MECH 303 Chapter 2 2.6 Strain at a point The strain components x, y , xy at a point completely determine the strain state ( deformation state ) of the material at this point. The normal strain N of any line element PN N l 2 x m 2 y lm xy The change of angle between PN and PN ' cos ll ' mm ' cos1 cos 1 N N ' 2ll ' x mm' y lm ' l ' m xy principal strain, principal plane. xy 0 y n x 2-8 MECH 303 Chapter 2 2.7 Boundary conditions (1) 8 basic equations for plane problems: x xy X 0 y x equilibrium y yx Y 0 y x geometrical physical u v , , x y x y xy u v , y x 1 x y x E 1 y y x E 1 xy xy G , (plane stress) Solve the 8 unknown functions under proper boundary conditions. stress (2) 3 kinds of boundary conditions : displaceme nt mixed displacement boundary problem (displacement boundary 2-9 MECH 303 Chapter 2 condition) u s= u , s= v , u and v are the prescribed functions of coordinates on the surface. Stress boundary problem ( stress boundary condition) Surface forces acting on the boundary of a body are prescribed, which can be expressed as a condition about the stress components at the boundary: l x m y s s m yx s XN l xy s YN s s X Y (relations between boundary stress components and the surface force) when the boundary is normal to a coordinate axis, the above boundary conditions are simplified Mixed boundary conditions 2-10 MECH 303 Chapter 2 2.8 Saint-Venant’s principal If a system of forces acting on a small portion of the surface of an elastic body is replaced by another statically equivalent system of forces acting on the same portion of the surface, the redistribution of loading produces substantial changes in the stresses only in the immediate neighborhood of the loading, and the stresses are essentially the same in the parts of the body which are at large distances in comparison with the linear dimension of the surface on which the forces are changed. By “ statically equivalent systems” we mean that the two systems have the same resultant force and the same resultant moment. 2-11 MECH 303 Chapter 2 2.9 Solution of plane problem in terms of displacements Equilibrium equation in terms of stresses: x xy X 0 x y express stress by strain via (physical equation) y yx Y 0 y x Equilibrium equation in terms of strain : express strain by displacement components (geometrical equations) : Equilibrium equation in terms of displacements: (plane stress) E 2u 1 2 1 2 x 2 E 2v 1 2 1 2 y 2 2u 1 u 2v X 0 2 2 xy y 2v 1 2 Y 0 2 2 xy x Solve the above equations of u(x,y) and v(x,y) with the following boundary conditions (1) displacement boundary condition : u sU , v s V (2) stress boundary condition ( in term of displacements) : E u v 1 u v l m X s s y 2 y x 1 2 x E v u 1 v u s Y m s l 2 x 2 x y 1 y Thus, to solve a plane stress problem in terms of displacements, we have to solve two differential equations simultaneously and the 2-12 MECH 303 Chapter 2 obtained u (x, y) and v ( x, y) must satisfy the displacement boundary condition or stress boundary condition or mixed boundary condition at the boundary. Once we have u (x, y) and v (x, y), then x (x, y) , y (x, y), x (x, y), y (x, y), xy (x, y) xy (x, y) are completely uniquely determined. For plane strain problem, the formulation is the same as before, only replace E by E and by . 2 1 1 The above method is suitable to solve any plane problems with any boundary conditions. y H Exercise 1 Exercise 2 (fixed boundary) (plane stress) x u0 L 2-13 MECH 303 Chapter 2 2.10 Solution of plane problem in terms of stress Take the 3 stress components x , y , xy (xy ) as the basic unknown functions. (1) 2 equilibrium equations contain 3 functions (2) The 3rd equation is the compatibility equation in terms of strain (or stress): The necessary and sufficient condition for the existence of single-valued continuous functions u ( x, y) and v (x, y) is simple x u x , y v , y xy u v y x 2 2 2 x y xy xy y 2 x 2 (by using physical equations x x ,……...) X Y 2 x y 1 x y (plane stress) 1 X Y 1 x y (plane strain) 2 x y (3) The obtained stress solution should satisfy stress boundary condition (B.C.) 2-14 MECH 303 Chapter 2 2.11 Case of constant body forces In the case of constant body forces: X Y 0 , 0 , then the 3 stress x y components x , y , xy are determined by equations x xy X 0 x y y yx Y 0 y x 2 0 x y l x s m xy s X and B. C. on S m y s l xy s Y x xy X 0 x y consists of: The general solution of y yx Y 0 x y x xy 0 x y (1) The general solution of y yx 0 x y (homogeneous system) x xy X 0 x y (2) A particular solution of y yx Y 0 x y (non-homogeneous system) When body forces are constant , the particular solution may be taken as 2-15 MECH 303 Chapter 2 x X x or: y Y y , y=0 , x 0 , xy yx 0 , xy X y Y x x xy 0 x y is The general solution of y yx 0 x y 2 x 2 y 2 y 2 always satisfy the homogeneous equations x 2 xy yx xy Airy’s stress function for plane problems. The complete solution of the equilibrium equations are 2 Xx x 2 y 2 Yy y 2 x 2 xy yx xy depends on To determine , substituting x, y, xy, into compatibility condition, we have 4 0 4 2 2 2 2 y x 2 2 2 2 y x 2-16 MECH 303 Chapter 2 Thus, in the solution of plane problems in terms of stress when body forces are constant, it is necessary only to solve for stress function from the single differential equation 4 =0, and x, y, xy satisfy stress boundary conditions. 2-17 MECH 303 Chapter 2 2.12 Airy’s stress function. Inverse method and semi-inverse method Direct method usually impossible Inverse method take some function satisfying the compatibility equation, then obtain stress components and find the surface force components, thus we know what problem they can solve. Semi-inverse method we assume the solution for stress or displacements in a given problem, then proceed to show that all the differential equations and boundary conditions are satisfied. If some of the boundary conditions are not satisfied, then we have to modify the assumptions made. (Numerical method ---- Finite element method, directly solve the equilibrium equation in terms of displacements.) 2-18