Rc Element Structures

IS 3370 (Part II) 1984 Determine the suitable dimension of the cantilever retaining wall which is required to support the bank of earth 4m height above. G.L. on the toe side of the wall considered the back fill surface to the inclined at an angle of 15o Assume good soil for foundation at a depth of 1.25m below G.L SBC – 160 KN/m2. Further assume the back fill to comprise of granular soil with unit wt 16KN/m3 and an ansle of shearing resistance of 30o. Assume co-efficient of friction b/w. soil and concrete is 0.5. Pa is the active earth pressure exerted by the retain earth on the wall. Solution : (both wan & the earth move in the same direction)  = 30o, SBC = 160 KN/m2,  = 16 KN/m3, μ= 0.5 Rankine’s min depth of foundation, dmin = SBC  1 − sin φ   γ  1 + sin φ 

2

= 160  1 − sin 30o    16  1 + sin 30o 

2

= 1.11m 1.25m Thickness of base slab h/12

14

IS 3370 (Part II) 1984 =

= 0.4375 5.25 12

Assume the top tk of stem as 150mm thickness of stem tappers from 450mm from bottom & 150mm to top. Lmin =  Ca  h 1    3  αR

Co-efficient of active earth pressure Ca

=  Cos θ − Cos 2θ − Cos 2 φ   Cos θ + Cos 2θ − Cos 2 φ

  Cos θ 

=  Cos 15o − Cos 2 15o − Cos 2 30o    Cos1 o 2 o 2 o  Cos 15 + Cos 15 − Cos 30 

= Ca = 0.373 Cp =

=3 1 + sin φ 1 + sin 30o = 1 − sin φ 1 − sin 30o

Assuming surcharge height of 0.4m at the end of heel slab. = 5.25 + 0.4 h′

15

IS 3370 (Part II) 1984 = 5.65 m h′

Assuming the trapezoidal below the base slab. α

R

= 0.67

Lmin = Lmin = 2.97 m Provide base slab width of 3m Width of heel slab is Xmin = Lmin x α

R

= 3 x 0.67 = 2m The preliminary proportion is shown in fissure For the assumed proportions the retaining wall is check for stability against overturning and sliding Force

Force (KN)

Distance from Moment

ID W1

heel (m) ½ x 1.85 x 0.5 x 16 = 7.44

KNm

1/3 x 1.85 = 4.583 0.616

W2

W3

1.85x 5 x 25 – 0.45 x 16 = ½

(1.85)

= 131.424

142.08

0.925

0.15 x 4.8 x 25 = 18

1.85 + 0.15/2 = 34.65 1.925 14

IS 3370 (Part II) 1984 W4

½ x (0.450 – 0.15) x 4.8 x 1.85 (25 – 16) = 6.48

–

1/3 11.34

(0.30) = 1.75

Pa sin 98.65 sin 15o = 25.53 Pa = Ca γ e x 4/2 = 0.373 x 16 x  5.75     2 

2

Pa = 98.65KN The resultant of the vertical of lies at a distance of Xw from the h end X4 =

= MW W

= 0.99m 232.62 283.28

Check for over turning moment FoSo = > 1.4 For retaining wall with sloping back fi Mo = Pa cos θ x h1/3 = 98.65 Cos 15o x 5.75 / 5 Mo = 182.63 KNm 14

IS 3370 (Part II) 1984 Mr = W (L. x W) + Pa Sin θ (L) = 233.28 (3 – 0.99) + 98.65 sin 15o = 545.5 Foso =

=2.68 > 1.4 0.9 x 545.5 182.63

Check for sliding Fos sliding = 0.9 F Pa Cos θ

F = μR = 0.5 x 232.28 = 116.64 KN Fos sliding =

= 1.10 < 1.4 0.9 x 116.64 Pa Cos 15o

Hence the section is not safe is sliding The shear key is required to resist sliding Assume shear key of size 300 x 300mm at a distance of 1.3m from the too as shown in figure. Figue***

15

IS 3370 (Part II) 1984 Tan 30o = x 1.3

x = 0.75m h1

= 1.25 - 0.3 + 0.8 = 1.25m

h2

= 1.25 + 0.75 = 2m

Ppe

= Cp pe

(h

2 2

− h 12

)

=3x 16((2 ) 2 − (1.25) 2 ) 2

Pps = 58.5KN Fos s = 0.9 F + Pps 98.65cos15o

= 0.9 F (116.64) + 58.5 98.65 cos15o

= 1.71 > 1.4 Hence the section is safe against sliding. Soil pressure from the supporting soil on base slab :

15

IS 3370 (Part II) 1984 The distribution of soil pressure from the base is trapezoida in nature. Maximum pressure at one end is q max = R L

 6e  1 +   L

and min-pressure is qmin = R 6e  1 −  L L

(where qmax is direct stress + Bending stress qmin is direct stress – bending stress) here the eccentricity ‘e’ is given by e = LR – L/2 where LR = The distance of R from the heel slab LR = (Mo + Mw) /R = (182.63 + 232.62 ) 233.38

LR = 1.78m E = 1.78 – (3/2) = 0.28m

14

IS 3370 (Part II) 1984 qmax = 233.28  6(0.28)  1 +  3  3 

= 121.30KN/m2 α SBC 34.21 > 0 KN/m2

qmin = 233.28  6(0.28)  1 −  3  3 

Hence pressure is within the limits. Figure*** Design of Toe slab : Fig*** The toe slab is design for a UD of 92.36 The Toe slab is design as cantilever Assume the clear cover as 75mm for base slab and bar dia of 16mm Eff. Depth, d = D – clear cover - φ /2 = 450 – 75 – 16/2 d

= 367mm

The max. B.M at the rear face of the stem is M = 92.36 x 1 x ½ + (½ x 1 x 28.96) 15

IS 3370 (Part II) 1984 M = 55.82 KNm The ma. Shear force occurs at the distance of ‘d’ from the face of the stear. V

= ½ x 1 (92.36 + 121.3) x (1-0.367)

V

= 67.62 KN

Factored B.M = 1.5 x 55.82 Mu = 83.73 KNm Factored shear force, Vu = 1.5 x 67.62 Vu = 101.43 KN Mu = 0.87fy Ast (d – 0.47

) 0.87fy Ast 0.36 fck

83.73 x 106 = 0.87 x 415 x Ast (367 - 0.42  0.87fy Ast     0.36 x 20 x1000 

Ast = 656.44mm2 S

= 1000 ast Ast

14

IS 3370 (Part II) 1984 = π x (16) 2 1000 x 4 Ast

S

= 306.19 mm

Refer table 2 in slab for M20 p fe 415 stal Provide 16mm @ 300mm c/c. Check for shear Design of heel slab : M = 34.3 x 1.55 x 1.55 2

½ x 1.55 x 47.89 x 1/3 x 1.55 M = 60.37 KNm V = ½ (82.19 + 34.3) (1.55 – 0.367) V = 68.90KN Mu = 1.5 x 60.37 = 90.55 KNm Vu = 1.5 x 68.90 = 103.35 KN Mu = 0.87fy Ast d  fy Ast  1 −   0.36 fck bd  15

IS 3370 (Part II) 1984 90.55 x 106 = 0.87 x 415 x Ast x 367   415 Ast 1 −   20 x 1000 x 367

Ast = 712.03mm2 S= π x 162 4 712.08

1000 x

S = 282.37mm Provide 16mm @ 280mm c/c Development length : The main reinforcement to be developed into the fixed support for a length of Ld

= φ σs 4 τ bd

= 16 x 0.87 x 415 4 x 1.2 x 1.6

Ld

= 752.18 mm

Check for shear

15

IS 3370 (Part II) 1984 τ

v

= 0.281 N/mm2

= vu 103.35 x 103 = bd 1000 x 367

Pt =

= 0.194% 100 Ast 100 x 712.03 = bd 1000x 367

c = 0.315 N/mm2 τ v<τ c Hence section is safe in shear Distribution steel : Ast min = 0.12 c/s is provided along the transverse direction of the base slab. Ast min = 0.12 bd = 0.12 x 1000 x 150 = 540.0 mm2 S= π x 102 1000 x 4 540

= 145.44mm Provide 10mm @ 140mm c/c Design of stem 15

IS 3370 (Part II) 1984 Stem is designed as a cantilever slab for a height of 4800mm (5250 - ) The max. moment on cantilever slab in head stem is M = C a h3/6 Assume clear cover of 50mm, bar of 16mm d = 450 – 50 – 16/2 = 392mm M = 0.373 x 16 x = 110KNm The maximum shear force at ‘d’ from compound base slab is V = Caγ Z2/2 where Z = 4.8 – 0.45 = 4.35m V = 0.373 x 16 x (4.35/2)2 V = 56.46. KN Mu = 1.5 x 110 = 165KNm Vu = 1.5 x 56.46 = 84.69 KN Mu = 0.87 fy Ast d  fy Ast  1 −   fck bd 

165 x 106 = 0.87 x 415 x Ast x 312   415 Ast 1 −   20 x 1000 x 312

Ast = 1248.30mm2

14

IS 3370 (Part II) 1984 S = 161.06mm Provide 16mm φ @ 160mm c/c. Check for shear v =

= 0.216 N/mm2 Vu 84.67 x 103 = bd 1000 x 312

Pt =

= 0.318 100 Ast 100 x 1248.3 = bd 1000 x 392

C = 0.392 N/mm2 v < c Hence section is safe in slab.

