Shear Wall to ACI Standards

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IntroductionShear walls play a critical role in the structural stability of buildings by providing stronger resistance to lateral loads. As trends in construction shifted towards high-rises and tall buildings in general, they have become even more important.
Lateral loads, such as wind and seismic forces, cause structures to sway and deflect (as shown in Figure 2). Without proper bracing and resistance against these loads, the structure's occupants may experience discomfort, or in the worst case, the structure may fail completely. One of the ways to limit the lateral sway and deflection is to increase the size of beams and columns - however, this method is costly and increases dimensions, which may not be permitted.
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 Lateral force diagram
Shear walls are an effective solution to this problem. They are thick blocks of reinforced concrete that extend from a structure’s foundation to the top, adding stiffness and rigidity. Shear walls can be constructed in many shapes, as shown in Figure 3.
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 different shaear wall shape
Arrangement (A) is common in structures as the shear walls are easier to construct and have no corners. Arrangement (B) has an aesthetic advantage as spaces between the shear wall allow the installation of decorative features such as large windows. In most tall structures, shear walls are placed in the center in a ‘c-shape’ (generally used as elevator shafts or stairways) to form a ‘core’. These cores prevent lateral torsion of the structure. Of course, there are other arrangements of shearwalls, and the design engineer must select one that suits the project at hand.
Often, openings are required in shear walls for access (e.g. elevator shafts), windows, or doorways, as seen in Figure 4. In these situations, the areas straight above the opening are significantly weaker than others. To counteract these internal stresses, ‘lintels’ are placed to act as beams to support additional height. For more, you can read this guide by StructuralGuide here.
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shear walls for guideline
Design Procedure
This design guide follows ACI 318-14 and ACI 318R-14 Chapter 11[1]. This chapter only applies to walls resisting vertical and lateral loads, and the ‘special walls’ design must follow Section 18 instead. A ‘special wall’ is a shear wall including coupling beams and wall piers forming part of a seismic-force-resisting system.
In-plane ShearStep 1. Determine the minimum shear wall thicknessACI 318-14 Table 11.3.1.1 provides the minimum shear wall thickness h for different types of walls. If justified with appropriate structural analysis, then thinner walls may be used.
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minimum shaer wall thickness value
Step 2. Determine factored axial force and momentShear walls are to be designed for the factored axial force, Pu, which produces the maximum factored moment Mu (if Pu is eccentric) for each applicable load combination. Pu must be smaller than ϕPn, max what is given in Table 22.4.2.1. Pn, max is a function of the nominal axial strength Po, at zero eccentricity, and varies depending on the prestressing of the member. Po can be calculated as:
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Po=0.85f′c(Ag−Ast)+fyAst If the member is non−prestressed or composite steel and concreteP_o=0.85f′_c(A_g−A_{st})+f_yA_{st} \ \\ \text{If the member is non−prestressed or composite steel and concrete} \\Po​=0.85f′c​(Ag​−Ast​)+fy​Ast​ If the member is non−prestressed or composite steel and concrete
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Po=0.85f′c(Ag−Ast−Apd)+fyAst−(fse−0.003Ep)AptP_o=0.85f′_c(A_g−A_{st}−A_{pd})+f_yA_{st}−(f_{se}−0.003E_p)A_{pt}Po​=0.85f′c​(Ag​−Ast​−Apd​)+fy​Ast​−(fse​−0.003Ep​)Apt​
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Table 2 - Maximum axial strength, ACI 318-14 Table 22.4.2.1
Step 3. Determine axial and flexural strength‍ACI 318-14 Clause 11.5.2 - Axial load and in-plane or out-of-plane flexure
