC, from Table -9, beam in.7 5. For twist of BC, from Eq. (-4), p., with T = () = 4 lbfin 45 C.48 rad = (.48) =.494 in. For bending of CD, from Table -9, beam.4 in.55 Chapter 4 - Rev B, Page 9/8

30 Summing the deflections results in in ns. D i i This problem is solved more easil using Castigliano s theorem. See Prob The deflection of D in the x direction due to F z is from:. The deflection due to the slope at B, B, due to the force and moment acting on B (x = BC B = 5 B ).. The deflection due to the moment acting on C (x ).. For B, I B = 4 /4 =.499 in 4. From Table -9, beams and 4 B d Fx MBx Fx MBx xl xl dx EI EI EI EI xl l Fl M B.7 rad EI.499 x = 5(.7) =.9 in. For BC, I BC = (.5)(.5) / =.95 in 4. From Table -9, beam 4 x C 5 M l.47 in EI.95 The deflection of D in the x direction due to F x is from:. The elongation of B due to the tension. For B, the area is = /4 =.7854 in x Fl 5 E B in 4. The deflection due to the slope at B, B, due to the moment acting on B (x = BC B = 5 B ). With I B =.497 in 4, B B Ml rad EI.499 xl Chapter 4 - Rev B, Page /8

31 x 4 = 5(.5) =.58 in 5. The deflection at C due to the bending force acting on C. With I BC =.95 in 4 x Fl in.95 5 EI BC. The elongation of CD due to the tension. For CD, the area is = (.75 )/4 =.448 in x Fl 5 E CD.448 Summing the deflections results in x D x i i 5. in in ns. 4-4 J O = J BC = (.5 4 )/ =.497 in 4, J B = ( 4 )/ =.987 in 4, I B = ( 4 )/4 =.499 in 4, and I CD = (.75 4 )/4 =.55 in 4. Tl Tl Tl T lo lb l BC GJ O GJ B GJ BC G JO J B J BC 5() 9. rad ns..5( ) Simplified Tl 5()() s GJ s.45 rad ns. Simplified is.45/. =. times greater ns. 5 Fl OC Fl CD D slcd.45() EI B EICD ().499 () D.847 in ns Reverse the deflection equation of beam 7 of Table -9. Using units in lbf, inches Chapter 4 - Rev B, Page /8

32 / x wx lx x l 5x x 5 4EI x x x ns The maximum height occurs at x = 5()/ = 5 in max in ns From Table -9-, Fbx L x b l EIl Fb L x b x l x EIl dl Fb x b l dx EIl d Fbb l L dx EIl x dl Let dx x 4 dl and set I. Thus, 4 d L Fb b l /4 El ns. For the other end view, observe beam of Table -9 from the back of the page, noting that a and b interchange as do x and x d R Fa l a /4 El ns. For a uniform diameter shaft the necessar diameter is the larger of dl and dr. 4-4 The maximum slope will occur at the left bearing. Incorporating a design factor into the solution for of Prob. 4-45, d L Chapter 4 - Rev B, Page /8

33 4 nfb l b d El d /4 (.8)()() d 8. mm ns. (7) ()(.) I.4 mm 4 From Table -9, beam, the maximum deflection will occur in BC where d BC /dx = d Fa l dx EIl x x a lx x lx a l x 4() 4.,.7 mm x x x x x =.7 mm is acceptable. x Fa l max x a lx EIl x.7mm ns mm I = (.5 4 )/4 =.98 in 4. From Table -9, beam ( ) ( x a lx ) ( x b l ) Fa l x Fb x EIl EIl 5(5)( 8) 8 5 ()(8) ().98 () 5()(8) 8 ().98 (). in ns. / Chapter 4 - Rev B, Page /8

34 4-48 I = (.5 4 )/4 =.98 in 4. For both forces use beam of Table -9. For F = 5 lbf: x 5 Fbx 55 x x b l x 5 EIl.98 xx () 5 x Fa lx 5 5 x x a lx x 5 x EIl.98 xx x () For F = 5 lbf: x Fbx 5 x z x b l EIl.98 x xx () x Fa lx 5 x z x a lx x x EIl.98 xx x (4) Plot Eqs. () to (4) for each. in using a spreadsheet. There are data points, too numerous to tabulate here but the plot is shown below, where the maximum deflection of =.55 in occurs at x = 9.9 in. ns. Chapter 4 - Rev B, Page 4/8

