ENFD_375_Midterm2_Solutions_Blue

ENFD_375_Midterm2_Solutions_Blue - ENFD 375 — Basic...

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Unformatted text preview: ENFD 375 — Basic Strength of Materials NAME: 1’5 Lu 6' 1431-2 BASIC STRENGTH OF MATERIALS ENFD 375 — WINTER 2008 MIDTERM #2 ' DR. MILLER FRIDAY, FEBRUARY 22, 2008; 4-5 PM NO BOOKS OR NOTES NOTE: There are multiple versions of this exam, all using the same figures. The numerical values are different. The various versions are randomly distributed to the class. Directions: 1) Put your name on each page of the exam. Exams without names receive “0”. Put youimame on each page in case the pages are accidently separated. 2) Do the problems on the test sheet. If you need additional sheets, put “continued” at the end of the problem sheet and staple the additional sheet to the end of the exam. Additional sheets MUST have your name and problem number. ONE PROBLEM per additional page. ‘ 3) Show ALL calculations. No calculation — no credit. 4) All answers must have proper value, units and sense. Sign (+ or -) is not enough unless the sign isproperly defined in the problem. 5) Be neat. Messy, unreadable or indecipherable calculations receive a grade of “0”. P UMAL =§=E£ 5 7% 3:— L V P Tavg = 21" 0' BEARING = A BEARING V = SHEAR FORCE FS ___ Gullimate 0' applied T 1' = J— ENFD 375 — Basic Strength of Materials NAME: 6 but: 1461 Problem 1 (15 points): For the hollow shaft shown, the inside diameter, di = 3 inches; the outside diameter, (10 = 5 inches. Determine the maximum shearing stress, rm“, and the shear stress on the inside of the S ._ hole. FIXED 50k-in :- ‘nuk is + - Tomcat? Av" I: 1 (Zlg’d)q’//I §r~ :- If 2. Z 5DW(2"") '3 237%“! $553“ a .__———-——--—'."7 mm! $3. in 7" CEL‘vévnw émw 3 2.3“¢SI< LG. a/Iyo psi 2.51:0) s.u.wwmmmmmwn-«-mnmmumww ENFD 375 —— Basic Strength of Materials NAME: 3L4; (5” l5 9: Problem 2 (35 points): The shaft shown has a diameter of 3 inches (J = 8 in4). It is fixed at A but free at C. Assume T is always in the direction shown. Assume + is defined by the internal torque sign convention. Gm! = 11200 ksi and Gamma.“ = 4000 ksi. a) If T = 110 k-in in the direction showngwhat is :1)? / b) Assume the aluminum has a maximum allowable shearing stress of raluminummx = +12 ksi. What is the maximum allowable value of T? This question is independent of part a. ' c) Assume the steel has a measured maximum shear stress of rsteelmx = +20 ksi. What is the value of T? This question is independent of parts a and b. - d) If the angle of twist in the direction of T is limited to +0.0015 radians, what is the maximum value of T for this case? This question is independent of parts a, b and c. "V1094"| ‘ a". ’4; p ¢-2n. - "‘l°""(s ) +M) ' 33' 'HZuoaou(€ab") "loavlcn(o,:,‘- .¢’.olw G 1:67?) xvé‘vfi‘” . M“; 60 ’ a u p 1- 1.» 5M6 w b) Fst Mn mm“ 5 a E , T(I.s.'.) T .1 (+3 1 ' M“ T 3,9? A 5y $14.! Milli: 7—: L-fe H) S0,,ztjzau 1.. 1Z3°°p F) 1;; I — Tc. (TM) C) “$0 5"“ f" 2" = ‘w'f " 9.2.7 T= 907 ""”’ A) THE“ W~ 77,00 0207:2441 0:077!» I . . T.300""”=T A ‘ I F60. ‘ fl A league“ Guam: : T—3oo"':‘ I T MW, + 7: is ‘5‘; w.