ME3180_HW2_Solutions_Spring2012

# ME3180_HW2_Solutions_Spring2012 - Georgia Institute of...

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Unformatted text preview: Georgia Institute of Technology ME 3180: Machine Design Spring 2012: Sections A, RCC, RPK, and RPY Homework #2 Assigned on Jan. 20, 2012 Due in class on Jan. 26, 2012 Problem 1 Using both the distortion energy and maximum shear stress theories, determine the factors of safety for a bar of AISI 1030 hot-rolled steel for the following plane stress states: a. ox = 25 kpsi, 0y = 15 kpsi b. ox =15 kpsi, O'y = -15 kpsi c. 6x = 20 kpsi, rxy = -10 kpsi (1. 6x = -12 kpsi, O'y =15 kpsi, txy = -9 kpsi e. ox = —24 kpsi, 6y = -24 kpsi, txy = -15 kpsi W If a steel shaft is to be designed with a maximum von Mises effective stress of 210 MPa and must transmit a steady torque of 340 N-m, what is the required diameter of the shaft? Problem 3 A countershaﬁ; carrying two V-belt pulleys is shown in the ﬁgure (see next page). Pulley A receives power from a motor through a belt with the belt tensions shown. The power is transmitted through the shaﬁ and delivered to the belt on pulley B. Assume the belt tension on the loose side at B is 15 percent of the tension on the tight side. a. Determine the tensions in the belt on pulley B, assuming the shaft is running at a constant speed. b. Find the magnitudes of the bearing reaction forces, assuming the bearings act as simple supports. c. Draw shear-force and bending moment diagrams for the shaft. If needed, make one set for the horizontal plane and another set for the vertical plane. (1. At the point of maximum bending moment, determine the bending stress and the torsional shear stress. e. At the point of maximum bending moment, determine the principal stresses and the maximum shear stress. Problem 4 The countershaft shown in the ﬁgure (see next page) is to be made from AISI 1035 CD steel. a. Select an appropriate static failure theory and determine the safety factor for the shaft as shown in the ﬁgure. b. Select an appropriate failure theory and determine the minimum shaft diameter to avoid yielding for a design factor of 2. mew l 419; {0&0 HIE Shad (page {st-203 / 3‘": 260 MFA: 3'15 law} 5” : 470 MP4: 68’ W96 4/\ :: ’lelfz‘jl‘ 20:9 in 2 ‘w _ \$ ’— w ' “B _ m ‘DE (Few eve ShessW V ,_- 1 g {/tégz {Riff-#4 +3rcx7> (er/v3 l3) “lb 2 i 2 K7 —- 25H? + 25—! L L a— h» (am a: O/I': LF\$I 5;: [5' Hm (I: IS L135: K: O [4,061 ("K3 - 25") - I25 . TML: K “*7/— \’ L \' k1” LA 4:15 k138i 6': ’/‘D— 12E 6": 29% LF\${—-’l/\: I‘M “12.3. {Malawi 5;: 01:93; 73W: l5kg>si~>n= [.25 {3: ’IS‘ laps: C) ({x": 20 W5] 5,= 0 L993" twat: '/0 [(15): 2‘; fl: Zbglcpsi ——> n= Hz ’12.. 6i: 24:! bps; 03: o \cps'. “(Mg ILL! \cpsa—v m= L33 {3'- '4JLI' laps, dvéﬁ’le/ps‘ 6: [5— Lars: {‘97: ’61 m; f/:2<5.| loPSi ’9 m: '33 V93 'f: Hf—Hzpﬁ 0/25 0 k103i Tm: no.2 lcPSi—7 n: We {a = 4%? léfsi 63 6);: —2L{ l=,>si 6,: ~l‘1‘ lapsi Tact: ”/§ J/x 069:4 7%6 5/)f£Ae/%€€/+ (140% page) bod/6w W‘Q Wﬁﬁe’ms qcmm WC «Li-fr?" page 09 W‘ﬂ W‘ws 71-0 gala: Arr E: 3- E: a- is. «N- :9. mxm wmwNNdH mNnéH- o wmthNH mNh.vH- meNhNH :9. m- as. 3 as. 3- E: msm mesmu mmEm_m NmEm_m HmEmﬁ nmEm_m mmEm_m VHNwHéH vH Nwaé- o VHNVHéN «HE‘Hé- VHNVH .VN we. 3. :9. o as. 8 :8. WE meDNH mmEm_m NmEm_m HmEmﬁ nmEmﬁ mmEmﬁ mm.m~ me. o as. 3- as. 3 39. WE as. o :9. 3 :8. 3 a8. msm an tma H E0305 ‘QSI I «SEE? NESAET m N admsmvzsugﬁ N Swoméwmsmvzamu 8d\$§£x<§u Anorﬁémrmaamam*E-~<wa [O in) — QED “39(28 it’s-l- PC (LID L) :‘D We in Jrﬁe +7.- madam) \$4; (95 lk€>( 125131315” ILA: ( AM ' 130' 213.25 Hop. (Mm: 14‘qu —% dire MA = (mg/L000 .113 .- 29325 mu; MB : (*l7\$,25159)(12ih\ —- «2/3? MM ...
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ME3180_HW2_Solutions_Spring2012 - Georgia Institute of...

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