1. Mechanics of Materials(ME-294)
Lecture 5:
Thermal Stress and Strain

2. Thermal Deflection From the graph, the slope α=εt /ΔT.
Since strain is a dimensionless quantity, the coefficient of thermal expansion has units equal to the reciprocal of temperature change.
In SI units the dimensions of α can be expressed as either 1/K (the reciprocal of kelvins) or 1/°C (the reciprocal of degrees Celsius). The value of α is the same in both cases because a change in temperature is numerically the same in both kelvins and degrees Celsius.
Substitute the definition of strain, ε=δ/L , and we have α= δ /L(ΔT) .
Rewrite the equation to solve for thermal deflection: δ=αL(ΔT) .

3. Thermal Expansion and Thermal Stress
Heat a piece of steel and it will expand. Cool the same piece, and it will shrink.
Plot the strain as a function of temperature change, and for most materials, you get a relatively straight line.
The slope of the line is called the coefficient of thermal expansion and is denoted by Greek letter (alpha) α.
It tells us how much strain we can expect for a given temperature change.

4. Coefficient of Thermal Expansion
The thermal expansion coefficient is a materials property; different materials expand at different rates.
For example, aluminum expands about twice as much as steel for a given temperature change, because α for Aluminum=23×10−6°C−1 and α for Steel=12×10−6°C−1 .
One reason we use steel as a reinforcement in concrete is α for Concrete=11×10−6°C−1 , so the steel and concrete expand and contract at roughly the same rate.
If the matrix and reinforcement in a composite expand at different rates, then the matrix and reinforcement may separate under repeated thermal cycles.
The change in temperature can be positive or negative, and is defined as the final temperature minus the original temperature: ΔT=Tf−To . If the material is cooled from 70°F to 40°F, the change in temperature is ΔT=40°F−70°F=−30°F . If the material is heated from 70°F to 90°F, the change in temperature is ΔT=90°F−70°F=+20°F

5. Example
A 5 m aluminum flagpole is installed at 20°C. Overnight, the temperature drops to -5°C. How much does the height change, in millimeters? What is the final height of the flagpole, in meters? Thermal expansion coefficient for aluminum is α for Aluminum=23×10−6°C−1 Solution : First, calculate the change in length using δ=αL(ΔT). Change in length δ=αL(ΔT)=−2.88mm . The negative sign indicates the flagpole is getting shorter. Final length Lf=L+δ=5m−2.88mm =4.997m

6. Example
Two cantilever beams made of different materials have a measurable gap between their ends. As the bars heat up, they grow towards each other and eventually meet if the temperature rises enough. Each bar has a different thermal coefficient of expansion. How do we calculate the temperature
T f at which they meet?
Consider that the gap between the two bars equals
δ total = δ steel + δ brass
Substitute the equation for thermal expansion, and we get
The change in temperature is the same for both materials, so Rewrite the equation to solve for temperature change:
final temperature:

7. Thermal Stress
If the material is restrained from expanding or contracting while the temperature changes, then stress builds within the part.
Consider a bar of steel embedded in two blocks of concrete. If the bar heats up, it will want to expand by an amount δ=α L (ΔT) .
However, the blocks of concrete prevent the bar from expanding, by exerting a force P on the bar. We know from that an axial load will cause a bar to deform an amount . However, the load P in this problem is compressive
Set the two deflections equal to each other
The thermal stress does not depend on the length of the bar; it depends only on materials constants α and E and the temperature change

8. Example
Two immovable concrete blocks are connected by a steel wire. At 72°F there is no stress in the wire. If the wire cools from 72°F to 55°F, what is the stress in the wire? α=6.5×10−6°F−1 and E=30×106psi Solution : Thermal stress σ=−αE(ΔT)=−6.5×10−6 °F 30×106lb. in.2 (55°F−72°F) =3,315psi The positive sign indicates the wire is under a tensile stress. The wire cooled and wanted to shrink, but the concrete blocks prevented it from shrinking, leaving the wire in tension

9. Example
Two immovable concrete blocks are connected with a 2 inch by 2 inch square bar. At 72°F there is no stress in the bar. If the bar heats from 72°F to 102°F, what is the stress in the bar? How much force do the blocks exert on the bar? α=6.5×10−6°F−1 and E=30×106psi .
Solution
Thermal stress σ=−αE(ΔT)=−6.5×10−6 °F 30×106lb. in.2 (102°F−72°F) =−5,850psi
The negative sign indicates the bar is under a compressive stress.
The bar heated up and wanted to expand, but the concrete blocks prevented it from expanding, leaving the bar in compression.
We can find the force exerted by the walls from the thermal stresss: σ thermal=P / A .Rewrite the equation to solve for force: P thermal=σ thermal /A=−5,850lb. in.2 2in.×2in.=−23,400lb. The negative sign shows that the force is compressive.

10. Thermal Stress
Some thermal expansion problems require both the deflection and the stress equations.
For example, if this cantilever beam heats up sufficiently, it will meet the right-hand wall.
If the temperature continues to rise, stress will build up in the beam.
If you know the initial length, material, and temperatures To and T2 (but not temperature T1), how do you find the thermal stress?
Use the thermal deflection equation and the temperature change ΔT=T1−To to figure out temperature T1, then use the thermal stress equation and the temperature change ΔT=T2−T1 to figure out σ thermal at temperature T2.