C

= distance between neutral axis and extreme fibre.

For breaking load of circular section replace bd2 by 0.59 d3.

For good wrought iron the value of R is about 80,000; for steel about 120,000. For cast iron the value of R varies greatly. Thurston found 45,740 for No. 2 and 67,980 for No. 1.

Beam supported at both ends, single unsymmetrical load.

Beam supported at both ends, single unsymmetrical load.

the modulus of elasticity or the moment of inertia. The following table gives the values of W, etc., without introducing

General Formulæ for Transverse Strength, Etc.

FORMULA FOR ROUND AND RECTANGULAR SOLID BEAMS

CHAPTER V

ACCELERATION OF FALLING BODIES

THE change in velocity of a falling body which occurs in a unit of time is its acceleration.

That due to gravity is 32.16 feet per second, in 'one second and is denoted by g.

Let

t

=

2=

number of seconds during which a body falls.

velocity acquired in feet per second at the expiration of t seconds.

u = space fallen through in each second.

The table below gives the values of h, v and u, for values of t up to ten seconds.

The graphical method of ascertaining the values of t, v, u and his easily remembered and is often of service.

1Seck

2

3"

4 "

5"

6"

In the triangle, Fig. 61, let the vertical divisions on the left of the perpendicular represent the number of seconds through which the body falls = t.

Let the base of each small triangle equal the velocity at the end of the first second = 32.16. Then the number of bases on each of the horizontal lines at 1, 2, 3, etc., multiplied by 32.16 will equal the acquired velocity for the corresponding time = v. Let the area of each small triangle = 16.08. Then the number of such areas between o and any horizontal line multiplied by 16.08 will equal the height in feet fallen through in the number of seconds corresponding to that line = h. And the number of small triangles between each pair of horizontal lines, multiplied by 16.08 will equal the number of feet fallen through in each second =u.

FIG. 61.

If two forces are applied to the same point, their resultant will be represented in intensity and direction by the diagonal of a parallelogram of which the adjacent sides represent the intensities and directions of the given forces.

Let AB and AC represent, in intensity and direction, any two forces applied to the point A; then AD will correspondingly represent their resultant.

FIG. 62.

Conversely, if AD be the known force acting at A, it may be resolved into two components, in any direction in the same plane; which components will be the adjacent sides of a parallelogram having AD for its diagonal.

Parallelopipedon of Forces

If three forces, not in the same plane, act on the same point, they may be represented by the edges of a parallelopipedon and the diagonal through the point of application is their resultant.