Euler Bernoulli Beam 4th order ODE -Balance of Units

[bugatti79]

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

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

$LHS \ne RHS$...?

 

[Mute]

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

 

[bugatti79]

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]

df/dx is defined as $\lim_{h\to 0} (f(x+h)- f(x))/h$. 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".