Timoshenko beam-element

beam theory

1. Gaziantep university Structure mechanic Comparison of Bernoulli-EULER and Timoshenko Beam Presented by Mohammed Muneam Mohammed Student number: 201623647

2. Content - Introduction to Classical “Euler- Bernoulli” Theory - Timoshenko Beam Theory - Finite Element of Timoshenko Beam and Euler- Bernoulli” Theory - References

3. Introduction to Classical “Euler- Bernoulli” Theory This is a well-Known the classical theory of Euler-Bernoulli beam assumes that: The cross- section plane perpendicular to the axis of the beam remains plane after deformation “assumption of a rigid cross-section plane . The deformed cross-section plane is still perpendicular to the axis after deformation. The classical theory of the beam neglects transverse shearing deformation where the transverse shear stress is determined by the equation of equilibrium. This is applicable to a thin beam.

4. Euler- Bernoulli beam element The strain in (x) direction And the stress :

5. And the resultant bending moment on the cross section area

6. For a beam with short effective length (length/depth < 5),( or composite beams , plates and shells , it is inapplicable to neglect the transverse shear deformation. In 1921 , Timoshenko presented a revised beam theory considering shear deformation which retains the first assumption and satisfies the stress-strain relations of shear. The difference between the Timoshenko beam and the technical, Bernoulli, beam is that the former includes the effect of the shear stresses on the deformation. A constant shear over the beam height is assumed.. Timoshenko beam element

7. Timoshenko Beam Theory Let the X axis be along the beam axis before deformation and the XZ plane be the deflection plane as shown in fig. above . The bending problem of a Timoshenko beam is considered the displacements û(x, z), ŵ(x, z) at any point (x , z) in the beam along X-axis and Z-axis ,respectively ,can be expressed in terms of two generalized displacements ,i.e. the deflection of beam axis w̃(x) and the rotation angle of the cross section Ѳ˜( x). û(x, z) = −z θ̃(x), ŵ(x, z) = w̃(x) Bending of Timoshenko beam

8. According to strain displacement є ᵪ = = Since Ѳ (x) is a function of (x) only, then the partial differential symbol can be replaced by an ordinary differential symbol            substituting equations above in stress – strain relationship get

9. ……….(1) ………(2) Multiplying both sides of equation (1) by (z) and integrating over the cross-section area . ……..(3) And further , integrating both sides of equation (2) directly over the cross-section area, we obtain:

10. ……..(4) When deducing the above equations , we assume constant shear stress on the cross section , which however are not true in actual situation. Hence a shear correction factor ( K ) of the cross section always introduced in equation (4) . Equation (3) and (4) becomes : M= EI K Fs= K GA ϓ Where K=

11. And ϓ =   Physically , they are referred as the relative rotation angle between two adjacent cross sections and the shear angle of section respectively . To determine the shear correction factor (K) it is usually to assume beforehand the type of shear stress distribution on various shape of cross sections. Different assumption results in different numerical values. For example K=1→1.2 for rectangular cross sections . An infinite shear correction factor K→ infinity , implies negligible effect of transverse shear deformation and the model degenerates to the classical theory of Euler- Bernoulli Beam The equilibrium equations of the Timoshenko beam theory are the same as those of the Euler – Bernoulli beam theory.

12. For Timoshenko beam And the strain

13. The equation of equilibrium: And we know that : Then, substitute the last two equations in the equilibrium equation get : These last two equations are called the Timoshenko beam equations

14. Finite Element of Timoshenko Beam In Euler-Bernoulli beam there are four degrees of freedom ( w₁ , w₂, Ѳ₁ , Ѳ₂ ) then the stiffness matrix is (4*4).which has the form : And the displacement w(x)=N₁(x) Z₁+ N₂(x) Ѳ₁ +N₃(x)Z₂+N₄(x) Ѳ₂ ……..(*)

15. The potential energy has a bending strain component and a shear strain component.

16. Then there are two types of bending: the first because of the moment and the second because of the shear, therefore; two equations for deflection shape will be assumed: Then, substituting it in the equation of potential energy we will get the shape function:

17. The transverse deformation of a beam with shear and bending strains may be separated into a portion related to shear deformation and a portion related to bending deformation It can be shown that the following shape functions satisfy the Timoshenko beam equations

18. The same steps used to derivative the stiffness of Bernoulli-Euler beam will be used to find the stiffness matrix for Timoshenko beam.

19. To explain effect of the shear stress i.e. the difference between Euler-Bernoulli and Timoshenko theories. we will take a simply supported beam and find the deflection at the mid span by using Timoshenko theory and Euler-Bernoulli theory . In each method we will find the deflection by Exact solution once and by numerical method (Finite Element Analysis) . And make a comparison between the results. We have a simply supported beam as shown below, subjected to distributed load (q) at all its length (L) L We will began with the Exact solution: q

20. Find the deflection at the mid span by using Timoshenko theory Now take the first Timoshenko equation and integrating it three times to get the equation of rotation ( Ѳ ) Then we will take the second Timoshenko equation , then substitute the equation of Ѳ on it

21. Use the boundary conditions: At x=0, w(x) = 0 At x=0, = 0 At x=L, w(x) = 0 At x=L, = 0

22. Applying the above boundary conditions get c ₂=0 , , c₄= 0 At x=L/2: Now will find the deflection at the mid span by using Euler-Bernoulli theory:

23. Integrating this equation two times with respect to x and applying the boundary conditions: At x = 0 , w = 0 At x = L , w = 0 We will get: At x = L/2 : If we return to Timoshenko displacement

24. Then the magnitude represented the effect of shear stress - Now we will find the displacement at the mid span by using the finite element analysis . For Euler-Bernoulli beam. Step 1- divide the beam in to two equal element Step 2- find the stiffness matrix of each element The stiffness matrix is the same for two elements

25. Make assembly for the two stiffness matrices to find the global stiffness matrix Step 3- find the displacement vector

26. Step 4 - find the force vector.

27. = Apply the equation F= K d, and find the displacement in the node 2. and knowing that u ₁= 0 ,u₃= 0 and Ѳ₂ = 0, then find u₂

28. Now use Timoshenko theory, we will make the same steps that used in Euler- Bernoulli beam, but the stiffness matrix will differ

29. Where The two elements have the same stiffness matrix. Now we will made assembly, to get the global stiffness matrix

30. Specify the displacement vector :

31. And the force vector And knowing that u ₁= 0 , u₃= 0 and Ѳ₂ = 0 , applying the equation : F= K d We will find that:

32. When we compare this result with the exact solution result We find there is small difference , that difference may be reduced if we used more elements. The value inside the bracket represent the effect of traverse shear strain

33. References 1- Bathe, K.J., Finite Element Procedures, Prentice Hall, 1996 2- Schäfer, M., Numerik im Maschinenbau, Springer, 1999 3- Henri P. Gavin , Spring, STRUCTURAL ELEMENT STIFFNESS MATRICES AND MASS MATRICES,2010 4- SYMPLECTIC ELASTICITY © World Scientific Publishing Co. Pte. Ltd. http://www.worldscibooks.com/engineering/6656.html 5- Dr Fehmi Cirak , Finite Element Formulation for Beams 6-Jose Palacios, Finite Element Method Beams, July 2008 7-Timoshenko beam theory , Wikipedia, the free encyclopedia

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