Amity University Noida B.Tech Admissions 2025
ApplyAmong Top 30 National Universities for Engineering (NIRF 2024) | 30+ Specializations | AI Powered Learning & State-of-the-Art Facilities
Thermal stress and thermal strain is considered one of the most asked concept.
25 Questions around this concept.
A compressive force, F is applied at the two ends of a long thin steel rod. It is heated, simultaneously, such that its temperature increases by T. The net change in its length is zero. Let be the length of the rod, A its cross-section area, Y its Young’s modulus, and α its linear expansion coefficient. Then, F is equal to :
When a rod is heated but prevented from expanding then the thermal stress produced is
When a rod whose ends are rigidly fixed such as to prevent expansion or contraction then the thermal stress produced is -
New: JEE Main 2025 Session 1 Result OUT; Check Now | Rank Predictor
JEE Main 2025: College Predictor | Marks vs Rank vs Percentile | Top NITs Cutoff
JEE Main 2025: January Session Official Question Paper for All Shifts | Exam Analysis
JEE Main 2025: High Scoring Topics | Sample Papers | Mock Tests | PYQs
When the rod is heated but prevent from expanding then the thermal strain is independent of :
When a rod is heated rigidly fixed at both ends, the force on the support will be -
The maximum stretching force of humans is $81 \times 10^4 \mathrm{~V}$, However the Young's modulus for stretch is $36 \times 10^9 \mathrm{~Pa}$. Then what is longitudinal strain $\left(\frac{\Delta_l}{l}=?\right) \longrightarrow \quad\left(A=9 \mathrm{~cm}^2\right)$
Thermal stress in a rod which is rigidly fixed : When a rod which rigidly fixed at ends such as to prevent expansion or contraction, when its temperature is increased or decreased. Due to preventing its thermal expansion or contraction, a compressive or tensile stress is developed in it. As the rod try to expand or contracts, then it apply a reaction force on the rigid support. If the change in temperature of a rod of length L is then -
Thermal strain $=\frac{\Delta L}{L}=\alpha \Delta \theta \quad\left[\right.$ As $\left.\alpha=\frac{\Delta L}{L} \times \frac{1}{\Delta \theta}\right]$
As, If we know the strain then with the help of Hooke's law, we can find the stress also. If we know the stress , then we can find the force by multiplying cross-sectional area with stress. Both stress and force can be written as -
So, Thermal stress $=Y \alpha \Delta \theta$
or, Force on the supports $F=Y A \alpha \Delta \theta$
"Stay in the loop. Receive exam news, study resources, and expert advice!"