[email protected] , 上海交通大学

2020年

万德成[email protected] , http://dcwan.sjtu.edu.cn/

上海交通大学船舶海洋与建筑工程学院海洋工程国家重点实验室

Class-3

NA26018

Finite Element Analysis of

Solids and Fluids

NA26018 Finite Element Analysis of Solids and Fluids

Timoshenko beam element

Governing EquationsRecall that the Euler-Bernoulli beam theory is based on the assumption that plane cross sections remain plane and normal to the longitudinal axis after bending This assumption results in zero transverse shear strain. When the normality assumption is not used, i.e.，plane sections remain plane but not necessarily normal to the longitudinal axis after deformation, the transverse shear strain 𝛾𝑥𝑧 = 2휀𝑥𝑧 is not zero. Therefore, the rotation of a transverse normal plane about the y-axis is not equal to −𝑑𝑤/𝑑𝑥

Beam theory based on these relaxed assumptions is known as a shear deformation beam theory, most commonly known as the Timoshenko beam theory



We denote the rotation about the y-axis by an independent function 𝛹(𝑥)

Kinematics of the Euler-Bernoulli beam theory

Kinematics of the Timoshenko beam theory



The equilibrium equations of the Timoshenko beam theory are the same as those of the Euler-Bernoulli beam theory, but the kinematic relations are different

Thus, the governing equations in terms of the deflection 𝑤 and rotation 𝛹(𝑥) become

where 𝐺 is the shear modulus and 𝐾, is the shear correction coefficient, which is introduced to account for the difference in the constant state of shear stress in this theory and the parabolic variation of the shear stress predicted by the elasticity theory through the beam thickness

−𝑑

𝑑𝑥𝐺𝐴𝐾𝑠 𝛹 +

𝑑𝑤

𝑑𝑥+ 𝑐𝑓𝑤 = 𝑞

−𝑑

𝑑𝑥𝐸𝐼

𝑑𝛹

𝑑𝑥+ 𝐺𝐴𝐾𝑠 𝛹 +

𝑑𝑤

𝑑𝑥= 0

−𝑑𝑉

𝑑𝑥+ 𝑐𝑓𝑤 = 𝑞, −

𝑑𝑀

𝑑𝑥+ 𝑉 = 0, 𝑀 = 𝐸𝐼

𝑑𝛹

𝑑𝑥, 𝑉 = 𝐺𝐴𝐾𝑠

𝑑𝑤

𝑑𝑥+ 𝛹



For chubby beams (i. e, length-to-height ratio less than 20), transverse shear strain 𝛹 + 𝑑𝑤/𝑑𝑥 ≠ 0

When the shear strain is zero, 𝛹 + 𝑑𝑤/𝑑𝑥 = 0(i.e, for long slender beam), substituting the second equation into the first, and replacing 𝛹 with 𝑑𝑤/𝑑𝑥, we obtain governing equation of the Euler-Bernoulli beam theory

𝑑2

𝑑𝑥2 𝐸𝐼𝑑2𝑤

𝑑𝑥2 + 𝑐𝑓𝑤 = 𝑞(𝑥)

−𝑑


𝑑𝑤

𝑑𝑥+ 𝑐𝑓𝑤 = 𝑞

−𝑑

𝑑𝑥𝐸𝐼

𝑑𝛹

𝑑𝑥+ 𝐺𝐴𝐾𝑠 𝛹 +

𝑑𝑤

𝑑𝑥= 0



Weak Form

The weak form over an element 𝛺𝑒=(𝑥𝑎, 𝑥𝑏) can be developed using the usual 3-step procedure, as already discussed. At the end of the second step (i. e, after integration by parts) we obtain

The coefficients of the weight functions 𝑣1 and 𝑣2 in the boundary integrals are, respectively.

0 =

𝑥𝑎

𝑥𝑏

𝑑𝑣1


𝑑𝑤

𝑑𝑥+ 𝑐𝑓𝑣1𝑤 − 𝑣1𝑞 𝑑𝑥 − 𝑣1𝐺𝐴𝐾𝑠 𝛹 +

𝑑𝑤

𝑑𝑥𝑥𝑎

𝑥0

0 =

𝑥𝑎

𝑥𝑏

𝑑𝑣2

𝑑𝑥𝐸𝐼

𝑑𝛹

𝑑𝑥+ 𝑣2𝐺𝐴𝐾𝑠 𝛹 +

𝑑𝑤

𝑑𝑥𝑑𝑥 − 𝑣2𝐸𝐼

𝑑𝛹

𝑑𝑥𝑥𝑎

𝑥𝑏

𝐺𝐴𝐾𝑠 𝛹 +𝑑𝑤

𝑑𝑥≡ 𝑉 𝑎𝑛𝑑 𝐸𝐼

𝑑𝛹

𝑑𝑥≡ 𝑀



where 𝑉 is the shear force and 𝑀 is the bending moment; these coefficients constitute the secondary variables of the weak form. The weight functions 𝑣1 and 𝑣2 must have the physical interpretations such that the products 𝑣1𝑉 and 𝑣2𝑀 have the units of work. Clearly, 𝑣1 must be equivalent to (the variation of) the transverse deflection 𝑤, and 𝑣2 must be ec quivalent to (the variation of) the rotation function 𝛹

