thrust anchor blocks for vertically inclined hrizontal bend

DESIGNED BY

DESIGN INFORMATION

Properties of Pipes

Internal diameter of pipe D = mm

Internal area of pipe A = πD2/4

= m2

Hydrostatic test pressure (Ultimate) = bar

Hydrostatic test pressure (SLS) = bar

Required maximum clearance in X dirn Lx = m

Required maximum clearance in Y dirn Ly = m

Required maximum clearance in Z dirn Lz = m

Required maximum length of inclined pipe L = √[Lz2 + Ly

2 + Lx2]

= m

Minimum required Bend rotation angle min = tan‐1(Lz/Ly)

= °

Let take Bend rotation angle as = °

Minimum required angle of bend min = tan‐1(√[Lz2 + Ly

2]/Lx)

= °

Let take angle of Bend as = °

Properties of Soil

Density of compacted soil (dry) d = kN/m3

Angle of internal friction = °

Soil cover above the base footing h1 = m

Passive earth pressure coefficient Kp = (1 + sin)/(1 ‐ sin)=

Active earth pressure coefficient Ka = (1 ‐ sin)/(1 + sin)=

Bearing capacity of soil = kN/m2

Factor of safety against thrust F.O.S =

Properties of Concrete

Density of concrete conc = kN/m3

Soil ‐ concrete interface friction angle = °

6.66667

45

24

20

REF

0.8

3

150.00

1.5

0.33333

30

SYSTEM REHABILITATION FOR NRW REDUCTION IN EAST PART OF THE COLOMBO CITY PAKAGE ‐ 02

18

1

1.5

BS 8110 & BS 8004 KUSHAN CHECKED BY ANRM 2016‐03‐17

Calculations Output

600

0.28274

10

45

1

45

43.3139

2.06

PROJECT

STRUCTURE SUPPORTS FOR CULVERT CROSSING TYPE E ELEMENT VERTICALLY INCLINED HRIZONTAL BEND

CODES

Reference

DESIGNED BY

REF



Calculations Output

PROJECT


CODES

Reference

ANALYSIS OF CULVERT CROSSING TYPE E

Force acting in X direction in bend Px0 = 2pA(sin/2)2

= kN

Force acting in Y direction in bend Py0 = pA(sin)(cosβ)= kN

Force acting in Z direction in bend Pz0 = pA(sin)(sinβ)= kN

Apparent Horizontal angle θH = tan‐1(Ly/Lx)

= °

Moment about Z1 axis of the base Mz1 = Py0Lx + Px0Ly

= kNm

Moment about Y1 axis of the base My1 = Pz0√[Lx2 + Ly

2]

+ [Px0(cosθH) ‐ Py0(sinθH)]Lz

= kNm

Moment about X1 axis of the base Mx1 = [Px0(sinθH) + Py0(cosθH)]Lz

= kNm

FORCES ACTING ON COLUMN SUPPORTS (SLS)

Px1 = Py0(sinθH) ‐ Px0(cosθH)

= kN

Py1 = Px0(sinθH) + Py0(cosθH)

= kN

Pz1 = Pz0

= kN

55.2091

196.581

33.6901

94.2478

94.2478

163.565

109.043

6.343

109.043

94.2478

DESIGNED BY

REF



Calculations Output

PROJECT


CODES

Reference

DESIGN OF COLUMN SUPPORTS

Properties of Pipes

Thickness of Pipe t = mm

Density of Pipe Material pipe = kN/m3

Effective pipe length over culvert crossing L0 = m

Effective pipe length for column support Leff = (L + L0)/2

= m

Total weight of pipe over effective length Wtotal = pipeD2/4 + waterD + t)t]Leff

= kN

Column typical dimensions

Depth of column section (Y1 dir) = mm

Width of column section (X1 dir) = mm

Height of short column = mm

Height of tall column = mm

Short Column (SLS)

Axial Force = kN

Bending moment about major axis (x‐x) = kNm

Bending moment about minor axis (y ‐y) = kNm

Tall Column (SLS)

Axial Force = kN

Bending moment about major axis (x‐x) = kNm

Bending moment about minor axis (y ‐y) = kNm

(Sign Convention : Compression +ve)

Short Column (ULS)

Axial Force = kN

Bending moment about major axis (x1 ‐ x1) = kNm

Bending moment about minor axis (y1 ‐ y1) = kNm

Shear force in x1 direction = kN

Shear force in y1 direction = kN

Tall Column (ULS)

