Final Exam Summer 07 Solutions

Final Exam Summer 07 Solutions - KEY AEM 201: Final Exam...

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Unformatted text preview: KEY AEM 201: Final Exam NAME: CWID: 1. (20 points) The 50-kg block rests on the horizontal surface, and a force P = 200 N, whose direction can be varied, is applied to the block. (a) If the block begins to slip when 6 is reduced to 30°, calculate the static coefficient of friction p3 between the block and the surface. (b) If P is applied with 6 = 45°, calculate the friction force F. W: m3 = (5°K3X‘IQIM/g‘) 1': 4‘1 0.5'N—5 \lo 0 2F=N“W“PS"“3O o ‘N:W-\'P$in3oo= +QOIS+ZOOgM30 (N) 0 Fees 30 Pcos‘lfl‘3 _ 200 COC‘JDO 5°: 0.5" fl 8 = .933 F = Prado .-.-.. man! Md? =,7fi33l9 5; M Z. I co 4 °—_—. 200 c09+§°=i+mzl 2. (20 points) Cafculate the moment of inertia and radius of gyration about the centroidal x-axis for the shaded area shown. a) Red-«nyk 80* 60 (+) ®fuaficv “C‘\rc\€ ) V: j 0 6—) ® +v§¢vfis‘e , 30’4ovm (‘3 (4‘; 90,40: +Boomm“ 1 '- git-11' (30}12370‘9.e§g WM: m 30. m 5‘an 3 3—11” 1 Y = m: ‘43’000mm :_ MM ZR 34°|3.\4Mm" V : EL h -.: L—L——o* 5"” '2 21°F“) "W1 in 3413 W V 3c—2.1.q4\3=- .0597W‘1 L3? = 4".ZL‘IL—l‘flfl‘lrl3-a \‘r.3u3...... “\T = 10- 131mm; -.-. ammo «m I ’ ____ gs'oluoS‘ m». 4. x ' 4- 1"“; ggo’ooomm z. I , 3 K): B" x 4 Ix! BSOIHoSMr-v - . h, M \<¥ ' A. ‘4 run" 2 l§.6007 MM 3. (10 points) If it is known that the center pin A supports 1/2 of the vertical loading shown, determine the force in truss member BF. State also whether BF is in compression or tension. The notation “6 panels at 12 m" means, for example, that CD = 12 m. Wm.....~—fi panels at 12 n1m—.._...:m.mw_..1 I V a.‘ £60“ ' o A A; “5 .. . . . ... . -- =_ 4x9+>9do . “MEL. ..... = s m 8 kN 1o MEISMN wim = SH.” 2 2. OK” C7 H Sx'wnme’rvw —-> A G, 3 H '1: ‘3 F35 TO\¢\+ P“ I ‘ (D AB a “‘9‘: $2Fx=EFAB”\)-‘EFAF= ‘ - ' Fm: = FM” .. 0 2M“) T+EF7= 7J0 ““ +1th“3=£‘=91=’ ‘ F 6-: Mb :-(Hom)F 3.: ‘L (43M)(‘3\<N)- (11m){\o\<~) "" (24m)(3kN)-—- (“Nam RN) + ham (19") x; 0 Wm, +0 A (48KB) -. (~11.(\,b)4(24)(8)- (36}(9)+{)\il3fi(7—9) W i Q ——- ’L $.13 H 4. (20 points) (a) Draw the shear and bending moment diagrams for the beam. (b) Determine the maximum absolute values of the shear and bending moment and their respective locations relative to point A. (c) Find the bending moment M at x = 3.5 m, i.e., 0.5 m from the leftiend. 8 kN 3 kN “‘5”- ZUkN-an ac+ S< 23.5'W1 3 "WE? ‘ ......... .mpL—ji...0‘75 1 111 mm... .._i.... 0.25 m gs; Mn: 49m)— zo «mm +3!) —35W= 0 (mm +1.0 «aim-i; s (is) b = 3 .5: 32.\667 \w 0‘ - - _ - 3mm T+ZFY .. 97 ‘3 ‘3 “3+ :LO fl = -\. WW kN ._‘L~_____. A‘s (44mm: —i.m n1: (-QJUCZfl = "2'73 93 -_ (—axrxns) =— - 6-93 fist—mum: -\‘1.i7 a“ 3; (isim -= 15' o KN- w‘ 5. (20 points) The welded tubular structure is secured to the horizontal x-y plane by a ball-and-socket joint at A and receives support from the loose-fitting ring at B (By = 0). Under the action of the 2-kN load in the y-direction, rotation about a line from A to B is prevented by the cable CD (this is a big hintt), where point D lies in the x-y plane, and the structure is stable in the position shown. Neglect the weight of the structure compared with the applied load and determine the tension T in the cable, the reaction components at the ring B, and the reaction components at A. F: 1 ~ k“ =mo°7+m x = 4's.- .- ~cb ‘cm— V‘rcls —— ‘1— A A “- I :1 Z‘c‘,“ fisawzs‘i—ck) K” :-.. t‘ . ‘5' ( A .. ~wn I +Z'5i h) KB/fl=4.53+cl< (M) ICE/9‘1'Stl—GK0") ZMAB: bAB’ 2(1'*.E\= O a: O 2:..31- (rt, *1: Surf»? _. T __ -| O. .4. 2.5’ '25 L 250 __ ’fifi: ['6‘1-Q‘f'gl‘1 2.sB+E'(°’0 Ol+'8!0 LU’O H ...— q%§E-.L)CG) 4—.3 {—2.563} +905) w-O T: —+’ 46.15 12.93364’ KN Zf-GKG) Jr ABC-1'3) + ' __ 0:32.333“ _ ¢ _(e)(2.e)m) " [+53% 348.15 ‘2] + l [6 Ex 546.25 1 ii-“ -(W = J“ [ 3" m *5] O <~ (61(23st ‘- 7'? Ba: fmw—ML '=- 4.05'5'5'6 Bzz4'ot’KM A =7 13 ._.. 2.93304 imam B =.‘H7 KN‘ ’Q +6.7: A (7.5)(23 u H) _ 5 - k '7 Bx': “"5 :AwWJ +~S 10 ; i ! ’azg: l-+_.\f3;4——*\"§%§=—"~—'4‘o‘sisé' 6. (10 points) Determine the resuitant force vector R of the three forces and two couples shown. Find the coordinate x of the point on the x-axis through which R passes. Find the coordinate y on the y_—axis through which R passes. Determine the equation for the line of action of R. ZkN y ' i . mo [email protected] ifimfl mm “#400 mmfl F . : R: i.53~33~1’§ (W) “V -- BkN g: M A = -,|01)-v:¢80 - (11.5).» (31(3) : R, *‘(LSXRA =— -- va (Wm) Shape Triangular urea Quarter-cirmllar area Semicirwlar area Quarter-viliptical ama Smilieliilfl ilflll area i» ; Smnipambniic ; :u'cn EWWWA”.M.7H.W......__..___._‘_A-“. l’a ralmiic area andufiic spandrcl General spandrel Circular sector 5.1m Camtoids of common shapes of areas‘ 13 Bectangie l. = MAN + 112} l Triangle Circle Stfl‘l‘ll't‘ircké Quarter birch: m fl¥7mb3 Eilipse '. 3 . “I.” m é‘amab [a = i—mlbmg +122) Fig. 9.12 Moments of inefl‘sa of common geememc sshapes. 14 ...
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This test prep was uploaded on 04/08/2008 for the course AEM 201 taught by Professor Freeman during the Spring '08 term at Alabama.

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Final Exam Summer 07 Solutions - KEY AEM 201: Final Exam...

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