197. Let us consider next a framework which is a little more complicated than the last.

Let ABCDE be the frame carrying a load at A and fixed in a wall at C and D.

First number each space so that each bar is represented by the numbers on its two sides. (It will generally be found convenient to number the spaces consecutively either clockwise or counter-clockwise.)

The point A is in equilibrium under the load W and the pulling or pushing forces in the bars 12 and 02. Set off then to scale 01 in the stress diagram vertically downwards and from its ends draw 02 and 21 parallel to the bars 02 and 12 and determine the point 2. Then 012 is the order of the Vector Triangle for equilibrium at A. The stress 12 is then from A to B, i.e. the bar 12 tends to shorten and is therefore in tension, the stress 20 is from E to A, i.e. the bar EA tends to lengthen and is in compression.

A fine line is drawn near and parallel to EA to indicate that it is in compression.

We now know the stress in one bar at B and in one bar at E. At B 3 bars, and at E 4 bars meet.

Knowing one force at a point, we can find the forces there in any two given directions which will be in equilibrium with the given force; to find such forces

in three given directions is however an indeterminate problem.

Hence at this stage we cannot find the stresses of the bars meeting at E, but we can for those meeting at B.

Starting then at 2 in the stress diagram, draw 23 and 13 parallel to the corresponding bars in the frames.

This determines the point 3. Now AB is in tension and pulls at B, i.e. for the point B, the direction in the stress diagram is 21, hence for equilibrium 213 is the order of the vectors. 13 pulls at B, therefore the bar 13 is in tension, 32 (upwards) pushes at B, the bar 32 is then in compression.

Returning to the point E, we now know the stresses in two of the bars and therefore those in the other two bars may be found. Draw 34 and 04 in the stress diagram parallel to bars 34 and 04 of the frame and we determine the point 4 of the diagram.

The bar 32 is in compression and therefore the force at E due to this bar is downwards, hence 23 is the corresponding sense in the stress diagram and the order is 23402 for the vectors. 34 slopes upwards, hence the bar 34 pulls at E and is in tension, while 40 pushes and is therefore in compression. 02 slopes downwards, hence the bar 02 pushes at A and is in compression, this agrees with the previous determination.

As in the previous example the lengths of the lines in the stress diagram give to scale the magnitudes of the stresses in the corresponding bars.

198. The Warren Girder. The figure shews a type of girder often used for bridges, named after its inventor Captain Warren. It consists of two parallel booms connected by slanting bars making (usually) an angle of 60° with one another.

We will suppose the booms freely jointed where the slanting bars join them. Let the lower joints be loaded with equal weights. Obviously the reactions at the supports must be equal and each half the total load.

To find the stresses in the bars we proceed as follows. Set the loads 12, 23, 34 to scale, bisect 1, 4 at 0, then 01, and 40 are the reactions at the supports.

Knowing now all the external forces we may proceed as before, taking as starting point A where there is one known and two unknown forces.

An instructive way of looking at the equilibrium of A is to imagine the bars meeting there to be cut through. We then have to find the forces which acting in the directions of the bars would keep any rigid body surrounding A in equilibrium.

Draw then in the stress diagram 1, 11 and 0, 11 meeting at 11. The sense of 01 is upwards hence 0, 1, 11, 0 is the order of the vector-triangle and 1, 11 is away from, while 11, 0 is towards A. The bar AC then pulls and the bar AB pushes at A and therefore AB is in compression, AC in tension.

Proceed now to the point B (not C where there are three unknowns) and draw 11, 10 and 0, 10 in the stress diagram. We know 0, 11 is in compression and pushes at B, its sense therefore is from 0 to 11. The vectortriangle is then 0, 11, 10, 0 and 11, 10 being away from B this bar is in tension while 10, 0 is in compression. Next take the point C (not D) and proceed as before, and finally D.

As the loading is symmetrical, the rest of the diagram may be drawn at once; if the loads had not been symmetrically placed it would have been necessary to have gone all along the girder determining the forces at each joint.

We may check our results as to tension or compression in any number of bars as follows. Draw through the bars any closed curve; this we may consider as a rigid body in equilibrium under the action of forces in the bars supposed cut. Thus for dotted curve in figure, suppose we know that 0, 11 is in compression, then taking the bars in order we get for the vector-polygon, 0, 11, 10, 9, 8, 7, 6, 5, 0, hence 11, 10 pulls, therefore tension, 10, 9 pushes, therefore compression, 9, 8 in tension, 8, 7 in tension, 7, 6 in compression, 6, 5 in tension, 5, 0 in compression.

199. In the girder just considered and in similar ones, the sloping bars constitute the Web of the girder and the horizontal ones the top and bottom Booms.

Practically it may be considered that the web resists the shearing force and the booms the bending moment.

200. Calculation of Stresses. The stresses in girders may also be determined by the calculation of shearing force and bending moment at various points of

the girder. For instance in the Warren girder, suppose we require the stress in the bar 76. Imagine a section xx

through the girder, then the sum of the vertical components of the stresses in 06, 76, 37 must equal the shearing force at the section, viz. the sum of the forces 40 and 34. But 06 and 37 are horizontal, hence if s is the stress in 76 and S the shearing force,

sx sin 0 = S.

To find the stress in a horizontal bar we must consider the bending moment where a vertical section cuts that bar, e.g. the line yy cuts the bar 29 at A and the top boom in B. Then since all the bars except 92 cut the sectional line at B we have, if s is the stress in 92,

8 × AB = bending moment.

The bending moment may be found either from the link-polygon or by multiplying the loads 23, 34, 40 by their known distances from A.

In complicated frames this process may become very laborious, so that except in special cases it will be found. easier to find the stresses by the stress-diagram.

201. Suspension Bridges. In these bridges the roadway is usually supported by a number of unequal vertical tie-rods which in turn are supported by the chains. These tie-rods are usually placed at equal distances and each may be supposed to bear an equal fraction of the whole weight of the roadway.

The problem presented by such chains may be looked at as follows:

Given the loads, the span and the dip of the chain, to determine the form of the chain, the stresses set up, and the reaction at the supports.

(1) In the figure AC and BD are the supporting pillars, CD the roadway, AE and BF tie-rods keeping the pillars vertical. The number of tie-rods is in this case even, so that the middle link of the chain is horizontal.

The loads in 12, 23, &c. being known the stress diagram would offer no difficulty, if the stress in 04 could be found.