This document discusses the design of foundations for a building located in Petrila, Hunedoara, Romania. It determines the depth of foundations as 5.35 meters based on soil conditions and frost depth. Isolated square foundations of reinforced concrete are selected, with dimensions of 2.0x2.0x0.7 meters. Loads and moments acting on the foundations are calculated. The document also calculates the pressures on the foundation base, designs the reinforcement for the concrete pad and pedestal, and checks settlement criteria.

1. 1
Calcul Fundatii
Stabilirea adancimii de fundare:
Construcţia este situată în Petrila, judetul Hunedoara, zonă în care adâncimea de
îngheţ este 100...110cm. Deasemenea, este o construcţie definitivă, iar terenul este supus
îngheţului. Conform studiului geotehnic nu s-a gasit apa subterana, acesta efectuandu-se
pe o adancime de 7 m.
Datorită sistemului structural de rezistenta – cadru spatial din beton armat – al
cladirii si a terenului bun de fundare, sistemul de fundare ales este cel cu fundatii izolate,
alcatuite dintr-un bloc de beton simplu si un cuzinet din beton armat.
Adâncimea de fundare stabiliă este de -5.35 m faţă de CTN.

2. 2
-incarcarile: N=1353 kN
Mz=21 kNm
Vy=8.88 kN
My=46.55 kNm
Vz=16.72 kN
Alegerea materialului:
-fundatie izolata din beton armat => minim C8/10
Predimensionare:
a =0.5 m
b = 0,5 m → dimensiunile stalpilor
-nisip argilos plastic consistent  −−−
convp =290 kN/m2
B =2.0
- B=L → dimensiunile blocului de beton simplu L = 2.0 m
3.1...0.10.2)65,0...5,0()65,0...5,0(65,0...5,0 ==== Ll
L
l
c
c
m
 lc = bc = 1,2 m

3. 3
30,0ch m
3,020,125,025,025,0  cccc
c
c
hhlh
l
h
m
22.035.0625,0625,0625,0625,0 1
1
 ccc
c
hhlh
l
h
tg mhc = 0,40
35,0
2
5,020,1
2
1 =
−
=
−
=
al
l c
m
375,011 1
1
 cc
c
hlh
l
h
tg m
-beton C12/15
3,1min =tg
- −−−
convp =290 kPa
65,05,03,12min
2
=== ltgH
l
H
tg  m  H = 0,7 m
5,0
2
0,10.2
2
2 =
−
=
−
= clL
l m
-dimensiunile: B = 2,0 m
L = 2,0 m dimensiunile blocului de fundare simplu
H = 0,7 m
bc = 1,0 m
lc = 1,0 m dimensiunile cuzinetului
hc = 0,40 m
 Df = 5.35 m
1.5 Calculul presiunii maxime acceptate de teren:
-strat de fundare : nisip argilos plastic consistent
 pconv = −−−
convp +Cb+Cd= 280+56+90.45=426.45 kN/m2
Cb=corectia de latime;
pt Df>5 m 𝐶 𝑏 = 0.2 ∗ 𝑝𝑐𝑜𝑛𝑣 = 0.2 ∗ 280 = 56
Cd=corectia de adancime;
-pt Df<2m:Cd= 𝑘2 ∗ 𝛾 ∗ 𝐷𝑓 − 2 = 1.5 ∗ 18 ∗ (5.35 − 2) = 90.45

4. 4
Calculul presiunii pe talpa fundatiei:
-solicitare excentrica pe 2 directii
)
66
1(4,3,2,1
B
e
L
e
BL
GN
W
M
W
M
BL
GN
p xyf
y
y
x
zf
ef 
+
=
+
=
2.772540,00,10,12470,00,20,2 =+=+= betcccbetf hlbBLHG  kN
03.0
2.771353
46
=
+
=
+
=
f
y
x
GN
M
e m
01,0
2.771353
21
=
+
=
+
=
f
z
y
GN
M
e m
400)
0,2
03,06
0,2
01,06
1(
0,20,2
2.771353
)
66
1(1 =

