than the actual sectional area of the stanchion itself. This is, of course, by reason of the bearing power of the soil or other natural foundation upon which it stands being limited to so many tons or lbs. per square foot according to circumstances, always less than the safe stress intensity of the material of which the stanchion is made. Practical illustrations of this are given in chapters on steel work. Between the base (usually of masonry, brickwork, or concrete) and the stanchion, in order to distribute the pressure evenly over the base, is interposed a base plate, which in cast-iron pillars is an integral part of the casting, the maximum fibre stress occurring either at the top or bottom of a plate of uniform thickness, or the bottom of a cast-iron plate of varying thickness. The ordinary formulæ for base plates are either empirical or based on the theory that this line of fracture of a base plate is a straight line tangent to the exterior of the foot of the column standing on it. The latest formulæ at the time of writing are those based on Ricker's experi ments, and for a circular plate,

We will now consider how shafting behaves when used to transmit power. While this is not intended to be a treatise on mechanical engineering, so much of this work comes under the province of the civil engineer that a few words on shafting will not be out of place, because the stresses set up there have a kind of relation to some of the stresses we have just discussed.

For

instance, the connecting rod of an engine transmitting the pressure of the steam on to the shafts sets up an enormous bending moment therein. Again, the rods of a large pump and the vertical shaft of a water turbine (unless it is balanced in a manner to be described under water power) are loaded in exactly the same way as a pillar of solid cylindrical section, besides the power they transmit.

FIG. 131.

Now a man turning a crank such as we described in fig. 99, Chapter VII., if he applied a force of 100 lbs. at the end of a crank 24 in. long, the torque or twisting moment which would be set up in that shaft would be 24 x 100 = 2400 in.-lbs. Now, in the same way as a beam having a bending moment, we must have a resisting moment in opposition to it, so we have in a shaft a resistance to torque or twisting moment supplied by the shaft itself. It is called the Torsional Resistance, and its value is, in in.-lbs.,

R= radius of shaft in inches. Fallowable stress on the material in lbs. (shear stress), or if F = tons, Tinch-tons.

Now the shaft will demonstrate its twist as in fig. 131, and the strain in any ring of fibres is proportional to the arc aB, which is called the arc of twist. Suppose any radius in question, and d and D the internal and external diameters of any hollow shaft in question, and ƒ the allowable shear stress (maximum),

Then the stress in the shaft

(D this time referring to the diameter of a solid shaft), while for a hollow shaft we have the torsional resistance

Therefore by comparison we see for the same length and weight a hollow shaft is 44% stronger than a solid one. Now if we know that in a certain shaft we are transmitting a certain horse-power (=33,000 ft.-lbs.) at a certain number of revolutions per minute, then we can find the twisting moment in that shaft by the formula

where N = revs. per minute, and the value of TM is in inch-lbs.

Now obviously in a gas or steam engine having a piston of "A" square inches in sec. area and a pressure (max.) of P lbs. per sq. in., the twisting moment on the shaft if the crank radius is R inches

But when the shaft is overhung we also have a bending moment in it (similar to a cantilever end-loaded) when l = distance in inches from the centre line of motion to the nearest bearing,

Now the equation for resistance to bending in any solid circular shaft when supported at both ends

that is, just one-half the resistance to torque (vide Formula 104), therefore 32 substituted for 16 in Equations 106 and 106a will give us the resistance to bending on those shafts.

When we have a bending moment as obtained by Equation 109 and a torque as in 108 we have to combine the two and get an equivalent twisting moment = TM1, and

In addition to the

the same as in beams.

TM1 = BM + √BM2 × TM2.

(111)

strength of shafts we also have stiffness to consider, But in this case the strength varies as D3 and stiffness as D. Referring to Equation 107, we have a shaft which we know will stand a certain torque of TM in.-lbs., then we can transpose the equation to give us the safe horse-power in any shaft, which we can transmit.

