Design of concrete frame building manual design

This report only covers the analysis and design of static forces. Static forces that took in the account in this report include self-dead load of structure, super dead load which include floor finishing load, ceiling load, duct and pipes loads and live loads only. The design building consists of five story. ACI coefficients are used to find moment and shear stresses in the beams and columns. Strength method is used to design all of these elements include beams and columns. ACI 318-19 and Abu Dhabi International Building Code is taken in mind while design the structure. The report shows the design of typical beam and slab. Similarly, the report covers the design of ground floor typical column and its foundation. Isolated footing is used to transfer the load of structure to the ground. AutoCAD is used for drafting purpose.?Excel sheets are used in beams and foundation for economical design.

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Figure 2 Layout of building

Figure 3 Layout of beams and column on AutoCAD

Load calculations for Slab

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Assume slab thickness 150mm

Dead load of Slab = 25 x 150/1000 = 3.75 KN/m2

Assume live load for minimum design load of building = 3 KN/m2

Assume super dead load include flooring marble, pipe ducts etc. = 2.5 KN/m2

Total dead load = 3.75 + 2.5 = 6.25 KN/m2

Total live load = 3 KN/m2

Load combination

1.4D = 1.4(6.25) = 8.75 KN/m2

1.2D + 1.6L = 1.2(6.25) + 1.6(3) = 12.3 KN/m2 Controls

Load Calculations for Beams

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Grid A

Assume size of Grid A beam = 300mmx 450mm

Self-weight of beam = 25 x 0.3 x 0.450 =3.375 KN/m

Dead load on Grid A beam = 6.25 KN/m2 x 3/2 = 9.375 KN/m

Live load on Grid A beam = 3 KN/m2 x 3/2 = 4.5 KN/m

Ultimate load on Grid A beam = 1.2(3.375 + 9.375) + 1.6(4.5)

??????????????????????????????????????????????????????= 22.5 KN/m

Grid B

Assume size of Grid A beam = 300mmx 450mm

Self-weight of beam = 25 x 0.3 x 0.450 =3.375 KN/m

Dead load on Grid A beam = 6.25 KN/m2 x (3/2 +4/2) = 21.875 KN/m

Live load on Grid A beam = 3 KN/m2 x (3/2 +4/2) = 10.5 KN/m

Ultimate load on Grid A beam = 1.2(3.375 + 21.875) + 1.6(10.5)

??????????????????????????????????????????????????????= 47.1 KN/m

Grid 2

Assume size of Grid A beam = 225mmx 350mm

Self-weight of beam = 25 x 0.225 x 0.350 =1.96 KN/m

Dead load on Grid A beam = 6.25 KN/m2 x (2.765/2) = 8.64 KN/m

Live load on Grid A beam = 3 KN/m2 x (2.765/2) = 4.14 KN/m

Ultimate load on Grid A beam = 1.2(1.96 + 8.64) + 1.6(4.14)

??????????????????????????????????????????????????????= 19.344 KN/m

Length of beam Grid 1 = 6.2m

Point load on Grid C beam = 19.344 KN/m x 6.2/2

????????????????????????????????????????????????= 59.96KN

Grid C

Assume size of Grid A beam = 300mmx 450mm

Self-weight of beam = 25 x 0.3 x 0.450 =3.375 KN/m

Dead load on Grid A beam = 6.25 KN/m2 x (6.2/2 +4/2) = 31.875 KN/m

Live load on Grid A beam = 3 KN/m2 x (6.2/2 +4/2) = 15.3 KN/m

Ultimate load on Grid A beam = 1.2(3.375 + 31.875) + 1.6(15.3)

??????????????????????????????????????????????????????= 66.78 KN/m

Load Calculation on Columns (Ground Floor)

C A-1 Ultimate Load

No of Floors x Ultimate load x Tributary area of column

5 x 12.3 x 6.6/2 x 3/2 = 304.425 KN

C B-1

5 x 12.3 x 6.6/2 x 3/2 = 304.425 KN

C C-1

5 x 12.3 x 4/2 = 123 KN

C A-3

5 x 12.3 x (3.3+3.15) x (3/2) = 595.01 KN

C B-3

5 x 12.3 x (3.4+3.2) x (3/2 + 4/2) = 1420.65 KN

C C-3

5 x 12.3 x (4/2+3.1) x (4.5325) = 1421.61 KN

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C A-4

5 x 12.3 x (3)/2 x (3.15) = 290.58 KN

C B-4

5 x 12.3 x (3/2+4/2) x (3.15) = 678.03 KN

C C-4

5 x 12.3 x (4/2+3.1) x (3.15) = 988 KN

C D-4

5 x 12.3 x (3.1) x (3.15) = 600.65 KN

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Beam Moments and Shear

Bending moment and shear coefficients of beams are found using ACI coefficients

Figure 4 Bending moment coefficients of beam

Figure 5 Shear coefficients of beam

Positive moment =

Span (6.6m) = ?

