Shear 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 occupants of the structure may experience discomfort, or in the worst case, the structure may completely fail. 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.

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.

Arrangement (A) is common instructures as the shear walls are easier to construct as they have no corners.Arrangement (B) has an aesthetic advantage as spaces between the shear wallsallow instalment of decorative features such as large windows. In most tallstructures, shear walls are placed in the centre in ‘c-shape’ (generally usedas elevator shafts or stairways) to form a ‘core’. These cores prevent lateraltorsion 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 inshear walls for access (e.g. elevator shafts), windows or doorways, as seen in Figure 4. In these situations, theareas straight above the opening are significantly weaker than others. Tocounteract these internal stresses, ‘lintels’ are placed to act as beams tosupport additional height. For more, you can read this guide byStructuralGuide here (https://www.structuralguide.com/concrete-lintel-beams/).

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 design of ‘special walls’ must follow Section 18 instead. A ‘special wall’is defined as a shear wall including coupling beams and wall piers that formpart of a seismic-force-resisting system.

In-plane Shear

Step 1. Determine the minimum wall thickness

ACI 318-14 Table 11.3.1.1 provides the minimum thickness h for different types of walls. If justified with appropriate structural analysis, then thinner walls may be used.

Step 2. Determine factored axial force and moment

Shear walls are to be designed for the factored axial force, Pwhich 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:

\begin{align*} \cdot P_o = 0.85f'_c(A_g-A_{st})+f_yA_{st} \ if\ the\ member\ is\ non-prestressed\ or\ composite\ steel\ and\ concrete\\ \cdot P_o = 0.85f'_c(A_g-A_{st}-A_{pd}) + f_yA_{st}-(f_{se} - 0.003E_p)A_{pt} \end{align*}
Table 2 - Maximum axial strength, ACI 318-14 Table 22.4.2.1

Step 3. Determine axial and flexural strength

ACI 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 behaviour.

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 which causes out-of-plane flexure, then Pn is calculated as:

\begin{align*} \cdot P_n=0.55f'_cA_g[1-(\frac{kl_c}{32h})^2]\\ Where,\\ l_c = length\ of\ the\ wall\ (in.)\\ k = effective\ length\ factor,\ which\ accounts\ for\ different\ boundary\ conditions\ (as\ shown\ in\ Figure\ 5)\ at\ the\ wall’s\ ends,\ determined\ using\ Table 11.5.3.2:\ \end{align*}

Table 3 - Effective length factor k, ACI 318-14 Table 11.5.3.2

Step 4. Determine the in-plane shear strength

ACI 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:

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.

ACI 318-14 Clause11.5.4.8 - Contribution from reinforcement

The portion of total in-plane shear strength provided by reinforcement  is calculated as:

\begin{align*} \cdot V_s=\frac{A_vf_{yt}d}{s} \\ Where,\\ Av= area\ of\ shear\ reinforcement\ (in.^2)\\ fyt= yield\ strength\ of\ transverse\ reinforcement\ (psi)\\ s= centre-to-centre\ spacing\ of\ transverse\ reinforcement\ \end{align*}

Once V_c and V_s have been calculated, check the total shear strength of the wall against in-plane design action effects:

\begin{align*} \cdot \phi V_n = \phi(V_c + V_s)> V_u \end{align*}

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

Identical 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.

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 on it, then the portion of the total out-of-plane shear strength provided by concrete V_c can be calculated as:

\begin{align*} \cdot V_c=2\lambda\sqrt{f'_c}b_wd\ (simplified)\ \end{align*}

Or, using Table 22.5.5.1 which provides more detailed formulae:

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 Vcan be calculated as:

\begin{align*} \cdot V_c=2(1+ \frac{N_u}{2000A_g})\lambda\sqrt{f'_c}b_wd \end{align*}

Or, using Table 22.5.6.1 which provides more detailed formulae:

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:

\begin{align*} \cdot V_c=2(1+ \frac{N_u}{500A_g})\lambda\sqrt{f'_c}b_wd \end{align*}

Nu is negative for tension

The equation above is similar to the equation from 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.

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:

\begin{align*} \cdot A_{ps}f_{se} \ge 0.4(A_{ps}f_{pu}+A_sf_y) \\ Where, \\ A_{ps}= area\ of\ prestressed\ longitudinal\ tension\ reinforcement\ (in.^2)\\ f_{se}= effective\ stress\ in\ prestressing\ reinforcement\ (psi)\\ f_{pu}= specified\ tensile\ strength\ of\ prestressing\ reinforcement\ (psi)\\ A_s= area\ of\ non-prestressed\ longitudinal\ tension\ reinforcement\ (in.^2)\\ f_y= specified\ yield\ strength\ of\ non-prestressed\ reinforcement\ (psi)\\ \end{align*}

                             

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:

\begin{align*} \cdot V_{ci}=lesser\ of\ [0.6\lambda\sqrt{f'_c}b_wd_p+V_d+\frac{V_iM_{cre}}{M_{max}} , 1.7\lambda\sqrt{f'_c}b_wd] \\ Where,\\ d_p= distance\ from\ the\ most\ extreme\ compressive\ fibre\ to\ the\ centroid\ of\ the\ prestressed\ tensile\ reinforcement\ (in.),\ must\ be\ greater\ than\ 0.8h\\ V_d= shear\ force\ at\ the\ section\ being\ designed\ for\ due\ to\ unfactored\ dead\ load\ (lb)\\ M_{max}= maximum\ factored\ bending\ moment\ at\ the\ section\ being\ designed\ for\ due\ to\ external\ applied\ loads\ (lb)\\ V_i= factored\ shear\ force\ at\ the\ section\ being\ designed\ for\ occurring\ simultaneously\ with\ M_{max}\\ M_{cre}= bending\ moment\ at\ the\ onset\ of\ flexural\ cracks\ =(\frac{I}{y_t})(6\lambda\sqrt{f'_c}+f_{pe}+f_d) \\ \cdot V_{cw}=(3.5\lambda\sqrt{f'_c}+0.3f_{pc})b_wd_p+V_p\\ V_p= vertical\ component\ of\ the\ effective\ prestress\ (lb)\\ The\ calculated\ V_c\ must\ satisfy:\\ V_u= \phi (V_c+8\sqrt{f'_c}b_wd \end{align*}

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:

\begin{align*} \cdot V_s \ge \frac{Vu}{\phi}-V_c \\ Where.\\ V_s=\frac{A_vf_{yt}d}{s} 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 \alpha + cos \alpha)d}{s} 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).\\ \end{align*}

Once Vc and Vs have been calculated, check the total shear strength of the wall against out-of-plane design action effects:

\begin{align*} \cdot \phi V_n= \phi(V_c+V_s)>V_u \end{align*}

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Published:
Jan 19, 2023
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January 19, 2023