Question

Small changes in the length of an object can be measured using a strain gauge sensor, which is a wire that when undeformed has length ℓ₀, cross-sectional area A₀, and resistance R₀. This sensor is rigidly affixed to the object’s surface, aligning its length in the direction in which length changes are to be measured. As the object deforms, the length of the wire sensor changes by Δℓ, and the resulting change ΔR in the sensor’s resistance is measured. Assuming that as the solid wire is deformed to a length ℓ, its density and volume remain constant (only approximately valid), show that the strain ( = Δℓ / ℓ₀ ) of the wire sensor, and thus of the object to which it is attached, is approximately ΔR / 2R₀.

Accepted Answer

Start by recalling the relationship between resistance and the physical properties of a wire. The resistance R of a wire is given by the formula: R = ρAℓ, where ρ is the resistivity, ℓ is the length, and A is the cross-sectional area of the wire. Since the problem states that the volume and density of the wire remain constant during deformation, we can use the relationship for volume: V = ℓ₀A₀ = ℓA. From this, we can express the new cross-sectional area A in terms of the original area A₀ and the change in length Δℓ: A = A₀ℓ₀ℓ. Substitute the expression for A into the resistance formula. The new resistance R becomes: R = ρA₀ℓ²ℓ₀. This shows that resistance depends on the square of the length of the wire. Now, calculate the change in resistance ΔR due to the change in length Δℓ. Using the approximation for small changes, expand the square term: R ≈ R₀(1 + 2Δℓ/ℓ₀), where R₀ is the original resistance. Finally, isolate the strain (Δℓ / ℓ₀) in terms of the change in resistance ΔR. From the previous step, we find: Δℓℓ₀ ≈ ΔR2R₀. This shows that the strain is approximately equal to ΔR / 2R₀, as required.

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