Question

question 7
2 pts
for a given material, the yield strength is 250 mpa and the ultimate tensile strength is 300
mpa. you need to use this material in a device with a factor of safety n = 2. what would be
the maximum working stress allowed for this material
v select
125 mpa
150 mpa
250 mpa
300 mpa

Explanation:

Step1: Determine the formula for working stress

The formula for working stress \(\sigma_{w}\) when using yield strength \(\sigma_{y}\) is \(\sigma_{w}=\frac{\sigma_{y}}{N}\), where \(N\) is the factor of safety.

Step2: Substitute the given values

Given \(\sigma_{y} = 250\space MPa\) and \(N = 2\). Substitute into the formula: \(\sigma_{w}=\frac{250}{2}\).

Step3: Calculate the working stress

\(\sigma_{w}=125\space MPa\)

Answer:

125 MPa