Beam and Frame Structures

1. Beam and Frame Structures • 5.1 Euler-Bernoulli beam theory • 5.2 1-D Euler-Bernoulli beam element • 5.3 2-D frame element • 5.4 3-D frame element • 5.5 Timoshenko beam theory • 5.6 Timoshenko beam element • 5.7 Shear deformation beam element

2. 5.5 Timoshenko beam theory It is equal to where is the transverse shear strain. The Timoshenko beam theory includes the effect of transverse shear deformation. As a result, a plane normal to the beam axis before deformation does not remain normal to the beam axis any longer after deformation. Consider a generic point located on the plane section at a distance y from the neutral axis. It is assumed that after deformation the plane section remains plane, but because of shear deformations its rotation is not equal to Thus, the displacement field is and

3. Using the strain-displacement equations, the nonzero strains are These strains are related to stresses as follows: where E is Young’s modulus, G is the shear modulus, and ks is a shear correction factor introduced to account for nonuniform shear stress distribution.

4. where ks is a shear correction factor introduced to account for nonuniform shear stress distribution. It is equal to the ratio of the effective area resisting shear deformation to the actual beam cross-sectional area. For a rectangular section For an I-beam, area of web/area of cross-section. More accurate values of ks account for the effect of Poisson’s ratio as follows: Rectangular section: Thin-walled square tube: Thin-walled circular tube: Solid-circular section:

5. The strain energy functional can be defined by writing the strain energy and the work done by the applied forces, the strain energy is where le is the length of the beam.

6. Substituting the strain-displacement relationship yields The potential energy is

7. 5.6 Timoshenko beam element Displacement and section rotation interpolate independently. For two-node element, linear interpolating functions are used:

8. Nodal displacement vector Define Displacement and rotation can be written into matrix form:

9. Curvature Shear strain

10. Element strain energy

11. Bending stiffness matrix Shear deformation stiffness matrix

12. The equations can be written in a more convenient form by introducing a non-dimensional relative stiffness parameter as follows:

13. Shear locking For thin-beam, shear strain is zero. Shear strain need to be zero in all beam. It requires is linear function, it yields It means beam can't bending that is not the actual case. The phenomenon is called as shear locking.

14. Consider the application of this element to a cantilever beam with moment applied at the free end. The beam is in the state of pure bending with no shear forces. The exact solution is constant bending strain and zero shear strain along the entire length of the beam. The interpolation functions used in the formulation of the two-node Timoshenko beam element are

15. The bending and the shear strains are as follows From the finite element solution the condition of zero shear strain implies that

16. For this condition to be true for the entire length. Clearly, the coefficient of the term multiplying x must be zero. That is, the only way that we can get a zero shear strain for the entire element length is when However, this implies that the bending strain must be zero. Thus, the shear strain term is preventing the element from giving reasonable results for the bending problem. This is exactly what is known as shear locking.

17. Remedies for shear locking: selective reduced integration For the two-node element, since the shear strain is linear, an exact integration requires two Gauss points. To under integrate this term, we use a single Gauss point integration. In 1*1 integrating rule, element center point is used to calculate Bs, it means linear change of is replaced by constant change in element. It makes and have the same order.

18. If using 1*1 integrating rule ks is The element stiffness matrix is

19. Function to calculate element stiffness matrix for 2D Timoshenko beam element function ke=beam2Te(ecoords,E,v,ks,A,I); b=[ ecoords(2,2)-ecoords(1,1); ecoords(2,2)-ecoords(1,2) ]; L=sqrt(b'*b); n=b/L; G=E/(2*(1+v)); %材料剪切模量 m=12*E*I/(ks*G*A*L^2) Kle=E[A/L 0 0 -A/L 0 0; 0 12*I/m^3 6*I*L/m^3 0 -12*I/m^3 6*I*L/m^3; 0 6*I*L/m^3 I*L^2*(m+3)/m^3 0 -6*I*L/m^3 -I*L^2*(m-3)/m^3; -A/L 0 0 A/L 0 0; 0 -12*I/m^3 -6*I/L^2 0 12*I/m^3 -6*I/L^2; 0 6*I*L/m^3 -I*L^2*(m-3)/m^3 0 -6*I*L/m^3 I*L^2*(m+3)/m^3]; T=[n(1) n(2) 0 0 0 0; -n(2) n(1) 0 0 0 0; 0 0 1 0 0 0; 0 0 0 n(1) n(2) 0; 0 0 0 -n(2) n(1) 0; 0 0 0 0 0 1]; ke=T'*Kle*T;