+

+

+
=++
+
=
B
e
L
e
BL
GN
p xyf
ef kN/m2
336)
0,2
03,06
0,2
01,06
1(
0,20,2
2.771353
)
66
1(2 =

−

+

+
=−+
+
=
B
e
L
e
BL
GN
p xyf
ef kN/m2
356)
0,2
03,06
0,2
01,06
1(
0,20,2
2.771353
)
66
1(3 =

+

−

+
=+−
+
=
B
e
L
e
BL
GN
p xyf
ef kN/m2
313)
0,2
03,06
0,2
01,06
1(
0,20,2
2.771353
)
66
1(4 =

−

−

+
=−−
+
=
B
e
L
e
BL
GN
p xyf
ef kN/m2
Calculul armaturii de rezistenta din cuzinet:
11902540,020,120,1117501 =+=+=+= betccccuz hblNGNN  kN
1164,0509601 =+=+= czyx hVMM kNm
1344,05811101 =+=+= cyzy hVMM kNm

5. 5
112,0
1190
134
01
01
===
N
M
e
y
x m
097,0
1190
116
01
===
N
M
e olx
y m
1289)
20,1
112,06
1(
20,120,1
1190
)
6
1(01
1 =

+

=+=
c
x
cc
cx
l
e
bl
N
p kN/m2
364)
20,1
112,06
1(
20,120,1
1190
)
6
1(01
2 =

−

=−=
c
x
cc
cx
l
e
bl
N
p kN/m2
1227)
20,1
097,06
1(
20,120,1
1190
)
6
1(01
1 =

+

=+=
c
y
cc
cy
l
e
bl
N
p kN/m2
426)
20,1
097,06
1(
20,120,1
1190
)
6
1(01
1 =

−

=−=
c
y
cc
cy
l
e
bl
N
p kN/m2
1000364)375,020,1(
20,1
3641289
)( 2
21
0 =+−
−
=+−
−
= cxxc
c
cxcx
cx pll
l
pp
p kN/m2
976426)375,020,1(
20,1
4261227
)( 2
21
0 =+−
−
=+−
−
= cyyc
c
cycy
cy pll
l
pp
p kN/m2
375,0
2
45,020,1
2
=
−
=
−
==
al
ll c
yx m
5.826
2
3641289
2
21
,, =
+
=
+
= cxcx
xmedc
pp
p kN/m2
5.826
2
4261227
2
21
,, =
+
=
+
=
cycy
ymedc
pp
p kN/m2
]375,0
3
2
2
375,0
)10001289(
2
375,0
1000[20,1]
3
2
2
)(
2
[
2
01
2
0 −+=−+= x
x
cxcx
x
cxcx l
l
pp
l
pbM
Mx = 185 kNm
]375,0
3
2
2
375,0
)9761227(
2
375,0
976[20,1]
3
2
2
)(
2
[
2
01
2
0 −+=−+= y
y
cycy
y
cycy l
l
pp
l
plM
My = 96.4 kNm
1,37)
2
8,1
2(40)
2
( =+−=+−=

nomc chd cm
201010min =+=+= devnom ccc mm
10}10;;max{min == mmccc durabad mm
10= devc mm

6. 6
Φmax = 18 mm
h0x = d = 37,1 cm
h0y = h0x-Φmax=37.1-1,1=36 cm
-armatura OB37 => Ra = 2100 kN/m2
13.27
2100371,0875,0
185
875,0 0
=

