Handbook of Civil Engineering Calculations

FIGURE 22 Thick-walled cylinder under internal and exter-

nal pressure.

3. Check the accuracy of the results
   Use the relation ^r 1 - s 0 r 2 = [(V 2 - ^ 1 X(V 2 + r)](p^ 1 + p 2 r 2 ). Substituting the known val-
   ues verifies the earlier calculations.

THERMAL STRESS RESULTING FROM

HEATING A MEMBER

A steel member 18 ft (5.5 m) long is set snugly between two walls and heated 8O^0 F
(44.4^0 C). If each wall yields 0.015 in (0.381 mm), what is the compressive stress in the
member? Use a coefficient of thermal expansion of 6.5 x 10-^6 /°F (1.17 x 10~^5 /°C) for
steel.

Calculation Procedure:

1. Compute the thermal expansion of the member without
   restraint
   Replace the true condition of partial restraint with the following equivalent conditions:
   The member is first allowed to expand freely under the temperature rise and is then com-
   pressed to its true final length.
   To compute the thermal expansion without restraint, use the relation AL = cLkT,
   where c - coefficient of thermal expansion, /^0 F (/^0 C); AJ = increase in temperature,^0 F
   (^0 C); L = original length of member, in (mm); AZ, = increase in length of the member, in
   (mm). Substituting gives M = 6.5(10-^6 XlS)(^)(SO) = 0.1123 in (2.852 mm).


2. Compute the linear restraint exerted by the walls
   The walls yield 2(0.015) = 0.030 in (0.762 mm). Thus, the restraint exerted by the walls is
   AZ,W = 0.1123 - 0.030 = 0.0823 in (2.090 mm).