LECTURE XVII.

CONTENTS.-Action of the Crank-Tangential and Radial Forces-Diagrams of Twisting Moments with Uniform and with Variable Steam Pressure on Piston, and neglecting as well as taking Account of the Obliquity of Connecting-rod-Effect of Inertia of Moving Parts-Case of a Horizontal Engine with Connecting-rod of Infinite Length-Indicator Diagrams as modified by Inertia - Graphic Representation of the Inertia-Case of a Horizontal Engine with Connecting-rod of Finite Length-Position of Instantaneous Axis of Connecting-rod.

Action of the Crank-Tangential and Radial Forces. In most steam engines the conversion of the reciprocating motion of the piston into circular motion is effected by means of the crank and connecting-rod.

The turning or tangential force exerted by the connecting-rod on the crank varies with the position of the crank itself. Thus, when the centre line of the crank coincides with a line drawn through the centre of the cylinder and the centre of the crank shaft, the crank is said to be at the "dead points," and the connecting-rod exerts no rotational effort on it. The crank arrives in those positions twice in one revolution, just when on the point of reversing the direction of motion of the piston. These positions are OA and O B in the next diagram. Again, when the crank is at an angle of about 90° to the centre line through the cylinder and crank shaft, the tangential force is a maximum.

Diagram of Twisting Moments-Neglecting Length of ConnectingRod. The simplest case is that in which the pressure on the piston is uniform throughout the stroke, and the obliquity of the connecting-rod is neglected. Then the pressure or thrust, Q, on the connecting-rod is equal to the total pressure, P, on the piston (see next figure).

Suppose the crank to be in the position, OC1, then by the parallelogram of forces the thrust, Q, on the connecting-rod may be resolved into two components, one, C1 R, acting along the line of the crank and representing a radial pressure, R, on the crank-shaft bearing; the other, CT, acting at right angles to OC1, and representing the tangential pressure, T, acting on the crank pin. Of course, the whole thrust on crank-shaft bearing is equal to the whole pressure on piston.

Let the angle AO C1 = 0, ... ▲ Q C1 0 = 0, • . • C1 Q is | to A O.
Then the radial pressure, R, or C1 R = Q. cos 0.

=

And the tangential pressure, T, or C1 T Q. sin 0.

These components may be plotted out separately for every position of the crank by curves in the following manner:-Let OC1 represent Q, to any convenient scale, and lay off to the same scale Ot1 =T=CT, the tangential component of Q. Then t1 is a point on the curve, and Ot1 measures to scale the tangential pressure on the crank pin for the position, OC1, of the crank. To find other points on this curve, take any other position of the crank and plot off along that line of the crank the tangential component of Q for that position. If we find a number of points and join them, they

POLAR CURVES OF TANGENTIAL FORCE (T) ON CRANK-PIN, AND RADIAL THRUST (R) THROUGH CRANK, WITH UNIFORM PRESSURE ON PISTON AND NEGLECTING OBLIQUITY OF CONNECTING-ROD.

will be found to lie on the circumference of two circles described with 0 to 90° and O to 270° as diameters. Similarly, if we lay off Ori on the position, OC1, of the crank, equal to the radial component of Q, for that position of the crank, and do the same for a number of other positions, we have, by joining the points, two complete circles described with OA and OB as diameters, representing the radial thrust on the crank-shaft bearing for any position of the crank. In the position of the crank taken (0 = 45°) the radial and tangential components are equal, and, therefore, r1 coincides with t1. These circles are known as Polar" curves. For any other position, OC, of the crank, the tangential or turning force is given by O tą, whilst the radial thrust on the crank shaft is given by Org.

The TWISTING or TURNING MOMENT or TORQUE on the crank shaft at any position is equal to the tangential pressure on the crank pin in that position multiplied by the length of the crank.

Let r = radius of crank-pin circle or the length of the crank.
= the angle made by the crank with the line of dead points.
Then the twisting moment = Qr sin 0.

Since the polar curves, 0 to 90° and O to 270°, represent the tangential forces (Psin 0), their values must be multiplied by r, the length of the crank, in order to find the TWISTING MOMENT at any point; but, seeing that, r, is constant, the polar curves may be taken to represent the relative values of the twisting moments.

The twisting moments may also be represented by the following diagram, in which the horizontal line represents the path of the crank, and the height of each vertical ordinate gives the tangential force or the twisting moment for that point. To draw the diagram for one stroke of the piston, or one half revolution of the crank, lay off a horizontal line equal to the semicircumference of the crank-pin circle, and divide it into 10 equal parts. Each division then represents a movement of 180 ÷ 10 18° of the crank.

=

Connecting-rod of Fixed Length

For Connecting-rod of Infinite Length.

