Answer:
![D_o=11.9inch](https://img.qammunity.org/2022/formulas/engineering/college/ma0nug4ebodrhize50gclfa7po1vgh2xac.png)
Step-by-step explanation:
From the question we are told that:
Thickness
![T=0.5](https://img.qammunity.org/2022/formulas/engineering/college/44ovm680k8pz2wftruzhoxa6ztp9sp6pgp.png)
Internal Pressure
![P=2.2Ksi](https://img.qammunity.org/2022/formulas/engineering/college/v6dy95lcekojdhcz4tdj1qib373o6o4gp2.png)
Shear stress
![\sigma=12ksi](https://img.qammunity.org/2022/formulas/engineering/college/m7byze0ntmsm9m1ppniuzdtslam4qrkoba.png)
Elastic modulus
![\gamma= 35000](https://img.qammunity.org/2022/formulas/engineering/college/gejlqag844c71mp5kz8w3d7njf1g6i23n9.png)
Generally the equation for shear stress is mathematically given by
![\sigma=(P*r_1)/(2*t)](https://img.qammunity.org/2022/formulas/engineering/college/j41hrx9gfiqfz3wofto1tiapewh0zdudn1.png)
Where
r_i=internal Radius
Therefore
![12=(2.2*r_1)/(2*0.5)](https://img.qammunity.org/2022/formulas/engineering/college/4z30mp8x2ctp9l9y0pogui0us4zd3owgsg.png)
![r_i=5.45](https://img.qammunity.org/2022/formulas/engineering/college/s08i061t6kdgl578zom2vjlckec7qfjbjt.png)
Generally
![r_o=r_1+t](https://img.qammunity.org/2022/formulas/engineering/college/h41d4qja43v3l664o1ennutyq1yxnt3ef9.png)
![r_o=5.45+0.5](https://img.qammunity.org/2022/formulas/engineering/college/uaw86s8auhfxqbftwazmktaarv6470jb2z.png)
![r_o=5.95](https://img.qammunity.org/2022/formulas/engineering/college/esv5h4o3ctfowba3y57w30s6u2xbt0n1qa.png)
Generally the equation for outer diameter is mathematically given by
![D_o=2r_o](https://img.qammunity.org/2022/formulas/engineering/college/6mtixerx39n2g232qc7a1lcgh7tosfet8s.png)
![D_o=11.9inch](https://img.qammunity.org/2022/formulas/engineering/college/ma0nug4ebodrhize50gclfa7po1vgh2xac.png)
Therefore
Assuming that the thin cylinder is subjected to integral Pressure
Outer Diameter is
![D_o=11.9inch](https://img.qammunity.org/2022/formulas/engineering/college/ma0nug4ebodrhize50gclfa7po1vgh2xac.png)