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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; 45 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
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¢ " 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
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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 kin 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“";
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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' mil9790‘“
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 Spring '08
 Miller

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