The equilibrium equations of the blocks in the spring-block system are 3(x2 – x1) – 2x1 = -80 3(x3 – x2) – 3(x2 – x1) = 0 3(x4 – x3) – 3(x3 – x2) = 0 - 3(x3 – x4) – 3(x4 – x3) = 60 - - - -2.x5 – 3(x5 – x4) = 0 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.

Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter4: Eigenvalues And Eigenvectors
Section4.6: Applications And The Perron-frobenius Theorem
Problem 65EQ
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3 kN/m
3 kN/m
2 kNm
2 kNm
ww80 N.
3 kN/m
3 kN/m
ww-60 N-www
The equilibrium equations of the blocks in the spring-block system are
3(x2 – x1) – 2.x1 = -80
3(x3 – x2) – 3(x2 – x1) = 0
3(x4 – x3) – 3(x3 – x2) = 0
%3D
3(x5 – x4) – 3(x4 – x3) = 60
%3D
-2.x5 – 3(x5 – x4) = 0
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.
Transcribed Image Text:3 kN/m 3 kN/m 2 kNm 2 kNm ww80 N. 3 kN/m 3 kN/m ww-60 N-www The equilibrium equations of the blocks in the spring-block system are 3(x2 – x1) – 2.x1 = -80 3(x3 – x2) – 3(x2 – x1) = 0 3(x4 – x3) – 3(x3 – x2) = 0 %3D 3(x5 – x4) – 3(x4 – x3) = 60 %3D -2.x5 – 3(x5 – x4) = 0 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.
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