Test 1 Solutions - th—dti—dlfll/JD KIT: iii-El...

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Unformatted text preview: th—dti—dlfll/JD KIT: iii-El L||"|'._. li_.1'~.x'1L I:NL'.I1N|:I:H1NL'.I D'KJ title.” 4'Kti-4 l" . Ell/Ell: ENGR 2200- Intermediate Strength ofMuterials WJ. Likes Test #1 Fall 2005 ENGR 2200 — Test #1 (100 points) Student Name #50 LU'T'! mg Student ID My signature verifies that l have completed these exam questions. independently and without the aid of additional notes, equations, or materials not provided by the instructor. I have not received assistance from other students. I understand that the use of prohibited items or assistance is considered an unfair advantage and will be documented and treated as academic dishonesty. Signature Equations for use: i 2" TL Chg anemia}: s E p ’r “ 1/6 th—dti—dlfllflb UT: DU LIITIL, L1U1L I:NL:I1N|:I:H1NL:I D'KJ titid 4'Kt14 1" . Eld/ Ell: ENGR 2200~ Intermediate Strength ofMerert'els WJ. Likes Test #1 Fall 2005 Question 1. (10 points). Sketch a typical o—e diagram for a ductile material (such as steel) when subjected to axial loading in tension. You must label the axes and denote the location of the following on the sketch: yield stress (o'y), ultimate stress (on), breaking stress (ob), Young‘s modulus (E), the elastic behavior region, and the plastic behavior region. ELM-hr. “Na-i“ ‘5 Question 2. (5 points). A steel member is loaded in tension from point A to point B. Sketch the expected stress—strain response if the member is then unloaded from point B back to o‘ = 0. “he Restit- Milan/“"103 Questions 3 s 7 (True or False; circle the correct answer; 2 points each) 3. If the design factor of safety for a member is 3.0, the ultimate stress is 8 ksi, and the allowable stress is 2.8 ksi, then this member is suitable for the design. p; g E: _ .- ZBE .1; 3 (T ® 2 3 5pm: 4. Machines typically operate at stresses that result in plastic deformation. (T ® 5. Axial stress distribution in an eccentrically loaded compression member is uniform. (T® 6. Shear modulus describes the relationship between axial strain and lateral strain. (T 7. If a ductile material is loaded in tension, we expect some stretching before failure. @ F) 2/6 SEP-EB-EE’IES [37:58 LIMC CIUIL ENGINEERING 573 882 4784 HEB/[216 ENGR 2200— Intermediate Strength efMeterictls WJ. Likes Test #1 F all 2005 Question 8 (25 points). Each of the four vertical links has an 8 X 36-mm unifcnn rectangular cross section and each of the four pins has a 164mm diameter. Determine the maximum value of average normal stress in the links connecting (a) points 13 and D, and 03) points C and E. 0.4 m4 ‘ C " 01.17ch We \\ 9 a; 1:5 It.) LbjfiynEfitwfi “a \\ see A : \Bm m)l3£,>uo a») '\ A: Zfieliolmz L+ 7 \\ 2 Lame, 2A : Smarts ml ’- " 3CD ('D'lelfw) _ {a lab “X mic N") \\ (y .. Fu: JZB' we?” N lav": 325W” “9 \ as “7:: We \ . b 1 fl film}: 1 \\ s“ : 2L7 XIO N’m : 3' _. O ‘-\ N " \\ (r 5"th ‘10 ' {O‘MWO m > \ «as: 2L7 Ml’a l7“; : - [7.5 Wag N \R on. Fw=ll.