20. 5.7 Shear deformation beam element While considering shear deformation, deflection of beam can express as two parts where is the deflection by bending deformation. is the additional deflection by shear deformation Rotation of section by bending deformation is Shear strain corresponding to

21. For two-node beam element, nodal displacement vector includes two parts where ，

22. For bending deflection ,three order polynomial function is used (Hermite interpolating function). For additional deflection cursed by shear deformation , linear interpolating function is used, which yields constant shear force in element. It is reasonable selection. is three order polynomial, from equilibrium equation, it also derives a constant shear force in element. 3 order polynomial 0 order polynomial 1 order polynomial 0 order polynomial

23. can be expressed as and

24. Element potential energy considering shear deformation is where is curvature is shear strain.

25. The curvature is The shear strain is

26. and are same as those of classic beam element. Element equations is derived from functional minimal condition (principle of minimal potential energy) respectively

27. Corresponding to shear strain, and are

28. Above two element equations are non-coupled. Each node has three DOF Using equilibrium equation, each node can reserve two independent DOF

29. Define According to , we have

30. Write in matrix form: Solve above two equations, we have

31. Substituting above two equations into element equations to merge nodal DOF:

34. Add above two equations and multiply , we have

35. Therefore

36. Add equations of shear deformation into first and third rows of bending deformation equation, we get:

37. For uniform distribution load q, it has The equivalent nodal load vector is same with that of classic beam element:

38. Function to calculate element stiffness matrix for 2D shear deformation beam element function ke=beam2Tse(ecoords,E,v,ks,A,I); b=[ ecoords(2,2)-ecoords(1,1); ecoords(2,2)-ecoords(1,2) ]; L=sqrt(b'*b); n=b/L; G=E/(2*(1+v)); %材料剪切模量 m=12*E*I/(ks*G*A*L^2) Kle=E/(1+m)*[A*(1+m)/L 0 0 -A*(1+m)/L 0 0; 0 12*I/L^3 6*I/L^2 0 -12*I/L^3 6*I/L^2; 0 6*I/L^2 4*I*(1+m/4)/L 0 -6*I/L^2 2*I*(1-m/2)/L; -A*(1+m)/L 0 0 A*(1+m)/L 0 0; 0 -12*I/L^3 -6*I/L^2 0 12*I/L^3 -6*I/L^2; 0 6*I/L^2 2*I*(1-m/2)/L 0 -6*I/L^2 4*I*(1+m/4)/L]; T=[n(1) n(2) 0 0 0 0; -n(2) n(1) 0 0 0 0; 0 0 1 0 0 0; 0 0 0 n(1) n(2) 0; 0 0 0 -n(2) n(1) 0; 0 0 0 0 0 1]; ke=T'*Kle*T;

39. Example 2 Consider the solution of a cantilever beam subjected to a point load at the free end as shown in following figure. Assume the section to be rectangular with width = 1, height = h, and ks = 5/6. Assume that G = E/3, E = 1, and L = 1. Obtain a solution using only one element. Compare solutions obtained by the Timosenko beam element, linked interpolation beam element and the conventional beam element.

40. Using the data given, we have ; Using the Timoshenko beam element, after imposing the essential boundary conditions, we have Solving this system of equations, we get the following tip displacement

41. Using the linked interpolation beam element, after imposing the essential boundary conditions, we have Solving this system of equations, we get the following tip displacement

42. Using conventional beam element, the equations after imposing the essential boundary conditions are as follows Solving this system of equations, we get the tip displacement

43. Comparison of solutions using a Timoshenko beam and a conventional beam element

44. Comparison of solutions using a linked interpolation TBT beam and a conventional beam element

45. The Timoshenko beam element solution is larger than the conventional beam element. However, as the beam becomes thin, say The Timoshenko beam element results do not make physical sense. The phenomenon demonstrated in this example is known as shear locking. The linked interpolation beam element does not exhibit shear locking and the solution approaches EBT as h decreases.

46. ANSYS beam elements Linked interpolation beam elements BEAM3/4 2-D/ 3-D Elastic Beam BEAM23 /24 2-D/3-D Plastic Beam BEAM54 /44 2-D/3-D Elastic Tapered Unsymmetric Beam Timoshenko beam elements BEAM188/189 3-D 2-Nodes/3-Nodes Linear Finite Strain Beam

47. BEAM3 Shape functions:

49. BEAM4 Shape functions: