Question

All members have constant El B.M. to be evaluated in terms of w or PL, H 2t/m solve by consistent deformation 7 m 1.8m 9m 1.8m

Answer 1

Given atim am ym ym DTTI ATU حي 1.8m 1.8m of the Given structure The static indeterminacy No of unknown Reaction - 3 is 6-3 - 3 24 m B. het, the unknown reactions be Qg, de 8 03 PT 92 A 33 02 know that the by castigliano's theorem deflection is given by, 6 da Ma.mx EI where, is the moment due to actual load & ma is the moment due to unit load applied

Consider Section at a between distance from 2 tim a -X x varies from 0 to 7 m. 82=1 unit 33=2 unit = 1 unit here, My = 0 KN-m (moment due to actual load) 1x0=o (due to unit load orf Ba ) en directic ma load en direction ma (due to unit of 82) in direction mz 1 (due to anit moment of B3) consider section at a distance Similarily, s from C. 2tim a varies from o tog m. А 8

3. Here, Ma? 2 3.24+x2 1 + 2x1.8 2 + 2x x3 2 ma 2 m2 - 7 tom mz 1 t-m. distance at a section a Now Consider section from Bi n 2 tlm a a varies from o to im DD A 2 Mas 2x9 81 tom. my - 9 t-m tom - (7x) mz 1 t-m

M3 Ma m2 mi - Member Origin Length 1 D o-7 -7 1 0-9 3.24+x² x С CB -(7x) 1 07 91 9 BA 12 Now, the flexiblity matrix F da. mi. mj € I ce 8l.da 810 Ст. a2 da ст EI -7x.dx Fr= Fal + 9 (-7+x) .da GI CI 7 -7% + 9 [-7x + 22 2 2 504 EI

F23 - For x.da t a da GG EI 7 9x a3 L - 1035 EI LEI 7 F22 x2 da t 49 da + (7x)? da. СТ СТ СТ 7 + 49847 + Í (70 BEI 3 - 669.66 EI F2z = F32 = 1 (-770.doc -x.dx t CI -7da GI 2 7 9 (-7) + [ -724 201 ET + (-49+245) - 2 4.5 EI ET EI EE १ F33 - I do + da t da EI G 23 EI

103.5 810 [f] = 504 669-66 12 -504 EI 23 103.5 -112 Now, the matrix due to actual loading Say COOL] DoLi - Ma. Mi. da. I 1 + DOLL - (3.24+x2) x da EI 8189 de ca 6874.47 EI 7 818 (-7+x) de t 13.24 +x²) C-7) da DOL2 - EI GI -388 9.62 - 1905-12 EI - 1984.5 ET + Dolz - (3.24+x2) da Si da E I EL 839.16 EI

[Dec] = 1 ET 6874-47 3889.62 839.16 at the There no deflection of Rotation Joint Di say cos Dis place Joint matrix [DS] = O [8]) Displacement matrix (say [8] = [08] - [DOL 6874-47 [8] - I GI - 3889.62 839-16 we know the compatiblity equation, EFI [0] - [8] [0]= - [F] -1 is We know, [ F] -1 = [ Ady E where, (Ady F] = adjoint of Matrix f If = Determinant of [F].

3 1035 {F} = 810 soy -504 669-66 -112 -112 EL 103.5 23 2858 18 - {ady f] - -12861.81 38556 28840 86 7917.75 38556 EI) 12861.81 (Fla $80x 2858.18 -103.5X12861.813 181)-3 IF = 983928.465 (EZ) [ F] -1 -0.01307 Oro0290 C & I 0.00805 0.03918 0.29312 -0.01307 0.03918 [o] [F] 2 [s [0]= EI 0.00290 -0.013 07 0.00805 0.03918 -6874.47 GI 3889-62 CI --839.16 EI -0.01307 003918 0.29312 - 8.968 KN t or -1.567 KN t -3.73 t-m

- 8.968t (1) O = 1.567t (e) 3.73 ton (5) Og - Net vertical force + Vp 2x 12.6 % 25.2- 8.968 V = 16.232 t (1) . Net horizontal force =0 HATH = 0 1.567€ (-) HAS Moment at A = (1.56787+2x92 – 8.968 x 9) 2? 11-257 t-m.
To solve using consistent deformation method, at first the static indeterminacy is determined. After that the flexiblity matrix and deflection matrix is developed. Hence, the compatiblity equation is used to get the unknowns.

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