Masonry Magazine August 1975 Page. 19
TABLE 5
Values of Load Constants S and T
| Arch k Value | | Uniform lead over the entire spon | Complementary parabolic lead over the entire spon |
|---|---|---|---|
| 3.00 | S | 1.0290 | 0.6095 |
| | T | 0.7000 | 0.4429 |
| 2.00 | S | 0.9143 | 0.5333 |
| | T | 0.6000 | 0.3714 |
| 1.40 | S | 0.8457 | 0.4876 |
| | T | 0.5400 | 0.3286 |
| 1.30 | S | 0.8343 | 0.4800 |
| | T | 0.5300 | 0.3214 |
| 1.20 | S | 0.8229 | 0.4724 |
| | T | 0.5200 | 0.3143 |
Note: From Table 18, "Frames and Arches." Intermediate valves may be obtained by interpolation.
TABLE 6
Arch Parameters a, B, y and 8
| Arch k Value | α | B | Y | 8 |
|---|---|---|---|---|
| 3.00 | 7.20 | 2.80 | 8.229 | 2.971 |
| 2.00 | 5.60 | 2.40 | 7.314 | 2.286 |
| 1.80 | 5.28 | 2.32 | 7.131 | 2.149 |
| 1.60 | 4.96 | 2.24 | 6.949 | 2.011 |
| 1.40 | 4.64 | 2.16 | 6.766 | 1.874 |
| 1.20 | 4.32 | 2.08 | 6.583 | 1.737 |
Note: From Table 13, "Frames and Arches." Intermediate valves may be obtained by interpolation,
When x L/2:
N. = H,cos + 1
sin
2
"Frames and Arches" also contains equations for other loading conditions; e.g. concentrated loads.
Notation. In these equations, the subscripts 1 and 2 denote the left and right supports respec- tively. The subscript x denotes values at any horizontal distance, x, from the origin. is the angle, at any point, whose tangent is the slope of the arch axis at that point. (See Table 4.)
M = moment
Naxial force
Q shearing force
frise of the arch
W total load under consideration.
horizontal thrust
H
V
vertical reaction
L= span of the arch
S and T are load constants (see Table 5). J, F and K are constants, determined by:
J = 1+(8/7)
K = Se/y
F=0-J
α, β, y and 8 are parameters (see Table 6).
0 = 2(α + β)
ILLUSTRATIVE EXAMPLE
Problem. Using the equations given in the book, "Frames and Arches," design a parabolic brick masonry arch to meet the following require- ments. The arch is integral with a loadbearing, brick-and-brick cavity wall. Wall weight is 80 psf. The arch dimensions are: span, 20 ft; rise, 12 ft; deptn, 16 in.; thickness, iz in.; totai wan neigia, 8 ft. The uniform roof load, bearing on the wall above the arch, is 1200 lb per ft. The arch is of solid brick (4000 psi) and type N mortar; allow- able compressive stress in the arch is 800 psi.
Solution. The arch is a constant-section, high- rise, symmetrical, parabolic, hingeless arch; there- fore, the equations previously given in this issue of Technical Notes are applicable. Each different loading condition must be analyzed separately. Similar loads, e.g. all uniform loads, may be added and treated as a single load. Moments, shears and thrusts resulting from each loading condition are combined to give total values. For symmetrically loaded symmetrical arches, only 1½ of the arch need be analyzed.
The loads carried by the arch are:
Uniform loads
Wall dead load
(80) (20/100)
Roof dead + live load
Arch dead load in excess of wall weight (approx.)
Total uniform load
W(1780) (20)
Complementary parabolic loading
Maximum ordinate
p = (80) (12)
Minimum ordinate
WpL/8960(20)/3
320 lb per ft
1200
260
=
1780 lb per ft
= 35,600 lb
960 lb per ft
0
6400 lb
Numbers before the following paragraphs refer to the outline of the recommended sequence.
1. The principal arch dimensions are:
d = 16 in.
t = 12 in.
L = 20 ft
f = 12 ft
f/L = 0.6
2. From Table 3, k = 2.80.
3a. For parabolic loading:
From Table 5,
S = 0.5943
From Table 6,
α = 6.88
β = 2.72
From the given relationships,
J = 1.3522
0 = 19.20
T = 0.4286
y = 8.046
8 = 2.834
F= 4.4884
K = 1.4182
3b. For vertical uniform loads:
From Table 5,
S = 1.0061
T=0.6800
From Table 6 (note that a, ß, y and 8 are the same for any given arch dimensions),
α = 6.88
β = 2.72
From the given relationships:
J = 1.3522
0 = 19.20
y = 8.046
8 = 2.834
F = 4.4884
K = 2.4008
4. The necessary substitution may now be made to evaluate the design moments and forces. In
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