Chapter 4 ( ) ( ) F Fl F y = = + Solving for k. k kt. y = = + +

Transcription

1 Chapter 4 4- For a torsion bar, k T T/ F/, and so F/k T. For a cantiever, k F/δ,δ F/k. For the assemby, k F/y, or, y F/k + δ Thus F F F y + k kt k Soving for k kkt k ns. k + kt + kt k 4- For a torsion bar, k T T/ F/, and so F/k T. For each cantiever, k F/δ, δ F/k, and,δ L F/k L. For the assemby, k F/y, or, y F/k + δ +δ L. Thus F F F F y + + k kt k kl Soving for k klkkt k ns. kkl + kt kl + kt k + + kt k kl 4- (a For a torsion bar, k T/ GJ/. Two springs in parae, with J πd 4 i /, and d d d, 4 4 JG JG π d d k + G + x x x x π 4 Gd + x x ns. ( Defection equation, T x JG ( T x JG ( T x resuts in T ( x From statics, T + T T 5. Substitute Eq. ( Shigey s MED, th edition Chapter 4 Soutions, Page /8

2 x x T + T 5 T 5 x ns. ( Substitute into Eq. ( resuting in x T 5 ns. (4 π 4 (b From Eq. (, k (.5.5( + 8.( bf in/rad ns From Eq. (4, T 5 75 bf in ns. 5 From Eq. (, T 5 75 bf in ns. T ( 5 i From either section, τ. ( psi. kpsi π d π.5 ns. i 4-4 Defection to be the same as Prob. 4- where T 75 bf in, / 5 in, and d.5 in Or, T T T 4 T T T 4 4 ( ( π 4 π 4 π 4 d G dg (.5 d d G 5 d ( 4 d ( 4 T T T T Equa stress, τ τ π d π d d d Divide Eq. (4 by the first two equations of Eq.( resuts in T T d d d.5 d (5 4T T 4 4 d d Statics, T + T 5 ( Substitute in Eqs. ( and (, with Eq. (5 gives 4 d 4 d Soving for d and substituting it back into Eq. (5 gives d.88 8 in, d.58 in ns. (4 Shigey s MED, th edition Chapter 4 Soutions, Page /8

3 From Eqs. ( and (, T 5( ( bf in T ( ( bf in ns. ns. Defection of T is T rad JG 4 ( π / ( ( T 5 8. bf in ns..5 8 Spring constant is k The stress in d is ( 4 (.88 8 T τ 9.7 psi 9.7 kpsi ns. π d π The stress in d is (.58 T 57 τ 9.7 psi 9.7 kpsi ns. π d π 4-5 (a Let the radii of the straight sections be r d / and r d /. Let the ange of the taper be α where tan α (r r /. Thus, the radius in the taper as a function of x is r r + x tan α, and the area is π (r + x tan α. The defection of the tapered portion is F F dx F δ dx E π E π E r x α tanα ( r + x tanα ( + tan F F π Er tanα tanα ( r + tanα π E tanα r r F r r F tanα F π E tanα r r π E tanα r r π r r E 4F π d d E (b For section, ns. F 4F 4(( δ (.5 (( E π d E π 4.4( in ns. For the tapered section, 4 F 4 ( 4 δ.( in ns. π dde π (.5(.75(( For section, Shigey s MED, th edition Chapter 4 Soutions, Page /8

4 F 4F 4(( 4 δ.5( in ns. E π d E π (.75 (( 4- (a Let the radii of the straight sections be r d / and r d /. Let the ange of the taper be α where tan α (r r /. Thus, the radius in the taper as a function of x is r r + x tan α, and the poar second area moment is J (π / (r + x tan α 4. The anguar defection of the tapered portion is T T dx T dx 4 GJ πg tanα G tanα tanα ( r + x π ( r + x T T π G r tanα tanα ( r + tanα π G tanα r r ( r + r r + r T r r T r r T π G tanα r r π G r r r r π G r r T π G ( d + dd + d d d (b The defections, in degrees, are: Section, ns. T 8 T 8 (5( 8.44 deg ns. π π π π (.5.5( π Tapered section, Section, 4 4 GJ d G T( d + d d + d 8 π π Gd d (5(.5 + (.5( deg ns. π.5( (.5 (.75 π T 8 T 8 (5( 8.48 deg ns. 4 4 GJ π π dg π π (.75.5( π 4-7 The area and the eastic moduus remain constant. However, the force changes with respect to x. From Tabe -5, the unit weight of stee is γ.8 bf/in and the eastic moduus is E Mpsi. Starting from the top of the cabe (i.e. x, at the top. F γ(( x Shigey s MED, th edition Chapter 4 Soutions, Page 4/8

5 [ ].8 5( ( x dx γ γ.9 in x x ( Fdx w δc o E E E E From the weight at the bottom of the cabe, [ ] W 4W 4(5 5( δ W E π d E π (.5 ( 5.9 in δ δc + δw in ns. The percentage of tota eongation due to the cabe s own weight.9 (.% ns ΣF y R F R F ΣM M Fa M Fa V B F, M B F (x a, V BC M BC Section B: F x B + EI EI B at x C F ( x a dx ax C ( ( F x F x x yb ax dx a + C EI EI y B at x C, and Fx yb ( x a ns. EI Section BC: BC dx + C EI ( From Eq. (, at x a (with C, F a a( a Fa C. Thus, EI EI Fa BC EI Shigey s MED, th edition Chapter 4 Soutions, Page 5/8

6 Fa Fa ybc dx x + C EI EI 4 ( From Eq. (, at x a (with C, F a a y a Fa. Thus, from Eq. ( EI EI Fa Fa Fa a + C4 C4 Substitute into Eq. (, obtaining EI EI EI Fa Fa Fa ybc x + ( a x ns. EI EI EI The maximum defection occurs at, Fa ymax ( a ns. EI 4-9 ΣM C F ( / R R F / ΣF y F / + R F R F / Break at x /: V B R F /, M B R x Fx / Break at / x : V BC R F R F /, M BC R x F ( x / F ( x / Section B: Fx F x B dx C EI EI 4 + From symmetry, B at x / F F + C C. Thus, 4EI EI F x F F ( 4x B ( EI 4 EI EI Shigey s MED, th edition Chapter 4 Soutions, Page /8

7 F F 4x yb ( 4x dx x C EI + EI y B at x C, and, Fx yb ( 4x ( 48EI y BC is not given, because with symmetry, Eq. ( can be used in this region. The maximum defection occurs at x /, F F ymax 4 ns. 48EI 48EI 4- From Tabe -, for each ange, I - 7 cm 4. Thus, I (7 ( 4 4.4( mm 4 From Tabe -9, use beam with F 5 N, a mm, and mm; and beam with w N/mm and mm. 4 Fa y max ( a EI w 8EI 4 5( (( [ ( ] (7 (4.4 8(7( (4.4( O 5.4 mm ns. M Fa w ( / 5( [( /] 9.5( N mm From Tabe -, from centroid to upper surface is y 9 mm. From centroid to bottom surface is y 9. 7 mm. The maximum stress is compressive at the bottom of the beam at the wa. This stress is My 9.5( ( 7 max σ MPa ns. I 4.4( Shigey s MED, th edition Chapter 4 Soutions, Page 7/8

8 4- R R O C 4 (45 + ( 45 bf (45 + ( 85 bf M 45(.48( bf in M.48( +5(4 4.( bf in σ M max 4. max 5 Z.8 in Z Z For defections, use beams 5 and of Tabe -9 F a[ ( / ] F y ft + a EI 48EI 45(7( (4 (( I(4 48(( I. in I /. in 4 4 Seect two 5 in-.7 bf/ft channes from Tabe -7, I ( in 4, Z (.. in. ymidspan.4 in σ max 5.7 kpsi. I 4- π 4 4 I ( in 4 From Tabe -9 by superposition of beams and 7, at x a 5 in, with b 4 in and 9 in Fba wa y [ a + b ] + ( a a EI 4EI y 4( ( (.4859 (5 /(5 + 4( (.485 (9( in ns. t x / 9.5 in Fa [ ( / ] ( / w y a + + EI 4EI Shigey s MED, th edition Chapter 4 Soutions, Page 8/8

9 4(5(9.5 y (( (.485(9 (5 /( (( (.485 (9( in ns % difference ( 5.% ns..978 ((.84 mm 4- I 4 From Tabe -9-, beam Fa yc ( + a EI Fax yb ( x EI dy B Fa ( x dx EI dyb t x, dx Fa Fa EI EI Fa yo a EI With both oads, Fa Fa yo ( + a EI EI Fa 4( ( + a [ (5 + ( ].7 mm ns. EI (7 (.84 t midspan, Fa( / Fa 4((5 ye. mm ns. EI 4 EI 4 7(.84( π I (.5.59 in 4 From Tabe -5, E.4 Mpsi From Tabe -9, beams and, by superposition Shigey s MED, th edition Chapter 4 Soutions, Page 9/8

10 [ ] [ ] F 4( ( B Fa (.4 (.59 (.4 (.59 [ ] yb + ( a + ( (4( EI EI yb.94 in ns. 4-5 From Tabe -7, I (.85.7 in 4 From Tabe -5, E. Mpsi From Tabe -9, beams and, by superposition 4 [ + ] 4 F ( w + w 5( 5 (5 / ( c y.8 in ns. EI 8EI ( (.7 8( (.7 π 4 4- I d 4 From Tabe -5, E 7( MPa From Tabe -9, beams 5 and 9, with F C F F, by superposition FB Fa y B (4a I FB Fa(4a 48EI + 4EI 48Ey + 48(7 { 55( ( 75 (5 4(5 ( } ( I mm 4 B (5.4. mm. d I ns π π 4-7 From Tabe -9, beams 8 (region BC for this beam with a and (with a a, by superposition M Fax yb ( x x + x + ( x EI EI M ( x x + x + Fax ( x ns. EI d M F( x ybc ( x x x ( x [( x a( x ] dx + EI + EI M F( x ( x + [( x a( x ] EI EI ( x { M + F ( x a( x } ns. EI Shigey s MED, th edition Chapter 4 Soutions, Page /8

11 4-8 Note to the instructor: Beams with discontinuous oading are better soved using singuarity functions. This eiminates matching the sopes and dispacements at the discontinuity as is done in this soution. a wa M C R w a a + R ( a ns. wa wa Fy ( a + R w a R ns. a V B R w ( a ( a x a ns. w w w a VBC R w ns. w x M B VBdx ax a x + C M B at x C M wx B a a x ns. wa wa M BC VBCdx dx x + C wa wa M BC at x C M BC ( x ns. M B wx w B dx ( a a x dx ax a x x C EI EI + EI w EI yb Bdx ax a x x + C dx w EI + + y at x C B 4 ax a x x Cx C4 4 BC B M a a EI EI w EI w + at x a BC BC dx ( x dx x x C5 w wa wa aa a a C a a C C C EI EI ( Shigey s MED, th edition Chapter 4 Soutions, Page /8

12 a a ybc BCdx x x C5 dx x x C5x C EI w EI w wa ybc at x C C5 wa ybc x x C5( x EI + y y at x a B BC w 5 4 wa aa a a + Ca a a + C5 ( a wa Ca ( a 4 + C5 ( a ( 4 wa Substituting ( into ( yieds C5 ( a 4. Substituting this back into ( gives 4 wa C ( 4a a 4. Thus, 4 w 4 4 yb ( 4ax a x x + 4a x a x 4a x 4EI wx y ( B ax a x a a ns. 4EI w 4 4 ybc ( a x a x a x 4 a x + a ns. 4EI This resut is sufficient for y BC. However, this can be shown to be equivaent to w w y ( BC ax a x x a x + a x a x + x a 4EI 4EI w 4 ybc yb + ( x a ns. 4EI by expanding this or by soving the probem using singuarity functions. 4-9 The beam can be broken up into a uniform oad w downward from points to C and a uniform oad w upward from points to B. Using the resuts of Prob. 4-8, with b a for to C and a a for to B, resuts in wx x y B bx ( b x b ( b w ax ( a x a ( a 4EI 4EI wx bx ( b b ( b ax ( a + a ( a ns. 4EI w 4 ybc bx ( b x b x ( b 4EI ( 4ax a x x 4a x + 4 a x a x ( x a ns. Shigey s MED, th edition Chapter 4 Soutions, Page /8

13 w ycd 4bx b x x 4b x 4 b x b x ( x b 4EI + + w ax a x x 4a x 4 a x a x ( x a 4EI + + w 4 4 ( x b ( x a + yb ns. 4EI 4- Note to the instructor: See the note in the soution for Probem 4-8. wa wa Fy RB w a RB ( + a ns. For region BC, isoate right-hand eement of ength ( + a x wa VB R, VBC w ( + a x ns. wa w M B Rx x, M BC ( + a x ns. wa EI B M Bdx x + C 4 wa EIyB x + Cx + C wa y B at x C EIyB x + Cx a y B at x C w wa wa wa x wa x EIyB x + x ( x yb ( x ns. EI w EI BC M BCdx + a x + C w EIy 4 BC + a x + Cx + C wa wa y BC at x + C + C4 C4 C ( 4 4 wa wa wa wa B BC at x + + C C ( + a 4 wa Substitute C into Eq. ( gives C4 a + 4 ( + a 4. Substitute back into y BC 4 w 4 wa wa wa ybc ( a x x ( a ( a EI w 4 4 ( + a x 4 a ( x( + a a ns. 4EI Shigey s MED, th edition Chapter 4 Soutions, Page /8

14 4- Tabe -9, beam 7, w ( R R 5 bf wx x yb ( x x ( x x 4EI 4.5 x( x x.7778 d y d x w 4EI Sope: B B ( x 4x w B 4EI w 4EI w ( ybc B ( x ( x x.7778 x 4EI 4( (.5 t x, ( 4 From Prob. 4-, ( wa wa R R ( a [ ] bf B ( 4 48 bf ( + ( + a x ( 4 x ( ( w yb ( x x x x EI.5 w 4 4 ybc ( + a x 4a ( x( + a a 4EI 4 4 ( + 4 x 4( 4 ( x( ( x x 9 Superposition, R bf R bf ns. B ( 4 y.7778 x x x x x ns. B ybc.7778 x x + 89x 9 ns. The defection equations can be simpified further. However, they are sufficient for potting. Using a spreadsheet, x y x y Shigey s MED, th edition Chapter 4 Soutions, Page 4/8

15 x y x y y (in Beam Defection, Prob x (in 4- (a Usefu reations F 48EI k y k 8 I.58 in 48E 48( 4 From I bh /, and b h, then I 5 h 4 /, or, I ( h in h is cose to / in and 9/ in, whie b is cose to 5.4 in. Changing the height drasticay changes the spring rate, so changing the base wi make finding a cose soution easier. Tria and error was appied to find the combination of vaues from Tabe -7 that yieded the coset desired spring rate. h (in b (in b/h k (bf/in / 5 8 / 5½ 78 / 5¾ / / Shigey s MED, th edition Chapter 4 Soutions, Page 5/8

16 h ½ in, b 5 ½ in shoud be seected because it resuts in a cose spring rate and b/h is sti reasonaby cose to. (b 4 I 5.5(.5 /.579 in ( / 4 4σ 4( (.579 Mc F c I σ F 58 bf I I c (.5 F (58 y.84 in ns. 48EI 48( ( From the soutions to Prob. -8, T bf and T 4 bf 4 4 π d π (.5 I.98 in 4 4 From Tabe -9, beam, F b x Fb x z ( x + b + ( x + b EI EI ( 575(( + ( ( (.98(4 ( (.98(4 ( 4 4 in 4(( in ns. y d z d F b x F b x ( x + b + ( x + b dx dx EI EI in in Fb Fb ( x + b + ( x + b EI EI (575( ( + 4 ( (.98(4 in 4( + 4 ( (.98(4 4.( rad ns. 4-4 From the soutions to Prob. -9, T 88 N and T 4 N 4 4 π d π ( I 9.7( mm Shigey s MED, th edition Chapter 4 Soutions, Page /8

17 The oad in between the supports suppies an ange to the overhanging end of the beam. That ange is found by taking the derivative of the defection from that oad. From Tabe -9, beams (subscript and (subscript, y a + y ( BC C beam beam BC C F a a ( x a x ( x x a d F a x F a + dx EI EI ( EI Equation ( is thus F a F a y ( a a ( + a EI EI ( 7( + (7 (9.7 (5 (7 ( mm ns. The sope at, reative to the z axis is ( z (7 (9.7 (5 ( 5 ( ( 5 F a d F ( x ( a + ( x a( x EI dx EI F a F ( a + ( x a( x a( x EI EI F a F ( a ( a + a EI EI ( ( (7 (9.7 ( (5( + a + a.4 rad ns. 4-5 From the soutions to Prob. -7, T 9. bf and T 58.8 bf 4 4 π d π ( I.49 9 in 4 4 From Tabe -9, beam, 4 Shigey s MED, th edition Chapter 4 Soutions, Page 7/8

18 F b x ( 5(4(8 y x + b + ns EI 8in ( (.49 9( in. F b x ( 45.98(( in. z x + b + ns EI 8in ( (.49 9( The dispacement magnitude is δ y + z in ns. d y d Fb x Fb x b ( a b d x dx + + EI EI ( z a ( (.499( a ( 5( rad ns. d z d F b x F b d x dx EI EI ( ( x + b ( a + b y a (45.98( + ( (.499( a 8.5 rad ns. Θ rad ns. The sope magnitude is 4- From the soutions to Prob. -7, T 5 N and T 7.5 N 4 4 π d π ( I mm 4 4 o ( F b x 45sin 45 (55( y ( x + b EI (7 (7 854(85 mm. mm ns. F zb x Fb x z ( x + b + ( x + b EI EI mm 4 y o 45cos 45 (55( (7 (7 854(85 (7 (7 854( (5( mm ns. δ y + z mm ns. The dispacement magnitude is Shigey s MED, th edition Chapter 4 Soutions, Page 8/8

19 d y d F b x F yb x + b ( a + b d x dx EI EI y ( z a a o 45sin 45 (55 + (7 (7 854( rad ns. d z d F b x F b x d x dx EI EI z x + b + ( x + b y a F b F b EI EI ( a b ( a b z + + o 45cos 45 (55 ( (7 (7 854(85 (7 (7 85 a 87.5( rad ns. 4(85 The sope magnitude is Θ rad ns. 4-7 From the soutions to Prob. -7, F B 75 bf 4 4 π d π (.5 I.98 in 4 4 From Tabe -9, beams (subscript and (subscript F b x F a x EI EI y ( y y x + b + x o ( 4 in o cos (4( 75sin (9( ( ( (.9 8( ( (.9 8(.85 in ns. F b x F a x EI EI z ( z z x + b + x o in o ( sin (4( 75 cos (9( ( ( (.9 8( ( (.9 8(.9 in ns. The dispacement magnitude is δ y + z in ns. Shigey s MED, th edition Chapter 4 Soutions, Page 9/8

20 d y d F b x F a x d x dx EI EI y ( y ( x + b + x z F b a a y ( a b ( a y + + EI EI o ( ( (.9 8( F a cos (4 ( + 4 o ( 75sin (9 5 + ( (.9 8( 8. rad ns. d z d F b x F a x d x dx EI EI z z x + b + ( x y a F b F a EI EI z ( a b ( a z + o a o ( ( 4 ( sin (4 75 cos (9 + ( (.9 8( ( (.9 8(.5 rad ns. The sope magnitude is 5 Θ rad ns. 4-8 From the soutions to Prob. -7, F B.8 ( N 4 ( 5 4 π d π I.8( mm 4 4 From Tabe -9, beam, F yb x F yb x y ( x b ( x b EI + + EI + x + (7 (.8 (5 4 4mm o sin (5( (7 (.8 (5 o.8 sin 5 (( mm ns. Shigey s MED, th edition Chapter 4 Soutions, Page /8

21 F zb x F zb x z ( x + b + ( x + b EI EI + (7 (.8 (5 4mm o cos (5( o.8 cos 5 ((4 + ( mm ns. (7 (.8 (5 The dispacement magnitude is x δ y + z mm ns. d y d F b x F b x d x dx EI EI z z x + b + ( x + b z F b a y ( a b ( a b y EI EI F b + (7 (.8 (5 o sin (5 ( o.8( sin 5 ( (7 (.8 (5.57 rad ns. z z x + b + ( x + b y a o cos (5 ( o.8( cos 5 ( a d z d F b x F b x d x dx EI EI F zb F zb ( a + b ( a + b EI EI + (7 (.8 ( (7 (.8 (5.489 rad ns. The sope magnitude is a Θ rad ns. 4-9 From the soutions to Prob. -8, T bf and T 4 bf, and Prob. 4-, I.9 8 in 4. From Tabe -9, beam, Shigey s MED, th edition Chapter 4 Soutions, Page /8

22 d z d F b x F b x y d x dx EI EI F zb F b 575( EI EI ( (.9 8(4 z z O x + b + ( x + b ( C z ( b ( b ( 4 4( 4.48 rad ns. ( (.9 8(4 ( ( z ( ( x a x x a x d z d F a x F a x z y d x dx EI EI F a F a EI EI F a F za ( a ( a EI EI z ( x x a ( x x a z + z ( 575( 4 4( rad ns. ( (.9 8(4 ( (.9 8(4 4- From the soutions to Prob. -9, T 88 N and T 4 N, and Prob. 4-4, I 9.7 ( mm 4. From Tabe -9, beams and d y d Fb x F ax ( O ( x b ( x z d x dx + + EI EI Fb F a Fb F a ( x + b + ( x ( b + EI EI EI EI (8 7( (5 ( (7 (9.7 ( C z (7 (9.7 (5. rad ns. d y d F a ( x ( Fa x x + a x + ( x d x dx EI EI F a F a F a F a (x x a + ( x ( a EI EI EI EI ( 7((5 (5 (7 (9.7 (5 (7 (9.7.9 rad ns. 4- From the soutions to Prob. -7, T 9.9 bf and T 58.8 bf, and Prob. 4-5, I.49 9 in 4. From Tabe -9, beam Shigey s MED, th edition Chapter 4 Soutions, Page /8

23 d y d F b x F yb ( z x + b b d x dx EI EI 5(4 4.7 rad ns. y ( O ( (.499( d z d F b x F b y d x dx EI EI 45.98( z z O x + b ( b ( (.499(.4 rad ns. ( Θ rad ns. The sope magnitude is O d y d F a ( x d x dx EI x y ( C ( x + a x z F ya F ya ( x x a ( a EI EI 5(8 8.5 rad ns. ( (.49( d z d F a ( x d x dx EI z ( C ( x + a x y F za F za ( x x a ( a EI EI 45.98(.84 rad ns. ( (.499( The sope magnitude is Θ rad ns. C 4- From the soutions to Prob. -7, T 5 N and T 7.5 N, and Prob. 4-, I mm 4. From Tabe -9, beam d y d F b x F yb x + b ( b d x dx EI EI y ( O z o 45sin 45 ( rad ns. (7 (7 854(85 Shigey s MED, th edition Chapter 4 Soutions, Page /8

24 d z d F b x F b x d x dx EI EI z z O x + b + ( x + b y o F 45cos 45 (55 zb F b EI EI (7 (7 854(85 z ( b ( b ( (5. rad ns. (7 (7 854(85 The sope magnitude is O ( 5 85 Θ rad ns. d y d F a ( x F a d x dx EI EI x y ( y ( C x + a x x x a z o F 45sin 45 ( ya ( a ( rad ns. EI (7 (7 854(85 d z d F a ( x F a ( x d x dx EI EI z z C x + a x + ( x + a x y o F 45cos 45 ( za F a EI EI (7 (7 854(85 z ( a ( a ( (7 5 ( (7 (7 854( rad ns. 5 The sope magnitude is Θ C rad ns. 4- From the soutions to Prob. -7, F B 75 bf, and Prob. 4-7, I.9 8 in 4. From Tabe -9, beams and d y d F yb x F ( ya x ( O x b z + + x d x dx EI EI x F yb F ya F yb F ya ( x + b + ( x ( b + EI EI EI EI ( (.9 8( o o cos (4 75sin (9( ( rad ns. ( (.9 8 Shigey s MED, th edition Chapter 4 Soutions, Page 4/8

25 d z d F b x F a x d x dx EI EI z z O x + b + ( x y F zb F za F zb F za ( x + b + ( x ( b EI EI EI EI o o sin (4 75 cos ( 4 (9(.4 rad ns. ( (.9 8( ( (.9 8 The sope magnitude is O Θ rad ns. d y d F a ( x F a x dx dx EI EI x y ( y ( C x + a x + x z F ya F ya F ya F ya ( x x a + ( x ( a EI EI EI EI o cos ( ( (.9 8( o 75sin (9(.9 rad ns. ( (.9 8 d z d F a ( x F a x d x dx EI EI z z C x + a x + ( x y F za F za F za F za ( x x a + ( x ( a + EI EI EI EI o sin ( ( (.9 8( ( ( (.9 8 The sope magnitude is o 75 cos (9( +.9 rad ns. Θ rad ns. C 4-4 From the soutions to Prob. -7, F B.8 kn, and Prob. 4-8, I.8 ( mm 4. From Tabe -9, beam d y d F b x F b x d x dx EI EI y ( y ( O x + b + x + b z F b + EI EI (7 (.8 (5 o sin (5 y F yb ( b ( b o.8( sin 5 ( + ( (7 ( ( rad ns..8 (5 Shigey s MED, th edition Chapter 4 Soutions, Page 5/8

26 d z d F b x F b x d x dx EI EI F zb F zb ( b ( b EI EI o ( cos (5 z z O x + b + ( x + b y (7 (.8 (5 ( 5 5 o.8( cos 5 ( 5.47 rad ns. (7 (.8 (5 The sope magnitude is Θ rad ns. O d y d F a ( x F a ( x d x dx EI EI x y ( y ( C x + a x + x + a x z F a F a (x x a + x x a EI EI y y o F ( sin (4 ya F ya ( a ( a + ( 5 4 EI EI (7 (.8 (5 (7 (.8 (5 o.8 sin 5 ( rad ns. d z d F a ( x F a ( x d x dx EI EI z z C x + a x + ( x + a x y F a F a EI EI z ( x x a ( x x a z + o F ( cos (4 za F za ( a ( a ( 5 4 EI EI (7 (.8 (5 (7 (.8 (5 o.8 cos 5 (75 ( rad ns. The sope magnitude is Θ C rad ns. 4-5 The required new sope in radians is new.(π /8.5 rad. In Prob. 4-9, I.9 8 in 4, and it was found that the greater ange occurs at the bearing at O where ( O y.48 rad. Since is inversey proportiona to I, Shigey s MED, th edition Chapter 4 Soutions, Page /8

27 new I new od I od I new π d 4 new /4 od I od / new or, d new 4 od Iod π new /4 The absoute sign is used as the od sope may be negative dnew in ns. π.5 4- The required new sope in radians is new.(π /8.5 rad. In Prob. 4-, I 9.7 ( mm 4, and it was found that the greater ange occurs at the bearing at C where ( C y.9 rad. See the soution to Prob. 4-5 for the deveopment of the equation /4 d new 4 od od π new I /4 4.9 dnew 9.7. mm ns. π.5 /4 4-7 The required new sope in radians is new.(π /8.5 rad. In Prob. 4-, I.49 in 4, and the maximum sope is Θ C.4 rad. See the soution to Prob. 4-5 for the deveopment of the equation d 4 od new I od π new /4 4.4 dnew in ns. π The required new sope in radians is new.(π /8.5 rad. In Prob. 4-, I mm 4, and the maximum sope is Θ O.75 rad. See the soution to Prob. 4-5 for the deveopment of the equation /4 Shigey s MED, th edition Chapter 4 Soutions, Page 7/8

28 d new 4 od I od π new / dnew mm ns. π The required new sope in radians is new.(π /8.5 rad. In Prob. 4-, I.9 8 in 4, and the maximum sope Θ. rad. See the soution to Prob. 4-5 for the deveopment of the equation /4 d new 4 od od π new I /4 4. dnew in ns. π The required new sope in radians is new.(π /8.5 rad. In Prob. 4-4, I.8 ( mm 4, and the maximum sope is Θ C.74 rad. See the soution to Prob. 4-5 for the deveopment of the equation /4 d new 4 od Iod π new / dnew.8.9 mm ns. π.5 /4 4-4 I B π 4 /4.499 in 4, J B I B.988 in 4, I BC (.5(.5 /.7 in 4, I CD π (/4 4 /4.55 in 4. For Eq. (-4, p., b/c.5/.5 β.99. The defection can be broken down into severa parts. The vertica defection of B due to force and moment acting on B (y.. The vertica defection due to the sope at B, B, due to the force and moment acting on B (y CD B B. Shigey s MED, th edition Chapter 4 Soutions, Page 8/8

29 . The vertica defection due to the rotation at B, B, due to the torsion acting at B (y BC B 5 B. 4. The vertica defection of C due to the force acting on C (y The rotation at C, C, due to the torsion acting at C (y CD C C.. The vertica defection of D due to the force acting on D (y 5.. From Tabe -9, beams and 4 with F bf and M B ( 4 bf in ( 4( y +.47 in ( (.499 ( (.499. From Tabe -9, beams and 4 d Fx M Bx Fx M Bx B ( x + ( x dx EI EI + EI EI [ F + M B ] ( ( + ( rad EI.499 y ( in. The torsion at B is T B 5( bf in. From Eq. (4-5 TL.54 rad B JG B y 5( in 4. For bending of BC, from Tabe -9, beam y 5.95 in For twist of BC, from Eq. (-4, p., with T ( 4 bf in 4( 5 C.48 rad y 5 ( in. For bending of CD, from Tabe -9, beam y.4 in.55 Shigey s MED, th edition Chapter 4 Soutions, Page 9/8

30 Summing the defections resuts in y y in ns. D i i This probem is soved more easiy using Castigiano s theorem. See Prob The defection of D in the x direction due to F z is from:. The defection due to the sope at B, B, due to the force and moment acting on B (x BC B 5 B.. The defection due to the moment acting on C (x.. For B, I B π 4 /4.499 in 4. From Tabe -9, beams and 4 B d Fx M Bx Fx M Bx ( x + ( x + dx EI EI EI EI [ F + M B ] ( ( + (.7 rad EI.499 x 5(.7.9 in. For BC, I BC (.5(.5 /.95 in 4. From Tabe -9, beam 4 x C ( 5 M.47 in EI.95 The defection of D in the x direction due to F x is from:. The eongation of B due to the tension. For B, the area is π / in x F 5 5.8( in E B The defection due to the sope at B, B, due to the moment acting on B (x BC B 5 B. With I B.497 in 4, ( M 5 5 B B.5 rad EI.499 Shigey s MED, th edition Chapter 4 Soutions, Page /8

31 x 4 5(.5.58 in 5. The defection at C due to the bending force acting on C. With I BC.95 in 4 x 5 EI BC F in.95. The eongation of CD due to the tension. For CD, the area is π (.75 /4.448 in F 5 x E CD in Summing the defections resuts in x D i 5 x i in ns. 4-4 J O J BC π (.5 4 /.497 in 4, J B π ( 4 /.987 in 4, I B π ( 4 /4.499 in 4, and I CD π (.75 4 /4.55 in 4. T T T T O B BC GJ O GJ B GJ BC G JO J B J BC 5( 9. rad. + + ns.5( Simpified T 5(( s GJ s.45 rad ns. Simpified is.45/.. times greater ns. 5 5 FyOC FyCD D + s ( CD ( + EI B EICD ( (.499 (.55 y yd.847 in ns Reverse the defection equation of beam 7 of Tabe -9. Using units in bf, inches Shigey s MED, th edition Chapter 4 Soutions, Page /8

32 ( / x { } wx y ( x x x x 4EI x x x + ns The maximum height occurs at x 5(/ 5 in ymax in ns From Tabe -9, beam, Fbx yl ( x + b EI Fb yl x + b x x EI dyl Fb x + b dx EI dyl dx ( Fb b EI Let ξ dy dx L and set I π d. Thus, 4 4 L d L Fb b ( /4 π Eξ ns. For the other end view, observe beam of Tabe -9 from the back of the page, noting that a and b interchange as do x and x d R Fa ( a /4 π Eξ ns. For a uniform diameter shaft the necessary diameter is the arger of dl and d R. 4-4 The maximum sope wi occur at the eft bearing. Incorporating a design factor into the soution for dl of Prob. 4-45, Shigey s MED, th edition Chapter 4 Soutions, Page /8

33 4 ( b nfb d π Eξ d /4 ( (.8(( d 8. mm ns. 4 ( 8. π (7 ((. π 4 I.4( mm 4 From Tabe -9, beam, the maximum defection wi occur in BC where dy BC /dx d Fa dx EI ( x ( ( x a x x x a x ± 4( 4.,.7 mm x x + + x x + x.7 mm is acceptabe. ( x ( Fa ymax x + a x EI (.7mm mm ns I π (.5 4 /4.98 in 4. From Tabe -9, beam ( ( x a x ( x b F a x F b x δ + EI + + EI 5(5( 8 ( ((8 ( (.98 ( 5((8 + ( 8 + ( (.98 (. in ns. / Shigey s MED, th edition Chapter 4 Soutions, Page /8

34 4-48 I π (.5 4 /4.98 in 4. For both forces use beam of Tabe -9. For F 5 bf: x 5 Fb x 5( 5 x y ( x + b x + 5 EI.98 ( x( x ( 5 x F a ( x ( 5 ( 5( x y x + a x x + 5 ( x EI.98 ( x( x x ( For F 5 bf: x Fb x 5( x z ( x + b EI.98 x + ( x( x ( x F a ( x ( 5 ( ( x z x + a x x + ( x EI.98 ( x( x x (4 Pot Eqs. ( to (4 for each. in using a spreadsheet. There are data points, too numerous to tabuate here but the pot is shown beow, where the maximum defection of δ.55 in occurs at x 9.9 in. ns Dispacement (in y (in z (in Tota (in x (in Shigey s MED, th edition Chapter 4 Soutions, Page 4/8

35 4-49 The arger sope wi occur at the eft end. From Tabe -9, beam 8 M Bx yb EI x + a a + dyb M B dx EI + + ( (x a a With I π d 4 /4, the sope at the eft bearing is dyb M B + dx E d / 4 Soving for d 4 ( π (a a M ( + + ( 4 (4 (4( B d 4 a a π E π ( (.(.4 in ns. 4-5 From Tabe -5, E.4 Mpsi ΣM O 8 F BC ( F BC. bf The cross sectiona area of rod BC is π (.5 /4.9 in. The defection at point B wi be equa to the eongation of the rod BC. y B FL.( 5.79( in ns. E BC (.9 ( 4-5 ΣM O F C ( F C 8. bf The defection at point in the negative y direction is equa to the eongation of the rod C. From Tabe -5, E s Mpsi. y FL ( in E C π (.5 / 4 ( By simiar trianges the defection at B due to the eongation of the rod C is y y 8 B 4 yb y (.75. in From Tabe -5, E a.4 Mpsi The bar can then be treated as a simpy supported beam with an overhang B. From Tabe -9, beam, Shigey s MED, th edition Chapter 4 Soutions, Page 5/8

36 dy BC Fa d F( x Fa yb BD ( + a 7 ( x a( x ( a dx + a EI dx EI + EI + a F Fa 7Fa Fa EI EI EI EI 7 ( x a( x a( x + a ( a ( a ( a [ ] ( 5 (.4 (.5 / 7 5 (.4.5 / ( + (5 ( in y B y B + y B in ns. 4-5 From Tabe -5, E 7 GPa, and G 79. GPa. T T FB FOCB FCB FB yb B + B GJ GJ EI G d / G d / E d / 4 FB OC C B π GdOC Gd C Ed B ( π ( π ( π OC C B OC C + + The spring rate is k F/ y B. Thus k + + B OC C B π GdOC Gd C Ed B ( π 79.( 8 79.( ( 7 ( 8 8. N/mm ns. 4-5 For the beam defection, use beam 5 of Tabe -9. F R R F F δ, and δ k k δ δ Fx δ 48EI k k x + + yb + x + (4x yb F x (4x ns. k kk 48EI Shigey s MED, th edition Chapter 4 Soutions, Page /8

37 For BC, since Tabe -9 does not have an equation (because of symmetry an equation wi need to be deveoped as the probem is no onger symmetric. This can be done easiy using beam of Tabe -9 with a / F Fk Fk F ( / ( x ybc + x + x + x k kk EI 4 ( x ( k k F + x + 4x + 8 x ns. k kk 48EI 4-54 Fa F R, and R ( + a Fa F δ, and δ ( + a k k δ δ Fax yb δ + x + ( x EI a x ax yb F + k a k ( a ( x ns. k kk + + EI δ δ F( x ybc δ + x + ( x a ( x EI a x ( x ybc F + k a k ( a ( x a ( x ns. k kk + + EI 4-55 Let the oad be at x /. The maximum defection wi be in Section B (Tabe -9, beam Fbx yb x + b EI dyb dx Fb ( x + b x + b EI b x, xmax.577 ns. For x /, xmin ns. Shigey s MED, th edition Chapter 4 Soutions, 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 9.5( + 55x 5 x - dy x EI 9.5( x 75x 5 x C dx + + dy x C dx at dy x EI 9.5( x + 75x 5 x dx 4 x EIy 4.75( x + 9.7x 4.7 x + C 4 y at x C, and therefore 4 y 4( x ( x x ( x 4EI + + y B 4 ( ( mm ns M O 9.5 ( N m. The maximum stress is compressive at the bottom of the beam where y 9. 7 mm My 9.5 ( 7 σ max ( Pa MPa ns. I 4.4( The soutions are the same as Prob See Prob. 4- for reactions: R O 45 bf and R C 85 bf. Using bf and inch units M 45 x 45 x 7 x Shigey s MED, th edition Chapter 4 Soutions, Page 8/8

39 dy EI.5x 5 x 7 5 x C dx + EIy 77.5 x 75 x 7 5 x C x y at x C y at x 4 in 77.5(4 75(4 7 5(4 + C x C.( bf in and, EIy 77.5 x 75 x 7 5 x.( x Substituting y.5 in at x in gives ( I ( ( 75( 7 5(.( ( I. in 4 Seect two 5 in.7 bf/ft channes; from Tabe -7, I ( in 4. ymidspan.4 in ns The maximum moment occurs at x in where M max 4.( bf in Mc 4.( (.5 σ max 5 7 psi O.K. I 4.98 The soutions are the same as Prob I π (.5 4 /4.485 in 4, and w 5/.5 bf/in. 4 R O ( (4 45. bf 9.5 M 45.x x 4 x 5 dy.5 EI.5x x 7 x 5 C dx EIy x x x + C x + C y at x C 4 at 9 in.85( bf in Thus, y x C y x.58x 5.7 x 5.85 ( x EI Evauating at x 5 in, Shigey s MED, th edition Chapter 4 Soutions, Page 9/8

40 y y midspan 75.5( 5.58( ( ( 4 (5 ( ( in ns. 75.5( ( ( ( 4 (9.5 ( ( in ns. 5 % difference ns. The soutions are the same as Prob I.5 in 4 ( 4 7( 4, R 4 bf and RB 98 bf M 4 x 5 x + 98 x dy dx + + EI x.7x 49 x C EIy x x + x + C x + C y at x C y at x in C 8 bf in. Thus, 4 y 7x 4.7x +. x 8x x 4.7x +. x 8 x ns. The tabuar resuts and pot are exacty the same as Prob R R B 4 N, and I ( / 84 mm 4. First haf of beam, M 4 x + 4 x dy EI x x C dx + + From symmetry, dy/dx at x 55 mm (55 + (55 + C C 48( N mm EIy.7 x +.7 x + 48( x + C Shigey s MED, th edition Chapter 4 Soutions, Page 4/8