14

IS 3370 (Part II) 1984

Design of heel slab : Wt from backfill = 16(5.25 – 0.45)

= 76.8

Self wt of heel slab 25 (0.45)

= 11.25 Kn = 88.05 Kn

M = 5.86 x 1.55 x 1.55/2 + ½ (47.89) x 1.55 x 2/3 (1.55) M = 45.39 KNm V = ½ (5.86 + 53.75) (1.53 – 0.367) 5.86 V = 35.25 KN Mu = 1.5 (45.39) = 68.08 KNm Vu = 1.5 x 56.463 = 84.69 KN Mu = 0.87 fy Ast. d (1-

) 415 Ast 20 x 1000x 3

S= π x 122 4 529.65

1000 x

S = 213.53mm 15

IS 3370 (Part II) 1984 Provide 12mm @ 210mm c/c. Provide distribution steel 10mm @ 160mm c/c. in the base slab as well as along the transverse direction for the steam in the rear face. Also provide distasted for the front face of the stem along both direction as 10mm @ 16mm c/c. Reinforcement Details : The main reinforcement in the stem can be curtailed at two places. At ½ height (4/3m) half the reinforcement is curtail. Provide 16mm @ 320mm c/c. At 2/8 h

the reinforcement is reduced to half. 2   x 4.8 8 

Provide 16mm @ 640mm c/c. The distribution steel is also curtail in the similar method.

Design the cantilever retaining wall to retain a level difference of 4m. Good soil is available at a depth of 1.25m below G.L. The unit wt of soil 16KN/m3 and SBC of soil is 160KN/m2. The backfill is leveled one with angle of internal friction φ = 30o. Soln : 14

IS 3370 (Part II) 1984 Height of wall above G.L= 4m Good soil depth below G.L. = 1.25m Unit wt of soil = 16 Kn/m3 = 160KN/m2

SBC

 = 30 Rankine’s min depth of foundation. dmin = SBC  1 − sin θ   γ  1 + sin θ

2

= 160  1 − sin 30o    16  1 + sin 30o 

2

= 1.11m  1.25 Earth pressure co-efficient Ca =

= 1 − sinθ 1 + sinθ

Cp =

= 0.333 1 − sin 30o 1 + sin 30o

= 1 − sinθ 1 + sinθ

=3 1 − sin 30o 1 + sin 30o

15

IS 3370 (Part II) 1984 Thickness of base slab =

= 0.4375 = 0.45m h 5.25 = 12 12

Assume top width of wall as 150mm & tk of stem tappers from 450mm to 150mm Lmin =  Ca  h   α  3 

R

Assuming the trapezoidal sters below the base slab 

R

= 0.67

Lmin =

= 2.610m  0.333   3  

 5.25    0.67

Provide base slab width of 3m = Lmin x α

Width of heel slab is Xmin

R

= 3.0 x 0.67 = 2.01 = 2m The preliminary proportions is shown in figure.

15

IS 3370 (Part II) 1984 For the assumed proportions the retaining is check for stability against overturning (s) Pa = Ca e x

= 0.333 x 16 x  (h)    2 

 (5.25) 2     2 

2

Pa = 73.43 Kn Force

Force (KN)

Distance from Moment

ID W1

heel (m)

½ x 1.85 x 0.45 x 16 = 1/2 x 1.85 = 181.42 142.08

W2

KNm

0.925

0.15 x (5.25 – 0.45) x 25 = 1.85 +

= 34.65  0.15    2 

18 1.925 W3

W4

½ x (0.45 – 0.15) x 4.8 x 1.85 – 1/3(0.3) 11.35 (25-16) = 6.48

= 1.75 3/2

3 x 0.45 x 25 = 33.75

3/2 = 1.5

W = 200.31KN

50.625

M = 228.045 KNm

The resistant of vertical forces lies at a the of x 01 from the heel end. X4 =

= 1.188m Mw 228 .045 = W 200 .31

14

IS 3370 (Part II) 1984 Check for overturning moment : Foso = 0.9 Mr Mo

Mo = Pa cos θ x h1/3 = 73.43  5.25    3 

Mo = 128.50 KNm Mr = W (L. x W) + Pa Sin (L) = 200.31 (3 – 1.138) = 372.97KNm Foso =

=2.61 > 1.4 0.9 x 372.97 128.50

Check for sliding Fos sliding = 0.9 F Pa Cos θ

F = μR = 0.5 x 200.31 = 100.155 KN

14

IS 3370 (Part II) 1984 Fos sliding =

= 1.22 < 1.4 0.9 x 100.155 73.43

Hence the section is not safe is sliding The shear key is required to resist sliding Assume shear key of size 300 x 300mm at a distance of 1.3m from the too as shown in figure. Figue*** Tan 30o = x 1.3

x = 0.75m h1

= 1.25 - 0.3 + 0.8 = 1.25m

h2

= 1.25 - 0.3 + 0.3 + 0.75 = 1.25m

Ppe

= Cp e

(h

2 2

− h 12

)

=3x 16((2 ) 2 − (1.25) 2 ) 2

Pps = 58.5KN

15

IS 3370 (Part II) 1984 Fos s = 0.9 F + Pps 98.65 cos θ

= 0.9 F (100 .155 ) + 58.5 73.43

= 2.02 > 1.4 Hence the section is safe against sliding. Soil pressure from the supporting soil on base slab : The distribution of soil pressure from the base is trapezoida in nature. Maximum pressure at one end is q max = R L

 6e  1 +   L

and min-pressure is qmin = R 6e  1 −  L L

here the eccentricity ‘e’ is given by e = LR – L/2 where LR = The distance of R from the heel slab

15

IS 3370 (Part II) 1984 LR = (Mo + Mw) /R = (128.5 + 228.04 ) 200.31

LR = 1.78m e = 1.78 – (3/2) = 0.28m qmax = 200.31  6(0.28)  1 +  3  3 

= 104.16KN/m2 α 160 qmin

29.37 Kn/m2 > 0

= 233.28  6(0.28)  1 −  3  3 

Hence pressure is within the limits. Figure*** Design of Toe slab : Fig*** 74 .79 x1 = 3 2

X = 49.86KN/m2

15

IS 3370 (Part II) 1984 74.79 x = 3 1.55

X = 38.64 KN/m2 The Toe slab is design for a UDL ofas cantilever beam Assume the clear cover as 75mm for base slab and bar dia of 16mm Eff. Depth, d = D – clear cover - φ /2 = 450 – 75 – 16/2 d

= 367mm

The max. B.M at the rear face of the stem is M = 79.23 x 1 x ½ + (½ x 1 x 21.93) x 2/3 M = 47.925 KNm The ma. Shear force occurs at the distance of ‘d’ from the face of the stear. V

= ½ x 1 (79.23 + 104.16) x (1-0.367)

V

= 58.04 KN

Factored B.M = 1.5 x 47.92 Mu = 71.85 KNm Factored shear force, Vu = 1.5 x 58.04 14

IS 3370 (Part II) 1984 Vu = 87.06 KN Mu = 0.87fy Ast (1 -

) fy Ast fck

71.88 x 106 = 0.87 x 415 x Ast (367)   415 Ast 1 −  20 x1000x367  

Ast = 560.21mm2 S

= 1000 ast Ast

= π x (12) 2 4 560.21

1000 x

S

= 201.80 mm

Provide 12mm @ 200mm c/c. Check for shear 

v

= 0.237 N/mm2

= Vu 87.06 x 103 = bd 1000x 367

16

IS 3370 (Part II) 1984 Pf =

= 0.152% 100Ast 100x 560.21 = bd 1000x 367

c = 0.2816N/mm2 τ

v

= c.

