{ "metadata": { "name": "", "signature": "sha256:b72b94f11100054eb36f3f6425e3380102d4e56c58db949d525795ac79669196" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "Euler Buckling" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "" ] }, { "cell_type": "heading", "level": 2, "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "Derivation" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "In 1759 the mathematician Leonhard Euler proved that buckling will occur in a column even when it is perfectly straight once the critical buckling load has been reached." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "\n", "\n", "\"Loaded\n", "\n", "\n", "\"Buckled\n" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "*(Click the blue down arrow for structural symbol review or the right arrow to continue.)*" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "subslide" } }, "source": [ "" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "For those proofreaders or students that have forgotten their idealized structural symbols, you can review the appropriate material [here (pdf)](http://www.sut.ac.th/engineering/civil/courseonline/430331/pdf/02_determinate_structures.pdf) and [here (MIT website)](http://web.mit.edu/4.441/1_lectures/1_lecture13/1_lecture13.html). You only need to look at pinned end connection, roller connections and notice that beams and columns are just indicated as lines." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "A little more about buckling can be found [here](http://en.wikipedia.org/wiki/Structural_engineering_theory#Buckling)." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "I am going to go through a simplified derivation to show what the critical buckling load is. \n", "\n", "First, drawing a column vertically takes up a lot of room so let's work with it on its side. We will neglect the effect of gravity since the beam is supposed to be a column." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "\"Loaded\n", "\n", "\"Buckled\n" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "To start with we need to cut the beam in half to look at the internal loading." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "\"Half\n" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "Because the internal moment must resist the rotation caused by the offset axial load force pair (P critical) the moment equals,\n", "\n", "$$M = -P_{cr} y$$" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "Using the [Euler beam equation](http://en.wikipedia.org/wiki/Structural_engineering_theory#The_Euler-Bernoulli_beam_equation),\n", "\n", "$$E I \\frac{d^4y}{dx^4} = w(x)$$\n", "\n", "We can show that,\n", "\n", "$$E I \\frac{d^2y}{dx^2} = M(x)$$\n", "\n", "or in our case,\n", "\n", "$$M = E I \\frac{d^2y}{dx^2}$$" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "With some substitution we get,\n", "\n", "$$E I \\frac{d^2y}{dx^2} = -P_{cr} y$$\n", "\n", "or,\n", "\n", "$$E I \\frac{d^2y}{dx^2} + P_{cr} y = 0$$\n", "\n", "Dividing through by $E I$\n", "\n", "$$\\frac{d^2y}{dx^2} + \\frac{P_{cr} y}{E I} = 0$$\n", "\n", "This is an [ordinary second order differential equation](http://en.wikipedia.org/wiki/Ordinary_differential_equation#Summary_of_exact_solutions) and one form of the general solution is,\n", "\n", "$$y = A \\sin(k x) + B \\cos(k x)$$" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "By evaluating the boundary conditions of the beam we can determine the constants.\n", "\n", "$$y = A \\sin(k x) + B \\cos(k x)$$\n", "\n", "\"Buckled\n", "\n", "If $x = 0$ or $x = L$ then $ y = 0$ because the beam is restrained by the pins.\n", "\n", "So at $x = 0$\n", "\n", "$$0 = A \\sin(k \\times 0) + B \\cos(k \\times 0)$$\n", "\n", "$$0 = A \\sin( 0) + B \\cos( 0) = A\\times 0 + B \\times 1$$\n", "\n", "$$0 = 0 + B$$\n", "\n", "Therefore $B=0$." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "Since $B=0$ our equation for $y$ reduces to,\n", "\n", "$$y = A \\sin(k x)$$\n", "\n", "At $x=L$\n", "\n", "$$0 = A \\sin(k \\times L)$$\n", "\n", "Now if $A=0$ then $y=0$ at every value for x and the columns would never buckle and we know that is not true.\n", "\n", "If $k\\times L=0$ then $k=0$ and $y=A\\sin(0\\times x) = 0$ for all values of $x$, which again we know is not true.\n", "\n", "If $k\\times L=n\\times\\pi$, for $n= 1, 2, 3, ...$ then $k=\\frac{n\\times \\pi}{L}$ and\n", "\n", "$$y = A\\sin\\left(\\frac{n \\pi x}{L}\\right)$$\n", "\n", "Let's check,\n", "\n", "|Location | Boundary Condition |Correct?|\n", "|--- | --- | ---|\n", "|$x=0$ | $y = A\\sin\\left(\\frac{n \\pi \\times 0}{L}\\right)=0$ | $\\checkmark$|\n", "|$x=L$ | $y = A\\sin\\left(\\frac{n \\pi \\times L}{L}\\right)=A\\sin(n \\pi )=0$ | $\\checkmark$|" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "This looks like it will work. Let's plug it into the differential equation.