the centre will be negative, and there will be two points of contrary flexure somewhere between B and D. In the diagrams e is taken at 3b, and the point of contrary flexure is then situated at t, half-way between E and C. The pontoon will now be sharply hogged in the middle, as shown in Fig. 476. From t to E it will be slightly sagged, but notwithstanding this curvature, the inclination will still be a downward inclination, even at the extreme end; for the upward or downward inclination at E will be proportional to the algebraical sum of the negative and positive areas of the moment diagram, and when the support is at D, it is evident, by mere inspection, that the negative area CtG is greater than the positive area EKt.

The curves of deflection shown in Figs. 476, 47c, and 47d, relate to the three cases in which e is taken to be,, and ğ of the half-length b, and are all drawn to the same (exaggerated) vertical scale. To trace the curves in detail, the simplest plan would be to set off calculated ordinates y1, Y2, etc., above or below a horizontal tangent drawn through the central point C; and at any section the ordinate y will be proportional to the algebraical sum of the moments of the positive and negative areas in the moment diagram, intercepted between that section and the centre line.

Με

Thus, to find the deflection at D in Fig. 476, let μ, denote the area CtG in Fig. 47a, and let 1 be the horizontal distance of its centre of gravity from the point D; also let με and 2 denote the same quantities for the positive area DKt; then the downward deflection at D, or the ordinate below the tangent, will μιξι Y2 – M151 — M252

be

ΕΙ

To obtain a general expression for the ordinates y1⁄2 and yз, the calculation may be simplified still further; for if we draw the horizontal line RK, and subtract the moment of the rectangle RKCD from the moment of the figure RKG, the difference will obviously be the same as the quantity μ11- 22 in the μεξ, equation just above written.

A little consideration will suffice to show that we can treat the diagram in the same way in all the three cases, and thus we may obtain, by successive integration, a general expression for the deflection ordinate at each point of support, viz.—

Y2 =

p(b − e)2
24EI

{6e2 - 5(be)2} .

(11)

and if we have

use z to denote the distance CD, or b e, we

In the same way, we can summate the moments of the positive and negative areas reckoned about E, and we then have for the deflection ordinate at either end of the pontoon

Of course, the deflections calculated in this way are the deflections that would be produced under the bending stresses represented in Fig. 47a; and these bending stresses are themselves calculated upon the assumption that the buoyant pressure will be uniformly distributed. If we turn the deflection curve upside down, and then build the pontoon with an initial camber represented by the inverted curve, the pontoon will come to a straight line when floating under the given load; the buoyant pressure will then be uniformly distributed, and the bending stresses will have exactly the values above calculated.

On the other hand, if the pontoon is built initially straight, the imposition of the load upon the two supports will have the effect of producing a certain curvature, and in each half of the pontoon a certain change in the position of the centre of buoyancy. The extent of these changes will evidently depend, not only on the stiffness of the pontoon, but very largely also upon the disposition of the two supports; and the changes will be very small if we make the distance e equal to about sevensixteenths of b.

Each individual case must, of course, be treated according to the actual dimensions and loads; and if it is found in any given case that the elastic deflection, as calculated by the formula, amounts to a considerable fraction of the mean draught, it may be desirable to introduce a correction for the altered distribution

of the buoyant pressure. This may be done in the manner already illustrated in the last article, but for such a purpose it would probably be more convenient to work by graphic construction of the several curves rather than by analytical methods.

Art. 59. A Series of Concentrated Loads imposed at Three or More Points at Known Distances.-When the floating pontoon is loaded with a series of weights placed at equal or unequal

distances, the diagram of moments may sometimes be similar in form to that which is so familiar to us in continuous girders; but the bending moments can be calculated without any of the difficulties which present themselves in the case of a continuous girder.

For if we know the actual distribution of the loads, we can find the distribution of the buoyant pressure within small limits of error; and when the vertical forces are all known, the bending moments can always be calculated by the general formula referred to in Art. 56.

Thus, for example, if the centre of gravity of the whole load coincides with the centre of the pontoon, we may assume that the buoyant pressure will be a constant quantity p per foot lineal, unless the elastic deflection is so great as to produce a sensible change in its distribution.

Then, to find the bending moment at any given section, whose distance from the right end of the pontoon is denoted by x, let W1, W2, etc., represent the weights of those loads which are imposed on the right-hand side of the section, and let X1, X2, etc., denote their horizontal distances from the given section; and we shall have for the bending moment

It is hardly necessary here to apply the formula to any particular case; but it may be noted that the curve of moments will not always intersect the datum line between the loaded. points, as it generally does between the piers of a continuous girder bridge.

Art. 60. Load uniformly distributed over a Portion of the Pontoon's Length. It is sufficiently evident that a flat rectangular pontoon, carrying a load of any given intensity per foot lineal, which is uniformly distributed along its whole length, will float with a level bottom, and will undergo no bending stress whatever in the longitudinal plane.

But we may sometimes have to consider the stresses that

will take effect when the load does not extend to the ends of

the pontoon.

We will suppose that the load extends over a length BD = 2z in the middle of a pontoon, whose total length AE = 2b, as sketched in Fig. 48.

Eliminating again the weight of the pontoon, and also that portion of the buoyant pressure which goes to support it, we may consider only the external load, and the uniformly distributed buoyant pressure by which the external load is supported. If q denotes the intensity of the external load per lineal foot, we shall have for the distributed buoyant pressure the uniform value, or, at all events, the mean value

q

e+

p = q = 142, in which e denotes the length of pontoon that b has no load upon it, or the length DE in Fig. 48 at each end of the pontoon.

Then, to find the bending moment and the vertical shearing force at any section, whose distance from the right end of the pontoon is denoted by x, we may proceed as follows:

For any distance x that is less than e, we have

At the point D, where x = e, the bending moment will have

the value MD

pe2
2

= ; and this moment is represented by the

ordinate DK in the diagram of Fig. 48a, the curve being a simple parabola from E to K. At the point D we shall have the greatest shearing force, its value being S pe. To complete the curve of moments beyond K, we have, when x is greater than e

=

; and S = px − q(x − e)

At the centre of the pontoon we find the greatest bending moment, whose value will be

The whole diagram of moments will consist of a pair of reverse parabolic curves, something like an admiral's cocked hat. The moments will everywhere be positive, and the pontoon

will be sagged throughout its whole length, the curvature being sharpest in the middle.

To trace the curvature that would theoretically be produced by these bending stresses, we may draw a horizontal tangent through the central point C, as in Fig. 486, and then we can calculate the ordinates at D and at E as follows:

In the diagram of moments, the figure GRK will have the area pez2, while its moment about K will be pez. The rectangle CRKD will have the area pe2z, and its moment about

K will be pe22. Adding these moments together and dividing by EI, we have for the upward ordinate at D, or the sagging deflection of the loaded portion

pez2
24EI (6e +52).

(16)

Then reckoning the moments of the same areas about the point E, we have for the first 24pez2(8e + 52), and for the second 24pe2z(12e +62), while the moment of the area DKE will be pe. Therefore, the upward ordinate at E, or the sagging deflection on the whole length of the pontoon, will be

pe

Y3 =
24EI

{3e3+6ez (2e + 2) + z2(8e + 52)} . (17)