Analysis of influence of imperfections on stiffness of fully anchored light-frame timber shear walls—elastic model

Abstract

In order to stabilize light-frame timber buildings against horizontal loads, the diaphragm or in-plane action of roofs, floors and walls is often used. This paper deals with the influence of imperfections such as gaps and uplift on the horizontal displacement of fully anchored shear walls. The significance of analyzing the effects of imperfections is evident when evaluating the stiffness of shear walls; tests of walls show that the horizontal displacement is underestimated in calculations using the stiffness of sheathing-to-framing joints as obtained from experiments. Also, in real structures where hold-downs are used according to the elastic design method, the influence of gaps and uplift should be included in order to obtain realistic displacements in the serviceability limit state. A new elastic model for the analysis, based on linear elastic behaviour of the mechanical sheathing-to-framing joints, is presented and the equations for the stiffness and the deflection versus the number of segments in the wall are derived. The fully anchored condition for the shear walls are modelled by applying a diagonal load to the wall. Three types of imperfections are evaluated: gaps at all studs, a gap only at the trailing stud, and gaps at all studs, except at the trailing stud. It is shown that the effect of imperfections on the stiffness of the wall in the initial stage is considerable. Depending on the distribution of the gaps and the number of segments included in the shear wall, the displacement of the shear wall is increased several times compared to that of a fully anchored shear wall with no gaps; e.g. for a single segment wall more than three times. However, for walls with more than six to ten segments, the effect of imperfections can be neglected. Finally, the theoretical model is experimentally verified.

Appendix A: Analysis of influence of imperfections—gaps or uplift

1.1 A.1 Shear walls with a single segment—without gaps or uplift

Consider a single segment wall with no gaps according to Fig. 5a, b. The diagonal force is applied on the frame work according to Fig. 5a and is transferred to the sheet according to Fig. 5b via the fasteners. The coordinate system for the frame is located in the fixed point of the lower right corner and is denoted (u _frame, v _stud). The coordinate system (u, v) is located in the CG of the sheet, which coincides with the CG of the equally spaced fasteners along the perimeter of the sheet.

For a fully anchored shear wall without gaps, there is no displacement (of the first order) in the vertical direction, i.e. v ₁ = v _stud,0 = v _stud,1 = 0. According to Källsner and Girhammar [1], the horizontal displacement, u _frame, for a wall without imperfections is given by

$$u_{\rm{frame}} =2\left[ {\frac{3}{1+3\frac{h}{b}}\frac{h^{2}}{b^{2}}+\frac{1}{1+\frac{5}{12}\frac{h}{b}}} \right]\frac{s_{\rm r} }{b}\frac{H}{k};\,\rightarrow\,u_{\rm{frame}} =4.52\frac{s_{\rm r} }{b}\frac{H}{k}\hbox{for}\frac{h}{b}=2$$

(23)

and the pertaining variables with reference to the sheet are given by

$$\theta_{\rm 1} =-\frac{6}{1+3\frac{h}{b}}\frac{h}{b}\frac{s_{\rm r} }{b}\frac{H}{kb};\,\rightarrow\,\theta_{\rm 1} =-1.71\frac{s_{\rm r} }{b}\frac{H}{kb}\quad \hbox{for}\quad \frac{h}{b}=2$$

(24)

$$u_{\rm 1} =\frac{1}{2}u_{\rm{frame}} +\frac{1}{1+\frac{5}{12}\frac{h}{b}}\frac{s_{\rm r} }{b}\frac{H}{k}; \quad \rightarrow \, u_{\rm 1} =2.80\frac{s_{\rm r} }{b}\frac{H}{k}\hbox{ for }\frac{h}{b}=2$$

(25)