\n", "\n", "$$\\frac{d^2y}{dx^2} + \\frac{P_{cr} y}{E I} = 0$$\n", "\n", "If, $y = A\\sin\\left(\\frac{n \\pi x}{L}\\right)$ then, *([calculus reference](http://www.eeweb.com/tools/math-sheets/images/calculus-derivatives-limits.png))*\n", "\n", "$\\frac{dy}{dx} = A \\frac{n \\pi }{L}\\cos\\left(\\frac{n \\pi x}{L}\\right)$\n", "\n", "and\n", "\n", "$\\frac{d^2y}{dx^2} = -A \\left(\\frac{n \\pi }{L}\\right)^2\\sin\\left(\\frac{n \\pi x}{L}\\right) = -\\left(\\frac{n \\pi }{L}\\right)^2 A \\sin\\left(\\frac{n \\pi x}{L}\\right)$\n", "\n", "so:\n", "\n", "$-\\left(\\frac{n \\pi }{L}\\right)^2 A \\sin\\left(\\frac{n \\pi x}{L}\\right) + \\frac{P_{cr}}{E I} A\\sin\\left(\\frac{n \\pi x}{L}\\right)= 0$\n", "\n", "Divide through by $ A\\sin\\left(\\frac{n \\pi x}{L}\\right)$,\n", "\n", "$- \\left(\\frac{n \\pi }{L}\\right)^2 + \\frac{P_{cr}}{E I} = 0$\n", "\n", "*(Continued on the next slide)*" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "$ \\frac{P_{cr}}{E I} = \\left(\\frac{n \\pi }{L}\\right)^2 $\n", "\n", "$P_{cr} =E I \\left(\\frac{n \\pi }{L}\\right)^2 $\n", "\n", "But we are looking for the lowest load that will cause buckling so use the lowest $n$, i.e. $n=1$\n", "\n", "$$P_{cr} =\\frac{ \\pi^2 E I}{L^2}$$\n", "\n", "This is the Euler critical buckling load.\n", "\n", "Note that the buckling load is independent of material strength $(F_y$ - for steel$ )$ and is only dependent on the flexural stiffness $E I$ and length $L$." ] }, { "cell_type": "heading", "level": 2, "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "Common Forms:" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "This Euler critical buckling load shows up in the steel code in the form,\n", "\n", "$$F_{cr} = 0.877 F_e = 0.877 \\left(\\frac{\\pi^2 E}{\\left(\\frac{K L}{r} \\right)^2} \\right)$$\n", "\n", "\n", "Note that $0.877$ accounts for an initial crookedness of $\\frac{1}{1,500}$ of the length $(L)$. For showing derivations we will assume that the column is perfectly straight, i.e. use $1$ instead of $0.877$.\n", "\n", "Also we will assume that $K=1$ for the [effective length factor](http://en.wikipedia.org/wiki/Column#Equilibrium.2C_instability.2C_and_loads) since we are using pinned end connections.\n", "\n", "$$F_{cr} =\\frac{\\pi^2 E}{\\left(\\frac{L^2}{r^2}\\right)} =\\frac{\\pi^2 r^2 E}{L^2} $$\n", "\n", "The value $r$ is the [radius of gyration](http://en.wikipedia.org/wiki/Radius_of_gyration#Applications_in_structural_engineering) which equals,\n", "\n", "$$r = \\sqrt{\\frac{I}{A}} \\Rightarrow r^2 = \\frac{I}{A}$$\n", "\n", "*(Continued next slide)*" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "$$F_{cr} =\\frac{\\pi^2 r^2 E}{L^2}=\\frac{\\pi^2 I E}{L^2 A} $$\n", "\n", "Since $P_{cr} =F_{cr} \\times A$,\n", "\n", "$$P_{cr} =F_{cr} \\times A=\\frac{\\pi^2 r^2 E}{L^2}=\\frac{\\pi^2 E I}{L^2 A} \\times A$$\n", "\n", "$$P_{cr} =\\frac{\\pi^2 E I}{L^2 } $$\n", "\n", "Which is the Euler critical buckling load." ] }, { "cell_type": "heading", "level": 2, "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "Buckling in Timber" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "In timber design we use rectangular sections.\n", "\n", "In a rectangular section the moment of inertia can be written as,\n", "\n", "$$I = \\frac{b d^3}{12} = \\frac{Ad^2}{12}$$\n", "\n", "Our Euler critical buckling load is,\n", "\n", "$$P_{cr} =\\frac{\\pi^2 E I}{L^2 } $$\n", "\n", "by substisuting for $I$, \n", "\n", "$P_{cr} =\\frac{\\pi^2 E A d^2}{12 L^2 } = \\frac{\\pi^2}{12}\\frac{E A d^2}{L^2 }$\n", "\n", "$P_{cr} = 0.822 \\frac{E A d^2}{L^2 } $\n", "\n", "*(Continued next slide)*" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "The factor $E_{min}$ is just $E$ with a factor of safety of $2.75$.\n", "\n", "$E_{min}=\\frac{E}{2.75}$\n", "\n", "$P_{cr} = 0.822 \\frac{E_{min} A d^2}{L^2 } $\n", "\n", "but we use allowable stress in timber, so we divide both sides by $A$ to get the critical buckling stress.\n", "\n", "$F_{cr} = 0.822 \\frac{E_{min} d^2}{L^2 } $\n", "\n", "In the Timber code we call it the critical Elastic buckling stress,\n", "\n", "$$F_{cE} = \\frac{0.822 E_{min}' }{\\left(\\frac{L}{d}\\right)^2 } \\text{ (see section 3.7.1 of the Timber NDS)}$$" ] }, { "cell_type": "heading", "level": 2, "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "References:" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "[Class website](https://elearning2.courses.ufl.edu/portal/site/UFL-BCN4423C-40908-52014) *(Use this link to if you are taking the course on e-learning.)*\n", "\n", "[Github.io version of course website](http://damontallen.github.io/Construction-Lectures/) *(Do not use this link if you are taking this course in Summer A or B.)*\n", "\n", "[IPython.org](http://ipython.org/) *(IPython is the opensource software used in the development of much of this course.)*\n", "\n", "[Complete Software List](http://nbviewer.ipython.org/github/damontallen/Construction-Lectures/blob/master/Resources.ipynb)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "skip" } }, "source": [ "CSS stylesheet\n", "\n", "\n", "\n", "\n", "\n", " /* class = \"max box\" for img tags\n", " \n", " \n", " \n", " */\n", " class = \"max box\" == style=\"max-width:300px; max-height:300px; border:1px solid blue; float:left; margin-right:3px;\"" ] } ], "metadata": {} } ] }