HW10 - 3.1 Determine the torque T which causes a maximum...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

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

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 3.1 Determine the torque T which causes a maximum shearin stress of 70 MPa in B M .1 8 PRO LE 3 the steel cylindrical shafl shown. 18 mm T SOLUTION K c, H [:1 n‘ 21;: - J T = g—c‘t = 3(O.0l8)3(70¥10‘) = 3.2 Determine the maximum shearing stress caused by a torque of magnitude T= 800 N-m SOLUTION 73..“ 2 3; J: 4C... ,c. _ 2T cameo» w _ _ ___.___._______.._._ we! PROBLEM 3.3 3.3 Knowing that the intemal diameter of the hollow shaft shown is d = 0.9 in, determine the maximum shearing stress caused by a torque of magnitude T= 9 kip-in. SOLUTION CZ : $.01: Aline) = 0.8 in. c--- 0.8 in. C. = id: : l"! J = yet"- CM: §(o.2’- o.?s"‘)= 0.5770 in“ 2;“! :: 2:;— _ (9 0.8) __ - J. — 0.5770 - 12.4416: ‘ PROBLEM 3.4 3.4 Knowing that d = 1.2 in., determine the torque T which causes a maximum shearing stress of 7.5 ksi in the hollow shaft shown. SOLUTION c2 : $442 =(—;—)(n.c3= 0.2 ‘m c = 0.2 s». c.=4;J, = (H02): 0.6 ;n J : %(c;’— cf‘) = amt- as“) = 0.5‘398 5:." :wflflfl -_— 4‘12 MIN" ‘ 0.8 3.11 Knowing that each portion of the shafl AD consists of a solid circular rod, determine (a) the portion of the shaft in which the maximum shearing stress occurs, (b) the magnitude of that stress. PROBLEM 3.11 30“ ll) -in. SOLUTION Shot-H AB 3 T = I400 Jb- in C: '3 0.30 in 1’th I; — \ lZOl) ll) - in 400 Hi ~ in‘ dCD = 0.9 in. 0 J‘ ' 1T0! dBC=O.75in. - - . a” ‘ “n (0.3033 ' 0'43, P“ SLQ‘H BC: T =—400 + I200 = 800 lbfim : 11:; = .21: _ (2)(2oo2 - c = {gal = 0.3751.n zw J was WHO‘S”? ‘ 9658’”. SM? CD: T= —Hoo+:200 +500 = I300 125.5" __ t - ~TZ:<:.__2_I_2130<>>- ‘ c -‘fiel 0.45m Ln», J. — TIC; -$TTJ<O.~q-$—ig - 90 32 P51 wost : (a) 51034 BC (in) ass us; 4 3. 12 Knowing that a 0.30-in.-diameter hole has been drilled through each portion of shafi AD, determine (a) the portion of the shafi in which the maximum shearing stress occurs, (b) the magnitude of that stress. PROBLEM 3.12 500 ll: - in. SOLUTION Hope: C. =7‘cl, = OJS in Sha‘F‘l’ A83 T: 900 Ila-O1 02: J-dz = 0.30 n, -'O.75in. LT: ¥<Czq— C". ) : ¥(O.3O4- 1200117 ~ in.\ dCD = 0.9 in. All) ll» - in. \ T = 0.0:)928 in”) 1-“ = C2 : ('400KO.30 _ . r J “~00”ng - '0600 PS: C Sha‘F‘i‘ 8C: T = -‘l‘00+1200 = 800 “win CZZ-édz: 0.375 in. J : Tflczl—cfl): 130.375"2 0.15“) : 0.030262 m 12”” - ‘T'cz - (800)(o.375)_ _ CH” PSI. ' ‘37" ’ 0.0309048 ' Sha'H' (:13: T = -400 +1200 +500 = 1300 )L-iu cl: i-alf 0.45 in J : mag—r") = 3; (0-45” - 0.15”) = 0.063617 .‘n’ thay ‘ TC; = = Pal. 0.0636 l7 AhSk/ews: (CL) sha‘F‘l‘ AB (b) IO. 06 VS? '4 PROBLEM 3.15 3.15 The allowable stress is 15 ksi in the 1.5-in.—diameter rod AB and 8 ksi in the 1.8-in.-diameter rod BC. Neglecting the effect of stress concentrations, determine the largest torque that may be applied at A. SOLUTION - TC- - " _ rm- ——T——, J—Iile T-ercsz’m v ) squ- AB: 21..., 2 1516; c, =Jz-d = 0.75;». T: 1g-(o.7.s)”(ls) = 9.9% kcpjih Squ BC: tn, = 23 ksi c. = lid = 030;" T = 1{(ofio)3(8)= cue. mm. T16 «Apow-‘MC +0"; U: is He end/er Vapue T = 9.16 HIP-in ‘ 3.16 The allowable stress is 15 ksi in the steel rod AB and 8 ksi in the brass rod PROBLEM 3.16 BC. Knowing that a torque T = 10 kip-in. is applied at A, determine the required diameter of (a) rod AB, (b) for BC. SOLUTION 2 T TF—Z’Tv Z’M,= l5 ks.“ [in d =2c 9 1.503 in. T; /o (0")th 11..., = 8 ksi 0. 77577 in Cl" 2C. = L853 in. ...
View Full Document

{[ snackBarMessage ]}

Page1 / 3

HW10 - 3.1 Determine the torque T which causes a maximum...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online