Superfluous members.

CE, HG and G B; and in strain diagram; be, downwards, therefore a vertical load; c h downwards towards joint, therefore compression ; h g to the left, towards joint, therefore compression ; and g b upwards towards the joint, therefore compression; as our Figure is a closed one, we having arrived back at the point of start b, the joint, is in equilibrium. We next examine similarly the joint at apex and at centre of horizontal tie. The joints to the right will be similar to the corresponding left-hand joints, as the truss is uniformly loaded. Figures 214 and 215 are similar to 213, the only difference being in the increased pitch of the rafter.

When we analyze Figure 216 we first take the joint A B at foot of rafter; next the joint B C immediately above; next the first joint along tie-rod; next the second joint along rafter and so on. Figures 217, 2181 and 210 are similar. Figures 220, 221 and 222 will present no difficulty; nor will Figures 226,227 and 228, but in the latter three we find that we cannot follow the above rule of passing from a joint along rafter to one on tie-rod; for we begin at joint A B (Figure 226) and find the strains B S and S 0; we next pass to joint B C and find the two unknown strains C R and R S ; were we now to pass to the joint at the tie-rod, we should find it impossible to continue, for there would be three unknown strains, namely, it N,

1 Angle of rafter (or angle Z) in Figure 218 is 26 30'.

NM and MO; but if we pass first to joint CD we find the two unknown strains D N and N R, and then going back to the joint at tie-rod, there remain only two unknown strains NM and MO which we can now find.

Figure 223, we will find, presents a difficulty at the joint C D and this truss, in effect, cannot be analyzed by the same method as the others.

We begin at joint A B and find the unknown strains B L and L O ; we pass to the joint B C and find the strains C M and M L ; we now pass to first joint along tie-rod and find the strains M N and NO; we now pass to joint CD but find three unknown strains D S, S R and R N; we try the second joint along tie-rod, but this, too, has three unknown strains N R, R U and U 0; so have all the other joints, we are, therefore, completely stopped. We reason, however, that the duties of L M and M N, are apparently the same as those of T S and S R, namely, to truss transverely the long (half) lengths of rafter. As the loads are the same on all joints we will assume that SR =M N and TS = L M.

Analysis of Figure 223.

We can now continue our work, for at joint CD there only remain two unknown strains D S and R N. Passing to the second joint along tie we can continue, for, NR being now known, there only remain the two unknown strains R U and U O.

This method of reasoning is correct, where loads are uniform and the rafter divided into four equal panels; were this not the case, we should have to find the strain on R N first, and by one of the methods shown in Figures 223a and 2236.

In the former we assume (temporarily) that the intermediate panels or joints do not exist.

We take care to reproportion our loads on the joints, as properly there are now only two panels to main rafter, where before there were four.

Fig 223 b.

We now find the strain on R N without any difficulty by drawing a separate strain diagram similar to the one shown in Figure 220 and having found the strain on R X we proceed with our original strain diagram annexed to Figure 223.

Or, we will assume (temporarily) that Figure 228 is altered to have three struts as shown in Figure 223b, taking care to keep R N the same, and the divisions of rafter the same, the loads on joints will therefore not be changed. We now find R N without difficulty by drawing a separate strain diagram on the principle of that shown in Figure 226; and having found R N proceed with our original strain diagram without difficulty.

Figures 224 and 225 are similar to Figure 223, excepting the inclination of main rafters, and angles between rafters and tie-rods.