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(1)

MINING INFLUENCE ON THE SURFACE (a 2D case)

Horizontal strain  causes:

horizontal forces in the beam

1. Design value 

d

(in ‰ thus in mm/m)

k wp p

d

ε k k k

ε     ,  = given characteristic value [‰]

k

p

= overloading factor (=1.1 usually)

k

wp

= working conditions factor (say 1.0 for beams) k

k

= directional factor (say 0.7 as an average)

2. Shearing stresses  [kPa] under the beam

For long beams, the tension ( > 0) is more dange- rous than the compression ( < 0), generally.

For prismatic beams, the plot of the function (x) is bilinear for 0  x  L/2:

 increases from 0 to an ultimate value 

d

in the interval [0;x

],

 reaches the ultimate value 

d

at x

and is constant in [x

;L/2],

 is the odd function in [-L/2;L/2].

Take 

d

= K(q

k

tg

k

+ c

k

)

f

[kPa] and:

x

= 0.3L/

d

if both the soil is cohesive and 

d

 6 (x

 1.52.0m usually), x

= 0 elsewhere.

d

is the design value of an ac- cidental action, so 

f

= 

A

=1.0 (

A

=0.0 is not considered here).

All values of the parameters are characteristic ones, i.e. c

k

, tg

k

and q

k

= [P

k

+ G

k

]/(LB).

Margin of safety is introduced by the experimental coefficient 0.5  K  1.0 for friction which depends on q

k

[kPa] – in Fig.

In very special cases, it could be justified to use some “lubricants” under foundation which can reduce the value of K, sometimes almost to zero. Note that for non-prismatic foundations, the center of tension CT should be used as x = 0 instead of the geometrical center (L/2).

3. Tensile and compressive cross-section forces Z(x), Z

b

(x) For a single (separated) prismatic foundation beam such forces can appear only on the contact of the foundation with

a natural soil, i.e. under foundation and/or on its vertical sides.

a) Z(x) under foundation beam: 

x

2 / L

dx ) x ( B )

x (

Z τ

b) Z

b

(x) on each vertical side, Z

b

(x) = Z(x)0.75h/B.

Clearly the resultant force is N = Z + 2Z

b

, generally.

Let h

s

denote the average depth of the contact zone from the ground surface – center of the shelf or center of the rectangular cross-section, respectively; Z

b

forces are applied at this points (centers of the contact zones).

If h/B  1/3, so take  =1 (a T-shaped cross-section, usually),

If h/B > 1/3, so take  = D/

d

where D = h

s

tg

k

tg

2

(45

o

-

k

/2) + c

k

[1-2tg

k

tg(45

o

-

k

/2)].

Note that the Z

b

forces will not appear on the contact with a backfill material like “Sand” in Fig. (applied bitumen or PVC isolation reduces the friction). Only natural undisturbed cohesive soils are of interest here,

so Z

b

= 0, frequently. All forces have already design values because such is 

d

- and , as a consequence.

Test yourself:

(1) Can you check that for the prismatic beam there is max{Z(x)} = B

d

[(L/2-x

)+x

/2] ? (ignore Z

b

).

(2) Can you calculate tensile Z

max

for a symmetrical beam if x

= 2m, L = 16m, 

d

= 30kPa but B is not constant: B = 2m for -4m  x  +4m and B = 1.5m elsewhere? (ignore Z

b

).

(3) Is it true that usually the tensile force Z can reduce bending in spans between columns on the beam?

(4) Can you see that for the lower reinforcement, the forces Z and Z

b

give some moments of opposite signs?

(5) How about for the upper reinforcement?

L/2 L/2

q

-

d

 (x)  

d

d

d

x

-x

Z

d

(x)

H w

max

r

m

r

m

h tg

tg

R>0,

>0 (convex)

0 Z

b

Z

b

Z

h

h

s

clay 0 Sand

x

K

qk [kPa]

R<0, <0 (concave)

W.Brząkała, WUST, Wroclaw/Poland

(2)

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