1. You have a Co-Cr alloy with Young's Modulus: 645 MPa, Poisson ratio 0.28, and yield strength 501 MPa for that alloy when used in a medical device. From these data estimate the following for a 1 x 2 cm cross-section bar of the alloy having length 12 cm: a. 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.

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