ENFD_375_Midterm2_Solutions_Red

ENFD_375_Midterm2_Solutions_Red - a a a a a a a 3 a a a a 3...

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Unformatted text preview: a a a a a a a 3 a a a a 3 3 a a 3 3 a a a a a a a a a a a NAME: [2'50 ENFD 375 — Basic Strength Of Materials BASIC STRENGTH OF MATERIALS V ENFD/375 - WINTER 2008 ' DR. MILLER MIDTERM #2 . N0 BOOKS 0R NOTES FRIDAY, FEBRUARY 22, 2008; 4-5 PM 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 your name 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 is properly defined in the problem. 5) Be neat. Messy, unreadable or indeeipherable calculations receive a grade of “0”. P a = — = E8 7 _ AXIAL A 6 8 = - L 1' - K 0' - —-———P M A BWNG ABEARING V = SHEAR FORCE FS = aultimate dapplied , = L9 J J = gr“ solid circular shaft TL ¢ " JG AL = aLAT ENFD 375 — Basic Strength of Materials NAME: my £91 Problem 1 (15 points): For the hollow shafi shown, the inside diameter, cl, = 2 inches; the outside diameter, do = 3 inches. Determine the maximum shearing stress, 1am, and the shear stress on the inside of the hole. " ‘ FIXED '45k-in é-Psy ’Sléu Cadvewmawl WIS 13Lo7 C‘) WI PUT: .. “5’ r ‘ ’IO'ELSIL')=_707 t—M— [at/WA» élwwflf ' I'C; gIC” WM Cad Va a a a 3 3 a a 3 3 a a a a a a 3 a 3 3 a a 3 a 3 3 3 3 a 3 3 3 ENFD 375 — Basic Strength of Materials NAME: I269 ' 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 = 100 k-in in the direction shown, what is (1)? b) Assume the aluminum has a maximum allowable shearing stress of mummmmx = +10 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 15mm“ = +16 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 +0.002 radians, what is the maximum value of T for this case? This question is independent of parts a, b and c. , ‘ . T +|oo“"; 0') \W‘ud ' Tongue-v. __ . ’zooh.‘d‘ unfit-m k____§L—_.l_____£.'_"_.__| . T S Iv ‘1‘: It. fl' _. _( ‘2 + ’90 sh ) =+’Oo7é /’ 2°” “43"” How”5’(6nfl tun fl: .ooum ,A‘ main“. oF"T". Vat/530 f A Tc. ‘3) FED A'L T; .11. g? /OL£I : . 'T mm T v Im d; BY 51w couv, 7": “lumped: 52,3; ’07» . Ivsgtflfioal WM C) FI‘SD $r‘nL: T-‘SW‘ 2-,. 7.2- fl . I, T. p”; . ) x“ T—‘SOO (IISI-v ‘ T—zoo f/é 'I- V I; i $15.: A I T: 3C3; guy gag IF WM“; '3 — ENFD 375 — Basic Strength of Materials NAME: 2&2 #6: z naval-J6 le T + I E ¢ 3 2 a: 10/19.; -: ,oaow/J— 0.13373 4— .oOIST‘ 1/50 = ,Oblfi'fiar 7’: 77'4"“) ,ZJ DI/LeT/flo") 3%” w F80 loo/WI" [‘DLQUC I £00 “ I, 200551-(8'7'1) T094906 “('3 . 775A») 4- : . 0002737— [Vlw JObr. ¢T : M 4.09;) (2)" l l ’4 cousma, "T' HUN“ (8’ n1” NJ F80 ‘ ".01197‘1+.000L‘/3T'= .oaz, . T L5 (4) ’r r T: 77|ffw I V I H3 'DIl/ELTM‘J a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a i 3 a a a a a a a a ENFD 375 — Basic Strength of Materials NAME: [6619 K51: Problem 3 (20 points): The shaft shown is 4 inches in diameter. It is made of steel (Gsm1= 11200 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 3 and T9 are both shown CCW). 'm _ \ v7 ENFD 375 — Basic Strength of Materials NAME: M— "+6 [grain/A1.— Sm}; cyan/wanna» gr war‘ (15499 53.91.. agorgflt— (rut/awgtwm! 621—: 0 = ‘73.?L"V+IS'DKq~«'/ao£'w+ T4 x"; » "y ;, ¢=I7zz3 ,u aim-“am '17 ‘3 I77I3zv'd 7D FLA~¢ To" 73'dizpm C49) max, Pan-«4r ow ENFD 375 — Basic Strength of Materials NAME: Keg ‘ Pia/5 3 mmeer ’1 £61, Law’s C l4 " (\\\’T" ’ fa ’ L ., . It- \\ 7fl=/7713 [-9 y/Le’Lfizu~ V ‘ 5 #w A3 G ZF=o £773,700 WW3]. 7; ~=‘73,7""” ,.. .nuwmuwww...“ ...._.. Wmm.a.~.ww.- a ’ . ENFD 375 — Basic Strength of Materials NAME: Kraft 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 735 inz. The girder length, L = 80 feet. To allow for thermal expansion, the engineer left a 1/8 inch gap on each end. Assume the girder is free to expand and contract and it expands or contracts equally 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 10°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 35° F. What is the STRESS in the girder due to this Emperature change? T118 inch gap 118 inch gap.".— Girder Length = L arr—“8 inch gap Detail ' ‘ ENFD 375 - Basic Strength of Materials NAMEMfl Flaws l/ 6N) 77% 4410 1'5 2 Hi); ‘9' '9 1' mil-9790‘“ «29‘ = W m (sari-MW»,— AT’+‘/3. 7 “F ' 1,) 1 5T = o<L AT = “IXXPF (fiaFr)(Iz'yn)/0°F =. 0576: "‘V t. L JAE .05“ 4'1» 3 u. ; = L = Wafimk 30’9’C'Z"‘/Fr 'coMP ‘9 L 57' ’“‘ “LAT ~= “M'é/‘F (50 Fr)(lz'“~/pr) 010% mar) 51- 3‘ . 752”! GAP : ,zs‘" - N ' A»: 2 Shaw 3;, /&2 77115 c e L 1/32,“ > . ‘ d3=fl£‘0’? :7 039:5“ ( )=o.8ss"" ‘- 80W(Iz..;/m~) COMP ...
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This note was uploaded on 03/27/2008 for the course ENGIN 375 taught by Professor Miller during the Spring '08 term at University of Cincinnati.

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ENFD_375_Midterm2_Solutions_Red - a a a a a a a 3 a a a a 3...

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