Constructional Iron and Steel Work As Applied to Public, Private, and Domestic Buildings

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General Books, 2013 - 110 стор.
This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1896 edition. Excerpt: ...twice as much as the extra weight in an 8 in. x 4 in. over that in a 5 in. x 2 in. joist. Of course, the joists cannot move laterally when fire-clay or terra-cotta lintels or permanent centres are placed between them and so determine their spacing. We now come to the trimmer joists D D; each of these has to carry its own normal load of 4-8 tons equally distributed and the concentrated load. brought upon it by the trimmer E E, which is one-half of the total load on the latter, or 2-3 tons at one foot from the bearing E. The amount of load thrown upon the bearing at the other end of joist D by the concentrated load will be 2-3 1 = 0-096 ton. 24 4-8 The load per lineal foot on the joist = = 02 ton. Let x equal the distance of any given point in the joist d from the bearing on the wall opposite A B, and M the moment of stress at that point, then, M= -w4-0-096 = 0-2 xa _ 0-2 x 24 x x _ 0.Q96 x = 0-1 Xx2-2-496 x The value of z which gives the maximum moment of stress must next be found. From the point of maximum stress it will commence decreasing in each direction, but we may imagine two points taken infinitesimally close together, so that no appreciable difference occurs between them; let a be an indefinitely small increment of x; then M = 0-1 (x + of-2-496 (x + a)-0-l(x-+ 2xa + or)--2-496 (x + a). As a is taken indefinitely small, or, which will be much smaller, may be disregarded, then 01 (2 + Ixa)-2-496 (x + a) = 0-1 3?-2-496x; subtracting 0-l x2 and--2-496 x from each side of tho equation, we have 0-2 x x.a--2-496 a; whence = 12-48 feet. Replacing x in the general equation by this value the maximum moment of stress is found, M--0-1 x 12-483-2-496 x 12-48 =--15-575 foot-tons. To find the equivalent distributed load this is...

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