An Elementary Treatise on the Calculus for Engineering Students; With Numerous Examples and Problems Worked Out

$17.75

This textbook provides instruction and practice problems for students learning calculus, specifically applied to engineering.

An Elementary Treatise on the Calculus for Engineering Students; With Numerous Examples and Problems Worked Out
An Elementary Treatise on the Calculus for Engineering Students; With Numerous Examples and Problems Worked Out
$17.75

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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1896 Excerpt: …Suppose one half of this girder to PlQ. 41. become embedded in a wall, then the shape of the beam will be unaltered, and the problem reduces to 202; but W I instead of W we have–, and instead of I we have–; therefore the equation of the deflection curve is where y is the height above and x the distance from the centre of the beam. The greatest deflection is equal to the greatest value of K-that is, where x =-. “48 EI” (2:’;”.) To find the shape and the deflection of a girder of uniform section supported at its ends and loaded uniformly. Let 1 denote the length of the girder in feet, w the load per foot ran. Taking the middle point 0 as origin, the soriKAtal line through 0 as the axis of X, and the vertical line thronsrh 0 as the axis of Y, we have the bending moment al a distance << from the origin, tlxwJWr Integrating, we have w/l2x x3 No constant is added, since-=-= 0 where x = 0. aa; Integrate again; therefore _ T w /Px2 X There is no constant added, since = 0 when a; = 0. The deflection is equal to the greatest value of y; that is, where =-; therefore a A =-+- =11? whereW = <<,z. 384 EI 384 EI' (206.) To find the shape and deflection of a girder of uniform section fixed at its ends and loaded at its centre. Let I denote the length of the girder in feet, W the load at its centre in lbs. Taking the middle point 0 as origin and axis, as in the preceding example, we have the bending moment at a distance x from the centre. (<<) M-5g-.)-M', where M' is the bending moment due to the stresses in the girder at its intersection with the wall. Therefore Integrating, we have,,.,-rdy W /lx x2,r, There is no constant to be added, since–= 0 where ax x=0. We have also-=-=-= 0 where a; =-, and this gives ax A 8 The bending moment at the…

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Dimensions 18.9 × 0.2 × 24.6 in

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