==
ax
x
ax
Rh
M
A cm2
=> 11Φ18 → 27.94 cm2
57.14
210036,0875,0
4.96
875,0 0
=

==
ay
y
ay
Rh
M
A cm2
=> 6Φ18 → 15.24 cm2
px
%
= 336.0100*
* 0
=
x
ax
hB
A
%
px
%
= 178.0100*
* 0
=
y
ay
hL
A
%
pmin=0.1 %
CALCULUL TASARILOR PROBABILE
𝐷𝑓 = −1,5 𝑚 ;
𝐵 = 1,2 𝑚 ;
𝑝 𝑒𝑓 =
𝑁 + 𝐺 𝑓
𝐴
=
1175 + 103
5,29
= 241,6 [𝑘𝑃𝑎]
1032540,020,120,12470,03,23,2 =+=+= betcccbetf hlbBLHG  kN
𝜎 𝑧𝑖 =∝0𝑖∗ 𝑝 𝑛𝑒𝑡;
∝0𝑖= 𝑓 (
𝑧𝑖
𝐵
,
𝐿
𝐵
) ;
𝜎 𝑔𝑧𝑖 = 𝛾 ∗ 𝐷𝑓 + ∑ ℎ 𝑖 ∗ 𝛾𝑖
ℎ 𝑖 ≤ 0,4 ∗ 𝐵 = 0,9 𝑚
𝑠 = 100 ∗ 0,8 ∗ ∑
𝜎 𝑧𝑚𝑒𝑑 ,𝑖 ∗ ℎ 𝑖
𝐸𝑖
≤ 𝑠𝑎𝑑𝑚
𝑠𝑎𝑑𝑚 = 8 𝑐𝑚 ;

7. 7
𝑝 𝑛𝑒𝑡 = 𝑝 𝑒𝑓 − 𝛾 ∗ 𝐷𝑓; 𝑝 𝑛𝑒𝑡 = 241,6 − 19,65 ∗ 1,5 = 212,1; 0,2z gz  
𝜎 𝑧𝑖 = 4 ∗ 𝐾 d*pnet;
Kd=f[L/B;2zi/2];
Pentru Ϭz1;
Ϭz1=4*Kd*pn=4*0.238*212,1=202;
2*z1/B=2*0,6/2,3=0,521;
-pentru L/B=1 =>0.25….0.027 =>x=0.0021→0.23
0.02…..x
0.5…...0.012
0.25…...x =>x=0.0072
Kd=0.23+0.0072=0.238;
Pentru Ϭz2:
Ϭz2=4*Kd*pn=4*0.1884*212,1=159.83;
2*z1/B=2*1,1/2,3=0,956;
-pentru L/B=1 =>0.25….0.031 =>x=0.0248→0.181
0.2…..x
0.5…...0.012

8. 8
0.25…...x =>x=0.0072
Kd=0.1812+0.0072=0.1884;
Pentru Ϭz3:
Ϭz3=4*Kd*pn=4*0.118*212,1=100.1;
2*z1/B=2*2/2,3=1.739;
-pentru L/B=1 =>0.25….0.037 =>x=0.017→0.104
0.23…..x
0.5…...0.024
0.25…...x =>x=0.0144
Kd=0.104+0.0144=0.118;
Pentru Ϭz4:
Ϭz4=4*Kd*pn=4*0.077*212,1=65.32;
2*z1/B=2*2,9/2,3=2.52;
-pentru L/B=1 =>1….0.039 =>x=0.012→0.063
0.521…..x
0.5…...0.023
0.25…...x =>x=0.0138
Kd=0.063+0.0138=0.077;
Pentru Ϭz5:
Ϭz5=4*Kd*pn=4*0.049*212,1=41.57;
2*z1/B=2*3.8/2,3=3.3;
-pentru L/B=1 =>1….0.018 =>x=0.005→0.039
0.304…..x
0.5…...0.016
0.25…...x =>x=0.00096
Kd=0.0039+0.0096=0.049;
Pentru Ϭz6:

