Question 8.1: A compressed-air tank having an inner diameter of 18 inches ...
A compressed-air tank having an inner diameter of 18 inches and a wall thickness of 1/4 inch is formed by welding two steel hemispheres (Fig. 8-5).
(a) If the allowable tensile stress in the steel is 14,000 psi, what is the maximum permissible air pressure p_{a} in the tank?
(b) If the allowable shear stress in the steel is 6,000 psi, what is the maximum permissible pressure p_{b}?
(c) If the normal strain at the outer surface of the tank is not to exceed 0.0003, what is the maximum permissible pressure p_{c}? (Assume that Hooke’s law is valid and that the modulus of elasticity for the steel is 29 × 10^{6} psi and Poisson’s ratio is 0.28.)
(d) Tests on the welded seam show that failure occurs when the tensile load on the welds exceeds 8.1 kips per inch of weld. If the required factor of safety against failure of the weld is 2.5, what is the maximum permissible pressure p_{d}?
(e) Considering the four preceding factors, what is the allowable pressure p_{allow} in the tank?
(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. 8-1). Solving this equation for the pressure in terms of the allowable stress, we get
p_{a}=\frac{2t\sigma_{allow}}{r}=\frac{2(0.25_\ in.)(14,000_\ psi)}{}=777.8_\ psiThus, the maximum allowable pressure based upon tension in the wall of the tank is p_{a} = 777 psi. (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. (8-3), from which we get the following equation for the pressure:
\tau_{\max}=\frac{\sigma}{2}=\frac{pr}{4t} (8-3)
p_{b}=\frac{4t\tau_{allow}}{r}=\frac{4(0.25_\ in.)(6,000_\ psi)}{}=666.7_\ psiTherefore, the allowable pressure based upon shear is p_{b} = 666 psi.
(c) Allowable pressure based upon the normal strain in the steel. The normal strain is obtained from Hooke’s law for biaxial stress (Eq. 7-39a):
\epsilon_{x}=\frac{1}{E}\left(\sigma_{x}-\nu \sigma_{y}\right) (h)
Substituting \sigma_{x} = \sigma_{y} = σ = pr/2t (see Fig. 8-4a), we obtain
\epsilon_{x}=\frac{\sigma}{E}\left(1-\nu\right)=\frac{pr}{2tE}\left(1-\nu\right) (8-4)
This equation can be solved for the pressure p_{c}:
p_{c}=\frac{2tE\epsilon_{allow}}{r(1-\nu)}=\frac{2(0.25_\ in.)(29 \times 10^{6}psi)(0.0003)}{(9.0_\ in.)(1-0.28)}=671.3_\ psiThus, the allowable pressure based upon the normal strain in the wall is p_{c} = 671 psi.
(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:
The corresponding allowable tensile stress is equal to the allowable load on a one-inch length of weld divided by the cross-sectional area of a one-inch length of weld:
\sigma_{allow}=\frac{T_{allow}(1.0_\ in.)}{(1.0_\ in.)(t)}=\frac{(3240_\ lb/in.)(1.0_\ in)}{}=12,960_\ psiFinally, we solve for the internal pressure by using Eq. (8-1):
p_{d}=\frac{2t\sigma_{allow}}{r}=\frac{2(0.25_\ in.)(12,960_\ psi)}{}=720.0_\ psiThis 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} = 666 psi
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 (666 psi), the tensile stresses in the shell are
\sigma=\frac{pr}{2t} (8-1)
=\frac{(666_\ psi)(9.0_\ in.)}{}=12,000_\ psiThus, at the inner surface of the shell (Fig. 8-4b), the ratio of the principal stress in the z direction (666 psi) to the in-plane principal stresses (12,000 psi) is only 0.056. 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.
