CN101576460B - 确定粒料基层lc非线性承载力的方法 - Google Patents

确定粒料基层lc非线性承载力的方法 Download PDF

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CN101576460B
CN101576460B CN2009101039239A CN200910103923A CN101576460B CN 101576460 B CN101576460 B CN 101576460B CN 2009101039239 A CN2009101039239 A CN 2009101039239A CN 200910103923 A CN200910103923 A CN 200910103923A CN 101576460 B CN101576460 B CN 101576460B
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梁乃兴
曹源文
赵毅
梅迎军
李志勇
冯五一
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Chongqing Jiaotong University
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Abstract

本发明公开了一种确定粒料基层LC非线性承载力的方法,该方法根据下式确定粒料材料回弹模量与大小主应力的关系:
Figure D2009101039239A00011
采用ANSYS软件对参数a、b、c进行求解。本发明的有益技术效果是:提出了一种确定粒料基层LC非线性承载力的模型,该模型具有参数少,计算简便,实验、分析成本低廉的优点,为进一步研究粒料基层材料模量与大小主应力的关系建立了基础。

Description

确定粒料基层LC非线性承载力的方法
技术领域
本发明涉及一种材料回弹模量计算模型,尤其涉及一种确定粒料基层LC非线性承载力的方法。
背景技术
粒料混合料是指砂砾石、砂卵石、碎石、石渣、石粉等不同粒径颗粒组成的无粘性混合料,不仅具有良好温度稳定性和湿度稳定性,粒料颗粒间嵌锁作用形成的非线性结构层具有较大的承载能力和较强的抵抗疲劳破坏能力,还可以有效吸收裂缝尖端应力,防止半刚性沥青路面早期开裂和反射裂缝的产生。因此,采用优质级配粒料混合料作为沥青路面基层得到了工程界的关注。
国内外许多研究表明:粒料类材料的回弹模量随作用应力状态而变化,其应力应变的非线性使得混合料的回弹模量在很大程度上受竖向和侧向应力大小的影响,众多研究采用基于重复三轴试验的方法,并建立粒料混合料回弹模量与应力状态相互关系模型。确定无粘性粒料混合料回弹模量的主要模型有:1)K-θ模型、2)Uzan模型、3)LADE和NELSON模型、4)Brown和Pappin模型、5)DAMA设计程序、6)德国学者F.Wellner提出的碎石材料的弹性模量公式: E = ( 8617.7 + 3064 σ 1 0.421 ) σ 3 0.33 + 30000 ; 前述种种模型及公式,有的参数较多计算复杂,有的理论还并不完善,有待进一步研究。
发明内容
本发明提出了一种确定粒料基层LC非线性承载力的方法,该方法根据下式确定粒料材料回弹模量与大小主应力的关系:
E = ( a + b σ 3 0.5 ) σ 1 0.5 + c
式中:σ1为单元大主应力,(MPa);
      σ3为单元小主应力,(MPa);
      E为粒料材料回弹模量;
      a、b、c为在试验结果的基础上进行理论分析回归的参数;
      a无量纲;b单位为[MPa]-1;c单位为MPa。
上述参数a、b、c的计算方法如下:
将参数a、b、c的关系表示为下列三式:
aA+bB+cC=D
aB+bE+cF=G
aC+bF+nc=K
式中, A = Σ i = 1 n σ 1 i , B = Σ i = 1 n σ 1 i σ 3 i , C = Σ i = 1 n σ 1 i , D = Σ i = 1 n E i σ 1 i
E = Σ i = 1 n σ 1 i σ 3 i , F = Σ i = 1 n σ 1 i σ 3 i , G = Σ i = 1 n E i σ 1 i σ 3 i , K = Σ i = 1 n E i ,
n为有限元模型划分单元数,n=90;
i为单元序数值,即i=1,2…90;
σ1i为第i个单元大主应力,(MPa);
σ3i为第i个单元小主应力,(MPa);
Ei为第i个单元粒料材料的回弹模量,(MPa)。
采用ANSYS软件建立尺寸为D×H=308mm×220mm的数字模型,将数字模型在高度方向划分为18等份,提取垂直方向的路径R=0mm,R=43.333mm,R=50mm,R=67.333mm,R=145.333mm,提取定义路径上的结点在单元形心处共90个单元的σ1、σ3,设压应力为正,拉应力为负且负值取为零;根据A、B、C、D、E、F、G、K的8个表达式,计算出各级荷载的A、B、C、D、E、F、G、K值,再分别求和,将求和结果代入参数a、b、c的三个关系式,最终求得参数a、b、c的值;其中,σ1为单元大主应力,(MPa);σ3为单元小主应力,(MPa);
本发明的有益技术效果是:提出了一种确定粒料基层LC非线性承载力的模型,该模型具有参数少,计算简便,实验、分析成本低廉的优点,为进一步研究粒料基层材料模量与大小主应力的关系建立了基础。
附图说明
图1、粒料混合料回弹模量测试装置实物照片;
图2、极限含水量情况下粒料混合料回弹模量测定装置实物照片;
图3、单元划分示意图。
具体实施方式
发明人针对现有技术的不足,进行了深入、细致的研究,提出了本发明的确定粒料基层LC非线性承载力的方法。
由于砂砾材料的应力应变关系具有明显的非线性(即随着单位压力的增大,回弹模量也增大),所以首先要进行回弹模量试验:通过回弹模量试验,建立单位应力与回弹模量的关系。实验过程简述如下:
采用室内承载板试验装置,取尺寸为D×H=308mm×220mm的试桶,承载板直径取100mm,试样成型用2000KN压力机静压成型,加载方式采用在承载板上放置千斤顶,通过反力架逐级加载卸载进行回弹模量试验,并用测力环控制加载量。磁性表架上安置两个千分表,读取每级荷载下的回弹变形。试验装置如图1、2所示。
根据实验数据,对参数a、b、c进行求解:
确定出的参数a、b、c应使LC模型确定的模量与回弹模量试验每一级荷载下对应的模量的均方差最小,可表示为下式:
H = Σ i = 1 n [ ( ( a + b σ 3 i 0.5 ) σ 1 i 0.5 + c ) - E i ] 2
式中,σ1i为第i级加荷下单元大主应力,(MPa);σ3i为第i级加荷下单元小主应力,(MPa);
当H最小时,回归得出参数a,b,c,进而得到模量E与实际试验所测模量最为接近。
为使H的值最小,将前式分别对a、b、c求导,得:
a Σ i = 1 n σ 1 i + b Σ i = 1 n σ 1 i σ 3 i + c Σ i = 1 n σ 1 i = Σ i = 1 n E i σ 1 i
a Σ i = 1 n σ 1 i σ 3 i + b Σ i = 1 n σ 1 i σ 3 i + c Σ i = 1 n σ 1 i σ 3 i = Σ i = 1 n E i σ 1 i σ 3 i
a Σ i = 1 n σ 1 i + b Σ i = 1 n σ 1 i σ 3 i + nc = Σ i = 1 n E i
为了便于分析,记
A = Σ i = 1 n σ 1 i , B = Σ i = 1 n σ 1 i σ 3 i , C = Σ i = 1 n σ 1 i , D = Σ i = 1 n E i σ 1 i , E = Σ i = 1 n σ 1 i σ 3 i , F = Σ i = 1 n σ 1 i σ 3 i ,
G = Σ i = 1 n E i σ 1 i σ 3 i , K = Σ i = 1 n E i ,
其中,n为有限元模型划分单元数,n=90;
i为单元序数值,即i=1,2…90;
σ1i为第i个单元大主应力,(MPa);
σ3i为第i个单元小主应力,(MPa);
Ei为第i个单元粒料材料的回弹模量,(MPa)。
则有
aA+bB+cC=D
aB+bE+cF=G
aC+bF+nc=K
采用ANSYS软件,建立尺寸为D×H=308mm×220mm的数字模型,将数字模型在高度方向划分为18等份,提取垂直方向的路径R=0mm,R=43.333mm,R=50mm,R=67.333mm,R=145.333mm,提取定义路径上的结点在单元形心处共90个单元的σ1、σ3,设压应力为正,拉应力为负且负值取为零;根据A、B、C、D、E、F、G、K的8个表达式,计算出各级荷载的A、B、C、D、E、F、G、K值,再分别求和,将求和结果代入参数a、b、c的三个关系式,最终求得参数a、b、c的值,即可得到本发明的粒料基层LC非线性承载力模型:
E = ( a + b σ 3 0.5 ) σ 1 0.5 + c
其中,n为有限元模型划分单元数,n=90;
i为单元序数值,即i=1,2…90;
σ1i为第i个单元大主应力,(MPa);
σ3i为第i个单元小主应力,(MPa);
Ei为第i个单元粒料材料的回弹模量,(MPa)。

Claims (2)

1.一种确定粒料基层LC非线性承载力的方法,采用室内承载板试验装置,取尺寸为D×H=308mm×220mm的试桶,承载板直径取100mm,试样成型用2000KN压力机静压成型,加载方式采用在承载板上放置千斤顶,通过反力架逐级加载卸载进行回弹模量试验,并用测力环控制加载量;磁性表架上安置两个千分表,读取每级荷载下的回弹变形;根据实验数据,对参数a、b、c进行求解;
其特征在于:根据下式确定粒料材料回弹模量与大小主应力的关系:
E = ( a + bσ 3 0.5 ) σ 1 0.5 + c
式中:σ1为单元大主应力,单位为MPa;
σ3为单元小主应力,单位为MPa;
E为粒料材料回弹模量;
a、b、c为在试验结果的基础上进行理论分析回归的参数;
a无量纲;b单位为[MPa]-1;c单位为MPa。
2.根据权利要求1所述的确定粒料基层LC非线性承载力的方法,其特征在于:参数a、b、c的计算方法包括:
将参数a、b、c的关系表示为下列三式:
aA+bB+cC=D
aB+bE+cF=G
aC+bF+nc=K
式中, A = Σ i = 1 n σ 1 i , B = Σ i = 1 n σ 1 i σ 3 i , C = Σ i = 1 n σ 1 i , D = Σ i = 1 n E i σ 1 i E = Σ i = 1 n σ 1 i σ 3 i , F = Σ i = 1 n σ 1 i σ 3 i , G = Σ i = 1 n E i σ 1 i σ 3 i , K = Σ i = 1 n E i ,
其中,n为有限元模型划分单元数,n=90;
i为单元序数值,即i=1,2…90;
σ1i为第i个单元大主应力,单位为MPa;
σ3i为第i个单元小主应力,单位为MPa;
Ei为第i个单元粒料材料的回弹模量,单位为Mpa;
采用ANSYS软件建立尺寸为D×H=308mm×220mm的数字模型,将数字模型在高度方向划分为18等份,提取垂直方向的路径R=0mm,R=43.333mm,R=50mm,R=67.333mm,R=145.333mm,提取定义路径上的结点在单元形心处共90个单元的σ1、σ3,设压应力为正,拉应力为负且负值取为零;根据A、B、C、D、E、F、G、K的8个表达式,计算出各级荷载的A、B、C、D、E、F、G、K值,再分别求和,将求和结果代入参数a、b、c的三个关系式,最终求得参数a、b、c的值;其中,σ1为单元大主应力,单位为MPa;σ3为单元小主应力,单位为MPa。
CN2009101039239A 2009-05-22 2009-05-22 确定粒料基层lc非线性承载力的方法 Expired - Fee Related CN101576460B (zh)

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