DESIGN OF R.C. TWO-WAY SOLID SLAB - HK CoP 2013
1.0 INPUT DATA
Material Properties
fcu = MPa Characteristic concrete cube strength
fy = MPa Characteristic steel yield strength [Clause 3.2.1.1]
fy,comp = N/mm² Characteristic steel yield strength for
compression steel
γc = Partial safety factor for concrete
γs = Partial safety factor for steel
Ratiobasic = Basic span/effective depth ratio [Table 3.9]
Section Dimensions
lx = mm Length of short span
ly = mm Length of long span
h =
mm Overall depth of slab
b =
mm Slab width for design
Loading
wult = kPa Design ultimate load (per area)
wservice = kPa Design service load (per area)
Boundary Condition
Select support condition [Table 6.6]:
Assumed Reinforcement Sizes and Cover
cnom = mm Nominal cover
φx = mm Short span main bar diameter
Sx = mm Short span main bar spacing
φy = mm Long span main bar diameter
Sy = mm Long span main bar spacing
φcomp = mm Compression bar diameter (if needed)
2.0 CALCULATIONS
2.1 Ultimate Limit State (ULS) - Bending
Aspect ratio
ly/lx = 7800/5600
= 1.393
Effective depths
dx = h − cnom −
φx/2 (Effective depth to short span steel)
dx = 250 - 35 - 12/2 = 209.00 mm
dy = h − cnom −
φx − φy/2 (Effective depth to long span
steel)
dy = 250 - 35 - 12 - 12/2 = 197.00 mm
dcomp = cnom +
φcomp/2 (Effective depth to compression steel)
dcomp = 35 + 12/2 = 41.00 mm
Step 1: Moment coefficients (βsx,
βsy)
Values from Table 6.6 with linear interpolation for intermediate aspect ratios. You can override
these values if needed.
For ly/lx = 1.393
βsx =
(Short span moment coefficient)
βsy =
(Long span moment coefficient)
Step 2: Calculate design moments
Short span moment
Msx = βsx · wult ·
lx² · 1m
Msx = 0.0313 × 12.6 × 5.6² × 1 = 12.39 kN·m
Long span moment
Msy = βsy · wult ·
lx² · 1m
Msy = 0.0429 × 12.6 × 5.6² × 1 = 16.99 kN·m
Step 3: Design reinforcement for short span (per 1m width)
Kx = Msx/(b ·
dx² · fcu)
Kx = 12.39×10⁶/(1000×209²×45) = 0.0063
Kbal = 0.156 (Maximum K for singly reinforced
section)
Compression steel NOT required in short span (Kx
< Kbal)
K'x = 0.0063
Lever arm for short span
zx = min(dx · (0.5 + √(0.25 −
K'x/0.9)), 0.95 · dx)
zx = min(209×(0.5 + √(0.25 - 0.0063/0.9)),
0.95×209)
zx = 198.55
mm
Tension steel required (short span)
Asx,req =
Msx/(fy/γs ·
zx)
Asx,req = 12.39×10⁶/(500/1.15×198.55)
= 143.51 mm²
Provide bars (e.g. T12@200mm)
Asx,prov = b · (φx/2)² ·
π/Sx
Asx,prov = 1000×(12/2)²×π/200 = 565.49 mm²
Compression steel required (short span)
Asxc,req = 0.00
mm² (Not required)
Check tension reinforcement (short span)
check_Asx = OK
(Asx,prov = 565.49 ≥
Asx,req = 143.51 mm²)
Step 4: Design reinforcement for long span (per 1m width)
Ky = Msy/(b ·
dy² · fcu)
Ky = 16.99×10⁶/(1000×197²×45) = 0.0097
Kbal = 0.156 (Maximum K for singly reinforced
section)
Compression steel NOT required in long span (Ky
< Kbal)
K'y = 0.0097
Lever arm for long span
zy = min(dy · (0.5 + √(0.25 −
K'y/0.9)), 0.95 · dy)
zy = min(197×(0.5 + √(0.25 - 0.0097/0.9)),
0.95×197)
zy = 187.15
mm
Tension steel required (long span)