35 4-49 The larger slope will occur at the left end. From Table -9, beam 8 MBx B ( x a al l ) EIl db M B x a al l dx EIl ( ) With I = d 4 /4, the slope at the left bearing is db M a al l dx E d l x Solving for d B 4 /4 ( ) M () 4 (4 ) (4)() B d 4 a al l E l () (.)().4 in ns. 4-5 From Table -5, E =.4 Mpsi M O = = 8 F BC () F BC =. lbf The cross sectional area of rod BC is = (.5 )/4 =.9 in. The deflection at point B will be equal to the elongation of the rod BC. B FL.() E BC in. 4-5 M O = = F C () F C = 8. lbf The deflection at point in the negative direction is equal to the elongation of the rod C. From Table -5, E s = Mpsi. FL 8. E C.5 / 4 ns 4.75 in B similar triangles the deflection at B due to the elongation of the rod C is 8 B 4 B (.75). in From Table -5, E a =.4 Mpsi The bar can then be treated as a simpl supported beam with an overhang B. From Table -9, beam Chapter 4 - Rev B, Page 5/8

36 d BC Fa d F( x l) Fa B BD ( la) 7 ( x l) a( x l) ( l dx x l a EI dx EI a) EI x l a F Fa 7Fa Fa EI EI EI EI 7 ( x l) a( x l) a( x l) x l a ( l a) (l a) ( l a) () (5) ( 5) (.4).5( ) / (.4).5( ) /.48 in B = B + B =..48 =.55 in ns. 4-5 From Table -5, E = 7 GPa, and G = 79. GPa. Tl Tl FlB FlOClB FlClB FlB B lb lb GJ GJ EI G d / G d / E d /4 FlB OC C lb GdOC GdC EdB OC C B OC C l l The spring rate is k = F/ B. Thus l l l k B OC C B GdOC GdC EdB l N/mm ns. 4-5 For the beam deflection, use beam 5 of Table -9. F R R F F, and k k l Fx 48EI B x (4x l ) k k x B F x (4x l ) ns. k kkl 48EI Chapter 4 - Rev B, Page /8

37 For BC, since Table -9 does not have an equation (because of smmetr) an equation will need to be developed as the problem is no longer smmetric. This can be done easil using beam of Table -9 with a = l / F Fk Fk Fl/ l x l BC x x lx k kkl EIl 4 k k l x F x 4x l 8 lx ns. k kkl 48EI 4-54 Fa F R, and R ( la) l l Fa F, and ( l a ) lk lk Fax B x ( l x ) l EIl a x ax B F k a kl a ( l x ). kl kkl ns EIl F( xl) BC x ( x l ) a ( x l ) l EI a x ( x l) BC F k a kl a ( x l ) a ( x l ) ns. kl kkl EI 4-55 Let the load be at x l/. The maximum deflection will be in Section B (Table -9, beam ) Fbx B x b l EIl db dx Fb EIl x b l x b l l b l x, xmax.577 l ns. For x l/, xmin l.577l.4 l ns. Chapter 4 - Rev B, Page 7/8

38 4-5 M R O O ()(5) 5() 9.5 N mm () N From Prob. 4-, I 4.4( ) mm 4 x M x 5 x - d x EI 9.5 x 75x 5 x C dx d x C dx at d x EI 9.5 x 75x 5 x dx 4 x EI 4.75 x 9.7x 4.7 x C 4 at x C, and therefore 4 4 x x x x 4EI B mm ns. 4 4 M O = 9.5 ( ) Nm. The maximum stress is compressive at the bottom of the beam where = 9. = 7 mm M 9.5 ( 7) max Pa MPa ns. I 4.4( ) The solutions are the same as Prob See Prob. 4- for reactions: R O = 45 lbf and R C = 85 lbf. Using lbf and inch units Chapter 4 - Rev B, Page 8/8

39 M = 45 x 45 x 7 x d EI.5x 5 x 7 5 x dx C EI = 77.5 x 75 x 7 5 x C x = at x = C = = at x = 4 in = 77.5(4 ) 75(4 7) 5(4 ) + C x C =.( ) lbfin and, EI = 77.5 x 75 x 7 5 x.( ) x Substituting =.5 in at x = in gives ( ) I (.5) = 77.5 ( ) 75( 7) 5( ).( )() I =. in 4 Select two 5 in.7 lbf/ft channels; from Table -7, I = (7.49) = 4.98 in 4. midspan.4 in ns The maximum moment occurs at x = in where M max = 4.( ) lbfin Mc 4.( )(.5) max 5 7 psi O.K. I 4.98 The solutions are the same as Prob I = (.5 4 )/4 =.485 in 4, and w = 5/ =.5 lbf/in. R.59 4 O (4) 45. lbf 9.5 M 45.x x 4 x 5 d.5 EI.5x x 7 x 5 dx C 4 EI 75.5x.58x 5.7 x5 C xc atx C at x 9 in C.85( ) lbf in 4 Thus, x.58x 5.7 x 5.85 x EI Evaluating at x = 5 in, Chapter 4 - Rev B, Page 9/8

