COMPUTATIONAL EXAMPLE
protection to building foundation framework
against impact of area horizontal deformations
1. Width of continuous footings
Fig. 1. Projection of building cellars.
1.1. Loading of load-bearing walls in continuous footings level
Load Continuous footings
No. Wall axes
go [kN/m] marking
1. 1/A-D 98,5 L1
2. 2/A-D 175,3 L2
3. 3/A-D 95,2 L3
4. A/1-3 85,6 L4
5. B/1-2 69,4 L5
6. C/1-2 69,4 L6
7. D/1-3 85,6 L7
1.2. Soil parameters on footing level
Kind of soil G, IL=0,05 (clays from medium cohesive to compact)
Limiting passive soil pressure qfn=200kPa
Internal friction angle Ću= 21°
Cohesion cu=35kPa
1
1.3. Width of continuous footings
go
Minimal width of continuous footings b* =
q
fn
go
Actual normal stress under continuous footings à =
b
Continuous footings Minimal width b* Employed width Normal stresses Ã
No.
[m] b [m] [MPa]
1. L1 0,4925 0,50 0,197
2. L2 0,8765 0,90 0,195
3. L3 0,476 0,50 0,190
4. L4 0,428 0,45 0,190
5. L5 0,347 0,35 0,198
6. L6 0,347 0,35 0,198
7. L7 0,428 0,45 0,190
Fig. 2. Projection of continuous footings.
2
2. Internal forces in continuous footings
2.1. Continuous footing L1
2.1.1. Tension
" Force Z
Coefficient K
KL1 = -0,5Å"Ã + 0,7 = -0,5Å"0,197 + 0,7 = 0,6015
L1
{value of coefficient K should be appointed according to graph or from the dependence:
K = -3,5Å"à +1 for à d" 0,1MPa , and K = -0,5Å"à + 0,7 for à > 0,1MPa }"
Shearing stress Åš
ÅšL1 = KL1(à Å"tg Ću + cu )= 0,6015(0,197Å"tg 21+ 0,035)= 0,0665MPa
L1
Force Z
ZL1 = 0,5Å" Ll1 Å"bL1 Å"ÅšL1 = 0,5Å"12,95Å"0,5Å"0,0665 = 0,2154MN = 215,4kN
{Due to lack of symmetry in continuous footing L1 against half of the length it is necessary to
appoint tensile forces in continuous footing on left and right side of the continuous footing}
" Shearing stresses J under adjacent continuous footings L4, L5, L6 and L7
KL4 = KL7 = -0,5Å"Ã + 0,7 = -0,5Å"0,19 + 0,7 = 0,605
L4
""
)
KL5 = KL6 = -0,5Å"Ã + 0,7 = -0,5Å"0,198 + 0,7 = 0,601
L5
ÅšL4 =ÅšL7 = KL4(à Å"tg Ću + cu )= 0,605Å"(0,19Å"tg 21+ 0,035)= 0,0653MPa
L4
ÅšL5 =ÅšL6 = KL5(à Å"tg Ću + cu )= 0,601Å"(0,198Å"tg 21+ 0,035)= 0,0667MPa
L5
JL4 = JL7 =ÅšL4 Å"bL4 = 0,0653Å"0,45 = 0,0294MN / m
JL5 = JL6 =ÅšL5 Å"bL5 = 0,0667Å"0,35 = 0,0233MN / m
" Normal stresses H to adjacent continuous footings L4, L5, L6 and L7
{the same height of continuous footings h was employed for the whole projection of
foundation}
h=0,3m po= 2,1kN/m2 gop=6,627kN/m2 Å‚g=18kN/m3
{loading of cellars floor po was employed as in residential buildings and floor deadweight gop
was appointed from actual thickness of layers and thickness of these layers was omitted in
further calculations, backfilling soil deadweight Å‚g was appointed based on soil standard}
Normal stress in fields between continuous footings Ão
à = po + gop +Å‚ Å"h = 2,1+ 6,627 +18Å"0,3 =19,317kPa
o g
"
Text provided in {...} is only a comment and it should not appear in the design.
