This section is from the book "The Building Trades Pocketbook", by International Correspondence Schools. Also available from Amazon: Building Trades Pocketbook: a Handy Manual of reference on Building Construction.

If a cord, fastened at each end, supports a number of loads, it will take a position of equilibrium.

depending on the amount and location of the loads, as in (a), Fig 51 In such a case, the cord lain tension. If the system is inverted, it will assume the position shown in (6), in which the forces are still in equilibrium; but. instead of a cord in tension, the lines op,pq, etc. must be. members capable of resisting compression. This latter case

Fig. 51.

represents what exists in an arch, and the broken line inters secting the vertical forces forms the line of pressure; the material in the arch must be of such strength and so disposed as to safely resist the compressive forces acting along this line.

To determine the line of pressure for any arch, points at the abutments and crown must be fixed upon; otherwise, an indefinite number of lines of pressure could be drawn. In metal arches, the abutments and crown are generally hinged, or pin-connected, and the line of pressure necessarily passes through these three points. In masonry arches the abutments and crown are generally not hinged, although there are exceptions; and, therefore, a point must be assumed at each abutment and at the crown, through which the line of pressure is to pass. The line of pressure should, for a masonry arch, be within the middle third of the arch ring, the depth of keystone of which is first assumed, or made equal to that of some existing arch of about the same span. Thus, with an arch 3 ft. deep, the line of pressure should be within a space 6 in. on either side of the center of the depth. If it lies without the middle third, the joints tend to open, which result, while the danger of failure of the arch may not be great, would be unsightly. Again, if the line of pressure passes through the middle third, the angle that it makes with any joint will not be such that the voussoirs are likely to slide on their surface of contact.

In taking the loads upon arches, all weights must be reduced to the same standard ; that is, in this case, the loads have been made equivalent to masonry weighing 140 lb. per cu. ft. The arch and loads are assumed to be 1 ft. in thickness, so that all superficial measurements also represent cubic contents.

In the arch shown in Fig. 52 (a), the pressure curve is considered as passing through points at the abutments j the depth of the voussoirs from the intrados, and through the center of depth at the crown. The theoretical rise of the arch is 10.75 ft., and the theoretical span, 51.32 ft. The arch and load is divided by dotted lines into sections, which, for convenience, are numbered.

If w be the width of any section, and h its height, then its area a is w X h. Also, if c is the distance from the crown to the center of gravity of a section, the moment m of any section about the crown is a X c. Call A the sum of all the a's from the crown up to and including the section considered; thus, A for section 3 is, in the following table,

Fig. 52.

31.25 + 63.75 + 70. Call M the total of the m's; thus, M for on 3 is 78.12 + 478.12 + 875. Then, the distance C from the crown to the center of gravity of the portion between the crown and the section considered is M / A of that section

(see Neutral Axis, page 75); thus, for section 3, the distance C for the portion including the sections 1, 2, and 3 is 1431.24 / 165

= 8.67 ft.

The above values may be tabulated as follows:

Section. | w | h | a = wXh | c | m= aXc | A = Σ a | M = Σm | C = M / A |

1 | 5 | 6.25 | 31.25 | 2.5 | 78.12 | 31.25 | 78.12 | 2.50 |

2 | 5 | 12.75 | 63.75 | 7.5 | 478.12 | 95.00 | 556.24 | 5.85 |

3 | 5 | 14.00 | 70.00 | 12.5 | 875.00 | 165.00 | 1,431.24 | 8.67 |

4 | 5 | 16.50 | 82.50 | 17.5 | 1,443.75 | 247.50 | 2,874.99 | 11.61 |

5 | 5 | 14.00 | 70.00 | 22.5 | 1,575.00 | 317.50 | 4,449.99 | 14.01 |

6 | 2 | 14.75 | 29.50 | 26.0 | 767.00 | 347.00 | 5,216.99 | 15.03 |

The horizontal thrust, Q = (x X p)/b in which x is equal to one-half the theoretical span, or 25.66 ft. minus the value of C for the 6th section, which is, in this case, 15.03, giving x = 10.63; P is equal to A for the last section, 347; and b equals the theoretical rise of the arch, 10.75 ft. Hence, taking moments about m, Q =(x X P)/b = (10.63 X 347)/10.75, or 343 cu. ft. Multiplying by 140 lb., the weight of masonry per cubic foot assumed in this case, the horizontal thrust is 48,020 lb.

The line of pressure may now be determined as follows: Draw through point p in Fig. 52 (6), the horizontal line yz; lay off to scale from p, in order, the distances C obtained from table. At these points lay off the vertical distance ef, g h', i j, etc., equal respectively to the values 31.25, 95,165, etc., from the column headed A. For instance, if the diagram is drawn to a scale of 1 in. to 100 cu. ft., the distance ef will be .31, or nearly 1/3 in. From f, h', j, etc., to the same scale, mark off the constant horizontal thrust Q, as at fq, h'r, js, etc. Thus, the vertical and horizontal forces at each section being given, the resultant of these two forces in each case is eq, g r, i s, etc Extending each until it intersects the joint beyond e, g, i, etc., the pressure curve may be drawn through these latter points of intersection, as shown by the heavy black line, and the thrust at the joints may be found by measuring e q, g r, is, etc. with the scale to which the diagram was drawn.

Since in this ease the pressure curve falls well within the middle third of the arch ring, the arch may be considered satisfactory, provided the safe crushing strength of the masonry is not exceeded.

The influence of the last oblique thrust, which is the resultant thrust of the arch upon the pier, or abutment, will be explained in the following method of determining the pressure curve. This method is somewhat simpler, as it requires practically no calculations.

Fig. 53.

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