My experience has been that many engineers have little or no concept of the design of arches, principally because they are so seldom occur in the design of commercial buildings. I hope that this example will give you a better feel for design.

The structure chosen is one that could have stability problems: a two hinged arch with a span of 100 feet and a rise of 15 ft. and a relatively light load. It is intended to show the importance of including nonlinear deflections and moments in the design of an arch. The analysis was made with the program FRAME-X.BAS previously described. The arches are on 20 ft. centers, and have a dead load of 20 psf and a live load of 15 psf on the slope (not on the horizontal projection). The loads at the nodes will the all be the same. Note that if the arch is curved, the loads should be concentrated at the nodes, otherwise there will be an erroneous moment at the nodes.

The arch is divided into 12 sections as shown in the following sketch.


First we will make a preliminary design. The maximum moment in the arch will occur at the quarter point for the live load on that half of the span, or at the opposite quarter point. It can be demonstrated that the moment due to live load is WL2/64, where W is the live load, and L is the span of the arch. On a parabolic arch with a uniform load, there will be no moment due to the dead load, so we will assume for a circular arch, the moment is low.

The horizontal thrust is, approximately, H=(D + L/2)xS²/8R where D is the dead load, L is the live load, S is the span, and R is the rise of the arch:

H = (20x20+20x15/2)x100²/8x15 = 45,833 pounds = 44.83 kips

The thrust at the quarter point is approximately equal to: H/cosA, where A is the angle with the horizontal. In this case it is 16.7 degrees and H = 50.46 kips

The moment at the quarter point is:

M = 20x15x100²/64 = 46,875 fp, =46.88 fk

The section chosen for investigation is a W16x26, with A=7.68 si, I=301, and S= 38.4. Using these values, and the equation f=P/A +/- M/(S), the maximum stress is 21.22 ksi.

Input for the program is shown below:


5010 ' Parameters
5020 DATA 12,13,4.17E+6
5025 ' Coordinates
5030 DATA 1,0,0, 2,8.33,4.88, 3,16.67,8.633, 4,25,11.494
5040 DATA 5,33.33,13.458, 6,41.67,14.617, 7,50,15
5050 DATA 8,58.33,14.617, 9,66.67,13.458, 10,75,11.492
5060 DATA 11,83.33,8.633, 12,91.67,4.880, 13,100,0
5065 ' Member properties
5070 DATA 1,.0533,.0145,0, 0
5075 ' Member incidences and types.
5080 DATA 1,1,2,1, 2,2,3,1, 3,3,4,1, 4,4,5,1, 5,5,6,1
5090 DATA 6,6,7,1, 7,7,8,1, 8,8,9,1, 9,9,10,1, 10,10,11,1
5095 ' Code for joint supports or releases. (0=fixed, 1=free to move.)
5100 DATA 11,11,12,1, 12,12,13,1
5110 DATA 1,0,0,1, 13,0,0,1, 0
5115 ' Loads (There are 11 joint loads
5120 DATA 1,DEAD + 1/2 LIVE , 11,0, 2,0,5.99,0, 3,0,5.98,0
5130 DATA 4,0,5.98,0, 5,0,5.98,0, 6,0,5.98,0, 7,0,4.15,0, 8,0,2.65,0
5140 DATA 9,0,2.65,0, 10,0,2.65,0, 11,0,2,65,0, 12,0,2.65,0

The results from running this program are as follows:

Defl. @ 4 = 0.373x12 = 4.48 in, Mom. = 76.71 fk
Defl @ 10 = 0.357x12 = 4.28 in., Mom = 97.09 fk,
Mom @ 11=112.12 fk, Thrust @ 4 = 41.42 k.

Defl @ 4 = 0.4739x12 = 5.69 in. Mom. = 98.09 fk
Defl @ 10 = 0.4996x12 = 6.00 in. Mom. = 115.56 fk.
Mom. @11 = 129.49 fk.

Note the great increase in deflections and moments, and that the maximum moment is on the right half of the arch. This is an extreme case but illustrates the stability calculations for an arch.

The final deflection, D(final), can be predicted from the initial deflection by assuming that the shape of the moment diagram is similar to the shape of the deflection diagram:

D(final)=D(initial) + (D(initial)/M)D(final)P

where P is the thrust at the section, and M is the moment.

Solve for D(final):

D(final) = D(initial)/(1 - PD(initial)/M)

Using the valuses obtained from the first pass:

D(final) = 0.373/(1-41.42*0.373/76.71) = 0.467 ft

This compares to 0.474 for the program.

If you would like to play around with this file (It will require a Zip program:)

Click HERE