A steel bar (shown in Fig. 1) has a cross-sectional area of 10−3 m2, Young’s modulusof 2 × 1011 Pa and Poisson’s ratio of 0.4. If the bar is subjected to a compressive forceof 105 N, find the corresponding longitudinal and transverse strains.b) Two strain gauges, each with an unstrained resistance of 120 Ω and a gauge factor of2.0, are bonded onto the steel bar (Fig. 1) so that one measures the longitudinal strainand the other the transverse strain. Determine the corresponding gauge resistancesfor the longitudinal and transverse strains
Question
A steel bar (shown in Fig. 1) has a cross-sectional area of 10−3 m2, Young’s modulusof 2 × 1011 Pa and Poisson’s ratio of 0.4. If the bar is subjected to a compressive forceof 105 N, find the corresponding longitudinal and transverse strains.b) Two strain gauges, each with an unstrained resistance of 120 Ω and a gauge factor of2.0, are bonded onto the steel bar (Fig. 1) so that one measures the longitudinal strainand the other the transverse strain. Determine the corresponding gauge resistancesfor the longitudinal and transverse strains
Solution
The problem can be solved in two parts:
a) To find the longitudinal and transverse strains, we need to use the definitions of stress, strain, and Poisson's ratio.
Stress (σ) is defined as the force (F) divided by the area (A) it's applied to. So, σ = F/A = 105 N / 10^-3 m^2 = 1 x 10^8 Pa.
Strain (ε) is defined as the stress (σ) divided by Young's modulus (E). So, ε = σ/E = 1 x 10^8 Pa / 2 x 10^11 Pa = 0.5 x 10^-3.
Poisson's ratio (ν) is the ratio of the transverse strain to the longitudinal strain. So, the transverse strain is ν * ε = 0.4 * 0.5 x 10^-3 = 0.2 x 10^-3.
b) The change in resistance of a strain gauge is given by the equation ΔR = GF * ε * R, where GF is the gauge factor, ε is the strain, and R is the unstrained resistance.
For the longitudinal strain gauge, ΔR = 2.0 * 0.5 x 10^-3 * 120 Ω = 0.12 Ω. So, the resistance of the gauge under strain is 120 Ω + 0.12 Ω = 120.12 Ω.
For the transverse strain gauge, ΔR = 2.0 * 0.2 x 10^-3 * 120 Ω = 0.048 Ω. So, the resistance of the gauge under strain is 120 Ω + 0.048 Ω = 120.048 Ω.
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