Hence section is safe in shear. Design of heel slab : Wt from back fill = 16(5.25 – 0.45) = 76.8 Kn Self wt of heel slab = 25 (0.45) = 11.25 KN/m2 88.05 Kn/m2 M = (20.04 x 1.55) x 1.55 2

½ x 1.55 x 38.68 x 2/3 x 1.55 M = 55.04 KNm Mu = 1.5 (55.04) = 82.56 KNm V = ½ (20.04 + 58.68) (1.55 – 0.367) V = 46.56KN Vu = 1.5 x (46.56) = 69.84 KNm

16

IS 3370 (Part II) 1984 Mu = 0.87fy Ast d  fy Ast  1 −   0.36 fck bd 

82.56 x 106 = 0.87 x 415 x Ast x 367   415 Ast 1 −   20 x 1000 x 367

Ast = 646.71mm2 S= π x 162 1000 x 4 646.71

S = 174.85mm Provide 12mm @ 170mm c/c Distribution steel Ast(min) = 0.12% bd = 0.12 x 1000 x 450 100

= 540mm2 S=

= 145.44mm π x (0) 2 4 540

1000 x

16

IS 3370 (Part II) 1984 Provide 10mm @ 140mm c/c. Design of Stem : Stem is designed as cantilever slab for a height of 5.25 – 0.45 = 4.8m The max. moment on cantilever slab in the head stem is M = Caγ h3/6 = 0.833 x 16 x ( 4.8) 3 4

M = 98.20 Knm Assume 50mm cover & 16mm  bar d = 450 – 50 – 16/2 = 392mm Max. shear force at ‘d’ from compound slab is V = Caγ Z2/2 Z = 4.8 – 0.45 – 4.3 V = 0.833 x 16 x ( 4.35) 2 2

V = 50.40 KN Mu = 1.5 x 98.20 = 147.3 Knm

14

IS 3370 (Part II) 1984 Vu = 1.5(50.40) = 75.6 KN 147.3 x 106 = 0.87 x 415 x Ast x 392 1−

415 Ast 20 x 1000

Ast = 1105.44mm2 S=

= 181.88mm π x 162 1000 x 4 1105.44

Provide 16mm @ 180mm c/c. Check for shear τ

v

= 0.192 N/mm2

= vu 75.6 x 103 = bd 1000 x 392

Pt =

= 0.282% 100 Ast 100 x 1105.44 = bd 1000x 392

τ

c

= 0.375 N/mm2

τ v<τ c Hence section is safe in shear Reinforcement details :

14

IS 3370 (Part II) 1984 Figure*** Ld =

= 752mm = 750mm φ σ0 16 x 0.87 x 415 = 4τ bd 4 x 1.2 x 1.6

Design a suitable counterfort retaining wall to support a leveled back fill of height 7.5m above g.l on the toe side. Assume good soil for the foundation at a design of 1.5m below G.L. The SBC of soil is 170KN/m2 with unit weight as 16KN/m3. The angle of internal friction is = 30o. The co-efficient of friction below the soil and concrete is 0.5 use M25 concrete and Fe415 steel. Figure **** Soln : Minimum depth of foundation = P  1 − sin θ    γ  1 + sin θ 

=

2

= 1.180m < 1.5m 170  1 − sin30o    16  1 + sin30o 

2

Depth of foundation = 1.5m Height of wall = 7.5 + 1.5 = 9m 14

IS 3370 (Part II) 1984 Thickness of heel & stem = 5% of 9m = 0.45m = 0.5m Thickness of toe slab = 6% of 9m = 0.54m Stability Conditions : Earth pressure calculations : Force

Force (KN)

Distance from Moment

ID W1

W2

heel (m) 16(7.5+1.5 – 0.5) x 2.5 = (3-0.5)/2

KNm = 425

340

1.25

25(0.5)(9-0.5) = 106.25

0.5/2 + 2.5 = 292.18 2.75

W3

25(0.5)x3 = 37.5

1.5

56.25

W4

25(1.5) (0.72) = 27

1.5/2+3=3.75

101.25

Total

W=510.75 KN

MH = 874.69

The resistant of vertical forces lies at a the of x 01 from the heel end. XW =

= 1.713m Mw 874 .69 = W 510 .75

Check for overturning moment : Foso = 0.9 Mr Mo

14

IS 3370 (Part II) 1984 Mo = Pa cos θ x h1/3 = 0.333 x 16 x  93     6

Mo = 647.35 KNm Mr = W (L. x W) = W = 510.75 (4.5 – 1.718) = 1423.6KNm Foso =

=1.98 > 1.4 0.9 x 1423.6 647.35

Hence, section is safe against overturning. Check for sliding Fos sliding = 0.9 F Pa Cos θ

F = μR = 0.5 x 510.75 = 255.87 KN

Pa = Ca e x

= 0.333 x 16 x  (h)    2 

2

 ( 9) 2     2 

Pa = 215.78 Kn 16

IS 3370 (Part II) 1984 Fos sliding =

= 1.065 < 1.4 0.9 x ( 255 .37 ) 215 .78

Hence the section is not safe is sliding The shear key is required to resist sliding Base pressure calculation: q max = R L

 6e  1 +   L

= 510.75  4.5

1 + 

6(73)   4.5 

= 223.97 KN/m2 > SBC un safe qmin = R L

 6e  1 +   L

= 510.75  4.5

1 + 

6(73)   4.5 

= 3.02 Kn/m2 > 0 Safe

16

IS 3370 (Part II) 1984 where the eccentricity ‘e’ is given by e = LR – L/2 where LR = The distance of R from the heel slab LR = (Mo + Mw) /R = (874.68 + 647 .352 ) 510 .75

LR = 2.98m e = 2.98 – (4.5/2) = 0.72 x 1/6 (0.75m) Since maximum earth pressure is greater than SBC of soil, the length of base slab has to increased preferably along the toe side. Increase the toe slab by 0.5m in length. EH = 510.75 + 0.5 x 25 x 0.72 = 519.75 KN LR =

= 3.01m (Mo + Mw) 917 .438 + 647 .352 = R 519 .75

∑M

Design of Toe slab :

16

IS 3370 (Part II) 1984 Fig*** 74 .79 x1 = 3 2

X = 49.86KN/m2

74.79 x = 3 1.55

X = 38.64 KN/m2 The Toe slab is design for a UDL ofas cantilever beam Assume the clear cover as 75mm for base slab and bar dia of 16mm Eff. Depth, d = D – clear cover - φ /2 = 450 – 75 – 16/2 d

= 367mm

The max. B.M at the rear face of the stem is M = 79.23 x 1 x ½ + (½ x 1 x 21.93) x 2/3 M = 47.925 KNm The ma. Shear force occurs at the distance of ‘d’ from the face of the stear. V

= ½ x 1 (79.23 + 104.16) x (1-0.367)

14

IS 3370 (Part II) 1984 V

= 58.04 KN

Factored B.M = 1.5 x 47.92 Mu = 71.85 KNm Factored shear force, Vu = 1.5 x 58.04 Vu = 87.06 KN Mu = 0.87fy Ast (1 -

) fy Ast fck

71.88 x 106 = 0.87 x 415 x Ast (367)   415 Ast 1 −  20 x1000x367  

Ast = 560.21mm2 S

= 1000 ast Ast

= π x (12) 2 4 560.21

1000 x

S

= 201.80 mm

Provide 12mm @ 200mm c/c. Check for shear 16

IS 3370 (Part II) 1984 v =

= 0.237 N/mm2 Vu 87.06 x 103 = bd 1000x 367

Pf =

= 0.152% 100Ast 100x 560.21 = bd 1000x 367

τ

c

= 0.2816N/mm2

τ

v

=  c.

