Locate and calculate the maximum normal tensile stress and t

Locate and calculate the maximum normal (tensile) stress and the maximum shear stress on me outside surface of the hollow propane tank shown below. These two stresses are likely not at the same location nor the same orientation. Calculate ALL stresses at these two points of max normal and max shear stress. The pressurized propane tank (120 psig) has an 18.000\" OD (outer diameter) and 16.000\" ID (inner diameter) and a length of 36.000\". The propane tank has its base rigidly secured to a brick BBQ so that it sticks horizontally out from the wall of the BBQ. The free end has the discharge/flow valve which is rusted stuck. Mr X needs to open the valve in order to keep cooking in the BBQ. He has tried a wrench and has finally resorted to a pipe at the end of the wrench and has decided he will jump on the lever arm. He is a large man at 300 pounds, assume his jump is 3 Gs down at the end of the three foot lever arm. Please ensure you take into account not only the internal pressure, but also the torque and the bending moment from the load (don\'t forget to add two inches for the valve). How would your answers change using thick walled vs thin walled calculations? Will Mr X join the ranks of other Darwin Award Winners or is his propane tank wrench jumping to loosen the valve/nozzle a sate move? How would your answers change if the Inner Diameter were 17.800\"? Yes, you should recalculate everything. Gas Tank Material Data: E = 28.5 Times 10^6 psi, v = 0.27, F_tu = 155 ksi, F_ty = 145 ksi, F_su = 95, alpha = 6 Times 10^-6 in/in/degree F

Solution

Locate and Calculate the Maximum normal stress , maximum shear stress in Propane tank.

Propane Tank Inner Diameter: Di = 16”

Propane Tank outer Diameter: Do = 18”

Tank height h = 36”

Wall thickness is t = (Do-Di)/2= 1”.

Pressure inside the cylinder: 120 psi

Do/t = 18. The tank may be considered as thick wall cylinder.

There two types of stress on the cylinder.

First is the hoop (tangential) stress . it is max at the inner end and slightly lower at outer end.

Magnitudes are

At the inner end

_ti= p x(Do² +Di²)/ (Do² - Di²) = 120 x 8.529 = 1023 psi

At the outer end

_to = 2p xDi²/ (Do² - Di²) = 120 x 7.529 = 903 psi

Second is the radial stress: It is max at the inner end and zero at the outer end.

At the inner end

_ri= p= 120 psi

At the outer end

_ro= 0 psi

The hoop and radial stresses are uniform all along the length of the tank.

NOW COMES Mr X trying to operate the valve thru a lever.

Weight of Mr X = 300 lbs.

Force applied by Mr X is 3G = 3 x 300 = 900 Lbs.

He is applying this force at the end of 3ft (36”) lever. The application force by Mr X has two fold effect. He is applying torque of 900 x 36 = 32,400Lb-in and bending load 900Lb at the end of the valve.

Let us calculate the maximum torsional stress in the tank.

Calculate J= Polar Moment of Inertia = PI x (Do⁴-Di⁴)/32 =3872 in⁴.

Max shear stress due to torsion on the outer diameter : _max = torque/(2J/Do) = 75 psi. This stress uniform across the length of the tank.

The propane tank may be taken as cantilever beam. That is 36+2 = 38” from the wall. This leads to a bending moment = 900 x 38” = 34,200Lb-in

Calculate the section Modulus : Z = PI x (Do⁴-Di⁴)/64 * 2/Do = 215 in³.

Bending stress : _x = Bending Moment/Z = 34,200/215 = 159 psi. This Maximum at the vertical end and minimum i.e = -159 psi at bottom end of the tank adjoining the wall.

Locate and Calculate the Maximum normal stress due to bending is as shown below. Maximum shear stress in Propane tank due to torsion is 75psi on the outer dia. Maximum tangential pressure is on the inner dia.

From the above it is clear that the loads due to Mr X action or internal pressure are do not pose any serious threat to the tank.

However, the valve at the end of the tank may give away due action of Mr X. It largely depends on it size. If it below 1”, will surely give away.

Now take the problem as Thin walled Cylinder.

Do = 18”, t= 0.1”

Then the tangential stress in the tank = PDo/(2t)= 10800 psi. This at the outer diameter.

Torsional stress due action of Mr X: 32,400/(PI/32 (Do⁴-Di⁴)*2/Do)= 648psi

Bending stress : 34,200/(PI/64 (Do⁴-Di⁴)*2/Do)=1368psi

All these stresses are very nominal and will not pose any threat to cylindrical tank design.

Practically, the stell properties for the tank are Ftu = 155 ksi, Fty = 145 ksi, Fsu= 95 ksivery low stresses were computed when wall thickness was more. The thin walled cylinder design is still safe and good to use.