Calculation of foundation membrane

1. Determining of internal forces in membrane

Membrane thickness:

10cm

Assumed slip layer made from two layers of building paper with powdered graphite

- friction coefficient

f

0.2

:=

2.1 Longitudinal direction

2.1.1 Continuous footing strip L1

- force Z

Z

L1

191.05kN

:=

- self weight of wall

g

oL1

74

kN

m

:=

- length of wall

l

L1

17.4m

:=

T

L1

g

oL1

0.5

⋅

l

L1

⋅

f

⋅

128.76 kN

⋅

=

:=

<

Z

L1

191.05 kN

⋅

=

2.1.2 Continuous footing strip L2

- force Z

Z

L2

292.42kN

:=

- self weight of wall

g

oL2

114

kN

m

:=

- length of wall

l

L2

17.4m

:=

T

L2

g

oL2

0.5

⋅

l

L2

⋅

f

⋅

198.36 kN

⋅

=

:=

<

Z

L2

292.42 kN

⋅

=

2.1.3 Continuous footing strip L3

- force Z

Z

L3

191.05kN

:=

- self weight of wall

g

oL3

74

kN

m

:=

- length of wall

l

L3

17.4m

:=

T

L3

g

oL3

0.5

⋅

l

L3

⋅

f

⋅

128.76 kN

⋅

=

:=

<

Z

L3

191.05 kN

⋅

=

2.1.4 Continuous footing strip L4

- force Z

Z

L4

191.05kN

:=

- self weight of wall

g

oL4

74

kN

m

:=

- length of wall

l

L4

17.4m

:=

T

L4

g

oL3

0.5

⋅

l

L3

⋅

f

⋅

128.76 kN

⋅

=

:=

<

Z

L4

191.05 kN

⋅

=

2.1.5 Strip between continuous footing L1/L2

- force H and J

J

L6

33.61

kN

m

:=

H

L6

13.32

kN

m

:=

J

L7

33.61

kN

m

:=

H

L7

13.32

kN

m

:=

- self weight of walls

g

oL6

114

kN

m

:=

g

oL7

114

kN

m

:=

- length of wall between L1 and L2

l

L6

3.5m

:=

l

L7

3.5m

:=

T

L1.2

g

oL6

g

oL7

+

(

)

l

L6

f

⋅

159.6 kN

⋅

=

:=

<

H

L6

J

L6

+

(

)

l

L6

⋅

H

L7

J

L7

+

(

)

l

L7

⋅

+

328.51 kN

⋅

=

Calculation of foundation membrane

2.1.6 Strip between continuous footing L2/L3

- force H and J

J

L6

33.61

kN

m

:=

H

L6

13.32

kN

m

:=

J

L7

33.61

kN

m

:=

H

L7

13.32

kN

m

:=

- self weight of walls

g

oL6

114

kN

m

:=

g

oL7

114

kN

m

:=

- length of wall between L2 and L3

l

L6

4.5m

:=

l

L7

4.5m

:=

T

L2.3

g

oL6

g

oL7

+

(

)

l

L6

f

⋅

205.2 kN

⋅

=

:=

<

H

L6

J

L6

+

(

)

l

L6

⋅

H

L7

J

L7

+

(

)

l

L7

⋅

+

422.37 kN

⋅

=

2.1.7 Strip between continuous footing L3/L4

- force H and J

J

L6

33.61

kN

m

:=

H

L6

13.32

kN

m

:=

J

L7

33.61

kN

m

:=

H

L7

13.32

kN

m

:=

- self weight of walls

g

oL6

114

kN

m

:=

g

oL7

114

kN

m

:=

- length of wall between L3 and L4

l

L6

2.6m

:=

l

L7

2.6m

:=

T

L3.4

g

oL6

g

oL7

+

(

)

l

L6

f

⋅

118.56 kN

⋅

=

:=

<

H

L6

J

L6

+

(

)

l

L6

⋅

H

L7

J

L7

+

(

)

l

L7

⋅

+

244.036 kN

⋅

=

2.2 Transverse direction

2.2.1 Continuous footing strip L5 and L8

- force Z

Z

L4

136.15kN

:=

- self weight of wall

g

oL4

74

kN

m

:=

- length of wall

l

L4

12.4m

:=

T

L4

g

oL4

0.5

⋅

l

L4

⋅

f

⋅

91.76 kN

⋅

=

:=

<

Z

L4

136.15 kN

⋅

=

2.2.2 Continuous footing strip L6 and L7

- force Z

Z

L5

208.39kN

:=

- self weight of wall

g

oL5

114

kN

m

:=

- length of wall

l

L5

12.4m

:=

T

L5

g

oL5

0.5

⋅

l

L5

⋅

f

⋅

141.36 kN

⋅

=

:=

<

Z

L5

208.39 kN

⋅

=

Calculation of foundation membrane

2.2.3 Strip between continuous footing L5/L6 and L7/L8

- force H and J

J

L1

21.96

kN

m

:=

H

L1

20.86

kN

m

:=

J

L2

33.61

kN

m

:=

H

L2

12.22

kN

m

:=

- self weight of walls

g

oL1

74

kN

m

:=

g

oL1

114

kN

m

:=

- length of wall between L5 and L6

l

L1

5.17m

:=

l

L1

5.17m

:=

T

L5.6

g

oL1

( )

l

L1

f

⋅

117.876 kN

⋅

=

:=

<

H

L1

J

L1

+

(

)

l

L1

⋅

221.379 kN

⋅

=

2..4 Strip between continuous footing L6/L7

J

L1

21.96

kN

m

:=

H

L1

20.86

kN

m

:=

J

L2

33.61

kN

m

:=

H

L2

12.22

kN

m

:=

- force H and J

- self weight of walls

g

oL1

74

kN

m

:=

g

oL1

114

kN

m

:=

- length of wall between L5 and L6

l

L1

5.1m

:=

l

L1

5.1m

:=

T

L6.7

g

oL1

( )

l

L1

f

⋅

116.28 kN

⋅

=

:=

<

H

L1

J

L1

+

(

)

l

L1

⋅

218.382 kN

⋅

=

2.3 Reinforcement in strip between continuous footing

- steel A-0

f

yd

190MPa

:=

- diameter of bar for
strip between constiuous footing

ϕ

1

16mm

:=

A

b1

ϕ

1

2

π

⋅

4

2.011 cm

2

⋅

=

:=

- diameter of bars for continuous
footing strip

ϕ

2

8mm

:=

A

b2

ϕ

2

2

π

⋅

4

0.503 cm

2

⋅

=

:=

Continuous

footing

Tensile

force [kN]

Minimum

reinfocement [cm2]

Employed

no. of rods

Employed

reinforcement [cm2]

L1, L3, L4

128,76

6,78

4

Φ

16

8,04

L2

198,36

10,44

6

Φ

16

12,07

L1/L2

159,60

8,40

17

Φ

8

8,55

L2/L3

205,20

10,80

22

Φ

8

11,07

L3/L4

118,56

6,24

13

Φ

8

6,54

L5, L8

91,76

4,83

3

Φ

16

6,03

L6, L7

141,36

7,44

4

Φ

16

8,04

L5/L6, L7/L8

117,88

6,20

13

Φ

8

6,54

L6/L7

116,28

6,12

13

Φ

8

6,54

2.4 Length of bar anchorage

f

bd

1.1MPa

:=

l

b

ϕ

1

f

yd

⋅

4 f

bd

⋅

0.691 m

=

:=

for bar 16mm

l

b

ϕ

2

f

yd

⋅

4 f

bd

⋅

0.345 m

=

:=

for bars 8mm