§wbsb/ CC.) a) ‘81:» die so “Twain/q) «so we? Sue.th cartel .. -3 ? , (gm) m1$t1\om)e (Mic MI 5w) W9 ‘ A : A ’ bluth 4; 3! LEI-tr A ,. l N5 W _. mp east. the We causal 50 As HAW) jsxs A —' 3 2 Xle W1 SEP-EB-EE’IES [37:58 LIMC CIUIL ENGINEERING 573 882 4784 REM/[216 ENGR 2200— Intermediate Strength of Materials WJ. Likes Test #1 Fall 2005 Question 9 (25 points). The 40-fi long steel rails on a train track are laid with a small gap between them to allow for thermal expansion. The coefficient of thermal expansion for steel is on = 6.6 X 10“5 1/ c“F and E = 29 X 103 psi. The cross—sectional area of each rail is A = 5.10 ml. (3.) Determine the required gap 5 so that the rails just touch one another when the temperature is increased from T1 = —20°F to T2 = 90°F. (.2 m": G)Q~(72_cu): \JD F a: 4mm pets 2 (Latte, NEW To Elm 07M) ems Em Moi-,7- S J . MM) 3/2 . “'30 WW ‘4’? use) ‘3 8 2 Mfr-TIL.) : “new? ‘MXHODDL Lime) g ; 55,029 it: g‘ : {£36qi (b) Using the gap from part (a), what would be the axial force in the rails if the temperature were to rise from —20°F to l 10"}? ? 355;” 5 HO *L‘ZD); {EDUF trite WWW. 6 : QMHCCTXL) : (es HELIIBDILiDDUZD : 0H: " 6D mmm m 42.5% eoMQMée'roA :59: Sf D-LH - 0343' = 9.06 g Ta .2: 1 0.09” i ‘ a " A (00" .o'm'i 2 so' ' I V : (0.0L: g I E) r 95‘ 1'. “105/ . . L (H0 gi‘l Jim/HQ __..-F-— 4/6 th‘dt‘—dlfllflb KIT: '3']. UITIL- L- l 'v' 1 L END 1 NIZIZH l NU CPI-(J Utid 4IKU4 I" . UD/ E": ENGR 2200- Intermediate Strength 0f Materials W1. LikOS Fall 2005 Tesr #1 Question 10 (25 points). The torques shown are acting at points B, C, and D. If the shaft is made of steel ((3 = 27 GPa), determine the angie of twist between (a) C and B, and (b) D and B. 3“ mm f [7]”) N - 1n 3“ mm (PH 11‘: m /' WM} N - l'l‘: I' (:36 mm Finis) MUN” Tommi" ’9) WI) ma CD m L, E inc) @D Fall «556194 5L Hi :Hl‘ two Ettéoo 90a The: SUD N m [C 900 (Lap : 313mm "7 AWL: 30mm (356 '5‘» j”;(amgmfiz LWYIOML} n. 1', W1 Gj'IJiOk-' lgwflm Qt jfiswo-Be LL93 LCM rfi -: ~ w, .5: a; J: raw—OW 6P; L # 0%,“ an as waoaltmm ) BC. the; : (3% {L I; (H‘OCJ N-M O- Bmze 75%" A (mm N/MX’HWIO mfl :(O.l”l9wmi 92—53 4 fisy4§fi : mam @mtm my" Egg”? SEP-EB-EEES 87:51 UMC CIUIL ENGINEERING 573 882 4784 P.86/EE ENGR 2200— Intermediate Strength aneierials WTJ. Likes Test #1 Fall 2005 Extra Credit (5 points). The change in diameter of a large steel bolt is carefully measured as the nut is tightened. Knowing that E = 200 GPa and Poisson’s Ratio v = 0,29, determine the tensile force in the bolt if the diameter is observed to decrease by 13 X 10'6 m. finmfiefi- ! (lie, mwa -!e 5 : Ham m (Mam I 3‘ 3 0.9(5m 5, w A‘ ’ ‘5' r' MAW * Jam” = aznm MM 3‘ 0.06m iacm , M” =_____——- W amassed emm F q A ate/“M "- K50 1 ,7 .r ZLWKL I" F- ? I." ego W ssfimxw c) A! as e = E2 : (ZOOM N/m1)L7,L+7 m AWL 13Kth in 5" : No fee "3 7‘ 1 1 2.???)th M fifljfl -: he a Li L. *3 ,_ ? ,.. :Kl‘t‘ifiilo m if: W27— “3 6X6 TEITFIL F' . BE: ...
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This note was uploaded on 04/03/2008 for the course CE ENGR 2200 taught by Professor Likos during the Spring '05 term at Missouri (Mizzou).

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Test 1 Solutions - th—dti—dlfll/JD KIT: iii-El...

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