Arches 100188

Fig. 108.

Therefore

Stress at C = 11250/240 + 6. 11250.3 1/2/240 = +96 pounds, and

Stress at D = 11250/240 - 6. 11250.3 1/2/240.20 = -3 pounds

The arch is, therefore, safe.

Example V.

A 12-inch brick semi-circular arch has 12 foot span. A solid brick wall is built over the arch to a level with one foot above the keystone.

The abutment piers are 5 feet high to the spring of arch and are each 3 feet wide, including, of course, the width of skew-backs. Are the arch and piers safe?

As before, we will assume arch, pier, and wall over arch, each one foot thick. We will divide the load over arch into seven equally wide slices. This will make uneven voussoirs, but this does not matter, as our joint lines (and voussoirs) are only imaginary anyhow, and not necessarily of the shape of the actual voussoirs, which in brick would, of course, be represented by each single brick. The amount of the sums of each voussoir and its load, and the vertical neutral axes of the different sets are given by the arrows and lines G1, G2, G3, etc. (in Fig. 110). When considering the safety of the abut-ment we treat it exactly the same as the voussoirs (and loads) of the arch; that is, we take the whole weight of the abutment, viz., C D E FIH C and find its neutral axis G8.

Arches 100189

Fig. 109.

Returning now to the arch, we go through the same process as before. We find the horizontal pressu (Fig. 109) g1 h1, g2 h1, etc. In this case we find that the last pressure g7 h7 is not as large as g6 h6 therefore we adopt the latter; it scales 1425 units or pounds. We now make (Fig. 110) a o = 1423 pounds; and a b = 251 pounds; b c = 280 pounds; c d = 373 pounds, etc.; g h is equal to the last section of arch or 1782 pounds. We continue, however, and make h i = 4600 pounds = the weight of abutment. Draw o a, o b, o c, etc., to o i. Then get the tangents to the curve of pressure, as before, viz.: a 1 i1 i2 i3 i4 i5 i6 K7; we now continue i6 K7, which is parallel with o h till it intersects the vertical axis G8 of the abutment at i7 and from thence we draw i7 K8 parallel with o i.

We will now examine the base joint I II of pier. I K8 scales 10 1/4", and as the pier is 36" wide, K8 is 7 3/4" from the centre of joint. The area is a = 12.36 = 432 and the pressure is p = o i = 9100 pounds. Therefore,

Thrust on abutment.

Stress at I = 9100/432 + 6. 9100.7 3/4/432.36 = + 48 pounds,= + 48 pounds. and

Stress at H = 9100/432 - 6. 9100.7 3/4/432.36 = - 6 pounds.

There is, therefore, a slight tendency for the pier to revolve around the point I, raising itself at H; still the tendency is so small, only 6 pounds per square inch, that we can safely pass the pier, so far as danger from thrust is concerned.

Joint C D at the spring of the arch looks rather dangerous, however, as i6 i7 cuts it so near its edge D. Let us examine it. D K7 measures 1 1/2", therefore K7 is 4 1/2" from the centre of joint, which is 12" wide. The area is, of course, a = 12. 12 = 144 and the pressure p = o h = 4600 pounds. Therefore,

Stress at D = 4600/144 + 6. 4600. 4 1/2/144.12 = + 104 pounds, and

Stress at C = 4600/144 - 6. 4600.4 1/2/144.12 = - 40 pounds.

It is evident, therefore, that the arch itself is not safe, and it should be designed deeper; that is, the joints should be made deeper (say, 16"), and a new calculation made.

If, instead of an abutment-pier, we had used an iron tie-rod, its sectional area would have to be sufficient to resist a tension equal to the greatest horizontal thrust o a; and care should be taken to proportion the washers at each end, large enough that they may have sufficient bearing-surface so as not to crush the material of the skew-backs.

Tie-rods to arches.

Tie rods to arches

Fig. 110.

Thus, in Example III, Fig. 105, if we should place the iron tie-rods to the beams 5 feet apart, they would resist a tension equal to five times the horizontal thrust o a, which, of course, was calculated for 1 foot only, or t = 5. 2040 = 10200 pounds. The safe resistance of wrought-iron to tension is from Table IV, 12000 pounds per square inch; we need, therefore:

10200/12000 = 0,85 square inches of area in the rod; or the rod should be 1 1/16" diameter. A 1" or even 7/8" rod would probably be strong enough, however, as such small iron is apt to be better welded, and, consequently, stronger, and the load on the arch would probably be a " dead " one.

As the end of rod will bear directly against the iron beam, the washers need have but about 1/4" bearing all around the end of the rod, so that the nut would probably be large enough, and no washer be needed.

Example VI.

A pier 28" wide and 10' high supports two abutting semi-circular arches; the right one a 20" brick arch of 8' span; the left one an 8" brick arch of 3' span. The loads on the arches are indicated in the Figure 111. Is the pier safe?

The loads are so heavy compared to the weight of the voussoirs, that we will neglect the latter, in this case, and consider the vertical neutral axis of and the amount of each load as covering the voussoirs also; except in the case of the lower two voussoirs, where the axes are considerably affected. "We find the curve of pressure of each arch as before.

For the large arch we would have the curve through a and i, for the small one through a1 and i1; the points i and i1 being the intersections of the curves with the last vertical neutral axis of each arch.

Now from i draw i x parallel with of and from i1 draw i1 x (backwards), but parallel with o1f till the two lines intersect at x. Now make f g parallel with and = o1 f1 and draw g h vertically = 2600 pounds = the weight of the pier from the springing line to the base (1' thick). Draw o g and o h. Now returning to x, draw x y parallel with g o till it intersects the neutral axis of the pier at y, and from y draw y z parallel with o h till it intersects the base joint C D of pier (or its prolongation) at K1. Continue also x y till it intersects the springing joint A B and K. Now, then, to get the stresses at joint A B we know that the width of joint is 28", therefore a = 28.12 = 336; further p = o g = 19250 pounds, and as B K scales one inch, K is distant 13" from the centre of joint, therefore