MEGR 3171 — Strain, Force & Temperature Sensing

Learning Objectives

Calculate strain from bridge output voltage using gauge factor and bridge configuration.
Design a rosette strain gauge arrangement and compute principal strains.
Compute thermocouple output voltage and apply cold-junction compensation.
Apply the Callendar-Van Dusen equation for RTD resistance-temperature conversion.
Use the Steinhart-Hart equation to convert thermistor resistance to temperature.
Explain emissivity correction for radiation pyrometers.

1. Strain Gauges

A foil strain gauge changes resistance proportionally to mechanical strain. The gauge factor G_f relates fractional resistance change to strain.

Gauge Factor and Strain

G_f = (ΔR/R) / ε     (typical: G_f ≈ 2.0 to 2.1 for metal foil)

Solving for strain from bridge output (quarter-bridge):
  ε = (4 · V_out) / (V_ex · G_f)    (linear approximation)

Rosette Configurations

To determine principal strains and their directions, three gauges at known angles must be used. The rectangular rosette (0°/45°/90°) is most common. Given strains ε_a, ε_b, ε_c at these angles:

Rectangular Rosette Principal Strains

ε_1,2 = (ε_a + ε_c)/2 ± (1/√2) · √[(ε_a - ε_c)² + (2ε_b - ε_a - ε_c)²]

tan(2θ) = (2ε_b - ε_a - ε_c) / (ε_a - ε_c)

2. Thermocouple Theory

The Seebeck effect: when two dissimilar metals form a junction at different temperatures, a voltage proportional to the temperature difference is generated. The Seebeck coefficient is specific to the metal pair.

Type K (Chromel/Alumel)

Most widely used. Range: −200 to 1350°C. Sensitivity: ~41 µV/°C. General purpose, good stability.

Type J (Iron/Constantan)

Range: −40 to 750°C. Sensitivity: ~52 µV/°C. Lower max temperature, susceptible to oxidation above 550°C.

Type T (Copper/Constantan)

Range: −200 to 350°C. Best for cryogenic and low-temperature measurements. Good accuracy.

Type E (Chromel/Constantan)

Highest Seebeck coefficient (~68 µV/°C). Best sensitivity. Non-magnetic.

Cold-Junction Compensation (CJC)

V_corrected = V_measured + V_CJC(T_ref)

where V_CJC(T_ref) is the thermocouple voltage at the
reference (ambient) temperature relative to 0°C.

T_junction = T_inverse(V_corrected)  (using NIST tables or polynomial)

Law of Intermediate Metals Inserting any third metal into a thermocouple circuit does not affect the net EMF as long as both junctions with the third metal are at the same temperature. This is why copper extension wires can be used if the terminal block is isothermal.

3. RTDs (Resistance Temperature Detectors)

Platinum RTDs (Pt100, Pt1000) offer excellent accuracy and stability. Resistance increases linearly with temperature over a wide range.

Callendar-Van Dusen Equation

R(T) = R_0 [1 + A·T + B·T² + C·T³(T-100)]

For Pt100: R_0 = 100 Ω at 0°C
  A = 3.9083×10&sup-₃ °C&sup-¹
  B = −5.775×10&sup-⁷ °C&sup-²
  C = −4.183×10&sup-¹² °C&sup-⁴ (only below 0°C; C=0 above 0°C)

4-wire connection eliminates lead resistance error.

4. Thermistors and the Steinhart-Hart Equation

Thermistors are semiconductor devices with a large, nonlinear resistance-temperature relationship. NTC (negative temperature coefficient) types decrease resistance with increasing temperature. High sensitivity makes them excellent for narrow-range precision measurements.

Steinhart-Hart Equation

1/T = A + B·ln(R) + C·[ln(R)]³

where T is in Kelvin, R in ohms
A, B, C are empirically determined constants

Simplified B-parameter equation:
1/T = 1/T_0 + (1/B)·ln(R/R_0)

Practice Problems

Problem 1 — Thermocouple CJC A Type K thermocouple reads 18.516 mV at its terminals. The reference junction (cold junction) is at 25°C, which corresponds to 1.000 mV for Type K relative to 0°C. What is the actual temperature at the measurement junction? (Use Type K Seebeck ≈ 41 µV/°C as linear approximation for this problem.)

V_corrected = V_measured + V_CJC = 18.516 + 1.000 = 19.516 mV

T ≈ 19.516 mV / 0.041 mV/°C ≈ 476°C (using linear approximation; precise answer requires NIST tables)

T_junction ≈ 476°C

Problem 2 — RTD A Pt100 RTD reads 138.50 Ω at an unknown temperature. Using the simplified linear approximation R(T) ≈ 100(1 + 0.00385·T), find the temperature.

138.50 = 100(1 + 0.00385·T) → 1.3850 = 1 + 0.00385·T → 0.3850 = 0.00385·T

T = 0.3850 / 0.00385 = 100.0°C

T = 100.0°C