Respuesta :
Solution :
i. Slip plane (1 1 0)
Slip direction -- [1 1 1]
Applied stress direction = ( 1 0 0 ]
τ = 50 MPa ( Here slip direction must be perpendicular to slip plane)
τ = σ cos Φ cos λ
[tex]$\cos \phi = \frac{(1,0,0) \cdot (1,1,0)}{1 \times \sqrt2}$[/tex]
[tex]$=\frac{1}{\sqrt2 }$[/tex]
[tex]$\cos \lambda = \frac{(1,0,0) \cdot (1,-1,1)}{1 \times \sqrt3}$[/tex]
[tex]$=\frac{1}{\sqrt3 }$[/tex]
τ = σ cos Φ cos λ
∴ [tex]$50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3} $[/tex]
σ = 122.47 MPa
ii. Slip plane --- (1 1 0)
Slip direction -- [1 1 1]
[tex]$\cos \phi = \frac{(1, 0, 0) \cdot (1, -1, 0)}{1 \times \sqrt2} =\frac{1}{\sqrt2}$[/tex]
[tex]$\cos \lambda = \frac{(1, 0, 0) \cdot (1, 1, -1)}{1 \times \sqrt3} =\frac{1}{\sqrt3}$[/tex]
τ = σ cos Φ cos λ
∴ [tex]$50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3} $[/tex]
σ = 122.47 MPa
iii. Slip plane --- (1 0 1)
Slip direction --- [1 1 1]
[tex]$\cos \phi = \frac{(1, 0, 0) \cdot (1, 0, 1)}{1 \times \sqrt2} =\frac{1}{\sqrt2}$[/tex]
[tex]$\cos \lambda = \frac{(1, 0, 0) \cdot (1, 1, -1)}{1 \times \sqrt3} =\frac{1}{\sqrt3}$[/tex]
τ = σ cos Φ cos λ
∴ [tex]$50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3} $[/tex]
σ = 122.47 MPa
iv. Slip plane -- (1 0 1)
Slip direction ---- [1 1 1]
[tex]$\cos \phi = \frac{(1, 0, 0) \cdot (1, 0, -1)}{1 \times \sqrt2}=\frac{1}{\sqrt2}$[/tex]
[tex]$\cos \lambda = \frac{(1, 0, 0) \cdot (1, -1, 1)}{1 \times \sqrt3} =\frac{1}{\sqrt3}$[/tex]
τ = σ cos Φ cos λ
∴ [tex]$50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3} $[/tex]
σ = 122.47 MPa
∴ (1, 0, -1). (1, -1, 1) = 1 + 0 - 1 = 0
