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under the rails. When any pairs are loaded we have certain stresses in the others. We want to find the value of such stresses. Using the lettering in the figure, the first thing is to find a coefficient

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The load on No. 4 will often prove negative, which shows that the member

is pulled upwards by the load. Fig. 143 shows what we call a trapezoidal

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truss.

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We want to find the BM in the horizontal members. It will be greatest when the load occurs at 4 or 1. Its value is

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Again, we have stresses in the diagonals. These are acting as struts, the verticals acting as ties, while the top member is the straining piece. The value of the stresses on the diagonals

C

=SD =

= w

2cK

(139)

Fig. 144 shows a diagram of a lattice girder. If the load were dead only, the diagonal members shown by thin lines would not be required. The vertical members are tension rods. First of all we want to find the bending moment in such a girder at any point (vide fig. 145).

=

Then having N number of divisions in the bottom chord, and n = the particular joint at which we require the BM denoted by its number, we have

M =

n(N-n) 110.

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So, having our line diagram, we have the following formula to give the stresses in each of the members :

:

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w= load per foot-run x distance between each joint

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(dead). w1=greatest live load on each joint=2 tons per foot-run on an average.

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Thus we can find the actual load in each member, and thereby obtain suitable scantlings, allowing good steel a safe load of 6.5 tons per square inch. We will take a practical example to illustrate our reasoning on lattice. girders.

Say the span was 60 feet divided into 8 divisions, as shown in fig. 145. The depth of proposed girder is 10 feet.

140.

1. The first thing to do is to find the length of the diagonals by Equation

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2. The Booms.—We are supposing our bridge is for an ordinary roadway, and are taking the load per foot-run as 2 tons 2 x 60 120 tons: =

=

=

120
8

=

15 tons dead load on each joint = w. We are also providing for a live load of 5 tons in each joint = 21°

Then by Equation 141 we find the stress on the boom at the centre

=

(15+5) x 60 4(8-4) 20 × 60

10

=

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x 1 = 120 tons.

In the same way we can find the others, and have our values so :-stress on the booms=

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Assuming our metal to stand 5 tons per square inch in tension or compression, we have for the main booms, say an I beam, 16′′ × 6′′ × 62 lbs., having a total sectional area of 18-23 sq. in. Then to take the extra load between 3 and 5 we add a 6′′ × 1⁄2" plate, and between 2 and 6 another 6" x 1" plate likewise, to provide for stresses of 120 and 114 tons.

Coming now to the diagonal bracing, we use Equation 144.
We have for the two diagonals at 3 and 5 a load of

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In a similar way we get the other calculation, and have

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Say at 4 we use two flat bars 31" x 1", kept apart by suitable distance pieces.

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Vertical Rods.-By Equation 142, at 3 and 5 we have

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22.5+ - 15 × 1:5+5x

=

2x8

=

75

=22·5+9.37 32.0 approximately.

.. in the vertical rods the stresses on the members are

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a negative result. And when Formula 145 gives us a negative result, we know that our live load is not large enough to warrant us putting on the thin line diagonals, assuming wind pressure at 40 lbs. per sq. ft. applied to onehalf the surface of the girder=60 × 40 × 512,000 lbs. ÷ 7 = 1700 lbs. at each joint, which is much less than requires a square inch of section, but we will use 4′ x 13′′ x 5′′ lbs. angles at each joint, placed between the main booms on top. They may be curved to give extra headway, for the flooring we shall use 60 x 25=1500 sq. ft. of troughing. We will have ample strength in the size known as "D. min." if used over the span of 26 feet width between booms. Its weight per sq. ft. is 28.75 lb... exclusive of the concrete it would be 1500 × 28.78 lbs. Then summing up the quantities obtained we have

=

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..total weight of bridge (steelwork only)

= 2 × 12,204 +44,850

= 24,408 +44,850 69,258 lbs.

=

÷2240-31 tons (nearly).

We shall give another illustrated example in Chapter XXXV.

Arched Ribs.-Instead of constructing our bridge on the lattice girder principle we can make it an arched rib, when the strains and general construction will of course be different.

In this case also, however, the first thing to do is to find the greatest moment of flexure of our proposed rib. Its value is when r=span

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It is the maximum bending moment, and it occurs at a point 30° above the springing, the angle being measured at the centre. This moment has the effect of tending to make the curvature sharper at the point where it acts.

W is the load per lineal foot of span live + dead when r is in feet also. Using the same notation, we have the horizontal thrust in the arch

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the greatest thrust. We now want to find the size of our proposed ribs, which we will assume to be of steel. We take any reasonable depth=d and allow a stress on the steel of 6 tons per sq. in., but if cast iron 2-3 is ample=ƒ= tons.

Then having d, we have

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Then bxd gives us the number of square inches of section that we require, which we can dispose of as we like, such as I beams, built-up girders (web or hollow type), cast-iron ribs, or cast-iron tubes. In Equations 146 and 147, w = of course dead + live load. Say we want to design a steel arch of 60 ft. radius, 7 ft. rise, and dead load 2 tons per foot-run and a live load ofton likewise, using Equation 146,

==

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using Equation 147. Rise of arch say 7 ft. = 160 = H. Assume d=10. Then

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=

Say width between parapets was 20 feet, we may use 4 ribs of 10′′ x 5′′ × 35 lb. I beam, each beam having a sectional area of 10.29 sq. in. × 4 41-16 sq. in., which is just sufficient. The arrangement is shown diagrammatically in fig. 153.

Or we may decide to build our bridge of cast iron, a practice which found

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