To draw the link-polygon proceed as follows:

Through the space 1 draw a line CE parallel to P1 cutting the line 12 at E, through the space 2 draw EF parallel to P2 cutting the line 23 at F, and so on until we have drawn through the space 5 a line GD parallel to P5.

Now since by supposition the reactions in 01 and 50 produce equilibrium with the given forces, the vector and the link-polygons must be closed. Hence the line joining C and D must be both the first and the last sides of the link-polygon. If then we draw in the vectorpolygon PO parallel to CD, 01 must be the reaction at A, and 50 that at B.

(In this and the succeeding examples the weight of the beam has been neglected.)

181. As the determination of the reactions is an important point, we will explain the matter from a

different point of view.

The resultant R of the forces 12, 23, &c. being found in the usual way, we may ask what forces at A and B are parallel and equivalent to R. If we can find them, then the reactions at A and B will be equal and opposite to these components.

Now the position of R is determined by the intersection of CE and ED, the first and last sides of the linkpolygon. Hence R can be resolved into two along CE and ED whose vectors are 1P and P5. The force in CE can be decomposed into two at C, one along CA, one along CD, and are given by 10 and OP in vector-polygon (where PO || CD); similarly the force in ED is equivalent to 05 in BD and PO in CD, the forces along CD cancel, hence R is equivalent to 10 at A and 05 at B, the reactions

at A and B being in consequence 01 and 50. The process therefore to obtain the reactions is to join CD in the linkpolygon, i.e. close the link-polygon, and to draw PO parallel to CD in the vector-polygon.

182. Bending Moments. If we require the sum of the moments of the forces in 12, 23, 34, 45 about any point Q, we draw through Q (see § 172) a line parallel to these and find the intercepts on this line of the links of the link-polygon. In the figure to paragraph 180, the sum of these moments is given by the length of the segment 1'5', if P is at unit distance from the vector 15. Similarly the moment of the force in 12 is given by 1′2′.

Now CD is the first and the last line of the linkpolygon for all the forces including 01 and 50. Hence the sum of the moments of all the forces about Q (and about all other points) is zero. The sum of the moments of 01 and 12 about Q is given by 0'2', the sum of the moments of the forces in 23, 34, 45, and 50 about Q is given by 2'3' + 3'4′+4'5' + 5'0'2'0'. The sum of 0'2' and 2'0', i.e. the sum of the moments to the left and to the right of Q, is zero, as of course it should be.

We see from this, that if we require the sum of the moments of all the forces on one side of a given point, it is given by the vertical distance between the sides of the link polygon; if this be upwards the moment is positive, if downwards it is negative (see § 171).

A

a

Fig. 118.

B

Thus if ACDEB be the linkpolygon, the sum of the moments of all the forces on one side of Q1 is given by the intercept a, for the forces on one side of Q2 it is given by b.

A beam loaded as in § 180 will not remain perfectly straight, it will bend, the lower side becoming slightly longer and the upper slightly shorter. This change of shape calls into play forces in the fibres of the beam, the

upper fibres resist being shortened, the lower ones resist being elongated. When the change of shape has taken place, the beam may be treated as a rigid body.

Now suppose the beam is cut across, what forces must be applied at the section to keep the part of the beam to the left in equilibrium? Clearly they must be equivalent to the reversed forces in the cut fibres at that section and they must also be equivalent to the loads and reaction to the right. By the theorem of § 175 the external forces may be replaced by a resultant force acting at some point of the section, together with a resultant couple, whose momental area is measured by the sum of the moments of the external forces about the same point.

The couple tends to bend the beam and since it is measured by the moments of the forces on one side of the section, it is called the Bending Moment.

The Resultant Force, since it is the force which tends to shear the beam at the section, is called the Shearing Force.

The beam being in equilibrium, the sum of the forces to the right of any section must be equal to minus the sum of the forces to the left.

Similarly

the left.

moments on the right

Σ moments on

The Bending Moment is defined as positive when at any section the part to the right tends to rotate counter-clockwise and the part to the left clockwise.

The Shearing Force is defined as positive when the part to the right tends to move upwards and the part to the left downwards.

As both shearing force and bending moment are of great importance in Engineering design, a diagram of shearing force should be drawn in most cases of loaded beam problems.

The diagram simply consists of vertical distances set up from a datum line to represent to scale the sum of the loads on one side of the section for every point. The

shearing force changes suddenly in magnitude when the

section passes through a point where a load is applied.

In the figure given the bending moment diagram is first drawn, this determines the point in the vectorP polygon.

From a datum line DD, drawn across the spaces of the link-polygon, are set up the sum of the forces to the right of any section, the forces being taken from the vector-polygon.

The shaded area is the Shearing Force diagram.

183. Travelling Load. It is sometimes required to find the bending moment at some particular point of a beam due to a load whose position changes.

The following simple but important theorem enables us to solve this problem :

The Bending Moment at any point of a beam, due to a single load at any other point, is equal to the Bending Moment at the second point due to the same load at the first point.

Let AB be the beam, D the position of the load, C the point where the bending is required to be known, then if W is the load and R the reaction at A, we have

and the Bending Moment at C is R× AC.

Hence the Bending Moment = W

AC × DB

AB

which since AC and DB are interchangeable is the Bending Moment at D due to the load W at C.

Hence to find the Bending Moment at C due to a load travelling from A to B, draw the Bending Moment diagram for the load at C; then the intercept on a line through E by the Bending Moment diagram will give the Bending Moment at C when the load is at E.

184. Bending Moments when the forces are not parallel. If a number of non-parallel forces act on a beam, then the components of the forces perpendicular to the beam alone contribute to the Bending Moment, the components along the beam simply tending to shorten or lengthen the beam.

Unless the horizontal components of the forces are in equilibrium among themselves, a horizontal beam will require some other method of fixing than that of being freely supported at the ends. A bolt or some such contrivance must be used at one end to prevent the beam bodily moving away.

Let AB be the beam supposed bolted at A and freely supported at B, and let the forces act on it as shewn. Suppose it is required to find the bending moment and shearing force diagrams, and the reactions at A and B.

First draw the vector-polygon 1, 2, 3, 4, 5, then σ = 15 is the resultant vector.

Project these vectors on the vertical and horizontal lines yy and xx. Then 1, 2, 3', 4', 5' is the vector-polygon