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
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
It is equal to
where
is the transverse shear strain.
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
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 A/L ;
*I/m^3 6*I*L/m^ *I/m^3 6*I*L/m^3;
*I*L/m^3 I*L^2*(m+3)/m^ *I*L/m^3 -I*L^2*(m-3)/m^3;
-A/L A/L ;
*I/m^3 -6*I/L^ *I/m^3 -6*I/L^2;
*I*L/m^3 -I*L^2*(m-3)/m^ *I*L/m^3 I*L^2*(m+3)/m^3];
T=[n(1) n(2) ;
-n(2) n(1) ; ;
n(1) n(2) 0;
n(2) n(1) 0; ];
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. and
can be expressed as

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

25. The curvature is
The shear strain is

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

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 A*(1+m)/L ;
0 12*I/L^3 6*I/L^ *I/L^3 6*I/L^2;
0 6*I/L^2 4*I*(1+m/4)/L *I/L^2 2*I*(1-m/2)/L;
-A*(1+m)/L A*(1+m)/L ;
0 -12*I/L^3 -6*I/L^ *I/L^3 -6*I/L^2;
0 6*I/L^2 2*I*(1-m/2)/L *I/L^2 4*I*(1+m/4)/L];
T=[n(1) n(2) ;
-n(2) n(1) ; ;
n(1) n(2) 0;
n(2) n(1) 0; ];
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:

53. BEAM44
The stiffness matrix is the same as for BEAM4 , except that an averaged area is used:
and all three moments of inertia use averages of the form:

54. BEAM188
Shape functions:

55. BEAM189
Shape functions:

56. Example 3. Define linear distribution loads
Analysis beam as following figure. Drawing moment diagram.
E=207e5, A=1, I= , h=0.5.

57. ANSYS Command lines
!Define linear distribution loads on beam elements.
/Clear
/prep7
ET,1,BEAM3
MP,EX,1,207e5
R,1,1, ,0.5
!define nodes
N,1,0,0
*DO, i,1,10
N,i+1,i*2,0
*ENDDO
N,100+i,20+i*2,0
!define elements
E,i,(i+1)

58. E,11,101
*DO, i,1,9
E,100+i,100+(i+1)
*ENDDO
!define displacement constraints
D,1,ALL
D,11,UY
D,110,UY
!define distribution loads
*DO, i,1,10
SFBEAM,i,1,PRES,(i-1)*50,i*50
SFBEAM,10+i,1,PRES,500-(i-1)*50,500-i*50
EPLOT
FINISH

59. /SOLVE
SOLVE
FINISH
/POST1
PLDISP,2
ETABLE,IMOMENT,SMISC,6
ETABLE,JMOMENT,SMISC,12
ETABLE,ISHEAR,SMISC,2
ETABLE,JSHEAR,SMISC,8
PRETAB
/TITLE,Shear force diagram
PLLS,ISHEAR,JSHEAR
/TITLE,Bending moment diagram
PLLS,IMOMENT,JMOMENT,-1

60. Example 4. Define tapered beam using BEAM54/44 and BEAM188/189
E=30GPa.

61. ANSYS Command lines (BEAM54)
!example for BEAM54 tapered beam.
!Although Beam54 can using SECTYPE,
!it can't be used in tapered beam.
!the only way is by defining Real constant.
/prep7
t=0.5
d=3
d0=1e-6
L=20
ET,1,BEAM54
MP,EX,1,30e6
MP,NUXY,1,0

62. *DO,i,1,10
xi=20/10*(i-1)
di=xi*3/20
Ai1=t*di
Iz1=xi*t*t*t/12.0
xi=20/10*i
Ai2=t*di
Iz2=xi*t*t*t/12.0
*SET,_RC_SET,i,
R,_RC_SET,Ai1,Iz1,t/2.0,t/2.0
RMODIF,_RC_SET,9,,,
RMODIF,_RC_SET,14,,
RMODIF,_RC_SET,5,Ai2,Iz2,t/2.0,t/2.0
RMODIF,_RC_SET,11,,,
RMODIF,_RC_SET,15, ,
RMODIF,_RC_SET,13, ,
RMODIF,_RC_SET,16, , , ,
*ENDDO
N,1,0,0
*DO,i,1,10
xi=20/10*i
N,i+1,xi,0
*ENDDO
REAL,i
E,i,i+1
/eshape,1
EPLOT
/VIEW, 1 ,1,1,1
/ANG, 1
/REP,FAST

