ANALYSIS OFSTRESSAND STRAIN Considering conjugate of shearing stresses and trigonometric identities we get 2、0_6c02c-zsin20 2 Similarly: O Fig 1 sin 2a+t cos 2a 2 y s yx O x Fig 2
Fig.1 x y sx txy sy O sy t yx sx s t x y O t n Fig.2 t s s s s s cos2 sin 2 2 2 xy x y x y − − + + = t s s t sin 2 cos2 2 xy x y + − = Considering conjugate of shearing stresses and trigonometric identities we get： Similarly：
考虑剪应力互等和三角变换,得: +.O.-0 coS 2a-T sin 2a 2 2 同理: O 图1 sin 2a+t cos 2a 2 y s yx O x 图2
x 图1 y sx txy sy O sy t yx sx s t x y O t n 图2 t s s s s s cos2 sin 2 2 2 xy x y x y − − + + = t s s t sin 2 cos2 2 xy x y + − = 考虑剪应力互等和三角变换，得： 同理：
ANALYSIS OFSTRESSAND STRAIN 2 The extreme values for the stress Let: dd \a=ao ao -or-o sin 2ao-2T, cos 20o=0 Thus we can get two stationary points (aoi+=) e 2 01 2 and two extremums =+±()2+2 2 2 u ry T=0 Extreme normal stresses are principal stresses O
: ( )sin 2 0 2 cos2 0 0 0 = − − − = = s s t s x y xy d d Let 2、The extreme values for the stress x y xy s s t − = − 2 、（ ）： tg2 0 2 01 01 + 0 0 t = ） 2 2 2 2 xy x y x y min max t s s s s s s + − ± + = （ x y sx txy sy O and two extremums Thus we can get two stationary points Extreme normal stresses are principal stresses
二、极值应力 -(o σ.kin2aa-2rcoS2cn=0 dd la=ao 由此得两个驻点: 2 a01、(aa1+)和两个极值:g2ao= 2 x max+可 ±{( )2+2 2 2 n ry x=0.极值正应力就是主应力! O
: ( )sin2 0 2 cos2 0 0 0 =− − − = = s s t s x y xy d d 令 二、极值应力 x y xy s s t − = − 2 、（ ）和两个极值： tg2 0 由此得两个驻点： 2 0 1 0 1 + t 0 =0极值正应力就是主应力！ x y sx txy sy O ） 2 2 2 2 xy x y x y min max t s s s s s s + − ± + = （
ANALYSIS OFSTRESSAND STRAIN O1=O, mx,O maX o is in the quadrant for the shearing stress to Main element point and lean to the larger of both o and o, y Exy oi dt Let. da 0tg2012r =( O X a=a x max G-6 =土1( )2+ min 2 I That is the angle between the planes in do- 4 which shearing stresses reach extremums and the principal planes is 450
s1 is in the quadrant for the shearing stress to point and lean to the larger of both sx and sy 0 d d et : 1 = = t L xy x y t s s 2 tg2 1 − = 2 2 2 x y x y min max t s s t t + − =± （ ） , 4 0 1 = + 1 max 2 max s =s ; s =s x y sx txy sy O Main elemeut s 2 s 1 That is the angle between the planes in which shearing stresses reach extremums and the principal planes is . 0 45