东南大学：《建筑力学 Architectural Mechanics》课程教学课件（英文讲稿）A16 Stress State

On the Damage Mechanism of Materials: Damage (fracture) of brittle materials is governed by the maximumtensilestress: Damage (yielding) of ductile materials is governed by the maximumshearing stress: The goal of stress state analysis is to determine thesurface/orientation on which normal/shearing stress achieves themaximum values.. Strength analysis can be subsequently performed in terms of themaximum normal/shearing stresses.11

• Damage (fracture) of brittle materials is governed by the maximum tensile stress On the Damage Mechanism of Materials • Damage (yielding) of ductile materials is governed by the maximum shearing stress • The goal of stress state analysis is to determine the surface/orientation on which normal/shearing stress achieves the maximum values. • Strength analysis can be subsequently performed in terms of the maximum normal/shearing stresses. 11

Plane Stress States. Plane Stress - state of stress in which twofaces of the cubic element are free of stressFor the illustrated example, the state ofstress is defined byOx, Oy, Txy and O, =tzx = tTy =0. State of plane stress occurs in a thin platesubjected to forces acting in the mid-planeof the plate.: State of plane stress also occurs on thefree surface of a structural element ormachine component, i.e., at any point ofthe surface not subjected to an externalforce.12

• Plane Stress - state of stress in which two faces of the cubic element are free of stress. For the illustrated example, the state of stress is defined by , , and 0.  x  y  xy  z  zx  zy  • State of plane stress also occurs on the free surface of a structural element or machine component, i.e., at any point of the surface not subjected to an external force. • State of plane stress occurs in a thin plate subjected to forces acting in the mid-plane of the plate. Plane Stress States 12

2-D Cauchy's Relation? Cauchy's relation:x1[0=ZFt.AA =o..AAcos0+ t...AAsin6n1肉ZFt,A = t.Acos0+o...AAsin0xx0=n+tunAA+oynyTn=Oβanp=OaβNp1yxt=n·o=o·n.Principal stresses & principal directionst=o·n=on=→(α-αl)·n=002TOxy13

0 cos sin 0 cos sin x x xx yx y y xy yy x xx x yx y y xy x yy y F t A A A F t A A A t n n t n n t n n                                                         t = n σ = σ n • Cauchy’s relation: • Principal stresses & principal directions   1 2 I 0 0 x xy xy y n n                                     t = σ n n σ n 0 n  t A  xx  yy yx  xy  t 2-D Cauchy’s Relation 13

Sample Problem. For the state of plane stress shown, determine the principal stressesand principal directions.. Solution1. Principal stresses10MPa40[50-α40MPa40-10-0n.4050-0=0=→ (50-α)(-10-)-1600= 050MPa40-10-α- 40g -2100 = 0 = 0, = 70,0, = -30>02. Principal directions40[50-70[0-20[n,4040040-10-70-80nV.5Ys40[0][50+3080n40071040-10+30400-2/520n2n14

• Solution    1 2 2 1 2 50 40 0 40 10 0 50 40 0 50 10 1600 0 40 10 40 2100 0 70, 30 n n                                                   • For the state of plane stress shown, determine the principal stresses and principal directions. 1 1 1 1 2 2 2 2 1 1 1 2 1 2 2 2 2 50 70 40 0 20 40 0 5 2 40 10 70 0 40 80 0 1 5 50 30 40 0 80 40 0 2 40 10 30 0 40 20 0 n n n n n n n n n n n n n n n n                                                                                                                   1 5 2 5               2. Principal directions 14 Sample Problem 1. Principal stresses

Plane Stress Transformation: Consider the conditions for equilibrium of aprismatic element with axis perpendicular toAAthe x, y, and x'.OAg,(AAcose)toZFx = 0 =αxA-x(Acos0)cos0- txy(Acos0)sin0Tuy(AAcose))-α,(Asino)sin-tx(Asin)cos0ZFy, = 0 = txsA+ox(Acos0)sin0 -Txy(AAcos0)cos0Txy(AA sin)-(Asin0)cosO + Tx(Asin0)sing, (AA sine): The equations may be rewritten to yieldax+aya.-01+cos20cos20+tr,sin2cos?Aor222ar01-cos20sin20+tx cos20sin?e22+元sin20=2sin0cos0cos2=cos?0-sin?0Gr+0a.1cos20-t..sin20a2215

                                          sin cos sin sin 0 cos sin cos cos sin sin sin cos 0 cos cos cos sin A A F A A A A A F A A A y xy y x y x xy y xy x x x xy                              • Consider the conditions for equilibrium of a prismatic element with axis perpendicular to the x, y, and x’. • The equations may be rewritten to yield   cos2 sin 2 2 2 sin 2 cos2 2 cos2 sin 2 2 2 2 x y x y x x y x xy x y x y x y y xy y                                              2 2 2 2 1 cos2 cos 2 1 cos2 sin 2 sin 2 2sin cos cos2 cos sin                  15 Plane Stress Transformation 