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Solutions 9 - 9-84 The pedal crank for a bicycle has the...

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Unformatted text preview: 9-84. The pedal crank for a bicycle has the cross section shown. If it is fixed to the gear at B and does not rotate while subjected to a force of 75 lb. determine the principal stresses in the material on the cross section at point C. Internal Forces and Ale-tent: As shown on FBD. S eetien Properties: l , , I= Eto.3)(0.s ) =0.0|23 n‘ a = i'A' = 0.3(02H03l = 0.0130 in’ V3750 lb Nor-tel Stress: Applying the (hut: formula. I? 3"“ ‘ i'05 0-2 :0. M 400402» ' - in. . . I 0.0l28 0.4 m Shear Stress: Applying the shear ionnuia. -=—--—=35L6 = , S ‘ _. 5-5516“. t‘ I: 00:230..» P" ”3 '6 "" i C unstrnetian of the Circle: in accordance with the sign conventmn. a, =4.6875 tn. 6, = 0. ad I" =0.35l6 ksi. Heme. O’ + 0, 4.687544) _ = — = 2.34315 ks: 2 1 a... 8 ‘lhccootdmates iorrelerenoeponlsa andC 81: A1168”. 0.35M) 02.34375. 0) 'l'herzllnisolthceitcleis n = (4.6875-134375): +0.3sn6: = 2.3670 ksi In . Plane Principal Stress: The coordinates of ports 8 and 0 represent a. and 03. respectively. a, = 2- 34375 + 2.3670 = 4.7: ksi Ans a2 = 2.34375 - 2.3670 = -0.0262 ksi Ans 9-87. The bent rod has a diameter of 15 mm and is subjected to the force of 600 N. Determine the principal stresses and the maximum in-plane shear stress that are developed at point A and point B. Show the results on elements located at these points. Section Properties.- 4 - x(aoovs‘) - $6.25n(10“) m’ 600a) 1-;(09015‘) - 2.4350( to") m‘ Q-Q-O 8 tree 1: a--t— A I 600 30.0(0.W75) ' 56.253004) * 2.4850( 10-9) 0‘ I 3.3953 -90.5414 I -87.14 MP: 0, I 3.39$3+90.$4|4 I 93.94 MP. 1" I f, IOsioaQ IQ I0 Contraction of the Circle: In accordance with the sign convention. a, I-87.l4 We. 6, I0. and t" I0lot 906nm. Home a, +0, 41144-0 a - I _ m 2 2 - «3.57 w; Theamtdinates for reference points A and C are M4114. 0) C l 43.57. 0). The Mitts olthecitcle is R I 87.!4-43.” I 43.57 MP: In-l'lone Principal Stresses: The cootdinateo of poinu B and A representa. and 0,. respectively. 0" .0 All! a, . 47.1 MP: Mu Mai-Inn: ln-Plnne Shear Stress: Reptaented by me coordinates of pow E on the circle r ... Ill-43.6w: Ans Orientation of the Plane for Maxi-inn. ln-Plnne Shear Street: From the dtde 20, I90' 0, I £5.0' (Conntereloekwiu) Ans Construction of the Circle: In accordance with the sign convenuon. a. I 93.94 MPa. 6, I 0. and r,, I Oiot point 8. Hence. a I a, +0, . 93.94“) "‘ 2 2 I 46.97 MP: 600N 600N 1mm 5‘ M" 6 44.5% +3-th 47.“. me 410». 450’ 45.9% 410% am} A Perm ln-Plone Principal Stresses: The coordimcs of poms A and 8 represent a. and 0:. respectively. a. I 93.9 MP; Am 0: 3 0 WI AM Moxitnmn ln-Plone Shear Stress: Represented by point E on the circle. f .“ ska-47.0w: All! ”-88. Draw the three Mohr‘s circles that describe each of the following states of stress. 6 ksi lOOpsi so MPa (b) 4? man) 1 TM) (.5) a) a... n6ksi a... so... -0 b) awasowh emu-0 ems-40w: c) a.