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ME383-Exam1-KEY - ’T’O‘H’rL PUTS 2 6 ME 383 Exam 1...

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Unformatted text preview: ’T’O‘H’rL. PUTS 2, 6) ME 383 Exam 1 Winter 2010 Closed Book and Notes Equation Sheet Provided and Calculator allowed Be neat and box all your answers Show all work and all units. Label all lots and sketches. Name kfifl Lab Section Tam“ (l9 1. Find the reactions at the supports and plot the shear-force and bendng diagrams (Hun mm) for beam shown in the figure below. Label all diagrams properly. mmye'ru mun-nu rm I... mum-“i“ m- M .. “pg. 8001M 27:3 3° ._ pl—FL'OD +Rz ”800 AC) _...... 1 u Rl'l'Q-L-a Ian-DD EH :7. o C) s . £th + ”awoo) 436%) 2’0 1mg; R1: nap—R, :llaa+l-}I.H -'-1m 0 Q. (UQILP‘C: L' _ .4 Vfi-U-FE mm am N’QUAF (moan. sue-M2— 3mm bananas memanfl' .thG I _ n——'—-——-‘ '——v--_ f— H. = (-12}: .qjcq] =\- 6895 “@414 m .-.-. -ess.s-(e=H.=fl Ce) H .-. _.. Moo + (89930;) (‘19in 3 ran—O .'.Cl+?='CV-SL Name K‘ES Lab Section 2. The following figure represents a stationary shaft and pulley system, subjected to 929%“ng @ a static load of value F=2000 lbf. For the l-in diameter shaft, with area A=0.785 mm ' inz, moment of inertia 1:0.049 in“, and polar moment of inertia J = 0.098 in4 ”strata: fiouwD 9M4" 1‘) :11 ”U - *50.1€S~ It)“. 1 .—_ moss m“ 3": 0.09? m” @ a. Draw a free body diagram with a cut-off point at the collar and compute the internal loads. b. Using the maximum torque value of T, the maximum shear force V, and maximum bending moment M as computed above (in part a): @ '1. Compute all direct normal and shear stresses associated with the loads. (9 ii. Where are these stresses at their maximum? Describe and show as points A, B, C, etc. on a sketch. pzzooolb; WEE” |~TM Lo'A'D‘S : "'06?ng 1 l:- RDOOIEDC'BIM)=~ 6000! (3rd _____..._...._....——-—- Brita-4L . _. F .1 o 0" Fear: ‘ lV" 299° '1' (E =- d atom-aft -. H = (29ml1¢]C&7~]:lLloool§-W l LEMD: OW Name K35 Lab Section @ 'T..-= Eooolkjuui H: qmtqupi v=2999 "4.0“ (9 .DH'EECT mama-L 2. 3m wages-$3 Jae: To THE? LOADS .moamm. fiaaa'sSJu-E To 6E~gp;me,; a; ”E! TI. @ mAx rm: TD? 09- 53991- ;J) (‘15.:th GUNS—- -————-—-—-_3 a+— (3:: (a [a] graham h: 1-49 z- 6‘z ; M C— : (amalgau) (05‘1”! p :I @.o£(°uu"q [0‘2 2-: Llafi’lqasil (C213 = (mow-”n1 Lesion) l/a- :-_ Basing-j 9. .‘mmvafif 45m 651115935 Jae—”m 36-20:!»36 : ”52% -‘—' W— CRQuan New) ”BA 1 @ mm: H- datum flM'S ...’.. 731$ 25" Lf CMOOIfi “7-31 ‘IT'JL%:~3 qk—P‘N ’bCo. WSW GD Lona-mob mamas was-£5 MLE'mflmmJlm 1 A AW“: SHAFT : - — C—fifiTMMMI‘E‘) 299%!!! 99‘5””? m “mum fiT W‘Du 'A—CTBJ‘SILE) ONUO-MIH. W3 $3; 115 M ‘? LeCJVI'IBN scampmlvu) L109? fio‘r'mm) @ X oTetLfia-NM entire. inn-23‘s ”Q3 M; «NH AT LoCfi‘Tl‘DAJ: A, 9, two 5. fiM—H%3 E'I'OP 99mm; Tm‘fme- ‘fll-Efl-{L— 411-04955 Q; I: max ail-'7' LUCA-“TIE“; (.1 — a Lm'mAgrAxn-s) Name Tfl‘Q? 3. The follo ing 2-D stress state is given: ox: 400 MPa, 6,: 160 MPa, I“: 120 MP (‘E 5 Lab Section a) (All other stresses may be assumed to be zero.) Construct the 2-D Mohr’s circle for this stress state. Label the x and y faces on the circle, the principal stresses, and the maximum 2—D shear stress. Compute the principal stresses and the maximum 2D shear stress. Show a stress element that has the two