Fa10_Exam3_key

Fa10_Exam3_key - NAME: [3 E1 E MCH 210H Statics and...

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Unformatted text preview: NAME: [3 E1 E MCH 210H Statics and Strength of Materials Section 1 (1:25-2:15) Fall 2010 Section 2 (2:30-3:20) Exam 3 This is a closed book exam. Show all of your work in a clear and logical progression. Partial credit will be based on the clarity of the work shown. If the instructor can’t read your work, you can’t get credit for it. Clearly mark your answers and include units. Check your work. The exam has four pages and a cover sheet. Let the instructor know immediately if your exam has less pages or if anything is illegible. Good luck! Fm = Irdm 1 AV=P+wangx Ix 2. of10 P=P0+Pgh AM=C+VAx W 3. of10 V=9J7A, A=6VL [y‘Ix 4 of 10 I = 1m, +Ad2 TOTAL of40 a = _& I --.h. X.-- 5-? “W h/2 <I>h/3 b b A=bh A:71;r2 A=bh/2 Ix = bh3/12 Ix = m4/4 Ix = bh3/36 Centroid 2nd Moment of Area NAME: 1. The structure shown below is in static equilibrium and consists of two rigid members AB and BC that are pin connected at B. The supports for the structure are pins at A and C. Determine the magnitude of the reaction at C. Neglect the mass of each member. 3 II 3 II 5” 2” '40 lb 2" 7MB; ? Bx Bx LEV H0 H3 6—, ._._9 C. A” TAV 4—» x lo, Cf EMA501’70(3\ +By(b\ + FAN-Q = 0 Lt Ema“? Mom + B\,(6\- BALL} =0 H'Bsk'i'ngl 7-940 :go ® ®~® ——+ Hat, = 130 (5%: we; no ital : Oi - BY 3 O '7 (A, = BY:\.\:Q iF¥:ot (_~,.+B,L-'+O=O Q a \a‘qa Cy‘ WD'Bx 0 n O x 4. fl —( \i C: mm; [b NAME: 2. Find (a) the resultant force component in the y-direction, Ry, from the water acting on the dam and (b) the distance from the y-axis to the line of action of Ry. The dam is 8 m long in the z-direction. Mass density of water: 1000 kg/m3 F1=P8VR=HDOOX Q¥ 5 ¥Z€j= $09008 F =Psq1= moo a TI‘A‘Sa/“i *zgzisuoofi 3 X\ =. :3“; m sea: 11.5.: mam R, = R + F; = saweog = 9.33 M” X Rx! : XIFI‘fXJ-FG- _ W = 3;": m X ‘ R’s’Hoo Ry: 8.95 m X NAME: 3. The force T causes the 300 kg reel of telephone cable to rotate counterclockwise while the reel rests on its hubs and bears against a vertical wall. The coefficient of kinetic friction for each pair of contacting surfaces is 0.55. (a) draw the free body diagram, (b) identify the unknown variables, (c) write the equations necessary to determine the force T. IF you have time, for an extra credit point, solve for the force T. FED F; mo) __ l g; N" Fl Tu, ___3>T Eqrchmns; F. = M M. 2mm: Tm — ram — F. (05% O 05F.+E T l\ T+N;-F|=O u @a® T = debiKT—E—Hiiyrum Q© T; 3 “out; : me) 036T = (0'5'UA'41 T + {HEN} = “m3 __ 9. \ + 0*“ \ (mg-HA 1. T W :MMCj 0(5‘EM \ +M+<DSAL1 _ T ®S+M ' yum} T, qqci N _________.————- NAME: 4. Draw the shear and moment diagrams for the beam in the space provided. Give values for shear and moment at all key points. Then compute the maximum tensile and compressive fiexural stresses. 2001b/ft 1400 lb " 3,, Beam cross section 2MA=01 Rfiz‘i— 400(io\-300($)LL+\=C R3 = l3OO \b 2R1 =03 R». + RB - Lino ~aoo($\=0 RA: 700 Ho I 700 — 2no¥ =0 -" >4: 35 A.= :5- (1oo\(3.e\ = was A5‘ = fiC-‘ioMQQ =~aoas - NW 3" f “ Q— : I <7 lh=3w30°=3593 + x I _ ‘— (gxgasm‘: 1.4m inl+ Mm” = \RRSMQ \b—m - '55 I 2 an: a 0.53%“ W = —%’Ho psi \lt a u (re = “ mm \lb = «ah/3 s—ma; _ W = «MN» psi Q—k’ : ——-—-____.._—-.r———-——-—/ MN?“ = - goo“; 003m _<300(m\§0$%.\ _—. +55%) psi Gt ’ ‘ mm (TB 5 _ —%ooua\(' L731“ = “1.15% st ...
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Fa10_Exam3_key - NAME: [3 E1 E MCH 210H Statics and...

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