Hence section is safe in shear. Design of heel slab : Wt from back fill = 16(5.25 – 0.45) = 76.8 Kn Self wt of heel slab = 25 (0.45) = 11.25 KN/m2 88.05 Kn/m2 M = (20.04 x 1.55) x 1.55 2

½ x 1.55 x 38.68 x 2/3 x 1.55 M = 55.04 KNm Mu = 1.5 (55.04) = 82.56 KNm V = ½ (20.04 + 58.68) (1.55 – 0.367) 16

IS 3370 (Part II) 1984 V = 46.56KN Vu = 1.5 x (46.56) = 69.84 KNm Mu = 0.87fy Ast d  fy Ast  1 −   0.36 fck bd 

82.56 x 106 = 0.87 x 415 x Ast x 367   415 Ast 1 −   20 x 1000 x 367

Ast = 646.71mm2 S= π x 162 4 646.71

1000 x

S = 174.85mm Provide 12mm @ 170mm c/c Distribution steel Ast(min) = 0.12% bd = 0.12 x 1000 x 450 100

= 540mm2

14

IS 3370 (Part II) 1984 S=

= 145.44mm π x (0) 4 540

2

1000 x

Provide 10mm @ 140mm c/c. Design of Stem : Stem is designed as cantilever slab for a height of 5.25 – 0.45 = 4.8m The max. moment on cantilever slab in the head stem is M = Caγ h3/6 = 0.833 x 16 x ( 4.8) 3 4

M = 98.20 KNm Assume 50mm cover & 16mm φ bar d = 450 – 50 – 16/2 = 392mm Max. shear force at ‘d’ from compound slab is V = Ca Z2/2 Z = 4.8 – 0.45 – 4.3 V = 0.833 x 16 x ( 4.35) 2 2 16

IS 3370 (Part II) 1984 V = 50.40 KN Mu = 1.5 x 98.20 = 147.3 Knm Vu = 1.5(50.40) = 75.6 KN 147.3 x 106 = 0.87 x 415 x Ast x 392 1−

415 Ast 20 x 1000

Ast = 1105.44mm2 S=

= 181.88mm π x 162 4 1105.44

1000 x

Provide 16mm @ 180mm c/c. Check for shear v =

= 0.192 N/mm2 vu 75.6 x 103 = bd 1000 x 392

Pt =

= 0.282% 100 Ast 100 x 1105.44 = bd 1000x 392

c = 0.375 N/mm2 v < τ c

14

IS 3370 (Part II) 1984 Hence section is safe in shear Reinforcement details : Figure*** Ld =

= 752mm = 750mm φ σ0 16 x 0.87 x 415 = 4τ bd 4 x 1.2 x 1.6

Design a suitable counterturn retaining way to support a leveled back fill of height 7.5 m above g. L on the toe side. Assume good soil for the foundation at a deim of 1.5 m below G. L. The SBC of soil is 170 KN/m2 with unit weight as 16 KN/m3. The angle of internal friction is φ = 300 the co-efficient of friction b/w the soil of concrete is 0.5 use M25 concrete of Fe 415 steel. Soln:Figure ********* Minimum depth of foundation = P γ

 1 − sin φ    1 + sin φ  

=

170  1 − Sin300    = 1.180 m < 1.5m 16  1 + sin 300  Depth of foundation = 1.5 m 14

2

IS 3370 (Part II) 1984 Height of wall = 7.5 + 1.5 = 9 m Thickness of heal & stem = 5% of 9m = Thickness of heal & stem = 5% of 9m = 0.45 m = 0.5 Thickness of toe slab – 8% of 9m = 0.72 m.

 Ca   X m in =   3  

h

=  0.333    < 9 = 3.0 m   3  

Lmin = 1.5 x 3 = 4.5 m Thickness of countertort = 6% of 9 m = 0.54 m Stabiling conditions: Earth pressure calculations: Force W1

IDforce (KN)

Distance from heal (m)

16 ( 7.5 + 1.5 – 0.5) x 2.5 (3 – 0.5)/2 = 1.25

Moment KNm 425

= 340 W2

25 (0.5) (9 – 0.5) = 0.5/2 + 2.5 = 2.75

292.18

106.25 W3

25(0.5)x 3 = 37.5

1.5 14

56.25

IS 3370 (Part II) 1984 W4

25 (1.5) (0.72) = 27

Total

W = 510.75 KN

1.5/2 + 3 = 3.75

101.25 M4 874.69

W = 25(0.3) x 0.75 = 9.00 XW =

= 1.713 m 874.69 = 1.713m 510.75

Fos(overturnis) = 0 .9 M M0

Mo = Pa. h.3 = Caγ e. h3/6 = 0.333 x 16 x (9)3/6 = 647. 35 KNm Mr = (L – Xw) w = 510.75 (4.5 – 1.713) = 1423.6 KNm Fos = 0.9 (1423.6) = 1.98 > 1.4 647.35 Hence section is safe against, overturnins. Silding:Fos(sliding) = 0 .9 F PaCos θ 14

=

IS 3370 (Part II) 1984 F = R = 0.5 x 510.75 = 255.37 Pa = Ca. γ e. h2/2 = 0.383 x 16 x (9)2/2 = 215.78 Fos sliding = 0.9 (255.37) / 215.75 = 1.065m < 1.4 Hence the section is not safe against sliding. Base pressure calculation: 9 mas = R 60  510 .75  6( 0.73)  1 +  = 1 +  L  L 4.5  4.5 

= 223.97 KN/m2 > SBC unsafe Qmin = R  6e  510 .75 14  = L L 4.5

6( 0.73)   1 +  4.5  

= 3.02 w/m2 > 0 safe Where, LR = Mh + M0 = C = LR − L / 2 where R LR = 874.688 + 647.352 = 2.98m 510.75 e = 2.98 – 4.5/2 = 0.78 < L/L (0.75mm) 16

IS 3370 (Part II) 1984 Since maximum earth pressure is stresses then SBC of sol, the length of base slab has to b. increased praberables along the toe side. Increase the toe slab by 0.5m in length. w = 510.75 + 0.5 x 25 x 0.72 = 519.75 KN Figure ********** Additional load due to increase in toe slab by 0.5. Moment = 0.5/2 + 4.5 = 4.75m Σ m = 874.69 + 42.75 = 917.44 KNm LR= ( M 0 + M W ) 417.438 x 647.352 = = 3.01m R 519.75 e = LR – L/2 = 30.11 – 5/2 = 0.511 m < L/6 (0.833m) q max = R  6e  519 .75  6 x 0.5 m  1 +  = 1 +  L  L 5. 0  5.0 

= 167.69 KN/m2 < SBC Qmin = R  6e  1 −  L  L

16

IS 3370 (Part II) 1984 = 519 .75  6(0.511)  1 −  5  5 

Qmin = 40.20 KN/m2 > 0 FOS siliding = 0.9 F 0.9 x 0.5 x 519.75 = Pa 215.784 = 1.082 < 1.4 Hence the section is not slab against sliding. Shear they is provided to resist sliding. Assume shear they of size 300 x 300 mm Figure ********* tan 300 = L 2400 x = 1.385 = 1.39 m PPb = Cp. Re  h 2 2 − h 12      2  

h2 = 1.39 + 1.2 = 2.89m h1 = 1.2 + 0 = 1.2m 15

IS 3370 (Part II) 1984 Meridonial thrust T = WR 1 + 1.30 = 4.5 x 9.06 1 + 0.724 T = 23.64 KN Meridonial stress T 23 .64 x 10 3 = c / s 1000 x 100

= 0.2364 N/mm2 x 7 N/mm2 Hence the section is safe against meridonial stress. Hoop stress:Hoop Stress = WR  1   Cos θ −  T  1 + cos ϑ 

= 4.5 x 9.06  1   0.724 −  23.64  1 + 0.721 

= 0.248 N/mm2 < 7 N/mm2 Since the section is safe to resist meridonial thrust of hoop stress the provided 100 mm thickness of sufficient. 15

IS 3370 (Part II) 1984 Ast min = 0.3 % of C/S = 0.3 x 1000 x 100 = 300 mm 2 100 For 8mm φ spacing = π x 82 4 = 167.5 mm 300

1000 x

∴

Provide 8 mm  @ 160 mm c/c both way.

