Question

A 20cm-long rod with a diameter of 0.250 cm is loaded with a 5000 N weight. If the diameter of the bar is 0.490 at this load, determine: I. the engineering stress and strain, and [2] II. the true stress and strain

Answer 1

If the diameter of the bar is 0.490 at this load, determine I. the engineering stress and strain, and [2] II. the true stress and strain is 1561. 84 MPa.

What is strain?

Strain is a unitless degree of ways a great deal an item receives larger or smaller from an implemented load. Normal stress happens while the elongation of an item is in reaction to an everyday pressure (i.e. perpendicular to a surface), and is denoted via way of means of the Greek letter epsilon.

1. L = 20 cm d x 1 = 0.21 cm
2. dx 2 = 0.25 cmF=5500 a) σ= F/A1= 5000/(π/4x(0.0025)^2)= 1018.5916 MPa lateral stress= Ad/d1= (0.0021-0.0025)/0.0025 = - 0.1 longitudinal stress (ɛ_l)= -lateral stress/v = -(-0.16)/0.3
3. (assuming a poisson's ration of 0.3) ε_l=0.16/0.3 = 0.5333
4. b) σ_true= σ(1+ ɛ_I)= 1018.5916(1+0.5333
5. = 1561.84 MPa.

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