HW10

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

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

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