Question

For the system shown below, (a) the global stiffness matrix   (b)displacements of nodes 2 and 3 (c)the reaction forces at nodes 1 and 4 (d)the force in the members

EA TRATAMI 70-400 = 100 x 10 kN/m 0.28 ATT L ( EA k, 100 - 200 = 200 x 10 kN/m 0.1 L (4 EA k, 200.70 =140x10 kN/m 0.1 I (4.2

Answer 1

The solution of the given problem is given below

Solution :- enษยา data ; 3 ki 100 V10 kalm ka 3 200 Y 10 Kalm к 3 lub No Kalm m. enlobal stiffness makin: 100 kn 50 kn ☺ (3) k 0.28 ool A 3 100 N10 3 -100 Y 10 ki 14 v trg 3 - 100 rio 3 10o NIO -19 14 100 -100 chy 3 & 8 [k loo 100 4 ราย 3 2.00 - 200 dar [K] 2007 lo .2oo 832 (3) luo 3 Tuo dzi [k] 10 -140 140 duen

enlobal stiffness matria [k] = [k + [ k) [k] + [k] du 4. น dza dur loc -100 0 3 - 100 300 -200 [K] 10 O -200 340 -luo -luo luo b. Displacements 아 nodes and @ 3:- X ello bol matin df] [K] da Boun dary condition's die - duz 20 modified global matriu is dza fax 300 200 3 10 d372 -200 fan 340 300 -200 42 -50 3 10 .] 200 340 dza 100 3 300 29 100 -50 = da } © re 3 10 -200 dan of 340 100 + 3uo dza] - © By solving & © ) we oft gef S 4 day = 4.838 710 3 j dan 3.22 58 710 m.

The al Peaction torce nodes and From global 3 Fi lo v { lot or 4.838 xios] R 4,838 ka Ru -4 to} i do to 140 3.22 58 y 10 -US. 162 Ry kn d Calculation member free cs :- from element Stitness matrinis Elemento- 3 {-1004 hox بها -100 & 4.838 10 -4.838 kn - 4.B 38 kn 3 ID B far 0 + 100 # 4.898 110 لم Element - @ -5 200 x 40838 710 3 10 200 Ý 3,258 * 18 of} fan ta 54.84 kal 3 10 -5 -200 x 4.838 X10 fr = . ? Į - tida} + 200 & 3.2258 410 f3r - 54.84 kr

Element 39 3 of نما 10 10 fax re 140 M 3.22 58 นม tan = u. 1612 3 fur luo & 3.22 58 rug fun 45. 1612 KN