problem1_sol

# problem1_sol - 22.314/1.56/2.084/13.14 Fall 2006 Problem...

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22.314/1.56/2.084/13.14 Fall 2006 Problem Set I Solution 1. Original dimension: D 0 = 12.8 mm, A 0 = πD 0 2 / 4 = 128.68 mm 2 ; and L 0 = 50.800 mm. (a) Stress at a load F = 22.2 kN: σ = F/A 0 = 172.5 MPa Strain: = ( L L 0 ) /L 0 = (50.848 - 50.8)/50.8 = 0.009448 Young’s Modulus: E = σ/� = 182.6 GPa (b) Maximum norimal strain is the strain when fracture occurs: ( L max L 0 ) /L 0 = (69.8 - 50.8)/50.8 = 0.374 (c) F max = 51.2 MPa. Tensile strength is: F max /A 0 = 397.9 MPa 2. This problem has a stress tensor: σ xx 0 0 σ = 0 σ yy τ yz 0 τ zy σ zz Note that the shear stresses on the plane normal to x direction are zero. Therefore, we can derive the two principal stresses on the yz plane using the solution for a plane stress condition. σ yy + σ zz σ yy σ zz ) 2 + τ 2 yz σ i,j ± ( = 2 2 σ i = 464.5 MPa and σ j = 40.5 MPa. Therefore, σ 1 = 464.5 MPa, σ 2 = 440MPa, and σ 3 = 40.5 MPa. The maximum normal stress is σ 1 = 464.5 MPa. The maximum shear stress is ( σ 1 σ 3 ) / 2 = 212 MPa. 3. (a) The principal stresses are the eigenvalues of the stress tensor. It’s solved by using MATLAB (See the code in the end). For 55 5 30 55 σ a = , 5 30 30 30 20 the principal stresses are found to be: σ 1 = 80 MPa, σ 2 = 60 MPa, and σ 3 = -10 MPa.

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