Assume a strain gage is bonded to the cylinder wall surface in the direction of the hoop strain. The strain gage has nominal resistance R0 and a Gage Factor GF. It is connected in a Wheatstone bridge configuration where all resistors have the same nominal resistance; the bridge has an input voltage, Vin. (The strain gage is bonded and the Wheatstone bridge balanced with the vessel already pressurized.)Calculate the voltage change ∆V across the Wheatstone bridge when the cylinder is pressurized to ∆P = 2.5 atm. Assume the vessel is made of 3004 aluminum with height h = 21 cm, diameter d = 9 cm, and thickness t = 65 µm. The Gage Factor is GF = 2 and the Wheatstone bridge has Vin = 6 V. The strain gage has nominal resistance R0 = 120 Ω.

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shahnoorazhar3 shahnoorazhar3 · Answer 1 · 2020-01-27T13:33:50+01:00

Answer:

5.994 V

Explanation:

The pressure as a function of hoop strain is given:

[tex]P = \frac{4*E*t}{D}*\frac{e_{h} }{2-v}[/tex]

[tex]e_{h} = \frac{D*P*(2-v)}{4*E*t} .... Eq1[/tex]

For wheat-stone bridge with equal nominal resistance of resistors:

[tex]V_{out} = \frac{GF*e*V_{in} }{4} .... Eq2[/tex]

Hence, input Eq1 into Eq2

[tex]V_{out} = \frac{GF*e*V_{in}*D*P*(2-v) }{16*E*t} .....Eq3\\[/tex]

Given data:

P = 253313 Pa

D = d + 2t = 0.09013 m

t = 65 um

GF = 2

E = 75 GPa

v = 0.33

Use the data above and compute Vout using Eq3

[tex]V_{out} = \frac{2*6*0.09013*253313*(2-0.33) }{16*75*10^9*65*10^-6} \\\\V_{out} = 0.006285 V\\\\change in V = 6 - 0.006285 = 5.994 V[/tex]