the formula, as shown by the dotted lines. If friction could be entirely eliminated, probably experiment and theory would agree.

In calculating the deflection of such springs, E should be taken at about 26,0c0,000, which is rather below the value for the steel plates themselves. Probably the deflection due to shear is partly responsible for the low modulus of elasticity, and

from the fact that the small central plate (No. 5) is always omitted in springs.

=

m.n.d. OH inch-lbs., using Then I inch on the vertical

M

CASE XII. Beam unevenly loaded. Let the beam be loaded as shown. Construct the bending-moment diagram shown below the beam by the method given on p. 416. Then the bending moment at any section is M the same notation as on p. 417. scale of the bending-moment d inch-lbs., and 1 inch on the horizontal scale = inches. Hence one square inch on the diagram m. n2OH inch-lbs. Then A = a. m. n'OH, where a = the shaded area measured in square inches.

diagram = = m. n. OH

=

The centre of gravity must be found by one of the methods

described in Chap. III. Then L = n., where is measured in inches, and the deflection

It is evident that the height of the supports above the tangent is the same at both ends. Hence the moment of the areas about the supports on either side of the tangent point must be the same. The point of maximum deflection must be found in this way by a series of trials and errors, which is very clumsy.

The deflection may be more conveniently found by a somewhat different process, as in Fig. 425.

L

FIG. 424.

We showed above that the deflection is numerically equal to the moment of each little element of area of the bendingmoment diagram about the free end EI. The moment of

d

R

Deflection curve

FIG. 425.

each portion of the bending-moment diagram may be found readily by a link-and-vector polygon, similar to that employed for the bending-moment diagram itself.

Treat the bending-moment diagram as a load diagram; split it up into narrow strips of width x, as shown by the dotted lines; draw the middle ordinate of each, as shown in full lines: then any given ordinate × by x is the area of the strip. Set down these ordinates on a vertical line as shown; choose a

pole O', and complete both polygons as in previous examples. The link polygon thus constructed gives the form of the bent beam; this is then reproduced to a horizontal base-line, and gives the bent beam shown in dark lines above. The only point remaining to be determined is the scale of the deflection

curve.

}

We have 1 inch on the load scale of the
first bending-moment diagram
also I inch on the length of the bending-)

= m lbs.

= n inches

and the bending moment at any point M = m. n.D.OH

M

where D is the depth of the bending-moment diagram at the point in inches, and OH is the polar distance, also expressed in inches. Hence 1 inch depth of the bending-moment diagram represents = mn. OH inch-lbs., and I square inch of the D bending-moment diagram represents mn2OH inch-inch-lbs. Hence the area xD represents xDmn2OH inch-inch-lbs. ; but as this area is represented on the second vector polygon by D, its scale is xmn2OH; hence

If it be found convenient to reduce the vertical ordinates of the bending-moment diagram when constructing the deflection vector polygon by say, then the above expression must be

multiplied by r.

I

Deflection of Built-in Beams.-When a beam is built in at one end only, it bends down with a convex curvature

upwards (Fig. 426); but when it is supported at both ends, it bends with a convex curvature downwards (Fig. 427);

and when a beam is built in at both ends (Fig. 428), we get a combined curvature, thus

Then considering the one kind of curvature as positive and the other kind as negative, the curvature will be zero at the points xx (Fig. 429), at which it changes sign; such are termed "points of contrary flexure." As the beam undergoes no

bending at these points, the bending moment is zero. Thus the beam may be regarded as a short central beam with free ends resting on short cantilevers, as shown in Fig. 430.

Hence, in order to determine the strength and deflection of built-in beams, we must calculate first the positions of the points x, x. It is evident that they occur at the points at which the upward slope of the beam is equal to the downward slope of the cantilever.

We showed above that the slope of a beam or cantilever at any point is given by the expression—

CASE XIII. Beam built in at both ends, with central load.

Hence, as the slope is the same at the point where the beam joins the cantilever, we have

This problem may be treated by another method, which, in some instances, is simpler to apply than the one just given.

When a beam is built in at both ends, the ends are necessarily level, or their slope is zero; hence the summation of the slope taken over the whole beam is zero, if downward slopes be