Concrete One-way Slab Designer to ACI318

Verified by the CalcTree engineering team on June 27, 2024
﻿
This calculator allows the user to assess the structural integrity of simply supported one-way slabs. The calculation will verify the reinforcement needs for concrete slabs ensuring compliance with flexural design criteria. Additionally, it checks the shear capacity of the slabs, employing ultimate limit state design (ULS) methods. 
All calculations are performed in accordance with ACI 318-19. 
﻿
Slab view - Geometric properties
﻿
﻿
Slab cross-section - Geometric properties
CalculationTechnical notes and assumptions
 Any top reinforcement is ignored and does not contribute to the slab's flexural capacity.
The slab is considered simply supported.  
The slab's self-weight is calculated automatically once the user sets the concrete weight and slab's depth.
Design forces and reinforcement are calculated and shall be distributed in 1-meter width.
An axial capacity check of the slab is not included.
If flexural or shear utilization surpasses 100%, the used should increase the assumed reinforcement or slab's depth.
The user has the flexibility to employ up to two distinct bar diameters and spacings for both principal and secondary reinforcement.
❗ NOTE: The user can choose between a manual calculation with the desired solicitations "Mu" and "Vu" or calculate them internally with the assumptions listed above. In case of checking reinforcement for negative moments calculated by hand or other software, the user should choose "MANUAL" mode and type the required solicitations.
InputsCalculation type
﻿﻿Calculation type
:AUTO​
﻿
﻿﻿User Vu
:{"mathjs":"Unit","value":3,"unit":"kN","fixPrefix":false}​
﻿
﻿﻿User Mu
:{"mathjs":"Unit","value":10,"unit":"kN m","fixPrefix":false}​
﻿
❗ Choose "MANUAL" to insert your desired "Mu" and "Vu" or "AUTO" to do it automatically. 
If the "AUTO" mode is set, users Mu and Vu won't work.
Loads
﻿﻿ DL: q
:{"mathjs":"Unit","value":2.5,"unit":"kN / m","fixPrefix":false}​
﻿
﻿﻿LL: q
:{"mathjs":"Unit","value":3.5,"unit":"kN / m","fixPrefix":false}​
﻿
Material Properties
﻿﻿Concrete weight
:{"mathjs":"Unit","value":24,"unit":"kN / m^3","fixPrefix":false}​
﻿
﻿﻿ f'c
:{"mathjs":"Unit","value":30,"unit":"MPa","fixPrefix":false}​
﻿
﻿﻿fsy
:{"mathjs":"Unit","value":420,"unit":"MPa","fixPrefix":false}​
﻿
Geometric Properties
﻿﻿D
:{"mathjs":"Unit","value":300,"unit":"mm","fixPrefix":false}​
﻿
﻿﻿ c
:{"mathjs":"Unit","value":20,"unit":"mm","fixPrefix":false}​
﻿
﻿﻿S
:{"mathjs":"Unit","value":4,"unit":"m","fixPrefix":false}​
﻿
Tensile (Bottom) Reinforcement:
﻿﻿db.st
:{"mathjs":"Unit","value":9,"unit":"mm","fixPrefix":false}​
﻿
﻿﻿ s
:{"mathjs":"Unit","value":150,"unit":"mm","fixPrefix":false}​
﻿
﻿﻿db.st 2
:{"mathjs":"Unit","value":0,"unit":"mm","fixPrefix":false}​
﻿
﻿﻿s
:{"mathjs":"Unit","value":150,"unit":"mm","fixPrefix":false}​
﻿
Design Properties
﻿﻿ϕf
:0.9​
﻿
﻿﻿ϕv
:0.75​
﻿
OutputGeometric Properties
﻿﻿ Ast
:424.11500823462154mm2​
﻿
﻿﻿Ag
:300000mm2​
﻿
﻿﻿Zx
:15000000mm3​
﻿
﻿﻿ Ixx
:2250000000mm4​
﻿
Design Forces by Load Case
﻿﻿SW: Self-Weight Bending Moment
:14.4kN m​
﻿
﻿﻿SW: Self-Weight Shear
:14.4kN​
﻿
﻿﻿DLBM
:5kN m​
﻿
﻿﻿DL: Uniform Load Shear 
:5kN​
﻿
﻿﻿LL: Uniform Load Bending Moment
:7kN m​
﻿
﻿﻿LL: Uniform Load Shear 
:7kN​
﻿
﻿
Ultimate Design Forces
﻿﻿Mu
:34.480000000000004kN m​
﻿
﻿﻿Mn
:38.31111111111111kN m​
﻿
﻿﻿Vu
:34.480000000000004kN​
﻿
﻿﻿ Vn
:45.973333333333336kN​
﻿
﻿
Design ChecksFlexural Checks
﻿﻿Minimum reinforcement
:540mm2​
﻿
Flexural capacity:
﻿﻿Required Reinforcement
:918.333333333333mm2​
﻿
﻿﻿Ast
:424.11500823462154mm2​
﻿
﻿﻿ϕadopted
:0.9​
﻿
﻿﻿Adopted Ka
:0.025355439800511144​
﻿
﻿﻿et adopted
:0.0975701348532159​
﻿
﻿﻿Check
:ERROR​
﻿
﻿﻿Mn adopted
:48.4521967696315kN m​
﻿
﻿﻿Mu adopted
:43.60697709266835kN m​
﻿
﻿﻿Flexural reinforcement utilization
:0.7906991563925005​
﻿
﻿
The required reinforcement checks for the maximum value between the minimum reinforcement in accordance with the ACI 318-19 Table 7.6.1.1 and the required reinforcement by flexural solicitations. The specified reinforcement should be distributed within a 1-meter width.
