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  1. ...material is its ability to deform under tensile stree. (b) Maximum strain = 6/250 = 0.024 (c) Cross-sectional area of rod = pi x 0.003^2 m^2 = 2.827x10...

  2. ...1.273 x 108 N/m2 Thinner bar = 3.537 x 108 N/m2 So strain in the thicker bar = 1.273 x 108/(2 x 1011) = 6.37 x 10-4 with extension...

  3. ...2 Hence, stress = 100/(7.854x10^-7) N/m^2 = 1.273x10^8 N/m^2 Strain = 2/1500 = 1.333 x 10^-3 Young's modulus = 1.273x10^8/(1.333x10^-3) ...

  4. stress and strain are different in various situation.... they depends on the setting...

  5. ...98 N / mm2 b) (i)Young modulus =(ii)Stress / (iii) Strain (i) Young modulus of (Al) , MUST ...

  6. ...'s modulus is Stress/ Strain which is Force/Area * length/ extension... should plot Stress in the y-axis and Strain in the x-axis. The shape of the graph really ...

  7. ...) = F / EA 250k / (200k * 1885) ~ 0.068%iv) increase in lenth = length * strain = 60 * 0.00068 ~ 0.041m ~ 41mm 2011-01-14 18:13:31 補充: i) cross section = 50/2 * 50...

  8. ...the same for both wires, hence stresses are not the same. 3. strain : strain = stress/Young's modulus, since Young's modulus is...

  9. Stress = F/A Strain = e/l Base surface area, A = pi(0.1/2)2 - pi(0.08/2)2 = 2.83...

  10. (delta L)/L = alpha (delta T) Stress=Young Modulus * Strain Strain =(delta L) / L You can first find out Strain , Then you can find out Stress.