ME3180_HW3_Solutions_Spring2012

ME3180_HW3_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 #3 Assigned on Jan. 27, 2012 Due in class on Feb. 2, 2012 Problem 1 Using both the Brittle Coulomb Mohr and Modified Mohr theories, determine the factors of safety for a bar of ASTM 30 cast iron for the following plane stress states: a. 6x =15 kpsi, 6y =10 kpsi b. ox = 15 kpsi, O'y = -50 kpsi c. ox =15 kpsi, txy = -10 kpsi d. e. ox = -10 kpsi, 6y — -25 kpsi, txy = ~10 kpsi ox = —35 kpsi, 0",, =13 kpsi, rxy = -10 kpsi Problem 2 What is the static stress concentration factor for a round shaft that changes from a diameter of 1.2 inches to a diameter of 1.0 inches with a transition radius of 0.1 inches if the loading is: 0.] in. Mag/us a) Tension? b) Torsion? 0) Bending? . O in. Problem 3 / in I A 1.875 inch diameter shaft machined from AISI 1040 CD steel supports a bending load during operation at 1750 rpm. If the reliability must be 90%, What is the endurance limit if the operating temperature is: a. Room temperature? b. 500 °F? Problem 4 Compute the factor of safety based upon the distortion energy theory for stress elements at locations “A” and “B” of the cantilevered bar shown in the figure. The bar is made of AISI 1006 CD steel and islanded'mtension (8.0 kN), torsion (30 Nm), and bending (550 N) as shown in the figur/ec' Y Axum ’50 caer irow “’9 ’fllfi,‘ 353%»; (Tame/“‘0 Sun: 106‘ #933 Qa‘mfim 67': [0 16PM x 6 aé-KZ ’2: 1 (£75: +( x7) [SHO + V5402 2 5"“ < L >' ( L ‘L'O Of~/5'L‘>$I *7 pig/254320 43:“)th -323-3‘h‘°$'~20¥ my ISI¢9$1~ ' ’( 5p. [SI/493‘. J) Or; ISLfsi DRgEMfioedfim 4': "mk'osl' {7,3 0 Since {KY 30/ > t ‘E ‘ 01/39?" ar'o/ 1:103! M g3z6; 89.4 K ’ 2 f --> .5. __ g 2 .L (A O 2 8 SM: Sun, h .L ; @i ._ (—Skas'B : 7:179: 0,?420 V‘ 3‘ 5‘ loci lopsi n: [00(0 w. cg: oz 63 (SwSuQ a, {2, L M Sucgut guc M 9:8. 7| 01‘ (ma—ayl’fl _’ (any a i 0033(3) 73‘1- V‘ firlgLPSi “(xy- ‘Iokfsl IS.z _ 7‘ 4,5: i +< /O> oat/lg = 75 i [215' (C—psi oz- 20 (337%! (“z 20 hPs.’ ~"5Lfsl ézokfsfi 6‘ 63': “5‘ ((103! M d BC : «2:. 5)‘ 42020? 4) m 51% ac, m “I m: 221;} r H5 BI [03 0:93 5420263 F S£_ ENC—psi [‘55- fi 4‘ 07‘ .. &\ 4: —-/o (V191?) i (—Io—('-Z§))2+(~/D>z Z l ’50,: '70 W5] ($45: 475 i ’25 "P35 6% : -5‘.0 ’3‘”; a: map} 6;: «30,0 bps; 67:: ~50 [99; g: ‘3543 has; $92 ‘24? o? a m: ‘2‘“: 4mm 0? _30 m: 3,405 “1’13 02 (£265 a VP- :m : 3,93 (g z I; (OPS, 5:191:91 4: '37 LPSI :12?“ 3~ "703' szo?6é J) :(%_ {ESL—Ehvflyt qt 94; 3i (:3? Mg in: L2‘ A202 {5 CS 4 30’ -1 I z m- “ ’9 g av‘ (51(ch ‘32:) H ((«o«,.__-m<w> .. z W ((61300 10‘! Piaf mt are, wes 6m fle/Kff‘dje- ME3180 HW3 Problem1 50 -100 -150 -150 -100 -50 0 50 Ga (kpsi) Se: Ltkbkck&kfik€ Se! : t hf“)! '2% 2m 1’ l ‘ _ . Se 2 (39 HM - L” '7 "P" H3: I70 nu A043) k; 2% ($35 0’2”: 0 $32 ( 9' kd= \ (WU/w he: 05m (70% mama/)3 [qt 1 (miseeaamms ewCFed‘s/ 4‘35qu :A Se = (OfBLXGfSZZXOQ )(o.%fi?)(\\(t}25 lfiosi) 5) T: 500°}: Cavmi Gar T: Sow: g,er 1%: M15“ (72% (H) (mks e11”. 944, I461=O,77(o> SLH7‘560”F : .ILA SthzT : 036,5 (3'3 L151) = 34.10 {ifs} Se, : 'éocdsut‘mB = "icé‘flo law” 3 42 3 1905' lam (be/41:33) 14d: | (CorreaLeJ {30V T: 950? agave.) ke: om} (70% mow/#5) : s (misc. eCCed‘s/ dsfumu’ 2 '3 PWm‘f AIS]: Iaoco (,1) stag) / P Lu k9 \ § : 230 M a = 53 “L Pm} 4’ $397330 M‘P4 : LN hos} of: 12+Mfc 20%Woniu23u. '\ W (mo (K @fiwsiennj ’5 level §D,+&e 1&6??? 6‘: 9492(7) is 23 laan .9” +th $77855 Mala/7w; a?” 24W B, ...
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ME3180_HW3_Solutions_Spring2012 - Georgia Institute of...

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