""
K coefficient for shearing stresses under perpendicular continuous footing Ji shall be calculated
according to the formula for K2
3
Fig. 3. Exemplary cross-section of continuous footing and floor.
Normal stresses H1
H1L4 = 0,7 Å"0,85Å"Ã Å"(x1,L4 - x2,L5)= 0,7 Å"0,85Å"19,317 Å" 4,1 = 47,12kN / m
o
H1L5 = 0,7 Å"0,85Å"Ã Å"(x1,L5 - 0)= 0,7 Å"0,85Å"19,317 Å"1,575 =18,10kN / m
o
H1L6 = 0,7 Å"0,85Å"Ã Å"(x1,L6 - 0)= 0,7 Å"0,85Å"19,317 Å"1,075 =12,36kN / m
o
H1L7 = 0,7 Å"0,85Å"Ã Å"(x1,L7 - x2,L6)= 0,7Å"0,85Å"19,317 Å" 4,6 = 52,87kN / m
o
{location of points x1,i and x2,i is appointed always from the half of calculated continuous
footing length in both directions to the end of the element}
Normal stresses H2 (from the condition of not exceeding shearing stresses in fields
between continuous footings)
H 2L4 = (à Å"tg Ću + cu )Å"(x1,L4 - x2,L5)= (19,317 Å"tg 21+ 35)Å" 4,1 =173,9kN / m
o
H 2L5 = (à Å"tg Ću + cu )Å"(x1,L5 - 0)= (19,317 Å"tg 21+ 35)Å"1,575 = 66,8kN / m
o
H 2L6 = (à Å"tg Ću + cu )Å"(x1,L6 - 0)= (19,317 Å"tg 21+ 35)Å"1,075 = 45,6kN / m
o
H 2L7 = (à Å"tg Ću + cu )Å"(x1,L7 - x2,L6)= (19,317 Å"tg 21+ 35)Å" 4,6 =195,11kN / m
o
{if employed backfilling soil will be other than virgin soil then internal friction angle and
cohesion should be changed}
Normal stresses H3 (from the condition of not exceeding limiting soil pressure)
îÅ‚ 1 Ć Ć Å‚Å‚
öÅ‚ ëÅ‚ öÅ‚ öÅ‚
2
Pb = hïÅ‚ëÅ‚ po + gop + h Å"Å‚ Å"tg 45° + + 2Å"c Å"tgëÅ‚45° + =
ìÅ‚ ÷Å‚ ìÅ‚ ÷Å‚ ìÅ‚ ÷łśł
g
2 2 2
íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚
ðÅ‚ ûÅ‚
îÅ‚ 1 21° 21° Å‚Å‚
ëÅ‚45° öÅ‚ öÅ‚
2
= 0,3ïÅ‚ëÅ‚ 2,1+ 6,627 + Å"0,3Å"18öÅ‚Å"tg + + 2Å"35Å"tgëÅ‚45° + = 37,81kN / m
ìÅ‚ ÷Å‚ ìÅ‚ ÷Å‚ ìÅ‚ ÷łśł
2 2 2
íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚
ðÅ‚ ûÅ‚
4
x1,L4 6,025
0,3
h
Kb,L4 = Kb,L7 =1,2 =1,2 = 0,867
x1,L4 6,025
7,7 +
7,7 +
0,3
h
x1,L5 1,575
0,3
h
Kb,L5 =1,2 =1,2 = 0,486
x1,L5 1,575
7,7 +
7,7 +
0,3
h
x1,L6 1,075
0,3
h
Kb,L6 =1,2 =1,2 = 0,381
x1,L6 1,075
7,7 +
7,7 +
0,3
h
H 3L4 = H 3L7 = Pb Å" Kb,L4 = 37,81Å"0,867 = 32,78kN / m
H 3L5 = Pb Å" Kb,L5 = 37,81Å"0,486 =18,38kN / m
H 3L6 = Pb Å" Kb,L6 = 37,81Å"0,381 =14,41kN / m
{in further calculations minimal value of normal stresses H is employed}
HL4 = H = 32,78kN / m
L7
HL5 =18,10kN / m
HL6 =12,36kN / m
" Tensile force in continuous footing on the left side NL
NL1,L = ZL1 + (JL4 + H )Å" sL1,L2 + (J + H )Å" sL1,L2 =
L4 L5 L5
