Circular Rings and Arches

This page provides tables of formulas for circular rings and arches from the "Stress Analysis Manual," Air Force Flight Dynamics Laboratory, October 1986.

Other related chapters from the Air Force "Stress Analysis Manual" can be seen to the right.

Frames & Rings

Table 5-6 gives formulas for the bending moments, tensions, shears, and deflections of closed circular rings and circular arches of uniform cross section loaded in various ways. Cases 1 through 21 treat closed rings, and cases 21 through 24 treat arches. By superposition, the formulas given by Table 5-6 can be combined to cover a wide variety of loading conditions.

These ring formulas are based on the following assumptions: (1) The ring is of such large radius in comparison with its radial thickness that the deflection theory for straight beams is applicable. (2) Its deflections are due solely to bending, the effect of direct axial tension or compression and that of shear being negligible. (3) It is nowhere stressed beyond the elastic limit. (4) It is not so severely deformed as to lose its essentially circular shape.

Since many of the formulas in Table 5-6 consist of a large number of terms, each of which may be large in comparison with the end result, calculations should be made with extreme care in order to ensure accurate results.

Table 5-6
Formulas for Closed Circular Rings of Uniform Cross Section

Closed Circular Ring Definitions
Case Illustration Formulas
1 Closed Circular Ring, Case 1 Closed Circular Ring Formulas, Case 1
2 Closed Circular Ring, Case 2 Closed Circular Ring Formulas, Case 2
3 Closed Circular Ring, Case 3 Closed Circular Ring Formulas, Case 3
4 Closed Circular Ring, Case 4 Closed Circular Ring Formulas, Case 4
5 Closed Circular Ring, Case 5 Closed Circular Ring Formulas, Case 5
6 Closed Circular Ring, Case 6 Closed Circular Ring Formulas, Case 6
7 Closed Circular Ring, Case 7 Closed Circular Ring Formulas, Case 7
8 Closed Circular Ring, Case 8 Closed Circular Ring Formulas, Case 8
9 Closed Circular Ring, Case 9 Closed Circular Ring Formulas, Case 9
10 Closed Circular Ring, Case 10 Closed Circular Ring Formulas, Case 10
11 Closed Circular Ring, Case 11 Closed Circular Ring Formulas, Case 11
12 Closed Circular Ring, Case 12 Closed Circular Ring Formulas, Case 12
13 Closed Circular Ring, Case 13 Closed Circular Ring Formulas, Case 13
14 Closed Circular Ring, Case 14 Closed Circular Ring Formulas, Case 14
15 Closed Circular Ring, Case 15 Closed Circular Ring Formulas, Case 15
16 Closed Circular Ring, Case 16 Closed Circular Ring Formulas, Case 16
17 Closed Circular Ring, Case 17 Closed Circular Ring Formulas, Case 17
18 Closed Circular Ring, Case 18 Closed Circular Ring Formulas, Case 18
19 Closed Circular Ring, Case 19 Closed Circular Ring Formulas, Case 19
20 Closed Circular Ring, Case 20 Closed Circular Ring Formulas, Case 20
21 Closed Circular Ring, Case 21 Closed Circular Ring Formulas, Case 21
22 Closed Circular Ring, Case 22 Closed Circular Ring Formulas, Case 22
23 Closed Circular Ring, Case 23 Closed Circular Ring Formulas, Case 23
24 Closed Circular Ring, Case 24 Closed Circular Ring Formulas, Case 24
25 Closed Circular Ring, Case 25 Closed Circular Ring Formulas, Case 25




5.10 Sample Problem - Circular Rings and Arches

Given: The circular ring shown in Figure 5-18.

Circular Ring

Find: The bending moments in the ring.

Solution: The bending moment in the ring may be obtained by superposing that due to the concentrated loads (Mp) and that due to the applied bending moments (Mm). From Table 5-6, case 1, the bending moment due to the concentrated loads is

$$ M_p = PR(0.3183 - b/2) = 70(10)(0.3183 - \sin{x}/2) = 223 - 350 \sin{x} $$

From Table 5-6, case 3, the bending moment due to the applied moment is

$$ M_m = M_A (0.6366 a - 1/2) = 800 (0.6366 \cos{x} - 1/2) = 510 \cos{x} - 400, ~~\text{if } (0 < x < \pi / 2) $$

or

$$ M_m = M_A (0.6366 a + 1/2) = 800 (0.6366 \cos{x} + 1/2) = 510 \cos{x} + 400, ~~\text{if } (\pi / 2 < x < \pi) $$

The bending moment due to the combined loading is

$$ M = M_p + M_m $$

Thus,

$$ \begin{eqnarray} M &=& (223 - 350 \sin{x}) + (510 \cos{x} - 400) \nonumber \\ &=& -177 - 350 \sin{x} + 510 \cos{x} \nonumber \\ &~& \text{if } (0 < x < \pi / 2) \end{eqnarray} $$

or

$$ \begin{eqnarray} M &=& (223 - 350 \sin{x}) + (510 \cos{x} + 400) \nonumber \\ &=& 623 - 350 \sin{x} + 510 \cos{x} \nonumber \\ &~& \text{if } (\pi / 2 < x < \pi) \end{eqnarray} $$

In the above expressions, x is the angular distance from the bottom of the ring.