principal stresses determined in part a. Show this element oriented from the X axis. Calculate and show the angle. Show a stress element with the maximum shear stress acting on it as determined from part a. Orient this element properly from the X axis. Calculate and show the angle. Show all stresses. In reality, all elements are in 3D stress state, with the stresses in one of the planes being equal to zero. The assumption of 2D stress state may cause serious errors when computing the maximum shear stresses the body will experience. In the situation described here, the first two principal stresses are as computed in Part a), and the third principal stress is equal to zero. Given these three principal stresses, com ute the maxim III hear stress for the 3-D stress state and com are to the maximum 2D shear stress. 5w Sic-{TN 2 (élrnlufifl- TL? , my Western} 622:; .— fl: 2830—1633- mm”: reaamf’a t‘ V/— Name \(EB Lab Section ( b. ) ‘fim's MW pPRthfi-L firms-5 9mm o 31> 61413—9 632 —o ImPLLE!) Mm GQLI‘S‘OHPH (Tenn HOG, ”TOTRL: Name K59 Lab Section @ 4. The figure below shows a crank loaded by a force F=300 Ibf that causes twisting and bending of a %-inch diameter shaft fixed to a support at the origin of the reference system. a) Draw separate free body diagrams of the shaft AB and the arm BC, and show (9 and describe all forces, moments, and torques that act them- Label the e @0 {Cupyughl fimMcan-lfifl Campnics,1m:. minim mind for «pm-cam “disfldyJ T : . . 3 m dia. directions of the coordinate axes on the diagrams. DO NQI LEQMPUTE VALUES OF LOADS —ONLY SHOW WHERE and WHAT. b) Locate the 2D stress element xz on the top surface of shaft at A, and calculate all the stress com onents that act upon these elements. (Use MA = 19501bfin, +z direction by right hand rule and T A=1200 lbf.in, +x directiou by right hand rule). Determine the maximum normal and shear stresses at A on the maximum stress planes. can m-Ji‘c E-mbi?‘ H‘s?“ 133:.“ i; __'i’}_ Name 1635 Lab Section (9 may @5095 DMMmS: (1 WT PDQ. bug—FCH } 1 'PM'? 99/?- ELK? LA MP'HDIU 3 4.?- ToflQuE’ JME "TD thfinué 94’: SW FIZ—Dm Rim RDWIUKJ w/cLflrdlc. I‘M—HOLE M515. Mme momw Just: '11:: 6W1“ BEL-’6 PLFd'ED DOM no 9551. =3: mam-mu Parade— Data? To Fencfi"fi‘r wmuacrmu B , 5-] C5 5, Racine») F935? #7 Can-’Hiscnnn C. out“ "to magma T; a“: ”rowan:- Due“ TO TM!6’T/u6 Fflmm WDLE '93, a Mel-1w React «r fivPPoL'r 6 — C63. =+F-) M9. = asthma MOM-act 1h E) ’T‘o CaumaLfl-cr Ccfi PuQHMJs amok, ’TtasI 2: Ta-{HDLLE— :zrp CDuMTE‘rLILcT mus-nae, m 141’ c; m»? To ‘F’Bflé‘e— 1L WSW/v16 {iii-MOLE— qDDNM . Name (:5 Lab Section 61060 '~ C 44: dden m) O 'TDflJQL-H: TA: : Q K513, :- E Nara .‘. FLd/Z d “3::qu - TA— C”?— -—_-, I437; ‘52" TEN/62 naa :- Wm :[m 145‘: S 17 Lopes-fix? ‘@ m Tumvexjrc“ 3% M 13mm flit—DQ‘HI‘DUQD (Q Q Name It; 2 Lab Section mRleum Mon—mm, 3 6%:er ‘JTIIE'S‘ES fi-r v4~ am Tue;- rnM:mum mares-5 67W 5Tfl‘e5‘5 W1"; :3 01 {Coca—3; [tr-M mm? Crud-91- Qaeanas) 3% 0—5,. ILL? o Equation Sheet: Mc Tr P P O'=— f:— O':— T:— I J A A (#23 rectangular section: A=bh,I=—,T=E 12 2A 0 ndsect'on A 7nd: 7rd4 J 750:4 0' 32M 1 16T T 4V ru 1 : =—, =—, =—, =_, = , =_ 4 64 32 m3 m3 3A , . ox + cry 0; — 0'3, 2 Mohr 3 ends: center : c = R = + 1' ...
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