40 midspan (5) ( )(.485).978 in ns (9.5) ( )(.485).7 in ns. 5 % difference ns. The solutions are the same as Prob I =.5 in , R 4 lbf and RB 98 lbf M = 4 x 5 x + 98 x d EI x x x.7 49 dx C 4 EI 7x 4.7x. x CxC = at x = C = = at x = in C = 8 lbfin. Thus, 4 7x 4.7x. x 8x x 4.7x. x 8 x ns. The tabular results and plot are exactl the same as Prob R = R B = 4 N, and I = ( ) / = 84 mm 4. First half of beam, M = 4 x + 4 x d EI x x C dx From smmetr, d/dx = at x = 55 mm = (55 ) + (55 ) + C C = 48( ) N mm EI =.7 x +.7 x + 48( ) x + C Chapter 4 - Rev B, Page 4/8

41 = at x = mm C =.( 9 ) N mm. The term (EI) = [7( ) 84] =.949 ( ) Thus =.949 ( ) [.7 x +.7 x + 48( ) x.( 9 )] O =.7 mm ns. x = 55 mm =.949 ( ) [.7 (55 ) +.7 (55 ) + 48( ) 55.( 9 )] =. mm ns. The solutions are the same as Prob MB Rl FaM R M Fa l M M R lf( la) R FlFaM l M RxM R x l d EI Rx M x R x l C dx EI Rx M x R x l C x C = at x = C = = at x = l C Rl M l. Thus, EI Rx M x R x l Rl M l x M Fa x M x l Fl Fa M x l Fal M l x EIl ns. In regions, B M Fax M x l Fal M l x EIl x M x lx l Fa l x EIl ns. Chapter 4 - Rev B, Page 4/8

42 BC M Fa x M x l FlFaM xl Fal M l x EIl M x x l x l xl F ax l ax l axl EIl M xll Flxlxl axl EIl x l MlFxl a xl ns. EI The solutions reduce to the same as Prob wb a 4- M D Rl w ba lb ba R lba l w w M Rx xa xb d EI Rx x a x b dx w w C w 4 w 4 EI Rx x a x b Cx C 4 4 = at x = C = = at x = l w w C Rl l a l b l 4 4 b a 4 4 w w w EI l l b a x x a x b b a b w w 4 w 4 x lba l la l l l 4 4 w 4 4 balbax l xa l xb 4EIl 4 4 x ba lba l la lb ns. The above answer is sufficient. In regions, Chapter 4 - Rev B, Page 4/8

43 w B ba lba x x ba lba l la lb 4EIl wx 4 4 balbax balbal la lb 4EIl 4 4 w BC ba lba x l xa 4EIl x ba lba l la lb w CD ba lba x l xa l x b 4EIl x ba lba l la lb These equations can be shown to be equivalent to the results found in Prob I = (.75 4 )/4 =.755 in 4, I = (.75 4 )/4 =.44 in 4, R =.5(8)() = 9 lbf Since the loading and geometr are smmetric, we will onl write the equations for half the beam For x 8 in M 9x9 x t x =, M = 7 lbfin Writing an equation for M / I, as seen in the figure, the magnitude and slope reduce since I > I. To reduce the magnitude at x = in, we add the term, 7(/I / I ) x. The slope of 9 at x = in is also reduced. We account for this with a ramp function, x. Thus, M 9x 9 7 x 9 x x I I I I I I I 58x95 x 7 x 95.5 x E d 54x 95 x 587 x 5.7 x C dx d Boundar Condition: at x 8 in dx Chapter 4 - Rev B, Page 4/8

44 C C = 8.7 ( ) lbf/in 4 E 854.7x 47 x 59 x.9 x 8.7( ) xc = at x = C = Thus, for x 8 in x 47 x 59 x.9 x 8.7( ) x ns. ( ) Using a spreadsheet, the following graph represents the deflection equation found above The maximum is max. in at x8 in ns. 4-4 The force and moment reactions at the left support are F and Fl respectivel. The bending moment equation is M = Fx Fl Plots for M and M /I are shown. M /I can be expressed using singularit functions M F Fl Fl l F l x x x I I I 4I I Chapter 4 - Rev B, Page 44/8

45 where the step down and increase in slope at x = l / are given b the last two terms. Integrate d F Fl Fl l F l E x x x x C dx 4I I 4I 4I d/dx = at x = C = F Fl Fl l F l E x x x x C I 4I 8I I = at x = C = F l l x lx l x x 4EI F l l 5Fl l l() (). xl/ ns 4EI 9EI F l l Fl l ll l l x xl 4EI EI The answers are identical to Ex Place a dumm force, Q, at the center. The reaction, R = wl / + Q / wl Q w x M x M x Q Integrating for half the beam and doubling the results ns. l/ l/ M wl wx x max M dx x EI Q EI dx Q Note, after differentiating with respect to Q, it can be set to zero l/ 4 w w xl x 5w max x l xdx ns. EI EI 4 84EI 4- Place a fictitious force Q pointing downwards at the end. Use the variable x originating at the free end at positive to the left wx M M Qx x Q l/ Chapter 4 - Rev B, Page 45/8

