Answer:
[tex]D_o=11.9inch[/tex]
Explanation:
From the question we are told that:
Thickness [tex]T=0.5[/tex]
Internal Pressure[tex]P=2.2Ksi[/tex]
Shear stress [tex]\sigma=12ksi[/tex]
Elastic modulus [tex]\gamma= 35000[/tex]
Generally the equation for shear stress is mathematically given by
[tex]\sigma=\frac{P*r_1}{2*t}[/tex]
Where
r_i=internal Radius
Therefore
[tex]12=\frac{2.2*r_1}{2*0.5}[/tex]
[tex]r_i=5.45[/tex]
Generally
[tex]r_o=r_1+t[/tex]
[tex]r_o=5.45+0.5[/tex]
[tex]r_o=5.95[/tex]
Generally the equation for outer diameter is mathematically given by
[tex]D_o=2r_o[/tex]
[tex]D_o=11.9inch[/tex]
Therefore
Assuming that the thin cylinder is subjected to integral Pressure
Outer Diameter is
[tex]D_o=11.9inch[/tex]
