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Contents The Stress State of a Point( 点的应力状态 ) Simple, General & Principal Stress State( 简单一般和主应力状态 ) Ordering of Principal Stresses( 主应力排序 ) On the Damage Mechanisms of Materials( 材料失效机制 ) Plane Stress States( 平面应力 ) -D Cauch s Relation( 平面柯西关系 ) Plane Stress Transformation( 平面应力转换 ) Mohr s Circle( 莫尔圆 ) Maimum Shearing Stress( 最大切应力 ) Construction of Mohr s Circle ( 莫尔圆的构建 )

Contents 3-D Cauch s Relation( 三维柯西关系 ) Stress Transformation of General 3-D Stresses( 空间应力转换 ) Application of Mohr s Circle in 3-D( 莫尔圆在三维应力状态中的应用 ) Generalied Hooke s Law( 广义胡克定律 ) The Relation among E, ν and G( 杨氏模量泊松系数及剪切模量间的关系 ) Fiber-reinforced Composites( 纤维增强复合材料 ) Strain Energ Densit under Generalied Stress States( 一般应力状态下的应变能 ) Volume Strain, Mean Stress and Bulk Modulus( 体应变平均应力和体积模量 ) Volumetric & Distortion Energ Densit( 体积应变能密度和畸变能密度 ) 3

The Stress State of a Point The stress state at a point - is defined as the collection of the stress distributions on all planes passing through the point. - can be analed via the stress distributions acting on a differential cube, arbitraril oriented with reference coordinates Method of differential cubes with: - infinitesimal side length - uniforml distributed stress on each of the si surfaces - equal and opposite stress on parallel sides 4

Stresses on Oblique Planes F F A F The stress distribution at a point depends on the plane orientation along which the point is eamined. F cos sin 5

Eamples of the Stress State of a Point F A F A M e M e A A A B F C A B C 6

Stress Under General Loadings A member subjected to a general combination of loads is cut into two segments b a plane passing through a point of interest Q The distribution of internal stress components ma be defined as, lim A0 lim A0 F A V A lim A0 V A For equilibrium, an equal and opposite internal force and stress distribution must be eerted on the other segment of the member. 7

General Stress State of a Point Stress components are defined for the planes cut parallel to the, and aes. For equilibrium, equal and opposite stresses are eerted on the hidden planes. The combination of forces generated b the stresses must satisf the conditions for equilibrium: F F F 0 M M 0 M 0 Consider the moments about the ais: similarl, It follows that onl 6 components of stress are required to define the complete state of stress M Aa A and a 8

Principal Stress State of a Point The most general state of stress at a point ma be represented b 6 components,,, normal stresses,, shearing stresses (Note:,, ) Same state of stress is represented b a different set of components if aes are rotated. There must eist a unique orientation (Principal Directions) of the differential cube, having onl normal stresses (Principal Stresses) acting on each of its si faces. 9

Ordering of Principal Stress State 3 ma inter min A 1 3 1 Principal aes & stresses Stress states classified based on the number of ero principal stresses: - Tri-aial stress state: none ero principal stresses; - Biaial stress state: one ero principal stresses; - Uniaial stress state: two ero principal stresses; 10

On the Damage Mechanism of Materials Damage (fracture) of brittle materials is governed b the maimum tensile stress Damage (ielding) of ductile materials is governed b the maimum shearing stress The goal of stress state analsis is to determine the surface/orientation on which normal/shearing stress achieves the maimum values. Strength analsis can be subsequentl performed in terms of the maimum normal/shearing stresses. 11

Plane Stress States Plane Stress - state of stress in which two faces of the cubic element are free of stress. For the illustrated eample, the state of stress is defined b,, and 0. State of plane stress occurs in a thin plate subjected to forces acting in the mid-plane of the plate. State of plane stress also occurs on the free surface of a structural element or machine component, i.e., at an point of the surface not subjected to an eternal force. 1

-D Cauch s Relation t n A t Cauch s relation: 0 F ta Acos Asin 0 F t A Acos Asin t n n t n n t n n t = n σ = σ n Principal stresses & principal directions t=σn n σ I n 0 n1 0 n 0 13

Sample Problem For the state of plane stress shown, determine the principal stresses and principal directions. Solution 1. Principal stresses. Principal directions 50 40 n1 0 40 10 n 0 50 40 0 50 10 1600 0 40 10 40 100 0 70, 30 1 50 70 40 n1 0 0 40 n1 0 n1 5 1 40 10 70 n n n 0 40 80 n 0 n 1 5 1 50 30 40 n1 0 80 40 n1 0 n1 5 n n1 40 10 30 n 0 40 0 n 0 n 5 14

Plane Stress Transformation 1 cos cos 1 cos sin sin sin cos cos cos sin Consider the conditions for equilibrium of a prismatic element with ais perpendicular to the,, and. F 0 A Acos cos Acos sin Asin sin Asin cos F 0 A Acos sin Acos cos Asin cos Asin sin The equations ma be rewritten to ield cos sin sin cos cos sin 15

Plane Stress Transformation Alternative Wa cos sin cos sin sin cos sin cos cos sin sin cos sin cos cos sin T 16

Mohr s Circle The previous equations are combined to ield parametric equations for a circle, cos sin sin cos ave R, ave ; R Principal stresses 0 sin cos R tan p o Note: defines two angles separated b 90 ave ma,min It is hard to predict which of the two angles results in the maimum/minimum principal stress. 17

Maimum Shearing Stress Maimum shearing stress occurs for cos sin sin cos ave R ave ma R tan s tan tan 1 s Note: defines two ang p offset from b 45 p o les separated b 90 an o d 18

Construction of Mohr s Circle Y, p X, cos sin sin cos ave R ma,min ave, p ave R, tan ; R Two options for plotting the diameter XY 0,, or X X,, Y,,, Y, The second wa is tpicall chosen, such that the direction of rotation of O to Oa is the same as CX to CA. 19

Construction of Mohr s Circle With Mohr s circle uniquel defined, the state of stress at other aes orientations ma be depicted. For the state of stress at an angle θ with respect to the aes, construct a new diameter X Y at an angle θ with respect to XY. Normal and shearing stresses are obtained from the coordinates X Y. 0

Construction of Mohr s Circle Mohr s circle for centric aial loading: P, 0 A P A Mohr s circle for torsional loading: T 0 T 0 W W P P 1

Sample Problem For the state of plane stress shown, (a) construct Mohr s circle, determine (b) the principal planes, (c) the principal stresses, (d) the maimum shearing stress and the corresponding normal stress. Solution: Construction of Mohr s circle 50 10 ave CF 50 0 30MPa R CX FX 40MPa 30 40 50MPa 0MPa

Principal planes and stresses ma OA OC CA 0 50 ma 70MPa OB OC BC 0 50 ma ma 30MPa FX 40 tan p CF 30 53.1, 33.1 p 6.6, 116.6 p 3

Maimum shearing stress s p 45 s 71.6, 161. 6 ma ma R 50 MPa ave 0 MPa 4

Sample Problem For the state of stress shown, determine (a) the principal planes and the principal stresses, (b) the stress components eerted on the element obtained b rotating the given element counterclockwise through 30 degrees. Solution: Construct Mohr s circle ave R 100 60 80MPa CF FX 0 48 5MPa 5

Principal planes and stresses tan p p p XF CF 67.4 33.7 48 0.4 clockwise ma OA OC CA 80 5 ma 13MPa ma OA OC 80 5 min 8 MPa BC 6

Stress components after rotation b 30 o Points X and Y on Mohr s circle that correspond to stress components on the rotated element are obtained b rotating XY counterclockwise through 60 o 180 60 67.4 5.6 OK OC KC OL OC CL KX 5sin 5.6 48.4 MPa 111.6 MPa 41.3MPa 80 5cos5.6 80 5cos5.6 7

3-D Cauch s Relation Consider the general 3D state of stress at a point and the transformation of stress from element rotation State of stress at Q defined b:,,,,, Consider tetrahedron with face perpendicular to the line QN with direction cosines: n cos n, n Cauch s relation: ta A A A 0 F t A A A A 0 F ta A A A ti jin j ijn j t = n σ = σ n 8

3-D Cauch s Relation t=σnn ( ) ( ) 0 ( ) 3 ( ) ( ) ( ) 0 3 I I I 0 1 3 Determine the normal and shearing stresses on the element in the,, and direction of a given reference; Calculate the three stress invariants; Solve the cubic equation for its three roots (principal stresses); Substitute the roots back into original characteristic equation to solve for the principal directions. I 1, I, and I 3 are known as stress invariants as the are independent of coordinate choices. 9

Sample Problem For the state of 3-D stress shown, determine the principal stresses and principal directions. Solution 1. Principal stresses I I I 1 3 0MPa 0 0 0 60 0 10 0 10 0 0 0 0 0 400 400 400 100 1100 8000 000 6000 3 60 1100 6000 0 40 300 0 30 0 10 0 30; 0; 10 1 1 1. Principal directions 0MPa 10MPa 0MPa 0 10 0 n 0 1 10 0 0 n 0 0 0 0 n 3 0 30

Stress Transformation of General 3-D Stresses Consider the conditions for equilibrium of the tetrahedron generated from an oblique plane with the elemental volume. The requirement F n 0 leads to n 31

Stress Transformation of General 3-D Stresses n o p n n n o o o p p p a n n n o o o p p p n n n n o p T o o o n o p a a p p p n o p Too complicated to find the principal stress states through stress transformation 3

