20--CE215MT#1 2

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Mechanical Engineering

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Oct 30, 2023

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13354} Tovo firces act on the hook, F; makes an angle of 1207, 887, " respectively w a . . Nzcesne® Pzosco’ Moty the x- v and z- axas. Find 1} {Opt} the angle ¥ which Fromishes with the v-axis ((,es 1*) ‘¢ (corea®)® ' o > . . . L . S = 1y {Spi) the umt vector in the direction of &), 7 l ' - z 7 e=sy)t Y= 45° tion (s vector!) of & along the direction of Fi, and 1 {1ph) the pr v} (Tipt) the angle @ between the two forves b, ney # 65 poon Cofy -9 [umotry-toe] - [, AT E:] [: - §0] [ 0.g '”"J‘b”fil d. £ -r, ¢ ’,F‘";’J 6 @ Ve | A1 Roasg f?.l..J_“x“”’ 2 - f-soo\ ¢ 300 ?'.‘5 t+z‘-zb‘.] 4 r |z°'l + QOJ -3 .KJ F;*“Ok[ogm‘, + Cot 60) 'HB“EI > + 9o D= M50 xfepe T EOOX T1ots 40t 40 rcany [chls" n.m',] Cos B = -300 Xlza 4 zg0x Y0 -229. 5, oty GO x 13 Fl ° 'a.;:’blj "‘u.ul P [SS.)[, 'm‘j "‘\L] &z 4 Ll el TET R DR ST YT fy 8 N e SN ey 8 oL & WY e o s o8 N \ Mg & & . Ly T oaD 3 B A “S ¥ §"3\‘;‘:’Q{> S a‘t\f a3 'i.s‘xf“‘\.}g 5“\’3‘»’?2 ) b d— & i PRI R wia g L] = o >-0 bos Ty res - . .’.’ 3 RS “ 4 L 1 r..‘.’ ! o K h “% g sn £l el i by Detarraine the trtemnad normal force, shear force, and moment at poird: Cin thee simply suppcred Heam. i 4 kN /m R e a DA TRIRTRTY. W) PROBO?_006.jpy N P I N 3 T H M Q 3 e S i B i : T “ . - X A& 3w b Im 5 AtC: (3ft from A) L +2=24 ¥, =04k M, +2(2)~24(3) =0 M, =72-4=32k-ft |-4] (tell me where you take momeggs!) A 3)(34pt)The simply-supported beam AB is loaded by 2 Point forces as shown. i} (12pt) Find the end reactions at A and B. ii) (22pt) Find the internat force and moment reactions at the pointsC and D: M, C-3ftfromd and D—]ftfromB. v [ KEN X, 2.4k H H § At D: (1t from B) . H V,+1.6=0 v, =-16k faan M — ‘ N 4 S 7 M, + L1617 =0 M, =1k-R 4] 2 Ly ~ - Lom M, BV\ T“ (ngain you nust tefl me where you {ake mnoments) * : “o4 1.6k o 38 Using the left side of the bar: 2k 2k V,+2+2=24 l‘ 4,8, v, =-16k L N My —V,(90-21)-X7TH=0 (where?) 7 P -20Y-2ATY=0 TZAk ¥, M, (longer A +B =4 M, =—1.609")+16 =16k -1t » T Dy eqns) A4,(10) = 2(9) +2(3)=24 A, =24k, B =16k 2]

< . o ¢ A - §‘\ PR w:}efi.w& af fess :‘.\y 20 e > - > 8 N N 2 AN A Y VAN RS W ATHRE SN VA hasy © Y T - T PR o 53 3 % WEEOR TS AT ASRIY Y arb 2 \a{{blcas 6 1 e iy Sy polled flew non N b 3 T Ky & O3 S L < P AR SAatle® YL s . R o . 3 NY ®a 3 SRR WA QPRAS T R FTN RN oy vy OPeY pin MLV dioimmven s o ¥ Y Con ov) SR Sl A o N RANTTA IS N sxsu“ kY WRARDTR N} . N TR \\ & Jtugstion: Wsing the mathed o sections, detarmine the forcez in members 49, 15, and 13 the following truss strecture: G ST M= N i L i L' : (35 m{TRIN} - (Bmify =0 s N Vv g oy == 22,5k ’ufi‘(‘iuo> W force acting in e t] boof the membe he truss bridge si i . . N JEMw cach member s in ! BN 1) Bolve for reantions using system Eg Eg'’s CATRIeRIETS RN - oo QUIMDBIession, i ~ PRFOFR=DIM =0 aum ’ % + Draw system FBD % G P sy oodens « SumforcesinX ¥ ‘L * Sum moments about point & _ with most unknown reaction forces & Ih= Tz Q s . ' s T aToy <bo -0 = 2} Solve for each joint summing forces in ¥y = FarFey bo-90 2 Q only ¥ and ¥ (no moment) » Draw a FBD at each joint « Sumforces in (f My = (Fey)(2e) ~{t)ie) - {83 20) = 0 \ FF‘{ - (L’ﬂ(ﬂ‘);’b"’ikl‘ﬂ 7) 33 hN ' ¥ at each joint » Piok e joint whers you have at least 1 known andg no 7 P s lew%a-T7333 = LG0T W mare than 2 uaknown forces » H you assume afl forces arz in lension, then if it somas aut negstive, IU's in cqilyrassion W Sy = W 4T + Fee on (39) = @] F\j = %G]*‘EficﬂFchV\L}o> Q RS 2 ’ Q (5 = T F 2y =0 \ F r"fi Fac e (e 3y F‘&D = -\§3.07 - (H.Su)matlﬂ = ~70\ A%l Vot 14 ZF\’ = b L7~ Faesm (1°} =0 " ST 7853 - Fepsmizn) = Q _ . vy EF . 5 Fre - [ \a% 33 e 3.3 \ Smize) . . Fer * Sugs = 2 N . i ' Fag = - o) (19993 = — 1837w e Tart VD X elitian C S¥y = 29431 Foe = O = =1 Venlze) * Fee = O ! - Eo= W80 & s g O £h me \ 2Fy =% -Tox = O Fs v ST, = MSds(E * Fee =0 < @ o ’& Fre = 129.9% W0 T o T * 7 { ' . , . - 4 F?sr, = bt w O v bee = 193,93 cos(ea) = ‘_:55 1 _k_h} ',Z'\,(_ \ F__i;‘w Fep~ zoviiwd & b T S Y CO R T Fap = Yol T Fog = - BokN Cog = 8300w T ) , s - . T - | 1)The truss is loaded as shown.Using both the method of Joint & Section: Fog = 32%¢ & i) (2.5pt) Find the reactions at the supports 4 and G. Foe = zolban C il) Find the forces in members CB, €D, EB, EC andEF (1.5pt per member) Ferp = 200 8V and (MUST) state if the members are in Tension or Compression i) ¥FBD 5 : E T, 0845 =10+5=15kN Meth ctioms o - SKN e D : if) Q= i " ; ZM@'b 7”(? 77777777777777 C E l E To#)-5H =0 [T, =5 1N 10KN ———> B ettt : Swase | /] cat REOMApssEsRS ey R AT +3)(4) = 3(3) - 10(4) =0 - Ty T Ty Ty : (T +5)=20 Ty =-25 ; I T Ay 4 le =5kN N Clp SM@G=0 ~A (4)~5(12)~10(8)~10(4) = 0 . l A, -—4Ay =604 80-+40=180 > A G, s A, ==45kN 14, =45k |G, = 506N T| A +5+10+10=0 . A, =-25KN | A, =25kN <

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