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
din the interval [0;x
],
reaches the ultimate value
dat x
and is constant in [x
;L/2],
is the odd function in [-L/2;L/2].
Take
d= K(q
ktg
k+ c
k)
f[kPa] and:
x
= 0.3L/
dif both the soil is cohesive and
d 6 (x
1.52.0m usually), x
= 0 elsewhere.
dis 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
kand q
k= [P
k+ G
k]/(LB).
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)0.75h/B.
Clearly the resultant force is N = Z + 2Z
b, generally.
Let h
sdenote the average depth of the contact zone from the ground surface – center of the shelf or center of the rectangular cross-section, respectively; Z
bforces 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/
dwhere D = h
stg
ktg
2(45
o-
k/2) + c
k[1-2tg
ktg(45
o-
k/2)].
Note that the Z
bforces 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
maxfor 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
bgive some moments of opposite signs?
(5) How about for the upper reinforcement?
L/2 L/2
q
-
d (x)
d
d
dx
-x
Z
d(x)
H w
maxr
mr
mh tg
tg
R>0,
>0 (convex)
0 Z
bZ
bZ
h
h
sclay 0 Sand
x
Kqk [kPa]
R<0, <0 (concave)
W.Brząkała, WUST, Wroclaw/Poland