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Wize University Physics Textbook (Master) > Heat and Temperature

Thermal Stress

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Thermal Stress and Strain

  • Inserting a force [ on the cross-sectional area AA of a rod causes a stress in the rod, which is expressed as:
Stress=FA=YΔLL0Stress=\dfrac{F}{A}=Y \dfrac{ \Delta L}{L_0}

  • ΔL\Delta L is the change in the length of the rod due to inserting force FF
  • YY is Young’s modulus which is a constant that depends on the material the rod is made of
  • Strain is the relative change in the length ΔLL0\frac{\Delta L}{L_0}
Stress=Y⋅StrainStress=Y \cdot Strain

  • The Young’s modulus of different material is shown in Table 3 below
MaterialY[109Nm2]Aluminum69Brass110Copper117Glass50−90Steel190\begin{array}{|l|c|}\hline Material & Y [10^9 \dfrac{N}{m^2} ]\\[7pt]\hline Aluminum & 69\\[7pt]\hline Brass & 110\\[7pt]\hline Copper & 117\\[7pt]\hline Glass & 50-90\\[7pt]\hline Steel & 190\\[7pt]\hline \end{array}
Young’s Modulus of Different Materials

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  • Preventing the expansion or contraction of a rod due to change in its temperature causes a stress in the rod which is called thermal stress
  • To calculate the thermal stress, we find the change in the length of the rod if it was free to expand or contract and find the stress needed to bring it back to its original length
ΔLthermal=αΔTL0ΔLtension=1YFAL0ΔLthermal+ΔLtension=0→αΔT=−1YFA or FA=−YαΔTthermal stress\begin{array}{c} \Delta L_{thermal}=\alpha \Delta TL_0\\[5pt] \Delta L_{tension}=\frac{1}{Y}\frac{F}{A} L_0\\[5pt] \Delta L_{thermal}+\Delta L_{tension}=0\\[5pt] \to \alpha \Delta T=-\frac{1}{Y}\frac{F}{A}\text{ or }\frac{F}{A}=-Y\alpha\Delta T \qquad thermal\ stress \end{array}
  • For a single rod fixed at both ends:
∣ΔLthermal∣=∣ΔLtension∣|\Delta L_{thermal} |=|\Delta L_{tension} |


  • The direction of FF is opposite to the change in the length
  • The inward force is negative and the outward force is positive

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  • For two rods in series and fixed at both ends:
∣ΔLthermalA∣+∣ΔLthermalB∣=∣ΔLtensionA∣+∣ΔLtensionB∣|\Delta L_{thermal}^A |+|\Delta L_{thermal}^B |=|\Delta L_{tension}^A |+|\Delta L_{tension}^B |


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Example: Thermal Stress

How much is the stress in each segment of the compound bar shown below if its temperature increases by 100°C100\degree C?
(αAl=2.4×10−5K−1,αSt=1.2×10−5K−1,YAl=69×109Pa,YSt=180×109Pa)\begin{array}{l} (\alpha_{Al}=2.4 \times10^{-5} K^{-1},\alpha_{St}=1.2 \times 10^{-5} K^{-1},\\ Y_{Al}=69 \times 10^9 Pa,Y_{St}=180 \times 10^9 Pa) \end{array}

∣ΔLthermalSt+ΔLthermalAl∣=∣ΔLtensionSt+ΔLtensionAl∣|\Delta L_{thermal}^{St}+\Delta L_{thermal}^{Al}|=|\Delta L_{tension}^{St}+\Delta L_{tension}^{Al}|
ΔLthermalSt=L0StαStΔT=9×10−4 m\Delta L_{thermal}^{St}=L_0^{St}\alpha_{St}\Delta T=9\times10^{-4}\ m
ΔLthermalAl=L0AlαAlΔT=1.44×10−3 m\Delta L_{thermal}^{Al}=L_0^{Al}\alpha_{Al}\Delta T=1.44\times10^{-3}\ m
→ΔLtensionSt+ΔLtensionAl=FStL0StAStYSt+FAlL0AlAAlYAl=2.34×10−3 m\to \Delta L_{tension}^{St}+\Delta L_{tension}^{Al}=\dfrac{F^{St}L_0^{St}}{A_{St}Y_{St}}+\dfrac{F^{Al}L_0^{Al}}{A_{Al}Y_{Al}}=2.34\times10^{-3}\ m
FSt=FAlF^{St}=F^{Al} (third law of Newton)
→F(0.0417×10−9+0.0435×10−9)=2.34×10−3 m\to F(0.0417\times10^{-9}+0.0435\times10^{-9})=2.34\times10^{-3}\ m →F=27.465×106 N\to F=27.465\times10^6\ N
Aluminum Stress = FAAl=27.465×1060.2=137.32 MPa\frac{F}{A_{Al}}=\frac{27.465\times10^6}{0.2}=137.32\ MPa
Steel Stress = FASt=274.65 MPa\frac{F}{A_{St}}=274.65\ MPa

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An aluminum rod is attached to a steel bar as shown in the figure below so that they expand or stretch together. The length of these bars is 2 m2\ m at 10°C10\degree C and the cross-sectional area of both bars are the same. How much is the stress in each bar at 100°C100\degree C?
αSt=1.2×10−51KYSt=20.0×1010Nm2αAl=2.4×10−51KYAl=69×109Nm2\begin{array}{ll} \alpha_{St}=1.2\times10^{-5} \frac{1}{K} & Y_{St}=20.0\times10^{10} \frac{N}{m^2}\\[10pt] \alpha_{Al}=2.4\times10^{-5} \frac{1}{K} & Y_{Al}=69\times10^9 \frac{N}{m^2} \end{array}




Extra Practice
Practice: Thermal Stress
An aluminum rod is attached to a steel bar as shown in the figure below so that they expand or stretch together. The length of these bars is 2 m2\ m at 10°C10\degree C and the cross-sectional area of both bars are the same. How much is the stress in each bar at 100°C100\degree C?
αSt=1.2×10−51KYSt=20.0×1010Nm2αAl=2.4×10−51KYAl=69×109Nm2\begin{array}{ll} \alpha_{St}=1.2\times10^{-5} \frac{1}{K} & Y_{St}=20.0\times10^{10} \frac{N}{m^2}\\[10pt] \alpha_{Al}=2.4\times10^{-5} \frac{1}{K} & Y_{Al}=69\times10^9 \frac{N}{m^2} \end{array}
Practice: Thermal Stress
An aluminum rod is attached to a steel bar as shown in the figure below so that they expand or stretch together. The length of these bars is 2 m2\ m at 10°C10\degree C and the cross-sectional area of both bars are the same. How much is the stress in each bar at 100°C100\degree C?
αSt=1.2×10−51KYSt=20.0×1010Nm2αAl=2.4×10−51KYAl=69×109Nm2\begin{array}{ll} \alpha_{St}=1.2\times10^{-5} \frac{1}{K} & Y_{St}=20.0\times10^{10} \frac{N}{m^2}\\[10pt] \alpha_{Al}=2.4\times10^{-5} \frac{1}{K} & Y_{Al}=69\times10^9 \frac{N}{m^2} \end{array}
When making cars companies cool aluminum bolts to −50° C-50\degree\ C so that they perfectly fit into the car when inserted. After being inserted into the car the bolts warm up to room temperature which is 20° C20\degree\ C.
Find the Stress caused from the bolt after it heats up to room temperature.
αAl=22⋅10−6 K−1α_{Al}=22\cdot10^{-6}\ K^{-1}
A 3 m3\ m bronze bar with a cross-sectional area of 2.5 cm22.5\ cm^2 is placed between two rigid walls as shown below. At T=−15°CT=-15\degree C, the gap is Δ=2 mm\Delta =2\ mm What temperature will make the stress in the bar to be 40 MPa40\ ΜPa?
(αbr=18.0×10−61K\alpha_{br}=18.0\times10^{-6} \frac{1}{K} and Ybr=80×109 PaY_{br}=80\times10^9\ Pa)
A copper sphere with a radius of 5 cm5\ cm at T=80°CT=80\degree C is placed into a 2 kg2\ kg steel vessel which is full of water and ice.
What is the final temperature of the system if the final radius of the copper sphere is 4.996 cm4.996\ cm?
If 1/3 of the initial ice-water mixture in the steel vessel is water, how much ice is melted in this process?(gr)
A compound bar made up of 15 cm15\ cm of Cu with Acu=25 cm2A_{cu}=25\ cm^2 and 20 cm20\ cm of steel with ASt=10 cm2A_{St}=10\ cm^2 and T=293 KT=293\ K is fixed at both ends as shown in the diagram below. How much is the stress in each metal if the system is heated uniformly by a 700 watt700\ watt heater for 5 minutes?