46 l l l M wx w max M dx xdx x dx EI Q EI EI Q 4 wl ns. 8EI 4-7 From Table -7, I - =.85 in 4. Thus, I = (.85) =.7 in 4 First treat the end force as a variable, F. dding weight of channels of (5)/ =.8 lbf/in. Using the variable x as shown in the figure 5.8 M Fx x Fx.97x M x F M M dx ( Fx.97 x )( x) dx EI F EI.8 in in the direction of the 5 lbf force 4 (5 / )( ) (.97 / 4)( ) ( )(.7).8 in ns. 4-8 The energ includes torsion in C, torsion in CO, and bending in B. Neglecting transverse shear in B M M Fx, x F In C and CO, T T FlB, l F B The total energ is l B Tl Tl M U GJ GJ EI C The deflection at the tip is CO B dx Chapter 4 - Rev B, Page 4/8

47 lb lb U TlC T TlCO T M M TlClB TlCOlB dx Fx dx F GJ F GJ F EI F GJ GJ EI C CO C CO B TlClB TlCOlB FlB FlClB FlCOlB FlB GJ GJ EI G d / G d / E d / 4 FlB C CO lb Gd C GdCO Ed B C CO B C CO B l l F l l l k l Gd Gd Ed C CO B B C CO B 8. N/mm ns I = (.75 4 )/4 =.755 in 4, I = (.75 4 )/4 =.44 in 4 Place a fictitious force Q pointing downwards at the midspan of the beam, x = 8 in R ()8 Q 9.5Q M M 9.5Q x.5x Q M M 9.5Q x9( x).5x Q For x in For x in B smmetr it is equivalent to use twice the integral from to EI Q EI EI 8 8 M M dx x dx x x x dx Q 8 4 x 9 x 9( x x x ) EI EI EI EI in ns. Chapter 4 - Rev B, Page 47/8

48 4-7 I = (.5 4 )/4 =.8 ( ) in 4, J = I =. ( ) in 4, = (.5 )/4 =.9 in. Consider x to be in the direction of O, verticall upward, and z in the direction of B. Resolve the force F into components in the x and directions obtaining. F in the horizontal direction and.8 F in the negative vertical direction. The. F force creates strain energ in the form of bending in B and O, and tension in O. The.8 F force creates strain energ in the form of bending in B and O, and torsion in O. Use the dumm variable x to originate at the end where the loads are applied on each segment, M. F: B M.Fx. x F M O M 4.F 4. F F a Fa.F. F M.8 F: B M.8Fx.8 x F M O M.8Fx.8 x F T T 5.F 5. F Once the derivatives are taken the value of F = 5 lbf can be substituted in. The deflection of B in the direction of F is* B F U FL a Fa TL T M M d x F E F JG F EI F O x d x d x O.8x d x in ns..8x dx Chapter 4 - Rev B, Page 48/8

49 *Note. This is not the actual deflection of point B. For this, dumm forces must be placed on B in the x,, and z directions. Determine the energ due to each, take derivatives, and then substitute the values of F x = 9 lbf, F = lbf, and F z =. This can be done separatel and then use superposition. The actual deflections of B are B =.8 i.8 j.77 k in From this, the deflection of B in the direction of F is B in F which agrees with our result. 4-7 Strain energ. B: Bending and torsion, BC: Bending and torsion, CD: Bending. I in 4 B = ( 4 )/4 =.499 in 4, J B = I B =.988 in 4, I BC =.5(.5 )/ =.7, I CD = (.75 4 )/4 =.55 in 4. For the torsion of bar BC, Eq. (-4) is in the form of =TL/(JG), where the equivalent of J is J eq = bc. With b/c =.5/.5 =, J BC = bc =.99(.5).5 = 7.8 ( ) in 4. Use the dumm variable x to originate at the end where the loads are applied on each segment, B: Bending M F x F M x F Torsion T 5F T 5 F BC: Bending M Fx M x F Torsion T F T F CD: Bending M Fx M x F U Tl T M D M d x F JG F EI F 5F F Fx dx.55fx dx F in ns..9 F.48 F.4 F.98 F 5.7 F 5.7 F x d x Chapter 4 - Rev B, Page 49/8

