2
$\begingroup$

For a rod of length $l$, breadth $b$, depth $d$, and Young's modulus $Y$, the rod's center of mass comes down by a distance $\delta$ when a weight $W$ is hanged from its center

$$\delta = \frac{Wl^3}{4bYd^3}.$$

How do I derive this formula? In what way should I take the element for integration, assuming it is derived using integration? (Please ignore the labelings in the diagram)

enter image description here

$\endgroup$
4
  • 1
    $\begingroup$ Diagram shows bending, not buckling. Buckling depends on how the ends are held. $\endgroup$
    – user137289
    Jan 2, 2017 at 18:21
  • 2
    $\begingroup$ take a look here (page 8) : freestudy.co.uk/statics/beams/beam%20tut3.pdf $\endgroup$
    – user98038
    Jan 2, 2017 at 21:09
  • $\begingroup$ What effort have you made to look for a derivation online (or elsewhere)? $\endgroup$ Jan 3, 2017 at 17:54
  • $\begingroup$ @sammygerbil I did not find this formula anywhere online. All the pages gave me some other formulas, just like the answer below $\endgroup$ Jan 5, 2017 at 5:55

1 Answer 1

6
$\begingroup$

A thin slender beam (rod) with length $\ell$, 2nd area moment $I = \frac{1}{12} b d^3$ and Young's modulus $E$ obeys the following differential equation

$$ M(x) = E I \frac{\partial ^2 y(x)}{\partial x^2} $$ where $y(x)$ is the deflection shape and $M(x)$ the internal moment function.

Using statics split the beam into two sections and do a free body diagram on each one

beam1

For the first section from A to C the internal moment is $$M_1(x) = -x\,A_y = - x \,\frac{F}{2} $$

For the second section from C to B the internal moment is $$M_2(x) = -x \, A_y+ F (x-\frac{\ell}{2}) = - (\ell -x) \frac{F}{2}$$

Now you can integrate the moment to get slope and deflection and make sure you use the constants of integration for fitting into the boundary conditions (at $y_1(0)=0$ and $y_2(\ell)=0$) and continuity ($y_1(\frac{\ell}{2})=y_2(\frac{\ell}{2})$ and $y_1'(\frac{\ell}{2}) = y_2'(\frac{\ell}{2})$)

The deflection shape is $$ y_1(x) = \frac{1}{E I} \int \int M_1(x)\,{\rm d}x{\rm d}x + K_1 + K_2 x = -\frac{F x (4 x^2-3 \ell^2)}{48 E I} \\ y_2(x) = \frac{1}{E I} \int \int M_2(x)\,{\rm d}x{\rm d}x + K_3 + K_4 x = -\frac{F ( \ell^3 -9 \ell^2 x+12 \ell x^2 -4 x^3)}{4 8 EI }$$

In the middle use $x=\frac{\ell}{2}$ for $$ \boxed{\delta = -\frac{F \ell^3}{48 E I} } $$

Here the result is negative because the deflection is along the negative y direction


There is also a shorter method using what is called an energy method. After finding the internal moments, find total internal energy of the system as a function of the applied load

$$U(F) = \int \limits_{0}^{\ell/2} \frac{M_1^2(x)}{2 E I}\,{\rm d}x + \int \limits_{\ell/2}^{\ell} \frac{M_2^2(x)}{2 E I}\,{\rm d}x = \frac{F^2 \ell^3}{96 E I}$$

The deflection at the point $F$ is applied (the middle) use the following partial derivative

$$\boxed{ \delta = \frac{\partial U(F)}{\partial F} = \frac{F \ell^3}{48 E I} } $$

Here the result is positive because the deflection is along the direction of the force

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.