would be in equilibrium, but not those represented by say

GE, KE, GK,

since here we do not go round in one direction only.

Now, to apply this, we note that the TAC which acts on the joint A is represented by EG-not by GE, GE representing the TAC which acts on C-we must go round then in the first way above, i.e.

EG, GK, KE,

which shows that TAB or KE is an upward force on the joint A, which we are considering, so that it is a thrust. Also TAD or GK acts to the left and is a pull, but this we could see at first, whereas the direction of TAB is not so easily seen, depending on the slopes of AC and AD, and it may in some instances be a pull.

We will now proceed to the joint at D.

TAD, W', TBD.

Forces

And TAD on D is represented by KG, so we draw GL parallel to BD, and KL parallel to W'.

Then

We have now found all the stresses, and considered the equilibrium of all the joints except B. If we examine our diagram, however, we shall find that a triangle showing the equilibrium of B already appears in it.

For the forces acting at B are TBD, TAB, ТCB, and the upward thrust Tp, say of the pivot at B, on which the crane turns.

Of these TAB and T, are in one line, so their resultant is Tp-TAB upward, and taking this as one force we now have only three forces.

But we have already a triangle, viz. LFG, with sides parallel to these three, and of which two sides, LG, GF, have already been seen to represent TBD and TBC; hence the third side FL must represent Tp - TAB

From this we have

T-TAB=FL,

.. T=TAB+ FL,
=KE+FL,

=KL+FE,
=W' + W,

and this must evidently be correct, since the pivot supports the whole weight resting on the crane, i.e. W + W'.

In practice, a large part, or the whole, of the counterbalance is provided by the weight of the boiler and engine used for driving the lifting winch.

Fixed Cranes.—In the crane just described, the jib, stay, and backstay are always in one plane, since the platform turns carrying the whole with it; there is then no tendency to turn the crane over sideways when the load hangs quietly; and although when the crane is turning there is a tendency to pull it over sideways, it is but small, and is provided against by the stiffness of the pivot and the resistance of its bearings.

YW

If, however, the crane post be simply pivoted in a bearing fixed to the ground, then if a single backstay only were provided, this could only balance the overturning effect of the weight when it was in the same plane as the jib and stay, as shown in plan by Fig. 336 (b), and then, when the jib was turned to a new position, (a) as (c), there would be an overturning moment Wx, which the (6) backstay could not prevent from overturning the crane. (c)

To overcome this difficulty it is usual to have two backstays, as Fig. 337, AE, AD being the

Fig. 336.

stays, always equal, and fastened to the ground at D

and E. The combined action of the two then renders overturning impossible in any direction.

along AF. There will then be no force causing overturning at right angles to FAC.

We may express this by saying that the two backstays must be equivalent to a single backstay, which lies along AF in plan.

But now the resultant of TAD TAE must be in the plane of AD and AE, so that it must be the line joining the top A of the crane post to the point F on the ground, this being the only line which is at once in the plane of jib and stay, and also in the plane of the two backstays. We say then that AD and AE are equivalent to the single backstay AF.

If then we replace AD and AE by AF, and then find the stress R in such a backstay, this stress R represents

the resultant of TAD, TAE, and we can then find its resolved parts, which will give us the required results.

We proceed then as follows: Draw a correct plan

take the length AF and set off BF equal to it at right angles to AB, i.e. along the ground.

Finally, join AF.

Fig. 339 is a correct side elevation of the crane and equivalent backstay.

Draw the diagram of forces (a), giving the value

of R.

TAE

plane (Fig. 340).

N

Fig. 340.

For this con

struction draw in

(a) AB the crane post, and make BD equal to AD in Fig. 338. Then joining AD we have the correct length

of each backstay. We take then DE from Fig. 338 in (b), and make DA, EA each equal to AD in (a). Also mark from Fig. 338 the position of F and join AF.

Then Fig. 340 (b) is a correct view of the backstays and the equivalent single backstay in their own plane. We now then draw in (c) MN equal to R from Fig. 339, and drawing MK, NK parallel to AD, AE respectively, we have finally

TAD=MK, TAE= KN,

which are the required results.

The point F (Fig. 338) may come anywhere in the line DE, and as AC revolves it will fall outside the backstays. In this case it can be easily seen that one of the stresses TAD, TAE must be compressive, and hence the stays must not be chains, but solid bars or spars. This is actually the case in practice. Sheer Legs. In this case (Fig. 341) there is no post, but the stay is prolonged

E

Fig. 341.

с

to meet the ground, thus havwing stay and backstay in one.

A nut is formed at the end A, which works on a screw, the rotation of the Screw running A in or out and so raising or lowering C.

Overturning sideways is now prevented by replacing

a single jib as CB by two legs, DC, EC, pivoted at D

and E respectively.

To find the stresses we proceed in an exactly similar manner to the preceding. The actual legs are imagined replaced by the single leg BC, which is now always midway between them, since the vertical plane through AC and W bisects DCE.

We then find the stress for AC and BC by drawing the triangle of forces for C: And, finally, resolve TBC