If the "Check" box gives an "ERROR" message, the user should check:  
Adopted reinforcement vs needed.
Et adopted > 0.004
Mu adopted > Mu 
 
Shear Checks
ExplanationSlabs are an important structural element in reinforced concrete (RC) structures. They are designed to provide resistance to external loads that cause shear forces and bending moments across their length. 
Concrete exhibits strength in compression but is relatively weak in tension. Therefore, reinforcement is incorporated to withstand the tensile stresses that arise when slabs are subjected to loads. In a simply supported slab, the tensile stresses primarily occur along the bottom of the section. Consequently, in theory, there is no necessity for top reinforcement.
General Steps One-way Slabs Design Generally, the limit state checks for RC elements design include the following:
Moment flexural capacity
Shear capacity
Deflection
Stability 
Crack control
This calculator covers the first two checks - flexural and shear capacity. Deflection, stability checks and crack control require a more detailed examination by the engineer.
Ultimate Flexural CapacityThe flexural (also referred to as 'bending' or 'moment') capacity of a slab cross-section is determined using the Rectangular Stress Block method. The stress distribution in concrete under bending is curved in reality, however, it can be converted to an equivalent rectangular stress block by the use of reduction factors shown in the following image.
﻿
The strain and equivalent rectangular stress block of a typical concrete section for flexural design
Factor ka is not defined explicitly in the standards and must be computed via mechanics solving the equation Cc=T as shown:
﻿
Cc=T  0.85fc′bd=Astfsyα=Astfsy0.85fc′bMn=Cc.(d−α2)=T.(d−α2)α=β1c=kadka=Astfsy0.85fc′bd=1−(1−2mn)mn=Mn0.85fc′bd2 C_{c}=T\   \\\ 0.85f'_cbd= A_{st}f_{sy}\\ \alpha=\frac{A_{st}f_{sy}}{0.85f'_cb}\\ M_n=C_{c}.(d-\frac{\alpha}{2})=T.(d-\frac{\alpha}{2})\\ \alpha=\beta_1 c=k_ad\\k_a = \frac{A_{st}f_{sy}}{0.85f'_cbd}=1-\sqrt{(1-2m_n)}\\m_n = \frac{M_n}{0.85f'_cbd^2} \\\ Cc​=T  0.85fc′​bd=Ast​fsy​α=0.85fc′​bAst​fsy​​Mn​=Cc​.(d−2α​)=T.(d−2α​)α=β1​c=ka​dka​=0.85fc′​bdAst​fsy​​=1−(1−2mn​)​mn​=0.85fc′​bd2Mn​​ 
The value of Beta is determined using ACI Table 22.2.2.4.3:
﻿
β1=0.85 if fc′≤ 30MPa β1=0.85−0.05(fc′−30)7 if 30MPa≤ fc′≤ 58MPaβ1=0.65 if fc′≥ 58MPa\beta_1=0.85\ if\ f'_c \leq\  30MPa \\\ \beta_1=0.85-0.05\frac{(f'_c-30)}{7}\ if\ 30MPa\leq\ f'_c \leq\  58MPa \\ \beta_1=0.65\ if\ f'_c \ge\  58MPa β1​=0.85 if fc′​≤ 30MPa β1​=0.85−0.057(fc′​−30)​ if 30MPa≤ fc′​≤ 58MPaβ1​=0.65 if fc′​≥ 58MPa
The ultimate flexural strength at critical sections should not be less than the minimum required strength in bending and by the minimum flexural reinforcement according to ACI Table 7.6.1.1:
﻿
