PROTECTION OF BUILDINGS IN THE MINING AREAS
DESIGNING EXERCISE
Impact of area horizontal deformations on building foundations
TOTAL TENSILE FORCE IN CONTINUOUS FOOTING:
N = Z + Zb + J + H
The following stresses influence foundation:
Z  shearing stresses in the base of calculated foundation;
Zb  shearing stresses on the foundation side surfaces;
J  shearing stresses in the base of adjacent continuous footings;
H  normal stresses (pressure forces) on the side surfaces of adjacent continuous
footings;
SHEARING STRESSES IN THE BASE OF CALCULATED FOUNDATION (Z)
Employed distribution of stresses under calculated continuous footing depends on
horizontal deformation value (µ):
o
Case no.1  µ > 4
oo
o o
Case no.2  2 d" µ d" 4
oo oo
1
o
Case no.3  µ < 2
oo
Value of stresses under calculated continuous footing
Åš = K1(Ã tgĆ + c)
à normal stresses under continuous footing;
Ć - soil internal friction angle;
c  cohesion of soil;
K1  coefficient according to the figure:
If there is deformable soil (or sand sub-crust), between foundation footing and rocky
subsoil, of thickness  t smaller than :
- for continuous footings and foundation frameworks
0,5m
Å„Å‚
t d"
òÅ‚1,5Å"b
ół
- for slabs
t d" 0,5m
then coefficient K=1
Value of tensile force Z on continuous footing length:
Zx = b Å"(0,5Å" L - x)Å"Åš
Maximal value of force Z for case no.1:
Zmax,1 = 0,5Å"b Å" L Å"Åš
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Maximal value of force Z for case no.2:
Zmax,2 = 0,5Å"b Å"Åš Å"(L - xÅš)
Maximal value of force Z for case no.3:
Zmax,3 = 0,25Å"b Å" L Å"Ä
SHEARING STRESSES ON SIDE SURFACES OF CALCULATED CONTINUOUS FOOTING (ZB)
In case if continuous footing was made in permanent shuttering and next after removal
of this shuttering the remaining spaces were backfilled (almost in 100% cases), then force Zb
equals 0.
SHEARING STRESSES IN THE BASE OF ADJACENT CONTINUOUS FOOTINGS (J)
n
J = si
"Ji
i=1
where: n  number of continuous footings adjacent to calculated continuous footing in
the section from particular cross-section to the end of continuous footing.
Value Ji from the formula
Ji = bi Å"Åši
where: bi  width of calculated adjacent continuous footing
Åši  shearing stresses under adjacent continuous footing
For adjacent continuous footing shearing stresses shall be calculated with the use of the
formula:
Åš = K2(Ã tgĆ + c)
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where: K2  coefficient according to the figure or formula:
s
b
K2 = 0,9Å"
s
17 +
b
where: s  length of considered continuous footing (between perpendicular continuous
footings  parallel to the ground deformation direction),
b  width of continuous footing.
si = si ,l + si,p
where: si,l and si,p are lengths of adjacent continuous footings employed for
calculations from the left and right side of calculated continuous footing.
FORCE FROM PRESSURE ON THE SURFACE OF ADJACENT CONTINUOUS FOOTINGS (H)
n
H = si
"Hi
i=1
Value of force Hi is appointed as the smallest of the following three (H1i, H2i and H3i):
H1i = 0,85Å"Ã Å"(x1,i - x2,i-1)
o
where: x1,i and x2,i-1  distance between continuous footings (size of the field between
continuous footings);
Ão  normal stress in fields between continuous footings appointed from the
formula:
à = p + ł1 h1 + ł h
o 2
where: p  cellars useful load;
h1  thickness of cellars floor;
Å‚1  bulk density of cellars floor:
h  continuous footing height;
Å‚2  bulk density of soil in field between continuous footings
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If soil in the field between continuous footings was replaced, then value of force H1i
can be decreased to 0.7.
Condition of not exceeding shearing stresses in the soil:
H 2i = (Ã tgĆ + c)(x1,i - x2,i-1)
o
Condition of limiting pressure force on continuous footing side surfaces:
H 3i = Pb Kb,i
where: Pb  passive soil pressure appointed from the formula:
îÅ‚ 1 Ć Ć Å‚Å‚
öÅ‚ ëÅ‚45 öÅ‚ öÅ‚
2 o
Pb = hïÅ‚ëÅ‚ p + h1 Å"Å‚ + Å" h Å"Å‚ Å"tg + + 2Å"c Å"tgëÅ‚45o +
ìÅ‚ ÷Å‚ ìÅ‚ ÷Å‚ ìÅ‚ ÷łśł
1 2
2 2 2
íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚
ðÅ‚ ûÅ‚
Kb,i  coefficient appointed from the formula or graph:
x1,i
h
Kb,i =1,2
x1,i
7,7 +
h
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ADDITIONAL COMMENTS:
1. In case if continuous footing loading is asymmetric on the length, axial forces
should be appointed on both sides (Nl and Np) and then calculate:
N = 0,5(Nl + N )
p
2. While tension is calculated for particular continuous footing, bending occurs in
adjacent ones. Bending moment should be appointed from the formula (as for the
fixed beam):
Ji + Hi
M = li2
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3. If exploitation direction is unknown values of tensile forces N and bending moments
M are calculated for each continuous footing assuming two perpendicular directions.
4. Oblique direction of exploitation should be employed, then values of continuous
footing loading are appointed as:
N' = 0,5Å" N
M' = 0,5Å" M
5. Reinforcement of each continuous footing should be appointed for three impacts:
" axial tension with force N
" bending with moment M
" eccentric tension with coupled pair of forces N and M
6. In case of big influences occurring at long continuous footings, elements facilitating
work of the construction  tie rods  should be used, as in the figure:
Value of tensile force in anchor tie rods:
Zk = (c + d)Å"(H + JL )
L
Value of tensile force in diagonal tie rods:
2 2
Z = [(HL + JL )Å"c] +[(H + JB )Å"b]
p B
tie rods are calculated only for axial force impact (tension)
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