Design of Rin beam The area of concrete received for the ring beam is bound for the hoop stress and area of steel required is bound for the horizontal component of meridmial thrust. HZ component of Meridonial thrust = T Cos θ x D/2 = 23.64 x 0.724 x 12.5/2 = 106.97 KN Ast = 106.97 x 103 = 713.13 mm 2 150

16

IS 3370 (Part II) 1984 For 16 mm φ , no. 08 bars =

713.13 π x 162 4 = 3.54 say 4 nos Size of rins beam is based on tensile stress of concrete PP8 = 3 x 16

 ( 2.89) 2 − (1.2 ) 2    2   PP8 = 165.89 KN/m2 FOS sliding =

 ( 2.89) 2 − (1.2 ) 2    2   = 1.852 > 1.4 Hence section is safe against sliding. Design of Toe slab: Eff.Cover = 75 + 20/2 = 85 mm Toe slab is designed similer to cantition and with maximum moment at trust face of the size a maximum shear at ‘d’ for face it beam. D = 720 – 85 = 635 mm

16

IS 3370 (Part II) 1984 M= 80.38 x

22 1 2 + x 2 x 49.94 x x 2 2 2 3

= 227.35 KNm S.F at 0.635 m = 49.94 x 0.635 = 15.85 KN 2 Area of tralizium = ½ x (a + b) h = ½ (130.82 + 95.98) (2 – 0.635) = 154.44 KN Factored S.F = 1.5 (154.44) S.F Vu = 231.66 KN Factored B.M, Mu = 1.5 (227.35) = 341.02 KNm K= M bd

u

2

=

341 .02 = 0.845 = 0.845 1000 x 635 2

Pt = 0.2445 Ct = 100 Ast 0.245 x 1000 x 635 = > Ast = bd 100 Ast = 1552.575 mm2

16

IS 3370 (Part II) 1984 Spacing = π x 16 2 1000 x 4 = 129.50 mm 1552.575

Provide 16 mm  @ 125 mm C/C Transerve Reinforcement = 0.12% of C/S = 0.12 x 1000 x 720 = 864 mm 2 100 Spacing = π x 10 2 4 = 90.90 mm 864

1000 x

Provide 10 mm  @ 100 mm c/C

∴

Check for shear :τ

v

= Vu 231 .66 x 10 3 = = 0.364 N / mm 2 bd 1000 x 635

τ

∴

c

\ 0.36 N/mm2 τ

v

~τ

c

The section is safe in shear

Design of heel slab 16

IS 3370 (Part II) 1984 The had slab is designed by countention at regular interval. The counterbant act as support and makes the had slab as one – way continuous slab. The heal slab is designed for a moment Wl2/l2 at the support & Wl2/16 at the mid span. The maximum shear at the support is W **** the maximum pressure at the best slab is consider for me design. Moment at the support, MSuf = Wl 2 106.92 x 3.2 = 12 12

= 111.65 KNm Moment at the mid span, Mmid = Wl 2 = 83.72 KNm 16

The max pressure acting on the had slab is the as ‘W’ for which the Ast required at mid span and support one found. Factored Msup = 167.47 KNm  Ast = 1172.69 mm2 Factored Mmidspan = 125.58  Ast = 868.27mm Using 16 mm φ bar, spacing = 1000 Ast = 170.02 Ast

∴

Provide 16 mm @ 170 mm c/c.

16

IS 3370 (Part II) 1984 At mid span, spacing = 231.72 mm  Provide 16mm @ 230 mm C/C. Transverse reinforcement = 0.12% of Bd = 0.12 x 1000 x 500 = 600 mm 2 100 For 8mm bar, spacing = 83.775 mm

∴

Provided 8 mm @ 80mm c/c.

Check for shear:Maximum shear l   2.5  W − d  = 107  − 0.415  2   2 

Factored S.F = 217.47 KN = Pt = 0.282 τ

v

= 0.624 N/mm2, τ

τ

v

> c

∴

c

= 0.376 N/mm2, τ

Depth has to be increased.

Design of stem

16

cmax

= 3.1 N/mm2

IS 3370 (Part II) 1984 The stem is also designed as one – way continuous slab with

support moment

and midspan moment Wl 2 12

Wl 2 16

For the negative moment at the support, reinforcement is provided at the rear side & for positive moment at mid span, reinforcement is provided at front base of the stem. The maximum moment various beam a base Intensitus of Ca γ e. h = 1/3 x 16 x (9 – 0.5) = 45.33 KN/m Mmid =

= Wl 2 16

1.5 x 45.33 x 3.54 2 16

= 53.25 KNm (or) 307.04 KNm Effective depth d = 500 – (50 + 20/2) = 440 mm M support = Wl 2 1.5 x 45.33 x 3.54 2 = = 71 KNm 12 12

Ast at support = 454.73 spacing = 69 mm

16

IS 3370 (Part II) 1984

∴

Provide 16 mm @ 300 mm C/C.

Ast at mid span = 339.54 mm2 For 16 mm , spacing = 592.54 mm Provide 16 mm @ 300 mm C.C Max S.F = W

= 60.28 KN l   − d 2 

Factored S.F = 90.42 KN Transverse reinforcement = 0.12 % of 6D = 0.12/100 x 1000 x 440 = 528 m For 8 mm , spacing = 95 8mm @ 90 mm C/C 8t = 0.134. τ

v

= 0.205 N/mm2, τ

v < τ

c

∴

c

= 0.29 N/mm2

Shear of safe.

Design of countertors:-

16

IS 3370 (Part II) 1984 The countertort is desined as a cautilever beam whose depth is equal to the length of the heal slab of the base is reduces to the thickness of the stem at the top. Maximum moment at the base of coutnerfort. Mmax = Ca γ e. h3/6 x Le Where Le  C/C distance from counterforb Mmax = 1932.5 KNm Factored Mmax = 2898.75 KNm Ast = 2755.5 mm2 No. of bars required =

Ast 2755.5 = 252 ast πx 4 = 5.61 Say n = 6 nos The main reinforcement is provided along the slanting face of the counterbort. Cartailment of reinforcement: Not all the 6 bars need to be taken to the for end Three bars are taken straight to the entire span of the beam upto me hop of the stem. 14

IS 3370 (Part II) 1984 One bar is Cut al a distance of ,

where n is the total no of

n −1 h = 12 n 8. 5 2

h1 distance from top . n = 6, γ = 6-1/6 = h12/8.52 h1 = 7.72th (from bottom ) n −2 h 6−2 h = 2 2 => = 2 2 = 6.94 m n 8 .5 6 8 .5 2

2

The third part is cut at a distance of n −3 h = 3 2 , h 3 = 6.01m ( From bottom ) n 8. 5 2

Vertical ties and horizontal ties are provided to connect the counter fort with the item is the heal slab. Design of horizontal ties:Closed stirrups are provided t, the vertical stem is the countertort. Considering 1m strip, the tension resisted by reinforcement is given by lateral pressure on the wall multiplied by contributing area. T = Cα.γ e.h

14

IS 3370 (Part II) 1984 Where Ast = T 0.87 fy

T = 1/3 x 16 x (9-0.5) x 3.54 = 160.48 KN Ast = 160 .48 x 10 3 x 1.5 = 1.5 ( 444 .48 ) = 666 .72 mm 2 0.87 x 415

For 10 mm φ spacing = 110 mm Provide 10 mm @ 110 mm c/c closed stirrups as horizontal ties. Design of vertical ties:The vertical stirrups connects the countertort and the heel slab considering 1m strip, the tensile force is the product of the average download pressure & the spacing between the counterforts. T = Avg 943.56 + 167) + Le = 266.49 KN Factored T = 399.74 KN Ast = 1107.15 mm2. For 10mm , spacing = 70.93 Provide 10mm @ 70m c/c. C/s of counter fort wall of midspan: Reinforcement details of stem, toe slab of heel slab.

15

IS 3370 (Part II) 1984 Isotropicall reinforced – Squave slab – fixed on all edges – udl:External work done = W. S W – load, S – virtual displacement Internal work done = Σ Mθ = Σ m.lθ = m.¬2L.2 2 L

I.W.D work done by positive yield line (ab, bc, cd, da) for

θ =

1 2 = L/2 L

Mθ = 4[W x L x 2/L] = 8m Total I.W.D = 16m E.W.D, Σ W.S = ½ x L x L/2 x W x ½ x (1) x 4 = Wl2/3 I.W.D = E.W.D 16m = WL2/3 M= WL2 4b

M – Moment per metre length alms the rived line.

14

IS 3370 (Part II) 1984 Design a square slab fixed along as the bour edges, with side 5m. the slab hase, to support a service load of hase to be support a service load of 4 KN.m2. use M20 concrete and Fe 415 steel. Soln:Side = L = 5m, All four edges fixed Given Service load = 4 KN/m2 Fcx = 20 N/mm2 Fy = 415 N/mm2 As per yield line theory for isotrophic reinforced square slab fixed on all four edges, M = ***** Step 1:As per IS 456 : 2000 L/d ratio for simply supported slab using Fe 415 steel. L/D = 0.8 (35) = 28 D = 5000/28 = 178.57m

∴

Provide effective depth d as 180 mm

15

IS 3370 (Part II) 1984 Solve the above problem using the co-efficient stam in IS 456:2000 Soln:Given Lens = l2 = ly = 5 m

∴

ly/l2 = 1

The slab is two way slab.