Axial Force = kN

Bending moment about major axis (x1 ‐ x1) = kNm

Bending moment about minor axis (y1 ‐ y1) = kNm

Shear force in x1 direction = kN

Shear force in y1 direction = kN

(Sign Convention : Compression +ve)

‐100.4

261.704

15.2221

10.1481

174.469

179.157

87.2347

5.07403

h 900

b 600

500

1500

25

5.03078

13.7777

8

7.95

163.565

54.5217

3.17127

‐61.03

9.51381

114.505

10.1481

174.469

DESIGNED BY

REF



Calculations Output

PROJECT


CODES

Reference

DESIGN INFORMATION OF SHORT COLUMN

Material Properties

Characteristic strength of concrete, = N/mm2

Characteristic strength of main r/f = N/mm2

Characteristic strength of links = N/mm3

Density of concrete, = kN/m3

Nominal maximum size of aggregate = mm

Exposure condition =

Fire resistance = hours

Exposed % of column =


Depth of column section = mm

Width of column section = mm

Floor ‐ to ‐ Floor hight = mm

For about X‐X(Major axis) and in Y‐Y direction

Condition of top end of column =

Top beam depth in Y‐Y direction = mm

Condition of bottom end of column =

Stability of Column =

Clear height between end restrains X‐X = mm

For about Y‐Y(Minor axis) and in X‐X direction


Top beam depth in X‐X direction = mm



Clear height between end restrains Y‐Y = mm

Reinforcement Properties

Diameter of main bars = T bars

Diameter of links = R bars

DESIGN FORCES

Applied direct load = kN

Larger moment about X1 axis = kNm

Smaller moment about X1 axis = kNm

Larger moment about Y1 axis = kNm

Smaller moment about Y1 axis = kNm

Shear Force in Y1 direction = kN

Shear Force in X1 direction = kN

M2y1 5.07403

M1y1 0

Vy1

174.469Vx1

10.1481

BS8110:1985

fcu 25

Table 3.1 fy 460

fyv 250

ρct 24

hagg 20

Durability Requirement

Mild

2

Fully exposed

h

Unbraced

lox 500

Monolithic Connection

0

Moment Connection to Foundation

Braced

900

b 600

500


0


loy 500

12

10

N 179.157

M2x1 87.2347

M1x1 0

DESIGNED BY

REF



Calculations Output

PROJECT


CODES

Reference

CHECK SLENDERNESS OF SHORT COLUMN

End Condition of Column

Top for Y‐Y direction =

Bottom for Y‐Y direction =

Top for X‐X direction =

Bottom for X‐X direction =

Values of β for both axes

For Y‐Y direction about X‐X axis βx =

For X‐X direction about Y‐Y axis βy =

= βxlox

= 1.2x500

= mm

= βyloy

= 0.75x500

= mm

For about X‐X axis lex/h = 600/900

=

< For Unbraced Column

→

For about Y‐Y axis ley/b = 375/600

=

< For Braced Column

→

DETERMINATION OF COVER

Grade of concrete =


Fire resistance = hours → Cover <

Nominal maximum size of aggregate = mm → Cover <

Maximum bar size = mm → Cover <

Minimum nominal cover = mm

Check for minimum dimension of column for fire resistance

→ OK

= h ‐ cover ‐ dia of link ‐ dia of bar/2

= 900 ‐ 25 ‐ 10 ‐ 12/2

= mm

= b ‐ cover ‐ dia of link ‐ dia of bar/2

= 600 ‐ 25 ‐ 10 ‐ 12/2

= mm

1

1

1

1.20

0.75

BS8110:1985

Part 1 1

Table 3.21 &

3.22

Clause 3.8.1.6 Effective height of column for about X‐

X axis

lex

600

Clause 3.8.1.6 Effective height of column for about Y‐

Y axis

ley

375

Clause 3.8.1.3

12

25

25

2

20

0.66667

10

Short

Clause 3.8.1.3

0.625

10

Short

Hence the column should design as Short column

Table 3.4 → Cover < 25Mild

Table 3.5 25

Clause 3.3.1.3 20

Clause 3.3.1.2 12

Cover = 25mm

Figure 3.2Requirement for minimum dimension of Fully

exposed column for 2 hours fire resistance= 300 mm

=<Minimum dimension of column

(=600mm)

