Verified by the CalcTree engineering team on October 31, 2024
This calculator performs capacity checks for a concrete unbraced shear wall. It checks axial and shear, calculates second-order effects in order to check in-plane bending and plots the interaction diagram to check flexural-compression combined actions.
This calculation has been written in accordance with ACI 318-19.
Results Summary
Summary
Design Check
Action
Resistance
Utilisation
Status
Shear Strength
V = 224.81 kips ΦVn = 237.21 kips
0.95 dimensionless
🟢
Axial Strength
N = 1124.04 kips ΦPn = 1999.19 kips
0.56 dimensionless
🟢
Flexural Strength
M = 1475.12 kips ΦMn = 2155.43 kips
0.56 dimensionless
🟢
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Calculation
Technical notes
Maximum strain at the extreme concrete compression fiber shall be assumed to be equal to 0.003 (ACI 318 22.2.2.1).
The tensile strength of concrete shall be neglected in flexural and axial strength calculations (ACI 318 22.2.2.2).
This calculator considers unbraced walls that cantilever from the ground, braced walls are not considered.
The user inputted design moment
Mu
shall be the first-order design moment. Second-order effects are automatically accounted for through the calculation of
Mc
.
1. Properties
1.1 Materials
fpc
:4,000 psi
wc
:145 lb / ft^3
Ec
:110,429 ksi
Ec=wc1.5×fpc0.5
Es
:29,000 ksi
fy
:60 ksi
1.2 Geometry
lw
:9.00 ft
hw
:50.00 ft
h
:8.00 in
1.3 Reinforcement
Vertical reinforcement:
db,v
:#4
s,v
:1.00 ft
layers
:2
cover
:1.00 in
Horizontal reinforcement:
db,h
:#4
s,h
:2.00 ft
2. Design Actions
Nu
:1,000.0 kips
Mu
:1,000.0 kips ft
Vu
:200.0 kips
Note, a positive values for
N
is compression, and a negative value is tension.
3. Design Checks
3.1 Shear Strength
According to ACI 318 Ch 11.5.4.2, the shear capacity
ϕVn
at any horizontal section shall not exceed
0.66fc′Acv
and shall be calculated by:
ϕVn=0.75(αcλfc′+ρtfyt)Acv
Where:
αc=⎩⎨⎧3.02.0varies linearly between 3.0 and 2.02.0(1+500AcvN)≥0.0forhw/lw≤1.5forhw/lw≥2.0for1.5<hw/lw<2.0for walls subject to a net axial tension
N
is the factored axial force normal to cross-section occurring simultaneously with
Vu
or
Tu
; to be taken as positive for compression and negative for tension.
ρt
is the ratio of the area of distributed transverse reinforcement to the gross concrete area perpendicular to that reinforcement.
Acv
is the gross area of the concrete section bounded by web thickness and length of the section in the direction of the shear force considered.
hw/lw
:5.56
αc
:2.00
αc=⎩⎨⎧3.02.03.0 to 2.02.0(1+500AcvN)≥0.0forhw/lw≤1.5forhw/lw≥2.0for1.5<hw/lw<2.0for tension walls
Ch 11.5.4.3
Acv
:864 in^2
Acv=lwh
ρt
:0.00
ρt=As,t/(svh)
Vn
:321.35 kip
Vn=(αcfc′+ρtfyt)Acv
Ch 11.5.4.3
ΦVn
:241.01 kip
ϕVn=0.75×Vn
Table 21.2.1
Shear utilization
:0.83 🟢
Vu≤ϕVn
3.2 Axial Strength
3.2.1 Maximum axial compressive strength
At a low load application rate, the accepted compressive strength for the concrete is 0.85 times the specified compressive strength that is
0.85fc′
.
The ultimate axial strength of the element is attained when the concrete undergoes crushing failure, and the reinforcement steel yields:
Pn=0.8Po
Po=0.85fc′(Ag−Ast)+fyAst
Where:
Pn
is the nominal axial compressive strength of the member.
Po
is the nominal axial strength at zero eccentricity.
Ag
is the gross area of the concrete section.
Ast
is the total area of longitudinal reinforcement.
3.2.2 Maximum axial tensile strength
The concrete is fully fissured when the ultimate axial tensile strength is reached, rendering its contribution null. Therefore, the ultimate strength is attained when the reinforcement reaches its yield stress.
Pnt=fyAs
Pn
:1,006.87 kip
Pn=0.8Po=0.8(0.85fc′(Ag−Ast)+fyAst)
Ch 22.4.2.2
ΦPn
:1,659.14 kip
ϕPn=0.65×Vn
Pnt
:-212.06 kip
Pnt=−fyAs
Ch 22.4.3.1
ΦPnt
:-190.85 kip
ϕPnt=0.90×Pnt
Axial utilization
:0.60 🟢
Nu≤ϕPn or ϕPnt
3.3 Second-order effects
As per Chapter 6.6.4.5.1, the second-order moment used for the design of columns and walls,
Mc
, shall be the first-order moment
Mu
amplified for the effects of the member curvature, given by:
Mc=δMu
Where:
Nu
is the axial force applied to the wall.
δ
is the magnification factor, given by
δ=1−0.75PcNu1
Pc
is the critical buckling load, given by
Pc=(khw)2π2(0.25EcIg)
Ec
is the concrete modulus of elasticity.
Ig
is the moment of inertia of the gross concrete section about the centroidal axis, neglecting reinforcement.
k
is the effective length factor, and for walls is recommended that a
k
value of
1.0
is used (R6.6.4.4.3)
Pc
:3.49 kilokip
Pc=(khw)2π2(0.25EcIg)
Ch 6.6.4.4.2
δ
:1.62
δ=1−0.75PcN1
Ch 6.6.4.5.2
Mc
:1,618.92 ft kip
Mc=δM
Ch 6.6.4.5.1
3.4 Flexural Strength
The nominal moment capacity
Mn
, is expressed as the sum of the moments from the internal forces, taking into account the stress distribution within the section. The equation for the nominal moment in a reinforced concrete column under eccentric axial loading is:
is the distance from the fiber of maximum compressive strain.
β1
is the equivalent rectangular concrete stress distribution.
c
is the distance from the fiber of maximum compressive strain to the neutral axis.
fs
is the strength for reinforcement in tension.
fs′
is the strength for reinforcement in compression.
This equation sums the moments produced by the concrete and steel in compression, along with the steel in tension, taking into account their respective distances from the neutral axis of the section.
The value for
Mu
is multiplied by the strength reduction factor
ϕ
, which varies from
0.65
in the compression behavior to
0.90
in tension. This can be better visualized in the interaction diagram. The value for
ϕMn
that is provided is the point on the interaction diagram that corresponds to the user inputted axial force demand
The interaction diagram for shear walls represents the relationship between nominal axial load
Pn
and nominal moment
Mn
, showing how the shear wall’s capacity changes under different axial force and bending moment combinations. In this diagram, the curve defines the safe limits of loading, where each point corresponds to a specific combination of
Pn
and
Mn
that the shear wall can resist without failure.
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