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CIV 207 - Midterm 2007

# CIV 207 - Midterm 2007 - McGill University Department of...

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Unformatted text preview: McGill University Department of Civil Engineering and Applied Mechanics CIVE—207A SOLID MECHANICS Midterm Exam Date: Oct. 29, 2007, Time: 3:30-5:30PM Notes: - 1) This is a closed book examination. No crib sheet is allowed. 2) Calculators that can store text in memory are not permitted. 3) This questionnaire contains 3 pages in total. 4) The examination will be marked out of 30. 5) You can keep the exam questionnaire. Useful section properties: Moments of Inertia of Common Geometric Shapes Rectangle Triangle Circle GIVE-207A Midterm Examination Oct. 29, 2007 Problem 1 110 marks): A rigid member ABC is pinned at A and supported by two metallic rods as shown in Fig. 1. The properties of the rods are as follows: Cross sectional area Maximum allowable stress (MPa) 20 150 100 -__J___10_*__J Calculate the following: a) The position of the load, “x”, (2 m<x<3 m), such that the allowable stresses are not exceeded in either rod. b) The average normal stress in each rod. c) The vertical displacement at joint C. Figure 1: GIVE-207A Midterm Examination Oct. '29, 2007 Problem 2 110 marks): The compound shaft shown in Fig. 2 is attached to rigid support at A and C. Segment AB is of aluminum (G = 28 GPa, ran = 40 ﬁlPa), and the segment BC is of steel (G = 80 GPa, Ta“ = 100 MPa). Calculate (a) the diameters (1a and d5 at which each material will simultaneously be stressed to its limit, and (b) the-angle of rotation at B. Figure 2: / ‘7 d3 ’/ I'z .~ ' / xx ,7 12/ NM i , , . B _ Steel c Problem 3 {10 marks}: The cross section of a composite beam made of aluminum and steel is shown in Figure} The moduli of elasticity are E31 = 75 GPa for aluminum and E5; = 200 GPa for steel. Under the action of a bending moment that produces a maxxmum normal stress of 40 MP2. in the aluminum, what is the maximum normal stress, 65:, in the steel? 1 3’ Figure}: I Aluminum Steel N Clx/EZCFH‘}l bow) MEGA , W200? goumms To YVHDTEIZM Test ' PELBLEM/L, QHWAMMa : 2: Pa“: @owlpq)(4oomnz’)= 8 m ® 97,: pa“: (\ONPQ>(%O\MW(\L) = ‘ aﬁﬁﬁ ZMA=O (8W)(2W)+(§m) (95m)=(lo) Z. x ‘ I ' ' B ‘Om X: 3.IM>%W\) MoTémm .. \ i \ < u‘ luz LN Uz l \ “L3 ———2— 2 '3— P‘um) _5_;_ “P2 (1.9») E:\A\ - a E'LA’L - : n 1 62/37, 31-. W3 (190* SDO K PVPA“ 3w! (Pfhs‘) Em?“ W) 150% \$6 (m) M10 FO‘Z— ?zzD4u2M, Pfam‘ ® \: L=_°r.SM0a \012—"=IOMPC\_ C. r: 3 ——————-~——‘—— _. ~ ‘ 4‘ o T“ T “(7 TL "CL .~ «3 @ ’C='—_=D—‘=-— =——-:———- l_LIP_ MW 1?: Iv VWMT'T‘T A g C “L TL 6 L =[ 6ET: KC =(ftm A“: 4'0 HQ , (6+:CZ‘g-3A“_ IOOWQ err GA 6* > 2 <33 “ 0&6 (ES?) WW) - WW) k 40—0 33—min 3.4 —@ Sam) 9;?) {‘20 (ﬂax; ) , “"1 -- I 3 ' 3 SUB @ NW® 3 GOXIS)%GT)(AA\) (moxugﬂﬁaﬁﬂ (Ask) . I / ‘OMDQJQ: T1. ‘5 3 I“ N : \$75ng CL“ 1‘ [(1.53 XmFO‘M‘L) : (\$39? LL33\3)¥‘8 (6‘; ﬁlm ‘ q 3 = sumo max-\$12). 3 d = 5 0.01qulo‘9 -= 0,00SXM= Shm é“: Hqu a\ ' >) _ Q L ' Ao\=g\8NM/\ (9 \$_ __ @OX10)M1,Y2W , 3”“ - ~ ~ “if—“'5'— : 2 - {5 (51A Rzgxmq ﬁh Limo m 0385*”? [203. Q85 20A. ’2. W; WITH (Eu)aa'=40Mva, fa”)q_=loomPa a (a) A B c Tm *JEZW A3 T 7‘413 3 3 - -- a a T . “- ECUUWWM 10= Tm Isﬁ @me + “(5% =\$8§AM+M.53J% @104“) L6) WVA‘hbiHH' GU". éem; (bslc lg _ TL A _ z A ‘ ( LANqu WT lava“) “WW, (Sp (00mm. an (190 g = C a a): 2% C 2 ) K“ T9 2.). (myoﬁbcia’) [email protected]+o® L 3 .. {a M - ll ‘3 N ) _. 4"“Dﬁﬁam) (DexacTFQf-A .I. , a _ m w n - 9 [DX'OWW‘ (b 1— “a I = éo‘quto (Add) 3;; A 3__ --3 3 _‘ . 4‘ -' yoﬂgxmg : OJQEX‘O [M I dmw‘g—g ND :o‘og'g M4 = “MM- [Asﬁ 3‘6 “MM. (km: gum 6N U) (P (POND M4,, (Du) q M = z o o " (5 Zﬁxtoqug x Sgwu" 88 (L 3 3 : TARA“ _ 40MPa~Tr-(3‘KW)_ _ T0.\ [[0 ' Ho “I552U‘M. [Elk/w - 2M 0 2 : O‘OQX’S‘IQAA. 9H, (a ’Zg’Wa- %'|T(L§W)4‘ a) Ev: l0000NM~Tm = 10,000“ '951334‘53NWL _ 3.5. = ZA‘ —~; (WW) (too) + (8°) (8“) (+0) . — 4. M — V: m = someway“) H(no~4‘l.§) ,__/_————- (QT/MK A: It H (43.?) (n) WWW k" It mm sf (4%?)(7-b4) (mood % Uu-L‘v‘ts‘) W (@HQA) mm" 3: Ute-48.9) = “Ema 4 ...
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