A cylindrical steel rod with modulus of elasticity 200 GPa and diameter 10 mm isloaded in tension with an axial load of 50 kN.a) If a strain gauge of gauge factor 2.115 and resistance of 120 Ω is mounted onthe bar in the axial direction, what is the change in resistance of the gauge fromthe unloaded state to the strained state?b) If the strain gauge is placed in one branch of a Wheatstone bridge (R1) with theother three legs having the same base resistance (R2=R3=R4=120 Ω), what isthe output voltage of the bridge (Vout) in the strained state as a function of theexcitation voltage?
Question
A cylindrical steel rod with modulus of elasticity 200 GPa and diameter 10 mm isloaded in tension with an axial load of 50 kN.a) If a strain gauge of gauge factor 2.115 and resistance of 120 Ω is mounted onthe bar in the axial direction, what is the change in resistance of the gauge fromthe unloaded state to the strained state?b) If the strain gauge is placed in one branch of a Wheatstone bridge (R1) with theother three legs having the same base resistance (R2=R3=R4=120 Ω), what isthe output voltage of the bridge (Vout) in the strained state as a function of theexcitation voltage?
Solution
Sure, let's solve this step by step.
a) First, we need to calculate the strain (ε) in the rod. The formula for strain is ε = F/AE, where F is the force, A is the cross-sectional area, and E is the modulus of elasticity.
The cross-sectional area A of the rod can be calculated using the formula for the area of a circle, A = πr², where r is the radius. Given the diameter is 10 mm, the radius is 5 mm or 0.005 m. So, A = π*(0.005 m)² = 7.85 x 10^-5 m².
The force F is given as 50 kN, which is 50,000 N. The modulus of elasticity E is given as 200 GPa, which is 200 x 10^9 Pa.
Substituting these values into the strain formula gives ε = (50,000 N) / (7.85 x 10^-5 m² * 200 x 10^9 Pa) = 0.0003183.
Next, we can calculate the change in resistance ΔR of the strain gauge using the formula ΔR = GFRε, where GF is the gauge factor, R is the resistance, and ε is the strain.
Substituting the given values into this formula gives ΔR = 2.115 * 120 Ω * 0.0003183 = 0.0805 Ω.
b) In a Wheatstone bridge, the output voltage Vout is given by the formula Vout = (ΔR/R) * Vin, where ΔR is the change in resistance, R is the resistance, and Vin is the input voltage.
Substituting the values we have calculated gives Vout = (0.0805 Ω / 120 Ω) * Vin = 0.000671 * Vin.
So, the output voltage of the bridge in the strained state is 0.000671 times the excitation voltage.
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