Question

A P-2-kN tensile load is applied to a test coupon made from 1.6-mm thick flat steel plate (E 220 GPa, v 0.30) Assuming a-45 mm and b-13 mm, (a) Determine the resulting change in the 45-mm gage length. (b) Determine the resulting change in the 13-mm width. (c) Determine the resulting change in the 1.6-mm thickness. (d) Determine the resulting change in the cross-sectional area value: 0.50 points The uniaxial state of stress in given as (MPa)

A P=2-kN tensile load is applied to a test coupon made from 1.6-mm thick flat steel plate (E = 220 GPa, v = 0.30). Assuming a=45 mm and b=13 mm,

(a) Determine the resulting change in the 45-mm gage length.

(b) Determine the resulting change in the 13-mm width.

(c) Determine the resulting change in the 1.6-mm thickness.

(d) Determine the resulting change in the cross-sectional area.

The uniaxial state of stress in given as ________ (MPa)

Answer 1

E = 200 Gpa mu = 0.30 = 200 times 10^3 N/mm^3 p = 2 KN Rightarrow uniaxial state y stress sigma_x = p/A = 2000/13 times 1.t = 96.15 N/mm^2 Rightarrow (a) change in gauge length strain, E_a = pa/AE = 2000 times 45/13 times 1.6 times 2 times 10^5 = 2.163 times 10^- 3 Therefore change in length Delta a = E_a times a = 2.163 times 10^- 3 times 45 Delta a = 0.9736 mm (increase) (b) change in 13-mm width E-2 = -mu sigma_x/E Rightarrow Delta b = -mu sigma_x b/E = -0.3 times 96.15/2 times 10^5 times 13 Delta b = -1.075 times 10^- 3 mm decreases \r\n (c) Resulting cgange in 1.6 mm E_y = -mu sigma_x/E Rightarrow Delta t = -0.3 times 96.15/2 times 10^5 times 1.6 Delta t = -2.3076 times 10^- 4 mm (decreases) (d) change in cross - sectional area b_f = b + Delta b = (13 - 1.275 times 10^- 3) mm t_f = t + Delta t