For bearing walls, Pn and in- or out-of-plane nominal flexural strength Mn is calculated using Pn, max obtained from Step 2.
For non-bearing walls, Mn can be calculated ignoring the effect of axial forces, as they have an insignificant effect on its flexural behavior.
ACI 318-14 Clause 11.5.3 - Axial load and out-of-plane flexure
If the location of the resultant forces acting on the wall lies within the middle third of the wall (as shown in Figure 4) and it is subject to an axial load that causes out-of-plane flexure, then Pn is calculated as:
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Pn=0.55f′cAg[1−(klc32h)2]wherelc=Length of wall (in.)k=effective length factor, which accounts for different boundary conditions (as shown in Figure 5) at the wall′s ends, determined using Table11.5.3.2:Pn=0.55f′_cA_g[1−(\frac{kl_c}{32h})^2] \\ \text{where} \\ l_c = \text{Length of wall } (in.)\\k=\text{effective length factor, which accounts for different boundary conditions (as shown in Figure 5) at the wall′s ends, determined using Table11.5.3.2:} Pn=0.55f′c​Ag​[1−(32hklc​​)2]wherelc​=Length of wall (in.)k=effective length factor, which accounts for different boundary conditions (as shown in Figure 5) at the wall′s ends, determined using Table11.5.3.2:
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Table 3 - Effective length factor k, ACI 318-14 Table 11.5.3.2
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 factored axial force and moment of shear wall
Step 4. Determine the in-plane shear strengthACI 318-14 Clause11.5.4.1 ~ 7 - Contribution from concrete
The total in plane shear strength of a wall Vn is the sum of the strengths of the concrete Vc and steel reinforcement Vs. The portion of total in-plane shear strength provided by the concrete section is calculated using Table 11.5.4.6:
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Table 4 - Vc of non-prestressed and prestressed walls, ACI 318-14 Table 11.5.4.6
It should also be noted that if the length of the wall is greater or equal to its height (i.e. hw <= 2lw), then it requires a more analytical design approach and must follow the strut-tie-method outlined in Chapter 23.
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ACI 318-14 Clause11.5.4.8 - Contribution from reinforcement
The portion of total in-plane shear strength provided by reinforcement  is calculated as:
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Vs=AvfytdswhereAv=area of shear reinforcement (in2)fyt=yield strength of transverse reinforcement (psi)s=centre−to−centre spacing of transverse reinforcement d=???V_s = \frac{A_vf_{yt}d}{s}  \\ \text{where} \\ A_v = \text{area of shear reinforcement } (in^2) \\ f_{yt} = \text{yield strength of transverse reinforcement } (psi) \\ s = \text{centre−to−centre spacing of transverse reinforcement } \\ d = ???Vs​=sAv​fyt​d​whereAv​=area of shear reinforcement (in2)fyt​=yield strength of transverse reinforcement (psi)s=centre−to−centre spacing of transverse reinforcement d=???
Once V_c and V_s have been calculated, check the total shear strength of the wall against in-plane design action effects:
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ϕVn=ϕ(V+c+Vs)>VuϕV_n=ϕ(V+_c+V_s)>V_uϕVn​=ϕ(V+c​+Vs​)>Vu​
Step 5. Determine the out-of-plane shear strengthIdentical to the calculation of in-plane shear strength, equations and design procedures for out-of-plane shear strength change depending on the presence of prestressing and axial forces.
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ACI 318-14 Clause 22.5.5 - Contribution from concrete (non-prestressed members without axial force)
If the wall is non-prestressed and there is no axial force applied to it, then the portion of the total out-of-plane shear strength provided by concrete V_c can be calculated as:
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Vc=2λf′cbwd(simplified)\large{V_c=2λ\sqrt{f′_c}b_wd \hspace{5cm} (simplified) }Vc​=2λf′c​​bw​d(simplified)
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Or, using Table 22.5.5.1, which provides more detailed formulae:
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Table 5 - Vc for non-prestressed members without axial force, ACI 318-14 Table 22.5.5.1