41 y at x mm C.( 9 N mm. The term (EI [7( 84].949 ( Thus y.949 ( [.7 x +.7 x + 48( x.( 9 ] y O.7 mm ns. y x 55 mm.949 ( [.7 ( ( ( 55.( 9 ]. mm ns. The soutions are the same as Prob M B R + Fa M R ( M Fa M M + R F( + a R F + Fa M M R x M + R x dy EI R x M x + R x + C dx EIy R x M x + R x + C x + C y at x C y at x C R + M. Thus, EIy R x M x R x + + R + M x y ( M ( Fa x M x F Fa M x Fa M x EI ns. In regions, yb ( M Fa x M x ( Fa M x EI + + x M ( x x Fa( x EI + + ns. Shigey s MED, th edition Chapter 4 Soutions, Page 4/8

42 y ( BC M Fa x M x + F + Fa M x + Fa + M x EI { M } x x x x F ax a x ax EI { M ( x F ( x ( x a( x } EI ( x M F ( x a ( x ns { } + EI The soutions reduce to the same as Prob w( b a 4- M D R w ( b a b + ( b a R ( b a w w M R x x a + x b dy EI R x x a x b C dx w + w + w 4 w 4 EIy R x x a + x b + Cx + C 4 4 y at x C y at x w w C R a b ( b a ( 4 4 w w w EI y b a x x a + x b ( b a w w 4 w 4 x b a a + b 4 4 w 4 4 { ( b a( b a x x a + x b 4EI 4 4 } x b a b a a + b ns. The above answer is sufficient. In regions, Shigey s MED, th edition Chapter 4 Soutions, Page 4/8

43 w yb b a b a x x b a b a a + b 4EI wx 4 4 ( b a( b a x ( b a( b a + ( a ( b 4EI 4 4 { ( ( } { w ybc b a b a x x a 4EI 4 4 } x b a b a a + b { w ycd b a b a x x a + x b 4EI } x b a b a a + b These equations can be shown to be equivaent to the resuts found in Prob I π (.75 4 /4.755 in 4, I π (.75 4 /4.44 in 4, R.5(8( 9 bf Since the oading and geometry are symmetric, we wi ony write the equations for the first haf of the beam For x 8 in t x, M 7 bf in M 9x 9 x Writing an equation for M / I, as seen in the figure, the magnitude and sope reduce since I > I. To reduce the magnitude at x in, we add the term, 7(/I / I x. The sope of 9 at x in is aso reduced. We account for this with a ramp function, x. Thus, 4 M 9x 9 7 x 9 x x I I I I I I I 58x 95 x 7 x 95.5 x dy E 54x 95 x 587 x 5.7 x C dx + Boundary Condition: at 8 in dy x dx Shigey s MED, th edition Chapter 4 Soutions, Page 4/8

44 C C 8.7 ( bf/in 4 Ey 854.7x 47 x 59 x.9 x 8.7( x + C y at x C Thus, for x 8 in 4 y 854.7x 47 x 59 x.9 x 8.7( x ns. ( Using a spreadsheet, the foowing graph represents the defection equation found above -. Beam Defection y (in x (in The maximum is ymax. in at x 8 in ns. 4-4 The force and moment reactions at the eft support are F and F respectivey. The bending moment equation is M Fx F Pots for M and M /I are shown. M /I can be expressed using singuarity functions Shigey s MED, th edition Chapter 4 Soutions, Page 44/8

45 M F F F F x x + x I I I 4I I where the step down and increase in sope at x / are given by the ast two terms. Integrate dy F F F F E x x x + x + C dx 4I I 4I 4I dy/dx at x C F F F F Ey x x x + x + C I 4I 8I I y at x C F y x x x + x 4EI F 5F y ( ( ns. / + 4EI 9EI F F y ( ( x ns. + 4EI EI The answers are identica to Ex Pace a dummy force, Q, at the center. The reaction, R w / + Q / w Q w x M + x M x Q Integrating for haf the beam and doubing the resuts / / M w wx x ymax M dx x dx EI Q EI Q Note, after differentiating with respect to Q, it can be set to zero / 4 w w x x 5w ymax x ( x dx ns. EI EI 4 84EI 4- Pace a fictitious force Q pointing downwards at the end. Use the variabe x originating at the free end and positive to the eft / Shigey s MED, th edition Chapter 4 Soutions, Page 45/8

46 wx M M Qx x Q M wx w ymax M dx ( x dx x dx EI δq EI EI Q 4 w ns. 8EI 4-7 From Tabe -7, I -.85 in 4. Thus, I (.85.7 in 4 First treat the end force as a variabe, F. dding weight of channes of (5/.8 bf/in. Using the variabe x as shown in the figure 5.8 M F x x F x.97x M x M δ M d x ( F x.97 x ( x d x ( Fx /.97 x / 4 EI EI EI 4 (5( / + (.97( / 4 ( (.7.8 in in the direction of the 5 bf force y.8 in ns. 4-8 The energy incudes torsion in C, torsion in CO, and bending in B. Negecting transverse shear in B M M Fx, x In C and CO, T T FB, B The tota energy is U B T T M + + GJ GJ EI C CO B dx Shigey s MED, th edition Chapter 4 Soutions, Page 4/8

47 The defection at the tip is B B U TC T TCO T M M TCB TCOB dx Fx dx δ GJ GJ EI GJ GJ EI C CO C CO B TCB TCOB FB FCB FCOB FB δ GJ GJ EI G d / G d / E d / 4 FB C CO B π Gd C GdCO Ed B ( π ( π ( π C CO B C CO B + + F π k + + δ Gd Gd Ed C CO B B C CO B π N/mm ns ( 79.( 8 79.( ( 7 ( I π (.75 4 /4.755 in 4, I π (.75 4 /4.44 in 4 Pace 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 x 9( x.5x Q For x in For x in By symmetry it is equivaent to use twice the integra from to 8 δ ( 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 ( 5.( + EI EI in ns. Shigey s MED, th edition Chapter 4 Soutions, Page 47/8

48 4-7 I π (.5 4 /4.8 ( in 4, J I. ( in 4, π (.5 /4.9 in. Consider x to be in the direction of O, y verticay upward, and z in the direction of B. Resove the force F into components in the x and y directions obtaining. F in the horizonta direction and.8 F in the negative vertica direction. The. F force creates strain energy in the form of bending in B and O, and tension in O. The.8 F force creates strain energy in the form of bending in B and O, and torsion in O. Use the dummy variabe x to originate at the end where the oads are appied on each segment, M. F: B M.F x. x M O M 4.F 4. F a a.f. M.8 F: B M.8F x.8 x M O M.8F x.8 x T T 5.F 5. Once the derivatives are taken the vaue of F 5 bf can be substituted in. The defection of B in the direction of F is* ( δ B F U Fa L a TL T M + + M d x E JG EI O ( ( (.x d x + d x O.8x d x 8( in ns. 5 (.8x d x Shigey s MED, th edition Chapter 4 Soutions, Page 48/8

49 *Note. Be carefu, this is not the actua defection of point B. For this, dummy forces must be paced on B in the x, y, and z directions. Determine the energy due to each, take derivatives, and then substitute the vaues of F x 9 bf, F y bf, and F z. This can be done separatey and then added by vector addition. The actua defections of B are found to be δ B.8 i.8 j.77 k in From this, the defection of B in the direction of F is δ in B F which agrees with our resut. 4-7 Strain energy. B: Bending and torsion, BC: Bending and torsion, CD: Bending. I B π ( 4 /4.499 in 4, J B I B.988 in 4, I BC.5(.5 /.7 in 4, I CD π (.75 4 /4.55 in 4. For the torsion of bar BC, Eq. (-4 is in the form of TL/(JG, where the equivaent of J is J eq βbc. With b/c.5/.5, J BC βbc.99( ( in 4. Use the dummy variabe x to originate at the end where the oads are appied on each segment, B: Bending M F x + F M x + Torsion T 5F T 5 BC: Bending M F x M x Torsion T F T CD: Bending M F x M x U T T M δ D + M d x JG EI 5F F ( F x (.7 F x d x (.55 F x d x ( 4 F 5.7 ( 4.4 in ns..9 F +.48 F +.4 F +.98 F F 5.7 d x Shigey s MED, th edition Chapter 4 Soutions, Page 49/8

