空間半無限體內(nèi)部作用集中力時的粘彈性沉降

空間半無限體內(nèi)部作用集中力時的粘彈性沉降

0關(guān)于樁基的粘彈性問題1936年，縫林提供了半自由體內(nèi)部功能的垂直直立力和水平向集中力的彈性解。徐志英根據(jù)mile理論原理，推斷出半自由體內(nèi)部垂直傾斜流量法中的土中垂直高度公式。從那時起，一些科學(xué)家將模糊理論應(yīng)用于樁基分析。袁云系統(tǒng)研究了垂直流量法中的負(fù)荷在半有限體內(nèi)和水平方向上的作用，并對半有限體內(nèi)的動態(tài)方程進(jìn)行了驗證。在軸向?qū)ΨQ條件下，文獻(xiàn)給出了keeli模型的答案。在文獻(xiàn)中，我們研究了半自由山體內(nèi)部或橫向固定力作用下的空間粘度。然而，作者在半自由空間內(nèi)的縱向激勵下對空間粘膜的回答還沒有系統(tǒng)完成。雖然工程實踐證明,以Mindlin解為依據(jù)導(dǎo)出的樁基礎(chǔ)以及其他深基礎(chǔ)的沉降計算公式,往往與實測值吻合較好,但Mindlin解并沒有考慮半無限體的粘彈性特征。建筑物的持續(xù)沉降通常與地基土的流變性質(zhì)有關(guān),因此,研究半無限體內(nèi)部作用有豎向集中力的粘彈性解,對于較準(zhǔn)確的計算樁基位移與建筑物的沉降,無疑具有非常重要的理論價值和實際應(yīng)用價值。1半無限空間的粘合彈性解1.1基于laplace變換的kelren模型為了研究問題的方便,對半無限空間體作如下假設(shè)(見圖1):(1)假定半無限體是均勻各向同性的連續(xù)變形體,在深度和水平方向上無限延伸;(2)半無限體在內(nèi)部集中力作用下的應(yīng)力為三維應(yīng)力狀態(tài),應(yīng)力球張量和應(yīng)變球張量之間符合彈性關(guān)系,應(yīng)力偏張量和應(yīng)變偏張量之間符合Kevin粘彈性模型。對時間t求Laplace變換后的復(fù)雜應(yīng)力狀態(tài)下本構(gòu)方程為:其中,sij、eij分別為應(yīng)力、應(yīng)變偏張量的Laplace變換;σkk、εkk分別為應(yīng)力、應(yīng)變球張量的Laplace變換;P1、Q1分別為與偏張量有關(guān)的微分算子的Laplace變換;P2、Q2分別為與球張量有關(guān)的微分算子的Laplace變換;P1、Q1和P2、Q2的表達(dá)式為ˉΡ1=m∑k=0p′ksk,ˉQ1=n∑k=0q′ksk,ˉΡ2=m∑k=0p″ksk,ˉQ2=n∑k=0q″ksk,p′k、q′k、p″k、q″k可由材料的粘彈性性質(zhì)確定。如圖2所示,對Kelvin模型,則有式中,K為體積彈性模量;Gk、ηk分別為Kelvin模型系數(shù)。由式(2)可得1.2土體剪切模量計算的計算假定粘彈性空間半無限體內(nèi)部深度h處受突加豎向集中力P(t)=P0H(t)作用,對P(t)求Laplace變換為P(s)=P0/s。(4)由文獻(xiàn),半無限彈性體內(nèi)部作用有豎直向集中力時,半無限體內(nèi)部任一點M(x,y,z)的Mindlin解答為σx=Ρ8π{-b1(z-h)R31+b23x2(z-h)R51-b13(z-h)R32+b44(z+h)R32+b230hx2z(z+h)R72+b33x2(z-h)R52-b16h(z+h)zR52+b512h2(z+h)R52+b6[4R2(R2+z+h)(1-x2R2(R2+z+h)-x2R22)]},(5a)σy=Ρ8π{-b1(z-h)R31+b23y2(z-h)R51-b13(z-h)R32+b44(z+h)R32+b230hy2z(z+h)R72+b33y2(z-h)R52-b16hz(z+h)R52+b512h2(z+h)R52+b64R2(R2+z+h)(1-y2R2(R2+z+h)-y2R22)},(5b)σz=Ρ8π{b1(z-h)R31+b23(z-h)3R51-b1(z-h)R32+b230hz(z+h)3R72+b33z(z+h