Its value is

Take now the case of a shaft in which we are given the twist in a certain

number of degrees =D; we have the strain also given = x. Then if we want a certain shaft which under the given strain will not have an angle of twist of more than that specified, its diameter, which is found from the circumference by dividing by = 3.1416,

Circumference of shaft =

360x
Ꭰ

there being 360° in a circle.

(113)

But if we want to find the size of a shaft to do a certain specific duty (e.g., the horse-power we want to transmit), find the torque by Equation 107 in in.-lbs. T.

=

Then ƒ being

=

10,000 for a steel shaft and 7000 for an iron shaft, we have the diameter of our proposed shaft in inches

For instance, required a solid shaft to transmit 1000 h.p. at 100 RPM.

A 7-in. shaft would no doubt be provided.

Coming again to the angle of twist, and considering stiffness proportional to D', it is obviously measured by the smallness of the angle of twist per unit length of shaft. The simple equation for this is

and 0, much used by engineers, will give us a fraction, say, that is, of a radian, which is equal to 57.3 degrees.

Reverting to Equations 108 and 109, having these values for BM and TM in a shaft, we can find the actual shear stress and tensile stress set up in any shaft. Their values are

CHAPTER X.

ON THE DESIGN OF ENGINEERING STRUCTURES.

THE following pages apply to a great extent to structures which concern the engineer in steel (or iron), but it is not intended that the treatment shall be exhaustive. In the first place it is a wide subject, and is fully treated in some very good text-books at the present day, while, in the second place, the municipal or county engineer does not now usually devote his time to going into designs for large bridges, etc., because there are specialists who devote their whole time to the work, and large firms now usually undertake their own designs as well as erection.

It is, however, quite essential that the engineer under whose supervision the above works will probably be undertaken should have a knowledge of the principles of such designs, because competition is now so keen that firms, in order to put in low tenders, cut the sections of their designs down to the smallest possible safe limit, and sometimes even exceed it. When the engineer has been instructed by his employers or clients to get in designs and tenders for a certain work, he therefore requires to decide which is the best tender. Naturally, a board of unprofessional men often want to take the lowest tender. If, however, they are wise, they consult their engineer first, and hold him responsible for the work. If, therefore, the engineer who looks over the tenders has the requisite knowledge of the design, and selects one as being on reasonable terms, he must carefully go over the drawings, and see whether or not the proposed design is in accordance with theoretical and practical principles. If he finds they are, well and good; if not, he is in a position to deter his employers from accepting such a tender, no matter what the price quoted. Of course, in out of the way places, small structures are usually left in the engineer's hands, and he orders his raw material and supervises the work done by direct labour or otherwise, the local blacksmith being able to do the smaller class of steelwork in a more or less satisfactory

manner.

Retaining-walls, earthworks, and foundations-in fact, masonry, brick, and concrete work in general—are left to the engineer entirely, together with his roads, sewerage, and waterworks. These subjects, therefore, will receive a much deeper consideration both from a theoretical and practical standpoint.

A way of analysing the stresses and strains in structures, called graphic statics, by means of diagrams upon the principle of the parallelogram and triangle of forces, is also much used, and is of great value to the engineer. We shall treat this subject in the next chapter.

Design of Suspension Bridges.—Dealing with the design of structures, we must note that

(1) Ties are always subject to tension, and the equilibrium of a movable tie is stable.

(2) Struts being subject to compression, their equilibrium is unstable.

(3) A brace is a stay bar on which there is permanent stress, and which does not alter the resultant of the forces applied to the joints which it connects, but only the distribution of the components. In fig. 132 we have a hanging chain similar to that in a suspension bridge. The supports are assumed to be at the same level.

while m, called the focal distance of the parabola (hanging chains always describe a parabolic curve) [vide fig. 154],

The inclinations at the points of support are equal to i,

(120)

And if the weight of the chain in lbs. per foot-run P, the horizontal tension in the chain

Hyp. log means that the hyperbolic or Napierian logarithms have to be used. If a table is not handy for finding their value, multiply the common log by 2.30258.

But if we had such a case where the supports were not level, the above reasoning would be altered; say, instead of y we had y1 and y2; then m, instead of the value attached to it in Formula 119, would be