50 4-7 B π ( / in, I B π ( 4 /4.499 in 4, I BC.5 (.5 /.95 ( in 4, CD π (.75 /4.448 in, I B π (.75 4 /4.55 in 4. For (δ D x et F F x 5 bf and F z bf. Use the dummy variabe x to originate at the end where the oads are appied on each segment, M y CD: M y Fz x a Fa F M y BC: M y F x + Fz x a Fa Fz M y B: M y 5F + Fz + Fz x 5 a Fa F 5 U FL a ( δ D ( F x Fz x d x x + + E EI CD + FL a ( 5F Fz Fz x ( 5 d x EI E F B F F ( Fz ( F F z F Fz F ( F ( Fz +.9( F z BC +.9 F +.54 F 8.5 F F Substituting F F x 5 bf and F z bf gives ( 4 4 δ D in ns. x 4-7 I O I BC π (.5 4 /4.485 in 4, J O J BC I O.497 in 4, I B π ( 4 /4.499 in 4, J B I B.988 in 4, I CD π (.75 4 /4.55 in 4 Let F y F, and use the dummy variabe x to originate at the end where the oads are appied on each segment, B z Shigey s MED, th edition Chapter 4 Soutions, Page 5/8

51 M T OC: M F x x, T F DC: M M F x x U TL T M ( δ D M d x y + JG OC EI The terms invoving the torsion and bending moments in OC must be spit up because of the changing second-area moments. ( δ D y F 4 F ( ( F x d x + F x d x + F x d x + F x d x F F F 4 5 F F F ( F ( ns in. For the simpified shaft OC, ( δ B y F ( F x d x 4 F F F F in ns. F x d x Simpified is.848/.7. times greater ns Pace a dummy force Q pointing downwards at point B. The reaction at C is R C Q + (/8 Q +. This is the axia force in member BC. Isoating the beam, we find that the moment is not a function of Q, and thus does not contribute to the strain energy. Thus, ony energy in the member BC needs to be considered. Let the axia force in BC be F, where Shigey s MED, th edition Chapter 4 Soutions, Page 5/8

52 F Q +. Q ( + U FL. ( 5 δ.79 B in ns. Q E Q BC Q Q π.5 / I OB.5( /.7 in 4 C π (.5 /4.9 in ΣM O R C ( 8 Q R C Q + 8. ΣM R O 5( Q R O Q + 8. Bending in OB. BD: Bending in BD is ony due to Q which when set to zero after differentiation gives no contribution. D: Using the variabe x as shown in the figure above M M x Q 7 + x 7 + x Q O: Using the variabe x as shown in the figure above M M ( Q + 8. x x Q xia in C: F Q + 8. Q Shigey s MED, th edition Chapter 4 Soutions, Page 5/8

53 U FL M δ B M dx Q + E Q EI Q Q Q Q ( ( x ( 7 x d x ( 8. x dx EI 5. + x ( 7 + x d x +.7 x dx.4( in ns. 4-7 There is no bending in B. Using the variabe, rotating countercockwise from B M M PR sin R sin P r Fr P cos cos P F Psin sin P MF PRsin P (4 4 mm, ro 4 + ( 4 mm, ri 4 ( 7 mm, From Tabe -4, p. 5, for a rectanguar cross section r n mm n(4 / 7 From Eq. (4-, the eccentricity is e R r n mm From Tabe -5, E 7( MPa, G 79.( MPa From Tabe 4-, C. From Eq. (4-8 π π π π M M F R F ( MF CFr R F d d d r δ d + + ee P E P E P G P ( sin ( sin PRsin ( cos π π π π P R PR CPR d + d d + d ee E E G π PR R ((4 4 (7 (. EC π E e G 4(4( δ.8 mm ns. Shigey s MED, th edition Chapter 4 Soutions, Page 5/8

54 4-77 Pace a dummy force Q pointing downwards at point. Bending in B is ony due to Q which when set to zero after differentiation gives no contribution. For section BC use the variabe, rotating countercockwise from B M M PR sin + Q( R + R sin R ( + sin Q r Fr ( P + Qcos cos Q F ( P + Qsin sin Q MF PR sin + QR + sin P + Q sin MF PR sin PR sin ( sin QR sin ( sin Q But after differentiation, we can set Q. Thus, MF PR sin ( sin Q + (4 4 mm, ro 4 + ( 4 mm, ri 4 ( 7 mm, From Tabe -4, p.5, for a rectanguar cross section r n mm n(4 / 7 From Eq. (4-, the eccentricity is e R r n mm From Tabe -5, E 7( MPa, G 79.( MPa From Tabe 4-, C. From Eq. (4-8, ( MF π π π π M M F R F CFr R F d d d r δ d + ee Q + E Q E Q G Q π π π PR PR PR sin + sin d + sin d sin + sin d ee E E π CPR + cos d G π PR π PR π PR π CPR PR π R π CE ee 4 E 4 E 4 G E e 4 G 4 π 4 π (.7 mm ns. Shigey s MED, th edition Chapter 4 Soutions, Page 54/8

55 4-78 (.5.5(.5.75 in ( +.5((.5 ( (.5(.5 R.5 in.75 Section is equivaent to the T section of Tabe -4, p. 5,.5( (.5 r n.79 in.5n[( +.75 /] +.75n[( + / ( +.75] e R r n in For the straight section I z (.5 ( +.5(( (.5 (.5 +.5( in 4 For x 4 in M V M Fx x, V F For π / r Fr F cos cos, F F sin sin M M F(4 +.5sin (4 +.5sin MF MF F(4 +.5sin F sin F(4 +.5sin sin 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 doube the resuts Shigey s MED, th edition Chapter 4 Soutions, Page 55/8

56 4 / F(4( π (4 +.5sin δ Fx dx F d E I ( G / E.75(.9 π / F sin (.5 π / F(4 +.5sin sin + d d π / ( F cos (.5 + d.75( G / E Substitute I.89 in 4, F 7 bf, E ( psi, G.5 ( psi ( π π δ 7( ( (.89.75(.5 /.75( π π.5 π + 4( (.5 / 4. in ns Since R/h 5/ use Eq. (4-8, integrate from to π, and doube the resuts M M FR ( cos R ( cos r Fr F sin sin F F cos cos MF F Rcos cos ( MF From Eq. (4-8, ( FRcos cos FR π FR π δ ( cos d cos d ee + E FR.FR ( + E G π π cos cos d sin d FR π R π E + +.π E e G 4.5(.5 mm, E 7 ( N/mm, G 79. ( N/mm, and from Tabe -4, p. 5, Shigey s MED, th edition Chapter 4 Soutions, Page 5/8

57 h 4.5 rn mm ro 7.5 n n ri.75 and e R r n mm. Thus, F ( 5 π 5 π 7 δ π F.5( where F is in N. For δ mm, F.5 N ns..858 Note: The first term in the equation for δ dominates and this is from the bending moment. Try Eq. (4-4, and compare the resuts. 4-8 R/h > so Eq. (4-4 can be used to determine defections. Consider the horizonta reaction to be appied at B and subject to the constraint ( δ. B H FR M π M ( cos HRsin Rsin < < H By symmetry, we may consider ony haf of the wire form and use twice the strain energy Eq. (4-4 then becomes, π / U M ( δ B H M Rd H EI H π / FR ( cos HRsin ( Rsin R d F 9.55 N. + F π F H H ns π π The reaction at is the same where H goes to the eft. Substituting H into the moment equation we get, M FR [ ( cos sin ] M R π π [ π ( cos sin ] < < π π Shigey s MED, th edition Chapter 4 Soutions, Page 57/8

58 U M π / FR δ P M Rd [ π ( cos sin ] R d P EI EI 4π FR π / ( π π cos 4sin π cos 4π sin 4π sin cos d π EI FR π π π π π 4 π 4π π π EI (π 8π 4 FR (π 8π 4 ((4.4 mm ns. 4 8π EI 8π 7( π ( / The radius is sufficienty arge compared to the wire diameter to use Eq. (4-4 for the curved beam portion. The shear and axia components wi be negigibe compared to bending. Pace a fictitious force Q pointing to the eft at point. M M PR sin + Q( Rsin + Rsin + Q Note that the strain energy in the straight portion is zero since there is no rea force in that section. From Eq. (4-4, π / M π / δ M Rd PR sin ( Rsin + Rd EI Q EI Q PR π / PR π (5 π ( R sin sin d R (5 4 4 EI EI 4 ( π (.5 / in ns. 4-8 Both the radius and the ength are sufficienty arge to use Eq. (4-4 for the curved beam portion and to negect transverse shear stress for the straight portion. Straight portion: M P B M B Px x Curved portion: M P[ R( cos + ] BC [ R( cos + ] BC M P From Eq. (4-4 with the addition of the bending strain energy in the straight portion of the wire, Shigey s MED, th edition Chapter 4 Soutions, Page 58/8