)2R52-b23h(z+h)(5z-h)R52}(5c)τyz=Ρ8π{b1yR31-b1yR32+b23y(z-h)2R51+b230hz(z+h)2yR72+b33z(z+h)yR52-b13h(3z+h)yR52},(5d)τxz=Ρ8π{b1xR31-b2xR32+b23(z-h)2xR51+b230hzx(z+h)2R72+b33z(z+h)xR52-b23hx(3z+h)R52},(5e)τxy=Ρ8π{b23xy(z-h)R51+b33xy(z-h)R52-b64R22(R2+z+h)(1R2(R2+z+h)-1R2)+b230hz(z+h)R72}(5f)ux=Ρ16π[b9(z-h)xR31+b7x(z-h)R32-b104xR2(R2+z+h)+b96xhz(z+h)R32],(6a)uy=Ρ16π[b9y(z-h)R31+b7y(z-h)R32-b104yR2(R2+z+h)+b96yhz(z+h)R32],(6b)uz=Ρ16π[b71R1+b88R2-b71R2+b9(z-h)2R31+b7(z+h)2R32-b92hzR32+b96hz(z+h)2R52],(6c)其中,r=√x2+y2;R1=√r2+(z-h)2;R2=√r2+(z+h)2;r為集中力的作用點到計算點的水平距離;μ為土體的泊松比;G為土體的剪切模量,G=E/2(1+μ);E為土的彈性模量;且有對式(7)進(jìn)行關(guān)于時間t的Laplace變換,并將(3)式代入式(8),可得對式(5a)～(6c)求關(guān)于時間t的Laplace變換,并將式(4)、(8)代入,再求Laplace逆變換,可得空間半無限粘彈性體的應(yīng)力和位移解答為σx=Ρ08π{[6Gk3Κ+4Gk(1-a1(t))+32a1(t)]×[-(z-h)R31-3(z-h)R32-6h(z+h)zR52]+[6Κ+2Gk3Κ+4Gk(1-a1(t))+12a1(t)]?[3x2(z-h)R51+30hx2z(z+h)R72]+[6Κ+14Gk3Κ+4Gk(1-a1(t))+72a1(t)]3x2(z-h)R52+[6Gk(3Κ-2Gk)6Κ+2Gk-9Κ3Κ+Gka2(t)+9Κ2(3Κ+4Gk)a1(t)]×4(z+h)R32+[3Κ-2Gk3Κ+4Gk(1-a1(t))+12a1(t)]?12h2(z+h)R52+[6Gk6Κ+2Gk(1-a2(t))+3a2(t)]?[4R2(R2+z+h)(1-x2R2(R2+z+h)-x2R22)]},(9a)σy=Ρ08π{[6Gk3Κ+4Gk(1-a1(t))+32a1(t)]?[-(z-h)R31-3(z-h)R32-6h(z+h)zR52]+[6Κ+2Gk3Κ+4Gk(1-a1(t))+12a1(t)]?[3y2(z-h)R51+30hy2z(z+h)R72]+[6Κ+14Gk3Κ+4Gk(1-a1(t))+72a1(t)]3y2(z-h)R52+[6Gk(3Κ-2Gk)6Κ+2Gk-9Κ3Κ+Gka2(t)+9Κ2(3Κ+4Gk)a1(t)]?4(z+h)R32+[3Κ-2Gk3Κ+4Gk(1-a1(t))-12a1(t)]?12h2(z+h)R52+[6Gk6Κ+2Gk(1-a2(t))+3a2(t)]?[4R2(R2+z+h)(1-y2R2(R2+z+h)-y2R22)]},(9b)σz=Ρ08π{[6Gk3Κ+4Gk(1-a1(t))+32a1(t)]?(z-hR31-z-hR32)+[6Κ+2Gk3Κ+4Gk(1-a1(t))+32a1(t)]?[3(z-h)3R51-3h(z+h)(5z-h)R52+30hz(z+h)3R72]+[(6Κ+14Gk)(3Κ+4Gk)(1-a1(t))+72a1(t)]3z(z+h)2R52},(9c)τyz=Ρ08π{[6Gk3Κ+4Gk(1-a1(t))+32a1(t)]?[yR31-yR32-3hy(3z+h)R52]+[6Κ+2Gk3Κ+4Gk(1-a1(t))+12a1(t)][3y(z-h)2R51+30hzy(z+h)2R72]+[6Κ+14Gk3Κ+4Gk(1-a1(t))+72a1(t)]3zy(z+h)R52},(9d)τxz=Ρ08π{[6Gk3Κ+4Gk(1-a1(t))+32a1(t)]?