Application of Mohr s Circle in 3-D 3 1 1 3 Points A, B, and C represent the principal stresses on the principal planes (shearing stress is ero) ma 1 ma min The three circles represent the normal and shearing stresses for rotation around each principal ais. Radius of the largest circle ields the maimum shearing stress. 33

Sample Problem For the state of 3-D stress shown below, determine the principal stresses and maimum shearing stress. Solution 1. Summarie the 3-D stress state. 0MPa 40 0 0 0 0 0 0 0 30 40MPa. Find the two principal stresses in - plane. i 40-0 40 0 46 MPa 0 j -6 MPa 3. Rearrange the three principal stresses. 0MPa 30MPa 1 46 MPa, 6 MPa, 3 30 MPa 4. Find the maimum shearing stress. 13 46 ( 30) ma 38 MPa 34

Generalied Hooke s Law For a bar under uniaial tension: Aial strain: E Transverse strain: E Uniaial stress along : Along :, E E, E E Along :, E E 35

Generalied Hooke s Law 1 E E E E E 1 E E E E E 1 E E E E E Shearing Stresses Shearing stress around : Around : Around : G G G

The Relation among E, ν and G L h 1 bd cos 1 sin L h h h h bd h 1 h 1 G 1 L h 1 h 1 bd h 1 h 1 E G 1 E 1 E Onl two of the three elastic constants are independent. 37

Sample Problem A circle of diameter d = 9 in. is scribed on an unstressed aluminum plate of thickness t = 3/4 in. Forces acting in the plane of the plate later cause normal stresses σ = 1 ksi and σ = 0 ksi. For E = 1010 6 psi and ν = 1/3, determine the change in: a) the length of diameter AB, b)the length of diameter CD, c) the thickness of the plate, and d)the volume of the plate. 38

SOLUTION: Appl the generalied Hooke s Law to find the three components of normal strain. E 1 1010 6 psi 0.53310 E 1.06710 E E E E 1.60010 E 1ksi 0 0ksi 3 3 3 in./in. E in./in. E in./in. 1 3 Evaluate the deformation components. 3 B d 0.53310 in./in. 9in. A C d D t t B A 4.810 3 in. 3 1.60010 in./in. 9in. C D 14.410 3 in. 3 1.06710 in./in. 0.75in. t 0.80010 Find the change in volume e V ev 1.06710 1.06710 V 3 in 3 3 /in in. 3 3 3 15150.75 in 3 0.187in 39

Sample Problem For the bar shown, find the change in length of line segment BC and in right angle ABC. Assume the Young s modulus and Poisson s ratio as E and ν respectivel. Solution A C 1. Stress state in reference F F coordinate sstem 60 o 30 o F bh All other stress components vanish.. Stress state in transformed coordinates: 3 0 cos sin 30 4 3 0 sin cos 30 4 1 0 cos sin 10 4 B 10 o 30 o 30 o 30 o b h 40

3. Longitudinal strain along BC 1 3 3 F 0 0 0 30 30 10 E 4E 4Ebh 4. The change in length of BC l l h BC 30 BC 30 0 0 5. The change in right angle ABC 3 Eb F 3 0 3(1 ) 3 (1 ) F G E bhe 30 4 E (1 ) 41

Fiber-reinforced Composites (Anisotrop) Fiber-reinforced composite materials are formed from lamina of fibers of graphite, glass, or polmers embedded in a resin matri. Normal stresses and strains are related b Hooke s Law but with directionall dependent moduli of elasticit, E E E Transverse contractions are related b directionall dependent values of Poisson s ratio, e.g., Materials with directionall dependent mechanical properties are anisotropic. 4

Strain Energ under Generalied Stress States Strain energ densit of nonlinearl elastic material under uniaial stress state du d Strain energ densit of linearl elastic material under uniaial stress state u 1 43

Strain Energ under Generalied Stress States Strain energ densit of nonlinearl elastic material under generalied 3-D stress states du d ij ij d d d d d d U Strain energ densit of linearl elastic material under generalied 3-D stress states 1 1 E 1 G d d d 44

Volume Strain, Mean Stress and Bulk Modulus Volume strain V0 ddd V d(1 ) d(1 ) d(1 ) 1 1- ddd 1 O V V 1 0 V V0 E For a given stress state, if no volume change happen, the first invariant of the stress tensor must be ero. 0 Bulk modulus can be defined as the ratio between mean stress and volume strain E K m V 3 V 0 1 3 1 d d d 45

Volumetric & Distortion Energ Densit m m d d d = + u u (a) Mean stress V ud tensor Volumetric energ densit: d m d d m d d d (b) Deviatoric stress tensor 1 3 1 1 E 6E Distortion energ densit: u V m V m 1 1 u d 6E G m m 46