OK

h'

h' = 859mm

b'

b' = 559mm

859

559

DESIGNED BY

REF



Calculations Output

PROJECT


CODES

Reference

DESIGN OF SHORT COLUMNS

Minimum eccentricity

emin,x = Lesser of (0.05h or 20mm)

= mm → Use 20 mm

emin,y = Lesser of (0.05b or 20mm)

= mm → Use 20 mm

Minimum moment X‐X = N x emin,x

= kNm ≤

Minimum moment Y‐Y = N x emin,y

= kNm ≤

For column subjected to moments and direct load

N/(bhfcu) = 179.16x10^3/(600x900x25)

=

β =

Mx1/h' = 87.23/859

=

My1/b' = 5.07/559

=

Hence Mx1/h' ≥ My1/b' Therefore Mx1' = Mx1 + β(h'/b')My1

= 87.23 + 0.98(859/559)x5.07

= kNm

Hence Mx1/h' < My1/b' Therefore My1' = My1 + β(b'/h')Mx1

= 5.07 + 0.98(559/859)x87.23

= kNm

N/(bh) = 179.16x10^3/(600x900)

=

M/(bh2) = 94.91x10^6/(600x900^2)

=

d/h = 859/900

=

100Asc/bh =

Asc = 0.4bh/100

= 0.4x600x900/100

= mm2

or 4 no of bars which ever greater

Minimum reinforcement

Required minimum compression r/f Asc, min = 0.4xbxh/100

= 0.4x600x900/100

= mm2

Clause 3.8.2.4

0.95

0.20

0.4

2160

94.9077

60.9386

Clause 3.8.4.50.33

Table 3.24

0.98

Clause 3.8.4.50.10155

0.00908

Clause 3.12.5

2160

Moment due maximum excentricity for

both axesemin = 0.05xh or 0.05xb

45

3.58314 Mxx

Significant moment present about x‐x axis

Myy

Significant moment present about x‐x axis

Column has significant moment at least in one axes

30

3.58314

0.01327

DESIGNED BY

REF



Calculations Output

PROJECT


CODES

Reference

Maximum reinforcement

For vertical cast column Asc, max = 6xbxh/100

= 6x600x900/100

= mm2

Hence required r/f area Asc = mm2

Required No of bars = bars

Other than 4 corner bars in column arrangement of main r/f for bars given bellow

No of bars on one side in short face of column =

No of bars on one side in long face of column =

Hence Provide no of bars = bars

= mm

Minimum C/C spacing of bars = (600 ‐ 25x2 ‐ 12x3 ‐ 10x2)/4

= mm → OK

Maximum allowable clear spacing = 47000/(5fy/8)

= mm

Maximum C/C spacing of bars = (900 ‐ 25x2 ‐ 12x5 ‐ 10x2)/6

= mm → OK

Design for shear

Mx1/N = (87.23/179.16)x1000

= mm < 0.75h

My1/N = (5.07/179.16)x1000

= mm < 0.75b

vx1 = Vx1/bh'

= (174.47x1000)/(600x859)

= N/mm2

vy1 = Vy1/b'h

= (10.15x1000)/(900x559)

= N/mm2

Maximum of 0.8√fcu or 5 N/mm2 = N/mm2

Containment of reinforcement

Minimum diameter of link = Maximum of (0.25xdia or 6mm)

= mm → OK

Maximum spacing of links = 12 x dia of bar

= mm

Hence provide shear links of @ mm

0.34

Minimum allowable clear distance between bars20

163.478

OK

3

5

OK

OK

4.00

Hence shear check is not required

20

486.92

Clause 3.12.6.2

32400

2160

20

128.333

123.50

BS8110:Part 3 :

1985 Clause

3.8.4.6

0.02

28.32

BS8110:Part 3 :

