中国地质大学：《构造地质学》课程PPT教学课件（英文版）Lecture 20 REVIEW OF STRESS

REVIEW OF STRESS

REVIEW OF STRESS

x y yx xy yZ Infinitesimal cube ↓σ Stress components in three dimensions

Stress components in three dimensions Infinitesimal cube xy =yx yz =zy zx =xz

(A) (A) General triaxial stress ellipsoid in perspective view.(B)Views normal to each of the principal planes of the ellipsoid. (After W.D. Means, 1976)

(A) General triaxial stress ellipsoid in perspective view.(B)Views normal to each of the principal planes of the ellipsoid. (After W.D. Means,1976) 1 1 1 2  2 3 3 3 2 (A) (B) 1> 2 > 3

STRESSES (2)

STRESSES (2)

Classes of Stress states

Classes of Stress States

Triaxial compressive stresses O10)23)0 o2 Biaxial compressive stresses G1G1O2O3=0

1 2 0 1 2 3   = 3 1 2 0 1 2 3    Triaxial compressive stresses Biaxial compressive stresses

Uniaxial stresses O1O2= 0 O2 Hydrostatic stresses O1=02=6

1  =1  =2 0 3   2 3 1  1  =2 0 3  = 1 Hydrostatic stresses Uniaxial stresses

O3 Plane stresses G112=0σ3 Pure shear stresses 1O1=O3:O2=0

3 1 1 2 3  = 0 3 1 1 3 2 Plane stresses Pure shear stresses

Stresses acting on a given plane 2 Equilibrium equations ∑Fa=0 ∑F;=0 1 Ae coSt coST o. w inosine=0 0 0 0-01 Ae coSe Sine+ o,A w sinecose=o 0 Ae cosO e Ao sine 2 A triangle element

Stresses acting on a given plane 1 2  t   1 2 A A cos A sin     A triangle element Equilibrium equations: F = 0 Ft = 0 -A + 1 A cos cos+ 2 A* *sinsin=0  A -1 A cos sin+ 2A * *sincos=0

σ=1cos2(0)+o2sin2(0) t=0 cossing o sinecose (2) Using the triangle formulae: c0s26=(1/2)(1+cos20) sin26=(1/2(1-cos20) We can rewrite equation(1)as follows: σ=12(σ1+∞2)+12(01-o2)c0s26 (3) Introduce the triangle formula sin20 =2cosesine into equation(2), we can obtain: T=1/2(01-2)sin26 (4) t reach a maximum value when 0 is 45 degrees and that is T=1/2(o1-02) (5)

 = 1 cos 2 ()+ 2 sin 2 () (1)  = 1 cossin － 2 sincos (2) Mohr’s Circle Stress Equations Using the triangle formulae: cos2  = (1/2)(1+cos2 ) sin 2 = (1/2)(1- cos2 ) We can rewrite equation (1) as follows:  = 1/2 (1 + 2 ) + 1/2 (1 - 2 ) cos2 (3) Introduce the triangle formula sin2 =2cossin into equation (2), we can obtain:  = 1/2 (1 - 2 ) sin2 (4)  reach a maximum value when  is 45 degrees and that is max= 1/2 (1 - 2 ) (5)