50 4-7 B = ( )/4 =.7854 in, I B = ( 4 )/4 =.499 in 4, I BC =.5 (.5 )/ =.95 ( ) in 4, CD = (.75 )/4 =.448 in, I B = (.75 4 )/4 =.55 in 4. For ( D ) x let F = F x = 5 lbf and F z = lbf. Use the dumm variable x to originate at the end where the loads are applied on each segment, M CD: M Fzx F Fa Fa F F M BC: M FxFz x F Fa Fa Fz F M B: M 5F Fz Fzx 5 F Fa Fa F F 5 U FL Fa D Fx Fzxdx x F E F EI CD FL Fa 5F Fz Fzx5d x EI E F B F F Fz F F z F Fz F F 4.7 Fz.9 F Fz F F BC F Substituting F = F x = 5 lbf and F z = lbf gives 4 4 D in ns. x 4-7 I O = I BC = (.5 4 )/4 =.485 in 4, J O = J BC = I O =.497 in 4, I B = ( 4 )/4 =.499 in 4, J B = I B =.988 in 4, I CD = (.75 4 )/4 =.55 in 4 Let F = F, and use the dumm variable x to originate at the end where the loads are applied on each segment, B z Chapter 4 - Rev B, Page 5/8

51 OC: M T M F x x, T F F F DC: M M Fx x F U TL T M D M dx F JGOC F EI F The terms involving the torsion and bending moments in OC must be split up because of the changing second-area moments. D F 4 F Fx dx F xdx Fxdx Fxdx F F F 4 5 F F F F ns in. For the simplified shaft OC, B F Fx dx 4 F F F F in ns. Fx dx Simplified is.848/.7 =. times greater ns Place a dumm force Q pointing downwards at point B. The reaction at C is R C = Q + (/8) = Q +. This is the axial force in member BC. Isolating the beam, we find that the moment is not a function of Q, and thus does not contribute to the strain energ. Thus, onl energ in the member BC needs to be considered. Let the axial force in BC be F, where Chapter 4 - Rev B, Page 5/8

52 F F Q. Q U FL F B in ns. Q E Q BC Q Q.5 / I OB =.5( )/ =.7 in 4 C = (.5 )/4 =.9 in M O = = R C () 8 Q R C = Q + 8. M = = R O 5() Q R O = Q + 8. Bending in OB. BD: Bending in BD is onl due to Q which when set to zero after differentiation gives no contribution. D: Using the variable x as shown in the figure above M M x Q 7 x 7 x Q O: Using the variable x as shown in the figure above M M Q8.x x Q xial in C: F F Q8. Q Chapter 4 - Rev B, Page 5/8

53 U FL F M B M dx Q E Q EI Q Q Q Q x 7 xd x 8.x dx EI 5. x7 xd x.7 x dx in ns. 4-7 There is no bending in B. Using the variable, rotating counterclockwise from B M M PRsin Rsin P Fr Fr Pcos cos P F F Psin sin P MF PR sin P (4) 4 mm, ro 4 () 4 mm, ri 4 () 7 mm, From Table -4, p., for a rectangular cross section rn mm ln(4 / 7) From Eq. (4-), the eccentricit is e = R r n = =.75 mm From Table -5, E = 7( ) MPa, G = 79.( ) MPa From Table 4-, C =. From Eq. (4-8) M M FR F MF CFrR F d d d r d ee P E P E P G P PRsin PRsin PRsin CPRcos d d d ee d E E G PR R ()(4) 4 (7 )(.) EC 4E e G 4(4)(7 ) mm ns. Chapter 4 - Rev B, Page 5/8

54 4-77 Place a dumm force Q pointing downwards at point. Bending in B is onl due to Q which when set to zero after differentiation gives no contribution. For section BC use the variable, rotating counterclockwise from B M M PRsin QRRsin Rsin Q Fr Fr PQcos cos Q F F PQsin sin Q MF PR sin QR sin P Q sin MF PRsin PRsin sin QR sin sin Q But after differentiation, we can set Q =. Thus, MF PRsin sin Q (4) 4 mm, ro 4 () 4 mm, ri 4 () 7 mm, From Table -4, p., for a rectangular cross section rn mm ln(4 / 7) From Eq. (4-), the eccentricit is e = R r n = =.75 mm From Table -5, E = 7( ) MPa, G = 79.( ) MPa From Table 4-, C =. From Eq. (4-8) M M FR F MF CFrR F d d d r d ee Q E Q E Q G Q PR PR PR sin sin d sin d sin sin d ee E E CPR cos d G PR PR PR CPR PR R CE 4 ee 4 E 4 E 4 G E 4 e 4 G mm ns. Chapter 4 - Rev B, Page 54/8

