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Author: Erik Bishop Class: CEE 379 Professor Eberhard Description HW#2, Problem #2a
a.) Solve for the Nodal Displacements and Internal Spring Forces for the following Spring System.
Spring Stiffnesses
25 -25 10 - -25 25 -10 10 35 -35 20 - -35 35 -20 20 50 -50 15 - -50 50 -15 15
Applied Loads
Q 1 -15 kip Q 2 25 kip Q 3 20 kip
Free Stiffness Matrix
k1+k2+k3 - k3 0 110 -50 0 K 11 = (^) - k3 k3+k4+k5 - k5 = -50 80 - 0 - k5 k5+k6 0 -20 35
Inverted Free Stiffness Matrix
0.0136 0.0099 0. (K 11 )-1^ = 0.0099 0.0218 0. 0.0057 0.0125 0.
Displacements At Free Nodes
Du Qk (kip) Du D 1 0.0136 0.0099 0.0057 -15 0. D 2 = 0.0099 0.0218 0.0125 25 = 0. D 3 0.0057 0.0125 0.0357 20 0.
kip/in
kip/in
(K 11 )-1^ (kip/in)
in
Spring 1, k` 1
Spring 2, k` 2
Spring 3, k` 3
kip/in
kip/in
kip/in
kip/in
kip/in
kip/in
Spring 4, k` 4
Spring 5, k` 5
Spring 6, k` 6
Compute Internal Spring Forces
qs Ds qN1 25 -25 0 -3. qF1 -25 25 0.157223796 3.
qN2 35 -35 0 -5. qF2 -35 35 0.157223796 5.
qN3 50 -50 0.157223796 -24. qF3 -50 50 0.645892351 24.
qN4 10 -10 0.645892351 6. qF4 -10 10 0 -6.
qN5 20 -20 0.645892351 -5. qF5 -20 20 0.940509915 5.
qN6 15 -15 0.940509915 14. qF6 -15 15 0 -14.
Spring Force Summary As expected, these forces are identical with those done in problem 1.
Spring 1 3.93 kip (T) Node 1 0.1572 in (R) Spring 2 5.50 kip (T) Node 2 0.6459 in (R) Spring 3 24.43 kip (T) Node 3 0.9405 in (R) Spring 4 6.46 kip (C) Node 4 0 in Spring 5 5.89 kip (T) Node 5 0 in Spring 6 14.11 kip (C) Node 6 0 in Node 7 0 in
Internal Forces Nodal Displacements
= x = kip (C)
= x = kip (T)
= x = kip (C)
= x = kip (T)
= x = kip (T)
k`-
= x = kip (T)
Compute Internal Spring Forces
qs Ds qN1 200 -200 0 37. qF1 -200 200 -0.187342833 -37.
qN2 35 -35 0 6. qF2 -35 35 -0.187342833 -6.
qN3 50 -50 -0.187342833 24. qF3 -50 50 -0.667854149 -24.
qN4 10 -10 -0.667854149 -6. qF4 -10 10 0 6.
qN5 20 -20 -0.667854149 5. qF5 -20 20 -0.953059514 -5.
qN6 15 -15 -0.953059514 -14. qF6 -15 15 0 14.
Spring Force Summary As expected, these forces are identical with those done in problem 1.
Spring 1 37.47 kip (C) Node 1 -0.1873 in (L) Spring 2 6.56 kip (C) Node 2 -0.6679 in (L) Spring 3 24.03 kip (C) Node 3 -0.9531 in (L) Spring 4 6.68 kip (T) Node 4 0 in Spring 5 5.70 kip (C) Node 5 0 in Spring 6 14.30 kip (T) Node 6 0 in Node 7 0 in
Internal Forces Nodal Displacements
k`-
x = kip (C)
= x = kip (C)
= x = kip (C)
= x = kip (T)
= x = kip (C)
= x = kip (T)
