#### Installation — business terrible - 1 part

September 8th, 2015

In the serpentine arrangement of the strain gage bonded on a calibration specimen in a longitudinal direction, the longitudinal grid elements and the inversion portions of the conductor are also stretched in a transversal direction, Fig.2.4 and the total resistance variation of the gage is also a function of these minor effects. Remembering that:

Ri + Rt — R

the total resistance variation for a strain gage is:

1 S • BiRi • + S • BtRt R |

AR — S • BiRi + S • BtRt

If:

Si = S ■ Ri /R is the longitudinal sensitivity St = S ■ Rt /R is the transversal sensitivity

= Si ■ єі + St ■ St = Si (ei + Kt St) (2.6)

R

with Kt = S.

If the strain gage is bonded in a longitudinal direction on a specimen with a known bending moment Fig.2.5, holds:

A R

= Si (ei + Kt St) = Si (ei – Kt V0Si) = Si ei (1 – V0 Kt) = K ■ e

R

K = Si (1 – v0Kt) is the gage factor, i. e. the proportionality factor between the fractional resistance change and the longitudinal strain. The factor also takes into account the backing and glue effect that partially shield the strain transition from the body to the grid. As the constant was defined, the calibration of the strain gage factor K depends on the ratio between longitudinal and transverse strains of the calibration bar. A systematic error arises in all the cases of strain gage applications in a biaxial stress field, with a ratio other than that used in the calibration specimen [16]. Even if errors in strain indication due to transverse sensitivity are generally small since the transverse sensitivity is small, (in common gages it varies from -9.2 to +1.8 %), these errors can become intolerable in biaxial strain fields with large differences between principal strains [17]. It can be taken into account and corrected in a simple way for two gages bonded in whatever perpendicular directions 1 and 2 of a biaxial state of stress, where the fractional gage resistances in both directions can be written as:

A R 1

= Si ■ S1 + St ■ S2 = K S1

R1 A R 2

= Si ■ S2 + St ■ S1 = K S2

R2

where: £1 and £2 are the indicated not corrected values with the gage factor K. Substituting its expression leads to the equations:

I Si (£1 + Kt Є2) = Si (1 – V0 Kt £

[ Si (£2 + Kt £1) = Si (1 – Vo Kt )£2

from which the true strains in both directions 1 and 2 are derived:

1 – v0 Kt ^ . ч

£1 = – J-Kf £ – Kt £2

1 – v0 Kt „ – ч

£2 = 1 _ K2 (£2 – Kt O)

## Leave a reply