= 215,4 + (29,4 + 32,78)3,3 + (23,3 +18,1)3,3 = 386,31kN
2 2
" Tensile force in continuous footing on the right side NP
NL1,P = ZL1 + (JL6 + H )Å" sL1,L2 + (J + H )Å" sL1,L2 =
L6 L7 L7
= 215,4 + (23,3 +12,36)3,3 + (29,4 + 32,78)3,3 = 376,84kN
2 2
" Tensile force N
NL1,L + NL1,P
386,31+ 376,84
NL1 = = = 381,58kN
2 2
2.1.2. Bending
{bending occurs when direction of horizontal deformations change, change of stresses
direction under continuous footings takes place in the middle of continuous footings A and D
length, along continuous footing 2 axis}
" Shearing stresses J under continuous footing L1
JL1 =ÅšL1 Å"bL1 = 66,5Å"0,5 = 33,25kN / m
" Normal stresses H to continuous footing L1
H1L1 = 0,7 Å"0,85Å"Ã Å"(x1,L1 - x2,L1)= 0,7 Å"0,85Å"19,317 Å"3,3 = 54,18kN / m
o
5
H 2L1 = (à Å"tg Ću + cu )Å"(x1,L1 - x2,L1)= (19,317 Å"tg 21+ 35)Å"3,3 =139,97kN / m
o
x1,L1 3,75
0,3
h
Kb,L1 =1,2 =1,2 = 0,743
x1,L1 3,75
7,7 +
7,7 +
0,3
h
H3L1 = Pb Å" Kb,L1 = 37,81Å"0,743 = 28,08kN / m
" Normal stresses were employed on continuous footing side surface
H = 28,08kN / m
L1
{to calculate continuous footing L1 bending, the biggest distance between supports
transverse continuous footings field C-D was employed, clear distance}
" Moment bending continuous footing L1
JL1 + HL1 33,25 + 28,08
2
M = (lL1/ C-D ) = Å" 4,62 = 81,11kNm
L1-CD
16 16
2.1.3. Eccentric tension
N'L1 = 0,5Å" NL1 = 0,5Å"381,58 =190,79kN
M'L1 = 0,5Å" M = 0,5Å"81,11 = 40,55kNm
L1
2.2. Continuous footing L2
2.2.1. Tension
" Force Z
Coefficient K
KL2 = -0,5Å"Ã + 0,7 = -0,5Å"0,195 + 0,7 = 0,6025
L2
Shearing stress Åš
ÅšL2 = KL2(à Å"tg Ću + cu )= 0,6025(0,195Å"tg 21+ 0,035)= 0,0662MPa
L2
Force Z
ZL2 = 0,5Å" LL2 Å"bL2 Å"ÅšL2 = 0,5Å"12,95Å"0,9Å"0,0662 = 0,3857MN = 385,7kN
" Shearing stresses J under adjacent continuous footings L4, L5, L6 and L7 (as in the
item 2.1.1)
" Normal stresses H to adjacent continuous footings L4, L5, L6 and L7 (between axes
1 and 2 as in the item 2.1.1)
" Normal stresses H to adjacent continuous footings L4 and L7 between axes 2 and 3
Particular normal stresses H
H1L4 = H1L7 = 0,7 Å"0,85Å"Ã Å"(x1,L4 - 0)= 0,7 Å"0,85Å"19,317 Å"6,025 = 69,25kN / m
o
H 2L4 = (à Å"tg Ću + cu )Å"(x1,L4 - 0)= (19,317 Å"tg 21+ 35)Å"6,025 = 255,55kN / m
o
6
x1,L4 6,025
0,3
h
Kb,L4 =1,2 =1,2 = 0,867
x1,L4 6,025
7,7 +
7,7 +
0,3
h
H3L4 = Pb Å" Kb,L4 = 37,81Å"0,867 = 32,78kN / m
" Normal stresses were employed on continuous footing L4 and L7 side surface in
axes 2 and 3.