63. ANSYS Command lines (BEAM188)
!example for BEAM188 tapered beam.
/prep7
t=0.5
d=3
d0=1e-6
L=20
ET,1,BEAM188
MP,EX,1,30e6
MP,NUXY,1,0
SECTYPE,1,BEAM,RECT
SECDATA,d0,t
SECTYPE,2,BEAM,RECT
SECDATA,d,t
SECTYPE,3,TAPER
SECDATA,1,0.0,0.0
SECDATA,2,L,0.0
N,1
N,8,L
FILL
N,10,,1,0
NGEN,8,1,10
SECNUM,3
E,1,2,10
EGEN,7,1,1,,,,,,1
/eshape,1
EPLOT

64. Example 5. Simulate pin joint by ANSYS
Analysis frame as following figure. E=1e6, I=1, A=1, h=0.1. All of them are using international system of units. Draw moment diagram.

65. ANSYS Command lines (BEAM3,share nodes)
/PREP7
!* define 2D elastic beam element
ET,1,BEAM3
!* define real constant
R,1,0.005,0.417e-5,0.1, , , ,
!* define material
MPTEMP,1,0
MPDATA,EX,1,,1e6
MPDATA,PRXY,1,,
!* define geometric points
K,1,0,0,,
K,2,1,0,,
K,3,1,1,,
K,4,0,1,,
K,5,1,1,,
!* define geometric lines
LSTR, ,
LSTR, ,
LSTR, ,
!* set element size of geometric lines
LESIZE,ALL,0.1, , , , , , ,1
!* mesh lines
LMESH,ALL
!* define nodes couples
CPINTF,UX,0.0001,
CPINTF,UY,0.0001,
FINISH
/SOL
!* define constraints and load
DK,1, , , ,0,ALL, , , , , ,
DK,2, , , ,0,ALL, , , , , ,
FK,4,FX,10
/STATUS,SOLU
SOLVE
/POST1
!* plot deformation
PLDISP,2
!* plot moment diagram
ETABLE,m1,SMISC, 6
ETABLE,m2,SMISC, 12
PLLS,M1,M2,-2,0

66. ANSYS Command lines (BEAM44, DOF release)
/PREP7
!* define 2D elastic beam element
ET,1,BEAM44
ET,2,BEAM44
keyopt,2,7,1
!* define real constant
R,1,1,1,1,0.1 ,0.1 , ,
!* define material
MPTEMP,1,0
MPDATA,EX,1,,1e6
MPDATA,PRXY,1,,
!* define geometric points
K,1,0,0,,
K,2,1,0,,
K,3,1,1,,
K,4,0,1,,
K,5,1,0.9,,
!* define geometric lines
LSTR, ,
LSTR, ,
LSTR, ,
LSTR, ,
!* set element size of geometric lines
LESIZE,ALL,0.1, , , , , , ,1
LSEL,S, , ,
latt,1,1,2
lsel,inve,,,all
latt,1,1,1
allsel,all
!* mesh lines
LMESH,ALL
FINISH

67. /SOL
!* define constraints and load
DK,1, , , ,0,ALL, , , , , ,
DK,2, , , ,0,ALL, , , , , ,
FK,4,FX,10
/STATUS,SOLU
SOLVE
FINISH
/POST1
!* plot deformation
PLDISP,2
!* plot moment diagram
ETABLE,m1,SMISC, 6
ETABLE,m2,SMISC, 12
PLLS,M1,M2,-2,0

68. ANSYS Command lines (BEAM189 , DOF release)
/PREP7
!* define 2D elastic beam element
ET,1,BEAM189
SECTYPE,1,BEAM,RECT
SECOFFSET,CENT
SECDATA,0.125,0.06
!* define real constant
!* define material
MPTEMP,1,0
MPDATA,EX,1,,1e6
MPDATA,PRXY,1,,
!* define geometric points
K,1,0,0,,
K,2,1,0,,
K,3,1,1,,
K,4,0,1,,
!* define geometric lines
LSTR, ,
LSTR, ,
LSTR, ,
!* set element size of geometric lines
LESIZE,ALL,0.1, , , , , , ,1
!* mesh lines
LMESH,ALL
FLST,5,2,2,ORDE,2
FITEM,5,20
FITEM,5,30
ESEL,S, , ,P51X
NSEL,S, , ,
endrelease,,-1,rotz
allsel,all
FINISH

69. /SOL
!* define constraints and load
DK,1, , , ,0,ALL, , , , , ,
DK,2, , , ,0,ALL, , , , , ,
FK,4,FX,10
/STATUS,SOLU
SOLVE
FINISH
/POST1
!* plot deformation
PLDISP,2
!* plot moment diagram
ETABLE,m1,SMISC, 3
ETABLE,m2,SMISC, 16
PLLS,M1,M2,-2,0