1985 Clause

3.12.7.1 6

144

R10 100

DESIGNED BY

REF



Calculations Output

PROJECT


CODES

Reference

DESIGN INFORMATION OF LONG COLUMN

Material Properties


Characteristic strength of main r/f = N/mm2

Characteristic strength of links = N/mm3


Nominal maximum size of aggregate = mm


Fire resistance = hours

Exposed % of column =


Depth of column section = mm

Width of column section = mm

Hight of column = mm

For about X‐X(Major axis) and in Y‐Y direction


Top beam depth in Y‐Y direction = mm



Clear height between end restrains X‐X = mm

For about Y‐Y(Minor axis) and in X‐X direction


Top beam depth in X‐X direction = mm



Clear height between end restrains Y‐Y = mm


Diameter of corner bars = T bars

Diameter of other main bars = T bars

Diameter of links = T bars

DESIGN FORCES

Applied direct load = kN

Larger moment about X1 axis = kNm

Smaller moment about X1 axis = kNm

Larger moment about Y1 axis = kNm

Smaller moment about Y1 axis = kNm

Shear Force in Y1 direction = kN

Shear Force in X1 direction = kN

16

Vx1

10.1481

M2x1 261.704

M1x1 0

M2y1 15.2221

M1y1 0

Vy1

174.469

0


Braced

loy 1500

20

10

N ‐100.4

600

1500


0


Unbraced

lox 1500


BS8110:1985

fcu 25

Table 3.1 fy 460

fyv 460

ρct 24

hagg 20


Mild

2

Fully exposed

h 900

b

DESIGNED BY

REF



Calculations Output

PROJECT


CODES

Reference

DETERMINATION OF COVER

Grade of concrete =


Fire resistance = hours → Cover <

Nominal maximum size of aggregate = mm → Cover <

Maximum bar size = mm → Cover <

Minimum nominal cover = mm

Check for minimum dimension of column for fire resistance

→ OK

h' = h ‐ cover ‐ dia of link ‐ dia of bar/2

= 900 ‐ 25 ‐ 10 ‐ 20/2

= mm

b' = b ‐ cover ‐ dia of link ‐ dia of bar/2

= 600 ‐ 25 ‐ 10 ‐ 20/2

= mm

ax = h ‐ 2(h ‐ h')

= 900 ‐ 2(900 ‐ 855)

= mm

ay = b ‐ 2(b ‐ b')

= 600 ‐ 2(600 ‐ 555)

= mm

For Short face of column Astx = M2xx/(0.87fyax)

= 261.7x10^6/(0.87x460x810)

= mm2

Provide 5 no of bars on each short face

Required area of each bar = mm2

For Long face of column Asty = M2yy/(0.87fyay)

= 15.22x10^6/(0.87x460x810)

= mm2

Provide 7 no of bars on each long face


Requiered area of steel for tension = N/0.87fy

= 100.4x10^3/0.87x460

= mm2

This area can be divided over the total no of 4 corner bars in the column.


Hence area of corner bar = mm2

Hence select 4 No of T20 bars corners and 16 No of T16 bars for all other

reinforcements. Arrangement of r/f is same as given in short column

161.465

10.6544

62.7209

810

74.5808

250.884

235

510

807.325

OK

855 h' = 855mm

555 b' = 555mm

25 Cover = 25mm

Figure 3.2Requirement for minimum dimension of Fully

exposed column for 2 hours fire resistance= 300 mm

=<Minimum dimension of column

(=600mm)

Table 3.5 2 25

Clause 3.3.1.3 20 20

Clause 3.3.1.2 20 20

Table 3.425

→ Cover < 25Mild

DESIGNED BY

REF



Calculations Output

PROJECT


CODES

Reference

Design for shear

vx1 = Vx1/bh'

= N/mm2

= (100x1232)/(600x855)

= < 3

→

= 400/855

= < 1

→ Not Ok use value 1

Design shear stress, vcx = 0.79[100As,x/bh']1/3[400/h']1/4[fcu/25]

1/3/γm= {0.79[0.2]^(1/3)x[1]^(1/4)x[25/25]^(1/3)}/1.25

= N/mm2

vy1 = Vy1/b'h

= N/mm2

= (100x1634)/(555x900)

= < 3

→

= 400/555

= < 1

→ Not Ok use value 1

Design shear stress, vcy = 0.79[100As,y/b'h]1/3[400/b']1/4[fcu/25]

1/3/γm= {0.79[0.3]^(1/3)x[1]^(1/4)x[25/25]^(1/3)}/1.25

= N/mm2

Vx1h/Mx1 = < 1

→ Ok

Vcx' = vcx + 0.75 (N/Acx)(Vx1h/Mx1)

= ≤ Min(0.8√fcu or 5 N/mm2)

= N/mm2

Vy1b/My1 = < 1

→ Ok

Vcy' = vcy + 0.75 (N/Acy)(Vy1b/My1)

= ≤ Min(0.8√fcu or 5 N/mm2)

= N/mm2

(vx/vcx') + (vy/vcy') =

> 1

vcx'' = vcx'vx1/(vx1 + vy1)

= N/mm2

vcy'' = vcy'vy1/(vx1 + vy1)

