Question

ft 101 -2 ft (equal to I- twoforce member) B₂C 86.1= Hsin 45 By 40 het B DI 2 11 30944 OG 11 E 40lbs. EMA- Icos 3016 tan 30) - 4(40) I - 160 53.3 lbs. .516) 311 Ay Ix< А. VIY Reaction The hanging load has a weight of 40 lbs. Determine: Magnitude x-component lbs y-component bs 90 lbs 26.7 lbs. 86. 1 53. 3lbs. al.7 lbs. 4. 121.8 lbs. 86.1 lbs. 53.3lbs 26.7 lbs. 46.lv B H 1 Determine: Ne Ve Me Internal Equilibrium FBD @ E (bottom): 86.1 lbs. C 26.7 lbs 80.1 lb ft Internal Equilibrium FBD @ E (top): Ne -86.1 2. Me Ve : 26.7 EMEF Ax(3) – Mg =0 ME - 80.1 lbf • Ax 26.7 Ay 86.1

Answer 1

Solution. F. BID of the Problem, aft eft Y(A) B + X(+) 2 ft 30 13t 40 lbs. E 3ft 30 2+1+3=6ft and to fan 30 50; I In Ax A b=storm 30° tay fig (a) 6 = 3.46uft b 160 3.464 According to fig (a) Taking Moment about A; considering 2 +(Ty xb) - (4x4) Ту= Ty = 46.189 lbs (1) Ify=0; - Iy it Ay - 40=0 Ay= 40 + 46.189 Ay = 86-189 lbs (1)

Joint I, Now dreawing F. B.D of along with cable for 'BI FBI Cosso у = FT Cos30° I FBI Efy = 0; & FBI FBI COS 30 - Ty=0 FBI singo FBI Ty Ty fig (6) cos30° 46.189 Cos36° FBI= 53.334 lbs Efx=0; – IX & FBI Simbo=0 Ix = 53.334 sin30° Ix = 26.667 lbs (4) Note: All the support reaction directions are assumed nitially so respect to positive sign of outcome awe can say assumed direction is alright else negative outcome will signify opposite of our assumption. again respect to fig (a) - IX + Ax=0 Ow2 Efa=0; Ax=IX Ax= 26 667 lbs (

30, Reaction at A: RA= A² iany? RA= √ 26.667²4 86-1892 RA= 90.22 lbs Ax Ay RA Reaction at I: so, 2 RI= 46-1892+ 26.667 RI = 53.334 lbs 46-189 RI 26.667 BI part Based on pin joint at B Separating the strincture into ABC part. and Bry By B 3 Bx H (G 40lbs fig (13) fin (d) fin (ABC) I A 26*667lb 26.667 lb 86.189 46.189 lbs ebs,

e) Efyzo In fig (c) If x=0 implies. - 26.667 Bx=0 Bx=26.667 lbs (in BiI part so must be m ABC part implies By 460189=0 By= 46.189 lbs Coin BI part So must be & in ABC part Reaction at Bi RB= V Bn²+ By? 26.667² +46.189 ?B= 53.334 lbs Bc) RB V

Considering fig (d); separating the figure into part BHC aft aft and BGA Brd Bx By in (Fort), fig (Bite) Bal Biol ESH 40lbs 8 Bix Fish co (Fon) y = For Soad (FGH), Fan Cosa A 2 tand= 2 / a=tant (1) Toning fiz (Bitc) a=45° Moment about Bi considering 26 fing (BGA) + (Tom) x 2 ~(rox 4) ao sina = 160 Fut- 80 Sinus 113. 137 lbs so (FGH) FGH Cosus = 113.137 cosus (Fort) x = 80 lbs and (Fony = F G H sinus = 113-1378, 245 (FH) y = And And GH 80 lbs,

Now concentrate on joint it B H& Hy HY FGH), يا (Ently (Hy) = (FG 4) x So Hy=80lbs in BHC so in CrH :) so, Hy=80lbs (in tim BHC. (Hy) = (FGH) y so & in bafet) hence RH = -√8024802 R4 = 113.137163 (Reaction atH)

Internal Equilibrium FBD E (bottom) B H +40lbs E 1 A * 26:667 lbs 26-6 67 وطا 86.189 lbs 46.189 ebs ME AA For Axial Forece; VE NE + 86-189=0 NE 86.189 lbs (so negative sign indicating its in & direction) 26.667lbs For Shear forece; VE+ 26.667 =0 186.189 VE= -26.667lbs ebs it must beint direction) for Moment, 24 ME + (26.667*3) = 0. ME = -80.001 dos-ft, (so it must be in a direction) Cso,

F.B.DE (top) Internal Equilibrium 25t B ++ 2ft 4006 E ME Fore Axial Force; 26-667 وفق 146.189 lbs #VE 3ft NE -40 - 46-189 zo NE= - 86.189 lbs. (so it must be 3.464 St in & direction) For Shear force; - 26.667 VE = 0 VE = -26.667 lbs (It must be in direction) For Moments 30 (46.189 x 3.464)- (26.667*3) - (40x4) - Me=0 ME= -80.000 lbs-It (So It must be in a direction)