ACI 318-14 Clause 22.5.6 - Contribution from concrete (non-prestressed members with axial compression)
If the wall is non-prestressed and axial compression is present, then the portion of the total out-of-plane shear strength provided by concrete Vc can be calculated as:
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Vc=2(1+Nu2000Ag)λf′cbwd\large{V_c=2(1+\frac{N_u}{2000A_g})λ\sqrt{f′_c}b_wd}Vc​=2(1+2000Ag​Nu​​)λf′c​​bw​d
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Or, using Table 22.5.6.1, which provides more detailed formulae:
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Table 6 -  Vc for non-prestressed members with axial compression, ACI 318-14 Table 22.5.6.1
ACI 318-14 Clause 22.5.7- Contribution from concrete (non-prestressed members with significant axial tension)
If the wall is non-prestressed and there is significant axial tension present, then the portion of the total out-of-plane shear strength provided by concrete Vc can be calculated as:
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Vc=2(1+Nu500Ag)λf′cbwdNU is negative for tension\large{V_c=2(1+\frac{N_u}{500A_g})λ\sqrt{f′_c}b_wd} \\ N_U\text{ is negative for tension}Vc​=2(1+500Ag​Nu​​)λf′c​​bw​dNU​ is negative for tension
The equation above is similar to Clause 22.5.6; however, as concrete is significantly weaker in tension than in compression, the provided shear strength is deemed to be lower. This is reflected in the equation as when Nu becomes negative, Vc is reduced.
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ACI 318-14 Clause 22.5.8- Contribution from concrete (prestressed members)
This clause is applicable for pre- and post-tensioned flexural members in regions of the structure where the effective stress in the prestressed reinforcement is fully transferred to the concrete. Effective stress If the member satisfies the following condition, then Vc can be calculated using Table 22.5.8.2:
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Apsfse≥0.4(Apsfpu+Asfy)whereAps=area of prestressed longitudinal tension reinforcement(in2)fse=effective stress in prestressing reinforcement(psi)fpu=specified tensile strength of prestressing reinforcement(psi)As=area of non−prestressed longitudinal tension reinforcement(in2)fy=specified yield strength of non−prestressed reinforcement(psi)A_{ps}f_{se}≥0.4(A_{ps}f_{pu}+A_sf_y) \\ \text{where} \\ A_{ps}= \text{area of prestressed longitudinal tension reinforcement} (in^2) \\ f_{se}= \text{effective stress in prestressing reinforcement} (psi) \\ f_{pu}= \text{specified tensile strength of prestressing reinforcement} (psi) \\ A_s= \text{area of non−prestressed longitudinal tension reinforcement} (in^2) \\ f_y= \text{specified yield strength of non−prestressed reinforcement} (psi)Aps​fse​≥0.4(Aps​fpu​+As​fy​)whereAps​=area of prestressed longitudinal tension reinforcement(in2)fse​=effective stress in prestressing reinforcement(psi)fpu​=specified tensile strength of prestressing reinforcement(psi)As​=area of non−prestressed longitudinal tension reinforcement(in2)fy​=specified yield strength of non−prestressed reinforcement(psi)
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Table 7 - Vc for prestressed members, ACI 318-41 Table 22.5.8.2
It should also be noted that equations in Table 7 may be applied to walls with prestressed reinforcements only as well as walls with a combination of prestressed and non-prestressed reinforcement.
Additionally, Vc for prestressed members can be the lesser of the out-of-plane flexural-shear strength Vci and web-shear strength Vw:
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Vci=min(0.6λf′cbwdp+Vd+ViMcreMmax,1.7λf′cbwd)wheredp=distance from the most extreme compressive fibre to the centroid of the prestressed tensile reinforcement (in.), must be greater than 0.8hVd=shear force at the section being designed for due to unfactored dead load (lb)Mmax=maximum factored bending moment at the section being designed for due to externally applied loads (lb)Vi=factored shear force at the section being designed for occurring simultaneously with MmaxMcre=bending moment at the onset of flexural cracks=(Iyt)(6λf′c+fpe+fd)Vcw=(3.5λf′c+0.3fpc)bwdp+VpVp=vertical component of the effective prestress (lb)The calculated Vc must satisfy: Vu=ϕ(Vc+8f′cbwd\large{V_{ci}=min(0.6λ\sqrt{f′_c}b_wd_p+V_d+\frac{V_iM_{cre}}{M_{max}},1.7λ\sqrt{f′_c}b_wd)} \\ \small{\text{where}} \\ \small{d_p=\text{distance from the most extreme compressive fibre to the centroid of the prestressed tensile reinforcement (in.), must be greater than 0.8h}} \\ V_d=\text{shear force at the section being designed for due to unfactored dead load } (lb) \\ M_{max} =\text{maximum factored bending moment at the section being designed for due to externally applied loads } (lb) \\ V_i= \text{factored shear force at the section being designed for occurring simultaneously with } M_{max} \\ M_{cre}=\text{bending moment at the onset of flexural cracks} =(\frac{I}{yt})(6λ\sqrt{f′_c}+f_{pe}+f_d) \\ \small{V_{cw}=(3.5λ\sqrt{f′_c}+0.3f_{pc})b_wd_p+V_p} \\ Vp=\text{vertical component of the effective prestress } (lb) \\ \text{The calculated } V_c \text{ must satisfy: }V_u=ϕ(V_c+8\sqrt{f′_c}b_wdVci​=min(0.6λf′c​​bw​dp​+Vd​+Mmax​Vi​Mcre​​,1.7λf′c​​bw​d)wheredp​=distance from the most extreme compressive fibre to the centroid of the prestressed tensile reinforcement (in.), must be greater than 0.8hVd​=shear force at the section being designed for due to unfactored dead load (lb)Mmax​=maximum factored bending moment at the section being designed for due to externally applied loads (lb)Vi​=factored shear force at the section being designed for occurring simultaneously with Mmax​Mcre​=bending moment at the onset of flexural cracks=(ytI​)(6λf′c​​+fpe​+fd​)Vcw​=(3.5λf′c​​+0.3fpc​)bw​dp​+Vp​Vp=vertical component of the effective prestress (lb)The calculated Vc​ must satisfy: Vu​=ϕ(Vc​+8f′c​​bw​d
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ACI 318-14 Clause22.5.10 - Contribution from reinforcement
In sections of the wall where concrete alone isn’t strong enough, the portion of total out-of-plane shear strength required from transverse reinforcement V_s is calculated such that:
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Vs≥Vu/ϕ−VcwhereVs=AvfytdsIf the shear reinforcement are stirrups, ties or hoops perpendicular to the longitudinal axis of the member or welded wire reinforcement located perpendicular to the axis of the member or spiral in shape.Vs=Avfyt(sin(α)+cos(α))dsIf the shear reinforcement are inclined stirrups making an angle of at least 45 degrees with the longitudinal axis of the member and cross the location of potential shear crack (only for non−prestressed members).V_s≥V_u/ϕ−V_c \\ \text{where} \\ V_s = \frac{A_vf_{yt}d}{s} \\\text{If the shear reinforcement are stirrups, ties or hoops perpendicular to the longitudinal axis of the member or welded wire reinforcement located perpendicular to the axis of the member or spiral in shape.} \\V_s = \frac{A_vf_{yt}(sin(α)+cos(α))d}{s} \\ \text{If the shear reinforcement are inclined stirrups making an angle of at least 45 degrees with the longitudinal axis of the member and cross the location of potential shear crack (only for non−prestressed members).}Vs​≥Vu​/ϕ−Vc​whereVs​=sAv​fyt​d​If the shear reinforcement are stirrups, ties or hoops perpendicular to the longitudinal axis of the member or welded wire reinforcement located perpendicular to the axis of the member or spiral in shape.Vs​=sAv​fyt​(sin(α)+cos(α))d​If the shear reinforcement are inclined stirrups making an angle of at least 45 degrees with the longitudinal axis of the member and cross the location of potential shear crack (only for non−prestressed members).
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Once Vc and Vs have been calculated, check the total shear strength of the wall against out-of-plane design action effects:
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ϕVn=ϕ(Vc+Vs)>VuϕV_n = ϕ(V_c+V_s)>V_uϕVn​=ϕ(Vc​+Vs​)>Vu​
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Design Guide: Shear Wall to ACI Standards

Introduction

In-plane Shear

Step 1. Determine the minimum shear wall thickness

Step 2. Determine factored axial force and moment

Step 3. Determine axial and flexural strength

Step 4. Determine the in-plane shear strength

Step 5. Determine the out-of-plane shear strength

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