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The figure shows a beam of length $ L $ embedded in concrete walls. If a constant load $ W $ is distributed evenly along its length, the beam takes the shape of the deflection curve$$ y = -\dfrac{W}{24EI} x^4 + \dfrac{WL}{12EI} x^3 - \dfrac{WL^2}{24EI} x^2 $$where $ E $ and $ I $ are positive constants. ($ E $ is Young's modulus of elasticity and $ I $ is the moment of inertia of a cross-section of the beam.) Sketch the graph of the deflection curve.

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Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 5

Summary of Curve Sketching

Derivatives

Differentiation

Volume

Oregon State University

Harvey Mudd College

Idaho State University

Lectures

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So first, what we want to do is, um, sketch this curve so we see that we'll have why equals w divided by e i Times X to the fourth. And that can be a minus. And we also have minus 24 minus 24 ei times x to the fourth, actually plus w l divided by 12 ei times X to the third or X Cube. And then lastly, we'll have minus W l squared. Yeah, divided by 24 e I X squared. You know, creation. Using these properties, we can. Now I'm just select constants for our values, and we end up getting this graph. Now we can change these values. But what we ultimately see is that we'll have this somewhat parabolic shape that's always concave down S O. This is the general shape of our graph, Andi. That is the deflection curve that we can sketch based on the values and plotting points

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