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EAS207-2nd-exam-sol-2011

# EAS207-2nd-exam-sol-2011 - 39293600 ’ EAS 207 STATICS...

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Unformatted text preview: 39293600 ’ EAS 207 - STATICS Second Exam Fall 2011 Total points: 100 Notes: 1. Closed book exam.‘ 2. Draw Free-body diagrams and reference axes for all problems Q.1(a) For the loaded truss shown in Fig. 1(a), use the method of sections to find the forces in member HJ and GI. The weight of the bucket is W = (’16 points) 1000 lbs. :20 7 QOQQXZ 2‘3)*Q0W>“ 03)‘ 3F on: an- *7 —2eooug;\ cowl? Z MEZO :7 goes ') (92\$) ”VLFHQB) («lion 0) [33+ a £63 3;“) 1:; o iii-Z" 64 ME] W @ Q.1(b) Fig. 2 (a) is a diagram of the bones and biceps muscle of a . person’s arm supporting a mass. Tension in the biceps muscle holds the forearm in the horizontal position, as illustrated in the simple mechanical =2kg. model in Fig. 2(b). The weight of the forearm is 9 N and the mass m (i) Determine the tension in the biceps muscle AB. (ii) Determine the magnitude of the to forearm at the elbow joint C. . (18 points) sadism i rce exerted on the upper arm by the ZPV/‘Cﬂ ' 1:) Q23 (9 8 3) (0/39) +C9) (O’IS) 200mm~4~150mm~+5 (b) (if \$933333,“ @) C005) ‘2: (D Q. 2) A beam is loaded and supported as shown in Fig. 3. i) Calculate the supportr eactions. (8 points) ii) Write the expressions for shear-force ('V) and bending- moment (M) for a beam section between points D and" I: as a function of x measured from the left— end A. Don’ t use integration of V to obtain the expression for bending moment. (10 points) iii) Draw complete shear and bending moment diagrams for the beam with the aid of areas of loading and shear-force diagrams (no integration). Be sure to label the values of shear force and bending moment on the diagram. ' (15 points) 130 \$3; 5642.} f: 900 - Bééré +3351; ,3?“ ~ 3 J 0.3) Determine the range of weight W for which the system shown in Figure 4 will remain in static equilibriUm. The coefficient of static friction between the disk D and the inextensible cable is 0.15 and between the iOO-lb block and the inclined surface is 0.3 . Neglect the friction between the wheels supporting weight W and the inclined plane. (33 points) D 4/jll\$=8,l§ #3 = 0.30 ‘l #k = 0.25l \ ——~/’\.’-»/-\ A”, ,— A K WW; W ‘ r7 Fig D TL if} // / x...’ ROWN F3391 _ “l EFUZO :2; N~too€os 2020 “F3951“ l (3Q §N=93y97 M E” > V 273:0 '3? 2T, ~MSN «1003mm :0 ER g) 17: =W3>(9387)+EOOSM20 . \/ a . r 3.»; §T=31..E3Mg§§ 3 :3 “E 231436 a»? BOE3’5)<%% ”ﬂ f; ‘ 1‘ If A g 1‘ ’ _. .m I __ E as». \ ”T 23(55mé) <1»3> »—' 40:53 p ’ 260:0 ' 4 WW Z—Tx JFMBN , LOO SW3 3 » \ I 1 ¢ C, I x ”‘ ““ \ E ©05m20 EEM‘L a: m .,.. -co»><E»Em «- E Mfg <71: 3 W35 ‘ / ,_ r“"’ __ f E “/glmce n77; Lg, :aXVWSW- 13 GE») 3 , J g .1, W ‘ E L, ,— “a Wm. W E227 3% a» W E »:E/EW WELL?” <9 M 3335155 w ...
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EAS207-2nd-exam-sol-2011 - 39293600 ’ EAS 207 STATICS...

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