3(x2 – x1) – 2.x1 = -80 3(x3 – x2) – 3(x2 – x1) = 0 - 3(x4 – x3) – 3(x3 – x2) = 0 3(x5 – X4) – 3(x4 – X3) = 60 -2x5 – 3(x5 – x4) = 0
3(x2 – x1) – 2.x1 = -80 3(x3 – x2) – 3(x2 – x1) = 0 - 3(x4 – x3) – 3(x3 – x2) = 0 3(x5 – X4) – 3(x4 – X3) = 60 -2x5 – 3(x5 – x4) = 0
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
where xi are the horizontal displacements of the blocks measured in mm.
Solve these equations by the Gauss-Seidel method.
Start with x = 0 and iterate until four-figure accuracy after the decimal point is achieved.
Start with x = 0 and iterate until four-figure accuracy after the decimal point is achieved.
Transcribed Image Text:3 kNm
2 kN/m
ww80 N-ww
3 kN/m
3 kN/m
ww
2
3 kN/m
ww60 Nww
3
2 kNm
ww
The equilibrium equations of the blocks in the spring-block system are
3(x2 – x1) – 2.xi
= -80
3(x3 – x2) – 3(x2 – x1) = 0
3(x4 – x3) – 3(x3 – x2) = 0
3(x, — ха) — 3(х, — хз) — 60
|
-2.x5 – 3(x5 – x4) = 0
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