Question

Figure 2

In Figure 2, a strain gage forms an angle, β, of 20° with a horizontal plane to measure the gage pressure in a cylindrical steel tank. The tank wall is 10mm thick, has a 520mm inner diameter, and is made of an iron alloy with E = 350 GPa and v = 0.42. Using this, find the pressure in the tank when the strain gage reads +245µ.

Answer 1

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day Date - we all know equation of shear strain at an angle o from the horizontal Ex = Entby + 6x - by cos 20 + Prey sin2o. 0 mm Given , B= 200, t= lomm, d= 520 mm i E=350 Gpa, ? Da 0.42, P=? Now, G= shear modulus- shearstren (Any) shear strain (Xry) But, tny =o . Vry=0 Now, On= Circum ferential ston = Pr by = Longitudinal ston = pr 1 at - En = Axial strain along the r-direction

day Date By 4 4 ④ Ey = Axial strain along y-direction Putting values in ① & 1 = Px 260x10 3 2 1 1 - 0.427 350X109 X 10x10 Er G = = px 0.2054 PX 0.2054 3.58109 - ③ Ey = Px260x103 350 X 108x10x10 3 - 0 42 - PX 0.0208 3.58109

By 0.0,80 245*10-6 0-0.5.868.3109P + 51.9428X10'2p. +((0.05868x100P) (5.9428x10t2p Cos(2420 + % sin (2420) 3,23114x10.1p +(2163686X10"p )* Cosco 24581000 = 245x100 = 3.23114X10-11p + 2,0199 5x10lp to 245x105 = 5.25109 1 p = 4066569 MPa Hrywer P= Prenme in the tank when the strain goge reads +2454
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