A cylindrical aluminum rod with a diameter of 10 mm is subjected to an axial tensile load of 5 kN. A foil strain gauge with the following properties is bonded axially along the surface of the rod:

Gauge factor (GF): 2.1

Unstrained resistance: 120.0 Ω

Measured change in resistance (ΔR): 0.189 Ω

Assume the strain gauge perfectly follows the deformation of the rod.

3. Is the strain value within the elastic range of aluminum (assume yield strain ≈ 0.0025). True or False?

Question

A cylindrical aluminum rod with a diameter of 10 mm is subjected to an axial tensile load of 5 kN. A foil strain gauge with the following properties is bonded axially along the surface of the rod:

Gauge factor (GF): 2.1

Unstrained resistance: 120.0 Ω

Measured change in resistance (ΔR): 0.189 Ω

Assume the strain gauge perfectly follows the deformation of the rod.

3. Is the strain value within the elastic range of aluminum (assume yield strain ≈ 0.0025). True or False?

BlackTom AI · Accepted Answer

The question asks whether the strain value is within the elastic range of aluminum given the rod under axial tension and the foil strain gauge reading.
First, examine the mechanical loading and material properties: the rod has a diameter of 10 mm, so its cross‑sectional area is A = π(0.005 m)^2 ≈ 7.85×10^-5 m^2. The applied axial load is F = 5 kN = 5000 N.
Next, compute the axial stress: σ = F / A ≈ 5000 / 7.85×10^-5 ≈ 6.37×10^7 Pa = 63.7 MPa.
To assess elastic range, we compare the stress to a typical aluminum yield stress. Given a yield strain ε_y ≈ 0.0025 and using a common Young’s modulus for aluminum E ≈ 70 GPa, the yield stress σ_y ≈ E·ε_y ≈ 70×10^9 × 0.0025 ≈ 175 MPa. Since 63.7 MPa < 175 MPa, the loading is within the elastic range, from a purely mechanical standpoint.
Now, determine the strain from the gauge reading: the foil gauge has GF = 2.1, initial resistance R0 = 120 Ω, and measured ΔR = 0.189 Ω.
Using the gauge factor definition, ε = ΔR / (GF · R0) = 0.189 / (2.1 × 120) = 0.189 / 252 ≈ 0.00075 (7.5×10^-4).
This strain value (7.5×10^-4) is well below the assumed yield strain (0.0025), and thus lies in the elastic regime for aluminum.
As a cross-check, one can also estimate strain from the mechanical stress: ε_mech ≈ σ / E ≈ 63.7 MPa / 70 GPa ≈ 0.00091 (9.1×10^-4), which is on the same order of magnitude as 0.00075 and also below ε_y. Given both independent approaches indicate elastic behavior, the statement that the strain value is within the elastic range is consistent.
In summary, the calculated strain from the gauge reading is small enough to be considered elastic for aluminum under these conditions, and the comparison with yield strain supports this conclusion.

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