CCu=390 Jkg⋅KCSt=470 Jkg⋅KYCu=11×1010 Nm2YSt=20×1010 Nm2αCu=1.7×10−51KαSt=1.2×10−51KρCu=9 kg/cm3ρSt=7.85 kg/cm3\begin{aligned} C_{Cu}=390\ \frac{J}{kg\cdot K} && C_{St}=470\ \frac{J}{kg\cdot K}\\ Y_{Cu}=11\times10^{10}\ \frac{N}{m^2} && Y_{St}=20\times10^{10}\ \frac{N}{m^2}\\ \alpha_{Cu}=1.7\times10^{-5} \frac{1}{K} && \alpha_{St}=1.2 \times10^{-5} \frac{1}{K}\\ \rho_{Cu}=9\ kg/cm^3 && \rho_{St}=7.85\ kg/cm^3 \end{aligned}
New Practice Question
At 80 degrees of Celsius, a steel tire 10 mm thick and 80 mm wide that is to be shrunk onto a locomotive driving wheel 2 m in diameter just fits over the wheel which is at temperature of 25°C. How much is the stress on the tire when the combination cools down to 25°C? Neglect the deformation of the wheel.
(αSt=11.7×10−6 1K and YSt=200×109 Pa)\left(\alpha_{St}=11.7\times10^{-6}\ \frac{1}{K}\ and\ Y_{St}=200\times10^9\ Pa\right)
Input your answer in Pa, to 4 sig. figs., do NOT include units.
New Practice Question
An aluminum rod is attached to a steel bar as shown in the figure below so that they expand or stretch together. The length of these bars is 2𝑚 at 10℃ and the cross-sectional area of both bars is the same. How much is the stress in each bar at 100℃?
αSt=1.2×10−5 1K          YSt=20.0×1010 Nm2\alpha_{St}=1.2\times10^{-5}\ \frac{1}{K}\ \ \ \ \ \ \ \ \ \ Y_{St}=20.0\times10^{10}\ \frac{N}{m^2}
αAl=2.4×10−5 1K          YAl=69.0×109 Nm2\alpha_{Al}=2.4\times10^{-5}\ \frac{1}{K}\ \ \ \ \ \ \ \ \ \ Y_{Al}=69.0\times10^9\ \frac{N}{m^2}
Thermal Stress
An aluminum rod is attached to a steel bar as shown in the figure below so that they expand or stretch together. The length of these bars is 2 m2\ m at 10°C10\degree C and the cross-sectional area of both bars are the same. How much is the stress in each bar at 100°C100\degree C?
αSt=1.2×10−51KYSt=20.0×1010Nm2αAl=2.4×10−51KYAl=69×109Nm2\begin{array}{ll} \alpha_{St}=1.2\times10^{-5} \frac{1}{K} & Y_{St}=20.0\times10^{10} \frac{N}{m^2}\\[10pt] \alpha_{Al}=2.4\times10^{-5} \frac{1}{K} & Y_{Al}=69\times10^9 \frac{N}{m^2} \end{array}
Thermal Expansion
How much taller does the Eiffel Tower become at the end of a day when the temperature has increased by 15oC. Its original height is 321m and you can assume it is made of steel.
Thermal Stress
An Aluminium rod of length of 4 meters and a cross sectional area of 10 cm210\ cm^2 weighs 1 kg. It is secured between two stable supports. There is no initial stress in the beam. If the temperature is initially 60 degrees and increases by 40 degrees. What is the stress in the beam. How much water at 20 degrees C must the 100 degree beam be put in to come to 50 degrees C.
αAl=22⋅10−6 K−1α_{Al}=22\cdot10^{-6}\ K^{-1}
YAl=65⋅109 PaY_{Al}=65\cdot10^9\ Pa
An Aluminium rod of length of 4 meters and a cross sectional area of 10 cm210\ cm^2. It is secured between two stable supports. There is an initial stress in the beam of 5⋅107 Nm25\cdot10^7\ \frac{N}{m^2}. If the temperature increases by 30 degrees. What is the new stress in the beam.
αAl=22⋅10−6 K−1α_{Al}=22\cdot10^{-6}\ K^{-1}
YAl=65⋅109 PaY_{Al}=65\cdot10^9\ Pa
Thermal Stress
We are planning to close a hole using an aluminum ball but the ball is too small to cover the entire hole. What could we do to make it fit the hole?