Assume all four edges discontinuous, rebearing annex d, take 20 αu = αy = 0.056 Mx = α x Wdx2 = 0.056 x 15 x 52 = 21 KN M2 = My = 21 KNm Mu =

 fy Ast   0.87 fy Ast d  1 − Fck bd   21 x 106 = 0.87 fy Ast d

  415 Ast  1 −  20 x 1000 x 180   Ast = 336.15 mm2 S= π x 82 1000 x 4 = 149 .59 mm 336

14

IS 3370 (Part II) 1984

∴

Provide 8 mm φ @ 140 mm C/C

Design the above problem for simply support condition. Soln:M= WL2 15 x 52 = = 15.625 KNm 24 24

Mu = 0.87 fy Ast d

 fy Ast   1 −  fck bd   15.625 x 106 = 0.87 x 415 x Ast x 180

  415 Ast 1 −  20 x 100 x 180   Ast = 247.48 mm2 S= π x 82 1000 x 4 = 203 .50 mm 247

∴

Provide 8 mm  bar @ 220 mm c/C

Check for shear:-

16

IS 3370 (Part II) 1984 τ

v

=

15 x 5 V 2 = = 0.208 N / mm2 d b 100 x 180 u

Pt = 100 Ast 100 x 247 = = 0.137 % bd 1000 x 180

τ

c

= 0.28 N/mm2

c = 1.2 x 0.28 = 0.336 N/mm2 τ

∴

< c

v

Hence the section is safe is shear.

Vu = Wu l x 13.5 x 4 = = 27 KN 2 2 τ

=

v

27 x 10 3 = 0.18 N / mm 2 1000 x 150

Pt = 100 x 277 = 0.184 % 1000 x 150

τ

c

= 0.307 N/mm2

τ

c

=K

c

= 1.26 x 0.307 = 0.386 N/mm2 16

IS 3370 (Part II) 1984

∴

Hence the section is safe is shear.

In the above problem design the slab if all support are fixed Soln:Based on yield line theory, for rectangular slab fixed on all four edges, subjects to UDL throushour M=

Wu L α  tan 2 φ    24  µ  2

tan  =

µ2   µ    1.5µ + −  2 4 α    2α  =

 (0.7 ) 2   0.7   1.5 (0.7 ) +  −   2  4 ( 0 . 67 ) 2 ( 0 . 67 )     tan φ = 0.8000 m= 13.5 x 6 2  0.8 x 0.8    48  0.7 

= 9.26 KNm

18

IS 3370 (Part II) 1984 M = 0.87 fy Ast d

 fy Ast   1 −  fck bd   9.26 x 106 = 0.87 x 415 x Ast x 150

  415 Ast  1 −   20 x 1000 x 150  Ast = 175 mm2 Ast min = 0.12 x 1000 x 170 = 204 mm 2 100

∴

Ast provided on shorter direction is 204 mm2

Assume 80 mm φ , S = π x 82 4 204

1000 x

S = 246.39 mm

∴

Provide 8 mm  @ 240 mm c/c both ways

Check for shear v = Vu 27 x 10 3 = = 0.18 N / mm 2 bd 1000 x 150

16

IS 3370 (Part II) 1984 Pt = 100 x 204 = 0.136 % 1000 x 150

τ

c

τ

c

τ

v

∴

= 0.28 N/mm2 1

= K c = 1.26 x 0.28 = 0.352 N/mm2

< c

Hence the section is sale in shear

Solve the above problem for two long edges fixed boundarks condition Soln:M=

Wu L x  tan 2 φ    24  µ  2

=

13.5 x 4 2  0.82    24  0.7  M = 8.28 Knm 8.23 x 106 = 0.87 x 415 x Ast x 150

  415 Ast  1 −  20 x 1000 x 150   Ast = 155.3 mm2 16

IS 3370 (Part II) 1984 Ast min = 0.12 x 1000 x 150 = 204 mm 2 100

∴

Ast provide on shorter direction is 204 mm2

Assume 8mm φ , S = π x 82 1000 x 4 = 246 .39 mm 204

∴

Provide 8 mm  @ 240 mm C/C bothways

Check for shear 

v

= 27 x 10 3 = 0.18 N / mm 2 1000 x 150

Pt = 100 x 204 = 0.136 % 1000 x 150

τ

c

= 0.28 N/mm2

c1 = Kc = 1.26 x 0.28 = 0.352 N/mm2 τ

v

< c

Hence the section is safe in shear.

16

IS 3370 (Part II) 1984 Design an equilateral triangular slab of side 5m, isotropically reinforced and simply supported along its edges. The span is subjected to a super imposed load of 3 KN/m2 use M20 concrete of Fe – 1415 Soln:Based on yield line theory, the ultimate moment for an etuilateral triangular slab simply supported alms all edges of subjected to VDL through out. M= Wu L2 72

Assume L/d = 28 D = 5000/28 = 178.54 mm Assume eff. Depth d = 180 mm Assume in eff.cover of 20 mm

∴

overall depth, D = 180 + 20 = 200 mm

Load calculation: Self wt of slab = 1 x 0.2 x 25 = 5 KN/m2 Live load = 3 KN/m2

16

IS 3370 (Part II) 1984 Floor finish = 1 KN/ m2 W = 9 KN/m2 Factored load, Wk = 1.5 x 9 = 43.5 KN/m for 1m strip. M= Wu L2 13.5 x 52 = = 4.69 KNm 72 72

4.69 x 106 = 0.87 x 415 x Ast x 180

 415 Ast   1 −  20 x 1000 x 10   Ast = 72. 77 mm2 8mm φ Transverse reinforce = 0.12 x 1000 x 200 = 240 mm 2 100 Spacing = π x 82 4 = 209 mm 240

1000 x

∴

Provide 8 mm  @ 200 mm C/C.

Check for shear τ

v

= Vu bd 16

IS 3370 (Part II) 1984 V= WL 13.5 x 5 = = 33.75 2 2 Vu = 33.75 = 33.75 τ

v

= 33 .75 x 10 3 = 0.187 N / mm 2 1000 x 180

Pt 100 x 240 = 0.133 % 1000 x 180

τ

c

τ

c

τ

v

∴

= 0.28 N/mm2 , τ 1

= K

c

c max

= 2.5 N/mm2

= 1.2 x 0.28 = 0.336 N/mm2

< c < τ

c max

Section is safe in shear

A right angled triangular slab simply support along all the edges it has sides AB = BC = 4m. the slab is isotrofically reinforced with 10 mm @ 100 mm C/C both way use M20 and Fe

415

(HYDS) bars. Find the safe

permissible service load (Live load) that can be apply on the slab. Soln:

16

IS 3370 (Part II) 1984 For A1m slab S.S along all the edges which is right angle isotuplication reinforced subjected to UDL through out. Mu = Wu x L2 (Based on yield line there ) 6

α=1 4    4 

S= 1000 ast Ast Ast = π x 10 2 1000 x 4 = 785 .4 mm 2 100

L/4 = 28  d =

D = 170 mm 4000 28

For Ast available moment resistance the section. Mu = 0.87 fy Ast d

 fy Ast   1 −  fck bd  

14

IS 3370 (Part II) 1984 Mu = 0.87 x 415 x 785.4 x 150

 415 x 785.4   1 −  20 x 1000 x 150   Mu = 37.9 KN m 37.9 x 106 = Wu x 1 x 4 2 6

Wu = 14.2 KN/m2 Factored load = 14.2 KN/m2

∴

Working load = 14.2 / 1.5 = 9.46 KN/m2

Self wt = 0.17 x 25 = 4.25 KN/m2 Floor finish = 0.71 KN / m 2 4.95

∴

The service live load that can be safly apply on slab is 9.46 – 4.96 =

4.5 KN/m2 Design a circular slab of diameter 5m, S.S along the edges and subjected to live load of 4 KN/m3 use M20 & Fe 415 steel. Soln:16

IS 3370 (Part II) 1984 Based on yield line theory the ultimate moment for circular slab isotopicus reinforcement Sr. S. along the edges subjectedto VDO through. M= Wu r 2 6

Assume L/d = 28 D = 5000 / 28 = 178.07mm

∴

eff. Depth, d = 180 mm

Adopt eff. Cover = 20 mm

∴

overall depth, D = 200 mm

Load calculation Self wt on slab = 0.2 x 25 = 5 KN/m2 L.L = 4 KN/m2 Floor finish = KN/m2 W = 10 KN/m2 Factored load, Wu = 1.5 x 10 = 15 KN/m

14

IS 3370 (Part II) 1984 M = Wu r 2 15 x ( 2.5) 2 = = 15.625KNm 6 6

15.625 x 106 = 0.87 x 415 x Ast x 180

  415 Ast  1 −   20 x 1000 x 10  Ast = 247.48 mm2 Transverse reinforcement = 0.12% of D = 0.12 x 1000 x 200 100 = 240 mm2 For 8 mm φ S = π x 82 4 = 203.10 mm 247.48

