Question

Consider a cylinder of height h, diameter d, and wall thickness t pressurized to an internal pressure P_0 (gauge pressure, relative to the external atmospheric pressure). The cylinder consists of material with Young's modulus E, Poisson's ratio v, and density rho. Derive expressions for the axial and hoop strains of the cylinder wall in terms of the can dimensions, properties, and internal pressure. You may assume plane stress conditions.
Continuing on Problem 1, assume a strain gage is bonded to the cylinder wall surface in the direction of the axial strain. The strain gage has nominal resistance R_0 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 V_in. (The strain gage is bonded and the Wheatstone bridge balanced with the vessel already pressurized.) Develop an expression for the voltage change delta V across the bridge if the cylinder pressure changes by delta P.

Repeat Problem 2, but now assuming the strain gage is bonded to the cylinder wall surface in the direction of the hoop strain. Does the voltage change more when the strain gage is oriented in the axial or hoop direction?

Continuing on Problem 3 (strain gage in the hoop direction), calculate the voltage change delta V across the Wheatstone bridge when the cylinder pressure increases by 1 atm. Assume the vessel is made of aluminum 3004 with height h = 10.5 cm, diameter d = 5.5 cm, and thickness t = 50 mu m. The Gage Factor is GF = 2 and the Wheatstone bridge has V_in = 6 V. The strain gage has nominal resistance R_0 = R_4 = 120 ohm.

Answer 1

Explanation:

Note: For equations refer the attached document!

The net upward pressure force per unit height p*D must be balanced by the downward tensile force per unit height 2T, a force that can also be expressed as a stress, σhoop, times area 2t. Equating and solving for σh gives:

 Eq 1

Similarly, the axial stress σaxial can be calculated by dividing the total force on the end of the can, pA=pπ(D/2)2 by the cross sectional area of the wall, πDt, giving:

Eq 2

For a flat sheet in biaxial tension, the strain in a given direction such as the ‘hoop’ tangential direction is given by the following constitutive relation - with Young’s modulus E and Poisson’s ratio ν:

 Eq 3

 

Finally, solving for unknown pressure as a function of hoop strain:

 Eq 4

 

Resistance of a conductor of length L, cross-sectional area A, and resistivity ρ is

 Eq 5

Consequently, a small differential change in ΔR/R can be expressed as

 Eq 6

Where ΔL/L is longitudinal strain ε, and ΔA/A is –2νε where ν is the Poisson’s ratio of the resistive material. Substitution and factoring out ε from the right hand side leaves

 Eq 7

Where Δρ/ρε can be considered nearly constant, and thus the parenthetical term effectively becomes a single constant, the gage factor, GF

 Eq 8

For Wheat stone bridge:

 Eq 9

Given that R1=R3=R4=Ro, and R2 (the strain gage) = Ro + ΔR, substituting into equation above:

Eq9

Substituting e with respective stress-strain relation

Eq 10

 

part b

Since, axial strain(1-2v) < hoop strain (2-v). V out axial < V out hoop.

Hence, dV hoop < dV axial.

part c  

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

Eq 11  

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