The note: when the body Substitute stress components into force is zero. the the compatibility equations which compatibility equations in ignoring the body force, we can see: terms of stress components the first three formulas and the last for spatial problems are are satisfied, the other two formulas(1+p)v2o2+ 02e =0 demand (1+)Vσy+ 0 V o=0 (1+)V2+2=0 V2=0 +u)VT =0 N amely ayaz VO=C (b) (+uVIr*O0 azar O 2⊙ (1+p)Vx+ 0 andy 11
11 (1 ) 0 (1 ) 0 (1 ) 0 2 2 2 2 2 2 = + + = + + = + + x y z x y z xy zx yz (1 ) 0 (1 ) 0 (1 ) 0 2 2 2 2 2 2 2 2 2 = + + = + + = + + z y x z y x The note: when the body force is zero, the compatibility equations in terms of stress components for spatial problems are Substitute stress components into the compatibility equations which ignoring the body force, we can see: the first three formulas and the last are satisfied, the other two formulas demand 0 0 2 2 = = y x Namely = C 2 (b)
注:体力为零时,空间冋: 题 将应力分量代入不计体力的应力分量表示的相容方程 相容方程,可见:前三式及最后 式得到满足,其余二式要求 (1 +)V2o + 0 0-9 V=0 (1+p)V =0 0 Vo=0 Oy a-e (1+)V2x+y=0 ayaz 即 Q VO=C (b) (+uVT aox 0 (1+n)V 0 Oxd 12
12 (1 ) 0 (1 ) 0 (1 ) 0 2 2 2 2 2 2 = + + = + + = + + x y z x y z xy zx yz (1 ) 0 (1 ) 0 (1 ) 0 2 2 2 2 2 2 2 2 2 = + + = + + = + + z y x z y x 注：体力为零时，空间问题 将应力分量代入不计体力的 应力分量表示的相容方程 相容方程，可见：前三式及最后 一式得到满足，其余二式要求 0 0 2 2 = = y x 即 = C 2 (b)
2. Boundary conditions The note the stress boundary On the profiles of the pole, substitute conditions for spatial problems n=0 and surface force components which are are zero into the boundary conditions. we get that the first two formulas can al ways G)+m(x)+)=X be satisfied. the third formula demands )+h)+1)= Namely(x: 2) s +m(tvs =o n()+(x-)+mr=)=z D a Being at the boundary d ds m=ax d s 13
13 2. Boundary conditions On the profiles of the pole, substitute n=0 and surface force components which are zero into the boundary conditions, we get that the first two formulas can always be satisfied, the third formula demands: The note:the stress boundary conditions for spatial problems are: ( ) ( ) ( ) ( ) ( ) ( ) n( ) l( ) m( ) Z m n l Y l m n X s z s xz s yz s xy s zy s y zx s s x s yx + + = + + = + + = l( x z ) s + m( y z ) s = 0 = 0 − s x s m y l Namely Being at the boundary: s x m s y l d d , d d = = −
扭装 二边界条件 注:空间问题应力边界条件 0)+m)+ Ⅹ 在杆的侧面上,将n=0.及面 m(n)2+n2)+1(n)= 力分量为零代入边界条件,可见前 两式总能满足,而第三式要求 n()+c=)+m()=z l(Ixa+m(t=o 即 =0 a 由于在边界上 d dx 二 d s d s 14
14 二 边界条件 在杆的侧面上，将 n=0，及面 力分量为零代入边界条件，可见前 两式总能满足，而第三式要求 注：空间问题应力边界条件 ( ) ( ) ( ) ( ) ( ) ( ) n( ) l( ) m( ) Z m n l Y l m n X s z s xz s yz s xy s zy s y zx s s x s yx + + = + + = + + = l( x z ) s + m( y z ) s = 0 = 0 − s x s m y l 即 由于在边界上 s x m s y l d d , d d = = −
Then we have ap dy. dodd ==0 av ds a dsd S S This illuminates that at the boundary of the cross-section, the stress function is a constant. Because the stress components dont change when the stress function subtracts a constant, we can suppose when it is a single successional section(solid pole) q。=0() At the arbitrary end of the pole, the shear stress composes torsion 15
15 0 d d d d d d = = + s s x s x y y s s Then we have This illuminates that at the boundary of the cross-section, the stress functionφ is a constant. Because the stress components don’t change when the stress function subtracts a constant, we can suppose when it is a single successional section(solid pole): s = 0 (c) At the arbitrary end of the pole, the shear stress composes torsion