9. 9
Ϭz6=4*Kd*pn=4*0.032*212,1=27.14;
2*z1/B=2*4.7/2,3=4.08;
-pentru L/B=1 =>2….0.014 =>x=0.0006→0.026
0.086…..x
0.5…...0.011
0.25…...x =>x=0.0066
Kd=0.026+0.0066=0.032;
Pentru Ϭz7:
Ϭz7=4*Kd*pn=4*0.028*212,1=23.75;
2*z1/B=2*5.6/2,3=4.782;
-pentru L/B=1 =>2….0.014 =>x=0.0054→0.0215
0.782…..x
0.5…...0.011
0.25…...x =>x=0.0066
Kd=0.0215+0.0066=0.028;
Pentru Ϭz8:
Ϭz8=4*Kd*pn=4*0.023*212,1=19.51;
2*z1/B=2*6.5/2,3=5.47;
-pentru L/B=1 =>2….0.014 =>x=0.0103→0.016
1.478…..x
2…...0.011
0.25…...x =>x=0.0066
Kd=0.016+0.0066=0.023;
Ϭgz0= γ1*Df=19.5*1.5=29.25;

10. 10
Ϭgz1= γ1*Df+ γ1*h1=19.5*1.5+19.5*0.6=41;
Ϭgz2= γ1*Df+ γ1*h1+ γ1*h2=19.5*1.5+19.5*0.6+19.5*0.5=50.75
Ϭgz3= γ1*Df+ γ1*h1+ γ1*h2+ γ1*h3=19.5*1.5+19.5*0.6+19.5*0.5+19.5*0.9=67.55
Ϭgz4= Ϭgz3+ γ2*h4=67.55+19.8*0.9=85.37
Ϭgz5= Ϭgz4+ γ2*h5=85.37+19.8*0.9=103.19
Ϭgz6= Ϭgz5+ γ2*h6=103.19+19.8*0.9=121
Ϭgz7= Ϭgz6+ γ2*h7=121+19.8*0.9=138.83
Ϭgz8= Ϭgz7+ γ2*h7=138.83+19.8*0.9=156.65
Fasia hi zi Hi γi pnet αoi Ei бzi бgzi
0,2
бgzi
бgzi
med s
[m] [m] [m] [KN/mc] [KN/mp] [kPa] [KN/mp] [KN/mp] [KN/mp] [KN/mp] [cm]
0 0.6 0.600 1.500 19.5 212 0.521 14000 110 29 6
1 0.5 1.100 2.100 19.5 212 0.956 14000 202 41 8 157 0.447
2 0.9 2.000 3.000 19.5 212 0.174 32000 160 59 10 120 0.270
3 0.9 2.900 3.900 19.8 212 0.252 32000 100 77 13 45 0.102
4 0.9 3.800 4.800 19 212 0.330 32000 65 95 17 62 0.139
5 0.9 4.700 5.700 19 212 0.408 32000 41 113 20 78 0.176
6 0.9 5.600 6.500 19 212 0.478 40000 27 129 24 94 0.169
7 0.9 6.500 7.300 19 212 0.547 40000 23 145 28 109 0.196
8 0.9 7.400 7.300 19 212 0.210 45000 19 145 31 80 0.128
𝑠𝑎𝑑𝑚 = 8 𝑐𝑚 ;
S < 𝑠𝑎𝑑𝑚
2. Fronton:
s= 1.615

11. 11
2.1 Incarcari de calcul:
-incarcarile:
N=1578,05 kN
Mz=145,14 kNm
Vy=87,07 kN
My=96,68 kNm
Vz=56,55 kN
2.2 Alegerea materialului:
-fundatie izolata din beton armat => minim C8/10
2.3 Predimensionare:

12. 12
a = b = 0,40 m → dimensiunile stalpilor
-roca stancoasa  −−−
convp =1000 kN/m2
 B = L = 1,50 m
- B=L → dimensiunile blocului de beton simplu
975,0...75,050,1)65,0...5,0()65,0...5,0(65,0...5,0 ==== Ll
L
l
c
c
m
 lc = bc = 0,80 m
30,0ch m
20,080,025,025,025,0  cccc
c
c
hhlh
l
h
m
13,02,065,065,065,065,0 1
1
 ccc
c
hhlh
l
h
tg m  hc = 0,50 m
20,0
2
40,080,0
2
1 =
−
=
−
=
al
l c
m
20,011 1
1
 cc
c
hlh
l
h
tg m
-beton C12/15
85,1min =tg
- −−−
convp =1000 kPa
65,035,085,12min
2
=== ltgH
l
H
tg  m  H = 1,00 m