55 4-78 Note to the Instructor. The cross section shown in the first printing is incorrect and the solution presented here reflects the correction which will be made in subsequent printings. The corrected cross section should appear as shown in this figure. We apologize for an inconvenience. = (.5).5(.5) =.75 in (.5)()(.5) (.75.5)(.5)(.5) R.5 in.75 Section is equivalent to the T section of Table -4, p.,.5(.75).75(.5) r n.79 in.5ln[(.75) /].75ln[( ) / (.75)] e Rr n in For the straight section I z (.5).5()(.5.5).5 (.5).5.5(.5) in For x 4 in M V M Fx x, V F F F For / F r Fr F Fcos cos, F Fsin sin F F M M F(4.5sin ) (4.5sin ) F MF MF F(4.5sin ) F sin F(4.5sin )sin F Chapter 4 - Rev B, Page 55/8

56 Use Eqs. (4-) and (4-4) (with C = ) for the straight part, and Eq. (4-8) for the curved part, integrating from to π/, and double the results 4 / F(4)() (4.5sin ) Fx dx F d E I.75( G/ E).75(.9) / / / () cos (.5) Fsin (.5) F(4.5sin )sin d d F d.75( G/ E) Substitute I =.89 in 4, F = 7 lbf, E = ( ) psi, G =.5 ( ) psi () (.5 / ).75(.9) / 4. in ns Since R/h = 5/4.5 = 7.78 use Eq. (4-8), integrate from to, and double the results M M FR cos R cos F Fr Fr Fsin sin F F F Fcos cos F MF F Rcos cos MF F From Eq. (4-8), FRcos cos FR FR ( cos ) d cos d ee E FR.FR E G cos cos d sin d FR R E. E e G = 4.5() =.5 mm, E = 7 ( ) N/mm, G = 79. ( ) N/mm, and from Table -4, p., Chapter 4 - Rev B, Page 5/8

57 h 4.5 rn mm ro 7.5 ln ln r.75 i and e = R r n = =.487 mm. Thus, F F where F is in N. For = mm, F.5 N ns..858 Note: The first term in the equation for dominates and this is from the bending moment. Tr Eq. (4-4), and compare the results. 4-8 R/h = > so Eq. (4-4) can be used to determine deflections. Consider the horizontal reaction, to applied at B, subject to the constraint ( ). B H FR M M ( cos ) HRsin Rsin H B smmetr, we ma consider onl half of the wire form and use twice the strain energ Eq. (4-4) then becomes, / U M ( B) H M Rd H EI H /FR ( cos ) HR sin ( R sin ) R d F F F H H 9.55 N ns. 4 4 Reaction at is the same where H goes to the left. Substituting H into the moment equation we get, M FR M R ( cos ) sin [ ( cos ) sin ] F Chapter 4 - Rev B, Page 57/8

58 P P EI F EI 4 U M / FR M Rd [ ( cos ) sin ] R d FR / ( cos 4sin cos 4 sin 4sincos ) d EI FR 4 4 EI 4 4 ( 8 4) FR ( 8 4) ()(4 ).4 mm ns. 4 8 EI 8 7 / The radius is sufficientl large compared to the wire diameter to use Eq. (4-4) for the curved beam portion. The shear and axial components will be negligible compared to bending. Place a fictitious force Q pointing to the left at point. M M PR sin Q( Rsin l) Rsin l Q Note that the strain energ in the straight portion is zero since there is no real force in that section. From Eq. (4-4), / M / M Rd PRsinRsin lrd EI Q EI Q PR / PR (5 ) Rsin lsin d R l (5) 4 4 EI EI 4.5 / in ns. 4-8 Both the radius and the length are sufficientl large to use Eq. (4-4) for the curved beam portion and to neglect transverse shear stress for the straight portion. Straight portion: M P B M B Px x Curved portion: M PR( cos ) l BC R(cos ) l BC M P From Eq. (4-4) with the addition of the bending strain energ in the straight portion of the wire, Chapter 4 - Rev B, Page 58/8

59 l M / B M BC M B dx MBC Rd EI P EI P P l / PR xdx R( cos ) l d EI EI Pl PR / R ( cos cos ) Rl( cos ) l d EI EI Pl PR / R cos R Rlcos ( R l) d EI EI Pl PR R R Rl ( R l) EI EI 4 P l R RR Rl R( R l) EI 4 4 (5 ) 5 (5 ) (5)(4) / in ns. 4-8 Both the radius and the length are sufficientl large to use Eq. (4-4) for the curved beam portion and to neglect transverse shear stress for the straight portion. Place a dumm force, Q, at verticall downward. The onl load in the straight section is the axial force, Q. Since this will be zero, there is no contribution. In the curved section M M PRsin QRcos Rcos Q From Eq. (4-4) / M / M Rd PRsin R cos Rd EI Q EI 5 4 Q PR / PR PR sin sincosd EI EI EI.74 in ns..5 / Both the radius and the length are sufficientl large to use Eq. (4-4) for the curved beam portion and to neglect transverse shear stress for the straight portion. Chapter 4 - Rev B, Page 59/8

