design of 2 way slab

8/12/2019 Design of 2 Way Slab

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Design of two-way slab over Abutments

Panel No = A

Input data: ACI Moment Coefficients

Case No = 1

la = 12 ft m = 0.75

lb = 16 ft Caneg = 0.000

f'c = 2.5 ksi Cbneg = 0.000

fy = 40 ksi Cadl = 0.061

DL add = 0.036 k/ft2 Cbdl = 0.019

LL = 1 k/ft2 Call = 0.061

h, cal = 3.73 in Cbll = 0.019

h, used = 9 in Wa = 0.760

Wb = 0.240

Loads Calculation: Moments Calculation:

DL wu = 0.208 k/ft2 Ma(-ve) = 0 k-in

LL wu = 1.7 k/ft2 Mb(-ve) = 0 k-in

TL wu = 1.908 k/ft2 Ma(+ve) = 201.12 k-in

Mb(+ve) = 111.37 k-in

Steel Calculation:

da = 8 in

db = 7.75 in

b1 = 0.85

Asmax = 2.158 in2

Asmin = 0.216 in2

Smax = 18

Moments Asreq #3(in c/c) #4(in c/c) #5(in c/c) #6(in c/c)

Ma(-ve) 0.216 6.1 11.1 17.2 18

Mb(-ve) 0.216 6.1 11.1 17.2 18

Ma(+ve) 0.754 1.8 3.2 4.9 7

Mb(+ve) 0.417 3.2 5.8 8.9 12.7



DESIGN OF STEP RETAINING WALL

Note: Change only Blue values.

0

gcon = 150 lb/ft3

2

gmat = 130 lb/ft3 1.25 2 1

gback = 120 lb/ft3

fsoil = 30 3

B.C = 3360 lb/ft2 1.25 3.25

TL 1120 lb/ft 3

Va 27 ft2 4.5

TL/ Area 41.4815 lb/ft2

Ka = 0.333

TOP STEP

Tli = 13.8133 0

P1 = 13.81

y1 = 0.5

BFLi = 39.96

P2 = 19.98 2y2 = 0.33 3

1

S.No W,lb x,ft M=w x

1 0 1 0 Over-Turning = [OK]2 780 1 780 Sliding = [OK]

Sum 780 780 Tension Check at the heel = [OK]

Over-Turning

Stabilizing Moment = 780

Over-Turning Moment = 13.4984

Factor of Safety = 57.78 [OK]

Sliding

Frictional Resistance = 390.14Sliding Force = 33.79


Tension Check at the heel

a1 = 0.98 [OK]

2

3

w2

w1



STEP#2

Tli = 13.8133 0

P3 = 55.25 2.5

y3 = 2 2

BFLi = 159.84 1.25 2 1

P4 = 319.68y4 = 1.33 4 3

3.25


1 450 2.5 1125 Over-Turning = [OK]2 780 1 780 Sliding = [OK]

3 1267.5 1.63 2066.03 Tension Check at the heel = [OK]

Sum 2497.5 3971.03

Over-Turning

Stabilizing Moment = 3971.03



Sliding

Frictional Resistance = 1249.21

Sliding Force = 374.93


Tension Check at the heela2 = 1.38 [OK]



STEP# 3

Tli = 13.8133 0

P5 = 96.69 2.5

y5 = 3.5 2

BFLi = 279.72 1.25 2 1

P6 = 979.02y6 = 2.33 3

7 1.25 3.25

4.5 3


1(a) 450 2.5 1125 Over-Turning = [OK]1(b) 900 2.75 24752 780 1 780 Sliding = [OK]3 1267.5 1.63 2066.03 Tension Check at the heel = [OK]

4 1755 2.25 3948.75

Sum 5152.5 10394.8

Over-Turning




Sliding





a3 = 1.51 [OK]

w1

w3



DESIGN OF RETAINING WALL

2

gmat = 150 lb/ft3

gback = 120 lb/ft3

fsoil = 30

B.C = 3306 lb/ft2

11

Ka = 0.333

P = 2417.58

y = 3.67 0 4.5 0


1 0 3.25 0 Over-Turning: [OK]

2 3300 1 3300 Sliding: [OK]

3 3712.5 3.5 12993.8 Tension: [OK]

4 2970 5 14850 Bearing: [OK]

5 0 6.5 0

Sum 9982.5 31143.8

Over-Turning



Factor of Safety = 3.51 2.5 [OK]

Sliding



Factor of Safety = 2.07 > 1.5 [OK]


a = 2.23

Bearing Capacity Check

Bearing Pressure at the = 2981.76 < than the given Bearing Capacity [OK]

0

6.5

as Resultant lies within the Middle

Third of the base width [OK]

w2

w3

w1

w4

w5

P



DESIGN OF DRY RETAINING WALL

1.5

gmat = 150 lb/ft3

gback = 120 lb/ft3

fsoil = 30

B.C = 3306 lb/ft2

11

Ka = 0.333

P = 2417.58

y = 3.67 0 2.75 0


1 0 2.13 0 Over-Turning: [NG]