Asmin ≥0.0018 × 420bdfy ≥ 0.0014 bdA_{smin}\ \ge\frac{0.0018\ \times\ 420\\bd}{f_y}\ \ge\ {0.0014\ bd}Asmin​ ≥fy​0.0018 × 420bd​ ≥ 0.0014 bd
A section will require compression reinforcement when the compressed concrete section is insufficient to balance the external moment. The maximum moment that concrete can balance is achieved when the maximum compression strain in concrete (0.003) is reached, and the minimum tensile strain in steel (0.005 for ductile failure) is maintained. This is deduced from the triangle relationship:
﻿
c=(ϵcϵc+ϵt)d c=kcdkcmaˊx≤ϵcϵc+ϵt=0.0030.003+0.005=0.375  ka=β1kckamaˊx=β1kcmaˊx=0.85×0.375=0.318c=(\frac{\epsilon_c}{\epsilon_c+\epsilon_t})d\\\ c=k_{c}d\\ k_{cmáx} \le\frac{\epsilon_c}{\epsilon_c+\epsilon_t}=\frac{0.003}{0.003+0.005}= 0.375\ \normalsize\small\ \\ k_{a}=\beta_1k_{c}\\ k_{amáx}=\beta_1k_{cmáx}=0.85\times0.375=0.318c=(ϵc​+ϵt​ϵc​​)d c=kc​dkcmaˊx​≤ϵc​+ϵt​ϵc​​=0.003+0.0050.003​=0.375  ka​=β1​kc​kamaˊx​=β1​kcmaˊx​=0.85×0.375=0.318
Once the reduction factor ϕ is determined from ACI Table 21.2.2, the reduced moment flexural capacity is thus given as:
﻿
ϕMu=ϕAstfsy×(d−α2)\phi  M_u = \phi A_{st}f_{sy} \times (d - \frac{\alpha}{2})ϕMu​=ϕAst​fsy​×(d−2α​)
Ultimate Shear CapacityThe total shear capacity of a section (Vu) is the combination of the shear strength contributed by concrete (Vuc) and shear reinforcement (Vus), limited by Vu.max.:
﻿
ϕ Vu = ϕ(Vuc+Vus) \phi\ V_u\ =\ \phi(V_{uc}+V_{us})\ ϕ Vu​ = ϕ(Vuc​+Vus​) 
Since slabs typically lack shear reinforcement, their capacity is assessed solely based on the contribution of concrete.
❗ NOTE: In the presence of significant point loads on the slab, such as those from columns, the engineer should verify punching shear.
﻿
Contribution to Shear Strength from Concrete (Vuc)
For non-prestressed members, the concrete capacity is determined by ACI Table 22.5.5.1:
﻿
Vc=(0.66λsλ(ρw)1/3fc′+Nu6Ag)bdV_{c} = (0.66\lambda_s\lambda (\rho_w)^{1/3}\sqrt{f'_c}+\frac{N_u}{6A_g}){bd}  Vc​=(0.66λs​λ(ρw​)1/3fc′​​+6Ag​Nu​​)bd
In this case, axial tension is neglected. 
﻿
Vc=0.66λsλ(ρw)1/3fc′bd≤ 0.42λfc′bdλs=21+0.004d≤1.0 fc′≤ 8.3MPaV_{c} = 0.66\lambda_s\lambda (\rho_w)^{1/3}\sqrt{f'_c}bd  \le\ 0.42\lambda\sqrt{f'_c}bd\\ \lambda_s=\sqrt{\frac{2}{1+0.004d}}\le1.0\\\ \sqrt{f'_c} \le\ 8.3MPaVc​=0.66λs​λ(ρw​)1/3fc′​​bd≤ 0.42λfc′​​bdλs​=1+0.004d2​​≤1.0 fc′​​≤ 8.3MPa
Lambda factor is set manually according to ACI Table 19.2.4.2.
The total shear capacity (Vu) is calculated below, with the reduction factor again taken from Table 21.2.1:
﻿
ϕVu=ϕVc\phi V_u = \phi V_{c} ϕVu​=ϕVc​
AcknowledgementsThis calculation was built in collaboration with Lucas Ghiglione. Learn more.
Related ResourcesBeam Analysis and Design Fundamentals
Concrete Column Design Calculator to AS 3600
Moment of Inertia Calculators - Various section shapes
Steel Beam and Column Designer to AISC
Steel Beam and Column Designer to AS4100
Wood Beam Calculator to AS 1720.1
﻿