This section is from the book "Safe Building", by Louis De Coppet Berg. Also available from Amazon: Code Check: An Illustrated Guide to Building a Safe House.

Thus, if x = 33° and y = 50°, formula (47) would become: p = w.L2/2. sin.2(17)o/sin2.50.o.sin.88o

1Above table of friction angles is taken from Klasen's "Hockbau and Bruck-etibau - Constructionen." As a rule it will do to assume the angle of friction at 33o and the weight of backing at 120 lbs. per cubic foot, except in the case of water.

The values of which, found in a table of natural sines, etc., is p = w.L2/2. 0,29242/0,7662.0,994

= w.L2/2.0,1458 = 0,729. w.L2 2

Similarly, in formula (48), we should have for the quantity:

Average Case. As already mentioned, however, the angle of friction - (except for water when it is = 0°, that is, normal to the back surface of wall) - is usually assumed at 33°; this would reduce above formulae to a very much more convenient form, viz.:

For the average angle of friction (33°)

If the backing is higher than the wall:

Backing higher than Wall.

(49) Or. if the backing is level with top of wall:

Backing level with Wall.

(50) Where p, w and L same as for formulas (47) and (48).

Where n = amount of slope or batter in inches (per foot height of wall) of rear surface of wall.

Thus, if the rear surface sloped towards the backing three inches (for each foot in height) we should have a positive quantity, or n = + 3.

If the rear surface sloped away from the backing three inches (per foot of height), n would become negative, or n = - 3.

When the rear surface of wall is vertical, there would be no slope, and we would have n=0.

The latter is the case generally for all cellar walls, which would still further simplify the formula, or, for cellar walls where weight of soil or backing varies materially from 120 pounds per cubic foot.

Cellar Walls.

p = W. L2. 0,138. (51)

Cellar Walls-general case.

For cellar walls, where the weight of soil or backing can be safely assumed to weigh 120 pounds per cubic foot p=16 2/3.L2. (52)

Cellar Walls-usual case.

Where p = the total amount of pressure, in pounds, per each running foot in length of wall.

Where w=the weight, in pounds, per cubic foot of backing.

Where L = the height, in feet, of ground line above cellar bottom.

For different slopes of the back surface of retaining walls (assuming friction angle at 33°) we should have the following table; + denoting slope towards backing, -denoting slope away from backing.

Slope of back surface of wall in inches per foot of height. | Value of p for backings of different weights per cubic foot. | Value of p for the average backing, assumed to weigh 120 lbs. per cubic foot. | ||||

+4" | p | = | 0,072. w. L2 | p | = | 8 2/3. L2 |

+3" | p | = | D,0S8. w.L2 | p | = | 11 . L2 |

+2" +1" | p | = | 0,098. w. L2 | p | = | 12 . L2 |

p | = | 0,112. to. L2 | p | = | 13i. L2 | |

0" | p | = | 0,138. w. L2 | p | = | 162/3. L2 |

- 1" | 1 | = | 0,157. to. L2 | p | = | 19 . L2 |

- 2" | p | = | 0,185. w. L2 | p | = | 22 1/2. L2 |

- 3" | p | = | 0,205. w. L2 | p | = | 24 2/3. L2 |

- 4" | p | = | 0,258. w. L2 | p | = | 31 . L2 |

Now having found the amount of pressure p from the most convenient formula, or from Table XI, and referring back to Figure 56, proceed as follows:

Find the centre of gravity G of the mass ABC D,1 from G draw the vertical axis G H, continue F O till it intersects G H at F. Make F H equal to the weight in pounds of the mass ABCD (one foot thick), at any convenient scale, and at same scale make H I = p and parallel to P O, then draw I F and it is the resultant of the pressure of the earth, and the resistance of the retaining wall. Its point of intersection K with the base D A is a point of the curve of pressure. To find the exact amount of pressure on the joint D A use formula (44) for the edge of joint nearest to the point K or edge D, and formula (45) for the edge of joint farthest, from the point K, or edge A.

To find Curve of Pressure.

Amount of Stress at Joint.

Fig. 57.

1To find the centre of gravity of a trapezoid A B C D.Fig. 57, prolong C B until BF = C I = D A and prolong D A until A E = D H = C B, draw E I and H Fand their point of intersection G is the centre of gravity of the whole.

Formula (44) was v= p/a+ 6.x.p/a.d

If M be the centre of D A, that is DM = MA = 1/2.DA, and remembering that the piece of wall we are calculating, is only one running foot (or one foot thick), we should have For x = EM; expressed in inches. For a = A D. 12; (AD expressed in inches). For d = AD; in inches, and

For p = F I, in lbs., measured at same scale as F II and H I; or, the stress at D, (the nearer edge of joint) would be v, in pounds, per square inch, v=FI/AD.12 + 6. KM.FI/12.AD2 Remembering to measure all parts in inches except F I, which must be measured at same scale as was used to lay out F H and II I. Similarly we should obtain the stress at A in pounds per square inch.

v = FI/AD.12 - 6. KM.FI/12.AD2 v should not exceed the safe crushing strength of the material if positive; or if v is negative, the safe tensile strength of the mortar. If we find the wall too weak, we must enlarge A D, or if too strong, we can diminish it; in either case, finding the new centre of gravity G of the new mass A B C D and repeating the operation from that point; the pressure, of course, remaining the same so long as the slope of back surface remains unaltered. If the wall is a very high one, it should be divided into several sections in height, and each section examined separately, the base of each section being treated the same as if it were the joint at the ground line, and the whole mass of wall in the section and above the section being taken in each time.

Fig. 58.

Thus in Figure 58, when examining the part A, BCD1, we should find O1 P11, for the part only, using L1, as its height, the point O1, being at one third the height of A, B or A1 O1 = 1/3. A1 B; G1 would be the centre of gravity of A, B C D1, while F1 II1 would be equal to the weight of its mass, one foot thick; this gives one point of the curve of pressure at K1, with the amount of pressure = F1 I1, so that we can examine the pressures on the fibres at D, and A,. Similarly when comparing the section A11 B C D11 we have the height L,„ and so find the amount of pressure O11 P11, applied at O11, where A11 O11 = 1/3 A11 B; G11 is centre of gravity of A11 B C D11 while F11 H11 is equal to the weight of A11 B C D11, one foot thick, and F11 I11 gives us the amount of pressure on the joint, and another point K11 of curve of pressure, so that we can examine the stress on the fibres at D11 and A„. For the whole mass A B C D we, of course, proceed as before.

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