This section is from the book "Cyclopedia Of Architecture, Carpentry, And Building", by James C. et al. Also available from Amazon: Cyclopedia Of Architecture, Carpentry And Building.

It is necessary to satisfy Equation 58. We shall consider in this case that the various points in the arch-rib curve and in the special equilibrium

y | z' | yz' | z'" | z" | yz" | |

1 | 1.55 | -9.04 | 14.01 | - 9.50 | -6.06 | ~ 9.39 |

2 | 5.15 | -5.44 | - 28.00 | - 7.15 | -3.71 | - 19.11 |

3 | 7.65 | -2.94 | - 22.50 | - 5.50 | -2.06 | - 15.76 |

4 | 9.70 | -0.89 | - 8.64 | - 4.20 | -0.76 | - 7.38 |

5 | 11.35 | + 0.76 | + 8.64 | - 3.20 | + 0.24 | + 2.72 |

6 | 12.70 | + 2.11 | + 26.80 | - 2.35 | + 1.09 | + 13.84 |

7 | 13.75 | + 3.16 | + 43.45 | - 1.70 | + 1.74 | + 23.93 |

8 | 14.45 | + 3.86 | + 55.80 | - 1.25 | + 2.19 | + 31.65 |

9 | 14.95 | + 4.36 | + 64.45 | - 0.90 | + 2.54 | + 37.95 |

10 | 15.20 | + 4.61 | + 70.07 | - 0.65 | + 2.79 | + 42.40 |

11 | 15.20 | - 0.55 | + 2.89 | + 43.90 | ||

12 | 14.95 | - 73.15 | - 0.60 | + 2.84 | + 42.45 | |

13 | 14.45 | + 269.21 | - 0.80 | + 2.64 | + 38.15 | |

14 | 13.75 | - 1.20 | + 2.24 | + 30.81 | ||

15 | 12.70 | + 196.06 | - 1.80 | + 1.44 | + 18.27 | |

16 | 11.35 | - 2.60 | + 0.84 | + 9.53 | ||

17 | 9.70 | - 3.65 | -0.21 | - 2.04 | ||

18 | 7.65 | - 5.05 | -1.61 | - 12.32 | ||

19 | 5.15 | - 6.85 | -3.41 | - 17.56 | ||

20 | 1.55 | - 9.35 | -5.91 | 9.16 | ||

+ 335.60 | ||||||

-68.85 | • 82.72 | |||||

212.90 | + 392.12 | - 3.44 | + 252.88 |

polygon are the points where these two lines are intersected by the load lines through the centers of the 20 sections. Therefore the points where these verticals intersect the center of the arch rib, give the points from which we measure down to the line OB, and obtain various values for y which are given in the tabular form above (Article 434). From Fig. 228 we may observe that z' equals the numerical difference between the value of y and the distance from OB up to the line vm. This distance has already been computed at 10.95 feet. Therefore, having scaled off as accurately as practicable the various values of y, it is unnecessary to scale off the values of z', but merely to take the numerical difference (carefully observing the algebraic sign) between 10.59 and the various values of y. We thus obtain the values of z' as given in the tabular form. Since z' measures the ordi-nates to points in the curve, and since the curve is symmetrical about its center, it is unnecessary, in this case, to set down the values of z' on both sides of the center; and therefore only the values from 1 to 10, inclusive, are written down in the tabular form. Multiplying the corresponding values of y and z', we find the products as given under the heading yz'. Adding these products for the half-span of the arch, we find an algebraic summation of + 19G.0G; multiplying this by 2, we find that the algebraic summation for the entire arch is + 392.12.

The ordinates z" are measured from vm to the trial equilibrium polygon when it has been shifted not only so that its closing line is horizontal, but also shifted vertically (up or down) so that its line v'"m'" corresponds with vm; but it is unnecessary to draw it in that way, since we may measure the ordinates from the transposed line v m', because we know that they are in each case the equal of the ordinates as they would be if the transposition had been actually made. But the lengths of these ordinates below vm' have already been determined; and since we know that v'" m" is 3.44 feet vertically below vm', we need only change the ordinates z'" by 3.44 (taking care of the algebraic sign), and we then obtain the values of z" as given in the tabular form. Multiplying each value of z" by the corresponding value of y, we obtain the various values (plus and minus) for yz" as given in the last column of the tabular form. The algebraic sum of these quantities is + 252.88.

Since this value is less than the value of the summation of yz', we must select a smaller value of H, so that the values of z" will be proportionately larger. We must therefore use a pole distance which shall be smaller than 02n in the ratio of 252.88 to 392.12. To solve this graphically, we must draw an indefinite line ns, and lay off the distance ns equal to 392.12, at any convenient scale. Laying off a distance nt at the same scale so that it equals 252.88; we may then join s and O2, and draw a line from t parallel to s02, obtaining the point O3 which is the required pole of the special equilibrium polygon.

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