DAR GROUP

Project

QF

Job Ref.

QF

Section

Sheet no./rev.

1

Calc. by

MS

Date

09-Nov-106

Chk'd by

AE

Date

App'd by

IC

Date

RC TWO WAY SLAB DESIGN TO ACI 318 – 05

½ Column strip Middle strip ½ Column strip

½ Column strip

Middle strip

½ Column strip

x

y

lx

ly

TWO-WAY SLAB PANEL

200 mmSection at middle strip along X axis

Slab definition

Slab thickness; h = 150 mm

Span along x – dir; lx = 6000 mm

Span along y – dir; ly = 4000 mm

Long span to short span ratio less than 2, hence two-way slab

Clear cover to positive reinforcement; cc = 20 mm

Clear cover to negative reinforcement; c’c = 20 mm

Effective flanged beam section

Flange width; bf = 1000 mm

Beam depth; hb = 550 mm

Beam width; bw = 300 mm

Materials

Specified compressive strength of concrete; f’c = 35 MPa

Specified yield strength of reinforcement; fy = 415 MPa

Modulus of elasticity; ESACI = 199948MPa

Loads

Total factored ultimate load on slab; wu = 18.00 kN/m2

DAR GROUP

Project

QF

Job Ref.

QF

Section

Sheet no./rev.

2

Calc. by

MS

Date

09-Nov-106

Chk'd by

AE

Date

App'd by

IC

Date

Check for the depth of slab provided

Clear span x – dir; lnx = lx - bw / 2 - bw / 2 = 5700 mm

Clear span y – dir; lny = ly - bw / 2 - bw / 2 = 3700 mm

Ratio; = max(lnx, lny) / min(lnx, lny) = 1.541

Moment of inertia of beam; Ib = 6922916667 mm4

Moment of inertia of slab along x – axis; Isx = 1125000000 mm4

Moment of inertia of slab along y – axis; Isy = 1687500000 mm4

Flexural stiffness ratio along x – axis; fx = Ib / Isx = 6.154

Flexural stiffness ratio along y – axis; fy = Ib / Isy = 4.102

Average flexural stiffness ratio; fm = (fx × 2 + fy × 2) / 4 = 5.128

Minimum depth required; hmin = 125 mm

PASS - slab thickness provided is sufficient for deflection

Moment computation (x – direction)

Width of column strip (half); lcol_stp = min(0.25 × lx, 0.25 × ly) = 1000 mm

Width of middle strip; lmid_stp_x = ly - 2 × lcol_stp = 2000 mm

Factored moment; Mox = wu × ly × lnx2 / 8 = 292.41 kNm

Moment distribution for interior panel; M’ux = 0.65 × Mox = 190.07 kNm

Mux = 0.35 × Mox = 102.34 kNm

Moment distribution in column and middle strips

ly / lx = 0.67

fx × ly / lx = 4.10

Moment distribution factors; fx_neg = 0.85

fx_pos = 0.85

Total column strip design moments; M’x_col_stp = fx_neg × M’ux = 161.56 kNm

Mx_col_stp = fx_pos × Mux = 86.99 kNm

Column strip beam moments; M’x_beam = 137.32 kNm

Mx_beam = 73.94 kNm

Column strip slab moments; M’x_slab = M’x_col_stp - M’x_beam = 24.23 kNm

Mx_slab = Mx_col_stp - Mx_beam = 13.05 kNm

Total middle strip design moments; M’x_mid_stp = (1 - fx_neg) × M’ux = 28.51 kNm

Mx_mid_stp = (1 - fx_pos) × Mux = 15.35 kNm

Required flexural nominal strength in column strip per 1 m width of slab

M’nx_col = M’x_slab / (0.9 × (2 × lcol_stp - bf)) = 26.93 kNm/m

Mnx_col = Mx_slab / (0.9 × (2 × lcol_stp - bf)) = 14.50 kNm/m

Required flexural nominal strength in middle strip per 1 m width of slab

M’nx_mid = M’x_mid_stp / (0.9 × lmid_stp_x) = 15.84 kNm/m

Mnx_mid = Mx_mid_stp / (0.9 × lmid_stp_x) = 8.53 kNm/m

Moment computation (y – direction)

Width of column strip (half); lcol_stp = min(0.25 × lx, 0.25 × ly) = 1000 mm

Width of middle strip; lmid_stp_y = lx - 2 × lcol_stp = 4000 mm

Factored moment; Moy = wu × lx × lny2 / 8 = 184.82 kNm

Moment distribution for interior panel; M’uy = 0.65 × Moy = 120.13 kNm

Muy = 0.35 × Moy = 64.69 kNm

Moment distribution in column and middle strips

lx / ly = 1.50

fy × lx / ly = 6.15

Moment distribution factors; fy_neg = 0.60

DAR GROUP

Project

QF

Job Ref.