13. 13
35,0
2
80,050,1
2
2 =
−
=
−
= clL
l m
-dimensiunile: B = 1,50 m
L = 1,50 m dimensiunile blocului de fundare simplu
H = 1,00 m
bc = 0,80 m
lc = 0,80 m dimensiunile cuzinetului
hc = 0,50 m
 Df = 3,70 m
2.4 Calculul presiunii maxime acceptate de teren:
-strat de fundare : roca stancoasa  pconv = −−−
convp = 1000 kN/m2
140010004,1 === convteren pp  kN/m2
Β = 1,4 → pentru gruparea fundamentala; incarcari cu excentricitate pe ambele
directii
2.5 Calculul presiunii pe talpa fundatiei:
-solicitare excentrica pe 2 directii
)
66
1(4,3,2,1
B
e
L
e
BL
GN
W
M
W
M
BL
GN
p xyf
y
y
x
zf
ef 
+
=
+
=
622550,080,080,02400,150,150,1 =+=+= betcccbetf hlbBLHG  kN
059,0
6205,1578
68,96
=
+
=
+
=
f
y
x
GN
M
e m
089,0
6205,1578
14,145
=
+
=
+
=
f
z
y
GN
M
e m
1161)
50,1
059,06
50,1
089,06
1(
50,150,1
6205,1578
)
66
1(1 =

+

+

+
=++
+
=
B
e
L
e
BL
GN
p xyf
ef kN/m2
817)
50,1
059,06
50,1
089,06
1(
50,150,1
6205,1578
)
66
1(2 =

−

+

+
=−+
+
=
B
e
L
e
BL
GN
p xyf
ef kN/m2
642)
50,1
059,06
50,1
089,06
1(
50,150,1
6205,1578
)
66
1(3 =

+

−

+
=+−
+
=
B
e
L
e
BL
GN
p xyf
ef kN/m2

14. 14
298)
50,1
059,06
50,1
089,06
1(
50,150,1
6205,1578
)
66
1(4 =

−

−

+
=−−
+
=
B
e
L
e
BL
GN
p xyf
ef kN/m2
2.6 Calculul armaturii de rezistenta din cuzinet:
15872550,080,080,005,157801 =+=+=+= betccccuz hblNGNN  kN
96,1245,055,5668,9601 =+=+= czyx hVMM kNm
68,1885,007,8714,14501 =+=+= cyzy hVMM kNm
119,0
1585
68,188
01
01
===
N
M
e
y
x m
079,0
1585
96,124
01
01
===
N
M
e z
y m
3366)
80,0
119,06
1(
80,080,0
1587
)
6
1(01
1 =