Asy,req =
Msy/(fy/γs ·
zy)
Asy,req = 16.99×10⁶/(500/1.15×187.15)
= 208.92 mm²
Provide bars (e.g. T12@200mm)
Asy,prov = b · (φy/2)² ·
π/Sy
Asy,prov = 1000×(12/2)²×π/200 = 565.49 mm²
Compression steel required (long span)
Asyc,req = 0.00
mm² (Not required)
Check tension reinforcement (long span)
check_Asy = OK
(Asy,prov = 565.49 ≥
Asy,req = 208.92 mm²)
2.2 Ultimate Limit State (ULS) - One-way Shear Check (short direction)
Maximum shear at support
Shear Factor = (Distribution factor, typically 0.5 for simply
supported)
Vx = wult · lx ·
Shear Factor · 1m
Vx = 12.6 × 5.6 × 0.5 × 1 = 35.28 kN
Applied shear stress
vx = Vx/(b ·
dx)
vx = 35.28×10³/(1000×209) = 0.169 MPa
Permissible shear stress in concrete
vc = 1 MPa · (0.79 · (100 ·
Asx,prov/(b · dx))1/3 · max(0.67,
(400/dx)1/4) · (fcu/25
MPa)1/3)/γc
vc = 1×(0.79×(100×565.49/(1000×209))¹ᐟ³×max(0.67,
(400/209)¹ᐟ⁴)×(45/25)¹ᐟ³)/1.25
vc = 0.475
MPa
Maximum shear stress [Clause 3.4.5.2]
vmax = min(0.8 · √(fcu/1 MPa), 5) × 1
MPa
vmax = min(0.8×√(45/1), 5)×1 = 5.000 MPa
Check shear stress
check_vx,max = OK
(vx = 0.169 ≤
vmax = 5.000 MPa)
check_vx,c = OK
(vx = 0.169 ≤
vc = 0.475 MPa)
Shear reinforcement is not required
2.3 Serviceability Limit State (SLS) - Deflection Check
Short span is critical for deflection
Service moment (short span)
Mservice,x = βsx ·
wservice · lx² · b
Mservice,x = 0.0313 × 12.6 × 5.6² × 1
= 12.39 kN·m
Required service steel (short span)
As,req,service,x =
Mservice,x/(fy/γs ·
zx)
As,req,service,x = 12.39×10⁶/(500/1.15×198.55) = 143.51 mm²
Modification factor for tension reinforcement
Factormod = min(0.55 + (477 MPa − 2 · fy ·
As,req,service,x/(3 · Asx,prov))/(120 · (0.9 MPa +
Mservice,x/(b · dx²))), 2)
Factormod = min(0.55 + (477 -
2×500×143.51/(3×565.49))/(120×(0.9 + 12.39×10⁶/(1000×209²))), 2)
Factormod = 1.632
Allowable span/effective depth ratio
Ratioallow = Ratiobasic ·
Factormod
Ratioallow = 20×1.632 = 32.64
Actual span/effective depth ratio
Ratioactual =
lx/dx
Ratioactual = 5600/209 = 26.79
Check deflection
check_deflection = OK
(Ratioactual = 26.79 ≤
Ratioallow = 32.64)
*** OUTPUT SUMMARY ***
Design Summary
Aspect ratio (ly/lx): 1.393
Short span moment, Msx = 12.39 kN·m
Long span moment, Msy = 16.99 kN·m
Required tension steel (short span), Asx,req = 143.51 mm²
Provided tension steel (short span), Asx,prov = 565.49 mm²
Tension reinforcement check (short span): OK
Required tension steel (long span), Asy,req = 208.92 mm²
Provided tension steel (long span), Asy,prov = 565.49 mm²
Tension reinforcement check (long span): OK
Required compression steel (short span): Asxc,req = 0.00 mm²
Required compression steel (long span): Asyc,req = 0.00 mm²
Design shear stress (short span), vx = 0.169 MPa
Allowable shear stress in concrete, vc = 0.475 MPa
Shear capacity check (short span): OK - Shear reinforcement not required
Actual span/depth ratio (short span), Ratioactual = 26.79
Allowable span/depth ratio, Ratioallow = 32.64
Deflection check: OK
RESULTS
ALL CHECKS PASSED - DESIGN IS ADEQUATE