= N/mm2

BS8110:1985

Part ‐ 1

Table 3.9

Conditions to

Satisfy

100As,y/b'h

Ok

400/b'

0.72072

BS8110:1985

Part ‐ 1

Table 3.9 0.44

0.40

0.37919

0.29146

0.021

1.15

Shear reinforcement in the form of links is required

BS8110:Part 3 :

1985 Clause

3.8.4.6

0.02

0.38

0.60

0.32705

BS8110:1985

Part ‐ 1

Table 3.9

Conditions to

Satisfy

100As,x/bh'

Ok

400/h'

0.46784

BS8110:1985

Part ‐ 1

Table 3.9 0.39

0.30887

0.31

0.34

0.24006

DESIGNED BY

REF



Calculations Output

PROJECT


CODES

Reference

Assume T10 mm dia bars used as links for the column

Area of two legs Asv = mm2

Required Maximum spacing in x1 dirn svx ≤ 0.87fyvAsv/[b(vx1 ‐ vcx'')]

≤ mm

Required maximum spacing in y1 dirn svy ≤ 0.87fyvAsv/[h(vy1 ‐ vcy'')]

≤ mm

Minimum reinforcement

Reinforcement provided = mm2

→ >

→

Maximum reinforcement

Reinforcement provided = mm2

→ <

→

Containment of reinforcement

All reinforcement in tension. Containment rules do not apply

Rules for minimum shear r/f in beams should apply.

S > 0.87fyvAsv/0.4b

>

Hence select link spacing as = mm

Check spacing of bars for crack width

Maximum size of aggregates = mm

Diameter of bar = mm

= mm

Minimum clear spacing of bars in column = (600 ‐ 16x3 ‐ 20x2 ‐ 2x10 ‐ 2x25)/4

= mm

→ OK

Assumed service stress fs = 5fy/8

Maximum allowable clear spacing of bars = 47000/(5fy/8)

= mm

Maximum clear spacing of bars in column = (900 ‐ 16x5 ‐ 20x2 ‐ 2x10‐ 2x25)/6

= mm

→ OK

Minimum alowable clear distance between bars

OK

OK

20

20

110.5

163.478

118.333

OK

4473.63

0.83% 6%

OK OK

261.93

250

hagg 20

157

2154.27

65993.4

4473.63

0.83%

OK

0.40%

DESIGNED BY

REF



Calculations Output

PROJECT


CODES

Reference

DESIGN INFORMATION OF BASE FOOTING

Material Properties


Characteristic strength of Main r/f = N/mm2


Elastic modulus of Soil = kN/m2

Elastic modulus of Concrete = kN/mm2

Maximum size of coarse aggregate = mm

Bearing Capacity Of Soil = kN/m2


Clear cover to top r/f, = mm

Clear cover to bottom r/f, = mm

Strip Footing Section Properties

b =

Neutral Axis depth from bottom face = mm

Second moment of area of the section = mm4

Maximum spacing of adjacent columns = mm

Required clearence from column faces = mm

Length of base a = mm


Reinforcement design width, = mm

Longitudinal direction top r/f = T bars

Longitudinal direction bottom r/f = T bars

Transverse direction r/f = T bars

DESIGN FORCES

For ULS (X1Z1 PLANE)

Max hogging moment = kNm

Max sagging moment = kNm

Max Shear force = kN

Max Soil pressure = kN/m2

For SLS (X1Z1 PLANE)

Max settlement = mm


For ULS (Y1Z1 PLANE)

Twisting moment due to column bending = kNm

For SLS (Y1Z1 PLANE)

Twisting moment from whole unit = kNm


h =

244.8

150

5.088

76.33

RESULTS FROM

PROKON

OUTPUT

100

2000

150


261.704

109.043

ØT 16

47.7

114

ØLT 20

ØLB 20

400

3500

BS8007:1987

Clause 2.7.6

Severe

50

75

Table 3.1 fy 460

BS8110:1985

fcu 35

ρct 24

Esoil 15000

Ec 26.5675

hagg 20

200

1.1E+10

1803

bt 1000

500

DESIGNED BY

REF



Calculations Output

PROJECT


CODES

Reference

DESIGN OF REINFORCEMENT

Reinforcement to Carry Bending Moment at ULS

Longitudinal direction Top R/F

Bending moment in ULS = kNm

Assume T20 reinforcement bars provided in longitudinal direction

Effective depth, = h ‐ top cover ‐ ØLT/2 ‐ ØT

= 400 ‐ 50 ‐ 20/2 ‐ 16

= mm

= Mu/bd2fcu

= (47.7x10^6)/(2000x324^2x35)

