【正文】 ?+?/2 代 . ? 3. Adding the results together, we obtain Eqs.()上述結(jié)果相加得本問題解答 () 彈性力學(xué) 第四章 74 C. A plate of any shape in plane stress or strain condition with a small circle hole located at some distance away from the boundary is subjected to any external forces. ? Assume that there is no hole, find the stress ponents and then the magnitudes and directions of the principal stresses, ?1 and ?2, at the point corresponding to the center of the hole. ? Place the origin of coordinates at the center of the hole, with x and y axes along ?1 and ?2 respectively, and apply Eqs.() with q1= ?1 and q2=?2 . 彈性力學(xué) 第四章 75 C. 任意平面問題的板中有一離邊界較遠(yuǎn)的半徑為 a的小圓孔，受到任意荷載 作用 。 ? 假定無(wú)洞，求出洞中心處的應(yīng)力，主應(yīng)力 ?1 和 ?2 及方向。 ? 置坐標(biāo)原點(diǎn)于洞中心， x 軸在 ?1 方向， y 軸在 ?2 方向，將 q1= ?1 和 q2=?2 代入式 ()得欲求的解答。 彈性力學(xué) 第四章 76 wedge loaded at the vertex or on the edges 契形體在契頂或契邊受力 ? P75(E) 彈性力學(xué) 第四章 77 A. a wedge is subjected to a concentrated load at its vertex. 契形體在契頂受集中力 ? 因次分析 dimensional analysis stress[force][length]2 P [force][length]1 r [length] ?,?,?dimensionless ? stressesP/r ? ? = r f(?) 彈性力學(xué) 第四章 78 ? ? ? = r f(?) (1) ? substitution of (1) into ?4? =[?/(r?r)+?2/(r2??2)+?2/?r2 ]2 ? =0 yields 1/r3[f(4)(?)+2f?? +f]=0 ? [f(4)(?)+2f?? +f]=0 ? f= [A cos? +B sin?+?(C cos?+Dsin?)] ? ?=r[A cos? +B sin?+?(C cos?+Dsin?)] ? ?=r?(C cos?+Dsin?) 彈性力學(xué) 第四章 79 ? stresses ? ?=r?(C cos?+Dsin?) ? ?r=??/(r?r)+?2?/(r2??2) =2/r (Dcos?Csin?) () ??= ?2?/?r2=0 () ?r?= (?/?r)[??/(r??)]=0 () 彈性力學(xué) 第四章 80 boundary condition ? ?r=2/r (Dcos?Csin?) ??= 0 ?r?=0 ? ??)?=??/2 = 0 ?r? ) ?=??/2 =0 are satisfied 彈性力學(xué) 第四章 81 boundary condition ?r=2/r (Dcos?Csin?) ? free body 0AB ?Fx=0 ? ?/2?/2 ?rcos? rd?+Pcos?=0 ?Fy=0 ? ?/2?/2 ?rsin? rd?+Psin?=0 彈性力學(xué) 第四章 82 boundary condition ?r=2/r (Dcos?Csin?) ? free body 0AB ?Fx=0 ? ?/2?/2 ?rcos? rd?+Pcos?=0 ?Fy=0 ? ?/2?/2 ?rsin? rd?+Psin?=0 ? solving for D C and substituting D C into Eq. (), we obtain: ?r=2P/r[(cos?cos?)/(?+sin?)+(sin?sin?)/(?sin?)] ??=0 ?r?=0 () 彈性力學(xué) 第四章 83 special case 1特例 1: ?=? ?=?/2 ? Fig. P y x ? ?r=2Psin?/(? r) ??=0 ?r?=0 彈性力學(xué) 第四章 84 Special case 2 特例 2: ?=? ?=0 concentrated normal load on a straight boundary(Flamant?s problem)直線邊界上作用法向集中荷載 (符拉芒問題） ? ?r=2P cos? /(? r) ??=0 ?r?=0 () 彈性力學(xué) 第四章 85 displacements 平面應(yīng)力問題的位移 ? ?r=2Psin?/(? r) ??=0 ?r?=0 ? ur=2Pcos?lnr/(?E)(1?)P?sin? /(?E) +Icos? u?=2Psin?lnr/(?E)+(1+?)Psin?/(?E)(1?) P?cos?/(?E)Isin? in which I is the rigidbody translation in the x direction. 其中 I為 x向剛體平動(dòng) 。 彈性力學(xué) 第四章 86 Settlementvertical displacement, positive downward. 沉陷 鉛直向位移，向下為正 ? u?=2Psin?lnr/(?E)+(1+?)Psin?/(?E)(1?) P?cos?/(?E)Isin? ? M(r, ?=?/2)arbitrary point on the surface M點(diǎn)沉陷 =u?)M=2Plnr/(?E)(1+?)P/(?E)+I ? M(r, ?=?/2)arbitrary point on the surface M點(diǎn)沉陷 = u?)M=2Plnr/(?E)(1+?)P/(?E)+I ? B(s, ?=??/2)base point B點(diǎn)沉陷 =2Plns/(?E)(1+?)P/(?E)+I ? M點(diǎn)相對(duì)沉陷 =M點(diǎn)沉陷 B點(diǎn)沉陷 = u?)M u?)B = 2P/(?E) ln(s/r) 彈性力學(xué) 第四章 87 P y x 中文書 410 B. Uniform normal loads on a straight boundary 直線邊界上作用法向分布荷載 彈性力學(xué) 第四章 88 ? 習(xí)題作業(yè)（英文書） 彈性力學(xué) 第四章 89 ? 4121 Derive the following equations for coordinate transformation of displacement ponents: ? 推導(dǎo)位移分量的坐標(biāo)變換式： ur=u cos?+v sin? u?=u sin?+v cos? u= ur cos? u? sin? v= ur sin?+ u? cos? 彈性力學(xué) 第四章 90 ? 4122 A hollow cylinder with inner radius a and outer radius b is subjected to an internal pressure of intensity q. Find the change of the inner radius, the outer radius and the thickness. ? 41 設(shè)有內(nèi)半徑為 a外半徑為 b的圓筒受內(nèi)壓 q，求內(nèi)半徑 外半徑和圓筒厚度的改變。 ur )r=a ur )r=b ur )r=b ur )r=a 彈性力學(xué) 第四章 91 42 設(shè)有一剛體，具有半徑為 b的圓柱形孔道，孔道內(nèi)放置內(nèi)半徑為 a外半徑為 b的圓筒，受圓筒內(nèi)壓 q，求圓筒的應(yīng)力。 平面應(yīng)變問題，軸對(duì)稱應(yīng)力 ?r=A/r2+2C ??=A/r2+2C ?r?=0 ?r )r=a =q ur )r=b=0 彈性力學(xué) 第四章 92 4125 A wedge of infinite length is subjected to uniform shearing forces of intensity q on its edges. Find stress ponents by using Eq. () 彈性力學(xué) 第四章 93 ? ?=r2(A cos2?+B sin2?+C?+D) ? ?r=??/(r?r)+?2?/(r2??2) =2A cos2?2B sin2? +2C?+2D ??= ?2?/?r2 = 2A cos2?+2B sin2? +2C?+2D ?r?= (?/?r)[??/(r??)] = 2A sin2?2B cos2? C ??)?=??/2 = 0 ?? r) ?=??/2 = ?q ABCD