Question 8.1: A compressed-air tank having an inner diameter of 5.5 m and ...
A compressed-air tank having an inner diameter of 5.5 m and a wall thickness of 45 mm is formed by welding two steel hemispheres (Fig. 8-5).
(a) If the allowable tensile stress in the steel is 93 MPa, what is the maximum permissible air pressure p_{a} in the tank?
(b) If the allowable shear stress in the steel is 42 MPa, 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 210 GPa 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 7.5 MN per meter of weld. If the required factor of safety against failure of the weld is 2.5, what is the maximum permissi-ble pressure p_{d} ?
(e) Considering the four preceding factors, what is the allowable pressure p_{\text {allow }} in the tank?
(a) Allowable pressure based upon the tensile stress in the steel. The maxi-mum tensile stress in the wall of the tank is given by the formula σ = pr/2t [see Eq. (8-5) \sigma=\frac{p r}{2 t}]. Solving this equation for the pressure in terms of the allowable stress, we get
p_{a}=\frac{2 t \sigma_{\text {allow }}}{r}=\frac{2(45 \mathrm{~mm})(93 \mathrm{~MPa})}{2.75 \mathrm{~m}}=3.04 \mathrm{~MPa}Thus, the maximum allowable pressure based upon tension in the wall of the tank is p_{a} = 3.04 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 maxi-mum shear stress in the wall of the tank is given by Eq. (8-7) \tau_{\max }=\frac{\sigma}{2}=\frac{p r}{4 t}, from which we get the following equation for the pressure:
p_{b}=\frac{4 t \tau_{\text {allow }}}{r}=\frac{4(45 \mathrm{~mm})(42 \mathrm{~MPa})}{2.75 \mathrm{~m}}=2.75 \mathrm{~MPa}Therefore, the allowable pressure based upon shear is p_{b} = 2.75MPa .
(c) Allowable pressure based upon the normal strain in the steel. The normal strain is obtained from Hooke’s law for biaxial stress [Eq. (7-40a)\varepsilon_{z}=-\frac{v}{E}\left(\sigma_{x}+\sigma_{y}\right)]:
\varepsilon_{x}=\frac{1}{E}\left(\sigma_{x}-v \sigma_{y}\right) (a)
Substituting \sigma_{x}=\sigma_{y}=\sigma=p r / 2 t (see Fig. 8-4a), we obtain
\varepsilon_{x}=\frac{\sigma}{E}(1-v)=\frac{p r}{2 t E^{-}}(1-v) (8-10)
This equation can be solved for the pressure p_{c}:
p_{c}=\frac{2 t E \varepsilon_{\text {allow }}}{r(1-v)}=\frac{2(45 \mathrm{~mm})(210 \mathrm{~GPa})(0.0003)}{2.75 \mathrm{~m}(1-0.28)}=2.86 \mathrm{~MPa}Thus, the allowable pressure based upon the normal strain in the wall is p_{c} = 2.86 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_{\text {allow }}=\frac{T_{\text {failure }}}{n}=\frac{7.5 \mathrm{~MN}/\mathrm{~m}{}}{2.5} = 3 \mathrm{~MN/m}The corresponding allowable tensile stress is equal to the allowable load on a one-meter length of weld divided by the cross-sectional area of a one-meter length of weld:
\sigma_{\text {allow }}=\frac{T_{\text {allow }}(1.0 \mathrm{~m})}{(1.0 \mathrm{~m})(t)}=\frac{3 \frac{\mathrm{~MN}}{\mathrm{~m}}(1 \mathrm{~m})}{(1 \mathrm{~m})(45 \mathrm{~mm})}=66.667 \mathrm{~MPa}Finally, we solve for the internal pressure by using Eq. (8-5):
p_{d}=\frac{2 t \sigma_{\text {allow }}}{r}=\frac{2(45 \mathrm{~mm})(66.67 \mathrm{~MPa})}{2.75 \mathrm{~m}}=2.18 \mathrm{~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 tension in the welded seam governs and the allowable pressure in the tank is
p_{\text {allow }}=2.18 \mathrm{~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 (2.18 MPa), the tensile stresses in the shell are
Thus, at the inner surface of the shell (Fig. 8-4b), the ratio of the princi-pal stress in the z direction (2.18 MPa) to the in-plane principal stresses (66.6 MPa) is only 0.033. Therefore, our earlier assumption that we can disregard the principal stress σ_3 in the z direction and consider the entire shell to be in biaxial stress is justified
