High School
SAT
ACT
University
MCAT
LSAT
DAT

Thermal Stress

Related Topics

Wize University Physics Textbook (Master) > Heat and Temperature

Thermal Stress

3 Activities

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

ΔL=L0αΔTΔL=L_0αΔT
ΔL=FALYΔL=\frac{F}{A}\frac{L}{Y}
Cwater=4.186 JgC_{water}=4.186\ \frac{J}{g}
CAl=0.9 JgC_{Al}=0.9\ \frac{J}{g}
More Thermal Stress Questions:
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.
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?