?????????????????????= 146.54KN – m

Negative moment =

(Interion support) =

?????????????????????????????????= 227.96KN – m

V’ =

?=

= 178.74KN

Slab Moments and Shear

Bending moment and shear coefficients of slabs are found using ACI coefficients

Figure 6 Slab Bending moment coefficients

Figure 7 Shear coefficients of slab

Design for one meter spam slab load = 12.3?x 1m

?????????????????????????????????= 12.3

Positive moment = ?=?????

??????????????????????= 12.3KN-m

Negative moment =

?= 12.3x

= 16.4 KN-m

Design Of beam for Flexure

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Fc’ =28Mpa

Fy = 420Mpa

Ultimate load on beam ‘’B’’ = 47.1KN/m

Initial size = (300 x 450) mm

Positive moment =

Span (6.6m) = ?

?????????????????????= 146.54KN – m

Negative moment =

(Interion support) =

?????????????????????????????????= 227.96KN – m

D = h -cover

= 450 – 40 = 410mm

Negative moment (edge support)

=

= ??

= 128.22Kn -m

Design for flexure

Positive moments = 146.54 KN – m

Row = 0.85?(1 –

Assume tension controlled = ? = 0.9

Row = ?(1-

Row = 0.00829

Row according to 9.6.1.2 ACI

Greater of

(A)

= ?= 0.0035

(b) = ?= ?= 0.0033

Amount of steel is in range

AST= Row x bx d

?????= 0.00829x300x410

???= 1019.67mm2

4 #19 gives 1136mm2

Check Nominal Strength

a =

??= ?= 66.82mn

B1 = 0.85

C = ?= 78.61mn

ET = 0.003[]

?= 0.003[] = 0.0037

Transition zone

Ф = 0.65 + (ET – 0.002)

?= 0.65 + (0.0037 – 0.002)

?= 0.7926

Ф Mn = ф Asfy ()

Ф Mn = 0.7926 x 1136 x 420[]

??= 142419845.9 N-mn

?= 142.42 KN – m < 146.54KN-m??

Increase the steel?

Let As = 3 # 22. As 1161mm2

As it is trail method use excel sheet

Try 4 # 22??As 154.8mm2

Selected 4#22

For negative steel

Interior = 4 # 25

?????Ast = 2040mm2

For negative steel edge support 3 # 22

????Ast = 1161mm2

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Figure 8 Design of beam for gird B Positive Steel

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Figure 9 Design of beam for gird B Interior Support

Figure 10 Design of beam for gird B Edge Support

Design of beam for shear

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Design for maximum shear

V’ =

?=

= 178.74KN

Vu = ф vc + ф vs

Ф vc = ??bwd

?= ?x300x410x10-3

=10.84KNx0.75 = 81.35KN

Vs’ =

Vs =

?= 129.85 KN

Provide 2 legged stirrups # 10

Vs = ?

S = ?

??= 188.31mn = 175mn

Vs max = ?bwd

?= x 300 x410 x 10-3

433.90 KN

Smax

If Vs ≤ ??bwd

?x 300 x 410 x10-3

129.85KN ≤ 216.95KN

Smax ? Smallest of

(1)??600

(2)??0.5d = 0.5x410 = 205mn

(3)???= ?=298.2mm

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Maximum spacing = 205mn ?200mn

Provided # 10 @175mn up to??from both side and rest on position & Provide # 10 @ 200mn c/c

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Detailing of beam

Figure 11 Long Section of Beam

Figure 12 Cross sections of beam

Design of slab

Spam (4m)

Fc’ = 28Mpa

Fy = 420 MPa

Design for one meter spam slab load = 12.3?x 1m

?????????????????????????????????= 12.3

Positive moment = ?=?????

