Appendix D.Load Cell Gauge Factor Recalculation

D.1Overview

This appendix describes how to recalculate the gauge factor for a load cell and then approximate the load where one or more strain gauges in the cell have failed after installation. This is not a foolproof method. For example, if the load distribution changes during monitoring, the calculations based on the method described above will be in error.

D.2Procedure

If the load is applied uniformly to the load cell then, as the load changes the change in reading on each gauge will be the same and, should one gauge fail, the gauge factor given on the calibration sheet can be applied to the average change of the remaining gauges.

Note the following example, where gauge number three in a six-gauge load cell has failed. The load cell gauge factor for the six gauges is 0.2439 tons/digit. If the load is uniformly applied to the load cell, then to calculate the load this gauge factor would be applied to the average reading change of the remaining five active gauges. In the example below the load on 7/1/02 would be calculated to be 0.2439(7298-6139) = 282.7 tons. However, in the field it rare to have the cell uniformly stressed, therefore, it may be more accurate to calculate a new gauge factor using only the active gauges.

In cases where the load is eccentric (in the present example the reading change on gauge number was higher than the other five gauges), the new gauge factor can be calculated for the remaining five active gauges as follows:

table 6: Gauge Factor for Remaining Gauges

1.Calculate a new zero load average using only the initial readings of the five remaining active gauges = 7308

2.Using only the readings of the active gauges: #1, #2, #4, #5, and #6 from the time of the last reading when all six gauges were active (6/1/02), calculate the average reading = 6432.

3.Calculate the new gauge factor for the remaining five active gauges by dividing the calculated load at the last time when all gauges were active, (6/1/02), by the change in the five gauge average readings calculated in steps one and two, = 220.2 /(7308-6432) = 0.2514. This is the new gauge factor to be applied to all subsequent changes of the remaining five active gauges.

4.Using the averages of the current and initial five-gauge readings, calculate the load on 7/1/02 by using the new gauge factor. Thus on 7/1/02: (7308 – 6139) x 0.2514 = 293.9 tons. This gives a better result than applying the old gauge factor for the six gauges to the average reading of the five active gauges. (The applied load was 291 tons).

5.Repeat step four for subsequent readings or repeat all steps if more gauges in the load cell fail.