Return to Answer

The usual simplification in structural engineering is to consider the sphere as infinitely thin. That is what is called a 'shell' as opposed to thicker 'plates'. Shells react to pressure by undergoing traction, but without bending. That makes things easier. The book by Timoshenko is the place to go for a detailed explanation of either case.

The easiest way to turn your problem into numbers is to consider only half of the sphere. The force exerted on that half sphere by the inner pressure will be $F=p \pi r^2$, and what keeps the sphere in place is the tension in the rim, exerted by the other half sphere. For a thickness $t$ of the shell means an area of $A = 2 \pi r t$, so that the tension is $\sigma = p r / (2 t)$.

So if you know the tensile strength of steel, you can calculate the thickness from $t = p r / (2 \sigma)$. This is another tricky question... steel has a lower yield limit, at which it starts to deform permanently, and a tensile limit, at which it snaps. These limits are also highly dependent on how the steel has been manufactured. You have some values here.

Being conservative and choosing $\sigma = 150\;\mathrm{MPa}$, and your pressure of $p= 300\;\mathrm{bar} = 30\; \mathrm{MPa}$, for a radius of $r=2\;\mathrm{cm}=0.02\;\mathrm{m}$ the resulting minimal thickness is $t=0.02\;\mathrm{m}=2\;\mathrm{cm}$.

Depending on the actual material you are using, and how much risk of your sphere blowing to pieces on your face you are willing to take, you may be able to get away with less, or want to avoid risks and go with even more. Normally using the thin shell approximation is deemed safe when $r/t > 10$, so for a much thicker plate things would be different.

If you plan to use this information in any real life project, be extremely careful with what you do: if your sphere suddenly breaks in half due to material failure, each half will be initially propelled by a force close to $40\;\mathrm{kN}$...

The usual simplification in structural engineering is to consider the sphere as infinitely thin. That is what is called a 'shell' as opposed to thicker 'plates'. Shells react to pressure by undergoing traction, but without bending. That makes things easier. The book by Timoshenko is the place to go for a detailed explanation of either case.

The easiest way to turn your problem into numbers is to consider only half of the sphere. The force exerted on that half sphere by the inner pressure will be $F=p \pi r^2$, and what keeps the sphere in place is the tension in the rim, exerted by the other half sphere. For a thickness $t$ of the shell means an area of $A = 2 \pi r t$, so that the tension is $\sigma = p r / (2 t)$.

So if you know the tensile strength of steel, you can calculate the thickness from $t = p r / (2 \sigma)$. This is another tricky question... steel has a lower yield limit, at which it starts to deform permanently, and a tensile limit, at which it snaps. These limits are also highly dependent on how the steel has been manufactured. You have some values here.

Being conservative and choosing $\sigma = 150\;\mathrm{MPa}$, and your pressure of $p= 300\;\mathrm{bar} = 30\; \mathrm{MPa}$, for a radius of $r=2\;\mathrm{cm}=0.02\;\mathrm{m}$ the resulting minimal thickness is $t=0.002\;\mathrm{m}=2\;\mathrm{mm}$.

Depending on the actual material you are using, and how much risk of your sphere blowing to pieces on your face you are willing to take, you may be able to get away with less, or want to avoid risks and go with even more. Normally using the thin shell approximation is deemed safe when $r/t > 10$, so for a much thicker plate things would be different.

If you plan to use this information in any real life project, be extremely careful with what you do: if your sphere suddenly breaks in half due to material failure, each half will be initially propelled by a force close to $40\;\mathrm{kN}$...

The usual simplification in structural engineering is to consider the sphere as infinitely thin. That is what is called a 'shell' as opposed to thicker 'plates'. Shells react to pressure by undergoing traction, but without bending. That makes things easier. The book by Timoshenko is the place to go for a detailed explanation of either case.

The easiest way to turn your problem into numbers is to consider only half of the sphere. The force exerted on that half sphere by the inner pressure will be $F=p \pi r^2$, and what keeps the sphere in place is the tension in the rim, exerted by the other half sphere. For a thickness $t$ of the shell means an area of $A = 2 \pi r t$, so that the tension is $\sigma = p r / (2 t)$.

So if you know the tensile strength of steel, you can calculate the thickness from $t = p r / (2 \sigma)$. This is another tricky question... steel has a lower yield limit, at which it starts to deform permanently, and a tensile limit, at which it snaps. These limits are also highly dependent on how the steel has been manufactured. You have some values here.

Being conservative and choosing $\sigma = 150MPa$, and your pressure of $p= 300bar = 30 MPa$, for a radius of $r=2cm=0.02m$ the resulting minimal thickness is $t=0.02m=2cm$.

Depending on the actual material you are using, and how much risk of your sphere blowing to pieces on your face you are willing to take, you may be able to get away with less, or want to avoid risks and go with even more. Normally using the thin shell approximation is deemed safe when $r/t > 10$, so for a much thicker plate things would be different.