=

= [0.5+(0.25‐K/0.9)1/2]d

= [0.5+(0.25‐0.006/0.9)^0.5]d

= d

= mm

Neutral axis depth = (d ‐ z)/0.45

= (324 ‐ 307.8)/0.45

= mm

As, reqd = M/0.87fyZ

= (47.7x10^6)/(0.87x460x308)

= mm2

Assume half of required steel area for torsion r/f = mm2

As, min = × bh/100

= 0.13x2000x400/100

= mm2

No of Bars Required = bars

Provided r/f area As, prov = π(ØLT/2)^2 × No of bars

= π(20/2)^2 × 15

= mm2

= mm

Minimum spacing of r/f = hagg + 5 mm

= 20 + 5

= mm

1

βb

1

1

= N/mm2

Maximum spacing of r/f = Min of (47000/fs or 300)

= mm

112

Required steel area to carry bending

moment,

4309

Top R/F for

Longitudinal

direction,

select

15/T20mm

diameter bars

23.6126

300

Provided Reinfocement and spacing are adequate

=5x460x387

×8x4712

BS8110:1985

Clause

3.12.11.1 25

Design service stress in tension r/f fs =5fyAs, req

4712

Clear Spacing =2000 ‐ (50x2) ‐ (20x15) ‐ (16x2)

14

387

×8As, prov

d

324

BS8110:1985

Clause 3.4.4.4

k

BS8110:1985

Table 3.27

Minimum r/f required in rectangular

beams for crack control

0.13

Limiting Z=0.95d, Z 308

x

36

0.006

K < 0.156, and compression r/f are not required

BS8110:1985

Clause 3.4.4.4

Z

0.99

Mu 47.7

1040

15

DESIGNED BY

REF



Calculations Output

PROJECT


CODES

Reference

Longitudinal direction Bottom R/F

Bending moment in ULS = kNm

Assume T reinforcement bars provided in longitudinal direction

Effective depth, = h ‐ bottom cover ‐ ØLB/2 ‐ ØT

= 400 ‐ 75 ‐ 20/2 ‐ 16

= mm

= Mu/bd2fcu

= (114x10^6)/(2000x299^2x35)

=

= [0.5+(0.25‐K/0.9)1/2]d

= [0.5+(0.25‐0.018/0.9)^0.5]d

= d

= mm


= (114x10^6)/(0.87x460x284)

= mm2

Assume half of required steel area for torsion r/f = mm2

As, min = × bh/100

= 0.13x2000x400/100

= mm2

No of Bars Required = bars

Provided r/f area As, prov = π(ØLT/2)^2 × No of bars

= π(20/2)^2 × 17

= mm2

= mm

Minimum spacing of r/f = hagg + 5 mm

= 20 + 5

= mm

1

βb

1

1

= N/mm2

Maximum spacing of r/f = Min of (47000/fs or 300)

= mm

×

8As, prov

=5x460x1003

8x5341

95.5


moment,

1003

d

299

4309

×

53.9904

BS8110:1985

Clause

3.12.11.2.4

300 Bottom R/F for

Longitudinal

direction,

select

17/T20mm

diameter bars

Provided Reinfocement and spacing are adequate

BS8110:1985

Clause

3.12.11.1 25

Design service stress in tension r/f fs =5fyAs, req

1040

17

5341

Clear Spacing between adjacent r/f =2000 ‐ (50x2) ‐ (20x17) ‐ (16x2)

16

BS8110:1985

Table 3.27

Minimum r/f required in rectangular

beams for crack control

0.13


BS8110:1985

Clause 3.4.4.4

Z

0.98


BS8110:1985

Clause 3.4.4.4

k

0.018

Mu 114

DESIGNED BY

REF



Calculations Output

PROJECT


CODES

Reference

Transverse direction Bottom R/F

= 150x(2000/2)^2/(2x10^6)

= kNm/m width

Assume T16 reinforcement bars provided in Transverse direction

Effective depth, =

= 400 ‐ 75 ‐ 16/2

= mm

= Mu/bd2fcu

= (75x10^6)/(1000x317^2x35)

=

= [0.5+(0.25‐K/0.9)1/2]d

= [0.5+(0.25‐0.021/0.9)^0.5]d

= d

= mm


= (75x10^6)/(0.87x460x301)

