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practice_exam1_solution

# practice_exam1_solution - Name Lab Group CE 231 Practice...

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Unformatted text preview: Name Lab. Group CE 231 - Practice Exam Closed books, 4 pages Equations you may ﬁnd useful: Q+g q-g _ q—%_ a. = + 00526+r 511126 T. = — sm26+ r cos23 .I: 2 xy X y’ ’0’ 21’ a" +0 0' —0' 2 _ Iy _ x y x y 2 tan29p — 0' —0' Omani“ —— 2 i ( 2 J + 11y 1: y NOTE: BE SURE TO INCLUDE UNITS WITH YOUR ANSWERS- (ANSWERS WITHOUT UNITS WILL NOT BE COUNTED) 1. (20 pt) Metal bar with a diameter of 0.505 in. is subjected to uniaxial compression force of 1200 lbs. For this state of loading determine the following: 1200 Db #0:)" a) Draw the Mohr’s circle and identify the pole 6—“ __=_ 2;;ng lit (0'30 D b) clearly mark the principal angle and draw a properly oriented element showing the a: C, u 9i stresses acting on principal planes c) clearly mark the angle that shows the orientation of planes of max. shearing stresses and draw a properly oriented element show".- , ..l - esses acting on these planes. ’5 IIIIIIIEIIIIIIIIIIIII IIIIIII Q Q IIIIIIIIIIIIIIIIIIIIIIIIIIII IIIIIIIIIIIIIIIIIIIIIIIIIIII IIIIIIIIIIIIIIIIIIIIIIIIIIII IIIIIIIIIIEIIIIIIIIIIIIIIIII IIIIIIIIEEIEIIIIIIIIIIIIIIII IIIIIIIHEIIIHIIIIIIIIIIIIIII c ,IIIIﬂEIINEIIILEIIIIIIIIIIIII III-IREVIIIIIIIIIIIIIIIIIIIEF o IIIIIII“IIIIHIIIIIIIIIIIIIII IIIIIIIIEIIEIIIIIII IIIIIIII cw IIIIIIIIIIﬂIIIIIIII IIIIIIIIIHiIIIIIIII 65:45 IIIIIIIIIIIIIIIIIII IIIIIIIIIIIIIIIIIII IIIIIIIIIIIIENIIIHI 2. (30 pt) A state of stress that occurs at a point on the free surface of a solid body is 63 = 110 MPa, 0‘, = 30 MPa, and 1;, = -30 MP3. (a) Evaluate the principal normal stresses and the associated shear stresses (b) Provide values of angles that determine the orientation of planes on which these stresses act (do not need to draw it, just provide a proper value). (c) What is the maximum shear stress? Solomon 30am} 3° a. Sin} [l0< i’al‘ozﬂ“ (CL) 6“: —.-. (Tine-ff‘ﬂ + G‘aL-o—LjDLq- 1133' 2—- 2 = tic-+30 + 0‘0’3031—1- (—30) '2_ 2. ._-. 70 +50 —_=. 120 MP?» 6"; = 10—50 a. 20 MPa. .. X’BU __ -—GC} :- Cb) +0”) 2.8) P _ 2. E1 =_ _g___————— - ‘90 0P1 = “‘9‘4’3e \j 0 CD "C'moor... = W; ‘5 CIG~§®L+C~3§ = 50 MPa,/ —————_ 3. (15 pt) The prism shown below is subjected to uniformly distributed forces. Determine the normal and shearing stresses that exist along diagonal AB. 3 —' l 40m _6 s 1333 NFC» GOXSOXIO ongo3 2. 5o Mpg. onsxuac’ Catt} =- O = i9. = o-étée 60 c0 = 334° V 0—19!) =- G'ﬂ—tG'DL‘G—‘j COSZQ+GLC3_ 51019 2. 2.. =~."l33'3+50 + ,13'3-3—5'0 605(67'1.) 2. '2. = ~41-c5 + C— 35-22) .s. -—'r£,-<a*1 Mpm a .JH-cgg —- 035-232;) —=. ~G°43MPm l i . Coma = .— @1163) Sun 2m —1- Tau} €052.03 ﬂ \ (P 63 DO bu I Ln 0 U E". O m 6\ .4 f) 4. (10 pt) The gage factor of an electrical resistance strain gage is about 2 . The coefficient of variation is deﬁned as the 5 D / ﬁVERAGi E The value of normal stress on planes of max shearing stress is equal to (select all that apply) om l/_2<3ave , 20m. The failure stress of the wood blocks that Gyou tested was about ID .A strain of 100% — I XIQC' microstrain (10 6) 5. (25 pt) Shown below is a deformed section of a body in a plane strain. It was originally a square with sides 10 mm in length. After deformation the line AC is still 10 mm long and AB is 10.008 mm long. The angle CAB decreased by 0.240x10'3 rad. (a) Draw Mohr’s Circle of strain (b) Locate the principal planes and the planes of max. shear. (0) Determine the magnitudes of the principal strains and that of the max. shear angle Ex. :1- o- 008 a O— 0003= 3‘300 Cm") 83 3 0‘, 073%,:- 0240Xt03 , 240Q0) 1:91: 120009 6* 106leJ Tm. CeolEa’ out rm. arch C ”‘0 ,9 mm. E was akEaUg .. goo-toga :400059 IIIIIIIIIIIIIIIII IIIIIIIIIIIIIIIII I IIIIIIIII I IIIIIIIII I IIIIIIIII I IIIIIIIII IIIIIIII IIIIIIII IIIIIIII IEIIIIII Haul-III ﬂ, IIIIIIII an. IIIIIIII liainiIIIIIIIIIIIII iIIIIIIIIIIIIIIIIII nIIIIIIIIIIIIIIIIIIIIII \511' M Co oﬁnal’es 05L cadre“! C (400 05\$) 0) \ TEL WeieTeﬁu— point A has Coordinator) A (80000) lZDC‘j thymusa,meannxm;ﬁ==K§E§F§;f «2—.- 441600)— IIIIEIII III'IIII I I I I I I Th. co Grahams at} points (3 ancﬂ D wepzresenﬂ— l‘ﬁi PII'Du‘poJ Shaka. Thug, 0m, '9‘ = Goo+ 411-0 Qty") _—. 811-61x15é £2. = QOOH‘HTQ Q59 = +9 \$11-\$10?) chnc. .1. LR =’- lXA-l'l-GCIEGD: 83S’2CH5C) \ocumoo a; m. anm g. .1251— : 2-40 20.3 ‘00 20P| gl— ﬁg, 900 O _—-—-———__ ...
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