H = 32,78kN / m
L4
{Continuous footing L2 asymmetrical calculating is required on the left and right}
" Tensile force in continuous footing on the left NL
NL2,L = ZL2 + (JL4 + HL4)Å"(sL1,L2 + sL2,L3)+ (JL5 + HL5)Å" sL1,L2 =
3,3
öÅ‚
= 385,7 + (29,4 + 32,78)ëÅ‚ 3,3 + + (23,3 +18,1)3,3 = 659,2kN
ìÅ‚ ÷Å‚
2 2 2
íÅ‚ Å‚Å‚
" Tensile force in continuous footing on the right NP
NL2,P = ZL2 + (JL6 + HL6)Å" sL1,L2 + (JL7 + HL7 )Å"(sL1,L2 + sL2,L3)=
3,3
öÅ‚
= 385,7 + (23,3 +12,36)3,3 + (29,4 + 32,78)ëÅ‚ 3,3 + = 649,73kN
ìÅ‚ ÷Å‚
2 2 2
íÅ‚ Å‚Å‚
" Tensile force N
NL2,L + NL2,P
659,2 + 649,73
NL2 = = = 654,47kN
2 2
2.2.2. Bending
{there are no impacts on continuous footing L2 perpendicular to axis, therefore there is
no bending and eccentric tension}
2.3. Continuous footing L3
2.3.1. Tension
" Force Z
Coefficient K
KL3 = -0,5Å"Ã + 0,7 = -0,5Å"0,19 + 0,7 = 0,605
L3
Shearing stress Åš
ÅšL3 = KL3(à Å"tg Ću + cu )= 0,605(0,19Å"tg 21+ 0,035)= 0,0653MPa
L3
Force Z
ZL3 = 0,5Å" LL3 Å"bL3 Å"ÅšL3 = 0,5Å"12,95Å"0,5Å"0,0653 = 0,2114MN = 211,4kN
7
" Shearing stresses J under adjacent continuous footings L4 and L7 as in the item
2.1.1
JL4 = JL7 = 29,4kN / m
" Normal stresses H to adjacent continuous footings L4 and L7 as in the item 2.2.1
HL4 = HL7 = 32,78kN / m
{Continuous footing L3 is symmetric can be calculated only on one side}
" Tensile force in continuous footing N
NL3 = ZL3 + (JL4 + HL4)Å" sL2,L3 = 211,4 + (29,4 + 32,78)3,3 = 314,0kN
2
2.3.2. Bending
{due to big distance between continuous footings perpendicular to L3 and too big
bending moment occurring because of it, anchorage tie rods were employed on continuous
footings extension in axes C and B, tie rods transfer only tensile force without bendings,
continuous footing maximal span decreases then to 4,775m tie rod width is not included}
" Shearing stresses J under continuous footing L3
JL3 =ÅšL3 Å"bL3 = 65,3Å"0,5 = 32,65kN / m
" Normal stresses H to continuous footing L3 (employed as in the item 2.1.2)
HL3 = 28,08kN / m
" Moment bending continuous footing L3
JL3 + HL3 32,65 + 28,08
2
M = (lL3 / C-D ) = Å" 4,7752 = 86,54kNm
L3-CD
16 16
2.3.3. Eccentric tension
N'L3 = 0,5Å" NL3 = 0,5Å"314,=157,0kN
M'L3 = 0,5Å" M = 0,5Å"86,54 = 43,27kNm
L3
2.4. Continuous footing L4
2.4.1. Tension
" Force Z
ZL4 = 0,5Å" LL4 Å"bL4 Å"ÅšL4 = 0,5Å"8,5Å"0,45Å"0,0653 = 0,1249MN =124,9kN
