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Wize University Physics Textbook (Master) > Equilibrium & Elasticity

Stress & Strain

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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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A mass of 22 kg is suspended from brass (Y=9.0×1010Y=9.0 \times 10^{10}) wire. The diameter of the wire is 0.15cm.
a) What is the stress on the brass wire?

r=di2=0.075 cmr=\dfrac{d_i}{2}=0.075\text{ cm}

FA=mgA=22×9.8π(0.075×10−2)2=1.22×108pa\frac{F}{A}=\frac{mg}{A}=\frac{22\times9.8}{\pi (0.075\times 10^{-2})^2}=1.22\times 10^8 pa
b) By what percent does the length of the wire increase?

ΔLLo=FA Y=Fπr2 Y=0.135%\frac{\Delta L}{L_{o}}=\frac{F}{A\ Y}=\frac{F}{\pi r^{2}\ Y}=0.135\%
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}