If a ball lying at the point A, Fig. 19, is propelled by a power sufficient to drive it in the direction of B, and as far as B in one minute, and at B is again propelled by a power sufficient to drive it in the direction of and as far as the point C in another minute, it will, of course, arrive at C at the end of two minutes, and by the route ABC.

If, on the other hand, both powers had been applied to the ball simultaneously, while lying at A, Fig. 20, it stands to reason that the ball would have reached C, but in one minute and by the route AC. AC (or E D), is, therefore, called the resultant of the forces A E and D A. If, now, we were to apply to the ball, while at A, simultaneously with the forces D A and A E, a third force (E D) sufficient to force the ball in the opposite direction to A C (that is, in the direction of C A), a distance equal to C A in one minute it stands to reason that the ball would remain perfectly motionless at A, as C A being the resultant (that is, the result) of the other two forces, if we oppose them with a power just equal to their own result, it stands to reason that they are completely neutralized. Now, applying this to a more practical case, if we had two sticks lying on A E and D A, Fig. 21, and holding the ball in place, and we apply to the ball a force ED=C A and in the direction C A, we can easily find how much each stick must resist or push against the ball. Draw a line e d, Fig. 22, parallel to E D, and of a length at any convenient scale equal in amount to. force E D; through e, Fig. 22, draw a e parallel to A E, and through d draw d a parallel to D A, then the triangle eda (not ead) is the strain diagram for the Fig. 21, and d a, measured by the same scale as e d, is the amount of force required for the stick D A to exert, while a e, measured by the same scale, is the amount of force required for the stick A E to exert. If in place of the force E D we had had a load, the same truths would hold good, but we should represent the load by a force acting downward in a vertical and plumb line.

Parallelogram Of Forces 10090

Fig. 19.

Parallelogram Of Forces 10091

Fig. 20.

Thus, if two sticks, B A and A C, Fig. 23, are supporting a load of ten pounds at their summit, and the inclination of each stick from a horizontal line is 45°, we proceed in the same manner. Draw c b, Fig. 24, at any scale equal to ten units, through b and c draw b a and a c at angles of 45° each, with c b, then measure the number of (scale measure) units in b a and a c, which, of course, we find to be a little over seven. Therefore, each stick must resist with a force equal to a little over seven pounds.

Now, to find the direction of the forces. In Fig. 23 we read CB,BAandAC,the corresponding parts in the strain diagram, Fig. 24, are c b, b a and a c. Now the direction of c b is downwards, therefore C B acts downwards, which is, of course, the effect of a weight.

The direction, however, of b a and a c is upwards, therefore B A and A C must be pushing upwards, or towards the weight, and therefore they are in compression. The same truths hold good no matter how many forces we have acting at any point; that is, if the point remains in equilibrium (all the forces neutralizing each other), we can construct a strain diagram which will always be a closed polygon with as many sides as there are forces, and each side equal and parallel to one of the forces, and the sides being in the same succession

to each other as the forces are. We can now proceed to dissect a simple truss. Take a roof truss with two rafters and a single tie-beam. The rafters are supposed to be loaded uniformly, and to be strong enough not to give way transversely, but to transfer safely one-half of the load on each rafter to be supported on each joint at the ends of the rafter. We consider each joint separately. Take joint No. 1, Fig. 25.

Parallelogram Of Forces 10092

Figs. 21 and 22.

Parallelogram Of Forces 10093

Figs. 23 and 24.

We have four forces, one O A (the left-hand reaction), being equal to half the load on the whole truss; next, A B, equal to half the load on the rafter B E. Then we have the force acting along B E, of which we do not as yet know amount or direction (up or down), but only know that it is parallel to B E; the same is all we know, as yet, of the force E O. Now draw, at any scale, Fig. 26, No. 1, o a =and parallel to O A, then from a draw a b = and parallel to A B (a b will, of course, lap over part of o a, but this does not affect anything). Then from b draw b e parallel to B E, and through o draw e o parallel to E O. Now, in reading off strains, begin at O A, then pass in succession to A B, B E and E O. Follow on the strain diagram Fig. 26, No. 1, the direction as read off, with the finger (that is, o a, a b, b e and e o), and we have the actual directions of the strains. Thus o a is up, therefore pushing up; ab is down, therefore pushing down; b e is downwards, therefore pushing against joint No. 1 (and we know it is compression); lastly, e o is pushing to the right, therefore pulling away from the joint No. 1, and we know it is' a tie-rod. In a similar manner we examine the joints 2 and 3, getting the strain diagrams No. 2 and No. 3 of Fig. 26. In Fig. 27, we get the same results exactly as in the above three diagrams of Fig. 26, only for simplicity they are combined into one diagram. If the single (combination) diagram, Fig. 27, should prove confusing to the student, let him make a separate diagram for each joint, if he will, as in Fig. 26. The above gives the principle of calculating the strength of trusses, graphically, and will be more fully used later on in practical examples.

Roof Trusses.

Roof Trusses

Fig. 25.

Parallelogram Of Forces 10095

Fig. 26.

Should the student desire a fuller knowledge of the subject, let him refer to "Greene's Analysis of Roof Trusses," which is simple, short, and by far the best manual on the subject.

In roof and other trusses the line of pressure or tension will always be co-incident with the central line or longitudinal axis of each piece. Each joint should, therefore, be so designed that the central lines or axes of all the pieces will go through one point. Thus, for instance, the foot of a king post should be designed as per Fig. 28.

In roof-trusses where the rafters support purlins, the rafters must not only be made strong enough to resist the compressive strain on them, but in addition to this enough material must be added to stand the transverse strain. Each part of the rafter is treated as a separate beam, supported at each joint, and the amount of reaction at each joint must be taken as the load at the joint. The same holds good of the tie-beam, when it has a ceiling or other weights suspended from it; of course these weights must all be shown by arrows on the drawing of the truss, so as to get their full allowance in the strain diagram. Strains in opposite directions, of course, counteract each other; the stress, therefore, to be exerted by the material need only be equal to the difference between the amounts of the opposing strains, and, of course, this stress will be directed against the larger strain.

Line of Pressure Central.

Line of Pressure Central

Fig. 27.

Parallelogram Of Forces 10097

Fig. 28.