Handbook of Civil Engineering Calculations

essary to load a body in a manner that produces identical boundary conditions and then to

ascertain the directions of the principal stresses.

5. Apply the principal-stress analogy
   Refer to Fig. 2a. Consider the surface directly below the dam to be subjected to a uniform
   pressure. Principal-stress trajectories may be readily constructed by applying the princi-
   ples of elasticity. In the flow net, flow lines correspond to the minor-stress trajectories
   and equipotential lines correspond to the major-stress trajectories. In this case, the flow
   lines are ellipses having their foci at the edges of the base of the dam, and the equipoten-
   tial lines are hyperbolas.
   A flow net may also be constructed by an approximate, trial-and-error procedure
   based on the method of relaxation. Consider that the area through which discharge occurs
   is covered with a grid of squares, a part of which is shown in Fig. 2b. If it is assumed that
   the hydraulic gradient is constant within each square, Eq. 5 leads to

hl+h 2 + h 3 + h 4 -4hQ = 0 (5)

Trial values are assigned to each node in the grid, and the values are adjusted until a

consistent set of values is obtained. With the approximate head at each node thus estab-

lished, it becomes a simple matter to draw equipotential lines. The flow lines are then

drawn normal thereto.

SOIL PRESSURE CAUSED BY POINT LOAD

A concentrated vertical load of 6 kips (26.7 IcN) is applied at the ground surface. Compute

the vertical pressure caused by this load at a point 3.5 ft (1.07 m) below the surface and 4

ft (1.2 m) from the action line of the force.

Calculation Procedure:

1. Sketch the load conditions
   Figure 3 shows the load conditions. In Fig. 3, O denotes the point
   at which the load is applied, and A denotes the point under consid-
   eration. Let R denote the length of OA and r and z denote the
   length of OA as projected on a horizontal and vertical plane, re-
   spectively.


2. Determine the vertical stress az at A
   Apply the Boussinesq equation:

3Pz

3

*' = 2^ (6)

Thus

'

with P = 600

°

lb

(

26

>

688

-° N), r = 4 ft (1.2 m), z = 3.5 ft

MGURE (^3) (1 0? m^ R = (42 + 352 )0. 5 = (^532) ft (1>62i m); then <rz =
3(6000)(3.5)^3 /[277(5.32)^5 ] = 28.8 lb/ft^2 (1.38 kPa).
Although the Boussinesq equation is derived by assuming an idealized homogeneous
mass, its results agree reasonably well with those obtained experimentally.
Surface