2 2475 0.75 1856.25 Sliding: [NG]3 2268.75 2.42 5490.38 Tension: [NG]

4 1815 3.33 6043.95 Bearing: [NG]

5 0 4.25 0

Sum 6558.75 13390.6

Over-Turning



Factor of Safety = 1.51 2.5 [NG]

Sliding

Frictional Resistance = 3280.57Sliding Force = 2417.58

Factor of Safety = 1.36 <1.5 [NG]


a = 0.69

Bearing Capacity Check

Bearing Pressure at the = 4669.65 > than the given Bearing Capacity [NG]

0

4.25

as Resultant does not lie within the

Middle Third of the base width [NG]

w2

w3

w1

w4

w5

P



Design of Footing

Footing Identification = C1

Input Data:

fy = 40 ksif'c = 2 ksi

Col. side,b = 12 in

S.Load, SL = 144 k

F.Load, FL = 208 k Use No.8 @ 12.2 c/c

B.Capacity,BC = 2 ksf

How deep ?h = 4 ft

Footing width,W = 9.5 ft Try( 9.8 )

Footing Depth,D = 16.5 in Try( 17.02 )

Enter Bar# Used = 8 [Ab = 0.79 in2]

SOLUTION:

Effective Bearing Capacity

qe = B.C. - 0.125 x h = 1.5 ksf

Area & sides of Footing:

A = S.L./ qe = 96 ksf

B = ( A )1/2 = 9.8 ft

Factored Soil Pressure:

qu = F.L./ A = 2.3 ksf

Minimum Depth of Footing Based on Punching Shear:

Depth = D = Quadratic Formula = 17.02 in

Punching Shear Check:

Effective Depth = d = D -3.5 = 13 in

bo = 4 ( b + d ) = 100 in

App. Shear Vup = qu [(W2-(b+d)/12)

2] 197.59 k

Capacity, Vcp = [0.85 x 4 (fc*1000)1/2

bo x d ]/1000 =

= 197.67 k [OK]

Beam Shear Check:

App.Shear = Vub = qu [(W/2-b/24-d/12) x W = 69.19 k

Capacity, Vcb = [0.85 x 2 (fc x 1000)1/2 (12 W) d]/1000 =

= 112.67 k [OK]

Flexural Design Per ft:

Mu = 12 [qu (W/2 - b/24)2/ 2] = 249.26 k-in

Asmin = 200 / (fy x 1000) 12 d = 0.78 in2

As = 0.85 12 (fc/fy) [d - (d2 - Mu/(0.3825 12 fc)

1/2)] =

= 0.43 in2

Asreq = Greater of above two = 0.78 in2

(Spacing of bars 12.2 in c/c)

Development Length Requirement:

Available Length = [12 W - b]/2 - 3 = 48 in

Required Length = 0.04 fy / (fc)1/2

db = 36 in [OK]



Material Quantity Nature of labour Labour

Required

Cement 9.94 Bags

sand 47.72 Cft

crush 95.44 Cft Labour 4.5 NoBricks 1450 Nos

cement 3.3 Bags

sand 23.76 cft Labour 3 No

Bricks 1450 Nos

cement 3.3 Bags

sand 23.76 cft Labour 5 No

Cement 0.96 Bags Mason 1 No

sand 6.94 Cft Labour 1 No


sand 3.10 Cft Labour 1.25 No

Cement 0.29 Bags Mason 3/4 No


Bricks 334 Nos

cement 1.12 Bags


Bricks 533 No

cement 1.5 Bags

sand 10.0 Cft Labour 2.25 No

Cement 3.06 Bags

sand 10.92 Cft

crush 10.34 Cft Labour 1 No


sand 5.92 Cft Labour 3/4 No

Cement 6.78 Bags

sand 22.93 Cft


Cement 17.23 Bags

sand 41.34 Cft

crush 82.62 Cft Labour 7 NoCement 17.23 Bags

sand 41.34 Cft


Cement 17.23 Bags

sand 41.34 Cft

crush 82.62 Cft

Pudlo 86.15 Lbs

Carpenter 3/4 No

Helper 1/4 No

Carpenter 1/4 No

Helper 1/8 No

Wood 0.10 Cft Carpenter 1/4 No 1.0 Sft Helper 1/8 No

Carpenter 1/10 No

Helper 1/30 No

Wood 0.13 Cft Carpenter 1/10 No

1 Sft Helper 1/8 No

Wood 0.06 Cft Carpenter 1/10 No

1.0 Sft Helper 1/10 No

Carpenter 1/10 No

Helper 1/8 No

Analysis of RatesBy: Atif Ghani

0.1 Cft22 Wooden Shelf 1" Thick 1 Sft Wood

21 G.I Wire Gauze Fitting 1 Sft

0.15 Cft

20Fly Proof Shutter 1.5"

thick1 Sft

19Battened Door Shutter

2" thick1 Sft

Wood

Wood 0.12 Cft

Wood 1.25 Cft

Mason 1 No

Labour

Steel 1.04 CWT Labour Cost per CWT

7 No

1 Cft

17Panalled Shutter 1.5"