1.2 A.2 Shear walls with a single segment—with gaps at all studs

Consider a single segment wall with gaps according to Fig. 5c, d. The forces arising due to the imperfections, R _V1 and R _V2 as given by Eqs. 2, 3, are applied on the frame work and transferred to the sheet via the fasteners. For this system of forces, there are no reactions in and no displacement (of the first order) of the shear wall in the horizontal direction, i.e. H = 0; u ₁ = 0. This shear wall system is analyzed by using the equations of equilibrium and compatibility together with the constitutive relation for the shear force per unit length of the fasteners, f = (k/s _r)δ, where δ denotes a displacement. The following equations must be valid

$$R_1 =R_{\rm{V1}} +R_{\rm{V2}}$$

$$M_1 =(R_{\rm{V2}} -R_{\rm{V1}} )\frac{b}{2}$$

(26)

$$M_1=-2\cdot \frac{1}{2}\frac{k}{s_{\rm r} }\frac{b}{2}\cdot \theta_1 \frac{b}{2}\cdot \frac{2}{3}\frac{b}{2}=-\frac{k}{s_{\rm r} }b\cdot \theta_1 \frac{b^{2}}{12}$$

Then, the initial horizontal displacement up to closure of the gap, i.e. 0 $\le v_{\rm{stud,1}} \le v_{\rm{trail}}^{\rm{imp}}$ or 0 ≤ H ≤ H _imp, is obtained as

$$\Updelta u_{\rm{frame}}^{\rm{imp}} =-\theta_1 \frac{h}{2}=-6(R_{\rm{V2}} -R_{\rm{V1}} )\frac{s_{\rm r} }{b}\frac{1}{kb}\frac{h}{2}=3(R_{\rm{V2}} -R_{\rm{V1}} )\frac{h}{b}\frac{s_{\rm r} }{b}\frac{1}{k}$$

(27)

$$u_{\rm{frame}}^{\rm{imp}} =u_{\rm{frame}} +\Updelta u_{\rm{frame}}^{\rm{imp}}$$

(28)

Using the fact that, R ₁ = −(k/s _r)bv ₁, R _{V1 = (k/s _r)h(−v ₁ + θ₁ b/2 + v _stud,0), and R _V2 = (k/s _r)h(v ₁ + θ₁ b/2−v _stud,1), the vertical displacement of the sheet and of the leading and trailing studs can be obtained as

$$v_{\rm 1} =-(R_{\rm{V1}} +R_{\rm{V2}} )\frac{s_{\rm r} }{b}\frac{1}{k}$$

(29)

$$v_{\rm{stud,0}} =\left[\left(\frac{b}{h}-4\right)R_{\rm{V1}} +2R_{\rm{V2}} \right]\frac{s_{\rm r} }{b}\frac{1}{k}$$

(30)

$$v_{\rm{stud,1}} =\left[2R_{\rm{V1}} -\left(\frac{b}{h}+4\right)R_{\rm{V2}} \right]\frac{s_{\rm r} }{b}\frac{1}{k}$$

(31)

1.3 A.3 Shear walls with arbitrary number of segments—with gaps at all studs

For a wood-frame shear wall with arbitrary number of segments and with gaps at all studs, all equations in the previous section are valid if only the force, R _V2, according to Eq. 3, is replaced by the general expression, $R_{{\rm V}2,n_{\rm{seg}}}$ as given by Eq. 5. Then, we arrive at the following general relationships for the representative segment of a shear wall with arbitrary number of segments,

$$\Updelta u_{\rm{frame}}^{\rm{imp}} =\frac{9}{n_{\rm{seg}} (1+3\frac{h}{b})}\frac{h^{3}}{b^{3}}\frac{s_{\rm r} }{b}\frac{H}{k};\,\rightarrow\,\Updelta u_{\rm{frame}}^{\rm{imp}} =\frac{72}{7n_{\rm{seg}} }\frac{s_{\rm r} }{b}\frac{H}{k}\hbox{ for}\;\;\frac{h}{b}=2$$

