This section is from the book "Building Construction And Superintendence", by F. E. Kidder. Also available from Amazon: Building Construction And Superintendence.

The supporting forces for cantilever beams or trusses may be found in the same way as for trusses supported at the ends, but to make the method perfectly plain we will give a few examples.

To find the respective loads on P1 and P2,

Fig. 257.

Ans. - We can take moments about any point in the beam, but in this particular case it will be simplest to take moments about a point directly over the center of P2. The weights to the left of P2 tend to turn the beam down to the left, and we will call their moments plus. The weight to the right of P2 tends to turn the beam in the opposite direction, and we will call its moment minus. The weight directly above P2 has no moment, as its arm is o. Multiplying each weight by its arm, we have:

Fig. 264.

10 | x | 22.................. | = | +220 |

8 | x | 13........................... | = | +104 |

7 | x | 7.................... | = | +49 |

9 | x | 3.......................... | = | -27 |

Total moment = +373 - 27 = 346. |

This moment must be resisted by P1 acting with an arm of 17 ft.

Consequently P1 must equal346/17 = 20.35 lbs. P2 must be the difference between the sum of the loads and P1, or 40 - 20.35 =

19.65 lbs.

To determine the supporting foces for the truss shown by Fig. 264, the loads being in tons.

In the case of cantilever trusses it is necessary to figure the loads at the end joints, because at one end they do not come over a support, and therefore they affect the stresses. In this example it will be simpler to take the moments about P2. Commencing with the load at the extreme left, and multiplying each load by its respective horizontal distance from P2, we have

.8 | X | 46.................................... | = | 36.8 |

1.6 | X | 40............................ | = | 64.0 |

1.6 | X | 34............................ | = | 54.4 |

1.7 | X | 28............................ | = | 47.6 |

1.8 | X | 21............................ | = | 37.8 |

1.8 | X | 14............................ | = | 25.2 |

1.8 | X | 7............................ | = | 12.6 |

.9 | X | 0............................ | = | .0 |

Sum of loads = 12.0 | Sum of moments | = | 278.4 |

Now the moments of the loads tending to rotate the truss to the left, about P2, must be balanced by the moment of P1, and as its moment must be equal to the sum of the moments of the loads, and its arm is 28, the reaction exerted by P1 must equal278.4/28 =

Fig. 265.

9.94 tons. As the sum of the reactions must equal the sum of the loads, P2 must equal 12 - 9.94, or 2.06 tons.

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