Handbook of Civil Engineering Calculations

kN); W 2 = 3(2O)(ISO) = 9000 Ib (40.03 kN); ^W= 18,000 + 9000 = 27,000 Ib (120.10
kN). ThCnJC 1 - (^2 /s)(12) = 8.0 ft (2.44m); x^2 = 12 + 1.5 = 13.5 ft (4.11 m).

2. Find the magnitude and location of the resultant of the
   hydrostatic pressure
   Calling the resultant H= Y 2 wh^2 =^1 / 2 (62.4)(20)^2 = 12,480 Ib (55.51 kN), where w = weight
   of water, lb/ft^3 (N/m^3 ), and h = water height, ft (m), thenj = (^1 /s)(20) = 6.67 ft (2.03 m).


3. Compute the moment of the loads with respect to the
   base centerline
   Thus, M= 18,000(8 - 7.5) + 9000(13.5 - 7.5) - 12,480(6.67) = 20,200 lb-ft (27,391 N-m)
   counterclockwise.


4. Compute the section modulus of the base
   Use the relation S = (^1 A)Zx/^2 = (^1 A)(I)(IS)^2 = 37.5 ft^3 (1.06 m^3 ).


5. Determine the soil pressure at the dam toe and heel
   Compute the soil pressure caused by the combined axial load and bending. Thus /j =
   ^WIA + MIS = 27,000/15 + 20,200/37.5 = 2339 lb/ft^2 (111.99 kPa);/ 2 = 1800 - 539 =
   126 lib/ft^2 (60.37 kPa).


6. Verify the computed results
   Locate the resultant R of the trapezoidal pressure prism, and take its moment with respect
   to the centerline of the base. Thus R = 27,000 Ib (120.10 kN); m = (15/3)((2 x 1261 +
   2339)7(1261 + 2339)] = 6.75 ft (2.05 m); MR = 27,000(7.50 - 6.75) = 20,200 lb-ft (27,391
   N-m). Since the applied and resisting moments are numerically equal, the computed re-
   sults are correct.

LOAD DISTRIBUTION IN PILE GROUP

A continuous wall is founded on three rows of piles spaced 3 ft (0.91 m) apart. The longi-

tudinal pile spacing is 4 ft (1.21 m) in the front and center rows and 6 ft (1.82 m) in the

rear row. The resultant of vertical loads on the wall is 20,000 Ib/lin ft (291.87 kN/m) and

lies 3 ft 3 in (99.06 cm) from the front row. Determine the pile load in each row.

(b) Plan (c) Pile reactions

piie group

(a) Elevation

FIGURE 35

Repeating

lgth