cylindrical tank

logistic_guy

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Apr 17, 2024
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588
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 1.5\displaystyle 1.5 m and a wall thickness of 18\displaystyle 18 mm. If the pressure in the tank is p=1.20\displaystyle p = 1.20 MPa, determine the force in each of the 16\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.

tank.png


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?😣
 
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 1.5\displaystyle 1.5 m and a wall thickness of 18\displaystyle 18 mm. If the pressure in the tank is p=1.20\displaystyle p = 1.20 MPa, determine the force in each of the 16\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....
 
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 10\displaystyle 10

rt>10\displaystyle \frac{r}{t} > 10

the radius isn't given☹️do i have to divide the diameter by 2\displaystyle 2?
 
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 10\displaystyle 10

rt>10\displaystyle \frac{r}{t} > 10

the radius isn't given☹️do i have to divide the diameter by 2\displaystyle 2?
Yes .... By definition of diameter and radius. That was taught to you in basic GEOMETRY!!!!
 
Yes .... By definition of diameter and radius. That was taught to you in basic GEOMETRY!!!!
rt=1.5218=1.536=0.042<10\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?
 
rt=1.5218=1.536=0.042<10\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.
 
Pressure = 1.2 MPa
Lid area = π×752\pi \times 75^2 cm²

Find the force acting on the lid. Divide by 16
 
You are correct that should be:

r = 1500/2 mm

Continue.....
thank

rt=1500218=41.7>10\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: ρh=prt\displaystyle \rho_h = \frac{pr}{t}

longitudinal stress: ρl=pr2t\displaystyle \rho_l = \frac{pr}{2t}

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

F1F2=0=pA16Fb\displaystyle F_1 - F_2 = 0 = pA - 16F_b

so the force in each bold is
Fb=pA16=1.2×106×π×0.75216=1.33×105=133\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?😣
 
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🙏
 
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
ρh=prt=1.2×106×75018=5×107=50\displaystyle \rho_{h} = \frac{pr}{t} = \frac{1.2 \times 10^6 \times 750}{18} = 5 \times 10^{7} = 50 MPa

longitudinal stress
ρl=pr2t=1.2×106×7502×18=2.5×107=25\displaystyle \rho_{l} = \frac{pr}{2t} = \frac{1.2 \times 10^{6} \times 750}{2 \times 18} = 2.5 \times 10^{7} = 25 MPa
 
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