[xR13-xR23]+[6Κ+2Gk3Κ+4Gk(1-a1(t))+12a1(t)]?[3x(z-h)2R15+30xhz(z+h)2R27-3hx(3z+h)R25]+[6Κ+14Gk3Κ+4Gk(1-a1(t))+72a1(t)]3zx(z+h)R25},(9e)τxy=Ρ08π{[6Κ+2Gk3Κ+4Gk(1-a1(t))+12a1(t)]×[3xy(z-h)R15+30hz(z+h)R27]+[6Κ+14Gk3Κ+4Gk(1-a1(t))+72a1(t)]3xy(z-h)R25-[6Gk6Κ+2Gk(1-a2(t))+32a2(t)]?4R22(R2+z+h)(1R2(R2+z+h)-1R2)},(9f)ux=Ρ016π{[6Κ+14GkGk(3Κ+4Gk)-2Gka3(t)-63Κ+4Gk×a1(t)]x(z-h)R23+[6Κ+2GkGk(3Κ+4Gk)-2Gka3(t)+63Κ+4Gka1(t)][(z-h)xR13+6xhz(z+h)R23]-[33Κ+Gk-33Κ+Gka2(t)]4xR2(R2+z+h)],(10a)uy=Ρ016π{[6Κ+14GkGk(3Κ+4Gk)-2Gka3(t)-63Κ+4Gk×a1(t)]?y(z-h)R23+[6Κ+2GkGk(3Κ+4Gk)-2Gka3(t)+63Κ+4Gka1(t)][(z-h)yR13+6yhz(z+h)R23]-33Κ+Gk[1-a2(t)]4yR2(R2+z+h)],(10b)uz=Ρ016π{(1R1-1R2+(z+h)2R23)[6Κ+14GkGk(3Κ+4Gk)-2Gka3(t)-63Κ+4Gka1(t)]+((z-h)2R13-2hzR23+6hz(z+h)2R25)[6Κ+2GkGk(3Κ+4Gk)-2Gka3(t)+63Κ+4Gka1(t)]+8R2[3Κ+4GkGk(6Κ+2Gk)-12Gka3(t)-36Κ+2Gka2(t)]},(10c)式(9a)～(10c)中,a1(t)=e3Κ+4Gk4ηkt,a2(t)=e-6Κ+2Gk2ηkt,a3(t)=e-Gkηkt。2垂直均勻布局的負(fù)荷在均質(zhì)材料的內(nèi)部的粘度炸彈沉降中起著作用2.1目標(biāo)函數(shù)如圖3所示,設(shè)半空間表面下深度h處有一均布荷載p作用在矩形面積上,矩形面積的長度和寬度分別為b和c,為計算矩形面積角點N處的粘彈性沉降,可在矩形面積范圍內(nèi)取單元面積dS=dξdη,作用在單元面積上的分布荷載可以以集中力dP=pdξdη來代替。由z粘彈性位移的表達(dá)式(10c),對其在矩形區(qū)域范圍內(nèi)積分,有w=∫0a∫0buz(x,y,z,t)dxdy=pb16π{[6Κ+14GkGk(3Κ+4Gk)-2Gka3(t)-63Κ+4Gka1(t)]×(A1-B1+D1)+8B1[3Κ+4GkGk(6Κ+2Gk)-12Gka3(t)-36Κ+2Gka2(t)]+[6Κ+2GkGk(3Κ+4Gk)-2Gka3(t)+63Κ+4Gka1(t)](C1+hzb2E1)},(11)式(11)中,A1=lnbc+(bc)2+(z-hc)2+1(z-hc)2+1+bcln1+(bc)2+(z-hc)2+1(bc)2+(z-hc)2+2z-hctan-1z-hcbc+(bc)2+(z-hc)2?[(bc)2+(z-hc)2+1-(bc)2+(z-hc)2]-(z-hc)tan-11z-hc,B1=lnbc+(bc)2+(z+hc)2+1(z+hc)2+1+bcln1+(bc)2+(z+hc)2+1(bc)2+(z+hc)2+2(z+hc)tan-1z+hcbc+(z+hc)2+(bc)2?[(bc)2+(z+hc)2+1-(bc)2+(z+hc)2]-(z+hc)tan-11z+hc,C1=(z+hc)tan-1bc(z-hc)(bc)2+(z-hc)2+1,D1=(z-hc)tan-1bc(z+hc)(bc)2+(z+hc)2+1,E1=2bc1(bc)2+(z+hc)2+1?