【正文】 be expressed as: 0?????? zy xzxy ??? If a stress function ?(y, z) exists such that: yz xzxy ????????? ??? Then the equilibrium equation is automatically satisfied and this stress function is referred to as Prantl’ s Stress Function. 57 ? When the deflections in terms of the angle of twist ? are considered. From the figure, it can be seen that for small ? : w = ?y v = ?z General Formulation (III) ? Since the rod is fixed at the origin of xyz, the angle of twist ? can be represented by the angle of twist per unit length ?? , ., ? = ?? x . ? w = ?? xy v = ?? xz 58 General Formulation (IV) zuyzuxwyuzyux xzxy ???????????????????????? ?????where, u = u(y,z) is the deflection in the xdirection and it is independent of x. ?????????????????????????????????????????????????vEyxeivEzyzuyvEGyuzvEGxyxzxzxzxyxy1..1)()1(2)()1(222222? It is called “ Prantl’ s Equation” , which is the governing equation for the torsional stress function ?(y, z). 59 General Formulation (V) .),(0),(00z e r obetoc h o s e nn o r ma l l yisCw h e r eCzyzyddzzdyydzdydzdyxyxzxzxy??????????????????? ????? To relate the torsional stress function ?(y, z) to the transmitted torque T, observe: ?? ???????d y d zzyTd y d zzyzd y d zyd y d zdTxyxzxyxzxyxz)()()()(??????60 dzdyyyydzyddzdyyyd y d zyyd y d zzzyyTyzxzxy? ??? ?????????????????????????????????????????)])[()()(:N o t e)(S i n c e12??? Since at the boundary, ?(y, z) = C is always chosen to be zero ? ????????? ???????????????????????????d y d zzyTd y d zd y d zzzd y d zdzdyd y d zyyyyy),(2S i m i l i a r l y)]0[000)(12 General Formulation (VI) 61 Example A solid circular shaft of radius r0 is transmitting a torque T. Determine the corresponding shearstress distribution. ? Solution: The boundary circle can be express as: y2 +z2 = r02 and note @ r = r0 = (y2 +z2)1/2 ? = 0 Therefore, try the following function: ? (y, z) = k (r2 – r02) = k (y2 + z2 – r02) where k is a constant to be determined. 62 4002020202ob t a i ntoS ol ve)(42)(2),(200rTkr drrrkr drrrkdy dzzyTrr???????????? ????The polar moment of inertia of a circular cross section is: )(2),(22 402240 rzyJTzyJTkrJ ?????????? ?Thus, the stresses can be determined as: JTyyJTzz xzxy ???????????? ?? Solution to Ex. 63 JTEvvEJTJTvEzy)1(2112222 ???????????????? ???? ?? ??? Solution (Continued) And at any given point, the shear stress is: JTrxzxyx ???22ne t )()()( ???For the angle of twist: GJTLJTLEv ??? )1(2?Assume the total length of the rod is L, then ? = ??L ? 64 Example Consider a equilateral cross section with sides of length a. Determine the shear distribution of the section which is transmitting a torque T. 65 Solution to Ex. ? Solution: Based on the boundary equations, assume the Prandtl’s Stress Function to be: )3)](3(32)][3(32[),( hyazahyazahyCzy ???????3:l i n eB o t t o m)3(32:l i n eL e f t)3(32:l i n eR i g h thyazahyazahy???????where C is a constant to be determined. The boundaries can be express as: 66 )(3)1(2)1(423a n d1S i n c e2222avaEvhECahvEzy?????????????????????????dyhydzzahhyzahhyCd y dzTA l s ohhzz????????????????????????????????????????????? ???? 3232322:23321)23(6isl i n er i g h tt h e)23(6isl i n el e f tt h e:21hyhazhyhazN o t e????? Solution to Ex. (II) 67 )(38023S i n c e15)3()3(34)3()23()6()2(31)23(6)3(4)3()2()3(254233323333222332221baTCahCahdyhyhyhaCdyhyhyhaahhyhahyCdyhydzzahhyCThhhhhhzz????????????????????????????????????????? ????)324()388(:b e c o m es t r e s sT h e)1(3160)(a n d)(C o n n e c t222223224yhyzahCyzahyzahCzEaTvbaxzxy?????????????????????? Solution to Ex. (III) 68 )333(380)36(3802/3ngS u b s t i t u t i2255 yayzaTazyzaTahxzxy ?????????? Solution to Ex. (IV) )33(38000w he rea xi st heA l ong 25 yaya Tzy xzxy ?????? ??03?? zxzTaNSe t ?This figure represents N as the function of y. One can see it is a parabolic curve. )](3 3)[(80 2 ayayN ????69 Topics of Today ? Chapter 4 Concepts from The Theory of Elasticity ? Plane Elastic Problems ? The Airy Stress Function ? Prandtl’ s Stress Function for Torsion ? General Formulation ? Tension on a Rectangular Cross Section 70 Torsion on a Rectangular Cross Section (I) ? The solution for the equilateral cross section was rather fortuitous. Normally, when the boundaries of the cross sections are not well defined by continuous functions, the determination of the stress functions is a more plicated matter. ? When this occurs, the stress functions employed are generally in the form of an infinite series. For the case of rectangular cross section, the technique employed for the equilateral cross section will not yield an accep