We have, then, from Formula (44) stress at D1111; v =10375/44.12 + 6.16.10375/44.12.44=+63pounds.

and from formula (45) stress at A1111; v =10375/44.12 - 6.16.10375/44.12.44= -23pounds.

The joint A111 D111, therefore, would be more than safe.

Let us try the bottom joint A D similarly. I K is parallel to a m now a m scales 480", therefore, the pressure at K is p = 480. 62 1/2 = 30000 pounds.

Now K is distant 53 inches from centre of joint, therefore, stress at D is v= 30000/56.12+6.53.3000/56.12.56= +298 pounds and stress at A is v =30000/56.12-6.53.3000/56.12.56= - 209 pounds

The wall would evidently have to be thickened at the base. If we could only brace the wall until the superimposed weight were on it, this might not be necessary. If we could do this we should lengthen b c an amount of inches equal to the amount of this load divided by 62 1/2 (the number of pounds per inch), or b c instead of being 36 inches long would be:

36 + 25000/62 1/2 = 436 inches long.

While this lengthening of b c would make the lines of pressure a c,ae, af, etc., very much longer, and consequently the actual pressure very much greater, it will also make them very much steeper and consequently bring this pressure so much nearer the centre of each joint,.

that the pressure will distribute itself over the joint much more evenly, and the worst danger (from tension) will probably be entirely removed.

Example IV.

A stone reservoir wall is plumb on the outside, 2 feet wide at the top and 5 feet wide at the bottom; the wall is 21 feet high, and the possible depth of water 20 feet. Is the wall safe?

Divide the wall into three parts in height; that is, D D1 = D1 D11 = D11 C. Find the weight of the parts from each joint to top, per running foot of length of wall, figuring the stonework at 150 pounds per cubic foot, and we have:

Weight of A11 BCD11 = 2975 pounds.

Weight of A, BCD, = 7140 pounds.

Weight of A B C D = 12495 pounds.

Find centres of gravity of the parts A B C D (at G), of A, B C D, (at G1) and of A11 B C D11 (at G11). Apply the pressures P O at 1/3 height of A E; P1 O1 at 1/3 height of A1 E and P11 O11 at a height of A11 E, where E top level of water.

The amount of pressures will be from formula (53).

For part A11 E P11 O11 = 31 1/4. L211 = 31 1/4. 62= 1125 pounds.

Reservoir

Wall.

For part

A1E;

P1O1

31 1/4.

L21

31 1/4.

132:

5281 ]

pounds.

For part

AE;

PO

31 1/4

L2:

311.

202

12500

pounds.

The pressures P O, P1 O11 etc., will be applied at right angles to A E, prolong these lines, till they intersect the vertical axes through (the centres of gravity) G, G1 and G11 at F, F1 and F11. Then make

Graphical Method 100143

Fig. 70.

F11 H11

=

2975,

weight of upper part.

F1 H1

=

7140,

weight of A1 B C D1 and

F H

=

12495,

weight of A B C D.

Draw through H, H1 and H11 the lines parallel to pressure lines making

H11I11

=

P11 O11

=

1125

H1 I1

=

P1 O1

=

5281

H I

=

P O

=

12500

Draw I11 F11 I1 F1 and I F, then will their lengths represent the amounts of pressure at points K11, K1 and K on the joints A11 D11 A, D, and A D.

F11I11 measures

3300

units or pounds.

F1 I1 "

9500

" "

F I "

18800

" "

By scaling we find that

K11 is 10 1/2 inches from centre of

A11 A11

K1 is 37 " "

D1 A1

K is 74 " "

D A

The stresses to he exerted by the wall will, therefore, be

at D11; v

=

3300/36.12+6. 10 1/2.3300/36.12.36 = +21 pounds

at A11; v

=

3300/36.12-6. 10 1/2.3300/36.12.36 = -6 pounds.

at D1; v

=

9500/48.12+6. 37.9500/48.12.48 = +93 pounds.

at A,; v

=

9500/48.12-6. 74.18800/60.12.60 = + 219 pounds.

at D ; v

=

18800/60.12+6. 74.18800/60.12.60 = +219 pounds.

at A ; v

=

18800/60.12-6. 74.18800/60.12.60 = - 167 pounds.

From the above it would appear that none of the joints are subject to excessive compression: further that joint D„A„ is more than safe, but that the joints D, A, and D A are subject to such severe tension that they cannot be passed as safe. The wall should, therefore, be redesigned, making the upper joint lighter and the lower two joints much wider.