= mm2 /m width

Ast, min = × hfl/100

= 0.15x400x1000/100

= mm2 /m width

Shear Resistance of base

Ultimate design shear force at support, = kN

Shear stress, = Vu/bvd

= (244.8x10^3)/(2000x299)

= N/mm2

Minimum of(0.8√fcu or 5 N/mm2) = N/mm3 >

→

= (100x5341)/(2000x299)

= < 3

→

= 400/299

= > 1

→ Ok

Design shear stress, vc = 0.79[100As/bd]1/3[400/d]1/4[fcu/25]

1/3/γm= {0.79[0.89]^(1/3)x[1.34]^(1/4)x[35/25]^(1/3)}/1.25

= /mm2

Shear enhancement factor for a section at a distance d from support,

= 2d/av

= 2

Ok

Vu 244.8


moment,

BS8110:1985

Table 3.90.73

BS8110:1985

Clause 3.4.5.10

Minimum required steel area for crack

control

0.15

600

Ok

BS8110:1985

Table 3.9

Conditions to

Satisfy

100As/bd

0.89314

Ok

400/d

1.33779

BS8110:1985

Clause 2.4.5.2

v

0.41

4.73 v

BS8110:1985

Clause 3.4.4.4

Z

0.98


317

BS8110:1985

Clause 3.4.4.4

k

0.021


Transverse direction Bending moment

in ULS

Mu

75

d h ‐ bottom cover ‐ ØT/2

622

DESIGNED BY

REF



Calculations Output

PROJECT


CODES

Reference

= 0.73 x 2

= N/mm2

= N/mm2

vc + 0.4 = N/mm3

Condition 0.5vc < v < (vc + 0.4) Satisfy

Assume T16 bars provided along the base

For shear ≥ 0.4bv/0.87fyv

≥ (0.4x2000x1000)/(0.87x460)

≥ mm2/ m width

Torsional Reinforcement

Torsional moment acting on base T = kNm

= N/mm2

Smaller C/C dimension of rectangular link x1 = h ‐ top cover ‐ bottom cover ‐ ØT

= mm

Larger C/C dimension of rectangular link y1 = b ‐ top cover x 2 ‐ ØT

= mm

> mm

v + vt = 0.41 + 1.75

= N/mm2

Limit to shear stress vtu = Minimum of 0.8√fcu or 5 N/mm2

= N/mm2

> v + vt → OK

vt,min = Minimum of 0.067√fcu or 0.4 N/mm2

= N/mm2

For Torsion Asv/sv ≥ T/[0.8 x1 y1 (0.95fyv)]

≥ 261.7x10^9/[0.8x259x1884(0.95x460)]

≥ mm2/ m width

→ vt > vt,min

v ≤ Vc + 0.4

Designed torsion reinforcement but not less than the minimum shear reinforcement

Hence required steel area for transverse r/f ≥ 2 x (As)bending

+ Max[(Asv)shear or (Asv)torsion]

≥ 2x622 + 1999

≥ mm2/ m width

2x261.7x10^6

400^2(2000 ‐ 400/3)=

1.75

1.86

BS8110:1985

Part 2

Clause 2.4.7

BS8110:1985

Part 2 Table 2.3

259

1884

2.16

550

4.73

BS8110:1985

Part 2

Table 2.4

3243

BS8110:1985

Table 3.8

Asv/sv

Enhanced design shear

stress,

vc

OK

BS8110:1985

Part 2

Clause 2.4.5

1.46

0.5vc 0.37

1999

261.704

2T

hmin2 (hmax ‐ hmin/3)

BS8110:1985

Part 2

Clause 2.4.4.1

Torsional shear stress vt =

0.40

1534.12

DESIGNED BY

REF



Calculations Output

PROJECT


CODES

Reference

Hence bar spacing for transverse direction sv = 1000/[3243/(2πx16^2/4) ‐ 1]

= mm

Minimum spacing of links for torsional r/f = Minimum of (x1, y1/2 or 200mm)

= mm

= mm → Ok

Hence provide @ C/C bars along the footing

Required longitudinal r/f for resist torsion

> mm2

→ Hence assumed area for torsion r/f is correct

Shear Resistance of Flange

Ultimate design shear force at flange = 150x2000/(2x1000)

= kN/m width

Shear stress, = Vu/bvd

= (150x10^3)/(1000x317)

= N/mm2

Minimum of(0.8√fcu or 5 N/mm2) = N/mm3 >

→

= (100x2211.68)/(1000x317)

