[1] A thin-walled spherical gas container (radius: r, thickness: t) is required

Question

[1] A thin-walled spherical gas container (radius: r, thickness: t) is required to be designed for an internal pressure of 1750 psi. Find the smallest mass of the container, if the factor of safety (SF) is 3 and the material used is structural steel (mass density,p=0.281bm/in2) The normal stress and shear stress of the container have to be less than 50 ksi and 45 ksi, respectively. The stress in this thin-walled structure can be evaluated as follows: pr 2t Note: The factor of safety (SF) needs to be multiplied to these stresses to compare with those limit stresses. (Identify design variables. -2 pts. (2) Identify design parameters. - 2 pts. (3) Set up the optimization problem in standard format. 6 pts Total: 10 pts Page 1 of 1

Explanation / Answer

Given Data:

A spherical gas container, radius = r , thickness = t

Material of the container is structural steel , mass density = 0.28 lbm /in3

To find : The smallest mass of the container when FOS = 3

Normal Stress < 50 ksi (Ultimate tensile stress)

Sheer Stress < 45 ksi (Ultimate sheer stress)

Internal pressure to be maintained inside the system = P = 1750 psi

Answer:

The design varaibles are radius r, thickness t, mass m, internal pressure P, density and factor of safety FOS

It is mentioned that the normal ans sheer stresses has to be less than the mentioned values.

Therefore, the given normal and sheer stresses are the ultimate tensile ans sheer stresses above which the material will go into a permanent deformation stage and below which the original dimensions of the material can be restored.

Given factor of safety = 3 = Ultimate stress /Permissible stress

Therefore,

The design parameters are obtained as follows:

Permissible tensile stress = 50/3 = 16.67 ksi

Permissible Sheer stress = 45/3 = 15 ksi

According to the given formula for normal stress ,

= Pr/2t = 16.67 ksi

r/t = (16.67*2) *1000/1750 = 19.048

According to the given formula for sheer stress,

= Pr/4t = 15 ksi

r/t = 15000*4/1750 = 34

The smallest mass can be obtained with the smaller ratio of r/t i.e. = 19

therefore,

mass = densilty *volume = 0.28 * [{4/3*22/7*(19t)3 } - {4/3*22/7*(19t-t)3 }]

                                        = 0.28 *4/3*22/7*1027t

                                        = 120.5t lb

Therefore the optimisation is also set up in the format above, in both the cases of obtained r/t ratios, the values of r and t if given, can give us the values of masses in both the cases.

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