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Section

Sheet no./rev.

3

Calc. by

MS

Date

09-Nov-106

Chk'd by

AE

Date

App'd by

IC

Date

fy_pos = 0.60

Total column strip design moments; M’y_col_stp = fy_neg × M’uy = 72.08 kNm

My_col_stp = fy_pos × Muy = 38.81 kNm

Column strip beam moments; M’y_beam = 61.27 kNm

My_beam = 32.99 kNm

Column strip slab moments; M’y_slab = M’y_col_stp - M’y_beam = 10.81 kNm

My_slab = My_col_stp - My_beam = 5.82 kNm

Total middle strip design moments; M’y_mid_stp = (1 - fy_neg) × M’uy = 48.05 kNm

My_mid_stp = (1 - fy_pos) × Muy = 25.87 kNm

Required flexural nominal strength in column strip per 1 m width of slab

M’ny_col = M’y_slab / (0.9 × (2 × lcol_stp - bf)) = 12.01 kNm/m

Mny_col = My_slab / (0.9 × (2 × lcol_stp - bf)) = 6.47 kNm/m

Required flexural nominal strength in middle strip per 1 m width of slab

M’ny_mid = M’y_mid_stp / (0.9 × lmid_stp_y) = 13.35 kNm/m

Mny_mid = My_mid_stp / (0.9 × lmid_stp_y) = 7.19 kNm/m

Reinforcement calculations

Stress block depth factor; 1 = max(0.65, 0.85 - 0.05 (f'c - 28 MPa) / 7 MPa = 0.85

Reinforcement percentage at strain of 0.004; b = 0.85 1 f’c / fy (0.003 / (0.003 + 0.004)) = 0.026

Maximum reinforcement percentage; max = b = 0.026

Column strip (x – direction)

Negative Reinforcement

Depth to negative steel; d’x_col = h - c’c - D’col_x / 2 = 124 mm

Flexural resistance factor; R’ux_col = M’nx_col / d’x_col2 = 1.77 MPa

Reinforcement required; ’reqx_col =0.85f’c/fy×[1-(abs(1-2×R’ux_col /(0.85×f’c)))] = 0.00439

Outer compression steel not required

Area of tension steel required; A’sx_req_col = ’reqx_col × d’x_col = 542 mm2/m

Minimum area of tension steel required; A’sx_min_col = max(1.4 MPa×d’x_col/fy,(f’c/1 MPa)×d’x_col1 MPa/(fy 4))

A’sx_min_col = 440 mm2/m

Maximum area of tension steel; A’sx_max_col = max × d’x_col = 3225 mm2/m

Maximum allowable spacing; smax = min(2 × h, 450 mm) = 300 mm

Reinforcement provided (negative), No. 4 @ 200 mm centers

Area of tension steel provided; A’sx_col = 664 mm2/m

PASS - area of steel provided is OK

Actual tensile bar spacing provided; s’x_col = 200 mm

PASS - spacing of bars less than allowable

Positive reinforcement

Depth to positive steel; dx_col = h - cc - Dcol_x / 2 = 124 mm

Flexural resistance factor; Rux_col = Mnx_col / dx_col2 = 0.95 MPa

Reinforcement required; reqx_col =0.85f’c/fy×[1-(abs(1-2×Rux_col/(0.85×f’c)))] = 0.00233

Outer compression steel not required

Area of tension steel required; Asx_req_col = reqx_col × dx_col = 288 mm2/m

Minimum area of tension steel required; Asx_min_col = max(1.4 MPa×dx_col/fy,(f’c/1 MPa)×dx_col1 MPa/(fy 4))

Asx_min_col = 440 mm2/m

Maximum area of tension steel; Asx_max_col = max × dx_col = 3225 mm2/m

Reinforcement provided (positive), No. 4 @ 200 mm centers

Area of tension steel provided; Asx_col = 664 mm2/m

PASS - area of steel provided is OK

DAR GROUP

Project

QF

Job Ref.

QF

Section

Sheet no./rev.

