How Beam Deflection Works? Beams are in every home and in one sense are quite simple. The weight placed on them causes them to bend (deflect), and you just need to make sure they don’t deflect too much by choosing a large enough beam. As shown in the diagram below, a beam that is deflecting takes on a curved shape that you might mistakenly assume to be circular. So how do we figure out what shape the curvature takes in order to predict how much a beam will deflect under load? After all you cannot wait until the house is built to find out the beam deflects too much. A beam is generally considered to fail if it deflects more than 1 inch. Solution to How Deflection Works: The model for SIMPLY SUPPORTED beam deflection Initial conditions: y(0) = y(L) = 0 Simply supported is engineer-speak for it just sits on supports without any further attachment that might help with the deflection we are trying to avoid. The dashed curve represents the deflected beam of length L with left side at the point (0,0) and right at (L,0). We consider a random point on the curve P = (x,y) where the y-coordinate would be the deflection at a distance of x from the left end. Moment in engineering is a beams desire to rotate and is equal to force times distance M = Fd. If the beam is in equilibrium (not breaking) then the moment must be equal on each side of every point on the beam. It all begins with a rather lovely formula for moment ultimately derived from the nature of curvature involving 풅ퟐ풚 a second derivative, which we will take at face value here: M = EI , where 풅풙ퟐ E is the modulus of elasticity (an expression of the flexibility of the material the beam is made of) I is the moment of inertia (an expression of a beams ability to resist bending based on its shape) which can be 푏푑3 calculated based on the beams width (b) and height (d) as 퐼 = . This formula also comes to us via an 12 integral that is rather beautiful … see the page on How Moment of Inertia Works. This moment then will be equal to the moment around any point (P) on our beam. Let’s consider the moment on the left side of P. Half the weight of the beam will be held by the force provided by the support at O. (1) The weight of the beam will be w lbs/ft * L ft = wL pounds which is a distance of x from point P. The support at O will be required to push up on the beam causing a negative moment around P. (2) The portion of the beam between O and P will have a weight of wx and will cause a positive moment which is x/2 away from P. Now we have what we need to start our formula for moment M around any point on the beam. ퟐ 퐝 퐲 퐰퐋 퐱 퐰퐋 퐰 ퟐ (3) M = EI = ∗ 퐱 − 퐰퐱 ∗ = ∗ 퐱 − 퐱 퐝퐱ퟐ ퟐ ퟐ ퟐ ퟐ Now we can integrate both sides of the equation to get rid of the derivatives. We can treat E, I, w and L as constants. 퐝ퟐ퐲 퐰퐋 퐰 ∫ 퐄퐈 = ∫ ∗ 퐱 − 퐱ퟐ 퐝퐱ퟐ ퟐ ퟐ 풅풚 풘푳 풘 (4) 퐄퐈 = ∗ 풙ퟐ − 풙ퟑ + 푪 풅풙 ퟒ ퟔ ퟏ 퐝퐲 퐰퐋 퐰 And now we can take the integral one more time ∫ 퐄퐈 = ∫ ∗ 퐱ퟐ − 퐱ퟑ + 퐂 퐝퐱 ퟒ ퟔ ퟏ 퐰퐋 ퟑ 퐰 ퟒ (5) 퐄퐈퐲 = ∗ 퐱 − 퐱 + 퐂 퐱 + 퐂 ퟏퟐ ퟐퟒ ퟏ ퟐ Now, recall that (0,0) has to be a solution which implies C2 = 0 And second that (L,0) has to be a solution which implies: ퟑ ퟑ 퐰퐋 ퟑ 퐰 ퟒ 퐰퐋 퐰 ퟑ 퐰퐋 ퟎ = ∗ 퐋 − 퐋 + 퐂 퐋 ... and solving for C1 implies 퐂 = − + 퐋 = − substituting into (5) we have: ퟏퟐ ퟐퟒ ퟏ ퟏ ퟏퟐ ퟐퟒ ퟐퟒ ퟑ 퐰퐋 ퟑ 퐰 ퟒ 퐰퐋 (6) 퐄퐈퐲 = ∗ 퐱 − 퐱 − 풙 ... now solving for y and cleaning up a bit we have the amazing result: ퟏퟐ ퟐퟒ ퟐퟒ 퐰 (7) 퐲 = (ퟐ퐋퐱ퟑ − 퐱ퟒ − 퐋ퟑ퐱) … Rad, so the curve is actually a fourth degree polynomial! ퟐퟒ퐄퐈 For the graph the weight (w) is 6 lbs/in, modulus of Elasticity (E) is 2million PSI, moment of inertia (I) is 320 in4, and the length of the beam is 288 in. Note that the maximum deflection of this beam is .84 inches which is safely less than an inch, so this beam design would be acceptable. 푳 Finally refer back to (7) and plug in x = since the maximum deflection occurs is the middle of the beam. ퟐ ퟓ풘푳ퟒ (8) ymax = … this is the classic simple algebra equation that allows an engineer to predict the ퟑퟖퟒ푬푰 deflection of a beam. It is interesting to note how often we are indebted to calculus for simple and useful algebra formulas. .