" Shearing stresses J under adjacent continuous footings L1 (2.1.2) and L3 (2.3.2)
JL1 = 33,25kN / m
JL3 = 32,65kN / m
8
" Normal stresses H to adjacent continuous footings L1 (2.1.2) and L3 (2.3.2)
HL1 = HL3 = 28,08kN / m
" Tensile force in continuous footing on the left side NL
NL4,L = ZL4 + (JL3 + H )Å" sLA,LB =124,9 + (32,65 + 28,08)4,275 = 254,7kN
L3
2
" Tensile force in continuous footing on the right side NP
NL4,P = ZL4 + (JL1 + HL1)Å" sLA,LB =124,9 + (33,25 + 28,08)4,1 = 250,6kN
2
" Tensile force N
NL4,L + NL4,P
254,7 + 250,6
NL4 = = = 252,7kN
2 2
2.4.2. Bending
" Shearing stresses J under continuous footing L4 (according to 2.1.1)
JL4 = 29,4kN / m
" Normal stresses H to continuous footing L4 (employed as in the item 2.1.1)
H = 32,78kN / m
L4
" Moment bending continuous footing L4
JL4 + HL4 29,4 + 32,78
2
M = (lL4 / 2-3) = Å"3,32 = 42,32kNm
L4-2,3
16 16
2.4.3. Eccentric tension
N'L4 = 0,5Å" NL4 = 0,5Å" 252,7 =126,35kN
M'L4 = 0,5Å" M = 0,5Å" 43,32 = 21,16kNm
L4
2.5. Continuous footing L5
2.5.1. Tension
" Force Z
ZL5 = 0,5Å" LL5 Å"bL5 Å"ÅšL5 = 0,5Å"8,5Å"0,35Å"0,0667 = 0,0992MN = 99,2kN
" Shearing stresses J under adjacent continuous footings L1 (2.1.2)
JL1 = 33,25kN / m
" Normal stresses H to adjacent continuous footings L1 (2.1.2)
H = 28,08kN / m
L1
" Tensile force in continuous footing N
2,65
öÅ‚
NL5 = ZL5 + (JL1 + HL1)Å"(sLA,LB + sLB ,LC )= 99,2 + (33,25 + 28,08)ëÅ‚ 4,1 + = 306,19kN
ìÅ‚ ÷Å‚
2 2
íÅ‚ Å‚Å‚
9
2.5.2. Bending
" Shearing stresses J under continuous footing L5 (according to 2.1.1)
JL5 = 23,3kN / m
" Normal stresses H to continuous footing L5 (employed as in the item 2.1.1)
HL5 =18,10kN / m
" Moment bending continuous footing L5
JL5 + HL5 23,3 +18,1
2
M = (lL5 / 1,2) = Å"3,32 = 28,18kNm
L5-1,2
16 16
2.5.3. Eccentric tension
N'L5 = 0,5Å" NL5 = 0,5Å"306,19 =153,1kN
M'L5 = 0,5Å" M = 0,5Å" 28,18 =14,09kNm
L5
2.6. Continuous footing L6
2.6.1. Tension
" Force Z
ZL6 = 0,5Å" LL6 Å"bL6 Å"ÅšL6 = 0,5Å"8,5Å"0,35Å"0,0667 = 0,0992MN = 99,2kN
" Shearing stresses J under adjacent continuous footings L1 (2.1.2)
JL1 = 33,25kN / m
" Normal stresses H to adjacent continuous footings L1 (2.1.2)
H = 28,08kN / m
L1
" Tensile force in continuous footing N
4,6
öÅ‚
NL6 = ZL6 + (JL1 + HL1)Å"(sLB ,LC + sLC ,LD )= 99,2 + (33,25 + 28,08)ëÅ‚ 2,65 + = 321,52kN
