and it pulls 5 down, so it is in tension. [This we should have expected, since at 4 it helps to support W, but it does not from that necessarily follow that it is in tension, and we must in all cases make sure by actual calculation.] We have now found all the stresses due to a load at one joint, and by the same method we can obtain them when any other joint or number of joints are loaded. Bridge Girder Loads-Permanent and Travelling.—We have shown that the reason why it was necessary to have two diagonals in one bay was to enable the bay to withstand either + or shear. This process of doubling the diagonal is called Counterbracing, and we are now about to inquire how it is that the shear on some bays is sometimes+, and at other times –, and so renders counterbracing necessary. For this purpose we must consider what loads a bridge has to bear. The first, and in long spans most important, load, is the weight of the bridge itself, and of the platforms, rails, etc., which it carries. This load is always on the bridge, and is hence called the Permanent or Dead Load. Secondly, there are the loads due to the passage of bodies across the bridge. These loads, being continually coming on and going off again, are called Live Loads or Travelling Loads. The permanent load will be in all cases a distributed load, and for parallel girders may be taken as uniformly distributed. It is then equivalent to a set of equal loads applied at the joints of whichever boom the platform rests on. We can then in a girder so place the diagonals that under the permanent load they shall be all in tension. Fig. 247 shows the proper arrangement; each joint is loaded with W, say, and there is one diagonal in each bay, which we can easily see is in tension, for— 66' There is symmetry, and S60 (page 326); therefore 56 and 67 are in tension, since they support W. Then it follows that 45 and 78 are in compression, since they hold up the ends of 56 and 67 respectively; and hence 43 and 89 are in tension, and so on. [We see here that although the diagonals are in tension, yet the uprights are in compression, and it may be asked, What is the gain? It is this: that the uprights are shorter than the diagonals-in some cases considerably so-and are hence better suited to withstand compression, since they bend less. general principle we try to keep short bars in compression and long bars in tension.] As a Effect of Travelling Load.-We now consider the effect produced when a load, such as a locomotive, crosses the bridge. Fig. 248. w' Let the weight of the locomotive be W', and we will treat it as a concentrated load, examining its effect as it reaches each joint in order. Suppose it to come on from the left. Then (Fig. 247) we have, before it comes on, S56=S34W' sec 0 compression, because they slope the wrong way for tension (see Fig. 237, page 325). Now 34 had originally 3 W sec tension, and hence if {W'>W, the total effect when W' is at 2 is that 34 is in compression, and S34(WW) compression. Therefore 56 is also in compression, for S56=(W' - W) sec compression, and if & W' – 3 W is plus, much more will W' - W be plus. But 56 will be in compression if W' be than W, so long as it is greater than not greater W. If then and we do not desire the braces to take compression, we must counterbrace both bays, 2354 and 456'6. however, If, we need, so far as we can see at present, only counterbrace 456'6. The effect of 'W' at 2 on the bars 1.2, 6.7, 8.9, and IO. II is to increase the tension in them; but this does not, for our present purpose, concern us, because we counterbrace not to decrease tension, but to prevent compression. Suppose finally that then the stresses in 23 and 56 are reduced to but they remain tensions, and thus so far no counterbracing at all would be required. S12= S343 W' sec tension, And we have only 56 with compression due to W', its } W'>{W, there would be compression in 56, and we should counterbrace. All the other bars are in tension, so only this one bay needs counterbracing. } W'< W, no counterbracing at all is required. Next, let W' come to 6. If Then the only effect is to increase the tension in all the diagonals, hence no counterbracing is required for this position of the travelling load. We have now traced the load half-way across, and we will summarise our results as follows :— When W' is at 2 it produces compression in 34, but not when at 4 or 6. When W' is at 2 or 4 it produces compression in 56, but its effect is greatest when it is at 4 directly under the end of 56, being then W' sec (0 against W' sec ◊ when at 2. The question whether 34 requires a counterbrace is then settled when W' is at 2, while for 56 it may be settled at 2 but must be settled at 4. If W> W, or W'>9 W, doth 34 and 56 require counterbraces; while if } W'>W, or W'> W but <9 W, only 56 needs it. We need not now go through the work while W' travels from 6 to 12, because of the symmetry. We shall have the counterbracing of 67 settled when W' is at 8, and it will be necessary if W'>3 W; while for 89 it is settled when W' is at 10, and is necessary only if W'> 9 W. In no circumstances can 12 and 10.11 require counterbraces (though practically they would be fitted in most cases), while if 56 and 67 do not require counterbracing, much less will 34 and 89. We can now then give a general method of proceeding for any number of bays based on what we have just seen, as follows: Commence at the centre bays, because if they do not need counterbraces, the outer ones will not do so. To decide whether the first bay from the centre needs the extra brace, consider the stress when the travelling load is at the outside extremity of that bay (from the centre outward), because it then produces its greatest compressive effect on that diagonal. If the first bay require counterbracing, go on to the second, taking the load at its extremity; and so on till a bay is found |