= < 3

→

= 400/317

= > 1

→ Ok

Design shear stress, vc = 0.79[100As/bd]1/3[400/d]1/4[fcu/25]

1/3/γm= {0.79[0.7]^(1/3)x[1.26]^(1/4)x[35/25]^(1/3)}/1.25

= /mm2

Shear enhancement factor for a section at a distance d from support,

= 2d/av

= 2

= 0.66 x 2

= N/mm2

= N/mm2

Hence v < 0.5vc , Therefore no shear links required

Vu

150

>402.12 x 460 x (259 + 1884)

100 x 460

8617.51

>As

Asv fyv (x1 + y1)

svfy

BS8110:1985

Part 1

Clause 3.4.5.5

200

Adopt T16 @

100 C/C bars

for transverese

direction of

footing

100

1.33

0.5vc 0.66

Ok

BS8110:1985

Table 3.90.66

BS8110:1985

Clause 3.4.5.10

Enhanced design shear

stress,

vc

5.00 v

Ok Ok

BS8110:1985

Table 3.9

Conditions to

Satisfy

100As/bd

0.69769

Ok

400/d

1.26183

BS8110:1985

Clause 2.4.5.2

v

0.47

224

T16 100

Check the Spacing of link not exceed

0.75d0.75d = 0.75x299

OK

DESIGNED BY

REF



Calculations Output

PROJECT


CODES

Reference

BEARING CAPACITY CHECK

Total load transferred to base P = Loads from columns + weight of soil

+ weight of base

= kN

Location of resultant force acting on base

Load excentricity in x direction ex = My1/P

= m

Load excentricity in y direction ey = Mx1/P

= m

= Region 4

309.675

528

352

Corresponding of region of base the resultant

force acting

a/6 a/6 a/6

a

a/4

b/4

b/6

b/6

b/6

b

DESIGNED BY

REF



Calculations Output

PROJECT


CODES

Reference

For resultan force acting at Region 4

k = ex/a + ey/b

=

P

ab

= kN/m2

Ultimate bearing Capacity of soil = kN/m2

Hence FOS for bearing =

Resistance for Horizontal Rotation

Ep ‐ Ea = 0.5(Kp ‐ Ka)d(h22 ‐ h1

2)

= kN/m

OK

= 0.5Kad(h22 ‐ h1

2)

0.33

Maximum bearing presser under SLS

ConditionPmax = k[12 ‐ 3.9(6k ‐ 1)(1 ‐ 2k)(2.3 ‐ 2k)

142.62

150

Bearing Pressure is adequate

1.05

Passive pressure at side of base footing

per unit lengthEp = 0.5Kpd(h2

2 ‐ h12)

19.2

Resultant horizontal earth pressure per

unit length

Active pressure at side of base footing per

unit lengthEa

DESIGNED BY

REF



Calculations Output

PROJECT


CODES

Reference

Torque developed from earth pressure = (Ep ‐ Ea)[(a/2)2 + (b/2)2]

= kNm

Friction force developed from sides of base = (Ep ‐ Ea)tan()= kN/m

Torque from Friction of sides of base = (Ep ‐ Ea)tan() x (a/2 x b/2 x 4)= kNm

F = (P/4)tan()= kN

Distance of froce from centroid of base = √[(a/4)2 + (b/4)2]

= m

Torque from Friction of bottom of base = kNm

= kNm

Moment acting about Z axis = kNm

FOS for Rotation = <

Rotational resistance of base about z axis is not adequate Not OK

28.1782

240.5

78.0

Friction force developed from 1/4 th area of

base

1.01

113.59

Total torque developed to resist rotation of

base

6.99

48.92

196.581

1.22 1.5

DESIGNED BY

REF



Calculations Output

PROJECT


CODES

Reference

SUMMARY OF REINFORCEMENT FOR SHORT COLUMN

Main reinforcement bars = x Nos

Shear links = @ mm

SUMMARY OF REINFORCEMENTS

Longitudinal direction top r/f = T × Nos

Longitudinal direction bottom r/f = T × Nos

Longitudinal direction bottom r/f for crack control = T @ C/C

Transverse direction r/f = T @ C/C

Shear links Required for T beam = ## @ C/C

16 #REF! #REF!

#REF! 100

20 17

#REF! #REF!

20 15

T12 20

R10 100

thrust anchor blocks for vertically inclined hrizontal bend

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