ìÅ‚ ÷Å‚
2 2
íÅ‚ Å‚Å‚
2.6.2. Bending
" Shearing stresses J under continuous footing L6 (according to 2.1.1)
JL6 = 23,3kN / m
" Normal stresses H to continuous footing L6 (employed as in the item 2.1.1)
HL6 =12,36kN / m
" Moment bending continuous footing L6
JL6 + HL6 23,3+12,36
2
M = (lL6 / 1,2) = Å"3,32 = 24,27kNm
L6-1,2
16 16
2.6.3. Eccentric tension
N'L6 = 0,5Å" NL6 = 0,5Å"321,52 =160,76kN
M'L6 = 0,5Å" M = 0,5Å" 24,27 =12,14kNm
L6
10
2.7. Continuous footing L7
2.7.1. Tension
" Force Z
ZL7 = 0,5Å" LL7 Å"bL7 Å"ÅšL7 = 0,5Å"8,5Å"0,45Å"0,0653 = 0,1249MN =124,9kN
" Shearing stresses J under adjacent continuous footings L1 (2.1.2) and L3 (2.3.2)
JL1 = 33,25kN / m
JL3 = 32,65kN / m
" Normal stresses H to adjacent continuous footings L1 (2.1.2) and L3 (2.3.2)
HL1 = HL3 = 28,08kN / m
" Tensile force in continuous footing on the left side NL
NL7,L = ZL7 + (JL3 + HL3)Å" sLC ,LD =124,9 + (32,65 + 28,08)4,775 = 269,89kN
2
" Tensile force in continuous footing on the right side NP
NL7,P = ZL7 + (JL1 + HL1)Å" sLC ,LD =124,9 + (33,25 + 28,08)4,6 = 265,96kN
2
" Tensile force N
NL7,L + NL7,P
269,89 + 265,96
NL7 = = = 267,93kN
2 2
2.7.2. Bending
" Shearing stresses J under continuous footing L7 (according to 2.1.1)
JL7 = 29,4kN / m
" Normal stresses H to continuous footing L7 (employed as in the item 2.1.1)
HL7 = 32,78kN / m
" Moment bending continuous footing L7
JL7 + HL7 29,4 + 32,78
2
M = (lL7 / 1,2) = Å"3,32 = 42,32kNm
L7-1,2
16 16
2.7.3. Eccentric tension
N'L7 = 0,5Å" NL7 = 0,5Å" 267,93 =133,97kN
M'L7 = 0,5Å" M = 0,5Å" 42,32 = 21,16kNm
L7
2.8. Tie rod in axis B SB
2.8.1. Tension
" Shearing stresses J under adjacent continuous footing L3 (2.1.2)
JL3 = 32,65kN / m
11
" Normal stresses H to adjacent continuous footing L3 (2.1.2)
HL3 = 28,08kN / m
" Tensile force ZKB
3
öÅ‚
ZKB = (JL3 + H )Å"(sLA,LB + sLB ,LC )= (32,65 + 28,08)ëÅ‚ 4,675 + = 233,05kN
ìÅ‚ ÷Å‚
L3
2 2
íÅ‚ Å‚Å‚
2.9. Tie rod in axis C SC
2.9.1. Tension
" Shearing stresses J under adjacent continuous footing L3 (2.1.2)
JL3 = 32,65kN / m
" Normal stresses H to adjacent continuous footing L3 (2.1.2)
HL3 = 28,08kN / m
" Tensile force ZKC
4,775
öÅ‚
ZKC = (JL3 + H )Å"(sLB ,LC + sLC ,LD )= (32,65 + 28,08)ëÅ‚ 3 + = 236,09kN
ìÅ‚ ÷Å‚
L3
2 2
íÅ‚ Å‚Å‚
2.10. List of internal impacts
Eccentric tension
Continuous Tension Bending
No.