18°

36°

54° 72°

90°

108° 126° 144° 162° 180 DIAGRAMS OF TWISTING MOMENTS FOR ONE HALF REVOLUTION OF CRANK. Both Curves are Drawn on the Assumption of Uniform Pressure on Piston.

Then calculate by the above formula, or plot out by the previous diagram of polar curves, tangential pressures for each of the 10 positions of the crank, and lay them off vertically at each division. Join these points, and we have the above full line curve which represents the twisting moments for one half revolution, neglecting the obliquity of the connecting-rod, and when the pressure on the piston is uniform throughout. A curve for the radial thrust through crank could be plotted out in the same way.

In the early days of the steam engine, it was imagined that the use of the crank and connecting rod involved a considerable loss of the work developed on the piston. The fallacy of this idea may now be made clear.

The

pressure on the crank-pin in the direction of rotation is = P sin 0;

therefore, in order to obtain the mean tangential pressure during a half revolution of the crank, we have only to find the mean value of sin 0, and multiply it by, P, the total mean pressure on the piston.

For an approximate result take the value of sin @ at every 10 degrees of the crank's movement and divide the total by 18, the number of divisions taken, thus

Sin 10°

173

20° .342

30°

500

40°

.643

50°

766

60°

.866

Hence, if L = length of stroke = 2 r.

The work done on the crank in one revolution Total mean pressure x distance passed through, 6349 x 2 r,

Which is practically the same thing as 2P L. But, the work done on the piston during one revolution is also equal to 2 PL. Consequently the employment of the crank and connecting-rod involves no loss of power if we neglect the power absorbed by friction due to bearing surfaces, &c.

Total 11-428

NOTE.-No such combination of mechanism as the crank and connecting-rod can involve a loss of power (neglecting friction), as it would be contrary to the "principle of the conservation of energy."

Diagram of Twisting Moments-Taking Account of Length of Connecting-Rod.-The next case is that in which the obliquity of the connecting-rod is taken into account,

and the pressure on the piston is supposed uniform. In this case, the twisting moment is equal to the total pressure on the piston,

NOTE TO FIGURE.-The pressure along connecting-rod and on the crosshead guides may be found graphically for any position, thus

Let POP, the pressure on piston to any convenient scale.

Then PV Q, the direction and pressure along connecting-rod to the same scale.

And OV=G, the direction and pressure on the lower crosshead guide to same scale

To prove this, let O be the centre of the crank shaft, and OP the centre line of the engine, passing through, O, and the centre of the cylinder. Let OC be the position of the crank, and PC the length of the connecting-rod. Produce PC to cut the vertical through O in the point V, and draw OE perpendicular to P V. Then VEO POV; also PVO is common: ... VOE OPV = 4, the inclination of the connecting-rod to centre line of engine.

=

=

Now the twisting moment = Q × OE=

P

× OV cos = P × OV.

cos

Knowing this, we can readily construct the polar curves. Suppose the crank in the position, O C, produce the centre line of the connecting-rod to cut the line ÖV in V. With centre, O, and radius, O V, describe the arc, Vt, cutting OC in t. Then, t, is a point on the tangential pressures or twisting moment's curve, and the twisting moment for the position, O C, of the crank is thus P x Ot. A similar construction for all the other positions of the crank gives all the other points, and the complete curve may then be described by joining them. We see that the curve is not a circle as in the last case, but differs therefrom in a marked degree. Now plot off the twisting moment at each of the 10 different points by this method on the rectangular diagram (page 157) as before, and we get the dotted line which shows the new diagram of twisting moments. It will be noticed that this curve rises more abruptly during the first quarter of a revolution, and falls flatter during the second quarter than when the obliquity of the connecting-rod is neglected, thus indicating a greater pressure on the crank pin during the first half of the stroke; also, the maximum pressure is reached before half stroke.

We can calculate the several twisting moments in this case without the aid of a scale diagram, thus

Since OCP+00V = 2 right angles, and OOP + 0 + = 2 right angles.

tangential pressure on crank

pin = T = Q'sin (8+4)} = (COP) s

... the twisting moment = Pr

sin (0 + $)

sin (0+) = P

cos

sin (0 + $).
cos ф

Where r sin = 1 sin e; r being crank radius, and I the length of connecting-rod.

It is, however, more tedious to work out the results by this formula than by the previous graphic method.

The effect of shortening the connecting-rod is to increase the effort upon the crank pin at the beginning of the stroke, and to decrease it towards the end, thus causing greater irregularity in the tangential pressure on the crank, and greater stress on the crosshead guides. The pressure on the latter is G P tan, as seen from the last figure and the footnote below it. The actual state of things which takes place in practice is, however, not so easily represented, for the pressure on the piston is never uniform, but falls away from the point of cut-off. In order, therefore, to construct a truer diagram of the twisting moments, we must find the positions of the