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1430_Mechanics Homework Mechanics of Materials Solution

1430_Mechanics Homework Mechanics of Materials Solution - f...

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Unformatted text preview: f.-. I 11 .01 A red consisting of n elements, each of which is homogeneous and of uniform cross section, is subjected to a load P applied at its free end. Elementn Elementi Element]. The length of element i is denoted by L, and its diameter by d.. (a) Denoting PROBLEM 1LC1 by E the modulus of elasticity of the material used in the rod, write a computer program that can be used to determine the strain energy acquired by the rod and the defamation measured at the free end. (b) Use this program to deter- mine the strain energy and deformation of the rods of Probs. 11.9 and 11.12. SOLUTION [Ax 7:54,; P g”; [ For? Fficfi‘ Farmer]? EN??? 1‘2 («my D; ,— P (maven Noe/wet sweets: u a .- T C 2 L - glee/N Lava/MY: U. = _E__f_ ‘ Z flit: 57mm» 5”"?de WEIW'TQ __ t: q“ z! 7B 74:. 5-7941»; 585%?6 ‘r OPP/6‘71? 77/120Udfi 7, 4.21:”; 1:”; U 7 u + U3. 7074; DE Fawn TIM I _ . =_.._2” EPA-U . A P Preteen/w ou7gu7’ Problem 11 . 9 Axial load a 8.000 kipe Modulus of elasticity = 29 it 10‘s psi Element Length delta L Stress Strain Energy Strain Energy Density in. in. kei in-lb 1b-in./in."3 1 24.000 0.022 26.08 86.32 11.72 2 36.000 0.022 18.11 89.92 5.65 Total Strain Energy :- 176.24 in-lb Total Detormation .. 0.0441 in. Problem 11 . 12 Axial loud :- 25.000 kN Modulus of elasticity = 200 GPa Element Length delta 1:. Stress Strain Energy Strain Energy Density in mm MPa J kJ/m"3 1 0.80 0.497 124.34 6.22 38.65 2 1.20 0.477 79.58 5.97 15.83 Total Strain Energy = 12.1853 J Total Defamation - 0.9748 mm ...
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