[1(bc)2+(z+hc)2+1(z+hc)2+1]。2.2a5t-32+3gka3t型若要計算圖4所示半空間內(nèi)部作用有豎向均布荷載所引起的矩形面積中點下某一深度O處的沉降,同樣可由式(10c)在矩形區(qū)域范圍內(nèi)積分求得,有w=4∫0a/2∫0b/2uz(x,y,z,t)dxdy=pb8π{[6Κ+14GkGk(3Κ+4Gk)-2Gka3(t)-63Κ+4Gka1(t)]×(A2-B2+D2)+8B2[3Κ+4GkGk(6Κ+2Gk)-12Gka1(t)-36Κ+2Gka2(t)]+[6Κ+2GkGk(3Κ+4Gk)-2Gka3(t)+63Κ+4Gka1(t)](C2+hzb2E2)},(12)其中,A2=lnbc+(bc)2+4(z-hc)2+14(z-hc)2+1-bcln1+(bc)2+4(z-hc)2+1(bc)2+4(z-hc)2+4z-hctan-12(z-hc)bc+(bc)2+4(z-hc)2?[(bc)2+4(z-hc)2+1-(bc)2+4(z-hc)2]-2(z-hc)tan-112(z-h)c,B2=lnbc+(bc)2+4(z+hc)2+14(z+hc)2+1+bcln1+(bc)2+4(z+hc)2+1(bc)2+4(z+hc)2+4(z+hc)tan-12(z+hc)bc+(bc)2+4(z+hc)2[(bc)2+4(z+hc)2+1-(bc)2+4(z+hc)2]-2(z+hc)tan-1c2(z+h),C2=2(z-hc)tan-1bc2(z-hc)(bc)2+4(z-hc)2+1,D2=2(z+hc)tan-1bc2(z+hc)(bc)2+4(z+hc)2+1,E2=8bc1(bc)2+4(z+hc)2+1?[1(bc)2+4(z+hc)2+14(z+hc)2+1]。2.3利用疊加原理求解形面上的荷載強(qiáng)度m點的粘彈性沉降公式簡稱“3+4gk”如圖5所示,若在半空間表面下深度處作用有沿軸方向線性增長的三角性分布荷載,為計算荷載強(qiáng)度為零邊角點下某一深度M處的粘彈性沉降,可在矩形范圍內(nèi)取面積元素dS=dξdη,作用在面積元素dS上的分布荷載可用集中力dP表示,dΡ=pξcdξdη,對表達(dá)式(10c)在矩形區(qū)域內(nèi)積分,有w=pb∫0a∫0buz(x,y,z,t)ydxdy=p32π{[6Κ+14GkGk(3Κ+4Gk)-2Gka3(t)-63Κ+4Gka1(t)]×[A3-B3+2(z+hb)2D3]+8B3[3Κ+4GkGk(6Κ+2Gk)-12Gka3(t)-36Κ+2Gka2(t)]+[6Κ+2GkGk(3Κ+4Gk)-2Gka3(t)+63Κ+4Gka1(t)](C3-4hzb2D3+hzb2E3)},(13)其中,A3=(bc)(bc)2+(z-hc)2+1+[1+(z-hc)2]×lnbc+(bc)2+(z-hc)2+1(z-hc)2+1-(bc)(bc)2+(z-hc)2-(z-hc)2lnbc+(bc)2+(z-hc)2z-hc?B3=bc(bc)2+(z+hc)2+1+[1+(z+hc)2]×lnbc+(bc)2+(z+hc)2+1(z+hc)2+1-(z+hc)2×lnbc+(bc)2+(z+hc)2z+hc-bc(bc)2+(z+hc)2?C3=2(z-hc)2[lnbc+(bc)2+(z-hc)2z-hc-lnbc+(bc)2+(z-hc)2+1(z-hc)2+11],D3=lnbc+(bc)2+(z+hc)2z+hc-lnbc+(bc)2+(z+hc)2+1(z+hc)2+1,E3=4(z+hc)2[bc(z+hc)2(bc)2+(z+hc)2-bc[1+(z+hc)2](bc)2+(z+hc)2+1]。公式(12)、(13)和(14)分別為矩形分布荷載作用在半空間內(nèi)部角點、中點以及矩形面積三角形分布荷載作用在半空間內(nèi)部時所引起的荷載強(qiáng)度為零邊角點下某一深度M點的粘彈性沉降計算公式,如果矩形面積上作用有梯形分布荷載或所求沉降點為地基中的任一點時,可以利用上述3個公式運用