Question

in the middle of the large 1. For the problem shown below, there is a hole of radius plate under uniform biaxial tension. a. Determine Or, 09, and tre- b. Is the stress amplified at the edges of the hole? Explain. S + S

Answer 1

a) As plate is under uniform biaxial tension, value of $\sigma _{r} = S$ ,
0g = S
and $\tau _{r\theta} = 0$ .

You can understand this from the equation or mohar circle.

$\sigma1,2 = \frac{ \sigma _{x} + \sigma _{y}}{2} \pm \frac{ \sigma _{x} - \sigma _{y}}{2} cos 2\theta - \tau _{xy} sin 2\theta$

$\tau _{\theta} = \frac{ \sigma _{x} - \sigma _{y}}{2} sin 2\theta - \tau _{xy} cos 2\theta$

Put value of $\sigma_{x}$ , $\sigma_{y}$ equal to S and $\tau$ equal to zero. For any angle value will remain same.

Mohar circle of this state of stress will be point on $\sigma$ axis (x axis) So shear stress will remain zero for any angle and normal stress will remain same for any angle.

b) I think, Yes. The stress will amplified at the edges of the hole. Whenever there are discontinuity or irregularities present in the component, stress will amplified around that region that is called stress concentration (also called stress riser). Stress will be higher around the hole by stress concentration factor.