1000 x

Provide 8 mm φ @ 200 mmC/C both ways Check for shear:τ

v

=

Wu l 15 x 5 x 103 Vu 2 = 2 = = 0.208 bd bd 1000 x 180

16

IS 3370 (Part II) 1984 Pt = 100 x 247.48 = 0.187 % 1000 x 180

τ c = 0.28 N/mm2 τ v<τ c Hence the section is safe in shear Design a simply supported rectangular slab of sie 5m x 8m which is orthotopilr reinforced with co-efficient of orthorooly µ

= 0.75 the

service live load on the slab is 5 KN/m2 use M20 concrete and Fe 415 steel Soln : L = 8m, α = 5m LL = 5 KN/m2 Based on yield line memory the rectangular slab orthotropically reinforced, S.S. along with edges of subjected to VDL through out. Α = 5/8 = 0.625 M= Wl 2 α2 24

[

3 + µ α2 − α µ

]

2

Assume L/D = 28 5000 / 28 = d, d = 178. 57 mm Provide d = 180 mm 15

IS 3370 (Part II) 1984 Eff. Cover = 20 mm, overall depth , D = 200 mm Load calculation Self wt on slab = 0.2 x 25 = 5 KN/m2 L. Load = 5 KN/m2 Floor finish = l KN/m2 Total load, W = 11 KN/m2 For 1 m strip , 4 = 11 KN/m Factored load, Wu = 1.5 x 11 = 16.5 KN/m Mu = Wl 2 α2 24

[

3 + µ α2 − α µ

= 16.5 x 0.625 2 x 82 24

[

]

2

3 + 0.75 x 0.6252 − 0.625 0.75

]

2

Mu = 27.86 KNm 27.86 x 106 = 0.87 x 415 x Ast x 180

  415 Ast  1 −  20 x 1000 x 180   Ast = 452.26 mm2 Transverse reinforcement ast min 14

IS 3370 (Part II) 1984 For 10mm φ mm bar Provide 10mm φ @ 170 mm C.C lonser direction Ast µ Ast = 0.75 x 452.26 = 339.19 mm2 Check for shear: provide 10 mm @ 230 mm C/C τ v = Vu/ bd Vu = WL/2 = 16.5 x 5 / 2 = 41.25 KN v = 41.25 x 103 / 1000 x 180 = 0.229 N/mm2 Pt = 100Ast / bd = 100 x 452.26/1000x 180 = 0.25 % τ c = 0.36 N/mm2 c1 = K τ c = 1.2 x 0.86 = 0.432 N/mm2 Hence section is safe in shear. Design a rectangular slab of size 5m x 4m which is fixed along the long edges of S.S along the two short edges. The slab is subjected to the distributed live load of 4 KN/m2 design the slab for orthotopically reinforced condition with co-efficient orthotropy µ = 0.7 Soln: Based on yield line theory for orthotropically reinforced slab fixed along long edges of S. S along the short edges of subjected to VDL through out, the ultimate mum M is 16

IS 3370 (Part II) 1984 M = Wulx2 / 24 =

 tan 2    µ   tan φ =

µ2   µ    1.5µ + −  4α 2   2α   Assume L/d = 28 4000 / d = 28 D = 142.85 Provide eff. Depth d = 150 mm Effective cover = 20 mm Overall depth D = 170 mm Load calculation: Self wt of slab = .0.17 x 25 = 4.25 KN/m2 L.L = 4 KN/m2 Floor finish = 0.75 KN/m2 W = 9 KN/m2 For 1 m strip, W = 9 KN/m Wu = 1.5 x 9 = 13.5 KN/m2 16

IS 3370 (Part II) 1984 Α = 4/5 = 0.8 Tan φ = 1.5 (0.7) Tan = 0.804 M = 8.311 KNm M = 0.87 fy Ast x d 8.311 x 103 = 0.87 x 415 x Ast x 150

  415 Ast 1 −   20 x 1000 x 150  Ast = 156.86 mm2 Ast provided in Longer direction = µ Ast = 0.7 (156.86) = 109.8 Ast min = 0.12% to D = 0.12 / 100 x 1000 x 170 Ast = 204 mm2 Provide Ast = 204 mm2 For 8 mm φ , spacking = π x 10 2 4 = 246 204

1000 x

Provide 8 mm  @ 240 mm c/c. Check for shear

14

IS 3370 (Part II) 1984 v = Vu/bd Vu = WuL/ 2 = 13.5 x 4 /2 = 27KN τ v= Vu bd Rf = WuL 13.5 x 4 = = 27 KN 2 2 τ c = 0.28 N/mm2 c1 = K τ c = 1.26 x 0.28 = 0.3528 N/mm2 v < c1 Hence section is safe in shear Design a doom for a cylindered water tank of diameter 12.5 m use M20 concrete of Fe 415 steel Soln: D = 12.5 m Rise = 12.5/5 = 2.5m R2 = (R – r)2 + (D/2)2 R2 = (R – 2.5)2 + (12.5/2)2

16

IS 3370 (Part II) 1984 R = 9.06 m Cos θ = 0.724 The done is subjected to meridonial thorust and hoof force for which the stress should be within permissible compressive strength of concrete Assume thickness of slab, t = 100 mm Loadings on slab: Self wt = 25 x 0.1 = 2.5 KN/m2 Live load = 2 W = 4.5 KN/m2 Ast =

HZo comp 90.5 x 103 = = 603.38mm3 σ st 150 Assume 16mm  , no of bars =

603.3 162 πx 4 Provide 4 nos of 16 mm  bar Size of ring beam is based on tensile stress of concrete  ct = Ft / Act (m-1) Ast 2.8 = 90.50 x 103 / AC + (13.33-1) 603.3 14

IS 3370 (Part II) 1984 Ac = 24886.4 mm2 Provide ring beam of size 160 x 160 mm Design a water tank for field base condition for a capacities of 4 lakh litres height of tanks is 4m. permissible stresses σ came σ

cbc

st

= 150 N/mm2 for M20

= 7 N/mm2, j = 0.84, R = 1.16

Soln:Capacity 4 x 105 lr = 400 m3 Ht = 4m Volume πd 2 x4 4 400 x 4 = d2 πx 4

d = 11.28 Provide diameter d = 115.m Thickness of tank based on emtiral relation. T min = 30 (4) + 50 = 170 mm Provide a thickness of 170 mm Non – dimensional parameter = 42 ( 4)2 = = 8.18 Dt 11 .5 x 0.79

Hoop tension devolor on the 10 all = T = co-efficient WH D/2 16

IS 3370 (Part II) 1984 BM develop along the W911, Co-eff x WH3 Reborins tasie 9 of SI = 3370 (Pert IV) Co-efficient for m x m hoof tension = 0.577 H2 = C.ett Dt

8 = 0.575 10 – 0.608 8.18 – 0.577 For

of 8.18 the co-efficient for hoof tension is actions it 0.64 H2 Dt

∴

mass – hoop tension, T = 0.577 x 9.81 x 4 x 11.5/2

T = 130.18 KN Co – efficient for max B.M is taken from table to of IS 3370 (Part IV) 8  0.0146 10  0.0123 8.18  0.0148 Co- efficient of max B.M – 0.0143 actions at 1.04 (at Base) Moment = Co. efficient x WH3 = -0.0143 x 9.81 x 43 B.M = -8.97 KNm Transverse Reinforcement:

16

IS 3370 (Part II) 1984 Ast =

H t 130.18 x 108 = σ st 150 Ast = 867.86 mm2 For 16 mm φ S = π x 16 2 1000 x 4 = 231 .69 mm 867.8

∴

Provide 16 mm φ @ 230 mm c/c

Adopt 12 mm φ @ 130 mm c.c Vertical Steel:- The vertical reinforcement tension for the max. BM 0.8 8.97 KN m by working shows mif Ast = M σst sd

Min – depth, d =

8.97 x 106 M = = 87.93 mm << 127 m Rb 1.16 x 1000 Overall depth of wan, thickness = 170 mm Adopiting a clear cover of 25 mm D = 170 – 25 – 12/2 – 12 D = 127 mm Ast = 8.97 x 10 6 = 560 .55 mm 2 150 x 0.84 x 127 15

IS 3370 (Part II) 1984 For 12mm, S = π x 12 2 1000 x 4 = 201.77 mm 560.5

∴

Provide 12 mm  @ 200 mm c/c.

