Solve Example 3.7 for the case where p = 689.4 kPa is applied externally and ΔT = 100°C is...

Solve Example 3.7 for the case where p = 689.4 kPa is applied externally and ΔT = 100°C is a decrease in temperature. Discuss the results.

Example 3.7

Composite ThinWall Cylinder Subjected to Pressure and Temperature Increase

Consider a composite cylinder of length L formed from an inner cylinder of aluminum with outer radius R andthickness tA and an outer cylinder of steel with inner radius R and thickness tS(Figure E3.7a); tA≪ R and ts ≪ R. The composite cylinder is supported snugly in an upright, unstressed state between rigid supports. An inner pressure p is applied to the cylinder (Figure E3.7b), and the entire assembly is subjected to a uniform temperature change ΔT. Determine the stresses in both the aluminum and the steel cylinders for the case tA = ts = t = 0.02R. For aluminum, EA = 69 GPa, νA = 0.333, and αA = 21.6 × 10−6/°C.For steel, ES = 207 GPa, vS = 0.280, and αS = 10.8 × 10−6/°C.

FIGURE E3.7 (a) Composite cylinder. (b)Cross section A–A. (c) Longitudinal section B–B. (d)Cylinder element.

Solution:

Since both cylinders are thin, we may assume that the stresses in the tangential direction θ, σθAand σθS in the aluminum and steel, respectively, are constant through the thicknesses tAand tS(Figure E3.7c). Also, it is sufficiently accurate to use the approximation R − t = R. From the free-body diagram of Figure E3.7c, we have ƩF = 2pRL − 2σθStL − 2σθAtL = 0. Hence,

Since ordinarily the radial stress σr in the cylinder is very small (of the order p) compared with both the tangential stress σθ and the longitudinal stress σL,we assume that σr is negligible. Therefore, the cylinder is subjected approximately to a state of plane stress (σL, σθ)(Figure E3.7d). Hence, for plane stress, the stress–strain–temperature relations for each cylinder are

Equations (b) hold for all points in the cylinder, provided that the ends are free to expand radially. The cylinder is restrained from expanding longitudinally, since the end walls are rigid. Then, ϵL = 0. Also, at radial distance R (the interface between the aluminum and the steel sleeves), the radial displacement is u and the tangential strain is ϵθ = [2π (R + u) − 2πR]/2πR = u/R. Assuming that t is so small that this strain is the same throughout the aluminum and the steel sleeves, we have by Eqs. (b)

Also, from the given data, 3EA = ESand αA = 2αS. Therefore, with Eqs. (a) and (c), we may write

By the first two of Eqs. (d) and with ESαS = 2.236 MPa/°C, we find that

Substitution of Eqs. (e) into the last of Eqs. (d) yields for the tangential stress in the steel cylinder

σθS = 37.16p + 0.8639(ΔT)    (f)

By Eqs. (a) and (f), we find the tangential stress in the aluminum cylinder to be

σθA = 50p −37.16p − 0.8639(ΔT) = 12.84p − 0.8639(ΔT)

and by Eqs. (e) and (f), we find the longitudinal stresses in the aluminum and steel cylinders, respectively,

σLA = 4.28p − l.779(ΔT)

σLS = 10.40p −1.994(ΔT)

Thus, for p = 689.4 kPa and ΔT = 100°C

σθA = −77.4MPa, σLA = −175 MPa

σθS = 112 MPa, σLS = −192 MPa