Question 5.11: An internal stainless tray is welded to the inside of a carb...
An internal stainless tray is welded to the inside of a carbon-steel vessel as shown in Figure 5.21a. If the coefficient of thermal expansion is 9.5 × 10^{−6} in./in.- °F for the tray and 6.7 × 10^{−6} in./in.- °F for the vessel, what is the stress in the weld due to a temperature increase of 400 °F? Use E =28 × 10^{6} psi and \mu=0.3 .
A conservative answer can be obtained by assuming the tray attachment to be rigid. Because the weld is subjected to both hoop and axial stresses, it can be treated as a biaxial condition. Hence, from Eq. (5.29b),
\sigma_{x}=\sigma_{y}=-\frac{\alpha \Delta T E}{1-\mu} for a biaxial case. (5.29b)
\sigma=\frac{-\left(28 \times 10^{6}\right)(9.5-6.7)\left(10^{-6}\right)(400)}{1-0.3}= −44,800 psi.
If a more accurate result is desired, then a discontinuitytype analysis can be performed. Referring to Figure 5.21b, and taking account of symmetry, the equations of compatibility and equilibrium can be written as
\delta_{3}=\delta_{1} (a)
\theta_{1}=0 (b)
\theta_{2}=0 (c)
\sum F=0 (d)
From the aforementioned four equations, the four unknowns F_{1}, F_{2}, F_{3}, and M_{0} can be obtained.
Equation (a) can be written as
deflection of tray due to temperature + F_{3}
= deflection of shell due to
temperature + F_{1} – M_{0}
\left(\alpha_{ ss }\right)(\Delta T)(r)+\frac{F_{3} r}{E T}(1-\mu)
=\left(\alpha_{c s}\right)(\Delta T)(r)+\frac{F_{1}}{2 \beta^{3} D}-\frac{M_{0}}{2 \beta^{2} D} (1)
Eqs. (b)–(d) can be written as
\frac{F_{1}}{2 \beta^{2} D}-\frac{M_{0}}{\beta D}=0 (2)
\frac{F_{2}}{2 \beta^{2} D}-\frac{M_{0}}{\beta D}=0 (3)
F_{1}+F_{2}+F_{3}=0 (4)
From Eq. (2),
F_{1}=2 \beta M_{0}From Eq. (3),
F_{2}=2 \beta M_{0}From Eq. (4),
F_{3}=-4 \beta M_{0}and Eq. (1) becomes
F_{3}\left[\frac{r(1-\mu)}{E T}+\frac{1}{8 \beta^{3} D}\right]=\left(\alpha_{ cs }-\alpha_{ ss }\right)(\Delta t)(r) (5)
Assuming the thickness of the cylinder is t =0.1875 in. and using the other given values, the value of F_{3} from Eq. (5) is
F_{3} = −618lb∕in. or
\sigma = 618 psi compression in weld.
The value 618 psi is significantly lower than the conservative value 44,800 psi obtained from Eq. (5.29b), because of the flexibility of the cylinder. If the thickness of the cylinder is t =3.0 in., then Eq. (5) gives
F_{3} = 21,200lb∕in. or
\sigma = 21,200 psi compression in weld.
This value indicates that as the cylinder gets thicker, the stress approaches that of Eq. (5.29b). In fact, if the cylinder is taken as infinitely rigid, then Eq. (1) becomes
\left(\alpha_{ ss }\right)(\Delta T)(r)+\frac{F_{3} r}{E t}(1-\mu)=\left(\alpha_{ cs }\right)(\Delta T)(r)and the equation yields F_{3} = −44,800 lb/in. or \sigma =44,800 psi compression in the weld, which is the same as that obtained from Eq. (5.29b).