Question

Use the stiffness method to analyse the elastic frame ABC shown below. Use a model made up of 2 the elements (AB and CB) and the axis indicated in the figure. All members have the following properties: E = 2 -10% kPa, A = 0.005 m², 1 = 1.5e - 4 m. Also the lengths of the elements are the same: AB = BC = L = 3.1 m and 6 = 45 kN/m. ות 0 B 3 2 x Structure, loading, member axes Degrees of freedom Calculate the the bending moment at Joint A, following the steps below: Part 1: Assemble the global structure stiffness matrix. Calculate: a) K11 = b) K22 = c) K33

Answer 1

338) FRIDAY W = 45kN/m 2 B 18 3 ă . DOF 2x10f 2 A - 0005m Given! E kla, .4 I m? 1.5 XI0 AB=BC = L 3.im, W =45kN/m. Fid F2d boss nd Mid -M₂d Element shi V 02 [ki] EI 12 matrix for o V2 6l -12 6d V 41² -6lze2 lol -bl -6l 2 212 -ol be - 12 12 6l 442 Jo, 2 EI 4 2x108 x 1.5x10 EI - 9.677 X10 31X3) *1*103 li l2² 3.13

रात्र ht= Би: та matrix EDAY from above 9-81 - 9-81 hn.se 9:31 - Ел 71 ?i : 2. 81- 38:2L - 12 2 аг b1 9. 31- 9-81 9-81 - 1 2 ht |-12 Іго 2.81 - hh. 8 € 2:31 19 11:19. 31- анх10 18. 0 12 3 а 9 81 1- 16 Кл 10 чл. (4)* L'я) = [я] mataidi TYPutiн ча, 9 hp - 82 2.У) - 14. 22 9:8) бл 2:31 - 12. 9 - 31- 래 82 xhxlos ісі. 2 2 2:31 hh: 8 18 - 9:31 "л 12. 2:5! = (Я) 1л 2:31 - 14:22 9-81 *о Гһһ• $ € 2-81- 5 0 1 X EE-E x'8) 12:01 2.31 9: 2:31 - 7- hh. 8 9.81 9.81 ܐܐ - [а] :* д

Shear force at node o fi d'= Fz d = wli 45x31 69.75 IN 2 Bending moment at node Mid Mzd we, - - 45 X 3.1 36.037 kNm 12 12 for Resultant Nodal force & moment is: fi F₂ M₂ fit fid + Mid Ft Fad M2 +12d F3 12 : Ma - fid 69.75 IN. Mid- - 36.037 INm Fed= 69. 75 kn M2 da 36.037 kNm: [fi-69.75 Mi-36.037 Fz 69.75 M2 + 36.037 F3 Mz F -

. [K] [w] leh ✓ , 9-81 14.2.2 0 Fi-69-7 M. 16037 69.75 2 + 26.057 . 24 180 41 186 381 V2 O2 02 18:6 19.12 100 180 1810 109 18.6 19.22 2.31- hn. si condition, Applying boundary V?=0 Oi=0 Oz = 0 FR M₂ = 0 7 4.00

103 24 D V2 02 -69.75 36.037 O 76.88 24x10²x W2 = -69.75 -2.91 X ID V2 -2.9 mm 10² x 76.88 O2 Oz 36.037 0.4687 x 10-3 rad 103 (-12 V2 +18.602) fi- 69.75 10? (12 x(-2.98.10^3) + 18.6 x0.4687610-3 ]: F1-69.75 fi= 113.268 CN 3 - 10 103 -18.6 V₂ + 19.220 2 ): M, - 36.037 [-18.66 (-2.9x1013) + 19.22*0.4687x10°3] : mi= 99.16 kNm Mi-36.037 10? [-12V2 -18.602], F; Li 103(-12* (-2.9 X10-3) 18.6 x0.4687x10-3] - Ez f3= 26.082 KN

- M3 w 103 10² [ 18.602 + 14.22 02] [18:6 x (-2.9 x103) 112 + 19.22 xo.u687x16 o 3] - My yu.93 cum PAS PB Pc 113.26 IN IN MA : 99.16 INm M1132 0 kNm Mc -uu.93 lerim. 26.08 IN u,a Om - 42 2.9 x 103 2 0 2 o rad 0.4687x10? rad o rad Uz = Om 3