An electrical resistance strain gauge of resistance 120Ω has a gauge ...
Given R = Resistance = 120Ω
Gf = Gauge factor = 2
E = 2 x 1011 N/m2
(a) Tensile stress = 60 x 106 N/m2
We know that,
where ΔR = change in resistance
∴ ΔR = Gf x ε x R
= 2 x 0.3 x 10-3 x 120
= 0.072Ω
(c) ΔT = increase of temperature = 40°C
AR1 = change in resistance due to temperature increase
= R (1 + αg ΔT) = 120 (1 + 20 x 106 x 40) = 120.096Ω
where αg = temperature coefficient of resistance of gauge.
View all questions of this test
An electrical resistance strain gauge of resistance 120Ω has a gauge ...
Given:
Resistance of strain gauge, R = 120Ω
Gauge factor, GF = 2
Modulus of elasticity of steel specimen, E = 2 * 10^11 N/m^2
Stress applied, σ = 60 * 10^6 N/m^2
Temperature coefficient of resistance of gauge, αg = 20 * 10^-6 per °C
Thermal coefficient of linear expansion of gauge, βg = 16 * 10^-6 per °C
Thermal coefficient of linear expansion of steel specimen, βs = 12 * 10^-6 per °C
Change in temperature, ΔT = 40°C
(a) Strain induced in the specimen:
Strain induced, ε = σ/E = (60 * 10^6 N/m^2)/(2 * 10^11 N/m^2) = 0.0003
(b) Change in the electrical resistance:
Change in resistance of the gauge, ΔR = GF * R * ε = 2 * 120Ω * 0.0003 = 0.072Ω
(c) Change in the electrical resistance due to temperature:
Change in resistance of the gauge due to temperature, ΔRt = R * αg * ΔT = 120Ω * 20 * 10^-6 per °C * 40°C = 0.096Ω
Change in resistance of the gauge due to thermal expansion, ΔRe = R * βg * ΔT = 120Ω * 16 * 10^-6 per °C * 40°C = 0.0768Ω
Change in resistance of steel specimen due to thermal expansion, ΔRs = R * βs * ΔT * ε = 120Ω * 12 * 10^-6 per °C * 40°C * 0.0003 = 0.0144Ω
Total change in resistance, ΔRtotal = ΔRt + ΔRe + ΔRs = 0.096Ω + 0.0768Ω + 0.0144Ω = 0.1872Ω
Final answer:
Change in resistance of the gauge due to tensile stress, ΔR = 0.072Ω
Change in resistance of the gauge due to temperature, ΔRtotal = 0.1872Ω
Total change in resistance, ΔRtotal = ΔR + ΔRtotal = 0.072Ω + 0.1872Ω = 0.2592Ω
Final strain induced in the specimen, ε = ΔRtotal/(GF * R) = 0.2592Ω/(2 * 120Ω) = 0.00136
Therefore, the final answer is 0.0014 (approximated to 3 significant figures).