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.

δ = 5/384.357500.7083/18000000.58000 = 1,62" Or, practically the same result, and showing how closely the different methods agree. Had we Figured the girder arithmetically we should have obtained practically the same results throughout. We should have considered our load as a uniform load of 357500 pounds, which would give us equal reactions, of 178750 pounds each, an error of hardly 2 per cent.

The bending-moment at the centre would be, Formula (21), m = 357500.708/8 = 31638750 pounds-inch.

The required moment of resistance therefore, would be, Formula

(18), r = 31638750/12000 = 2636,5

Deflection by Table XLI.

Deflection found arithmetically.

Bending-moment arithmetically.

The actual moment of resistance (by section No. 14, Table I) will be found to be at the centre r =2740 or considerably more than required. The load, however, is not strictly equal to a uniform load, hence the discrepancy, we should, of course, use the result found graphically, which was based on the actual conditions.

In Figuring the girder arithmetically the required moments of resistance at different points along the girder should be ascertained; after winch the curve of moments of resistance can be laid out and the flanges, web, rivets, etc., of girder, calculated the same as already explained. If our girder were not braced sideways we should have to calculate for lateral flexure, using Formula (5). For the area a we should take the area of top flange at centre, plus two angles and the part of web between angles. For the square of the radius of gyration we should take the same parts around

Lateral flexure, an axiss M. . . N at right angles to flange, or as top flange, shown in Figure 211. We omit rivet-holes in this case, for ease of calculation, and as all parts are in compression.

We have then a = 3.21 + 2.9,78 + 6.5/8

= 86,21 square inches.

Now for not finding the exact section in Table I, we must find the moment of inertia i and divide this by the area. We have then

i = 3.213/12 + 7/8.12 5/83/12 + 5 1/8.2 3/83/12 = 2468

Therefore = 2468/86,21=28,6 and from Formula (5), l being, of course, the span of girder in inches: w = 3.86,21.12000/1+4.7082.0,000025/9.

= 1149600 pounds.

One-third of this would be safe, therefore w/3 = 383200 pounds would be the safe stress in flanges not to cause lateral flexure, or the safe stress per square inch should not exceed

383200/86,21 = 4445 pounds.

The actual stress will, of course, equal the area 86,21 multiplied by the average fibre stress, per square inch. To find the average fibre stress, use the following Formula :

Average fibre stress in flanges.

v = 2.x (k/f)/d

(128)

Where v = the fibre stress, per square inch, in flanges of plate girders.

Where x = the distance of the centre of gravity of part of flange being strained - (flange, angles and part of web between them) - from the centre of depth of web, in inches.

Where (k/f)=safe modulus of rupture, in pounds, per square inch, or stress on extreme fibres, in pounds per square inch.

In our case this distance x would be found by rule given on p. 7, (Vol. I), and would be (see Figure 211) x= 21.3.19 1/2 + 12 5/8.7/8.17 9/16+2 3/8. 5 1/8.14 9/16

86,21 = 18,63 Therefore the average fibre stress from Formula (128) v = 2. 18,63.12000/42

= 10645 pounds.

The actual total compressive stress on flange will therefore be

= 86,21 .10645

= 917705 pounds.

This result should, of course, be the same as our horizontal flange stress, previously found, and by referring back, we see that this was practically the same (920000 pounds.)

Our actual compressive stress we see therefore is about two and one-half times larger than the safe stress to resist lateral flexure as found above (383200 pounds.)

We should therefore, either brace the girder sideways - which was done by the beams in our case- or we should have to broaden the top flange.

We can readily see that there is no danger of wrinkling in so heavy a flange, but did we wish to calculate it, we would do so by Formula (4) or Table III.

Since the publication of Table XX, the Homestead Steel Works of Pittsburgh (E) have begun rolling 24" deep steel beams from 240 to 300 pounds per yard in weight.

The data in regard to these beams is as follows :

Wrinklingof flange.

A new deep beam.

Depth of beam (d).... | 24" | 24" | |

Weight per yard....... | 300 | 240 | |

Width of flanges(b).... | 7,20 | 6,95 | |

Thickness of web.. | 0,75 | 0,50 | |

Area of each flange.. | 6,83 | 6,55 | |

Area of each web.... | 16,34 | 10,90 | |

Total area (a)... | 30,00 | 24,00 | |

Neutral axis normal to web. | Moment of intertia.... | 2349,00 | 2061,00 |

Moment of resistance,(r) | 195,75 | 171,75 | |

sq. of rad. of gyration . | 78,30 | 85,88 | |

Transverse value (steel)... | 1958000 | 1718000 | |

Neutral axis parallel to web. | Moment of inertia (i)... | 47,13 | 41,65 |

Moment of resistance (r)..... | 13,10 | 12,00 | |

Sq. of rad. of gyration . | 1,57 | 1,74 | |

Transverse value (steel).... | 131000 | 120000 |

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