MATERIALS ENGINEERING CONCEPTS - Materials for Civil and Construction Engineers - page 24

L = original length of the specimen

in the volume of the specimen
volume of the specimen

dV = change

V = original

For isotropic materials,
The coefficient of thermal expansion is very important in the design of
structures. Generally, structures are composed of many materials that are
bound together. If the coefficients of thermal expansion are different, the
materials will strain at different rates. The material with the lesser expan-
sion will restrict the straining of other materials. This constraining effect
will cause stresses in the materials that can lead directly to fracture.
Stresses can also be developed as a result of a thermal gradient in the
structure. As the temperature outside the structure changes and the tem-
perature inside remains constant, a thermal gradient develops. When the
structure is restrained from straining, stress develops in the material. This
mechanism has caused brick facades on buildings to fracture and, in some
cases, fall off the structure. Also, since concrete pavements are restrained
from movement, they may crack in the winter due to a drop in temperature
and may “blow up” in the summer due to an increase in temperature. Joints
are, therefore, used in buildings, bridges, concrete pavements, and various
structures to accommodate this thermal effect.

a V = 3a L .

Sample Problem 1.5

A steel bar with a length of 3 m, diameter of 25 mm, modulus of elasticity of 207 GPa,
and linear coefficient of thermal expansion of 0.000009 m/m/°C is fixed at both ends
when the ambient temperature is 40°C. If the ambient temperature is decreased to
15°C, what internal stress will develop due to this temperature change? Is this stress
tension or compression? Why?

Solution
If the bar was fixed at one end and free at the other end, the bar would have contract-
ed and no stresses would have developed. In that case, the change in length can be
calculated by using Equation 1.13 as follows:

d L = a L * d T * L = 0.000009 * 1 - 25 2 * 3 =- 0.000675 m

e = d L / L =- 0.000675/3 =- 0.000225 m/m

Since the bar is fixed at both ends, the length of the bar will not change. Therefore, a
tensile stress will develop in the bar as follows:

s = e E = 0.000225 * 207000 = 46.575 MPa

The stress will be tension; in effect, the length of the bar at 15°C without restraint
would be 2.999325 m and the stress would be zero. Restraining the bar into a longer
condition requires a tensile force.