
MECHANICS  EXAMPLE



Example

Gas Tank under Pressure


During testing, a thinwalled tank was pressurized with gas and the stresses in the x and y direction were found to be σ_{x} = 150 MPa, σ_{y} = 75 MPa, and τ_{xy} = 0. However, in actual use, the tank must also withstand a torque on the cap that will introduce a possible shear stress in the tank walls. This shear stress is in addition to the stresses due to the tank pressure. What is the maximum torque that can be applied if the vessel material can only withstand a shear stress of 100 MPa? The wall thickness is 15 mm and the outside diameter is 35 cm. Also, the maximum allowable shear stress for the vessel material is 100 MPa.






Solution

Mohr's Circle 

Mohr's circle can be used to understand the solve for this stress state by first finding the largest radius R for which the shearing stress does not exceed 100 MPa and then determine the allowable torque.
To draw the Mohr's circle, first the center should be determined. Since, (σ_{x} + σ_{y})/2 = (150 + 75)/2 = 112.5, this will be at (112.5 , 0).
Assuming a radius R, the Mohr's circle is plotted on the left. From the diagram principal stresses are,
σ_{1} = (σ_{x} + σ_{y})/2  R
= 112.5  R
σ_{2} = (σ_{x} + σ_{y})/2 + R
= 112.5 + R
where R =  (σ_{1}  σ_{2})/2 






The goal is to determine the largest radius R for which the shear stress does not exceed 100 MPa. This can be found using "Maximum Shear Stress Theory" that states that for any combination of loading for σ_{1} and σ_{2}, the shear stress cannot exceed σ_{yld}/2 = τ_{max}. This condition gives three separate possible situations that need to be checked,
In this problem, the maximum allowable shear stress is given, and not the yield stress. However, the yield stress is simply two times the yield stress for a uniaxial test. The three conditions become,
1st:  σ_{1}  = σ_{yld} = 2τ_{max} = 2(100)
 112.5  R  = 200
R = 312.5 MPa
2nd:  σ_{2}  = σ_{yld} = 2τ_{max} = 2(100)
 112.5 + R  = 200 MPa
R = 87.5 MPa
3rd:  σ_{1} σ_{2}  = σ_{yld} = 2τ_{max} = 2(100)
R =  (σ_{1}  σ_{2})/2  = 100 MPa




Mohr's Circle for τ 

Comparing these three situations, the maximum value of R is the minimum of above three.
R = 87.5 MPa
From the Mohr's Circle diagram corresponding torsional shear stress is found as,
τ = ( R^{2}  (112.5  75) ^{2} )^{1/2} = 79.06 MPa
Hence, the torque that will cause this torsional shearing stress,
T = τJ / r
= [(79.06) (π/2) (r_{o}^{4}  r_{i}^{4})] / (0.35/2)
= 124.2 [(0.35/2)^{4}  (0.35/2  0.015)^{4}] / 0.175
= 0.2005 MNm = 200.5 kNm 