60 Place a dumm force, Q, at verticall downward. The load in the straight section is the axial force, Q, whereas the bending moment is onl a function of P and is not a function of Q. When setting Q =, there is no axial or bending contribution. In the curved section M M PRcos lqrsin Rsin Q From Eq. (4-4) / M / M Rd PRcoslRsinRd EI Q EI Q / PR PR PR EI EI EI Rsin Rsincos lsind R l R R l in.5 / 4 Since the deflection is negative, is in the opposite direction of Q. Thus the deflection is.45 in ns Consider the force of the mass to be F, where F = 9.8() = 9.8 N. The load in B is tension FB FB F F For the curved section, the radius is sufficientl large to use Eq. (4-4). There is no bending in section DE. For section BCD, let be counterclockwise originating at D M FRsin M Rsin F Using Eqs. (4-9) and (4-4) Fl F B M Fl FR sin M Rd d E B F EI F E EI Fl FR F l R E EI E I 7 /4 /4.7 mm ns. Chapter 4 - Rev B, Page /8

61 4-8 O = (.5) =.5 in, I OB =.5( )/ =.7 in 4, I C = (.5 4 )/4 =.8 ( - ) in 4 ppling a force F at point B, using statics, the reaction forces at O and C are as shown. FO O: xial FO F F M O Bending M O Fx x F B: Bending M F B M B Fx x C: Isolating the upper curved section M M C FR sin cos R sin cos F Fl FO 4Fx dx F x d x E O F EI EI OB OB / 9FR sin cos d EI C C F 4F F / 9F sin sin cos sin cos cos d.8.7 F 7.9 F.58 F.9778F 4 4.F.. in ns O = (.5) =.5 in, I OB =.5( )/ =.7 in 4, I C = (.5 4 )/4 =.8 ( - ) in 4 ppling a vertical dumm force, Q, at, from statics the reactions are as shown. The dumm force is transmitted through section Chapter 4 - Rev B, Page /8

62 O and member C. FO O: FO F Q Q C: M F QRsin F QRcos C Rsin cos C Fl F M E Q EI Q E EI O / O C M C Rd O C Q / FlO FR sin cos d C M Q.4 in. ns I = ( 4 )/4 =. mm 4 / M M FRsin Rsin F T T FR( cos ) R( cos ) F ccording to Castigliano s theorem, a positive U/ F will ield a deflection of in the negative direction. Thus the deflection in the positive direction is U / / ( ) F( Rsin ) R d F[ R(cos )] R d F EI GJ Integrating and substituting J I and G E/ FR FR ( ) ( ) 4 8 ( 8) EI 4 4 4EI (5)(8) [4 8 ( 8)(.9)].5 mm ns. 4() The force applied to the copper and steel wire assembl is Fc Fs 4 lbf () Since the deflections are equal, c s Chapter 4 - Rev B, Page /8

63 Fl Fl E E Fl c c s ( / 4)(.9) (7.) ( / 4)(.55) () s Fl Yields, Fc.4F s. Substituting this into Eq. () gives.4fs Fs.4Fs 4 Fs 5. lbf Fc.4Fs 4.5 lbf Fc 4.5 c 75 psi. kpsi ns. c ( / 4)(.9) Fs 5. s 7 57 psi 7. kpsi ns. s ( / 4)(.55 ) Fl 5.()().7 in ns. E s ( / 4)(.55) () 4-9 (a) Bolt stress.75(5) 48.8 kpsi ns. b Total bolt force Fb bb (48.8) (.5 ) 57.5 kips 4 Clinder stress Fb 57.4 c.9 kpsi ( / 4)(5.5 5 ) ns. (b) Force from pressure D (5 ) P p (5) 987 lbf 9.8 kip 4 4 F x = P b + P c = 9.8 () Since c b, Pl c Pl b ( / 4)(5.5 5 ) E ( / 4)(.5 ) E c P c =.5 P b () Substituting this into Eq. () P b +.5 P b = 4.5 P b = 9.8 P b =.8 kip. From Eq. (), P c = 7.8 kip Using the results of (a) above, the total bolt and clinder stresses are.8 b kpsi ns. ( / 4)(.5 ) Chapter 4 - Rev B, Page /8

64 7.8 c.9. kpsi ns. ( / 4)(5.5 5 ) 4-9 T c + T s = T () c = s Tl Tl JG T JG JG JG c s c c c s s T s () Substitute this into Eq. () JG JG c s Ts Ts T Ts T JG JG JG s The percentage of the total torque carried b the shell is JGs JG JG % Torque ns. s 4-9 R O + R B = W () O = B Fl Fl E E O B c s c 4RO RB RO R E E Substitute this unto Eq. () B () R R 4 R. kn ns. B B B From Eq. () RO..4 kn ns. Fl 4(4). mm ns. E O ()(7.7)( ) 4-9 See figure in Prob. 4-9 solution. Procedure :. Let R B be the redundant reaction. Chapter 4 - Rev B, Page 4/8