(32)

$$u_{\rm{frame}}^{\rm{imp}} =u_{\rm{frame}} +\Updelta u_{\rm{frame}}^{\rm{imp}} ;\,\rightarrow\,u_{\rm{frame}}^{\rm{imp}} =(4.52+\frac{10.3}{n_{\rm{seg}} })\frac{s_{\rm r} }{b}\frac{H}{k}\hbox{ for}\;\;\frac{h}{b}=2$$

(33)

It is evident from Eqs. 23 and 33 that the stiffness of the wall is decreased more than three times in case of gaps in a single segment wall (n _seg = 1). For a wall with infinite number of segments, the stiffness approaches that of a perfect single segment shear wall. The other quantities in the imperfect wall system are given by

$$R_{\rm 1}=\frac{n_{\rm{seg}} +3\frac{h}{b}}{n_{\rm{seg}} (1+3\frac{h}{b})}\frac{h}{b}H;\,\rightarrow\,R_{\rm 1} =\frac{2(n_{\rm{seg}} +6)}{7n_{\rm{seg}} }H \quad \hbox{ for}\;\;\frac{h}{b}=2$$

(34)

$$M_{\rm 1} =\frac{3}{2}\frac{1}{n_{\rm{seg}} (1+3\frac{h}{b})}\frac{h}{b}Hh;\,\rightarrow\,M_{\rm 1} =\frac{3}{7n_{\rm{seg}} }Hh \quad \hbox{ for}\;\;\frac{h}{b}=2$$

(35)

$$\theta_{\rm 1} =-\frac{18}{n_{\rm{seg}} (1+3\frac{h}{b})}\frac{h^{2}}{b^{2}}\frac{s_{\rm r} }{b}\frac{H}{kb};\,\rightarrow\,\theta_{\rm 1} =-\frac{72}{7n_{\rm{seg}} }\frac{s_{\rm r} }{b}\frac{H}{kb} \quad \hbox{for}\;\;\frac{h}{b}=2$$

(36)

$$v_{\rm 1} =-\frac{n_{\rm{seg}} +3\frac{h}{b}}{n_{\rm{seg}} (1+3\frac{h}{b})}\frac{h}{b}\frac{s_{\rm r} }{b}\frac{H}{k};\,\rightarrow\,v_{\rm 1} =-\frac{2(n_{\rm{seg}} +6)}{7n_{\rm{seg}} }\frac{s_{\rm r} }{b}\frac{H}{k} \quad \hbox{for}\;\;\frac{h}{b}=2$$

(37)

$$\begin{aligned} v_{\rm{stud,0}}&=\frac{1}{2}\frac{n_{\rm{seg}} -2(n_{\rm{seg}} -2)\frac{h}{b}+12\frac{h^{2}}{b^{2}}}{n_{\rm{seg}} (1+3\frac{h}{b})}\frac{s_{\rm r} }{b}\frac{H}{k}; \\ &\rightarrow v_{\rm{stud,0}} =\frac{n_{\rm{seg}} -4(n_{\rm{seg}} -2)+48}{14n_{\rm{seg}} }\frac{s_{\rm r} }{b}\frac{H}{k} \quad \hbox{for}\;\;\frac{h}{b}=2 \end{aligned}$$

(38)

$$\begin{aligned} v_{\rm{stud,1}}&=-\frac{1}{2}\frac{n_{\rm{seg}} +2(n_{\rm{seg}} +3)\frac{h}{b}+12\frac{h^{2}}{b^{2}}}{n_{\rm{seg}} (1+3\frac{h}{b})}\frac{s_{\rm r} }{b}\frac{H}{k};\\ & \rightarrow v_{\rm{stud,1}} =-\frac{n_{\rm{seg}} +4(n_{\rm{seg}} +3)+48}{14n_{\rm{seg}} }\frac{s_{\rm r} }{b}\frac{H}{k} \quad \hbox{for}\;\;\frac{h}{b}=2\\ \end{aligned}$$