“ = 6!!) psi a,“ I Mp3: a... 8 IN psi a... - 6 ksi a... = 50 MP: 0.“ = 600 psi an: ' can - 0 an: 8 0 a... = -40 MPa out ' MP“ on. = 'w PSI (e) CW) 12-6. Determine the equations of the elastic curve for the beam using the x. and x3 coordinates Specify the beam’s maximum deflection. E] is constant. Support Reactions and Blade Corn: As shown on FED“). Mellon: Function: As shown on F300) aid (c) . Slope all Elam: Curve: 4% a? ”((3) P Pablo.) I -§x. . (‘0, P a? ---2’" do P adj-74w. m P 1 Eu. - -3“ +C,x, +6, [2] F“M(x’) 3P3, -3—':L , J’v, 3PL ”T; ”’3 ’7 du P 3PL 53-:- - 5:: -Tx, +c, [3) H u, I: gt: - $13.} Qt, +C‘ [4] Boundary Condition: 0' .0 "1' .0. PM EQJZI. C3 . 0 ”'30‘3"L melZ] PL’ Pl} 08 -1_2 +CIL C. 3 '17 0,3031’.L me [4], PL’ 3PL’ 03 T-T+C,L+C‘ 7PL’ 0=-—+C,L+C. [5] Douala) Conditions: v. =0 at. 80. From Eq.[2]. C; 8 0 PL’ PL’ os-l—2+C.L c. . T2- 0=-—+C,L+C. [5] 1." Elastic CU"... Substitute the m 0! C. . C3.C, .IndC. Continuity Coalition: in: Eqs.[2] aid [4] respectively do. 40, A! x I: I L. — g _. . P: ' , 4.. 4., memm- v. =—m;,(-x:+v) m _PL= P13 P13 3m m} u, . 9,1. , ¢ = fl -2 +1} g ”32"" T‘H'T‘T *C’ C"? ‘7‘ ‘2” 3 ‘7 PL, 0 g P 2‘, 2 3 , From Eq. [5]. C‘ ._T 3 ~IZE( 3 - 91:, +101. 33 -31. ) All: nu Slopc: Substime the value etc. into law 1. ”c = ”3 I.,.;L = L 2(3L)’ 9L 31..)24-10L2 3 L ’ ‘11. P (13-341) 12a 2 (2 (i )‘3” 4:, 12a ' pp 40 P 1, aka I 2 3 — .0. _ - — 4:. 12a“ 3") " ' J3 PL’ Hence! v..‘ 3 0c . — A” SE! *12—16. A torque wrench is used to tighten the nut on a bolt. if the dial indicates that a torque of 60 lb ' ft is applied when the bolt is fully tightened. determine the force P acting at the handle and the distance 3 the needle moves along the scale. Assume only the portion A8 of the beam distorts. The cross section is square having dimensions of 0.5 in. by 0.5 in. E = 29(10‘) ksi. C©_- A lain. 72o 1m V‘" C I) Moosmx-Zw K 400% Equation: of Equilibrium: Fm FUD“). (am-o. mama-o P-40.0Ib Ans «(Tug-0. A,-d0.0-0 5-40.05 Moment Function: N shown on FED“). Slope and Elastic Curve: a—-20.ox'-1mx+c. m a» -6.661x’-360x'+c.x +c, [2] Boundary Conditions: g 8 0 at x a Oandu U 0 a x I 0. From Eq.[l] 0:0-0+C, C. 80 From aq.(2| Ono—mow, c, so The Elastic Curve: Substitute the vain: of C. and C, in» Eq.[2|. u- -£'—I(6.667x’-360:’) tll Atxle‘uL. v=-:. FromEq.[l]. t (29)( IO‘)( ,5) (0.5)(05’) [6.667(12’) -360( t2’)] 3 a 0.267 it. All: *12—28. Determine the elastic curve for the cantilevered beam using the x coordinate. Also determine the maximum slope and maximum deflection. E] is constant. 4'0 d’o__3£ .mfiE--§§L+q (0 fl m 310- _wr’ 9-,: tau ” ’ ‘) o —w. * mm. 5L " l)MflF' d—B-0atx-L ”WW 4: "3 3mm % I On: I- L Froan.(l). 0- --;f-L(L‘)+c.: C. - 3%?- 0-308an Fmfiqom. _ We s We!) , 0 - _120L“' )+ TA (0+ Ca . “can”: FmBq.(l). 4” _ V0 _ 4 a a 24311.‘ ’ H" a 0... . .2...“ - %-;I Nautical-vs: Emma). W0 4 s nosu.‘ ’ ’ o (2) 30 All: All: All ...
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