Question

Q1: For the frame shown in the Fig. (1), check the determinacy and stability then draw the axial force, shearing force, and bending moment diagrams for the frame. 1.2 k/Ft Pin 25k 15 15k B 20 ft 15 ft 30 ft

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Answer 1

Solution 1.2kft 25k 15 ft 13 15k 15h HA HD RA Ro into CD so If number of unknown number of equations reactions then stincture is determinale as well as stable. Based on that hinge we can separate the strencture parts. So, the parts will be ABC member and member. since there are two members each member will hold 3 equilibrium equations de 3x2=6 nos, of equation are there. a vertical and Now due to hinge horizontal reaction will be developed on both ABC and CD after separation Hence there are 2 reations + 2 reactions in support A + 2renctions in D hinge = 6 nos of reaction. Cunknown)

thereforce nos of reeaction = nos. of equation implying the structure is stable and determinate, Support Reaction calculation tcy Moment about e; 15ft Considering 2 -(15x15) + (H *30)=0 ca ca 15K Ho= 7.5k () If x=0; Cx - 15+ Ho=o C = 15-75 (n=754 in CD so ABC) HD mnet be in Ro 1.2klift Ben A Cx (30-20) = loft chco 25k B Moment about A; considering 26 - (1-2*30*39) - (25*20) +((x+30) + (Cyx3=0 30cy= 540+500 (7.5x30) 2oft HA cy= 815 30 3oft Ca= 27.167k (A in ABC so & inco) IFx=0; HA +25-(x=0 HA=25-75 Ify=0; RX-(1.2X30) + Cy=0 HA= 17.5K (4) RA= 36-27-167 RA= 8.833K (*)

Now in main strencture; { Fy = 0; RA - (1.2830) + Ro=0 Ro=36-8-833 Rb= 27.167K (A) Note: At hinges moment is always zero. Now we will separate each member and analyse part by part. since rigid joint so that a moment will transferred from AB end to be end. B is be Member AB MB Bx 25k B 2oft_o A Ef=0; 25-Bx-17-5=0 By = 7.5k (& in AB in BC) {fy=0; -By +8.833=0 By=8.833k (tim AB f in BC) Moment about A; considoing ? • (25*20) + (Bx*26) + MB=0 8.833K = 500 - (7.5720) 350 h-SE (4 in AB 2 in Be) Taking section 0-O; considering Left (lower) portion (Osu220) (xis taken from A) & 17.5K MB MB

Ma Va A my 17.su For axial force; Nx+8.833 =0 Nx = -8.833 at x=0; NA= -8-8334 at a= 20; No = - 8-8334 For shear forces Vx17.5=0 Vx lzs at x=0 VA= IZSK at 2:20; VB = 17.gu For Moment; ? Mx - (17.5x2) =0 Mrx = 17-52 at MA=0 u.ft at x=20; MB = 350 u.ft S 8.833k 2=0 12.5k B -8.836 B 3500-ft B (+) C +) on-ft A -8.833k A A 17.5k SFD AFD BMD

Member BC: (102x30)=36k x 1.24lift - righe 36sino 1. 27-1674 350u.ft 75k. B loft @ 8.833K +36coso 30 COLO BC 30 BC= COSE BCF 30 Cos 18-435 3oft Along forme = 10 horizontal Direction total forice= 36 sin 18.435 O= fant (1880) O=18.435° in udl; total force length 36 Sin 18.435 30 Cos18-435 1.2 sin (8-43SC0318.435 102 sin (2x18.435) = 0.6sin 36-87° Along Vertical Direction total. forece = 36 cos 18.435 In ude, total force 36 CosI8-435 30 length Cos 18.45 1.2 Cos² 18.435

30% Cos 18.45 7-scose 2 -7.54 zssino | 1.2003²18.435 27-164 sino Z.sino 350 k-SY 0.6sin 36.87° 27.167 Z.5k e e 27.164 coso B 2 75cost 8-833 sino 8-833 18.833 coso Tahing section - Considering Left portion of it ( okaz 30 (a is taken from B) Cos18-435 Mx 7.5sine -Na (+) 1.2 Cos²18.435 G x(f) Vx 350 +t 4 B 0.6 Sin3687 7.5 Cose 8.833 sino i 8833 Cose

The =0 Nx= . For Axial forece; Z5 Co s 18-435 +8.833 Sin 18.435 - (0.65 in 36-87 **) = (0.6 sin 36-87x) 7.5cos 18-435 - 8.833 sin 18.435° So, at x=0 NB -9.908k at x= 30 Cos 18.435 ; Ne = 1.47.6.k for shear force; 75 Sim/Q.435' + 8-8 33 C0518.435° – (.2008218.435°x2) - Vy=0 8.833 cos 18.435 75 sin 18.435-(1.2008 18.438x) So, VB= 6.008k Ve = - 28. 145k Vn - s at x=0 30 at x = cos18.us For Bending Moment; -350 +(75 Sin 18435*2) - (8 833 cos 18-435x2) +(1.2 co3? 18.435** *22) + More to Mx = (8.833 (0518.435-7.5 in 18435)2 1.2 cos? 18.435.22 So, at x =0 MB = 350 u-ft ме ok-fe + 350 at 2 = 30 Cos18.435°

sam, at nax'; Vx=0 8.833 cos 18.435 - 7.5 sin 18435 - 1.2005218.4352' so, ,وہ or 1.2003²18-435n = 8.83320s 18-435-7.5sim 18-435 x = 5.563ft at n=5.563 cie a=5.563 ft Mx=(8.833 cos 18.435 - 7.5 sim 18.458) 5.563 1.2.0032 18.435 ( 515424) +350 Mr = 366.711 koft.

чти AFD - 9.9084 6.ook - 28. {usu 5.563e SFD 366 tuk-fe 350 и-де Louift БНР 5.5% 3

member: 27 16th 7.5k c 15ft 3 Elt4 15k 15 SE 75k 27.167k Taking section ② - 3 ; considering Left (upper) portion of it (ozne 15) (x is taken from )

2 Vx Max Nx so, 2=0 at xão; Mc = o unft at 2=15; ME= 112.5 n-It Mr = Fore Axial Force; 27-167k -27167 - Nx=0 7.5k Nx=-27167 So, at 2=0, Ne= -27.1674 at x=15 NE = -27.1674 For shear forrce; 7.5-Vx=0 Vr = 7.5 at Ve= 7.54 at 2=15 Ve= 754 For Moment; Mx (75xx) =( 7.50 Taking section 6 G Considering right portion of it (osna15) (x is taken from E) For Axial Force; Na +27.16750 (15-x) Ne = -27.1674 So, at x=0 Ne s. - 270167K 1 27. 167k at x=15, ND = -27.167k Fort Shear Force; Vx+7520 Vx= -75 so, at 2=0 VE - 7.5k at ze=15, Vo = 7.5k Nxi Mawar 1.5k > >

For Moment; ? -Mx +7.5(15-x) =0 Mr = 7.5(150) so, at n=0 ME = 112.5 u-ft at =IS, MD = ou-ft -2746tk c 7.ru ) - 7.5h (+) / 112.5 unft 7.5u A - 7.54 D -27.16IK SEO BMD AFD

For full frame: 1-4764 C -24167 u B 17-8-8334 1-9.9ogu AFD -8-8334 А -27167 Z.ru + 6.0084 175u 1-28.1454 B -754 7.gu SED A 17.5n 366.11 -7.50 350 unft (+) C BMD 350 u-ft B + (+) 112.5 noft A D (Answer)