??????????????????????= 12.3KN-m

Negative moment =

?= 12.3x

= 16.4 KN-m

Design of positive moment

B = 1000mm

H = height of slab

?????????= ?for both end continues

??????????= ?= 142.85 = 142.85 ? 150mm

D = h – cover

?????????????= 150 – 20 = 130mm

Row = 0.85x?(1-

Row = 0.0019

Ast = 0.0019 x 1000 x 130

= 254.70 mn2

# 10 @ 250mm

Minimum spacing 3h = 3x150 = 450mm

Design for negative steel

M = 16.4 KN – m

Row = 0.0026

Ast = 0.0026x1000x130

??????= 341.66mm2

Spacing = # 10 @ 200mn

Temperature and shrinkage reinforcement

Ast = 0.0018 bH

Ast = 0.0018x100x150

?????????= 270mm2?????????????????????????????

Maximum spacing = 5h

????????????????= 5x150mn??= 750mm

Detailing of Slab

Figure 13 Design of 4m Span Slab

Column design?

Design of column B – 4

Pu = 678.03KN

Mu = negative moment of gride B beam

Mu = 128.22KN -m

Assume column dimension = 400x600

?=

?= 0.8

F’c = 28Mpa

Fy = 420 Mpa

Use interaction diagram

Kn =

???????=

= 0.15

Rn =??

?=

?= 0.048

Plot the values in interaction diagram

Row = less than 0.01

Row min = 0.01bH

Ast = 0.01x400x600

= 2400mm2

Provide 8#22 gives 3096mm2

deign for shear rings spacing

(1) 16 Times dia of main bar

(2) 48 Times dia of ring

(3) minimum dimension of column

(1) 16x22 = 352 = 250m

(2) 48x10 = 480mn

(3) 400

Provide # 10??@ 250mn

Column detailing

Figure 14 Column Cross section

Design of foundation

F’c = 28MPa

Fy = 420Mpa

Column size = 400mm x 600mm

Service dead load = 5x6.25x[] x (3.15)

??????????????????= 344.53KN

Service live load = 5x3x ()(3.15)

?????????????????= 165.375KN

Bearing capacity =

Concrete unit weight =

Soil unit weight =

Assume depth of foundation = 1.5m

Effective bearing Capacity = Qa – rh

Qe = 180 – ()1.5 = 146.25 ????????????????????

Area required =

?????????????=??????

?????????????=???3.48m2

Size selected for foundation = 2m x 2n

?Qu = ?

???=

?= 169.509

Assume depth of foundation = 400mm

. d = 400 – 75 = 325mm

Check for one way Shear

?????????????Ф vc =0.75 x???bwd

?????????????????????=0.75 x???x 2000 x 325 x10-3

?????????????????????= 429.93 KN

????????????Vu = Qu x tributary area

???????????Tributary area = (2000/2 – 600/2 – 325) x 325

??????????????????????????????????????= 121875 mm2

??????????Vu = 169 x 121875/1000^2 = 20.59 KN

Ф vc > Vu Depth is satisfactory

Check for Two-way Shear / Punching Shear

?Ф vp =0.75 x???bod

??????????????????Bo = 2(400 + 325) + 2(600 + 325)

????????????????????????= 3300mm

?????????????????????=0.75 x???x 3300 x 325 x10-3

??????????????????????????????????= 354.69 KN

????????????Vu = Qu x tributary area

???????????Tributary area = 4m2 – (0.725 x 0.925)

??????????????????????????????????????= 3.32m2

??????????Vu = 169 x 3.32 = 518.59 KN

Ф vc < Vu Depth is not satisfactory, Increase the depth and trail using Excel sheet

Final d = 475mm, H = 550mm

Design for Reinforcement

Mu = Qu x B x l2/2

??????=??169 x 2 x 2^2/2 = 676 KN.m

Row = 0.85?(1 –

Assume tension controlled = ? = 0.9

E = ?(1-

?= 0.00411

Row minimum = 0.003

Ast =0.00411 x 2000 x 475 = 3904.5 mm2

Provide 8 #25 gives = 3925 mm^2m in Both directions

Detailing of foundation

Figure 15 Footing Details

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References

o??Hibbler, R.C. and Kiang, T., 2015.?Structural analysis. Upper Saddle River: Pearson Prentice Hall.

o??Darwin, D., Dolan, C.W. and Nilson, A.H., 2016.?Design of concrete structures. New York, NY, USA:: McGraw-Hill Education.

o??ACI Committee and International Organization for Standardization, 2008. Building code requirements for structural concrete (ACI 318-08) and commentary. American Concrete Institute.

o??Hassoun, M.N. and Al-Manaseer, A., 2020.?Structural concrete: theory and design. John wiley & sons.

o??DMA, 2013. Abu Dhabi International Building Code.