Question

Question 3. Using the Stress Transformation Equations, find the state of stress at 30 deg, 60 deg and the Principal Stresses (o and 02) and their orientation (@pi and 0.2) and the maximum in-plane shear stress (Tmax).for the following; 20 15 MPa

Answer 1

SOLUTION:

A state of stress is shown in the figure 1 below.

Given That:

У 10 MPa 9 oy у(1) х(1) о(х1), s(x1y1) 2 с ху 20 MPa а(y1) 9 OX OX 15 MPа х х (y1) с ху a(x1) s(x1y1) ду figure. 1

0,= -15M Pa

Oy 10MPa

and

TH 20MPa.

Analysis :

we know by stress transformation Equations

011 or + Oy 2 + Ycos20 + Trysin 20 ----(i) 2

$\tau _{x_{1}y_{1}}=-\frac{\sigma _{x}-\sigma _{y}}{2}sin2\theta+\tau _{xy}cos2\theta----(ii)$

$\sigma _{x_{1}}+\sigma _{y_{1}}=\sigma _{x}+\sigma _{y}-----(iii)$

1) Find state of stress when,

When $\theta=30^o$ ,

by equation (i),(ii) and (iii)

$\sigma_{x_{1}}=\frac{-15+10}{2}+\frac{-15-10}{2}cos60^o+20sin60^o={\color{Red} 8.57 MPa}...answer$

T1191 - 15 – 10 -sin60° + 20cos60° = 20.82 M Pa.....answer 2

and $8.57+\sigma _{y_{1}}=-15+10$

$\sigma _{y_{1}}={\color{Red} -13.57 MPa} .....answer$

When $\theta=60^o$ ,

by equation (i),(ii) and (iii)

$\sigma_{x_{1}}=\frac{-15+10}{2}+\frac{-15-10}{2}cos120^o+20sin120^o={\color{Red}21.07 MPa}...answer$

$\tau _{x_{1}y_{1}}=-\frac{-15-10}{2}sin120^o+20cos120^o={\color{Red} 0.825 MPa}.....answer$

and $21.07+\sigma _{y_{1}}=-15+10$

$\sigma _{y_{1}}={\color{Red} -26.07 MPa} .....answer$

2) For principal stress:

$\sigma _{1,2}=\frac{\sigma _{x}+\sigma _{y}}{2}\pm \sqrt{\left ( \frac{\sigma _{x}-\sigma _{y}}{2} \right )^{2}+\tau _{xy}^2}-----------(iv)$

so, $\sigma _{1,2}=\frac{-15+10}{2}\pm \sqrt{\left ( \frac{-15-10}{2} \right )^{2}+20^2}=-2.5\pm 23.58$

so,$\sigma _{1}={\color{Red} 21.08 MPa}...answer$

and $\sigma _{2}={\color{Red}- 26.08 MPa}...answer$

now We know that, $tan2\theta_{p}=\frac{2\tau_{xy}}{\sigma _{x}-\sigma _{y}}----(v)$ ,

where $\theta_{p}$ , defined as the orientation of the Principal Planes.

so, $tan2\theta_{p}=\frac{2*20}{-15-10}=-1.6$

or, $\theta_{p}=-28.99^o ....or.....28.99^o..clockwise$

so, $\theta_{p_{1}}={\color{Red} 28.99}^oclockwise...answer$

and $\theta_{p_{2}}=90-28.99^o={\color{Red} 61.01^o anticlockwise}...answer$

3) Maximum In-Plane shear stress, $\tau _{max}$

we know that

$\tau _{max}= \sqrt{\left ( \frac{\sigma _{x}-\sigma _{y}}{2} \right )^{2}+\tau _{xy}^2}-----------(vi)$

so, $\tau _{max}= \sqrt{\left ( \frac{-15-10}{2} \right )^{2}+20^2}= {\color{Red} 23.58 MPa}...answer$