1.16 Experimental Measurement of Surface Strains 39TheradiusofthecircleisCQand
CQ=√
CN^2 +QN^2
Hence,
CQ=√[
1
2 (εa−εc)] 2
+
[
εb−^12 (εa+εc)] 2
whichsimplifiesto
CQ=1
√
2
√
(εa−εb)^2 +(εc−εb)^2Therefore,εIisgivenby
εI=OC+radiusofcircleis
εI=^12 (εa+εc)+1
√
2
√
(εa−εb)^2 +(εc−εb)^2 (1.69)Also,
εII=OC−radiusofcirclethatis,
εII=^12 (εa+εc)−1
√
2
√
(εa−εb)^2 +(εc−εb)^2 (1.70)Finally,theangleθisgivenby
tan2θ=QN
CN
=
εb−^12 (εa+εc)
1
2 (εa−εc)thatis,
tan2θ=2 εb−εa−εc
εa−εc(1.71)
Asimilarapproachmaybeadoptedfora60◦rosette.
Example 1.7
A bar of solid circular cross section has a diameter of 50mm and carries a torque, T, together
with an axial tensile load,P. A rectangular strain gauge rosette attached to the surface of the bar
gavethefollowingstrainreadings:εa= 1000 × 10 −^6 ,εb=− 200 × 10 −^6 ,andεc=− 300 × 10 −^6 ,where
the gauges ‘a’ and ‘c’ are in line with, and perpendicular to, the axis of the bar, respectively. If
Young’smodulus,E,forthebaris70000N/mm^2 andPoisson’sratio,ν,is0.3,calculatethevaluesof
TandP.