- #1

Rusty789

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Firstly I have no idea how to tackle this one, im not sure if we've been taught it yet or if its sort of a lead on to the next set of topics..

The shear stress t in a shaft of diameter D under torque T is given by t = KT / [tex]\Pi[/tex]D

Determine the approximate error in calculating t if T is measure 3% too small and D is measured 1.5% to large.

None

As previously stated i don't know how to handle this one, so any advice will be great.

Also i don't want to know the answer, i would like to know how to work it out..

EDIT:

My first theory was to put numbers in do the actual calculation twice with physical number and compare the difference but that didnt work. e.g.

Completely random values.

t = 3*10/3.14*10

then do:

t = 3*9.7/3.14*10.15

This gives a percentage difference of 7.243%

t = (3*154)/(3.14*56

t = (3*149.38)/(3.14*56.84

this gives a percentage difference of 7.233%

Is this right?

## Homework Statement

The shear stress t in a shaft of diameter D under torque T is given by t = KT / [tex]\Pi[/tex]D

^{3}.Determine the approximate error in calculating t if T is measure 3% too small and D is measured 1.5% to large.

## Homework Equations

None

## The Attempt at a Solution

As previously stated i don't know how to handle this one, so any advice will be great.

Also i don't want to know the answer, i would like to know how to work it out..

EDIT:

My first theory was to put numbers in do the actual calculation twice with physical number and compare the difference but that didnt work. e.g.

**(although it did this time around, so im now completely confused)**Completely random values.

t = 3*10/3.14*10

^{3}= 9.554*10^{-3}then do:

t = 3*9.7/3.14*10.15

^{3}= 8.862*10^{-3}This gives a percentage difference of 7.243%

t = (3*154)/(3.14*56

^{3}) = 8.378*10^{-3}t = (3*149.38)/(3.14*56.84

^{3}) ) =7.772*10^{-3}this gives a percentage difference of 7.233%

Is this right?

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