Answer:
![\sigma _1=10.33MPa](https://img.qammunity.org/2020/formulas/engineering/college/30yv4tgyl0za4avuy4tlee0p8dwtngghio.png)
![\sigma _2=5.16MPa](https://img.qammunity.org/2020/formulas/engineering/college/5y6tdmxaqilub2ej7yzyts8hmhwti7lers.png)
Step-by-step explanation:
Given that
Diameter(d)=62 mm
Thickness(t)= 300 μm=0.3 mm
Internal pressure(P)=100 KPa
Actually there is no any shear stress so normal stress will become principle stress.This is the case of thin cylinder.The stress in thin cylinder are hoop stress and longitudinal stress .
The hoop stress
![\sigma _h=(Pd)/(2t)](https://img.qammunity.org/2020/formulas/engineering/college/5bnnrqiol4oy8ko3eha5apaqi7gt7aqikg.png)
Longitudinal stress
![\sigma _l=(Pd)/(4t)](https://img.qammunity.org/2020/formulas/engineering/college/if46x2xrhiwcy308t3wmsi99l11arqk5kp.png)
Now by putting the values
![\sigma _h=(Pd)/(2t)](https://img.qammunity.org/2020/formulas/engineering/college/5bnnrqiol4oy8ko3eha5apaqi7gt7aqikg.png)
![\sigma _h=(100* 62)/(2* 0.3)](https://img.qammunity.org/2020/formulas/engineering/college/u3k067eceujt90601iv134e6zynecpi5i0.png)
![\sigma _h=10.33MPa](https://img.qammunity.org/2020/formulas/engineering/college/yhtzfk9izopibelqi3oxjm2tvem8uj4kzx.png)
![\sigma _l=5.16MPa](https://img.qammunity.org/2020/formulas/engineering/college/htug2sypk944pgmew03gryvocmr950wbgh.png)
So the principle stress are
![\sigma _1=10.33MPa](https://img.qammunity.org/2020/formulas/engineering/college/30yv4tgyl0za4avuy4tlee0p8dwtngghio.png)
![\sigma _2=5.16MPa](https://img.qammunity.org/2020/formulas/engineering/college/5y6tdmxaqilub2ej7yzyts8hmhwti7lers.png)