{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Solutions 9

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

This preview shows pages 1–9. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern