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 哈尔滨工程大学学报  2019, Vol. 40 Issue (8): 1448-1454  DOI: 10.11990/jheu.201812024 0

### 引用本文

XIONG Ergang, XI Yanghong, SONG Liangying, et al. Method for calculating peak displacement of reinforced concrete shear wall[J]. Journal of Harbin Engineering University, 2019, 40(8), 1448-1454. DOI: 10.11990/jheu.201812024.

### 文章历史

1. 长安大学 建筑工程学院, 陕西 西安 710061;
2. 西安建筑科技大学 土木工程学院, 陕西 西安 710055

Method for calculating peak displacement of reinforced concrete shear wall
XIONG Ergang 1, XI Yanghong 1, SONG Liangying 1, LIANG Xingwen 2
1. School of Civil Engineering, Chang'an University, Xi'an 710061, China;
2. School of Civil Engineering, Xi'an University of Architecture & Technology, Xi'an 710055, China
Abstract: To study the peak displacement of reinforced concrete shear walls (the displacement corresponding to the peak load), we established a calculation model and formula for the peak displacement of medium-and high-rise I-section shear walls with boundary columns walls based on a truss model for the longitudinal elongation of the vertical cantilever beam. Then, using this formula, we determined the peak displacements of 34 I-shaped shear walls with boundary columns. The results show that the calculated and experimental results basically agree. The proposed formula enables the estimation of the peak displacement for reinforced concrete shear walls, which can be directly obtained by solving the vertical elongation strain of the shear wall using the calculation model introduced in this paper.
Keywords: reinforced concrete    shear wall    peak displacement    vertical cantilever beam    truss theory    plastic hinge    plastic deformation    elastic deformation

1 竖向悬臂梁纵向伸长量计算桁架模型

 Download: 图 1 竖向悬臂梁纵向伸长量计算 Fig. 1 Longitudinal elongation calculation of vertical cantilever beam
1.1 位移协调条件

 $\Delta_{t}=\Delta_{p}+\Delta_{\theta}+\Delta_{e f}$ (1)

 $-\Delta_{p} \sin \alpha+\varepsilon_{l t} l_{p} \cos \alpha \approx 0$ (2)

 $e_{l}=\left(\varepsilon_{l t}+\varepsilon_{l c}\right) l_{p} / 2$ (3)

 $\Delta_{p}=\varepsilon_{l t} l_{p} \cot \alpha=\left(2 e_{l}-\varepsilon_{l c} l_{p}\right) l_{p} / h_{s}$ (4)

 $\Delta_{\theta}=\theta_{p}\left(H-l_{p}\right)=\left(\varepsilon_{u}-\varepsilon_{l c}\right) l_{p}\left(H-l_{p}\right) / h_{s}$

 $\Delta_{\theta}=2\left(e_{l}-\varepsilon_{l c} l_{p}\right)\left(H-l_{p}\right) / h_{s}$ (5)

 $e_{l}=\frac{\left(\Delta_{t}-\Delta_{e f}\right) h_{s}}{2 H}+\varepsilon_{l c} l_{p}\left(1-\frac{l_{p}}{2 H}\right)$ (6)

Δef可采用Δef=ϕy(Hlp)2/3计算。根据Priestley[15]，梁截面的屈服曲率近似估计为ϕy=1.7εyield/h，式中εyield为纵筋的屈服应变，h为梁截面高度。由式(6)可知，el为Δtεlc的函数。

1.2 力的平衡条件

 $\left\{\begin{array}{l}{F_{D} \sin \alpha+V=0} \\ {F_{D} \cos \alpha+F_{C}+F_{T}+N=0} \\ {V H-N h_{s} / 2=F_{T} h_{s}}\end{array}\right.$

 $\left\{ \begin{array}{l} {F_C} = - {F_T}\left( {1 - \frac{{{h_s}\cot \alpha }}{H}} \right) - N\left( {1 - \frac{{{h_s}\cot \alpha }}{{2H}}} \right) = \\ - {F_T}\left( {1 - \frac{{{l_p}}}{H}} \right) - N\left( {1 - \frac{{{l_p}}}{{2H}}} \right)\\ {\sigma _{lc}} = - {\sigma _{lt}}\left( {\frac{{{A_s}}}{{{{A'}_s}}}} \right)\left( {1 - \frac{{{l_p}}}{H}} \right) - \frac{N}{{{{A'}_s}}}\left( {1 - \frac{{{l_p}}}{{2H}}} \right) \approx \\ - {f_y}\left( {\frac{{{A_s}}}{{{{A'}_s}}}} \right)\left( {1 - \frac{{{l_p}}}{H}} \right) - \frac{N}{{{{A'}_s}}}\left( {1 - \frac{{{l_p}}}{{2H}}} \right) \end{array} \right.$ (7)

1.3 钢筋的本构关系

 $\varepsilon_{l c}=\varepsilon_{1}-\varepsilon_{\text { yield }}-\left(-\eta \frac{\sigma_{l c}}{f_{y}}\right) \varepsilon_{1}=\varepsilon_{1}\left(1+\eta \frac{\sigma_{l c}}{f_{y}}\right)-\varepsilon_{\text { yield }}$ (8)

1.4 构件纵向伸长的分步求解

 Download: 图 3 构件纵向伸长分步计算的流程 Fig. 3 Flow chart of the stepwise calculation of the longitudinal elongation of the component

 $e_{l(n)}=\frac{\left(\Delta_{t}-\Delta_{e f}\right) h_{s}}{2 H}+\varepsilon_{l c(n)} l_{p}\left(1-\frac{l_{p}}{2 H}\right)$ (9)

 $\varepsilon_{l t(n)}=\frac{2\left(H-l_{p}\right)}{l_{p}\left(2 H-l_{p}\right)} e_{l(n)}+\frac{\left(\Delta_{t}-\Delta_{e f}\right) h_{s}}{l_{p}\left(2 H-l_{p}\right)}$ (10)

 $e_{l(n+1)}=\frac{\left(\Delta_{t}-\Delta_{e f}\right) h_{s}}{2 H}+\varepsilon_{l c(n+1)} l_{p}\left(1-\frac{l_{p}}{2 H}\right)$ (11)

 $\begin{array}{*{20}{c}} {{\varepsilon _{lc(n + 1)}} = {\varepsilon _1}\left( {1 + \eta \frac{{{\sigma _{lc}}}}{{{f_y}}}} \right) - {\varepsilon _{{\rm{yield}}}} = }\\ {{\varepsilon _{lt(n)}}\left( {1 + \eta \frac{{{\sigma _{lc}}}}{{{f_y}}}} \right) - {\varepsilon _{{\rm{ yield }}}}} \end{array}$ (12)

 $\begin{array}{*{20}{c}} {{e_{l(n + 1)}} = \left( {1 - \frac{{{l_p}}}{H}} \right)\left( {1 + \eta \frac{{{\sigma _{lc}}}}{{{f_y}}}} \right){e_{l(n)}} + }\\ {\frac{{\left( {{\Delta _t} - {\Delta _{ef}}} \right){h_s}}}{H}\left( {1 + \eta \frac{{{\sigma _{lc}}}}{{2{f_y}}}} \right) - {\varepsilon _y}{l_p}\left( {1 - \frac{{{l_p}}}{{2H}}} \right)} \end{array}$ (13)

 $e_{l(n)}=f\left(\Delta_{t}\right)+\beta^{n-1}\left[e_{l(1)}-f\left(\Delta_{t}\right)\right]$ (14)
 $\left\{ \begin{array}{l} f\left( {{\Delta _t}} \right) = \frac{{\frac{{\left( {{\Delta _t} - {\Delta _{ef}}} \right){h_s}}}{H}\left( {1 + \eta \frac{{{\sigma _{lc}}}}{{2{f_y}}}} \right) - {\varepsilon _{{\rm{yield}}}}{l_p}\left( {1 - \frac{{{l_p}}}{{2H}}} \right)}}{{1 - \left( {1 - \frac{{{l_p}}}{H}} \right)\left( {1 + \eta \frac{{{\sigma _{lc}}}}{{{f_y}}}} \right)}}\\ \beta = \left( {1 - \frac{{{l_p}}}{H}} \right)\left( {1 + \eta \frac{{{\sigma _{lc}}}}{{{f_y}}}} \right)\\ {e_l} = \frac{{\frac{{\left( {{\Delta _t} - {\Delta _{ef}}} \right){h_s}}}{H}\left( {1 + \eta \frac{{{\sigma _{lc}}}}{{2{f_y}}}} \right) - {\varepsilon _{{\rm{yield}}}}{l_p}\left( {1 - \frac{{{l_p}}}{{2H}}} \right)}}{{1 - \left( {1 - \frac{{{l_p}}}{H}} \right)\left( {1 + \eta \frac{{{\sigma _{lc}}}}{{{f_y}}}} \right)}} \end{array} \right.$

 $\varepsilon_{l}=e_{l} / l_{p}$

 ${\varepsilon _l} = \frac{{\frac{{\left( {{\Delta _t} - {\Delta _{ef}}} \right){h_s}}}{{{l_p}H}}\left( {1 + \eta \frac{{{\sigma _{lc}}}}{{2{f_y}}}} \right) - {\varepsilon _{{\rm{yield}}}}\left( {1 - \frac{{{l_p}}}{{2H}}} \right)}}{{1 - \left( {1 - \frac{{{l_p}}}{H}} \right)\left( {1 + \eta \frac{{{\sigma _{lc}}}}{{{f_y}}}} \right)}}$ (15)
 ${\varepsilon _l} = \frac{{\left( {{\Delta _t} - {\Delta _{ef}}} \right){h_s}}}{{2{l_p}H}}\;当\;{\sigma _{lc}} < - {f_y}\;或\;{\varepsilon _{lc}} = 0$ (16)
2 剪力墙构件峰值位移计算

 Download: 图 4 钢筋混凝土剪力墙桁架模型的内力及荷载作用 Fig. 4 Internal force and load of reinforced concrete shear wall truss model

 $\Delta_{m}=\Delta_{p}+\Delta_{\theta}+\Delta_{e f}+\Delta_{e s}$

 $\Delta_{e s}=\mu V H / G A$

 ${\varepsilon _l} = \frac{{\frac{{\left( {{\Delta _m} - {\Delta _{ef}} - {\Delta _{es}}} \right){h_s}}}{{{l_p}H}}\left( {1 + \eta \frac{{{\sigma _{lc}}}}{{2{f_y}}}} \right) - {\varepsilon _{{\rm{yield}}}}\left( {1 - \frac{{{l_p}}}{{2H}}} \right)}}{{1 - \left( {1 - \frac{{{l_p}}}{H}} \right)\left( {1 + \eta \frac{{{\sigma _{lc}}}}{{{f_y}}}} \right)}}$ (17)
 ${\varepsilon _l} = \frac{{\left( {{\Delta _t} - {\Delta _{ef}} - {\Delta _{es}}} \right){h_s}}}{{2{l_p}H}},{\sigma _{lc}} < - {f_y}\;或\;{\varepsilon _{lc}} = 0$ (18)

 $F_{D} \cos \alpha+F_{C}+F_{T}+F_{W}+N=0$
 $F_{D} \sin \alpha+V=0$
 $V H-\frac{N h_{s}}{2}=F_{T} h_{s}+F_{W}\left(\frac{h_{s}}{2}\right)$

 $F_{C}=-F_{T}\left(1-\frac{l_{p}}{H}\right)-\left(F_{W}+N\right)\left(1-\frac{l_{p}}{2 H}\right)$

 $\begin{array}{*{20}{c}} {{\sigma _{lc}} = - {\sigma _{lt}}\left( {\frac{{{A_s}}}{{{{A'}_s}}}} \right)\left( {1 - \frac{{{l_p}}}{H}} \right) - \left( {{\sigma _{lw}}\frac{{{A_w}}}{{{{A'}_s}}} + \frac{N}{{{{A'}_s}}}} \right)\left( {1 - \frac{{{l_p}}}{H}} \right) \approx }\\ { - {f_y}\left( {\frac{{{A_s}}}{{{{A'}_s}}}} \right)\left( {1 - \frac{{{l_p}}}{H}} \right) - \left( {{f_{yw}}\frac{{{A_w}}}{{{{A'}_s}}} + \frac{N}{{{{A'}_s}}}} \right)\left( {1 - \frac{{{l_p}}}{H}} \right)} \end{array}$ (19)

 ${A_{Dc}} = {h_d}{b_w} = \left( {3{l_p} + 2{l_c}} \right){b_w}\sin \alpha /4$
 $l_{c}=\frac{\left(x_{c}-\alpha_{s}^{\prime}\right) l_{p}}{2\left(h_{s}-x_{c}+\alpha_{s}^{\prime}\right)} \geqslant 0$

 Download: 图 6 有效抗压强度fce-主拉应变εt相关关系 Fig. 6 Effective compressive strength fce-primary tensile strain εt correlation
 $F_{D}=-f_{c e} A_{D c}=-\frac{f_{c} A_{D c}}{0.8+0.34\left(\varepsilon_{t} / \varepsilon_{c o}\right)} \geqslant-f_{c} A_{D_{c}}$ (20)

 ${V_{wc}} = \frac{{{f_{ce}}{A_{Dc}}\sin \alpha }}{{0.8 + 0.34\left( {{\varepsilon _t}/{\varepsilon _{co}}} \right)}} \le {f_c}{A_{Dc}}\sin \alpha$ (21)

 $\varepsilon_{x}=\varepsilon_{v} / 2, \varepsilon_{y}=\left(\varepsilon_{l t}+\varepsilon_{l c}\right) / 2$

 $\varepsilon_{x}=\varepsilon_{v y} / 2, \varepsilon_{y}=\varepsilon_{l}$ (22)

 Download: 图 7 开裂混凝土平均应变及平均应变莫尔圆 Fig. 7 Average strain and average strain of cracked concrete
 $\boldsymbol{\varepsilon}_{x}+\boldsymbol{\varepsilon}_{y}=\boldsymbol{\varepsilon}_{t}+\boldsymbol{\varepsilon}_{c}$ (23)

 $\varepsilon_{t}=\varepsilon_{l}+\varepsilon_{v y} / 2-\varepsilon_{c}$

 $\varepsilon_{t} \approx \varepsilon_{l}+3 \varepsilon_{c o} / 2$ (24)

 ${V_{wc}} = \frac{{{f_c}{A_{Dc}}\sin \alpha }}{{1.31 + 0.34\left( {{\varepsilon _l}/{\varepsilon _{co}}} \right)}} \le {f_c}{A_{Dc}}\sin \alpha$ (25)

 $\frac{{{f_{ck}}{A_{Dc}}\sin \alpha }}{{1.31 + 0.34\left( {{\varepsilon _l}/{\varepsilon _{co}}} \right)}} = {F_m}$ (26)

 Download: 图 8 钢筋混凝土剪力墙峰值位移的计算流程 Fig. 8 Calculation process of peak displacement of reinforced concrete shear wall
3 本文计算方法有效性验证

4 结论

1) 与文献[1-3]介绍的剪力墙构件变形指标计算方法相比该模型只要给定剪力墙的截面设计信息、剪力墙的高度以及轴力和剪力就可以直接求解得到与该剪力对应的峰值位移。无需再计算受压区高度或相对受压区高度。

2) 采用本文提出的方法，对收集的34片I形截面剪力墙的峰值位移进行了计算，计算值与试验值相差20%以内的有15组，计算值与试验值相差30%以内的有20组，基本验证了本文所提方法的有效性。

3) 本文所建立模型是基于单片剪力墙，且塑性铰区部位是规则的，可对塑性铰部位不规则的剪力墙做进一步研究。