If you plan to use this information in any real life project, be extremely careful with what you do: if your sphere suddenly breaks in half due to material failure, each half will be initially propelled by a force close to $40kN$...

The usual simplification in structural engineering is to consider the sphere as infinitely thin. That is what is called a 'shell' as opposed to thicker 'plates'. Shells react to pressure by undergoing traction, but without bending. That makes things easier. The book by Timoshenko is the place to go for a detailed explanation of either case.

The easiest way to turn your problem into numbers is to consider only half of the sphere. The force exerted on that half sphere by the inner pressure will be $F=p \pi r^2$, and what keeps the sphere in place is the tension in the rim, exerted by the other half sphere. For a thickness $t$ of the shell means an area of $A = 2 \pi r t$, so that the tension is $\sigma = p r / (2 t)$.

So if you know the tensile strength of steel, you can calculate the thickness from $t = p r / (2 \sigma)$. This is another tricky question... steel has a lower yield limit, at which it starts to deform permanently, and a tensile limit, at which it snaps. These limits are also highly dependent on how the steel has been manufactured. You have some values here.

Being conservative and choosing $\sigma = 150\;\mathrm{MPa}$, and your pressure of $p= 300\;\mathrm{bar} = 30\; \mathrm{MPa}$, for a radius of $r=2\;\mathrm{cm}=0.02\;\mathrm{m}$ the resulting minimal thickness is $t=0.02\;\mathrm{m}=2\;\mathrm{cm}$.

Depending on the actual material you are using, and how much risk of your sphere blowing to pieces on your face you are willing to take, you may be able to get away with less, or want to avoid risks and go with even more. Normally using the thin shell approximation is deemed safe when $r/t > 10$, so for a much thicker plate things would be different.

If you plan to use this information in any real life project, be extremely careful with what you do: if your sphere suddenly breaks in half due to material failure, each half will be initially propelled by a force close to $40\;\mathrm{kN}$...

The usual simplification in structural engineering is to consider the sphere as infinitely thin. That is what is called a 'shell' as opposed to thicker 'plates'. Shells react to pressure by undergoing traction, but without bending. That makes things easier. The book by Timoshenko is the place to go for a detailed explanation of either case.

The easiest way to turn your problem into numbers is to consider only half of the sphere. The force exerted on that half sphere by the inner pressure will be $F=p \pi r^2$, and what keeps the sphere in place is the tension in the rim, exerted by the other half sphere. For a thickness $t$ of the shell means an area of $A = 2 \pi r t$, so that the tension is $\sigma = p r / (2 t)$.

So if you know the tensile strength of steel, you can calculate the thickness from $t = p r / (2 \sigma)$. This is another tricky question... Steel has a lower yield limit, at which it starts to deform permanently, and a tensile limit, at which it snaps. These limits are also highly dependent on how the steel has been manufactured, you have some values here.

Being conservative and choosing $\sigma = 150MPa$, and your pressure of $p= 300bar = 30 MPa$, for a radius of $r=2cm=0.02m$ the resulting minimal thickness is $t=0.02m=2mm$.

Depending on the actual material you are using, and how much risk of your sphere blowing to pieces on your face you are willing to take, you may be able to get away with less, or want to avoid risks and go with even more. Normally using the thin shell approximation is deemed safe when $r/t > 10$, so for a much thicker plate things would be different.

If you plan to use this information in any real life project, be extremely careful with what you do: if your sphere suddenly breaks in half due to material failure, each half will be initially propelled by a force close to $40kN$...

The usual simplification in structural engineering is to consider the sphere as infinitely thin. That is what is called a 'shell' as opposed to thicker 'plates'. Shells react to pressure by undergoing traction, but without bending. That makes things easier. The book by Timoshenko is the place to go for a detailed explanation of either case.

The easiest way to turn your problem into numbers is to consider only half of the sphere. The force exerted on that half sphere by the inner pressure will be $F=p \pi r^2$, and what keeps the sphere in place is the tension in the rim, exerted by the other half sphere. For a thickness $t$ of the shell means an area of $A = 2 \pi r t$, so that the tension is $\sigma = p r / (2 t)$.

So if you know the tensile strength of steel, you can calculate the thickness from $t = p r / (2 \sigma)$. This is another tricky question... steel has a lower yield limit, at which it starts to deform permanently, and a tensile limit, at which it snaps. These limits are also highly dependent on how the steel has been manufactured. You have some values here.

Being conservative and choosing $\sigma = 150MPa$, and your pressure of $p= 300bar = 30 MPa$, for a radius of $r=2cm=0.02m$ the resulting minimal thickness is $t=0.02m=2cm$.

Depending on the actual material you are using, and how much risk of your sphere blowing to pieces on your face you are willing to take, you may be able to get away with less, or want to avoid risks and go with even more. Normally using the thin shell approximation is deemed safe when $r/t > 10$, so for a much thicker plate things would be different.

If you plan to use this information in any real life project, be extremely careful with what you do: if your sphere suddenly breaks in half due to material failure, each half will be initially propelled by a force close to $40kN$...