Andrew Sansome
New member
- Joined
- May 3, 2020
- Messages
- 1
Please excuse me but my maths only reached “O” level standard and that was over 40 years ago!
The problem:
Imagine a short (in relation to its radius) section of round tube.
The wall of the tube being infinitely distensible with a modulus of elasticity of zero.
The wall being subject to an internal pressure P (assume the external pressure is zero)
The wall of the tube is supported by a finite number of radial "spokes" acting through the centre of the circle.
The spokes are elastic and obey Hooke's law.
Starting pressure: P
Change in pressure: δP
Number of spokes: n
Starting radius: r0
Length of tube (constant): L
Then the force on a spoke must be given by: F=2Pπr0L/n
Increasing the pressure by δP must result in a force: F= 2(P+δP)πr0L/n But only if the length of the spoke remains constant, i.e its stiffness is infinite.
So increasing pressure by δP would increase the force supported by the spoke: δF=2(δP)πr0L/n.
But where the spoke is extensible (since for a spring F=kx, where k is spring constant and x is change in length)), δr0=δF/k.
So the radius of the tube (length of spoke) would increase by δr0=2(δP)πr0L/nk.
But this increase in radius will by increasing the circumference, further increase the force supported by the spoke and result in further extension.
Intuitively (?) if the spring constant K is less than 2π, then the extension will continue without restriction, if less than 2π, an equilibrium will be reached.
Is this correct and if so, what is the proof?
The problem:
Imagine a short (in relation to its radius) section of round tube.
The wall of the tube being infinitely distensible with a modulus of elasticity of zero.
The wall being subject to an internal pressure P (assume the external pressure is zero)
The wall of the tube is supported by a finite number of radial "spokes" acting through the centre of the circle.
The spokes are elastic and obey Hooke's law.
Starting pressure: P
Change in pressure: δP
Number of spokes: n
Starting radius: r0
Length of tube (constant): L
Then the force on a spoke must be given by: F=2Pπr0L/n
Increasing the pressure by δP must result in a force: F= 2(P+δP)πr0L/n But only if the length of the spoke remains constant, i.e its stiffness is infinite.
So increasing pressure by δP would increase the force supported by the spoke: δF=2(δP)πr0L/n.
But where the spoke is extensible (since for a spring F=kx, where k is spring constant and x is change in length)), δr0=δF/k.
So the radius of the tube (length of spoke) would increase by δr0=2(δP)πr0L/nk.
But this increase in radius will by increasing the circumference, further increase the force supported by the spoke and result in further extension.
Intuitively (?) if the spring constant K is less than 2π, then the extension will continue without restriction, if less than 2π, an equilibrium will be reached.
Is this correct and if so, what is the proof?