Question 7.1: A compressed-air tank having an inner diameter of 450 mm and...

(a) Allowable pressure based upon the tensile stress in the steel. The maximum tensile stress in the wall of the tank is given by the formula σ = pr/2t (see Eq. 7-1). Solving this equation for the pressure in terms of the allowable stress, we get

$frac{pr}{2t}$                 (7-1)

$p_a=\frac{2t\sigma _{allow}}{r}  =  \frac{2(7  mm)(115  MPa)}{225  mm}$ = 7.16 MPa

Thus, the maximum allowable pressure based upon tension in the wall of the tank is $p_a$ = 7.1 MPa. (Note that in a calculation of this kind, we round downward, not upward.)

(b) Allowable pressure based upon the shear stress in the steel. The maximum shear stress in the wall of the tank is given by Eq. (7-3), from which we get the following equation for the pressure:

$\tau _{max}=\frac{\sigma }{2}=\frac{pr}{4t}$                   (7-3)

$p_b=\frac{4t\tau _{allow}}{r} = \frac{4(7  mm)(40  MPa)}{225  mm}$ = 4.98 MPa

Therefore, the allowable pressure based upon shear is $p_b$ = 4.9 MPa.

(c) Allowable pressure based upon the normal strain in the steel. The normal strain is obtained from Hooke’s law for biaxial stress (Eq. 6 39a):

$\epsilon _x=\frac{1}{E} (\sigma _x-\nu \sigma _y)$                (6 39a)    (h)

Substituting $\sigma _x = \sigma _y =\sigma$ =pr/2t (see Fig. 7-4a), we obtain

$\epsilon _x=\frac{\sigma}{E} (1-\nu )=\frac{pr}{2tE} (1-\nu )$                (7-4)

This equation can be solved for the pressure $p_c$:

$p_c=\frac{2tE\epsilon _{allow}}{r(1-\nu )} =\frac{2(7  mm)(210  GPa)(0.0003)}{(225  mm)(1  –  0.28)} = 5.44  MPa$

Thus, the allowable pressure based upon the normal strain in the wall is $p_c$ = 5.4 MPa.

(d) Allowable pressure based upon the tension in the welded seam. The allowable tensile load on the welded seam is equal to the failure load divided by the factor of safety:

$T_{allow}=\frac{T_{failure}}{n} =\frac{1.5  MN/m}{2.5} =0.6  MN/m = 600  N/mm$

The corresponding allowable tensile stress is equal to the allowable load on a one millimeter length of weld divided by the cross-sectional area of a one millimeter length of weld:

$\sigma _{allow}=\frac{T_{allow}(1  mm)}{(1  mm)(t)} =\frac{(600  N/mm)(1  mm)}{(1  mm)(7  mm)}$ = 85.7 MPa

Finally, we solve for the internal pressure by using Eq. (7-1):

$p_d=\frac{2t\sigma _{allow}}{r} =\frac{2(7  mm)(85.7  MPa)}{225  mm}$ = 5.3 MPa

This result gives the allowable pressure based upon tension in the welded seam.

(e) Allowable pressure. Comparing the preceding results for $p_a$ , $p_b$ , $p_c$ , and $p_d$ , we see that shear stress in the wall governs and the allowable pressure in the tank is

$p_{allow}$ = 4.9 MPa

This example illustrates how various stresses and strains enter into the design of a spherical pressure vessel.

Note: When the internal pressure is at its maximum allowable value (4.9 MPa), the tensile stresses in the shell are

$\sigma =\frac{pr}{2t} =\frac{(4.9  MPa)(225  mm)}{2(7  mm)}$ = 78.8 MPa

Thus, at the inner surface of the shell (Fig. 7-4b), the ratio of the principal stress in the z direction (4.9 MPa) to the in-plane principal stresses (78.8 MPa) is only 0.062. Therefore, our earlier assumption that we can disregard the principal stress $\sigma_3$ in the z direction and consider the entire shell to be in biaxial stress is justified.