Question

A material is deformed by slip along a plane with a critical resolved shear stress (CRSS) of 150 MPa. What is the applied stress needed if the orientation factor is 0.45?

• A. 75 MPa
• B. 333 MPa
• C. 150 MPa
• D. 675 MPa

Hint:

Schmid's Law. Resolved shear stress \(\tau_R = \sigma \cos\phi \cos\lambda = \sigma \times m\). Slip starts when \(\tau_R\) reaches the Critical Resolved Shear Stress (\(\tau_{CRSS\)). Required applied stress \(\sigma = \tau_{CRSS / m\).

Answer 1

The Correct Option is B

Schmid's Law relates the applied tensile or compressive stress (\(\sigma\)) to the resolved shear stress (\(\tau_R\)) acting on a specific slip plane in a specific slip direction.
Slip occurs when the resolved shear stress reaches the critical resolved shear stress (CRSS or \(\tau_{CRSS}\)).
The relationship is: $$ \tau_R = \sigma \cos\phi \cos\lambda = \sigma \times m $$ where \(\phi\) is the angle between the applied stress direction and the normal to the slip plane, \(\lambda\) is the angle between the applied stress direction and the slip direction, and \(m = \cos\phi \cos\lambda\) is the Schmid factor (or orientation factor).
Slip occurs when \(\tau_R = \tau_{CRSS}\).
$$ \tau_{CRSS} = \sigma_{applied} \times m $$ We need to find the applied stress (\(\sigma_{applied}\)) required to initiate slip.
$$ \sigma_{applied} = \frac{\tau_{CRSS}}{m} $$ Given: CRSS (\(\tau_{CRSS}\)) = 150 MPa Orientation factor (Schmid factor, \(m\)) = 0.
45 $$ \sigma_{applied} = \frac{150 \, \text{MPa}}{0.
45} $$ $$ \sigma_{applied} = \frac{150}{45/100} = \frac{150 \times 100}{45} = \frac{10 \times 100}{3} = \frac{1000}{3} \approx 33(3)33 \, \text{MPa} $$ The required applied stress is approximately 333 MPa.