+

=+=
c
x
cc
cx
l
e
bl
N
p kN/m2
1595)
80,0
119,06
1(
80,080,0
1587
)
6
1(01
2 =

−

=−=
c
x
cc
cx
l
e
bl
N
p kN/m2
3068)
80,0
079,06
1(
80,080,0
1587
)
6
1(01
1 =

+

=+=
c
y
cc
cy
l
e
bl
N
p kN/m2
1893)
80,0
079,06
1(
80,080,0
1587
)
6
1(01
1 =

−

=−=
c
y
cc
cy
l
e
bl
N
p kN/m2

15. 15
29241595)20,080,0(
80,0
15953366
)( 2
21
0 =+−
−
=+−
−
= cxxc
c
cxcx
cx pll
l
pp
p kN/m2
27751893)20,080,0(
80,0
18933068
)( 2
21
0 =+−
−
=+−
−
= cyyc
c
cycy
cy pll
l
pp
p kN/m2
20,0
2
40,080,0
2
=
−
=
−
==
al
ll c
yx m
2481
2
15953366
2
21
,, =
+
=
+
= cxcx
xmedc
pp
p kN/m2
2481
2
18933068
2
21
,, =
+
=
+
=
cycy
ymedc
pp
p kN/m2
]20,0
3
2
2
2,0
)29243366(
2
20,0
2924[80,0]
3
2
2
)(
2
[
2
01
2
0 −+=−+= x
x
cxcx
x
cxcx l
l
pp
l
pbM
Mx = 51,5 kNm
]20,0
3
2
2
2,0
)27753068(
2
20,0
2775[80,0]
3
2
2
)(
2
[
2
01
2
0 −+=−+= y
y
cycy
y
cycy l
l
pp
l
plM
Mx = 47,5 kNm
3,47)
2
4,1
2(50)
2
( =+−=+−=

nomc chd cm
201010min =+=+= devnom ccc mm
10}10;;max{min == mmccc durabad mm
10= devc mm
Φmax = 14 mm
h0x = d = 47,3 cm
h0y = h0x-Φmax=47,3-1,4=45,9 cm
-armatura OB37 => Ra = 2100 kN/m2
93,5
2100473,0875,0
5,51
875,0 0
=

==
ax
x
ax
Rh
M
A cm2
=> 5Φ14 → 7,70 cm2
64,5
2100459,0875,0
5,47
875,0 0
=

==
ay
y
ay
Rh
M
A cm2
=> 5Φ14 → 7,70 cm2
3.Grinda rigidizare curenta:
2.1 Incarcari de calcul:

16. 16
80,32
2
20,1
00,52
2
00.5 =−=−= x
l
l c
m
6,32,12540,030,0* === xchbg betgrgrgr  kN/m
5,92510,0)2
2
80,3
( ==plg kN/m
1120,3)2
2
30,3
( ==plq kN/m
8,1668,2 === zidetajzid hg  kN/m
9,408,16115,96,3 =+++=+++= zidplplgrtotal gqggQ kN/m
-talpa fundatiei se gaseste in stratul: praf argilos vartos
 pconv = −−−
convp +Cb+Cd= 275 kN/m2
Cb=corectia de latime;
B=latimea fundatiei in m;
K1=coef.care pt pamanturile coezive=0.05;
Cb= −−−
convp *K1*(B-1)=23,25
Cd=corectia de adancime;
-pt Df<2m:Cd= −−−
convp *
4
2−Df
=-38,75
167,0
275
4115,115,1
=

==
conv
nec
p
Q
B m =17 cm < bp => bgr = 40 cm
5,102
40,0
41
===
gr
ef
b
Q
p kN/m2
< pconv => armare constructiva
=> 3Φ14 jos si 3Φ14 sus - bare longitudinale
=> Φ10/15 - etrieri

17. 17
3.Grinda rigidizare curenta:
2.1 Incarcari de calcul:
80,52
2
80,0
60,62
2
60,6 =−=−= cl
l m
52550,040,0 === betgrgrgr hbg  kN/m
2,62515,0
2
30,3
==plg kN/m
3,520,3
2
30,3
==plq kN/m
6,47178,2 === zidetajzid hg  kN/m
1,646,473,52,65 =+++=+++= zidplplgrtotal gqggQ kN/m
-talpa fundatiei se gaseste in stratul: nisip mare si mijlociu cu fragmente de roca
 −−−
convp =600 kN/m2
56436600 =−=++= −−−
DBconvconv CCpp kN/m2
-pamant necoeziv => k1=0,1
- 36)14,0(1,0600)1(1 −=−=−= −−−
BKpC convB kN/m2
- 0)2(2 =−= fD DKC 
-pamant necoeziv => K2=2,5
-γ = 0
131,0
564
1,6415,115,1
=

==
conv
nec
p
Q
B m = 13,1 cm < bp => bgr = 40 cm
160
40,0
1,64
===
gr
ef
b
Q
p kN/m2
< pconv => armare constructiva
=> 3Φ14 jos si 3Φ14 sus - bare longitudinale

18. 18
=> Φ10/15 - etrieri