The moment of inertia about this axis must now be computed. The relation used is that the moment of inertia about any axis is equal to the moment of inertia about some other axis, plus the product of the square of the distance between the two axes by the area of the section whose moment of inertia is desired. The moments of inertia of the various parts of the section (see "Steel Construction," Part IV, pp. 292 and 293) are computed and are as follows: The radius of gyration is equal to the square root of the quotient obtained by dividing the moment of inertia by the area. It is Using this value of the radius of gyration in the formula for the pressive stress, there is obtained 13 800 pounds as the unit allowable stress in compression, and this requires an area of 449 500 = 32.5 13 800 square inches. Since this is considerably less than the actual area of the section, the section will not be changed but will be taken as first assumed.
In order that the section should be safe about both axes, the moment of inertia about the axis perpendicular to the cover-plate should be equal to or greater than that as above computed. By com puting the moment of inertia about the axis perpendicular to the cover-plate, it is found to be 3 256.3, which gives a radius of gyration of 8.11; and since both of these are greater than those first computed, it is seen that the section is safer about the axis perpendicular to the cover-plate than it is about an axis perpendicular to the web plates.
There are small stresses in this member due to its own weight and to the fact that the pins are not placed directly upon the neutral axis (see "Strength of Materials," p. 82). These stresses are seldom more than 1 000 pounds per square inch in the extreme fibre; and since the section has such an excess of area, they will not be computed, as it is evident that there is sufficient strength in the member to with stand them.
The section just designed is that for the top chord having the greatest stress; and since this is the minimum section allowed by the Specifications, it must be used in all the sections of the top chord.
The section as finally designed is: • A pin 6; inches in diameter will be used at the point The stress in the is 378 200 pounds, and the bearing area 00 required is 24 000 = 15.75 square inches, or 7.875 for each side.
This makes a total required thickness of 7.875 = 1.265 inches for one 6.25 side. Since the thickness of the web plate is ./ inch, it will be necessary to provide pin-plates whose total thickness must be 1.265 — 0.5 =
0.765 inch. Two *-inch plates will give a thickness of 0.75 inch; and since this is less than the required thickness by an amount not over 21 per cent, they may be used. The total thickness of the bearing area is now 1.265 inches. The stress transferred to the two *-inch plates is: s = X X 189 100 = 113 500 pounds. 1.25 The rivets required to keep the outer plate from shearing off the other are 113 2X7220 — 8 shop rivets, and the rivets required to keep both of the i-inch plates from shearing off the web of the chord section are 113500 — 16 shop rivets in single shear. The bearing of a j-inch 7 220 shop rivet on a 1-inch plate is 10 500 pounds, and therefore the num ber of rivets required to keep these pin-plates from tearing the rivets out of the 5-inch web plates is 500 = 11 shop rivets in bearing. Fig. 1S5 shows the detail of this end of the top chord section. The pin plates should extend well back on the member, and at least one pin plate should go over the angle, and enough rivets, as computed above, should go through the angles and this pin-plate. Experiments on full sized bridge members go to show that unless the pin-plates cover the angles and extend well down on the member, the member will fail before the unit-stress reaches that value computed by the formula for compression.
Since the ends of the chord are milled at the splices, and therefore butt up against each other and allow the stress to be transmitted directly, only sufficient rivets need be placed in the splice to keep the top chord sections in line (55).
At the point it is not necessary to put in a pin-plate to take the stress in the upper chord; but it is only necessary to provide a pin-plate to take up the difference in stress between the two chord sections. This difference in stress is equal to the horizontal com ponent of the maximum stress in the member This is 110 000 pounds, and the area required on each side for bearing is 2.3 square inches; and as a 5-inch pin is used here, the thickness of the bearing area is = 0.46 inch. As this thickness is less than the thick ness of the web plate, no pin-plates will be required.
At the point a bearing area will be required to withstand the horizontal component of the member This is 56 300, and the inch pin is used here also. As this thickness is less than the thickness of the web plate, no pin-plate will be required.