Answer:
F_x = 100,200 N
x' = 21.321 cm ... Length
y' =0.7825 cm
z' = 1.565 cm
A' = ( 0.783 x 1.565 ) cm
Explanation:
Given:
- The Modulus of Elasticity E = 645 MPa
- The poisson ratio v = 0.28
- The Yield Strength Y = 501 MPa
- The Length along x-direction x = 12 cm
- The length along y-direction y = 1 cm
- The length along z--direction z = 2 cm
Find:
The maximum tensile load that can be applied in the longitudinal direction of the bar without inducing plastic deformation. b. The length and cross-sectional area of the bar at its tensile elastic limit.
Solution:
- The Tensile forces within the limit of proportionality is given as:
F_i = б_i*A_jk
- A maximum tensile Force F_x along x direction can be given as:
F_x = Y*A_yz
F_x = 501*( 0.01*0.02)*10^6
F_x = 100,200 N
- The corresponding strains in x, y and z direction due to F_x are:
ξ_x = Y / E
ξ_x = 501 / 645 = 0.7767
ξ_y = ξ_z = -v*Y / E
ξ_y = ξ_z = -0.28*501 / 645 = - 0.2175
- The corresponding change in lengths at tensile elastic stress are:
Δx = x*ξ_x = 12*0.7767 = 9.321 cm
Δy = y*ξ_y = - 1*0.2175 = -0.2175 cm
Δz = z*ξ_z = - 2*0.2175 = -0.435 cm
- The final lengths are:
x' = x + Δx = 12 + 9.321 = 21.321 cm
y' = y + Δy = 1 - 0.2175 = 0.7825 cm
z' = z + Δz = 2 - 0.435 = 1.565 cm
