wheatstone bridge

Thread Starter

lukus08

Joined Mar 14, 2009
34
sorry i didnt see your post got it now thanks. yes im happy with the other equation. but that example on post 14 is an example question given. r1 nd r2 - the black box is just a strain guage but i drew it.

so does this not workout ?
 

studiot

Joined Nov 9, 2007
4,998
Here is a derivation of the output voltage calculation.

I will leave it to you to fill in the numbers.

Be careful when following formulae about (wheatstone) bridge circuits as not everybody labels R1, R2, R3 and R4 in the same order so the formula can look different and therefore confusing.
I have followed your labelling.
 

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Thread Starter

lukus08

Joined Mar 14, 2009
34
great thanks so much. i will work through it this morning and post my result.

ok i had a go. cross sectional area = 19.63m^2

1/ stress = FA/Cross sectional area = 40000 / 19.63 = 2038m^2

2/ modulus = stress/strain
1.61x10^11 = 2038/ strain
strain = 1.27x10^-8

3/ (change in r1)/r1 = gf x strain
(change in r1) / 100 = 2 x 1.27x10^-8
(change in r1) = 2.53 x 10^-6

4/ filing it into Va-Vb i get vout to be 1.2375x10^-8 which surely cannot be correct?
 
Last edited:

studiot

Joined Nov 9, 2007
4,998
Well I make it 777 microvolts.

Not bad for a relatively low steel stress of 20MN/sqm, which has a strength of 250 MN/sqm


A tip for exams

Don't work out intermediate results.

Keep the expressions as I have done and just collect powers of 10.

That way you will get a mark for each correct line and only loose one if you make an arithmetic booboo at the end.
 

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Thread Starter

lukus08

Joined Mar 14, 2009
34
thanks

im not following yours...

why is your example have you got x4 beside force applied?
is the cross sectional area not jus pie x 25^2 / 10000? (/10000) to get to meters.
 

Thread Starter

lukus08

Joined Mar 14, 2009
34
it can be either pi d^2/4 or pi x r^2. both give same result.

but in your example first line should it not be...... stress = FA/area
stres = 4 x 10^3 / pi d^2/4???


also in post 7 you said:
'strain = gauge factor x change of resistance/original resistance'

but in your example on post 24 in your notes you said:
'change of resistance/original resistance = gauge factor x strain'

but they are contradicting each other.....
 
Last edited:

studiot

Joined Nov 9, 2007
4,998
it can be either pi d^2/4 or pi x r^2. both give same result.

but in your example first line should it not be...... stress = FA/area
stres = 4 x 10^3 / pi d^2/4???
Force = 40x10^3 Newtons

Diameter = 5 x 10^-2 metres

Sectional Area = π(5 x 10^-2)^2 / 4

Stress = Force x 1/area = 40 x 10^3 4/ π(5 x 10^-2)^2

which is what I had.


also in post 7 you said:
'strain = gauge factor x change of resistance/original resistance'

but in your example on post 24 in your notes you said:
'change of resistance/original resistance = gauge factor x strain'
Yes you have spotted something. Post 7 was an error, copying for the Wiki article. Sorry. The later post has the correct expression.
 
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