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Euler Bernoulli Beam 4th order ODE Balance of Units 
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#1
Nov1512, 02:23 PM

P: 660

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##....? 


#2
Nov1512, 04:40 PM

HW Helper
P: 1,391

The derivatives ##\frac{d^2}{dx^2}## have units of ##1/m^2##.



#3
Nov1612, 05:45 AM

P: 660

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


#4
Nov1612, 07:39 AM

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P: 39,564

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 d^{2}f/dt^{2} is an acceleration with units of "meters per second squared". 


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