59 M / B π M BC δ M B dx + M BC Rd EI P EI P P / PR π x dx [ R( cos ] d EI + EI + P PR π / + R ( cos cos R( cos d EI EI P PR π / + R cos ( R R cos ( R d EI EI P PR π π + R ( R R ( R EI EI P π π R R( R R R( R EI π π / π (5 5 (5 (5( in ns. 4-8 Both the radius and the ength are sufficienty arge to use Eq. (4-4 for the curved beam portion and to negect transverse shear stress for the straight portion. Pace a dummy force, Q, at verticay downward. The ony oad in the straight section is the axia force, Q. Since this wi be zero, there is no contribution. In the curved section M M PRsin + QR( cos R ( cos Q From Eq. (4-4 π / M π / δ M Rd PR sin R ( cos Rd EI Q EI ( 5 π ( 4 Q PR PR PR ( sin sin cos d EI EI EI π /.74 in ns..5 / Both the radius and the ength are sufficienty arge to use Eq. (4-4 for the curved beam portion and to negect transverse shear stress for the straight portion. Shigey s MED, th edition Chapter 4 Soutions, Page 59/8

60 Pace a dummy force, Q, at verticay downward. The oad in the straight section is the axia force, Q, whereas the bending moment is ony a function of P and is not a function of Q. When setting Q, there is no axia or bending contribution. In the curved section M M P R ( cos + QRsin R sin Q From Eq. (4-4 π / M π / δ M Rd P R( cos + ( R sin Rd EI Q EI Q π / PR PR PR EI EI EI ( Rsin Rsin cos sin d R R ( R ( 5 π ( in.5 / 4 Since the defection is negative, δ is in the opposite direction of Q. Thus the defection is δ.45 in ns Consider the force of the mass to be F, where F 9.8( 9.8 N. The oad in B is tension B FB F For the curved section, the radius is sufficienty arge to use Eq. (4-4. There is no bending in section DE. For section BCD, et be countercockwise originating at D M FR sin M R sin π Using Eqs. (4-9 and (4-4 F π B M F π FR δ + M Rd ( sin d E F + EI E EI B ( 4 F π FR F π R π E EI E I 7( π ( / 4 π / 4.7 mm ns. Shigey s MED, th edition Chapter 4 Soutions, Page /8

61 4-8 O (.5.5 in, I OB.5( /.7 in 4, I C π (.5 4 /4.8 ( - in 4 ppying a force F at point B, using statics (C is a two-force member, the reaction forces at O and C are as shown. O O: xia FO F M O Bending M O Fx x B: Bending M B M B F x x C: Isoating the upper curved section M M C FR sin + cos R sin + cos F O δ + 4Fx dx + F x d x E EI EI ( EI F C ( ( O OB OB π / 9FR + + d C ( sin cos ( 9F ( 4F F π / ( sin sin cos sin cos cos d π π π.7( F + 7.9( F +.58( F F F.. in ns O (.5.5 in, I OB.5( /.7 in 4, I C π (.5 4 /4.8 ( - in 4 ppying a vertica dummy force, Q, at, from statics the reactions are as shown. The dummy force is transmitted through section O and member C. Shigey s MED, th edition Chapter 4 Soutions, Page /8

62 O O: FO F + Q Q C: M ( F + Q R sin ( F + Q R( cos C R( sin + cos C F M δ + E Q EI Q π / O C M C Rd O C Q π / FO FR + + d ( E ( EI O C ( sin cos M Q π π π +.4 in ns I π ( 4 /4. mm 4 π / M M FR sin Rsin T T FR( cos R( cos ccording to Castigiano s theorem, a positive U/ F wi yied a defection of in the negative y direction. Thus the defection in the positive y direction is U ( δ ( sin + [ ( cos ] y π / π / F R R d F R R d EI GJ Integrating and substituting J I and G E / ( + ν FR π π FR ( δ y ( ν [ 4π 8 (π 8 ν ] EI EI (5(8 [4π 8 + (π 8(.9].5 mm ns. 4( The force appied to the copper and stee wire assemby is F + F 4 bf ( c Since the defections are equa, δc δ s s Shigey s MED, th edition Chapter 4 Soutions, Page /8

63 F F E E c Fc F ( π / 4(.9 (7. ( π / 4(.55 ( s s Yieds, F.4F. Substituting this into Eq. ( gives c s.4f + F.4F 4 F 5. bf s s s s Fc.4Fs 4.5 bf Fc 4.5 σ c 75 psi. kpsi ns. c ( π / 4(.9 Fs 5. σ s 7 57 psi 7. kpsi ns. ( π / 4(.55 s F 5.(( δ.7 in ns. E s ( π / 4(.55 ( 4-9 (a Bot stress σ.75( kpsi ns. b Tota bot force π Fb σ b b (48.8 ( kips 4 Cyinder stress Fb 57.4 σ c.9 kpsi c ( π / 4(5.5 5 ns. (b Force from pressure π D π (5 P p (5 987 bf 9.8 kip 4 4 ΣF x P b + P c 9.8 ( Since δ δ, c b P P c b ( / 4(5.5 5 E ( / 4(.5 π P c.5 P b ( π E 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 resuts of (a above, the tota bot and cyinder stresses are.8 σ b kpsi ns. ( π / 4(.5 Shigey s MED, th edition Chapter 4 Soutions, Page /8

64 7.8 σ c.9 +. kpsi ns. ( π / 4( T c + T s T ( c s ( JG T T 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 s c The percentage of the tota torque carried by the she is ( JG s ( JG + ( JG % Torque ns. s 4-9 R O + R B W ( δ O δ B F F E E O B 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. F 4(4 δ. mm ns. E O ((7.7( 4-9 See figure in Prob. 4-9 soution. Procedure :. Let R B be the redundant reaction. Shigey s MED, th edition Chapter 4 Soutions, Page 4/8

65 . Statics. R O + R B 4 N R O 4 R B (. Defection of point B. δ B ( ( R 4( 4 RB B + ( E E 4. From Eq. (, E cances and R B N ns. and from Eq. (, R O 4 4 N ns. F 4(4 δ. mm ns. E O ((7.7( 4-94 (a Without the right-hand wa, the defection of point C woud be δ C ( π F + E / / ( π. in >.5 in Hits wa ns. (b Let R C be the reaction of the wa at C acting to the eft (. Thus, the defection of point C is now ( π 5 R 8 C R C 5 δc + / / ( π 4RC π ( or,. 4.9 R.5 R 5 bf.5 kip ns. C Σ F x 5 + R 5 R 947 bf, R.95 kip ns. Defection. B is 947 bf in tension. Thus ( 8 947( 8 E ( π R δ B δ B 5. ( in ns. B / Since O B, C TO (4 TB ( TO T JG JG B ( Shigey s MED, th edition Chapter 4 Soutions, Page 5/8

66 Statics. T O + T B ( Substitute Eq. ( into Eq. (, 5 TB + TB TB TB 8 bf in ns. From Eq. ( TO TB 8 bf in ns. TL 8( 8.9 ns. 4 JG π /.5.5 π B τ T τ 489 psi 4.89 kpsi ns. π d π (.5 max O 8 τ B psi. kpsi ns. π ( Since O B, TO (4 (.9 ( /.5 TB T 4 /.75 4 O TB G ( π G ( π Statics. T O + T B ( Substitute Eq. ( into Eq. (,.9T + T.9T T 54. bf in ns. B B B B T.9T bf in ns. From Eq. ( O B ( π /.75.5 π ns τ (.5 max O.48. T 45.7 τ 8 psi.8 kpsi ns. π d π 54. τ B 8 psi.8 kpsi ns. π (.75 Shigey s MED, th edition Chapter 4 Soutions, Page /8