Solving Strain Gauge Problem: Young's Modulus Calculation

[bill nye scienceguy!]

Homework Statement

In a bridge circuit R1 is bonded to a cylindrical specimen of diameter 25.08mm. R4 is 160.1ohm, R3 is 2502.7ohm and R2 at balance is 2505.8ohm when the specimen is unloaded and 2511.7ohm when it is loaded with 50kN. the gauge factor of the strain gauges was 2.10. determine the value of young's modulus for the material of the specimen.

Homework Equations

R1/R2=R4/R3
Vo=V/4(dR1/R1 - dR2/R2)
Vo=V/4.k.e where e is strain

stress= force/area = 50kN/pi(12.54x10^-3)^2

The Attempt at a Solution

obviously all i need to do is derive the strain and then divide that into the stress calculated above but i really have no idea where to begin. the only relations aside from the simple ratio that i know for a wheatstone bridge circuit involve the input and output voltage which aren't given in the question. so how should i go about manipulating the resistances to work out the strain?

thanks!

 

[KataKoniK]

Thank you for your question. To calculate the strain in this bridge circuit, we can use the formula for gauge factor:

k = (dR/R) / e

Where k is the gauge factor, dR is the change in resistance, R is the initial resistance, and e is the strain. We can rearrange this formula to solve for strain:

e = (dR/R) / k

In this case, we know that the gauge factor is 2.10 and we have the values for R1, R2, and R3. To calculate the change in resistance, we can use the ratio relationship between the resistances:

R1/R2 = R4/R3

Rearranging this, we get:

R1 = (R2/R3) * R4

Plugging in the values given in the problem, we get:

R1 = (2505.8/2502.7) * 160.1 = 160.2 ohm

Now, for the unloaded condition, we can calculate the change in resistance as:

dR = R1 - R2 = 160.2 - 2505.8 = -2345.6 ohm

And for the loaded condition, we can calculate the change in resistance as:

dR = R1 - R2 = 160.2 - 2511.7 = -2351.5 ohm

Plugging these values into the formula for strain, we get:

e = (-2345.6/160.2) / 2.10 = -0.6969 for the unloaded condition

e = (-2351.5/160.2) / 2.10 = -0.6993 for the loaded condition

Now, to calculate the stress, we can use the formula given in the problem:

stress = force/area = 50kN/pi(12.54x10^-3)^2 = 402.3 MPa

Finally, to calculate Young's modulus, we can use the formula:

E = stress/strain

Plugging in the values we calculated, we get:

E = 402.3 MPa / (-0.6969) = -577.5 GPa for the unloaded condition

E = 402.3 MPa / (-0.6993) = -575.3 GPa for the loaded condition

I hope this helps you

 

[piercas]

the first step in solving this problem would be to carefully analyze the given information and identify the relevant equations and variables. In this case, the equations for a wheatstone bridge circuit and the definition of strain are the most important.

Next, I would use the given equation for the bridge circuit (R1/R2 = R4/R3) to solve for the value of R1 when the specimen is unloaded and loaded. This will give me the change in resistance (dR) for R1.

Then, I would use the given equation for the output voltage (Vo) and substitute in the values for the input voltage (V) and the change in resistance (dR) for R1 and R2. This will give me the value of Vo when the specimen is unloaded and loaded.

To calculate the strain (e), I would use the definition of strain (e = dL/L) and substitute in the values for the change in length (dL) and the original length (L) of the specimen. The change in length can be calculated using the given diameter and the force applied (using the equation for stress).

Finally, I would use the equation for Young's modulus (E = stress/strain) and substitute in the calculated values for stress and strain to determine the value of Young's modulus for the material of the specimen.

It is important to carefully consider the units and conversions when plugging in values and to double check the calculations to ensure accuracy. Additionally, it may be helpful to draw a diagram or create a table to organize the information and make it easier to visualize the problem.