cylindrical tank

[logistic_guy]

here is the question

The cap on the cylindrical tank is bolted to the tank along the flanges. The tank has an inner diameter of \(\displaystyle 1.5\) m and a wall thickness of \(\displaystyle 18\) mm. If the pressure in the tank is \(\displaystyle p = 1.20\) MPa, determine the force in each of the \(\displaystyle 16\) bolts that are used to attach the cap to the tank. Also, specify the state of stress in the wall of the tank.




my attemb
this question can be solved by the method of equilibrium
i think there is two stress on the wall of the tank
hoop stress and longitudinal stress
to preceed further in my calculations, i need to know if the tank can be considered as a thin walled structure or not
how to do that?

 

[khansaheb]

here is the question

The cap on the cylindrical tank is bolted to the tank along the flanges. The tank has an inner diameter of \(\displaystyle 1.5\) m and a wall thickness of \(\displaystyle 18\) mm. If the pressure in the tank is \(\displaystyle p = 1.20\) MPa, determine the force in each of the \(\displaystyle 16\) bolts that are used to attach the cap to the tank. Also, specify the state of stress in the wall of the tank.

View attachment 38963


my attemb
this question can be solved by the method of equilibrium
i think there is two stress on the wall of the tank
hoop stress and longitudinal stress
to preceed further in my calculations, i need to know if the tank can be considered as a thin walled structure or not
how to do that?

Do a Google search of "thin walled pressure vessel" and tell us what you find....

 

[logistic_guy]

Do a Google search of "thin walled pressure vessel" and tell us what you find....

what will happen if i blindfold assume it's thin walled structure?
do it give incorrect result?

 

[khansaheb]

what will happen if i blindfold assume it's thin walled structure?
do it give incorrect result?

You claim to be studying Engineering and you want to make blind-assumptions ?

 

[logistic_guy]

You claim to be studying Engineering and you want to make blind-assumptions ?

you're right. i'll be a bad engineer if i design base on my assumption

Do a Google search of "thin walled pressure vessel" and tell us what you find....

i do
it say the ratio between radius and thickness must bigger than \(\displaystyle 10\)

\(\displaystyle \frac{r}{t} > 10\)

the radius isn't givendo i have to divide the diameter by \(\displaystyle 2\)?

 

[khansaheb]

you're right. i'll be a bad engineer if i design base on my assumption


i do
it say the ratio between radius and thickness must bigger than \(\displaystyle 10\)

\(\displaystyle \frac{r}{t} > 10\)

the radius isn't givendo i have to divide the diameter by \(\displaystyle 2\)?

Yes .... By definition of diameter and radius. That was taught to you in basic GEOMETRY!!!!

 

[logistic_guy]

Yes .... By definition of diameter and radius. That was taught to you in basic GEOMETRY!!!!

\(\displaystyle \frac{r}{t} = \frac{\frac{1.5}{2}}{18} = \frac{1.5}{36} = 0.042 < 10\)

i don't know how to solve the question when it's not thin walled structure
can you guide me?

 

[khansaheb]

\(\displaystyle \frac{r}{t} = \frac{\frac{1.5}{2}}{18} = \frac{1.5}{36} = 0.042 < 10\)

i don't know how to solve the question when it's not thin walled structure
can you guide me?

Incorrect....

Use consistent unit/s.

 

[logistic_guy]

Incorrect....

Use consistent unit/s.

what do you mean? i don't understand

 

[khansaheb]

what do you mean? i don't understand

r = I.5 m = 1500 mm
t = 18 mm
r/t = ??

 

[Agent Smith]

Pressure = 1.2 MPa
Lid area = [imath]\pi \times 75^2[/imath] cm²

Find the force acting on the lid. Divide by 16

 

[khansaheb]

Pressure = 1.2 MPa
Lid area = [imath]\pi \times 75^2[/imath] cm²

Find the force acting on the lid. Divide by 16

Be careful about consistent units.

 

[logistic_guy]

r = I.5 m = 1500 mm
t = 18 mm
r/t = ??

you tell me by definition radius is half diameter. now i'm confused

Pressure = 1.2 MPa
Lid area = [imath]\pi \times 75^2[/imath] cm²

Find the force acting on the lid. Divide by 16

what is lid? one bolt you mean?

 

[khansaheb]

you tell me by definition radius is half diameter. now i'm confused

You are correct that should be:

r = 1500/2 mm

Continue.....

 

[logistic_guy]

You are correct that should be:

r = 1500/2 mm

Continue.....

thank

\(\displaystyle \frac{r}{t} = \frac{\frac{1500}{2}}{18} = 41.7 > 10\)

this mean i can treat the tank as a thin walled structure

the properties of thin walled structure give me this two formulas

hoop stress: \(\displaystyle \rho_h = \frac{pr}{t}\)

longitudinal stress: \(\displaystyle \rho_l = \frac{pr}{2t}\)

the \(\displaystyle 16\) bolts making force downward. an equal and opposite force created in the tank so

\(\displaystyle F_1 - F_2 = 0 = pA - 16F_b\)

so the force in each bold is
\(\displaystyle F_b = \frac{pA}{16} = \frac{1.2 \times 10^6 \times \pi \times 0.75^2}{16} = 1.33 \times 10^5 = 133 \) kN

is my analize correct?

 

[Agent Smith]

you tell me by definition radius is half diameter. now i'm confused


what is lid? one bolt you mean?

The cap is the lid, held down by 16 bolts. Other people have better answers

 

[logistic_guy]

The cap is the lid, held down by 16 bolts. Other people have better answers

i'll be so glad if you show me your calculations. what's one bolt force according to your calculation?
show me numericals please smith

thank very much
appreciate it

 

[topsquark]

i'll be so glad if you show me your calculations. what's one bolt force according to your calculation?
show me numericals please smith

thank very much
appreciate it

No. It is up to you to provide that.

-Dan

 

[logistic_guy]

No. It is up to you to provide that.

-Dan

i'm not sure if my calculation of the bolt force is correct
that's the tough task. now i can calculate the stress in the tank wall easily

hoop stress
\(\displaystyle \rho_{h} = \frac{pr}{t} = \frac{1.2 \times 10^6 \times 750}{18} = 5 \times 10^{7} = 50 \) MPa

longitudinal stress
\(\displaystyle \rho_{l} = \frac{pr}{2t} = \frac{1.2 \times 10^{6} \times 750}{2 \times 18} = 2.5 \times 10^{7} = 25\) MPa