Min. reinforcement Ast min = 0.3 % C.S = ***** Ast min = 510 mm2 < Ast rerd. Design of base slab: Assume 150 mm thick base slab, provide 0.3% of C/S as reinforcement along both the faces. Ast = 0.3 x 1000 x 150 = 450 mm 2 100 For 8 mm, S = π x 82 1000 x 4 =111.70 mm 450

Provide 8 mm 

@ 200 mm c/c both base along both faces.

Provide haunch of side 150 x 150 mm with min. steel of 8 mm @ 200 mm C/C along the face of the haunch Figure **** TUTORIAL Design a rectangular water tanks of size 4m x 7m with height 3.5m use M20 concrete of Fe – 415 steel. Design constant j = 0.053 R = 1.32 15

IS 3370 (Part II) 1984 Soln:L x B  7m x 4m L/B = 7/4 = 1.75 x2 H = 3.5 m h = H/4 or 1 =

3.5 = 0.875 or 1m 4

h=1m P = p (H-h) = 9.81 (3.5 – 7) P = 24.53 KN/m2 To find the final moment at the junction of lons will & short will based on the FEM & D.F. Moment distribution is done For any joint, D.F =

D.F =

I 1 / L1 I1 I 2 + L1 L 2

FEM MFAB = PB 2 24.53 x 4 2 = = 32.70 KNm 12 12

MFAD = 82 2 24.53 x 7 2 = = 100.16 KNm 12 12

D.F Jo int

A

Member AB

14

AD

IS 3370 (Part II) 1984

I1 / L1 I1 I 2 + L1 L2

D.F

1/ 4 1 1 + 4 7

I 2 / L2 I1 I 2 + L1 L2

1/ 7 = 0.36 1 1 + 4 7

= 0.611

Joint

Member

A

AB

AD

D.F

0.64

0.36

FIM

-32.70

100.16

B.M

-43.17

-24.28

Total

-75.87 KNm

75.88 KNm

Fixed B.M is 75.88 KNm Free B.M (For B.M) a) Along shorter direction = PB 2 24.53 x 4 2 = = 49.06 KNm 8 8

b) Longer direction = PL2 24.53 x 7 2 = = 150.24 KNm 8 8

resultant B.M at the mid span is a) Short will == -75.88 to 49.06 = -26.82 KNm b) Long will = -75.88 + 150.24 = 74.36 KNm At support = 75.88 KNm The will is design for me max B.M of for 14

IS 3370 (Part II) 1984 M max = 75.88 KNm Reinforcement details: Ast L = M − PL x P + 2 σst id σst

Ast B = M − Pb x P + B σst id σst

PL = P x B/2 = 24.53 x 4/2 = 49.06 KN PB = P x L/2 = 24.53 x 7/2 = 85.86 KN X = D/2 – eff.cover Providing a clear cover of 25 mm of bar 10 φ mm Eff.cover = 25 x 10/2 = 30 mm d=

75.88 106 M = = 239.75mm Rb 1.32 x 1000 provide d = 240 mm overall depth, D = 240 + 30 = 270 mm

∴

thickness of will is 270 mm

X= 270 − 30 2 X = 105 mm

14

IS 3370 (Part II) 1984 Ast L = 75.88 x 10 6 − 85.86 x 10 3 x 105 49.06 x 10 3 + 150 x 0.853 x 240 150

Ast L = 2630.33 mm2 Ast B = 75.88 x 10 6 − 85.86 x 10 3 x 105 85.86 x 10 3 + 150 x 0.853 x 240 150

Ast 8 = 2749.83 mm2 Provide 20mm φ bar, S = π x 20 2 1000 x 4 = 119 .43 mm 2630 .33

∴

Provide 20 mm φ @ 110 mm c/c

Since the thickness is greater than 200mm, the reinforcement is placed alogn both forces.

∴

Along transverse direction (HZ) provide

20 mm φ @ 220mm c/c along both faces in the short well & long wall. Vertical Reinforcement For

the

cantilever

action,

PHh 2 9.21 x 3.5 x 2 2 = 6 6

M = 5.723 KN-m

16

moment

develop

is

IS 3370 (Part II) 1984 Ast min:0.3 % c/s for 100 mm of 0.2% C/S for 450 mm

∴

for 270 mm  0.251% C/S

Ast min = 0.251 x 1000 270 100 = 677.7 mm2 For 10mm φ , S = π x 10 2 1000 x 4 = 115.87 mm 677.7

∴

Provide 10mm  @ 220 mm C/C as vertical reinforcement along both

the faces the taks for long will and short will. Design of Base slab:Assuming a thickness of 150 mm min. slab is provided for the base slab. Since it is resting on firm ground. = 450mm2

Ast min = 0.3 x 100 x 150 = 450mm 2 1000 For 8mm φ S =

π x 82 1000 x 4 = 111 .7 mm 450

14

IS 3370 (Part II) 1984

∴

Provide 8 mm φ @ 220 mm C/C along both ways along both faces

Provide a layer of lean mix (M10) for a thickness of 75 mm below base slab. Figure ************** A straight staircase is made of structures independent tread slab cantilevered from the R.C wall given riser is 150 mm and trend is 300mm. width if flight is 1.5m. design a typical cantilever tread slab. Any live load for over crowing. Use M20 concrete of Fe. 250 steel. Soln:Given R = 159 mm, T= 300 mm, w = 1.5 m M20 9 Fe 250 Loading on tread slab (0.3m) i) self wt = 0.15 x 25 x 0.3 = 1.125 KN/m ii) floor finish = 0.6 x 0.3 = 1.125 KN/m dead load = 1.305 KN/m B.M D.L = 13.5 x 1.52 = 1.46 KNm 2

Live load moment is a minimum of i) UDL due to live load on stair (for over providing case) WL = 5 KN/m2 ii) For cantilever tread slab the live load condition, is check for it load of 1.3 KN at tread.

16

IS 3370 (Part II) 1984 Figure ************* B.M LL1 = 1.52 x 1.5 2

B.M LL2 = 1.3 x 1.5 = 1.95 KN Maximum of B.M LL = 1.9 KN m Total BM = BM DL + B.D LL = 1.46 + 19.5 = 3.41 KNm. Factored B.M = 1.5 x 3.41 = 5.12 KNm Assume clear covet of 15 mm & 12mm  bar. d = 150 – 15 – 12/2 = 129 mm Mu = 0.87 fy Ast d

f Ast    1 − y  Fck td   5.12 x 106 – 0.87 x 250 x Ast x 129

  250 Ast  1 −  20 x 300 x 129   Ast = 194.73 mm2 n=

Ast 194373 = 2 = 2.47 say 3 nos ast π x 10 4 provide 3 nos of 10 mm  bar Distribution steel:-

14

IS 3370 (Part II) 1984 Ast min = 0.15% & D = 0.15 x 300 x 150 100 = 67.5 mm2 S= π x 82 4 = 744 mm 67.5

1000 x

Provide 8 mm  @ 300 mm C/C Check for shear:τv =

Vu bd

Dead load V DL = 1.3.5 x 1.5 = 1.959 KN V LL1 = 1.5 x 1.5 = 2.25 KN VLL2 = 1.3 KN Max VLL 2.25 KN Total Vu = 2.25 + 1.958 = 4.208 KN Factored Vu = 1.5 x 4.208 = 6.31 KN 

v

= 6.31 x 10 = 0..168 N / mm 2 300 x 129

∴

τ

c

= 0.48 N/mm2

K = 1.3 from table 16

IS 3370 (Part II) 1984 Tc1 = K Tc = 1.3 x 0.48 = 0.624 N/mm2 τ

c

2.8/2 = 1.4 N/mm2

τ

v

x c x τ

c

mm

Hence safe in shear Ld = φ σ3 10 x 0.87 x 250 = = 453 .15 mm 4 τ bd 4 x 1.2

Design a suitable countertor reintaining way to support a leveled back till of height 7.5m above g.L on the toe side. Assume good soil for the foundation at a desm of 1.5m below G.L the SBC of soil is 170 KN/m2 with unit weight as 16KN/m3. The angle of internal friction is = 300 the co-efficient of friction b/w the soil and concrete is 0.5 use M25 concrete and Fe 415 steel. Soln:Figure ************** Minimum depth of foundation = *** = *** Depth of foundation = 1.5 m Height of wall = 7.5 + 1.5 = 9m Thickness of had and stem = 5% of 9m = 0.45m ** 0.5. Thickness of toe slab = 8% of 9 m = 0.72 m.

16