Question: 17. A tapered bar with circular cross-section is fixed at ,r = 0, and an axial force of 0.3 x 106 N is applied at

17. A tapered bar with circular cross-section is fixed at ,r = 0, and an axial force of 0.3 x 106 N is applied at the other end. The length of the bar (L) is 0.3 m, and the radius varies as r(x) = 0.03 — O.OTx, where r and x are in meters. Use three equal-length finite elements to determine the displacements, axial force resultants and support reactions. Compare your FE solutions with the exact solution by plotting u vs. x and P(element force) vs. x. Use E= 1010 Pa.

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. 0.3x10 N 0.3 m

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