CHAPTER XVI

OPENWORK BEAMS-WARREN AND N GIRDERS

THE beams which carry the platforms of bridges, etc., may be solid, or, more generally, built up of bars and plates connected together by pins or rivets. They are then usually called Girders, or, when consisting of a network of bars, Trusses; and since the manner in which the B. M. and S. F. are resisted, and the nature of these actions, is shown more clearly in open work than in solid beams, we will take them first in order. Fig. 220 shows such a girder in skeleton.

There

are two horizontal members, AB and CD, called the Booms or Flanges. These booms are connected by cross bars AC, CE, etc., called the Diagonals, or collectively constituting the Web.

In the figure the top and bottom booms are parallel, and hence the girder is called a Parallel Girder; in many cases either CD or AB may not be straight and parallel, but these cases we shall not discuss at present.

In the figure also the diagonals do not cross each other, or we say the web is a simple triangulation; in

many cases this is not so (see next chap.), but in the present chapter we shall deal only with such a simple triangulation. There are two main forms which the simple parallel girder takes, viz. the form of Fig. 220, the diagonals being inclined at 60° to the horizontal, known as the Warren Girder; or the form shown in Fig. 221, with diagonals alternately upright and sloping at 45°, known as the Linville or N Girder. In this case the uprights are not usually called diagonals but verticals; diagonals," however, may be used when we wish to denote generally any bar of the web.

Pin Joints and Riveted Joints.-There are two distinct ways in which a girder may be constructed :

A

и

B

Fig. 221.

First, the booms may consist (Fig. 220) of separate bars—CF, FH, etc., and AE, EG, etc.—jointed by pin joints to each other and to the diagonals, no bar being continuous through a joint. Or

Second, the booms may be continuous bars, or the joints all riveted so as to make the structure practically one continuous whole.

For reasons explained fully in chap. xxii. (page 437) we shall confine our attention to the first case. Our girder then is supposed made up of separate bars united by pin joints; the joints being supposed frictionless, so that they can offer no resistance to turning.

Loads at the Joints.-In practically all cases the loaded platform carried by the girder or girders rests not on the bars of the girder but on cross beams, which cross beams rest on the boom or booms at the joints.

Thus Fig. 222 (a) represents the plan of a bridge,

this is carried by a pair of girders shown in plan by AB, A'B', one under each side of the bridge. Then 11', 22,'

as distinguished from the platform or the bridge as a whole, is loaded, not with a distributed load, but with loads concentrated at the joints. The loaded joints may be those of the top boom, as Fig. 222, or of the bottom boom, or of both; also the girder may be supported at the two ends, or as in the case of a balcony, fixed at one end only. We will proceed to examine some of these cases.

Warren Girder fixed to a Wall. Taking a general case, let it be loaded as shown at the joints 1,

10 9

8

YW 3

Fig. 223.

2, and 7, with the loads W1, W2, and W. The joints are denoted by numbers, which is found more convenient than using letters.

We want now to find the stresses in the bars, and we must begin by finding for any given bar what duty it performs. The simplest way of doing this is to imagine the bar removed, and see what effect would be produced.

Action of Booms. -Consider then the bar 34.

3

Imagine it removed, then evidently (Fig. 224) the part to the right swings downward round the joint 8, i.e. the joint opposite to 34. It appears then that the duty of 34 is to prevent rotation round the joint 8. This it effects by preventing the separation of 3 and 4, which is

10 9

Fig. 224.

necessary if the rotation take place. Since, when 34 is removed, 3 and 4 would separate, or the distance 34 would increase, it follows that when 34 is in place there is a tendency to elongate 34. Now we know from chap. xiii. that such a tendency will be resisted by the internal stresses on the cross-sections of the bar, or that 34 tries to regain its original length; and in so doing it exerts pulls on the pins of the joints at 3 and 4, these pulls being equal and opposite, and each representing the total stress on any section (page 262). Let us denote this total stress in 34 by H4, then 34 pulls at each of the points 3 and 4 with a force H34, and this is represented in Fig. 223 by placing the arrows on the bar, these arrows pointing inward, and thus representing the force exerted by the bar on the joints. This is a point of the first importance, and the arrows should always be put in this manner; if we wished for any reason to denote the forces acting on the bar, they would be put thus

3

It is advisable to put the arrow near

the joint on which the force it represents acts; and always to put in the pair of arrows, not one only; then the action of each bar is plain, e.g. 34 in Fig. 223 evidently pulls 3 to the left and 4 to the right.

Next, consider the

effect of removing a bar of the bottom boom, say 89. Then the right-hand part swings down round 4, the joint opposite to 89.

It appears then that the function of a bar in either boom is to resist rotation round the opposite joint.

10 9

8

7

Fig. 225.

There is, however, a difference in the nature of the stresses in the two cases, for whereas 34 elongated, we now find that 8 and 9 tend to approach and are thrust apart by the forces in 89; hence the arrows

point outward, and H ̧ is the resistance of 89 to compression. We see then that in this instance the bars of the top boom are extended, and of the bottom boom compressed.

34

349

Next, we want to find the magnitudes of H, Hg9, etc. Take now H34, then to find a connection between it and the loading, we must consider the equilibrium of some body on which H4 acts. We have (Fig. 223) a choice of two bodies, one being the piece 8, 3, 2, 1, 6, 7, which H34 pulls to the left, and the other the body 5, 4, 8, 9, 10, which it pulls to the right. We select the first because we know

Moment of H34 about 8=moments of W1, W2, and W3 about 8.

But the quantity on the right-hand side is what we have defined as the B. M. at 8, and, moreover, this