For end stiffeners we use Formula (125) s = 182500 - 12000.5/8.16/1+0,0003.362/(5/8)2

= 122500 pounds. Or we should need

122500/6562 = 18,6

Or we should need some 19 rivets in the end stiffener, we therefore decide to use a filler plate 24" x 23 5/8" x 3/4" each side, which will not

Number of rivets in end stiffeners.

only help stiffen the web, but affords us the room to get in the necessary number of rivets, without cutting more than six rivet-holes on any one vertical line. Figure 208 gives a plan of arrangement of web at end. We next take the stiffener located some 4' 6" from the end. The vertical shearing here (see Figure 200) is 163000 pounds.

From Formula (126) we have therefore strain on this stiffener

Number of rivets in central stiffeners.

s = 163000 - 12000.5/8.36/1+0,0003.362/(5/8)

= 28000 pounds.

Or we should need

28000/6562 = 4,3 or say five rivets, we must, however, locate them oftener, see

Formula (107).

We next decide to splice the web at the point shown in Figure 206, and as shown in plan Figure

210. The vertical shearing at this point (see Figure 200) is 163000 pounds, we need, therefore, each side of joint

163000/6562 = 24,8 or say 25 rivets. Including those in the angles, which, of course, help splice the joint, we have 26 each side.

We next settle the size of splice plate by Formula (114). We shall have for its neat breadth b = 24" - 6.7/8 = 18 3/4" and have for thickness of splice plates h= 163000/18 3/4.12000 = 0,72

As there are two plates, one each side of the web, we shall make each one-half the above, or say 3/8" thick, remembering, however, to fill out behind the angle with an additional

1/2" thick (filler) plate.

We must next settle the number of rivets connecting the angle and the web.

Splicing the web.

Numberof rivets connecting web and angles.

Fig 209.

The vertical through moment of resistance curve (Figure 199) at the centre measures 276" and the axis x y in (Figure 202) we made previously found to be 6562 pounds, gives the total number of rivets required, or

= 120000 pounds, therefore, from Formula (93) the bending-moment at centre, or: mcentre =276. 120000

= 33120000 pounds-inch.

This divided by the depth will give the horizontal flange strain from centre to end of girder, see Formula (121), or

s = 33120000/36 = 920000 pounds.

This again divided by the least value of web rivets, which we

Fig. 210.

920000/6562 = 140

In reality we have placed 143 rivets from centre to end, so as not to place the central ones too far apart. Again take a point just under the column (or wix) say fourteen feet from the left reaction. The vertical (Figure 199) measures 225", therefore bending-moment at wIX, or mw =220.120000

IX

= 27000000 pounds-inch and horizontal flange strain s = 27000000/36 = 750000 and required number of rivets 750000/6562 = 114

In reality there are only 113 between this point and end, but that is near enough, as it spaces more evenly so. Again, take a point at the first load to the left, or wXI which is four feet from left reaction. The vertical (Figure 199) measures 75", therefore bending-moment, mw =75.120000

XI

= 9000000 pounds-inch and horizontal flange strain to end s = 9000000/36= 250000

Therefore number of rivets required, 250000 = 39 6562

In reality there are 42 rivets.

It will be noticed that in allowing for horizontal flange stress we take all the rivets to the very end of girder, this, of course, is right; although before right through for convenience we have considered the end as at the reaction. The amount the girder will run over the reaction will be determined by the crusbing strength of the wall, or pier, or column it is supported by.

In our case we have a bearing 21" x 16 = 336 square inches, and therefore load per square inch on masonry

= 182500/336 = 543 pounds per square inch. This was distributed onto the brickwork by heavy, ribbed, enlarged cast-iron plates.

The only thing remaining to be done now is to Figure the deflection.

In Figure 199 we draw the vertical lines, 1, 2, 3, 5, 6, etc., through the moment of resistance curve, the distance between them (l,) being practically 60", the first one being a half distance. In Figure 203 we carry down these lengths in succession on line m 1, 2, 3, etc., to a. We select our pole z arbitrarily at a distance zj = 1000". In Figure 201 we now construct the deflection curve. The longest vertical is in the centre of girder and = v1=243". The momenl of inertia of the section of the girder at the centre will approximate very closely to 58000 (for exact amount see Table I, section No. 14) and remembering that for built-up plate girders of wrougbt-iron we must use a modulus of elasticity equal to only 18000000 pounds-inch, we have the central deflection, in inches, of the girder (see Formula 97)

Length of bearing on reactions-

Deflection found graphically.

δ = 243.60. 1000. 120000/18000000.59000 = 1,66" The safe deflection, not to crack plastering, would be, (see Formula 28)

δ = 59.0,03 = 1,77"

Or, our girder is amply stiff. Were we to consider our load as equal to a uniform load of 357500 pounds we could use the approximate formula for deflection given in Table XLI, and should have had

δ = 592/75.42 = 1,105

We must add one-half to this for a modulus of elasticity of only 18000000 (the approximate formula being based on 27000000) and would have

δ = 1,105 + 0,552 = 1,657" or the same as by the graphical method.

Or, we might have calculated the deflection by Formula (39) again considering the load as a uniform load, and should have had