# Pressure Vessel Design Query

## Main Question or Discussion Point

Design a railway cistern for 120,000 kg of Liquid. It has to be a cylindrical shell on two supports.

The cistern is
18 m long
3 m in diameter (inner).
It is made of steel with the yield strength of SY= 240MPa.
Safety factor of FY= 1.9;
Total corrosion allowance, c= 3.0 mm;
Welded joint efficiency = 0.85
Shell thickness, t - 32mm

Check whether this thickness is sufficient to withstand combined action of the internal pressure, t
he weight of liquid, and the action of longitudinal external forces (tensile or compressive) .
Assume the mass of the train as m= 12 x 10^6kg, deceleration during braking of a = 3m/sec
, and during climbing uphill assume the inclination angle of 10 degrees

I assume there will be 3 design cases for this problem:
Case 1 Internal Pressure+ Weight of Liquid
Case 2Internal Pressure+ Weight of Liquid + Inertia during breaking of the railway cistern
Case 3 Forces acting on the pressure vessel: Internal Pressure + Weight of liquid+ inertia of breaking + inclination angle (locomotive applied to the first cistern in the train)

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I don't see any problem with your method of reasoning, but my question is why a railway cistern needs to be pressurized. AFAIK, a cistern just a big container to hold water or some other liquid, isn't it?

So for each case I work out the total longitudinal stress?

For case three I am unsure if I am going about it correctly:

Determine mgsinx and Ff(friction).
mgsinx pushes block down the incline and Ff(friction) opposes this.

Write the equation
Fnet = mgsinx - FfSince Ff=uFn and we know Fn=mgcosx and we also know Fnet=ma, we write

ma = mgsinx- u(mgcosx)

Divide by m throughout, so we don't need mass a=gsinx - ugcosx

How would I work out u ?

I think in your equations, "u" is really mu, the coefficient of friction. You should just look it up or experimentally determine it.

AlephZero