Euler Bernoulli Beam 4th order ODE -Balance of Units


by bugatti79
Tags: balance, beam, bernoulli, euler, order, units
bugatti79
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#1
Nov15-12, 02:23 PM
P: 652
Folks,

I am trying to understand the balance of units for this eqn

## \displaystyle \frac{d^2}{dx^2}(E(x)I(x) \frac{d^2 w(x)}{dx^2})+c_f(x)w(x)=q(x)##

where ##E## is the modulus of Elasticity, ##I## is the second moment of area, ##c_f## is the elastic foundation modulus, ##w## is deflection and ##q## is the distributed transverse load.

Based on the above I calculate the units to be

## \displaystyle \frac{d^2}{dx^2}[\frac{N}{m^2} m^4 \frac{d^2 m}{dx^2}]+\frac{N}{m^2} m=\frac{N}{m}##

gives

##\displaystyle {Nm^3} +\frac{N}{m}=\frac{N}{m}##

##LHS \ne RHS##....?
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Mute
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#2
Nov15-12, 04:40 PM
HW Helper
P: 1,391
The derivatives ##\frac{d^2}{dx^2}## have units of ##1/m^2##.
bugatti79
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#3
Nov16-12, 05:45 AM
P: 652
Quote Quote by Mute View Post
The derivatives ##\frac{d^2}{dx^2}## have units of ##1/m^2##.
Ok, I see how they balance now. The question I have is how is this shown mathematically that the 2nd derivatives have ##1/m^2## units?

##f(x)= f(units in meters)##
##f'(x)= f(units in meters)##
##f''(x)= f(units in meters)##....?

HallsofIvy
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#4
Nov16-12, 07:39 AM
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Euler Bernoulli Beam 4th order ODE -Balance of Units


df/dx is defined as [itex]\lim_{h\to 0} (f(x+h)- f(x))/h[/itex]. The numerator is in what ever units h has. The denominator is in whatever unis x has- "meters" in your case- so the derivative has the units of f divided by the units of x and the second derivative has units of units of f divided by the units of x, squared.

Surely you learned this in basic Calculus? if f(t) is a distance function, with units "meters" and t is time, in "seconds", then df/dt is a speed with units "meters per second" and d2f/dt2 is an acceleration with units of "meters per second squared".


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