4

Calc. by

MS

Date

09-Nov-106

Chk'd by

AE

Date

App'd by

IC

Date

Actual tensile bar spacing provided; sx_col = 200 mm

PASS - spacing of bars less than allowable

Middle strip (x – direction)

Negative Reinforcement:

Depth to negative steel; d’x_mid = h - c’c - D’mid_x / 2 = 124 mm

Flexural resistance factor; R’ux_mid = M’nx_mid / d’x_mid2 = 1.04 MPa

Reinforcement required; ’reqx_mid = 0.85f’c/fy×[1-(abs(1-2×R’ux_mid/(0.85×f’c)))] =0.00255

Outer compression steel not required

Area of tension steel required; A’sx_req_mid = ’reqx_mid × d’x_mid = 315 mm2/m

Minimum area of tension steel required; A’sx_min_mid = max(1.4MPa× d’x_mid /fy,(f’c/1 MPa)× d’x_mid1MPa/(fy 4))

A’sx_min_mid = 440 mm2/m

Maximum area of tension steel; A’sx_max_mid = max × d’x_mid = 3225 mm2/m

Maximum allowable spacing; smax = min(2 × h, 450 mm) = 300 mm

Reinforcement provided (negative), No. 4 @ 200 mm centers

Area of tension steel provided; A’sx_mid = 664 mm2/m

PASS - area of steel provided is OK

Actual tensile bar spacing provided; s’x_mid = 200 mm

PASS - spacing of bars less than allowable

Positive reinforcement

Depth to positive steel; dx_mid = h - cc - Dmid_x / 2 = 124 mm

Flexural resistance factor; Rux_mid = Mnx_mid / (dx_mid2) = 0.56 MPa

Reinforcement required; reqx_mid = 0.85 f’c/fy×[1-(abs(1-2×Rux_mid/(0.85×f’c)))] = 0.00136

Outer compression steel not required

Area of tension steel required; Asx_req_mid = reqx_mid × dx_mid = 168.00 mm2/m

Minimum area of tension steel required; Asx_min_mid = max(1.4 MPa× dx_mid /fy,(f’c/1 MPa)× dx_mid1MPa/(fy 4))

Asx_min_mid = 440 mm2/m

Maximum area of tension steel; Asx_max_mid = max × dx_mid = 3225 mm2/m

Reinforcement provided (positive), No. 4 @ 200 mm centers

Area of tension steel provided; Asx_mid= 664 mm2/m

PASS - area of steel provided is OK

Actual tensile bar spacing provided; sx_mid = 200 mm

PASS - spacing of bars less than allowable

Column strip (y – direction)

Negative Reinforcement:

Depth to negative steel; d’y_col = h - c’c - D’col_x - D’col_y / 2 = 111 mm

Flexural resistance factor; R’uy_col = M’ny_col / d’y_col2 = 0.98 MPa

Reinforcement required; ’reqy_col =0.85f’c/fy× [1 -(abs(1-2×R’uy_col/(0.85×f’c)))] = 0.00241

Outer compression steel not required

Area of tension steel required; A’sy_req_col = ’reqy_col × d’y_col = 266 mm2/m

Minimum area of tension steel required; A’sy_min_col = max(1.4 MPa×d’y_col/fy,(f’c/1 MPa)×d’y_col1 MPa/(4fy))

A’sy_min_col = 394 mm2/m

Maximum area of tension steel; A’sy_max_col = max × d’y_col = 2886 mm2/m

Maximum allowable spacing; smax = min(2 × h, 450 mm) = 300 mm

Reinforcement provided (negative), No. 4 @ 200 mm centers

Area of tension steel provided; A’sy_col = 664 mm2/m

PASS - area of steel provided is OK

Actual tensile bar spacing provided; s’y_col = 200 mm

DAR GROUP

Project

QF

Job Ref.

QF

Section

Sheet no./rev.

5

Calc. by

MS

Date

09-Nov-106

Chk'd by

AE

Date

App'd by

IC

Date

PASS - spacing of bars less than allowable

Positive reinforcement

Depth to positive steel; dy_col = h - cc - Dcol_x - Dcol_y / 2 = 111 mm

Flexural resistance factor; Ruy_col = Mnx_col / dy_col2 = 1.19 MPa

Reinforcement required; reqy_col =0.85f’c/fy×[1-(abs(1-2×Ruy_col/(0.85 × f’c)))] = 0.00292

Outer compression steel not required

Area of tension steel required; Asy_req_col = reqy_col × dy_col = 322.74 mm2/m

Minimum area of tension steel required; Asy_min_col = max(1.4 MPa×dy_col/fy,(f’c/1 MPa)×dy_col1 MPa/(4fy))

