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In an experiment used to find the Young's modulus of a steel bar, the formula for change in height is given as follows: $$\delta_x=\frac{Fx^2}{6EI}(3L-x)$$ $$I=\frac{bh^3}{12}$$ where: $δ$- the change in height of the steel bar under load at a distance L

$δ_x$- the change in height of the steel bar under load at the point where deflection is measured

$L$- the distance between the point of application of the load and the end support

$x$- the distance between the point at which deflection is measured and the end support

$E$- Young’s modulus

$F$- applied load

$I$- second moment of area of the cross- section

$b$- the width of the beam

$h$- the height of the beam

$w$- the weight of each value in the weighted mean

$m$- the mass hung from the bar

The problem is outlined below: formula background

The problem is that I do not understand how they have derived this top formula. Thanks in advance.

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    $\begingroup$ Look up derivation of Euler-Bernoulli beam theory. Deflection is the solution to a fourth-order differential equation with respect to load distribution. Hope that gets you started! $\endgroup$ – wwarriner Dec 9 '18 at 1:14
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    $\begingroup$ @starrise is right but you can solve this specific case a bit more easily. Draw a moment diagram for the beam, use the formula that relates moment to curvature, and integrate the curvature twice to get the displacements. $\endgroup$ – alephzero Dec 9 '18 at 10:17
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I try to tackel the problem in two different ways, as suggested in the comments.

The first approach is based on Bernoulli beam theory, a set of differential equations, (if you want to know how drive them, then you can ask it here or in physics SE):

$$q(x)=-\frac{dv}{dx}=\frac{d^2M}{dx^2}=-EI_{yy}\frac{d^3\alpha}{dx^3}=EI_{yy}\frac{d^4u}{dx^4}$$

I respect the right handed coordinate system, the $x$ axis lies along the beam, $q$, $v$, $M$, $\alpha$ and $u$ represent the vertical force, shear force, moment, curvature and displacement respectively.

I assume you know how to drive the moment, if you don't, then take a look at the second method:

$$M=F(L-x)$$

Integrating the equation twice results in:

$$EI_{yy}u(x)=-F(L\frac{x^2}{2}-\frac{x^3}{6})-C_1x+C_2$$.

The boundary conditions in this case are:

$$EI_{yy}\alpha(x=0)=0 \rightarrow C_1=0$$ $$EI_{yy}u(x=0)=0 \rightarrow C_2=0$$

So:

$$EI_{yy}u(x)=-F(L\frac{x^2}{2}-\frac{x^3}{6})$$.

The second way is a bit more lengthy but it helps you to develop your insight:

enter image description here

Image: Mechanics of materials Prof. Wim VAN PAEPEGEM

First we find the reaction forces:

$$R_A+F=0 \rightarrow R_A=F$$

$$RM_A=-F.L$$

Now we wish to find the shear forces and the moment, you can choose between the right hand side of the beam or the left hand side, as you can see in the photo, i find it easier to write the equilibrium equations for the right hand side (BC).

$$-V+F=0 \rightarrow V=-F $$

$$M+F(L-x)=0 \rightarrow M=F(L-x)$$

you can now integrate the moment as i did before.

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    $\begingroup$ Thank you so much, this is a very detailed and thorough answer that solves my question nicely! $\endgroup$ – Henry Lee Dec 9 '18 at 22:49

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