This section is from the book "Safe Building", by Louis De Coppet Berg. Also available from Amazon: Code Check: An Illustrated Guide to Building a Safe House.

This, however, is probably an extreme case. The writer would recommend that the following be used, where no experiments can be made: For wrought-iron plate girders e= 18000000 pounds-inch.

For mild-steel plate girders e = 20000000 pounds-inch.

Where e = the modulus of elasticity, in pounds-inch, to be used in calculating the deflection of plate girders.

Before giving an example, Tables XLI, XLII and XLIII should be explained. They have been calculated to enable architects to lay out the required size of plate girders by their use, and without elaborate calculations. They will be found to be very accurate and valuable for preliminary estimates, quick designing of girders, and checking of final calculations.

Deflection of plate girders.

Decreased modulus of elasticity.

Table XLI gives the value of the web in resisting the bending-moment. It should be remarked here, that some engineers do not include the web at all; others include only one-sixth of the web at top and bottom. This is practically reducing the web to the same level as if the top and bottom flanges were merely latticed together. The writer believes, that in house-work at least, it can and should be safely included, particularly as it does not greatly affect the final result anyhow. In box girders the two webs should be considered as one web of thickness equal to the sum of the two; except when calculating for buckling, when, of course, each web is taken separately.

Table XLII gives the value of the four angle-irons, for six different sizes of angles, and Table XLIII the value of each inch of effective width of flange. In all of the Tables the horizontal column of Figures at the top indicates the length of span of girder, in feet; the vertical columns of Figures to the left indicate the respective values in tons (of 2000 pounds each) ; while the Figures on the curves indicate the depth of web of the plate girder.

The tables are calculated for a safe modulus of rupture (k/f ) or extreme fibre strain of 12000 pounds per square inch, and intended, of course, for wrought-iron.

For those desiring to use a smaller or greater strain it will only be necessary to increase or reduce the actual load (respectively) in the calculation. Thus, if it is desired to use a fibre strain of 15000 pounds, this will be one-quarter more than allowed for in Tables. We therefore use but four-fifths of our load in the calculation and find by the Tables, what sized girder will safely carry four-fifths of our load with an extreme fibre strain of 12000 pounds. When we then add one-fifth of the original load (or a quarter additional to the calculated load) it will, of course, add also one-quarter or 3000 to the extreme fibre strain. Or, we wish to use an extreme fibre strain, of only 10000 pounds. We add one-fifth to our load making it 6/5 (or one and one-fifth) and find from Tables the size of plate girder to carry this increased load at 12000 pounds fibre strain; when now we deduct one-fifth of the original load (or one-sixth of the calculated load), we will, of course, at the same time diminish our extreme fibre strain one-sixth to 10000 pounds.

Use of Tables XLI, XLII and XLIII.

For different fibre strains.

The use of the Tables is very simple and easy. For loads other than uniform, and for steel, the data at bottom of Table XLI shows their respective values as compared to those given in Tables.

It should be noted that the "greatest deflection" has been calculated for the most perfect work. For ordinary work this deflection will be increased, according to the quality of the workmanship, to one-half more than for perfect work.

In using the Tables, first settle the size of web, then of the angles, and finally the size of flange plates.

Example I. We will suppose that we have a wrought-iron plate girder of 60 feet span, which is to carry a uniform load of 89§ tons and two loads of 44 11/16 tons each, one concentrated near each end, and one quarter span from reactions. We are to use a web 36" deep and flange 21" wide. Design the girder parts by use of Tables.

From the arrangement of loads Wl11 and WlV (at bottom of Table XLI) we see their sum is equal to a uniform load, or

44 11/16 + 44 11/16 = 89 3/8 that is, our two concentrated loads will have the same effect on the girder as a uniform load of 89 3/8 tons ; our total load on the girder, therefore, will be equal to 178 3/4 tons uniform load, which we will assume includes the weight of the girder itself. We will decide to use four 6" x 6" x 7/8" angles, as the loads are very heavy, and 7/8" rivets throughout.

We now settle the thickness b of web, from formula given in right-hand corner of Table XLI, namely: b = U/8.d = 178 3/4/8.36 = 0,621 or, say, web should be 5/8" thick.

We now find the value of web in resisting the bending-moment from Table XLI. Pass down the verticil span line marked 60' 0" till we strike the curve marked 36", this is two-thirds way between the horizontal lines marked on the left - (in the 5/8" thick vertical column, the second from left)-7,5 and 10,0 respectively, or our web would safely carry about 9 tons. We next take Table XLI I, remembering that we selected the 6"x6"x7/8" angles, the values for which are in the extreme left column.

Designing girders by Tables.

Value of web.

We again pass down the 60' 0" vertical line till we strike the 36" curve, which is a little more than half-way between the horizontal lines marked in the extreme left - (6" x 6" x 7/8") - column 28,2 and 32,9 respectively; or, our four angles together will take care of about 30 3/4 tons, this added to the web value (9 tons; makes 39 3/4 tons of the 178 3/4 tons to be cared for, or a balance of 139 tons to come on the flange.

The flange is to be 21" wide, from this we must deduct the two 7/8" rivet holes,1 or our effective width of flange would be = 19 1/4", therefore each inch must carry

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