RelativeResourceManager-1

RelativeResourceManager-1 - Name ’I‘K Instructor...

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Unformatted text preview: Name: ’I‘K Instructor: Siegmund Sarkar Susilo (Print) (Last) (First) (Circle one) W ME 323 EXAM #1 FALL SEMESTER 2010 8:00 PM - 9:30 PM Sep. 29, 2010 Instructions 1. Begin each problem in the space provided on the examination sheets. If additional space is required, use the paper provided. Work on one side of each sheet only, with only one problem on a sheet. 2. Each problem is of value as indicated below. 3. To obtain maximum credit for a problem, you must present your solution clearly. Accordingly: a. Identify coordinate systems b. Sketch free body diagrams 0. State units explicitly d. Clarify your approach to the problem including assumptions 4. If your solution cannot be followed, it will be assumed that it is in error. Prob. 1 _(33) Prob. 2 _(34) Prob. 3 __(33) Total (100) 1 ME 323 Exam 1, Fall 2010 Name: Instructor: Siegmund Sarkar Susilo ' (Circle one) 0’ = F" /A Tavg = V / A F.S. = FM /Fa,,0w F.S. = 0/0,, /0'a,,0w F.S. = rm], Mann“, Em=(A~9"‘AS)/As=5/L0 7=(7r/2)—6' 0'=E£ V=—8/a,/8,Mg T=G}/ 1. 5m =a(AT)L 5”, = ja(AT)dx 0 1 5x =E[0'x —v(0'y +az)]+aAT 8y =%[0y —v(0'x +az)]+aAT 1 £2 = E[0’Z —v(0'x +O'y)]+aAT _ 1 _ 1 _ 1 yxy—ery 7xz‘Esz 7yz Gryz A «5 L £44 AL=IEE<x> dx AB 0 (x)A(x) Tp TL T,(x) = L 1=Gc— r=’ =—'— = ’ dx = + CW 7 ¢ G] ‘15 JG(x)J(x) ¢B ¢A ¢AB 4 4 4 J = 7m bar = MC" ' ) tube 2 2 2 ME 323 Exam 1, Fall 2010 Name: lnstructor: Siegmund Sarkar Susilo ' ' (Circle one) PROBLEM #1 {33 points) When thin sheets of material, like the outer skin of an airplane are subjected to stress, they are said to be in a state of plane stress (0': = 1x: = 1y: = 0). On an airplane (as shown below) it was determined that a biaxial state of stress exists such that rxy = 0, Le. no shear stresses occur. The strains in the x and y direction were measured as ex = 200x106 and 5y =140x1o—6 .. Determine the stresses ax and cry. ircraft is made of an aluminum alloy. The relevant For the analysis consider that the a n addition, it is known that Poisson's ratio is v = 0.3. stress strain curve is given below. I 0 0 002 0.004 0.005 0.003 0.01 Strain a 3 ME 323 Exam 1, Fall 2010 Name: Instructor: Siegmund Sarkar Susilo (Print) (Last) (First) (Circle one) PROBLEM #2 (34 points) An assembly consists of three identical bars (L = 1.0 m, A = 5.0 mm: ). The bars are made of steel (E = 200 GPa, v = 0.3, a = 16 x10‘6 /° C). The assembly is fixed at walls on its two ends. The bar EF is heated from room temperature (T = 7O C) to an elevated temperature (T = 200C). (a) Calculate the stresses in the bars after heating to T = 200C. (b) If the yield strength of steel is Q = 750 MPa (independent of temperature), determine to which temperature the structure can be heated such that the bars do not yield. 4 ME 323 Exam 1, Fall 2010 6": Ac 3 \G\ =_éE:mAT ;__584 l‘17z EF 3 .———,——-——‘_—/_‘-_r#“ 6‘ >G“ Name: Instructor: Siegmund Sarkar Susilo (Print) (Last) (First) (Circle one) PROBLEM #3 (33 goints) The bar shown below as a solid circular cross-section over a segment of length L1 :70 in. and a hollow circular cross section over a segment of length L2 =14.0in.. The shear modulus of the material of the bar is G=3.8><106 psi. The outer radius is co $40 in. and the inner radius is c, =20 in. The bar is loaded by two torques. A torque of magnitude 150 Kip.in. acts at the location where the solid section transitions to the hollow section, and a torque of magnitude 100 kip.in acts at the free end. The bar is attached to a rigid wall. ((4) Determine the rotation of the free end of the bar. Provide magnitude and direction of rotation. ' ‘ 4’ 3C- ) 9&1” In £44 - l, ‘ Q” J =£(2.;)=25‘./3,..4 So/I'd 11350 K' ‘ J ’7 - " (117-23 54$ "0"“ yea/M a: (2") * -"‘ ' C. 100 Kip.in / \>><“‘ __ 7:31) 2Hx:0 —TBC+ IDO M’ m— D 1 ’36.: MO ‘4‘!” "‘ a) l00 KPH“ T R \ Br— ZHA; J — 1’43— IIOKJ‘nM +lopgfggo 44—] » m K7,”; 115—- »50 Krr m M {AB /5’0K?4I~ ¢ _ 7;: 13¢ ‘ max/03 - /‘/ 0.0 /5”7’ rad ” ’ .2" /é .;2 BC 5" 3,, 23-56 ‘3 3"” ' fl ME 323 Exam1,Fall2010 "t 5/”? {/06 ,.: A3 ,/ ,. Co é 7: (A ).~ I“ M = 3‘73” ’/~‘ ’5’ qu’ 35 5%“, lu & T Q _;/~; c .- 32r/é L ’Z(€)= 32,, '2 (a L’ % ...
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RelativeResourceManager-1 - Name ’I‘K Instructor...

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