ADVANCED STRUCTURAL MECHANICS

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Ηιονη Λύτρας
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1 VSB TECHNICAL UNIVERSITY OF OSTRAVA FACULTY OF CIVIL ENGINEERING ADVANCED STRUCTURAL MECHANICS Lecture 1 Jiří Brožovský Office: LP H 406/3 Phone: jiri.brozovsky@vsb.cz WWW:

2 Topics 1. Basic equations 2. Plane structures 3. Energetical principles, variational methods 4. Finite element method 2

3 Recommended books Timoshenko. S. Gere, J.: Mechanics of Materials, Van Nostrand Company, New York, 1972 (or newer edition) Boresi, A. Schmidt, R.: Advanced Mechanics of Materials, John Wiley & Sons, 2003 Cook, R. D., Malkus, D. S., Plesha, M. E., Witt, R. J.: Concepts and Applications of Finite Element Analysis, John Wiley and Sons,

4 Topic 1: Basic equations of elasticity Basic unknowns Geometrical relations Equilibrium equations Material law Compatibility equations 4

5 Basic unknowns y τ zy τ σ y yz τ yx τ xz τ xy σ x Stress vector σ = {σ x, σ y, σ z, τ xy, τ yz, τ zx } T (1) Strain vector ε = {ε x, ε y, ε z, γ xy, γ yz, γ zx } T (2) σ z τ zx x z Displacement vector u = {u x, u y, u z } T (3) 5

6 Geometrical relations (1) 0 y, v x, u A B D C β α A D u v dx u x dx dy x v B dx x y v u dy dy C 6

7 Geometrical relations (2) 0 y, v x, u A B D C β α A D u v dx u x dx dy x v B dx x y v u dy dy C εx = A B AB AB = (x + dx + u + u x dx) (x + u) dx dx = u x 7

8 Geometrical relations (3) Normal deformations ε x = u x ε y = v y ε z = w z, (4) Shear deformations γ yz γ zx γ xy = γ zy = v z + w y (5) = γ xz = w x + u (6) z = γ yx = u y + v x. (7) 8

9 Geometrical relations (4) Uvedené vztahy obecně neplatí: γ yz = γ zy, γ zx = γ xz, γ xy = γ yx. The the assumption of equality of shear stresses is based on approximative fulfillment of moment equilibrium equations of differential element. Shear deformations are assumed to be equal in the same way. 9

10 Differential equilibrium conditions (1) σ y σ x dy τyx τzx τ yz σz τzy dx τ zy τ xy τyz σz τ τ xz xz τxy σy τyx τzx dz σ x σ x = σ x + σ x x dx, τ xy = τ xy + τ xy dy,... (8) x 10

11 Diff. equilibrium conditions (2) σ x = σ x + σ x x dx, τ xy = τ xy + τ xy dy,... y Fi,y = (σ x σ x ) dx dy+(τ xy τ xy) dx dz+(τ xz τ xz ) dx dy = 0 (σ x σ x σ x x dx) dy dz+(τ xy τ xy τ xy y dy) dx dz+(τ xz τ xz τ xz z dz) dx dy = 0 After simplification: σ x x + τ xy y + τ xz z = 0 (9) 11

12 Diff. equilibrium conditions (3) σ x x + τ xy y τ xy + τ xz z + X = 0 x + σ y y + τ yz z + Y = 0 (10) τ zx x + τ zy y + σ z z + Z = 0 where X,Y, Z are volume forces. 12

13 Material law (1) Defines relations between stresses and strains. Hooke law for 1D problem (simple tension/compression): ε x = σ x E x Α F ε x = L L = L x L L σ x = F A = E ε x 13

14 Material law (2) Hooke law for 3D problems: ε x = 1 E [σ x ν (σ y + σ z )], γ yz = τ yz 2 G ε y = 1 E [σ y ν (σ x + σ z )], γ xz = τ xz 2 G ε z = 1 E [σ z ν (σ x + σ y )], γ xy = τ xy 2 G (11) 14

15 Conclusion 15 unknowns: 3 displacements u 6 stresses ε 6 strains σ 15 available equations: 6 geometrical relations 6 matrial law equations 3 equilibrium conditions 15

16 Compatibility conditions (1) They define continuity of deformations the continuous volume must remain to be continuous also after deformation. They can be obtained from geometrical relations by elimination of displacements. 2 ε x y ε y x 2 + = 2 ε xy x y 2 ε y z ε z y 2 + = 2 ε yz y z 2 ε z x ε x z 2 + = 2 ε zx z x (12) 16

17 Compatibility conditions (2) 2 ε x y z = ε y z x = ε z x y = 1 2 γ yz x + γ yz x + γ yz x + γ xz y γ xz y + γ xz y + γ xy z + γ xy z γ xy z (13) For plane problems (2D, to be discussed later) it can be written: 2 ε x y ε y x 2 = 2 γ xy x y (14) 17

18 Topic 2: Plane problems 2D plane structures supported and loaded in their middle plane will be discussed. Two main variants of plane problem: plane stress plane strain 18

19 Plane strain 1 All deformations (strains) must lie in the x y plane: ε = {ε x, ε y, γ xy } T (15) The structure can not freely move in z direction. As a result, there is σ z stress. σ z : σ = {σ x, σ y, σ z, τ xy } T (16) 19

20 Plane strain y σ y τ xy σ x All stresses lie in the x y plane: σ = {σ x, σ y, τ xy } T (17) z x The wall can freely move in z direction, therefore there is on σ z stress but the ε z strain is nonzero: ε = {ε x, ε y, ε z, γ xy } T (18) 20

21 Plane problem conclusions (1) Equilibrium equations σ x x + τ xy y + X = 0 (19) τ xy x + σ y y + Y = 0 Geometrical relations ε x = u x, ε y = v y, γ xy = γ yx = u y + v x. (20) 21

22 Plane problem conclusions (2) Material law: plane stress σ x = σ y = τ x = E 1 µ 2 (ε x + µ ε y ) E 1 µ 2 (ε y + µ ε x ) (21) E 2(1 µ) γ xy (22) 22

23 Plane problem conclusions (3) Material law: plane strain σ x = σ y = τ x = E (1 + µ)(1 2µ) [(1 µ) ε x + µ ε y ] E (1 + µ)(1 2µ) [µ ε x + (1 µ) ε y ] (23) E (1 + µ)(1 2µ) γ 1 xy (1 µ) 2 23

24 Airy s equation for plane problem Airy function F describes stress state of plane problem if: σ x = 2 F x 2, σ y = 2 F x 2, τ xz = 2 F x y. (24) Airy equation compatibility relation is (14) writtnen with use of Airy function: 4 F x F x 2 y F y 4 = 0 (25) 24

25 Stress distribution on wall: N S N... beam theory, S... plane stress theory. 25

26 Forces in wall Nx = σx h [ N m ] Ny = σy h [ N m ] Nxy = τxy h [ N m ] x yx n xy z n n n y τ xy yx h τ σ σ x y y x 26

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