Pressure design thickness and how relates to hoop stress

Question

Stress = Force / Area

Lets see with static force... what happens to stress output when decrease the area.

# Stress = Force / Area 
F = 50000
A = [15.0,10.0,5.0]

S = zero(A) # Initialize the output 
for i in 1:length(A)
    S[i] = F / A[i]
end

m = DataFrame(hcat(A,S))
names = ["area","stress"]
names!(m, Symbol.(names))
print(m)

3×2 DataFrame
│ Row │ area    │ stress  │
│     │ Float64 │ Float64 │
├─────┼─────────┼─────────┤
│ 1   │ 15.0    │ 3333.33 │
│ 2   │ 10.0    │ 5000.0  │
│ 3   │ 5.0     │ 10000.0 │

With decreasing area we increase the stress.

So in thinking about hoop stress then how the minimum required thickness is obtained. Using the ASME B31.3 pressure tmin to find t

304.1.2 Straight Pipe Under Internal Pressure
t = PD / 2(SEW + PY)

The numerator has P * D so anything in the denominator will dilute the numerator...

So to see how P and D affect the t output we can iterate through a range of static pressure and changing Diameter... so lets see what this looks like:

# Define numerator inputs 
P = 200 # static P
D = collect(1.0:1.0:24.0) # varying D

# define the denominator static inputs 
S = 20000
E = 1.0
W = 1.0 
Y = 0.4 # temp coefficient 

# initialize output 
out = zero(D)
for i in 1:length(D)
    out[i] = P * D[i] / (2*(S*E*W) + (P*Y))
end

# plot the output
plot(D,out,seriestype = :bar, title="t = static P in relation to Diameter")

So a big factor in the minimum required thickness for pressure containment is the P vs diameter.. at say static P increase the D and we need a thicker wall for pressure containment.

So in thinking about S = F / A

How does this relate to the pressure design thickness calculation.... S in the tmin calculation is in the denominator and S is static in this equation governed by the 3:1 ASME safety factor in relation to materials tensile stress.

Intuitively i can see how at a static F over varying Area - how the stress INCREASES... (as we saw top of the post) because there is less area to distribute the stress over... so stress would increase pounds per sq inch...

But doesn't that mean for smaller D in a pipe, we have less A so that the stress would increase... but running pressure design thickness calculation.. at smaller D we need a thinner wall for pressure containment.

For sure im missing something here! (then again i see no area in the pressure tmin calculation)

Anyone can help me connect the dots?

Thanks

Answer 1

The "area" terms in the equations all cancel out. Think about a section of pipe of internal diameter $d$, length $l$, thickness $t$ with internal pressure $P$.

Now imagine you slice the section of pipe (and the fluid causing the internal pressure) into two semicircular parts, and think about the forces acting one of the parts.

The "area" of the cut through the fluid is d wide by $l$ long, and the fluid pressure of $P$. So the force exerted by the fluid is pressure $\times$ area = $P\,d\,l$.

This is balanced by the hoop stress in the pipe wall. The area of the cut through the walls is $2\,t\,l$ (2, because there are two cuts, on either side of the pipe).

If the average hoop stress is $S$, the force on the cut surfaces of the pipe is stress $\times$ area = $2S\,t\,l$.

So $2Stl = Pdl$ or $$S = \frac{Pd}{2t}.$$

The actual hoop stress varies between the outside and inside of the pipe wall, and this formula doesn't include any safety factors, so in the ASME formula the $2t$ term is replaced by something more realistic for design purposes.