Arched Trusses.

Fig. 235a.

Fig. 235 b.

Fig. 235 a1.

Fig 235 b1.

In separating the truss we note that in Figure 239, we have the additional tension member JI (see Figure 229) ; and in Figure 238 the members KJ and IH. are in tension instead of in compression as they were in Figure 230. These changes are due to the fact that the upper and lower chords are broken instead of straight lines, Had the truss been a Warren latticed truss, we should in separating find that we had only half the number of joints and hence would use the full load on each.

Where there is a horizontal tie or abutment to take up the horizontal thrust of an arched truss, we must find the amount of this thrust by the rules given at the end of Chapter I (Strength Of Materials) and at the beginning of Chapter V (Arches) (both in Vol. I). Having found this we draw an arrow at the foot of the truss in the proper direction to represent this tie (if a rod) or thrust (if an abutment) and then proceed to analyze the truss without difficulty.

In Figure 240 we have what is known as a "scissors" truss. It can be built in either wood, or iron, or a combination of both. For small spans it is a cheap truss where a vaulted ceiling is needed. If we attempt to analyze this truss in one strain diagram, as is done in Figure 243, we first take the lower joint A B and find the two unknown strains B G and G 0. At the next joint B C there remain three unknown strains, we therefore abandon it for the present and pass to the joint at the apex, here there are but two unknown strains DJ and JC and we find them by drawing in our strain diagram Figure 243 the lines dj and j c. parallel to them.

Scissors Truss.

Fig.237.

Having now found the point j in the strain diagram there is no difficulty with the rest.

This truss is really a combination of two trusses; we might build it by dividing it horizontally along the line JH as shown in Figure 240. We should then have a bottom truss with two inclined struts B G and E K, and one horizontal strut J H and two ties GO and K O the loads on the joints B.J and J E will each be increased by half of the apex load. Over this truss we should have another truss, with two inclined struts JC and D J and a horizontal tie JH. This truss transfers its load to the lower truss. Now the remarkable thing that we have discovered is that the member JH is at one and the same time both in compression and tension. If we analyze the top truss separately (Figure 241) and also the bottom truss (Figure 242) we find that the compression greatly exceeds the tension; the member, however, should be designed to resist both, having straps at the ends, sufficient to take up the tension. The actual stress in the member itself will, of course, be the difference between the two, and by referring to our single strain diagram (Figure 243) we see this clearly, for hj is the difference between h o the total compression and j o the total tension. Usually there is a bolt at the joint H 0 connecting the two ties. This is done, as, when the wind blows from one side only, the tie pointing to that side becomes a strut and is stiffened by being reinforced at this joint.

Fig. 238 a.

Fig. 239.

Fig.239 a.

Figure 244 represents what is known in Gothic architecture as a "hammer-beam" truss. We can analyze this truss in two ways, first (Figure 244) assuming that the wall takes up the thrust of the truss, due to the absence of any horizontal tie; or second (Figure 246) we can consider the truss as composed of two inclined trussed rafters united at their upper joints and so taking up their own thrust.

The latter method will require a very much heavier truss. In the latter case the semicircular member in the central panel becomes a tie, but the two lower quarter-circle members do not act at all, while in the first assumption the reverse is the case, the two lower quarter circle members acting as struts, while the semicircular member has no duty. In either case, however, these members come into play under wind-pressure calculated from one side only; it should be remarked here that when drawing the strain diagram all curved members must be assumed to be straight - (as shown at L 0 in Figure 244 and at Y 0 and TO in Figure 246). Of course, the stress to resist the strain thus found will be greatly increased when the curved member is called upon to do the same duty as a straight member. The increase in stress is equal to that produced on a beam of the length of the curved member with a bending-moment at its centre equal to the original strain multiplied by the (longest) versed sine of the circle, or :