Question

Use the stiffness method to analyse the structure shown below. For the beam ABC, E = 2 -108 kPa, A = 0,1 = 1.2e - 4 mº.. For the truss member DB, E = 200000000 kPa, A = 0.002 m². Also, take L = 6.5 m and o = 41 kN/m. 00 2 B с TIIL TE 3 Degrees of freedom D 2L Calculate the the bending moment at Joint B following the steps below: Part 1: Assemble the global structure stiffness matrix. Note that ABC is infinitely rigid in the axial direction. Calculate: a) Kui b) K22 = c) K33 d) K44 = Part 2: Specify the load vector. Drovide your anewore with two decimal lange

Answer 1

w c B VEL 2L L k Fig(1) given structure with loading تول A D Degree of Freedom stiffnen Matrix - Apply unit rotation at A ( 09=1) to get the Ist column of stiffnen Matsin 2E I _ k31 Daz1 3* L HEIKH L GEL okai

-GEI 2² K3 =2€ I, kuio ku = HEZI: Veysel get Apply unit displacement at B (461) to the Selond column of stiffness Mätsin. =k334 бет GEI. kze (242 رح B K GEI 2 .=kig GEI IEI B .kzz ILEI (2233 (242 K42 "AE sinto =kzz ↑ 122 I upal o=45 val Sino alla Sinto=%. 00 K12=-6EI AE Kzz = 12EI L² + + -Sino; IEI CE (24)3 EL K23=k32 = 6E I LEI, Kzy z Kyz = GEI (24² (262 sotation at B (86=1) to get stiffness Matrix. k13 ik33 (22) =kzz Apply unit the thoud column of GEL, K23 -K23 3

LEI LET + k13 = 2E 1, K23 (zipi kzz 20 Kus = 2E I ; Apply (621) to get the unit sotation at Pourth column of stiffness Matsin. UEI = kyy 2L 2013 kau 22 DAT 6E1 (22= kay kizu ZEI kyy LEI ZE ZL kuto; Kzu = 6 EI (26) 2 یا ple 4 B As اما WL 2 132 P 1 ALL 2-1912 2 w(24) 2 W(242 12 12 قا 21 с 27 B lig W(21) 19(21) P" 1012); P3 5+ W124) 2 P =-w24? 12

- P=PP اوا +WL 번 3 760. لایا ㅍ 3 @3 to L2 뺨 given the For the beam ABS, E=24108 kolm? I= 1.2x16" , For truss Member E= 2x1 kWh A = 0002m) L = 6.5m; w = 411 kN/m. (ay ki = UEI uy 101215 LL - 45 192000) 13 476 23077 2 (①) k22 12EI +3EI 2.13 + AE - 따 2. 구 X2X108 12.3 2259 37 28\2x16. 6.co2) 1.53 2x (653 2. = 90 155 2306 Sol 02 605 처 10 K3 =4년 0220는, 62세로 하리해 133 - 92153.15. ES

(0) Куч 2х 2-4.8 ) 2x 1.9 с чET 2. 96ooo 13 C5 2 1 384:5335 </х «5- - 144 •354 (2. ~ 91- Р: -399,759 (2- — 3 List 2. L933 ,ос25 یا -3х1х4.5 2 - Их Бs- Ч +41xt •5- 3 ч tur +511-417 kg = kir=-6E I — х 2-ке (2x) Ч 65- = -39c8, 2gy ks = <13 = 2E1 2х2х16, 2к) L С.5 23 89.bl589 Кузгkz GEI LE I ЧЕТ 2 - 2 L2 (292 2 — 9х2х).в.2х). Ч -2556, 213 2x (cs)- 3х2х10 | 2xt.Ч 32 x 6.5 - 8524071 Куча L2 = GEI К34 - 43 2ък 12, 3072 - 368, - ПЕТ 9. 65 2х2х10 1.2k/s } <чң ? ҶEL с +384 . 15 2 С. 5

global Equation - Paku -144.3547 514769.231 -3408.28 738446150 - 399.75! -34081284 23068.61 -255621 852071 -433,063 738416154 -2556.21 22153185 3692,308 3692.308 7384.615] $52.071 577.4167/ Och 0.004988 sad -0.02489 m lu Oy oc -0.04101 rad 0.11569 rad Momentum at jant Mabaz z mobbat 2E I (Ba+200-3€) 2EI (ab 1.243 -342 Tz + ZEI = -41x 6.5² 2x 12 6.5 624000 0.001988-2X0.0410) +380.62489 =-628.387kwim Mbe=mbe + 2EI (206+20c +386) 2 xestores - Ooy lol 6.5 21- = fool? 3 la + 2x2400 2x6.5 +0.101569 ~3x0.024899 2x65 - +628,387kNin Me = 628.387kNim.

	1	2	3	4
1	14769.231	-3408.28	7384.615	0
2	-3408.284	23068.01	-2556.21	852.071
3	7384.6154	-2556.21	22153.85	3692.308
4	0	852.071	3692.308	7384.615

-144.354

-399.75

-433.063

577.4167

0.004988

-0.02489

-0.04101

0.101569