1
$\begingroup$

I just thought about this problem recently but am not sure where to find a solution. I am not certain which parameters are important in this problem, so please bear with me.

Assume that a beam of length $L$ with rectangular cross section of width $W$ and height $H$ is pinned on the wall horizontally so that the width of the beam's cross section is parallel to the ground and the height of the beam's cross section is almost parallel to the normal to the ground (almost parallel because the beam bends, causing the height at each point to tilt a little). Suppose that the beam has mass $M$ which is distributed uniformly. Under gravity, the beam bends (see the picture below). What is the shape of this bent beam? (It suffices to determine the shape of the central line that runs the length of the beam.)

I assume that Young's modulus $E$ and the shear modulus $G$ of the beam are needed. Furthermore, the material that makes up the beam is assumed to be isotropic. With the parameters $L$, $W$, $H$, $M$, $E$, $G$, and the gravitational acceleration $g$, is this problem complete? Do you know how to solve it or any reference towards a solution? I would like the exact shape, rather than an approximated one. I have seen an approximation by a circular arc. Thank you very much in advance.

enter image description here

Improve this question
$\endgroup$
1
  • $\begingroup$ @Qmechanic This is a request for resources. I couldn't care less about homework of this type. Homework related to this question usually asks only for approximated shapes. I want exact shapes, and I am looking for research papers in this topic. $\endgroup$
    – Batominovski
    Commented Jun 18, 2018 at 18:02

2 Answers 2

Reset to default
2
$\begingroup$

The thing you are describing is called a cantilevered beam, and there are countless resources showing the derivation of its shape. Many of the derivation appear to be copied from books, and I'm not sure what the copyright situation is. However, I found a youtube video that shows the derivation.

In general, the curvature at a given point depends on the weight of the beam to the right of the point, and the bending moment it produces. The mass is proportional to the length of beam to the right, and the distance over which the force acts is equal to the distance divided by two (for a uniform beam).

That means the curvature is quadratic with distance from the wall; after integrating twice, you can see that the height will be proportional to the fourth power.

All this is shown in the link provided. With all the right scaling factors...

If you are interested, I wrote a more detailed answer about deflection of a center-loaded beam:

beam supported at both ends, with a weight in the middle

Improve this answer
$\endgroup$
2
  • $\begingroup$ What is the reference of this book pages of your link? $\endgroup$
    – freecharly
    Commented May 16, 2018 at 16:24
  • $\begingroup$ @freecharly it appears to be "Mechanics of Materials" by Hibbeler. Looking closely I realize this may be an illicit copy (although it's one of the top hits on Google) and I will find a better reference. $\endgroup$
    – Floris
    Commented May 16, 2018 at 16:26
1
$\begingroup$

This is a typical cantilever beam with the length L, under uniform load q , its own weight.

Assuming the beam is slender L/H >20 the shear deflection will be of tertiary order and will not have meaningful impact on bending.

then the bending or deflection will be $$\delta_x = qL^4 /8EI \space ,while\space I = Wh^3/12 $$

And $$\theta_b =qL^3/6EI $$

Moment at support is $ = qL^2/2$ And shear at support $ V= qL $

Improve this answer
$\endgroup$

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.