triangle, the resolution of the resultant in these three lines can be uniquely determined.

explained in § 174 (3).

The process is

First draw the vectors representing the forces 12, 23, and 01, choose some convenient pole P, and complete the vector-polygon. Then draw the link-polygon a through the space 0 to PO, then b through space I || to P1, then c through space 2, and d through space 3 to cut a at Q. The resultant σ = 03 is thus determined.

The next step is to find the components of σ along the three bars. Let A be the intersection of σ and bar 3, 12. Resolve o along the bar 3, 12 and along AB, where B is the point of intersection of the other two bars. The components are a and e in figure. Now decompose e along the bars 12, 11, and 11, 0, the components are B and y. Hence a gives the stress in 3, 12, and since it presses against the rigid body, 3, 12 is in compression. Similarly, and y give the stresses in 12, 11, and 11, 0.

If the drawing has been performed carefully, a, B, and should have the same lengths as the lines 3, 12; 11, 0; and 11, 12 of the stress diagram in first method.

γ

210. 3rd Method. By the Bending Moment diagram. If we had completed the link-polygon in the last method, the line a would have been the first and last of

the closed link-polygon. If then the lines a and d be produced to cut the line of action of the force at the vertex of the frame, the intercept on this line would give us the bending moment at the vertex.

In Fig. 135, if u is the distance of P from the line of forces 13, e the intercept of the bending moment diagram, h the distance of the vertex of the frame from the bar 0, 11, and f the force in that bar, then

fxh=uxe.

The reason for this equation is obvious, the bending moment ex u is the sum of the moments of all the forces to the left of the vertex about that point, and is therefore equal to the sum of the moments of the forces in the three bars. Two of the bars pass through the vertex and have therefore no moment, and the moment of the third is fx h.

The moments may also be found by calculation without graphics.

The sum of the moments of the external forces about the vertex can be easily calculated.

and if h1, h2, h, be the horizontal distances of the vertex from the lines of action of the forces (see Fig. 135), then

211. The frames we have considered were all such that there were just enough bars and no more to enable each frame to maintain its shape under all kinds of loading. Such frames are said to be rigid. The simplest example of such a frame is a triangular one.

If the number of bars is insufficient to maintain the shape of the frame under all circumstances, it is said to be non-rigid.

Many frames are in common use in which the number of bars is in excess of that required for rigidity. In this case there are redundant bars and the frame is said to be over-rigid.

212. A quadrilateral of jointed bars is a non-rigid structure, if we apply forces at the joints it would in

Fig. 136.

general change its shape. By joining two vertices by a fifth bar, the structure becomes rigid and when we know the forces applied at the joints the stresses in the bars can be calculated.

If the remaining vertices be joined by a sixth bar then the frame is over-rigid. For such frames we cannot calculate the stresses unless we know the stress in one of them. ABCD being the frame and AC being joined by a bar, then the points B and D are at a perfectly definite distance apart; it would be however impossible to make a bar fit exactly in BD, it would either be too long (perhaps by ever so little-a 1/10,000th of an inch say), or it would be a little too short. However little short of the proper length it might be, the points B and D would have to be squeezed nearer each other in order to joint the new piece to B and D. This would set up stresses in all the bars and it is quite impossible to say how large these would be. Even supposing the extra bar could be accurately fitted in, any force applied to the frame or a change of temperature would at once destroy the balance of the lengths.

In complicated frameworks it is as well, before attempting to find the stresses, to see whether it is a rigid frame or otherwise.

There are certain considerations which enable us to determine with comparative ease whether any given frame

is rigid, over-rigid, or non-rigid; these considerations we will now explain.

213. A point free to move in a line straight or curved is said to have one degree of freedom, it can only move to and fro in that line.

Suppose now the line itself can move by translation without rotation, then this line has also one degree of freedom, for any given point on the line can move to and fro along some straight line fixed in direction.

The line and the fixed direction determine a surface and any point in that surface is fixed when the position. of the moving line and the position of the moving point in the line are known (in other words the position of a point is fixed when its coordinates are known). A point free to move in a plane has therefore two degrees of freedom.

Consider now the degrees of freedom of a rigid body constrained to move in a plane. Any particular point of the body has two degrees of freedom; if that point be fixed the body can still turn about the point as axis and any second point, since it can move only in a line, has one degree of freedom. Hence the rigid body in a plane has three degrees of freedom.

214. To return to our frame.

Let n the number of joints in the frame,

=

m = the number of bars.

For n points free to move anywhere in a plane there are 2n degrees of freedom, each point having two degrees.

If two points be joined by a bar one degree of freedom is lost, and hence every bar may reduce the degrees of freedom by one.

But if the whole frame is rigid there are 3 degrees of freedom and therefore

2n 3 = m

and the number of bars must be odd.

215. A point where 2 bars are jointed will be called a single joint.

If 3 bars are jointed at one point, they form a double joint.

If 4 bars are jointed at one point, they form a triple joint, and so on.

We see therefore that every rigid frame must have single or double joints or both. If there are no double joints, there must at least be three single joints, and if there are no single joints, there must at least be six double joints.

216. Suppose we have any frame A, B, C, ... and we want to connect some point D rigidly to it. This can be done by joining D to any two joints, say A and B, of the frame by two bars, provided that D does not lie in the straight line AB. Conversely we can take away any single joint of a frame without altering its rigidity. Thus to see if a frame be rigid or not we may at once simplify it by removing all single joints. If the remaining frame is rigid the original was rigid.