Question #293705

A cylinder contains pressurized gas at 500 MPa and has the outer and inner diameters of 600 mm and 450 mm respectively. Determine the maximum permissible radial crack dimension on the outer surface of the cylinder in order to avoid further propagation of crack based on safety factor of 2.0. The cylinder material is BS 816 M40 of yield strength 1515 MPa.


1
Expert's answer
2022-02-04T14:51:02-0500

Solution;

For thin cylinders under internal pressure only;

σt=σr=ri2piro2ri2(1+ro2r2)\sigma_t=\sigma_r=\frac{r_i^2p_i}{r_o^2-r_i^2}(1+\frac{r_o^2}{r^2})

Where subscript i is for internal and o for outer

ri=225mm=0.225mr_i=225mm=0.225m

ro=300mm=0.3mr_o=300mm=0.3m

r=ri+ro2=0.2625mr=\frac{r_i+r_o}{2}=0.2625m

By substitution;

σ=0.2252×5000.320.2252×(1+0.320.26252)\sigma=\frac{0.225^2×500}{0.3^2-0.225^2}×(1+\frac{0.3^2}{0.2625^2})

σ=1482.5MPa\sigma=1482.5MPa

The factor of safety is 2;

σallowable=1482.52=741.25MPa\sigma_{allowable}=\frac{1482.5}{2}=741.25MPa

But we know;

E=σϵE=\frac{\sigma}{\epsilon}

ϵ=σE=741.251515=0.4893\epsilon=\frac{\sigma}{E}=\frac{741.25}{1515}=0.4893

But;

ϵ=ΔCCo\epsilon=\frac{\Delta C}{C_o}

C is the circumference;

ΔC=π×0.6×0.4893=0.922m\Delta C=π×0.6×0.4893=0.922m


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