Question

Given condition: k1=k2=k3=k4=k5=k,F3=P, and nodes 1,2,5 are fixed
Decision
a.Global stiffness matrix
b. Displacement of nodes 3 and 4
c. Reaction force at nodes 1,2,5
d.Internal force of spring 2,4,5
1.7 한 스프링계가 아래와 같다. k2 ki -F3 Amino k4 K3 X 주어진 조건: k = k = kg = k 4 = ks = k, F = P, 그리고 절점 1, 2, 5는 고정됨; 결정할 사항: a. 전역 강성행렬 b. 절점 3과 4의 변위 C. 절점 1, 2, 5에서 반력 d. 스프링 2, 4, 5의 내력

Oh, and I would appreciate it if you could tell the answer using computer typing instead of handwriting.

Because I don't know the cursive

If it is difficult, I would appreciate it if you could respond in a convenient way.

Answer 1

Given : k2 M F3 K3 ko 3 3. А UT 4 КА 2 2 Ki = K2=kz=Khak5 =k F = P. solution : for element ☺ KI -ki A ki U3 Fз -ki for element U2 F2 KH -ky 1 U4 -kh ky for element ③ F2 Katk3 ~K2-K3 JE に [-k2-k3 Katk3 uy

for element 4 KE _k5 -k5 ko us Global stiffness matix : ki OK k4 0 -K4 o o -ki kitkatk3 -k2-k3 0 -ky -k2-k3 Katkzt kytk5 -k5 0 0 K5 -k5 O K -k 0 (k - 0 K 0 -K 6 - k O 3k -2k o -k - Qk 4K -k b k ak O u=0 node I @ node a U2=0 Boundary conditions: fixed fixed @ node 3 F3 =P no de 5 fraed Eisite element equation {F} = [k]{u} s 1520

0 UD A K (FO U12.00 6 -k 6 0 K F2=0 uz F3 =P 3k -2k o -k - ak 4k -k 14 6 FHEO -F5=0 ak k u5=0 6 0 o the rows and columns of By cancelling zero displacements. (36) U₂ – 2k(44) Р -(ak) U3 + 4K (U4) = 0 2P 0 x 2 6k Uz - 4K U4 -2kU3 +4KUL. - 0 =&P 4kuz displacement at node ③ u3 &P 4k. Р Qk. sub is een 2nd has B - ZK uz + 4k U4 = 0 -&k AK & 4K44 4kU4 = P U4 = displacement at rode 6 Р 4K 13

di Reaction force : {R} - [K] {u} _{f} ~KU3 -0 => -Klak R,= ald O-k R2 = kuz - ku4 = ,4K R2= - P 4 R5 -* (*) -0 = -kly-ku5= -KI R5= - P 4 (h) internal force : 3 - -kui 43 kug- & ku4 - P --P =o+3k 4k (5)-2x( 3 - 1 / -P F3. O F4 -KU2 - QKUz + 4K 44- K45 11 0.2k * () +44() - 4 = 0