Thick1 Sft

13

18

16 Wooden Frames

2 No

1 No

15 Reinforcement 1 Cwt

1 No

14R.C.C 1:2:4 Water

Tank100 Cft

Mason

12R.C.C 1:2:4 Lintel,

Column, Beam,Shelf

Etc

100 CftMason

Mason

3/4 No

10Skirting 1/2" thick in 1:3

Cement Mortar 100 Cft

91.5" Thick 1:2:4 Tiles

on Roof with 1:6

cement Mortar &

100 CftMason

1.25 No

8Ist class Burnt Bricks

on Edge in 1:6 cement

mortar & pointion with

100 CftMason 1.25 No

7Ist class Burnt Bricks

Flate in court yard in

1:6 cement mortar &

100 CftMason

6P.C Pointing in 1:3

Cement Mortar 100 Sft

Glazed Shutter 1.5"Thick 1 Sft

11P.C.C Floor 3" Thick

1:2:4 3" thick 1"4:8100 Cft

R.C.C 1:2:4 Slab 100 Cft

5Celling Plaster 1/2"

thick in 1:3 Cement100 Sft

4 No

4Wall Plaster 1/2" Thick

in 1:6 Cement Mortar 100 Sft

3Ist class Burnt Bricks in

Bricks Masnory in

Foundation in 1:6

100 CftMason

1/2 No

2Ist class Burnt Bricks in

Bricks Masnory in

super structure in 1:6

100 CftMason 2.5 No

1 1.4.8 100 CftMason

S. No Description of item Unit

Material Cost Labour Cost



Design of Singly Reinforced Beams

Input data:

Mu(+or-) = 2256 k-in MainBar# = 6

Critical,Vu = 70 k Stirrup# = 3

f'c = 2 ksi No.ofRows = 2

fy = 40 ksi Ln = 10

R = 16 no c/c b/w = 10

hmin = 6 in 1.Rectangulr, 2.Tee, 3.L-Beam

h, used = 24 in Choice = 1

bw = 12 in d = 20.875

hf = 8 in

As,used = 3.53 in2, [Try 3.62 in2]

be

hf

h

Flexural Design:

bw

be = 22 in

b1 0.85

Check for Rectangular or Tee Beam Analysis, For a=hf

As = abs(Mu)/(0.9fy(d - hf/2) = 3.714 in2

a = As fy/(0.85 fc be) = 3.972 < hf [Rect. Analysis]

b = 12 in

Asmax = 5.47 in2 (for b=bw)

Asmax = 10.03 in2 (for b=be)

Asmin = 1.253 in2

From Quadrtic Equation

Asreq = 3.62 in2

[OK]

Design Moment (Only Singly):

a = As fy/(0.85 fc be) = 6.922 in [OK]

Md = 0.9 As fy (d - As fy/(1.7 fc b) = 2213 k-in [NG]

Shear Design:

Shear Capacity of Beam:

fVc =0.85 [2 (fc)1/2bw d ] = 19.04 k

Maximum Spacing of 2-Leg Stirrups: Av = 0.22 in2

Smax = [Minimum of Avfy/(50bw), d/2, 24]= 10.4 in

Shear Taken by Stirrups:

Vs = (fVu - fVc)/f = 47.6 k

Required Spacing of Stirrups:

S = [Minimum of Smax & (Av fy d / Vs)] = 3.9



Design of two-way slab

Panel No = A

Input data: ACI Moment Coeffici

Case No = 9

la = 15 ft m =

lb = 20 ft Caneg =

f'c = 4 ksi Cbneg =

fy = 60 ksi Cadl =

DL add = 0.036 k/ft2 Cbdl =

LL = 0.1 k/ft2 Call =

h, cal = 4.67 in Cbll =

h, used = 5 in Wa =

Wb =

Loads Calculation: Moments Calculation:

DL wu = 0.138 k/ft2 Ma(-ve) = 64.86

LL wu = 0.17 k/ft2 Mb(-ve) = 20.7

TL wu = 0.308 k/ft2 Ma(+ve) = 32.66

Mb(+ve) = 15.24

Steel Calculation:

da = 4 in

db = 3.75 in

b1 = 0.85

Asmax = 0.962 in2

Asmin = 0.108 in2

Smax = 10

Moments Asreq #3(in c/c) #4(in c/c) #5(in c/c) #6(in c/c)

Ma(-ve) 0.319 4.1 7.5 10 10

Mb(-ve) 0.108 10 10 10 10

Ma(+ve) 0.156 8.5 10 10 10

Mb(+ve) 0.108 10 10 10 10

Load Transfer:

Load Transfer To Longer Beam = 1.987 k/ft

Load Transfer To Shorter Beam = 0.431 k/ft



ents

0.75

0.078

0.014

0.031

0.007

0.046

0.013

0.860

0.140

k-in

k-in

k-in

k-in

design of 2 way slab

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