Asy_min_col = 394 mm2/m

Maximum area of tension steel; Asy_max_col = max × dy_col = 2886 mm2/m

Reinforcement provided (positive), No. 4 @ 200 mm centers

Area of tension steel provided; Asy_col= 664 mm2/m

PASS - area of steel provided is OK

Actual tensile bar spacing provided; sy_col = 200 mm

PASS - spacing of bars less than allowable

Middle strip (y – direction)

Negative Reinforcement:

Depth to negative steel; d’y_mid = h - c’c - D’mid_x - D’mid_y / 2 = 111 mm

Flexural resistance factor; R’uy_mid = M’ny_mid / d’y_mid2 = 1.09 MPa

Reinforcement required; ’reqy_mid = 0.85f’c/fy×[1-(abs(1-2×R’uy_mid/(0.85×f’c)))] =0.00268

Outer compression steel not required

Area of tension steel required; A’sy_req_mid = ’reqy_mid × d’y_mid = 297 mm2/m

Minimum area of tension steel required; A’sy_min_mid = max(1.4 MPa× d’y_mid/fy,(f’c/1 MPa)× d’y_mid1 MPa/(4fy))

A’sy_min_mid = 394 mm2/m

Maximum area of tension steel; A’sy_max_mid = max × d’y_mid = 2886 mm2/m

Maximum allowable spacing; smax = min(2 × h, 450 mm) = 300 mm

Reinforcement provided (negative), No. 4 @ 200 mm centers

Area of tension steel provided; A’sy_mid = 664 mm2/m

PASS - area of steel provided is OK

Actual tensile bar spacing provided; s’y_mid = 200 mm

PASS - spacing of bars less than allowable

Positive reinforcement

Depth to positive steel; dy_mid = h - cc - Dmid_x - Dmid_y / 2 = 111 mm

Flexural resistance factor; Ruy_mid = Mny_mid / dy_mid2 = 0.59 MPa

Reinforcement required; reqy_mid = 0.85 f’c/fy×[1-(abs(1-2×Ruy_mid/(0.85×f’c)))] =0.00143

Outer compression steel not required

Area of tension steel required; Asy_req_mid = reqy_mid × dy_mid = 158 mm2/m

Minimum area of tension steel required; Asy_min_mid = max(1.4 MPa× dy_mid/fy,(f’c/1 MPa)× dy_mid1 MPa/(4fy))

Asy_min_mid = 394 mm2/m

Maximum area of tension steel; Asy_max_mid = max × dy_mid = 2886 mm2/m

Reinforcement provided, No. (positive) 4 @ 200 mm centers

Area of tension steel provided; Asy_mid= 664 mm2/m

PASS - area of steel provided is OK

Actual tensile bar spacing provided; sy_mid = 200 mm

PASS - spacing of bars less than allowable

Distribution reinforcement (for shrinkage and temperature)

Minimum ratio of distribution reinforcement; t = 0.0018

DAR GROUP

Project

QF

Job Ref.

QF

Section

Sheet no./rev.

6

Calc. by

MS

Date

09-Nov-106

Chk'd by

AE

Date

App'd by

IC

Date

Minimum area of distribution reinforcement required; At_req = t × h = 270 mm2/m

Maximum allowable spacing of bars; smax_t = min(5 × h, 450 mm) = 450 mm

Distribution steel provided , No. 4 @ 400 mm centers

Distribution reinforcement provided; At_prov = 332 mm2/m

PASS - area of inner steel provided (Distribution) OK

Actual bar spacing provided; st = 400 mm

PASS - spacing of distribution bars is less than allowable

Check for shear

Shear force on the slab (along long beam); Vu = wu × (ly / 2 - bw / 2 - dx_mid) × 1 m = 31.08 kN

Design shear strength of slab; Vc = 0.75 × (f’c / 1 MPa) / 6 1 MPa × 1 m × dx_mid = 91.33 kN

PASS - one-way shear capacity

Check for clear cover (ACI 318 – 05 cl. 7.7.1)

Permissible min nominal cover to all reinforcement; cmin = 20 mm

Clear cover to tension reinforcement (negative); cc_neg = 20 mm

PASS - cover to steel resisting hogging exceeds allowable minimum cover

Clear cover to tension reinforcement (positive); cc_pos = 20 mm

PASS - cover to steel resisting sagging exceeds allowable minimum cover

Schematic reinforcement distribution.

½ C

olu

mn

str

ipM

iddl

e s

trip

½ C

olu

mn

str

ipx

y

lx

A’sx_col

A’sy_col

Asy_col

A’sy_mid

Asx_mid

Asy_mid A’sx_mid

Asx_col

TWO-WAY SLAB PANEL(Reinforcement Layout)

½ Column strip Middle strip ½ Column strip

ly

;