Civl113Final95

Civl113Final95 - {gait 113 LI 159 128 NOV199§ /,...

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Unformatted text preview: {gait 113 LI 159 128 NOV199§ /, Ebol'dfdm’. f p,{ {(3 ET 2"“; EM“ “M CIVL 113 STATICS and DYNAMICS Dept. of Civil and Structural Engineering, HKUST Instructor: Dr. X. S. Li FINAL EXAMINATION 16 December, i995 Emblem l l 10% 1: / The truss is composed of equilateral triangles of sides a and is loaded and supported as shown. Determine the forces in members EF and DE. {Lexie-mi".-- l Problem 2 1 19% )2 The homogeneous block of density p and height h is floating between two liquids of densities p, < p and p2 } 9. Determine an expression for the distance b that the block protrudes into the top liquid. .'.?~:-.‘:‘ 1- "m mm‘s‘a ' “ ,tk‘925§s§3§§::;§{?;:W-Jgm Fig. 2 160 Emblem 3 (10%1: The uniform links each of mass m are held in vertical alignment by the two springs each of stiffness k and unstretched in the position shown. Determine the minimum value of k which will ensure stable equilibrium in this position. Problem 4 1 152% 1: Each of the four uniform movable bars has a mass m, and their equilibrium position in the vertical plane is controlled by the force F applied to the end of the lower bar. For a given value of P, determine the equilibrium angle 6. Is it possible for the equilibrium position shown to be maintained by replacing the force P by a couple M applied to the end of the lower horizontal bar? Explain why. ‘ Fig. 4 161 e 0%: At the instant represented the truck rounds a l 10-meter curve with a constant speed of 60 km/h. and the car traveling on the straight road has a speed of 80 kmlh which is increasing at the rate of 1.5 m/sz. Determine the velocity and acceleration of the truck as abacn'ed from the car. Fig. 5 Pr blem 10% : A IG~kg block is released from rest 150 mm abo e two stiffness 5 KN/m and negligible mass. The e! tic Determine the maximum additional deflection x ecompressed springs each of ng are precompressed 50 mm. the springs dfifltfififiiflm 162 r \_ (f ‘ Emblerfi7 iggmz Two masses m, = 2 kg and m; = 1 kg are mounted on a carriage of mass mo = 10 kg and rotate about point 0 at radii r, = 10 cm and r; = 15 cm, respectively, with a constant angular velocity 9. = 60 rev/min. The angular positions of m, and m; are kept 90" apart as shown. The assembly is constrained to move horizontally and held in position with a spring k = 1000 Nlrn and a dashpot c = 25 N—s/m. Calculate the maximum displacement of the resulting steady-state vibration at O. k=1000Nlm Proble\m 5 (20%): / x The uniform solid cylinder of mass m and radius r rolls without slipping during its oscillation on the circular surface of radius R. If the motion is confined to small amplitudes 9 = 60, determine the period T of oscillations. Also determine the maximum angular velocity to of the cylinder about its mass center 6. The energy loss caused by rolling friction is negligible. (hint: for small 9, sine = 9) 9’ 74.31 EM :8 :> Za-C’os30°_ {.3q..s‘";’;oo:0 K: ZL/E ; ‘\ \;u’\\, EF {1' $3 Zwa é. apnea": K: 31:; 60° \vx' ‘2 2 EL; CC) . DE 2%:0 é DE=EF_QS€O ‘ _ L. y 2.2, Mwa_ “135—52” “.513— (T) Dmérz - PA EFY=O 79 ~ “5.5% ML r if”? .u fafi- ‘ Llf+ fiL+ Paid—[07A / J R "PA’FAfi—ZO oV (F‘_FL\H°+€L€=FQ 1‘ PL-ffiv .40 A E \ )7 b: (MW F—ufir v r‘ “3 ° T73 =m7€1+mm= tans-zoom nag—3mm 2 cost? I 1. Va 43% zQ<—x)‘-— fix = a; ska-gin 7‘ 17EH78. = 2M3!) easel + tabla-LL19 a; = -2m3L‘S‘hlé- + 7315.96039- =, éazxw—zwxfi m9 [ 60:0; PJ/ZQwsa) + m3 (572;: sing f 2[%57>a5]~ amt?) ,0 -_2Pa5/}75 66’ ytzmya. c056’d’6’ = o (t W F's/r76? :my 6050 &:/vn'/C"_9 . I ___/> // ,0 /: ftp/06644 A} a (cap/8 M no war/r by -M Car; 5: Jam? J/I/ytc #1: /ow~cr éor- remains banish/2M d cannoz‘ r‘of‘ar‘é, f/‘iaa MCou/q’ no)‘ product flfL/II/J'éf/lym‘ M“ .3: F, J 5’ '. ‘ Hal-CM, 15' = 8000 .3 some :6 1X / c . césgoo A + = I7.7.S‘f+ 1W? (“fr”) 0% = l-Q‘ 6053a°f+15<430°j = 1,277,? +' o.7§j (Ia/5e?) 133 .h .3. A '3 vb: Uf-Vc, = (a —[7_z§? ,; ’4 (IQC7 —- ILH)J :1 ~123§j+ ESE? \l-Tr/ok ’ (7.1:‘+ m—c,‘ =— 20.0%(Mlac) Tit/c. . ~'r.sc. , ‘fi,’— 19‘” 77775:]6'” ’5 _> _.\ I A A 7‘ at/b:at:—4L =- (7.518- L177)‘ -+ (0—05)”; = [.1164 ~o.7§) Vie/Ci = [3.163; 0.789” =V (“B7 (IA/“3) _. w. -1 K A [X :H 078 a ?l%° A”, [7-16, (WM; Ugo! 47:49, Diana/l 4V=o 77;: /0.€.8/-o.1§+ 2. 21.53.90.0'031: 17.11? (3') .. .. 1 ‘ ‘ 7 —~ (0 f X) + Z. E-fooo.(o.a&'+x)1= Shoo/(7; Qiflifl'k’i-‘iz‘s 0/- ‘L AV=o é V;V,= foooxisw/ex—mrsw _ sac/.7 I / W15: GWW'WK , ~ W'7 i L?" X2" . ’ aSooo 2- 5.000 Dark, 69 “PP/“‘9, X; 0.0273 (H) ZUnCMc-ffi) :— "(-45.7743 a d(r+v) no 167 77:3 1Lrwns/a‘l'c5 f. ra'lzf‘esl $0. 124; bite-4Lth (Saar/41"}. ~ mg; 4‘- 1L aylléa/cr 1.3} _,.L L. I '1 v m I Y ‘ T~ lam/5* sza'fi' ='€’“Vc+ Riff FedCJé-Qezwz a0. :‘Lw‘vl J(rnxk)y1w~ 1 c+£- H’ a: .=;mvc+?mrw 74.8 Wham V admi'. It... da’hwv 4; V: - my} ( 2*) case— (Tbi V) z d (ij V) dc axial 7W: : éCR~V)-é:+51§.;fi:0 , 6L n 22— _ _‘ _ '2? . EWWLQ , 19+ 3C£_y)‘6'-0 a Mu~/3(R_r>/Tzlfl/§j 19-: 610$ng+ % 51:904.),caswhf- 9 15¢qu = 6:50., vwfii' “2:: f?“ arr X r amax“ 3(E‘r> :' T ZéK—V) 2_ .u 0M" :- gfin (2"é'};’?6'4€“+ mgcrfi-rfi. ‘ALGw‘nfLL—"éo “WNW—WM“ w—m. W ...
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Civl113Final95 - {gait 113 LI 159 128 NOV199§ /,...

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