2D Problem in rectangular coordinate.pptx

2D Problem in rectangular coordinate

1. TWO DIMENSIONAL PROBLEMS IN RECTANGULAR COORDINATES LECTURE # 4 By Dr S Basack PhD, MIGS, MISRMTT, CE Associate Professor of Applied Mechanics Bengal Engineering and Science University Shibpur , Howrah Theory of Elasticity: Mechanics of Solids Department of Applied Mechanics BESU, Howrah August 2013

2. Governing Differential Equations 1

3. Solution Φ should be expressed as : A Polynomial Function of x and y A Fourier Series of x and y Second degree polynomial Third degree polynomial Fourth degree polynomial Fifth degree polynomial An E igen Solution

4. Solution by Polynomial Function : Second Degree Putting Φ = Φ 2 , the differential equation 1 is satisfied, hence this is a solution. Hence

5. Physical Interpretation D h h L L O X Y X’ Y’ X Y X’ Y’ O σ x = c 2 σ y = a 2 σ x = c 2 σ y = a 2 τ x y = - b 2 [opposite direction due to negative sign]

6. Third Degree Polynomial a 3 = b 3 = c 3 = 0 and d 3 ≠ 0 : Pure bending about X axis a 3 ≠ 0 and b 3 = c 3 = d 3 = 0 : Pure bending about Y axis a 3 = d 3 = 0 and b 3 ≠ 0, c 3 ≠ 0 : σ x α x ; σ y α y ; τ xy varies with x and y; a 3 = c 3 = d 3 = 0 and b 3 ≠ 0 : σ x = 0 ; σ y = b 3 y and τ xy = - b 3 x A B C D

7. Case - A X Y X’ Y’ O σ x = d 3 h σ x = - d 3 h σ x = d 3 h σ x = - d 3 h σ y = τ xy = 0

8. Case - B X Y X’ Y’ O σ y = a 3 L σ y = a 3 L σ y = - a 3 L σ y = - a 3 L σ x = τ xy = 0

9. Case - C X Y O σ x = c 3 .2 L σ x = 0 σ y = b 3 .2 h σ y = 0 τ xy = - b 3 .2L – c 3 .2h τ xy = - b 3 .2L τ xy = - c 3 .2h τ xy = 0 a 3 = d 3 = 0 and b 3 ≠ 0, c 3 ≠ 0 : σ x = c 3 x σ y = b 3 y τ xy = - b 3 x – c 3 y

10. Case - D X Y O Y’ σ y = b 3 . h σ y = - b 3 . h σ x = 0 σ x = 0 τ xy = - b 3 .2L τ xy = - b 3 .2L τ xy = 0 τ xy = 0 a 3 = c 3 = d 3 = 0 and b 3 ≠ 0 : σ x = 0 σ y = b 3 y τ xy = - b 3 x

11. Fourth Degree Polynomial a 4 + e 4 = - 2 c 4 Adjusting the arbitrary constants, various stress conditions can be obtained. Fifth Degree Polynomial

12. Displacements

13. Practical Examples P 1 Cantilever loaded at free end w per unit run 2 Simply supported beam under u d l

14. 1. Cantilever Loaded at End P Y Y ’ X L h h 1 - - P Shear force diagram PL Bending moment diagram Stress Conditions: 1. σ x α x (see BM diagram) 2. σ y = 0 (no normal loading on Y-plane) 4. τ xy does not vary with x (see SF diagram)

15. Stress Function Second degree polynomial a 2 = c 2 = 0 and b 2 ≠ 0 Fourth degree polynomial a 4 = b 4 = c 4 = e 4 = 0 and d 4 ≠ 0 Add

16. Boundary conditions and solutions I = m.i . of beam section = 2h 3 /3 By integration By substitution Back- substutution

17. 2. Simply supported beam under u d l wL Y Y ’ X L L h h 1 w per unit run wL X’ Stress Conditions: At y = h , σ y = - w At y = - h , σ y = 0 3. At y = + h , τ xy = 0 4. At x = + L , ∫ σ x dy = 0 At x = + L , ∫ σ x y dy = 0 6. At x = + L , ∫ τ xy dy = + wL h -h h -h h -h

18. Solution by Fourier Series α 4 f (y) -2 α 2 f”(y) + f IV (y) = 0 f (y) = C 1 cosh α y +C 2 sinh α y +C 3 y cosh α y+C 4 y sinh α y Φ = sin α x ( C 1 cosh α y +C 2 sinh α y +C 3 y cosh α y+C 4 y sinh α y )

19. Example : Beam under Sinusoidal Load X 1 X’ Y Y’ A h h B At y = h , τ xy = 0 ; σ y = - A sin α x At y = - h , τ xy = 0 ; σ y = - B sin α x L

20. Beam Loaded with Distributed Load of Varying Intensity X 1 X’ Y Y’ h h L