DESIGN OF R.C. SOLID SLAB (ONE-WAY) - 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

L = mm   Effective span of slab

h = mm   Overall depth of slab

b = mm   Width of slab

Loading

M = kN·m   Design ultimate moment

V = kN   Design ultimate shear force

Mservice = kN·m   Service moment for deflection check

Assumed Reinforcement Sizes and Cover

cnom = mm   Nominal cover

φmain = mm   Main bar diameter

Smain = mm   Main bar spacing

φcomp = mm   Compression bar diameter (if needed)

φdist = mm   Distribution bar diameter

2.0 CALCULATIONS

2.1 Ultimate Limit State (ULS) - Bending

Effective depth to tension steel

d = hcnomφmain/2
d = 150 - 30 - 12/2 = 114.00 mm

Effective depth to compression steel

dcomp = cnom + φcomp/2
dcomp = 30 + 10/2 = 35.00 mm

Step 1: Check if compression steel is required

K = M/(b·d²·fcu)
K = 26.9×10⁶/(1000×114²×45) = 0.0461
Kbal = 0.156   (Maximum K for singly reinforced section)
Compression steel is NOT required (K < Kbal)
K' = 0.0461

Step 2: Calculate lever arm (z)

z = min(d·(0.5 + √(0.25 − K'/0.9)), 0.95·d)
z = min(114×(0.5 + √(0.25 - 0.0461/0.9)), 0.95×114)
z = 108.30 mm

Step 3: Calculate tension reinforcement (As)

As,req = M/(fy/γs·z)
As,req = 26.9×10⁶/(500/1.15×108.30) = 571.48 mm²
Provide main bars (e.g. T12@150mm)
As,prov = b·(φmain/2)²·π/Smain
As,prov = 1000×(12/2)²×π/150 = 753.98 mm²   [in 1m width]

No compression steel required

As',req = 0 mm²
As',prov = 0 mm²

Check if tension reinforcement is sufficient

check_As = OK   (As,prov = 753.98As,req = 571.48 mm²)

Check if compression reinforcement is sufficient

check_As' = OK   (Not required)

2.2 Ultimate Limit State (ULS) - Shear Check

Permissible shear stress in concrete

vc = 1 MPa·(0.79·(100·As,prov/(b·d))1/3·max(0.67, (400/d·1 mm)1/4)·(fcu/25 MPa)1/3)/γc
vc = 1×(0.79×(100×753.98/(1000×114))¹ᐝ³·max(0.67, (400/114)¹ᐝ⁴)·(45/25)¹ᐝ³)/1.25
vc = 0.583 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

Applied shear stress

v = V/(b·d)
v = 69.5×10³/(1000×114) = 0.610 MPa

Check shear stress

check_vmax = OK   (v = 0.610vmax = 5.000 MPa)
check_vc = CHECK   (v = 0.610 > vc = 0.583 MPa)
Shear reinforcement is required

2.3 Serviceability Limit State (SLS) - Deflection Check

Modification factor for tension reinforcement

Factormod = min(0.55 + (477 MPa − 2·fy·As,req/(3·As,prov))/(120·(0.9 MPa + Mservice/(b·d²))), 2)
Factormod = min(0.55 + (477 - 2×500×571.48/(3×753.98))/(120×(0.9 + 26.9×10⁶/(1000×114²))), 2)
Factormod = 1.216

Allowable span/effective depth ratio

Ratioallow = Ratiobasic·Factormod
Ratioallow = 23×1.216 = 27.97

Actual span/effective depth ratio

Ratioactual = L/d
Ratioactual = 2900/114 = 25.44

Check deflection

check_deflection = OK   (Ratioactual = 25.44Ratioallow = 27.97)

*** OUTPUT SUMMARY ***

Design Summary

Effective depth, d = 114.00 mm

Lever arm, z = 108.30 mm

Required tension steel, As,req = 571.48 mm²

Provided tension steel, As,prov = 753.98 mm²

Tension reinforcement check: OK

Required compression steel (if needed), As',req = 0.00 mm²

Provided compression steel (if needed), As',prov = 0.00 mm²

Compression reinforcement check: OK

Design shear stress, v = 0.610 MPa

Allowable shear stress in concrete, vc = 0.583 MPa

Shear capacity check: CHECK - Shear reinforcement is required

Actual span/depth ratio, Ratioactual = 25.44

Allowable span/depth ratio, Ratioallow = 27.97

Deflection check: OK

RESULTS

DESIGN FAILS - SHEAR REINFORCEMENT REQUIRED