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The maximum vertical displacement takes place in the trailing stud. Using Eqs. 31, 2 and 6, this displacement is given by

$$\begin{aligned} v_{{\rm stud},n_{\rm seg}}&=-\frac{1}{2}\frac{1+2(n_{\rm{seg}} +5)\frac{h}{b}+12(n_{\rm{seg}} +1)\frac{h^{2}}{b^{2}}}{n_{\rm{seg}} (1+3\frac{h}{b})}\frac{s_{\rm r} }{b}\frac{H}{k};\\ &\rightarrow v_{\rm{stud,}n_{\rm{seg}}}=-\frac{52n_{\rm{seg}} +69}{14n_{\rm{seg}} }\frac{s_{\rm r} }{b}\frac{H}{k} \quad \hbox{for}\;\;\frac{h}{b}=2 \end{aligned}$$

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1.4 A.4 Shear walls with a single segment—with a gap only at trailing stud

To evaluate a fully anchored shear wall without gaps except at the trailing stud, the vertical force in and the displacement of the leading stud, R _V1 = 0 and v _stud,0 = 0, respectively, are introduced in Eqs. 26–31. The horizontal displacement, Δu ^imp_frame , for a single segment shear wall (n _seg = 1) with this imperfection then becomes

$$\Updelta u_{\rm{frame,1}}^{\rm{imp}} =\frac{3}{2}\frac{1+6\frac{h}{b}}{1+3\frac{h}{b}}\frac{h^{2}}{b^{2}}\frac{s_{\rm r} }{b}\frac{H}{k};\,\rightarrow\,\Updelta u_{\rm{frame,1}}^{\rm{imp}} =\alpha_{u_{\rm{frame}} ,1}^{\rm{imp}} \frac{s_{\rm r} }{b}\frac{H}{k}=\frac{78}{7}\frac{s_{\rm r} }{b}\frac{H}{k} \quad \hbox{for}\;\;\frac{h}{b}=2$$

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For a shear wall with arbitrary number of segments, the horizontal displacement is evaluated as shown in Sect. 4.4. The vertical displacement of the trailing stud is, according Eq. 31 together with R _V1 = 0 and Eq. 3, then given by

$$v_{\rm{stud,}} n_{\rm{seg}}=-\frac{(1+4\frac{h}{b})(1+6\frac{h}{b})}{2(1+3\frac{h}{b})}\frac{s_{\rm r} }{b}\frac{H}{k};\,\rightarrow \,v_{\rm{stud,}} n_{\rm{seg}}=-\frac{117}{14}\frac{s_{\rm r} }{b}\frac{H}{k}\quad \hbox{for}\quad \frac{h}{b}=2$$

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i.e. the displacement is independent of the number of segments of the wall.

1.5 A.5 Shear walls with arbitrary number of segments—with gaps at all studs, except at trailing stud

A shear wall with gaps at all studs, except at the trailing stud is a situation that occurs after the gap at the trailing stud is closed as looked at in Sect. A.3. In this case, the vertical force in and the displacement of the trailing stud is R _V2 = 0 and v _stud,1 = 0, respectively, which can be introduced in Eqs. 26–31 to obtain the results for this type of imperfect shear wall. The horizontal displacement, Δu ^imp_frame , for a wall with these imperfections then becomes

$$\Updelta u_{\rm{frame}}^{\rm{imp}} =-\frac{3}{2n_{\rm{seg}} }\frac{1}{1+3\frac{h}{b}}\frac{h^{2}}{b^{2}}\frac{s_{\rm r} }{b}\frac{H}{k};\quad \rightarrow \, \Updelta u_{\rm{frame}}^{\rm{imp}} =-\frac{6}{7n_{\rm{seg}} }\frac{s_{\rm r} }{b}\frac{H}{k} \quad \hbox{for}\;\;\frac{h}{b}=2$$

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