footings N [kN] M [kNm]
N [kN] M [kNm]
1. L1 381,58 81,11 190,79 40,55
2. L2 654,47 0 327,24 0
3. L3 314,00 86,54 157,00 43,27
4. L4 252,70 42,32 126,35 21,16
5. L5 306,19 28,18 153,10 14,09
6. L6 321,52 24,27 160,76 12,14
7. L7 267,93 42,32 133,97 21,16
8. SB 233,05 - - -
9. SC 236,09 - - -
3. Reinforcement dimensioning
Material parameters steel A-II fyd=310MPa
concrete B20 fcd=10,6MPa Ä…=0,85
N
3.1. Reinforcement for continuous footings tension AsR AsR = rods of diameter 16mm
f
yd
were employed
12
Minimal Employed Employed
Continuous
No. reinforcement number of reinforcement
footings
AsR [cm2] rods AsRr [cm2]
1. L1 12,309 7Ć16 14,074
2. L2 21,112 11Ć16 22,116
3. L3 10,129 6Ć16 12,064
4. L4 8,152 5Ć16 10,053
5. L5 9,877 5Ć16 10,053
6. L6 10,372 6Ć16 12,064
7. L7 8,643 5Ć16 10,053
8. SB 7,518 4Ć16 8,042
9. SC 7,616 4Ć16 8,042
3.2. Reinforcement for continuous footings bending AsM
Employed a=0,065m d=b-a
M
Designing reinforcement sc =
2
Ä… Å" fcd Å" h Å" d
¾ =1- 1- 0,5Å" sc
 =1- 0,5Å"¾
M
AsM =
f Å"Â Å" d
yd
Minimal Employed Employed
Continuous
No. sc [-] ¾ [-] Å› [-] reinforcement number of reinforcement
footing
AsM [cm2] rods AsMr [cm2]
1. L1 0,159 0,040 0,980 6,139 4Ć16 8,042
2. L2 0 0 0 0 0Ć16 0
3. L3 0,169 0,043 0,978 6,559 4Ć16 8,042
4. L4 0,106 0,027 0,987 3,594 2Ć16 4,021
5. L5 0,128 0,033 0,984 3,242 2Ć16 4,021
6. L6 0,111 0,028 0,986 2,786 2Ć16 4,021
7. L7 0,106 0,027 0,987 3,594 2Ć16 4,021
3.3. Reinforcement for eccentric tension AsMR (symmetric reinforcement)
M'
Static eccentric es =
N'
d - a
Distance from the middle of reinforcement eo =
2
13
Distance from the
Continuous Static eccentric middle to
No.
footing es [m] reinforcement eo
[m]
1. L1 0,2125 0,185 Big eccentric
2. L2 0 0,385 -
3. L3 0,2756 0,185 Big eccentric
4. L4 0,1675 0,160 Big eccentric
5. L5 0,0920 0,110 Small eccentric
6. L6 0,0755 0,110 Small eccentric
7. L7 0,1579 0,160 Small eccentric
3.3.1. Case of big eccentric (continuous footings L1, L3 and L4)
Compression zone height
xL1,eff = 0,0706m üÅ‚
ôÅ‚
N'
xeff = xL3,eff = 0,0581m < 2Å" a = 2Å"0,065 = 0,13m
żł
Ä… Å" fcd Å"h
xL4,eff = 0,0467môÅ‚
þÅ‚
Tensile reinforcement
N'Å"(es - eo )
AsMR =
1
f Å"(d - a)
yd
AsMR = 0,457cm2 1Ć16 AsMRr = 2,011cm2
1,L1 1,L1
AsMR =1,240cm2 employed 1Ć16 AsMRr = 2,011cm2
1,L3 1,L3
AsMR = 0,096cm2 1Ć16 AsMRr = 2,011cm2
1,L4 1,L4
3.3.2. Case of small eccentric (continuous footings L5, L6 and L7)
- N'Å"(es - eo )
AsMR =
1
f Å"(d - a)
yd
AsMR = 0,404cm2 1Ć16 AsMRr = 2,011cm2
1,L5 1,L5
AsMR = 0,813cm2 employed 1Ć16 AsMRr = 2,011cm2
1,L6 1,L6
AsMR = 0,028cm2 1Ć16 AsMRr = 2,011cm2
1,L7 1,L7
3.4. List of reinforcement
Continuous
No. Tension Bending Eccentric tension
footing
1. L1
7Ć16 4Ć16 1Ć16
2. L2
11Ć16 0Ć16 0Ć16
3. L3
6Ć16 4Ć16 1Ć16
4. L4
5Ć16 2Ć16 1Ć16
5. L5
5Ć16 2Ć16 1Ć16
6. L6
6Ć16 2Ć16 1Ć16
7. L7
5Ć16 2Ć16 1Ć16
8. SB - -
4Ć16
9. SC - -
4Ć16
14
3.5. Length of rods anchorage
fbd =2,0 according to the table 26 Polish Standard
f
Ć 0,016 310
yd
lb = = = 0,62m
4 fbd 4 2
{Tie rods of square cross-section with reinforcement distributed evenly on the
perimeter}
15
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