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Diagonals. We write the table as follows:-

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not put in, because for the sign of S we look not to F but to the figure.

The case can also be treated directly from the general formulæ, and the student should in this way verify the results we have obtained by superposition.

The N Girder.-The general treatment is identical with that of the Warren type, but there are one or two points which require examination. The figure shows a

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girder in six divisions, the sloping bars leaning away from the centre, as is always the case, and inclined, as is usual, at 45°. Then for the diagonals,

SF sec 45°F√2,

and for the uprights

SF seco° = F.

Let now the girder be loaded at, say, joint 4 of the lower boom. Then

P=W, Q}W.

Now the question is, What is S45? 45 being directly over W, F there changes from P to Q; but 45 must have a definite value, and so we have to inquire what value of F is there to be taken.

We shall

principles.

now have to examine 45 from first Its duty is to keep up the end 4 of the body

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So it appears that F is to be taken as Q, and it is

tension.

If, however, W had been at the top joint 5, then we should have simply

S45= P,

from the equilibrium of 0134, as before (Fig. 239), and S would be compression.

This special examination must be applied to all vertical bars, which are directly under or over loaded joints.

There is another bar which requires special consideration, viz. 66'.

If we remove 66', then without there be a load at 6', no effect at all is produced, for in theory the bar 57 can withstand either pull or thrust, even though it be jointed at 6'. Practically, however, we know that a jointed bar cannot take compression, because some slight accidental circumstance bends the joint a little, and then the strength is gone. This, then, is the reason for fitting 66', that is, to keep 57 straight, and so enable it to take thrust; and then 57 being straight, there can, from the equilibrium of the joint at 6', be no stress in 66' (see page 439). In any case we cannot calculate what stress is on 66', because it depends on accidental causes; but we know it will, in the absence of a load at 6', be small. If there be a load at 6' then 66′ must wholly support it, because the jointed bar 57 has no sideways strength (see page 439).

EXAMPLES.

1. A Warren girder 9 ft. long on top projects from a wall, the top boom being in three divisions. A load of two tons is placed at the end. Find the stresses in all the bars.

Ans. Number as in Fig. 223, then the stresses are, in tons

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2. Obtain the above results when either of the other joints of the top boom is loaded, and hence deduce the result when a distributed load of tons per foot run is carried.

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The load at I will be only 1 ton, therefore halve the first results and add the last two for the distributed load.

3. A Linville girder in 8 divisions carries a load of 25 tons at the joint next on the left to the centre of the lower boom. Find the stresses in all the bars.

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4. What would be the effect on the results of the preceding if the load were placed at the top joint?

Ans. The stress in 6.7 would be 15 tons tension.

5. A bridge 100 feet span is supported by a pair of Warren girders under the platform, which rests on the joints of the upper boom, which is in 12 divisions, and also, by struts, on the joints of the lower boom. The bridge is loaded with 1 tons per ft. Find the stresses on the bars. Ans. 23 joints are loaded with 2 end joint with 12 tons. forces are, therefore, 30 tons almost exactly; and apply the tabular method.

run.

NOTE.

tons at each, and each The virtual supporting

The truss described in this chapter as an N girder is also known as a Linville, a Pratt, or a Whipple-Murphy girder or truss, after various designers, who have made small changes in details of construction.

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