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

To determine the supporting forces for the truss shown by Fig. 265.

Take moments about P2, the moment of the load at the extreme right being marked minus ( - ) because it acts in the opposite direction from the other moments.

.6 | x | 36........................... | = | + 21.6 |

1.2 | X | 30........................... | = | + 36.0 |

1.4 | X | 24........................... | = | + 33.6 |

1.6 | x | 16........................... | = | + 25.6 |

1.6 | X | 8........................... | = | + 12.8 |

1.6 | X | 0........................... | = | .0 |

.8 | X | 8........................... | = | - 6.4 |

Sum of loads = 8.8 | Sum of moments | = | +123.2 |

P1 =123.2/24= 5.133 tons. P2 = 8.8 - 5.13 = 3 2/3 tons.

To prove the correctness of the answer, the sum of the moments tending to turn the truss to the left about P1 must just equal the sum of the moments tending to turn the truss to the right.

The forces tending to turn the truss to the left are P2 and the two loads on the left of P1. The four loads to the right of P1 tend to turn the truss to the right. The load directly above P1 has no moment. The sum of the moments tending to turn the truss to the left about P1 is 102.4, and of those tending to turn the truss to the right is also 102.4; therefore the values obtained for P1 and P2 are correct.

Fig. 266.

To find the supporting forces for the truss shown by Fig. 266, which is similar to that shown in Fig. 109, the overhanging end supporting a center king rod truss. In this example we will take the moments of the loads about P1. Doing so, we have

.6 | X | 0............................. | = | .0 |

1.2 | X | 6............................. | = | 7.2 |

1.2 | X | 12............................. | = | 14.4 |

1.2 | X | 18............................. | = | 21.6 |

3.0 | X | 24............................. | = | 72.0 |

Sum of loads = 7.2 | Sum of moments | = | 115.2 |

P2 = 115.2/12 = 9.6 tons. P1 = 7.2 - 9.6 = - 2.4, showing that P1 must pull down, to form an anchorage for the overhanging arm. If now we take the sum of all the moments about P2, we find them to be + 43.2 and - 43.2, showing that our solution is correct. After working these examples, the reader should have no difficulty in finding the supporting forces for any truss or beam under vertical loads only.

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