ME260_Final_exam_2007 - Instructor: THE UNIVERSITY OF...

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Unformatted text preview: Instructor: THE UNIVERSITY OF BRITISH COLUMBIA Department of Mechanical Engineering Sessional Examination —— December 7, 2007 V Mech 260 - Introduction to Mechanics of Materials - Section 10] J. Mikkelsen Time: 2.5 hours~ One sheet (double-sided) of hand written notes allowed. ’ Draw a box around your answers and clearly indicate units. All questions carry marks as indicated. Show all work, including any equations used. 1. (10 points total) a) In plane stress there are two principal stresses and in plane strain there are three principal stresses. True or False? Explain. b) The best place to drill a hole in a beam is ‘ through the centroid. True or False? Explain c) If you know the geometry of the cross section and the maximum bending stress on a cross section, then the bending normal stress at any point on the cross section can be found. True or False? Explain. (20 points) The engine of a helicopter delivers 700 horsepower to the rotor shaft, AB, at a rotational speed of 1400 rpm. The shaft is 2 feet long with an allowable shear stress of Ianow = 8 ksi. In addition, the angle of twist for the shaft must not exceed 10°.(l hp = 550 ft-lb/sec) (a) (5 pts) Determine the torque delivered to the shaft. (b) (15 pts) Determine the diameter of the shaft to'the nearest 1/8 inch for this application. c) Associate the stress cube with the appropriate Mohr circle given: Circle C ‘ Circle D Circle E Circle F 60 MPa 30 MPa 60 MP3 30 MPa 20 MPa Mech 260 (sections 101) - December 2007 examination Page 1 of 4 3. (25 pts) The assembly consists of a steel rod (A), a bronze bar (B), and a rigid bearing plate (C) fastened to the steel rod. A 0.015inch gap separates the plate and bronze bar BEFORE the assembly is loaded. ' Ame; = 2.5m? Abm = 3.75 inz. Esteel = x106 pSi, astee] = 6.6 x10-6/OF. Ebronze = x106 pSi, “bronze = "10'6/0F. (a) (5 pts) Determine the load required to close the gap. (b) (10 pts) Determine the temperature required to close the gap. (0) (10 pts) If a load of? = 5000 lbs is used and the temperature is raised 60 °F, Determine the final position of the steel rod and the stress in the steel rod. 4. (20 pts) Beam AC rests on a cantilever beam DE as shown. Two steel W410 * 39 Wide Flange Section are used for each beam (Esteel = 2‘00GPa). a) (5 pts) The reactions at A, C, and E. b) (10 pts) The deflection of point B. c) (5 pts) The deflection of point D. Mech 260 (sections 101) - December 2007 examination Page 2 of 4 5. (25 pts.) The solid circular shaft has a yield strength, cyield = 64 ksi. For R = 8.5 kip, determine the following: a) (5 pts.) Determine the reactions at Point A. b) (10 pts.) The maximum normal stress and maximum shear stress at point A. c) (10 pts) Determine if the shaft will fail according to the i) maximum shear stress theory and ii) maximum distortion energy method. Recall that for a circle, I x = i717”4 2?: . Web _M.~Flél§: ...... . i . thickness width thickness §.W_.__._.:;:K axis .............................................................. .. 2w l b! tr 3 I E S 1 ' 7" ~ - l mm l mm mm 3 10" mm“ £103 mm3 . 171800le 12.70 324.0 19.0 a; 1 200 4220 W610 x 140 179001 ‘617 13.10 i 230.0 ’ 22.2 1 1120 1' .3630 W610 x 125 15 .9001 612 11.90 § 229.0 ‘ 19.6 ‘ 985 2 3220 1 W610 x 113 144001 608 11.20 1 228.0 1 17.3 875 1 2880 W610 x 101 12 900} 603 10.50 1 223.0 14.9 764 2530 W610 >< 92 11800; 603 10.90 1 179.0 315.0 646 2140 W610 x 32 10500 599 10.00 E 178.0 12.3 ' 560 1570 1 W460 x 97 12300 466 11.40 1 193.0 19.0 445 1910 W460 x 89 911400; 463 10.50 192.0 ' 17.7 410 1770 5 W460 x 82 10400; 460 9.91 5 191.0 16.0 370 1610 1 W460 x 74 9460:i 457 9.02 § 190.0 14.5 333 - 1460 W460 x 68 8730 1 459 9.14 E 1540 15.4 297 1290 1 W460 x 60 7590:. 455 8.00. 1 163.0 13.3 255 1 120 1 W460 >< 52 6640i «:50 7.62 § 152.0 10.8 21,2 942 W410>< 85 10800 417 i' 10.90 181.0 118.2 315 1510 E W410 x 74 95101 413 9.65 180.0 16.0 275 1330 W410 x 67 8560 410 8.76 1 179.0 14.4 245 1200 W410 >< 53 (1820 403 7.49 3 177.0 10.9 186 923 ‘1 W410 x 46 5890 t 40:; 6.99 1 140.0 11.2 156 774 W410 x 39 49ml 399' 6.35 1 140.0 8.8 126 632 1 1 1 W360 >< 79 101.00] 354 9.40 1 205.0 16.8 227 .1280 1 W360 x 64 3150‘ 347 7.75 1 203.0 13.5 179 1030 y 1 £33221??? 23.231 33? 3'31 {33:3 it}. if: 332" , 9.63 l _ 2 E W360 x 45 57101 352 6.86 1 1.71.0 9.13 12.1. 683 8.1.6 i 95.4 37.3 W360 >< 39 4960 E 353 6.48 l 128.0 10.7 . 102 578 3.75 i 58.6 I 27.5 1W360><33 41901 349 5.84 1 127.0 . 3.5 82.9 475 2.91 1 458 g 26.4 1 1 1 Mech 260 (sections 101) - December 2007 examination Page 3 of 4 ~>Pnb(l. + b) ~ 6E1L 6E“ v = 4mm - 1% ~ n1) Pub(L + a) .r‘=u 615.11, 0 t < a” ‘1 "—“M'” .. x .n [I ' 6E". -MOLZ VEEEI “mm: = v ,wL/g a. 7685/ um“, = "0.006561 I at .\' = 0.45981". _ r~~w.\' _3 _ p 3 ' v—MEIQ ’2Lr2+L) 'v ""“ (m3 — 24L:z + 91.1) 3841:! ' , 0 5 .r s L/z- i 8.1" 24sz + 17sz ~L3) U2 5 x < L E] a! .r = 0.5193L ..PL2 - __ PLJ 9m“ Pm; ~ *g—ET' w L." um —0.00052-—"—~ v=— 3x4 _ ‘2 J. Al” 7L4 36DEIL( ‘ W * ) Elastic Curve _ ~sz 12 = Wm ~ .\') 1 D E 1;":er 1 —w(,l.: 7+4 I 0mm a 5 ' .x, ,..J 1 It»me .4 Wm ' ........ "’m’“ 3841-7 ~ WOL'1 vm'“ : 30131 | i l i ru 1;: I 1 _ *sz ; I W“ ” 8131 "m 4815] PL: _ “ a l- : w—zAEl(nx-3L) L/2<rsL _ . 4 .. . 2‘ Um“ = “L 1; z 7;?! (x2 ~ 4Lx + 61.2) ~wl.3 16:5wa ‘4 14/2) u: s ,\' s I. < (101.3 — 10sz + 5m w R) wwm" IZOEIL Mech 260 (sections 101) - December 2007 exami nation Page 4 of 4 ...
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This note was uploaded on 03/28/2012 for the course MECH 260 taught by Professor Ross during the Spring '12 term at The University of British Columbia.

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ME260_Final_exam_2007 - Instructor: THE UNIVERSITY OF...

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