Okay, let's continue our lesson on stresses in beams. So let's look at this example. I have a simply supported beam which is bent by two couples, M zero, at the ends. And the length is eight feet and the height, h, of the beam is six inches. Question is, if the deflection at the midpoint is delta as shown here after deflection, so, if the deflection of the center here is delta, and we're given that the longitudinal of normal strain on the bottom surface is .00125, The distance from the neutral surface by symmetry is 3 inches half the height. The question is find the radius of curvature row, the curvature of kappa and the defamation delta. So here's our basic relationship. The longitudinal strain, epsilon x, is equal to minus kappa times y. And also by definition, kappa is 1 over the radius of curvature. So the radius of curvature is minus y over epsilon x. And I can substitute in the numbers. The distance from the nutrax is three inches. The strain is given as .00125 so calculating I find that is equal to positive 24 inches or 200 feet. So the radius of curvature is a big number because the curvature is small, 200 feet, and the curvature of kappa is the inverse of the radius of curvature is 0.005 feet to the minus one. And the last portion to get the deflection delta we have to apply a little geometry here, and from the geometry of this triangle here, much exaggerated, from the geometry of the triangle, we have delta is equal to roe, one minus cosine theta, where theta is this angle here, the half angle of defamation. So rearranging l over 2 is approximately equal to row sin theta for small angles or sin theta is equal to l over 2 row. Substitute the numbers who've all ready calculated is equal to .02. And that of course is in radians. So converting to degrees, the angle there is 1.15 degrees.So finally the defamation delta is 2,400 into 1 minus co sign 1.5 degrees. Which is approximately half an inch. So, obviously, as we would expect, the deflection of this beam at the center is very small, it's only about half an inch. Now this example illustrates the geometric relationships. We have a copper wire bent around a tube of radius r. The maximum normal strain in the wire is most nearly which of these alternatives. Given that the radius of the tube is .6 inches, and the diameter of the copper wire 1.5 mm. So, here's our basic relationship again. Epsilon X is equal to minus kappa Y, and kappa is one over row by definition. So, we're arranging, we have the longitudinal strain, the normal strain, epsilon X, is equal to minus Y over row. So rearranging y, the maximum distance from the neutral surface. So in this case our wire looks like this, and the neutral surface passes through the centroid in the middle here. And the maximum distance y from the neutral axis is plus or minus d over two, half the diameter. So epsilon X is equal to Y is D over two and the radius of curvature here is R plus D over two plus half of the diameter here to go to the center of the wire. Substituting in the numbers that is equal to 1.25 times ten to the minus three. So the answer is A. And this completes our preliminary discussion of torsion.