65 . Statics. R O + R B = 4 N R O = 4 R B (). Deflection of point B. B R 44 RB B () E E 4. From Eq. (), E cancels and R B = N ns. and from Eq. (), R O = 4 = 4 N ns. Fl 4(4). mm ns. E O ()(7.7)( ) 4-94 (a) Without the right-hand wall the deflection of point C would be C Fl E / / in.5 in Hits wall ns. (b) Let R C be the reaction of the wall at C acting to the left (). Thus, the deflection of point C is now C 5 RC 8 RC 5 / / RC or,. 4.9 R.5 R 5 lbf.5 kip ns. C Statics. Considering +, 5 R 5 = R = 947 lbf =.95 kip ns. Deflection. B is 947 lbf in tension. Thus R B B 5. in ns. BE / Since O = B, C TO(4) TB () TO T JG JG B () Chapter 4 - Rev B, Page 5/8

66 Statics. T O + T B = () Substitute Eq. () into Eq. (), 5 TB TB TB TB 8 lbf in ns. From Eq. () TO TB 8 lbf in ns ns. 4 /.5.5 max O T 489 psi 4.89 kpsi ns. d.5 8 B psi. kpsi ns Since O = B, G Statics. T O + T B = () TO(4) ().9 () /.5 TB T 4 /.75 4 O TB G Substitute Eq. () into Eq. (),.9T T.9T T 54. lbf in ns. B B B B From Eq. () TO TB / lbf in ns. ns max O.48. T psi.8 kpsi ns. d B 8 psi.8 kpsi ns..75 Chapter 4 - Rev B, Page /8

67 4-97 Procedure.. rbitraril, choose R C as a redundant reaction.. Statics. F x =, ( ) ( ) R O R C = R O = ( ) R C (). The deflection of point C. ( ) ( ) R C () ( ) R C () RC (5) C E E E 4. The deflection equation simplifies to 45 R C + ( ) = R C = lbf. kip ns. From Eq. (), R O = ( ) = 4 7 lbf 4.7 kip ns. F B = F B + R C = +. = 7. kips compression FB 7. B 4.7 kpsi (.5)() ns. Deflection of. Since O is in tension, Rl OO 47() O E (.5)()(). in ns Procedure.. Choose R B as redundant reaction.. Statics. R C = wl R B () MC wl RBla (). Deflection equation for point B. Superposition of beams and of Table -9, RB la wla B 4llala l EI 4EI 4. Solving for R B. w RB l 4llala 8l a w l ala ns. 8 l a Substituting this into Eqs. () and () gives Chapter 4 - Rev B, Page 7/8

68 w R w lr l ala ns 8 5 l a C B. w M C w l RBla l ala ns See figure in Prob solution. Procedure.. Choose R B as redundant reaction.. Statics. R C = wl R B () MC w l RBla (). Deflection equation for point B. Let the variable x start at point and to the right. Using singularit functions, the bending moment as a function of x is M M w x RB xa xa RB U M B M dx R EI R B l l B l x dx x RB x a x adx EI w EI w a or, 4 4 a R w l a l a B l a a a 4 Solving for R B gives 4 4 RB l a a l a l al a ns. 8 w l a From Eqs. () and () w 4 8 w R w lr l ala ns 8 5 l a C B. w M C w l RBla l ala ns. 8 l a Chapter 4 - Rev B, Page 8/8

69 4- Note: When setting up the equations for this problem, no rounding of numbers was made. It turns out that the deflection equation is ver sensitive to rounding. Procedure.. Statics. R + R = wl () Rl M w l. Bending moment equation. () M Rx wx M d EI Rx wx Mx C dx 4 EI Rx wx Mx Cx C 4 () (4) EI = ( )(.85) = 5.5( ) lbfin.. Boundar condition. t x =, = R /k = R /[.5( )]. Substitute into Eq. (4) with value of EI ields C = 7 R. Boundar condition. t x =, d /dx = M /k = M /[.5( )]. Substitute into Eq. () with value of EI ields C =. M. Boundar condition. t x = l, = R /k = R /[.( )]. Substitute into Eq. (4) with value of EI ields 4.75R Rl wl Ml.Ml7 R (5) 4 Equations (), (), and (5), written in matrix form with w = 5/ lbf/in and l = 4 in, are R 4 R M 57 Solving, the simultaneous equations ields R = lbf, R = lbf, M =. lbfin ns. For the deflection at x = l / = in, Eq. (4) gives Chapter 4 - Rev B, Page 9/8