- (+) ¢: 2 3% J“:- ¢o~stf= 8w" J’¢= 2% 8.."( ems”)? U-—-—-———->'3°°W)(s“ A+ Tm“) 7 L llZoo HI Vow” ‘ 0.2-1» 3. ‘ .DOOV‘IGT- 0.133%; 1-.00IS T" .l‘15'93 = .001?” Tr.- 7 SL""’ ,2 Digger»; ‘ SHowu 0107764 2 :- Coustom Jusr THE 300 “'W WWI“; law ‘ p3“, w.» ¢ = 23, {non-“(533) ~ . GI I/ufi"(8,;‘l) ’ _’ 0’67 m T 35 \ . L.) w 5"» Cant/- T «~de WI<T+ ‘waKU (3.9”) “200 (8") + alum. .om’” - 'T :0 (“Bl .ooow37'=-0’922 . K) 7’:- 79. 8 ""“' DIFFMMK I S (awe OFF ENFD 375 — Basic Strength of Materials NAME: gout: Problem 3 (20 points): The shaft shown is 4 inches in diameter. It is made of aluminum (deinum = 4000 ksi). Both ends (A and D) are fixed against rotation. Determine TA and TD. Are the directions correct? Specify the directions as clockwise or counterclockwise as if you are looking down the z axis (As shown, TA and T9 are both shown CCW). ENFD 375 — Basic Strength of Materials NAME: 'SW ts" ’4 ‘7 Few 3 GOA/7’: (. N0"); musi‘ T‘ng?" - PL}; 1’ +2350 0 CsJ" “,1, z Tom-'0: 63' 4f . (1—1 T915Zzza7p . 1:1 OILW‘IIW SHAW” 5) gram; HID-Vac +22L7 7:) =17,31W~ 7: ~. z73"“." “‘4 n- ,2. = 2,2. 7 Cab) TD 2 g 3 3 3 3 3 3 3 3 a 3 3 3 mmm“an”mmy»wummmmumnumumwyuewmmmmmmnmm mummmmm “mum” mummmmmm 'ENFD 375 — Basic Strength of Materials NAME: 61,03” Ié 0/ m. Wm... WM... .mww» ....w-.......mmm.‘.mwumyu.m~~—-m»~mmn~m WHOLE S444?!" ENFD 375 — Basic Strength of Materials NAME: ELUE I461 Problem 4 (30 points): A bridge is made of concrete girders (E = 4500 ksi). The coefficient of thermal expansion is 6 x 10'6 /°F. The girder area‘is 785 inz. The girder length, L = 70 feet. To allow for thermal expansion, the engineer left a 1/8 inc gap on each end. Assume the girder is free to expand and contract and it expands or contracts e ually in both directions. NOTE: Parts a, b and c can be solved independently. a) What change in temperature will cause the gaps to close? . b) Assume the girder has expanded enough to just close the gaps. If the temperature ' increases another 12°F, what is the FORCE in the girder? c) Ohio Department of Transportation inspectors call the engineer and tell her the gap is closed and there is some cracking at the end of the bridge. The cracking occurred when the temperature of the girders was measured at 110°F. The engineer checked the records and found that the contractor had built the bridge in late October. The engineer estimates the girder temperature during construction was 40° F. What is the STRESS in the girder due to this temperature change? T118 inch gap 1/8 inch gapT Girder Length = L TVS inch gap Detail ENFD 375 — Basic Strength of Materials NAME: 6 L416 '4’» 9 Z 910$ 7 /' $111 0 A/ s 2 J WL/Z F é X/p'é/‘F ( /Z 7 l w 2, o t ’ (905W fitoupf=éwo 75(70PrX/Z'flv>(/Z‘F) ‘0 - 7-1.; 7900002” m 0405’” Jr ’ 7:? 78550500”) P212573“ 15W; 3””1’0 . 0 . C) AT=7O F , ,fi _ - Iw’. ‘ ‘ :. 5 3 * (g f “LN'séx/o V’FK7O'FJOZ fl‘r'XM F) ’g M "I:' ;,/03/ a? 7 L W'tijfleLc : 7E5: ) xi 6;“; (E ,; 70"”(IZ’W' :7 0’; «3317" " ’ ’ “ “0’03 3: 0’ VS—Dafiffi 3 ,SS’I W W ...
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ENFD_375_Midterm2_Solutions_Blue - ENFD 375 — Basic...

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