O1=O, mx,O maX O 2 G在剪应力相对的象限内, 主 单元体 且偏向于∝.及σ较大的一侧。 y Exy oi dt 0 tga, O X da 2T max G-6 )2+ lIn m 2 a0=a1+,即极值剪应力面与主平面成45 4
s1在剪应力相对的象限内， 且偏向于sx 及sy较大的一侧。 0 d d : 1 = = t 令 xy x y t s s 2 tg2 1 − = 0 0 1 , 45 4 即极值剪应力面与主平 面成 = + 2 2 2 x y x y min max t s s t t + − =± （ ） x y sx txy sy O 主 单元体 s 2 s 1 1 max 2 max s =s ; s =s
ANALYSIS OFSTRESSAND STRAIN Example 2 Analyze the failu ure of the circular shaft in torsion Solution: Determine the critical point and plot the original element O=0=0 M T=T Xy 2 Determine the extreme -value stress R+o y ±( 2 2 2 O x =士√z2=士z
Example 2 Analyze the failure of the circular shaft in torsion. Solution：Determine the critical point and plot the original element. Determine the extreme-value stress s x =s y =0 P n xy W M t =t= 2 2 2 1 2 2 xy x y x y t s s s s s s + − + = （ ） = t =t 2 xy C t xy t yx M C x y O txy t yx
「例2分析受扭构件的破坏规律。 解:◎确定危险点并画其原 始单元体 O=0=0 M T=T Xy 2求极值应力 R+o y ±( 2 2 2 O x =士√z2=士z
[例2] 分析受扭构件的破坏规律。 解：确定危险点并画其原 始单元体 求极值应力 s x =s y =0 P n xy W M t =t= = t =t 2 xy C t xy t yx M C x y O txy t yx 2 2 2 1 2 2 xy x y x y t s s s s s s + − + = （ ）
ANALYSIS OFSTRESSAND STRAIN O1=O2=03= 27 tg2ao o:C=45° o-0 工m=±1()2+zx=土rtg2a1=s 2 y=0.a1=0 min 3Analysis of failure LoW-carbon steel Low-carbon steel O.=240 MPa:T=200 MPa Cast iron Cast iron OLb=98280 MPa 0640~960MPa;h=198~300MPa
Analysis of failure: t t s s t t + = − = 2 2 min max 2 xy x y （ ） s =t s = s =−t 1 2 3 ; 0; 45 2 tg2 0 = 0 = − =− s s t x y xy 0 1 0 1 0 2 tg2 = = − = t s s xy x y s s = 240MPa;t s = 200MPa 640 ~ 960MPa; 198 ~ 300MPa 98 ~ 280MPa = = = yb b Lb s t s Low-carbon steel Cast iron Low-carbon steel: Cast iron:
O1=O2=03= 27 tg2ao o:C=45° o-0 工m=土1(xy)2+zx=trtg2a,、、-0y=0:aO0 2 min 2T 6破坏分析 低碳钢σ240MPa;zs=20MPa 低碳钢 灰口铸铁ab=98~280MPa Ob=640960MPa;zb=198300MPa 铸铁
破坏分析 s =t s = s = −t 1 2 3 ; 0 ; 45 2 tg 2 0 = 0 = − = − s s t x y xy 0 1 0 1 0 2 tg 2 = = − = t s s xy x y 低碳钢:s s =240MPa;t s =200MPa 640~960MPa; 198~300MPa : 98~280MPa = = = yb b Lb s t 灰口铸铁 s 低碳钢 铸铁 t t s s tt + = − = 2 2 min max 2 xy x y （ ）