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.

It would be:

Cross shearing from | A | to | w1 | = | 01H | = | 10000 | pounds. |

Cross shearing from | w1 | to | w11 | = | T I | = | 7500 | pounds. |

Cross shearing from | w11 | to | w111 | = | V J | = | 5000 | pounds. |

Cross shearing from | W111 | to | wIv | = | L K | = | 2500 | pounds. |

Cross shearing from | WIv | to | wv | = | 0 | = | 0 | pounds. |

Cross shearing from | wv | to | wvI | = | MN | = | 2500 | pounds. |

Cross sheering from | wVl | to | wvII | = | PP1 | = | 5000 | pounds. |

Cross shearing from | wVll | to | wv111, | = | RR1 | = | 7500 | pounds. |

Cross shearing from | WvIII | to | B | = | SO | = | 10000 | pounds. |

The area of web of a 15" - 150 pounds beam (Table XX) is 7,59 square inches; the safe resistance of wrought-iron to cross-shearing per square inch being (g/f) = 8000 pounds, we need not worry any further on that score.

To find the deflection we now make the lower load line g c equal to the sum of the lengths of verticals wVlll, wvlII, wVl, etc., through parabola C E G, beginning at top g with length of right vertical wVIII. We select 2 at random, scale z j = 246" (inch scale), draw z g,zc, etc., and figure

1 Had we taken more parts, the steps in shearing figure would become smaller and smaller till they would finally assume the straight line H S, which is the real outline of shearing figure.

c1f1 g1 We now daw z o parallel c1 g1 and find it bisects g c, or greatest deflection will be at centre of beam, which we know is the case. We scale ff1 = 62" (inch scale); find from Table XX for our 15" - 150 pounds beam i = 523,5 and from Table IV for wrought-iron e = 27000000, therefore, Formula (95):

δ = 62.37,5.246.12000 = 0,486 27000000.523,5

Had we figured arithmetically, Formula (39), we should have had

δ = 5.20000.3003/ 384.27000000.523,5 = 0,497" or practically the same result.

The safe deflection for plastering should not exceed (28)

8 = 25.0,03 = 0,75" bo that we are perfectly safe, providing our beam is well braced sideways.

Example VI.

A wrought-iron beam, braced sideways, of 30-foot span, Figure 157, carries a uniform load of 200 pounds per foot, including weight of beam. It carries also a concentrated load wl = 10000 pounds ten feet from the right-hand support. What beam should be used?

We draw beam A B = 360" at inch scale, we divide uniform load into, say, six equal parts, each 5 feet long, or l1 = 60". The total uniform load will be u = 30.200 = 6000 pounds, therefore each part u/6 = 6000/6 =1000 pounds. We draw arrows at the centre of each uniform part, so that the end arrows will be one-half part from supports. These arrows will therefore answer for our verticals, when drawing deflection figure.

At 120" from right hand support we locate the load w1 = 10000 pounds.

We now make load line b a = 16000 pounds the total load and divide it, so that

Uniform and Concentrated Load.

bl | = | wvll | = | 1000 | pounds. |

1h | = | Wv1 | = | 1000 | pounds. |

hf | = | w1 | = | 10000 | pounds. |

fe | = | Wv | = | 1000 | pounds. |

ef | = | w1v | = | 1000 | pounds. |

dc | = | w111 | = | 1000 | pounds. |

ca | = | w11 | = | 1000 | pounds. |

Continue to: