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Exam 3 solutions

# Exam 3 solutions - AEM 201 Exam 3 NAME CWID 1(20 points...

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Unformatted text preview: AEM 201: Exam 3 NAME: CWID: 1. (20 points) Determine 2 of the center of mass of the bracket, which is made from - a thin plate of uniform thickness. 3, CD +~riauaie l60*)00 __ ii B 165,557 Z ‘-'- 1 A 7.5,92c6 2. (20 points) Determine the resultant force and its location (with respect to point A) due to the distributed load acting on the top of the beam and the reactions at A and B. l.¢_ XR (500 N/m - _I-- _-- _I-- If-III- _I-- TOTAL 20 3. (30 points) The truss is supported by a pin at A and a short link (force in a known direction) at G. Determine the force (in kips) in member CG and and whether it is in tension or compression. c, Identify all zero-force members and illustrate through joint FBDs that each is a zero-force member. __- —I-- _I-- 1-1.!!- —I-- —I-- TOTAL 30 - 5 4. (30 points) Neglecting the weight of the frame, determine the forces acting on all 4 members. Show your ﬁnal results on an FBD for each of the 4 members. +EMC: A, (39-30(40) -5’0 (V2.3 a O h f30)(4°')+(§0)(\z) _ Qv ' 3o " 60 Fir =- 60, 0 No \L li'ZFY ‘='— CW“ “160)": O . 1 5 C7 ~.—.. nw+ gram): £0440=l00 C =loo.olL ’l‘ j . \$3sz- Rxth +30 4‘33-(90):O . - -- ‘ Cx‘nx_= 304-30 Cx'nx :GOO‘LG No-l'e J'ka'i— EF :5 a. +wc*-Covce M?M\o~ev . Procedure 18 - _I-- Signiﬁcant Digits(3) - Results 3 TOTAL 30 - 7 n, 6: Mﬁs z-é-(sovaon time) 2 0 Ex 3M5 A : 3.1(so)(2.o) -:. IS' X 40 4. ﬁx 7: '5'. 001% '9 9X 3 so“) NOw G 9 Cx: (,o—x-nx -: 7S ‘0“? Q: 75.0 n» e— : Fx" 30 “3x *%(5o)+ls .—. o 3k=30-33_—(50)+I5 :: [5 “Ex: 1510011. <— EFY ‘-‘- B), + ;(So)——'C,O= 0 13‘, = co-§(so)=zo ‘BY= ~2.0.0 “o T an-d Reed-+9 2.0.0“: IE”: 30.0“:- I}, up 50.0“ 80.0 G- 6°"’“° 5'0 onD F E D ' 50.0% E p . _ I skate +0 \‘We +ke 30—“, 4mm ari- 3 4.6" o“ AFB . QnaWema-Hve woo” Le +9 ckooge -Hq..’r' 4M; 30 H. ad‘s on 35C“ ”Jake“ HFB B): "37 62MB: ‘%(3"°)(20)+Rx(‘to) -.-.—~ 0 ﬁx;- 1510M, 9 3 s ‘50 z F;< ‘5'- Bx-‘rprx “31(50): 0 9" Bx = gang—13 .: 15 “‘5 13;: 1510011, ——> ... 4- __ __ 2137— 13y +3=-(so) 4,0- 0 13‘, == ~10.0 IL T W Fined ReSuH-s \$50M; :0“. 9°‘°“’ G— 503“; 60.01%: F .b fem; F E; I500“ 50.0“, E a sum, 60.0“> ...
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Exam 3 solutions - AEM 201 Exam 3 NAME CWID 1(20 points...

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