Hence, the primary variable of the formulation are 𝑤 and 𝛹

𝑣1~𝑤, 𝑣2~𝛹

𝐺𝐴𝐾𝑠 𝛹 +𝑑𝑤

𝑑𝑥≡ 𝑉 𝑎𝑛𝑑 𝐸𝐼

𝑑𝛹

𝑑𝑥≡ 𝑀



Denoting the shear force bending moments at the endpoints of the element by the expressions

we arrive at the final weak statements

𝑄1𝑒 ≡ − 𝐺𝐴𝐾𝑠 𝛹 +

𝑑𝑤

𝑑𝑥|𝑥𝑎

= −𝑉 𝑥𝑎

𝑄2𝑒 ≡ − 𝐸𝐼

𝑑𝛹

𝑑𝑥|𝑥𝑎

= −𝑀 𝑥𝑎

𝑄3𝑒 ≡ 𝐺𝐴𝐾𝑠 𝛹 +

𝑑𝑤

𝑑𝑥|𝑥𝑏

= 𝑉 𝑥𝑏

𝑄4𝑒 ≡ 𝐸𝐼

𝑑𝛹

𝑑𝑥|𝑥𝑏

= 𝑀 𝑥𝑏

0 =

𝑥𝑜

𝑥𝑏

𝐺𝐴𝐾𝑠

𝑑𝑣1

𝑑𝑥𝛹 +

𝑑𝑤

𝑑𝑥+ 𝑐𝑓𝑣1𝑤 − 𝑣1𝑞 𝑑𝑥 − 𝑣1 𝑥𝑎 𝑄1

𝑒 − 𝑣1 𝑥𝑏 𝑄3𝑒

0 =

𝑥𝑎

𝑥𝑏

𝐸𝐼𝑑𝑣2

𝑑𝑥

𝑑𝛹

𝑑𝑥+ 𝐺𝐴𝐾𝑠𝑣2 𝛹 +

𝑑𝑤

𝑑𝑥𝑑𝑥 − 𝑣2 𝑥𝑎 𝑄2




We note that 𝑄𝑖𝑒 have the same meaning as well as sense as in

the Euler-Bernoulli beam. We can identify the bilinear and linear forms from the weak forms as

Weak form are equivalent to the statement of the principle of virtual displacements for the Timoshenko beam theory. The total potential energy functional of the isolated beam finite element is given by

𝐵 )𝑣1, 𝑣2 , (𝑤,𝛹 =

𝑥𝑜

𝑥𝑏

𝐺𝐴𝐾𝑠

𝑑𝑣1

𝑑𝑥+ 𝑣2 𝛹 +

𝑑𝑤

𝑑𝑥+ 𝐸𝐼

𝑑𝑣2

𝑑𝑥

𝑑𝛹

𝑑𝑥𝑑𝑥

𝑙 𝑣1, 𝑣2 = 𝑥𝑎

𝑥𝑏

𝑣1 𝑞𝑑𝑥 + 𝑣1 𝑥𝑎 𝑄1𝑒 + 𝑣2 𝑥𝑎 𝑄2

𝑒 + 𝑣1 𝑥𝑏 𝑄3𝑒 + 𝑣2 𝑥𝑏 𝑄4

𝑒

𝛱𝑒(𝑤,𝛹) =

𝑥𝑒

𝑥𝑟−1

𝐸𝐼

2

𝑑𝛹

𝑑𝑥

2

+𝐺𝐴𝐾𝑠

2

𝑑𝑤

𝑑𝑥+ 𝛹

2

+1

2𝑐𝑓𝑤

2 − 𝑤𝑞 𝑑𝑥

− 𝑤 𝑥𝑎 𝑄1𝑒 − 𝛹 𝑥𝑎 𝑄2

𝑒 − 𝑤 𝑥𝑏 𝑄3𝑒 − 𝛹 𝑥𝑏 𝑄4

𝑒



The first term in the square brackets represents the elastic strain energy due to bending

The second term represents the elastic energy due to the transverse shear deformation

The third is the strain energy stored in the elastic foundation, The fourth is the work done by the distributed load

The remaining terms account for the work done by the generalized forces in moving through the respective generalized displacements (𝑤, 𝛹) at the ends of the element

Once again, the principle of minimum total potential energy, 𝛿Π = 0. gives the weak forms

𝒆(𝑤,𝛹) =

𝑥𝑒

𝑥𝑟−1

𝐸𝐼

2

𝑑𝛹

𝑑𝑥

2

+𝐺𝐴𝐾𝑠

2

𝑑𝑤

𝑑𝑥+ 𝛹

2

+1

2𝑐𝑓𝑤

2 − 𝑤𝑞 𝑑𝑥

− 𝑤 𝑥𝑎 𝑄1𝑒 − 𝛹 𝑥𝑎 𝑄2

𝑒 − 𝑤 𝑥𝑏 𝑄3𝑒 − 𝛹 𝑥𝑏 𝑄4

𝑒



General Finite Element Model

A close examination of the terms shows that both 𝑤 and 𝛹 are differentiated only once with respect to 𝑥. Since the primary variables are the dependent unknowns themselves (and do not include their derivatives), the Lagrange interpolation of both 𝑤and 𝛹 is admissible here. The minimum admissible degree of interpolation is linear, so that 𝑑𝑤/𝑑𝑥 ≠ 0 and 𝑑𝛹/𝑑𝑥 ≠ 0. The variables 𝑤 and 𝛹 do not have the same physical units they can be interpolated, in general, with different degrees of interpolation

0 =

𝑥𝑜

𝑥𝑏

𝐺𝐴𝐾𝑠

𝑑𝑣1

𝑑𝑥𝛹 +

𝑑𝑤



0 =

𝑥𝑎

𝑥𝑏

𝐸𝐼𝑑𝑣2

𝑑𝑥

𝑑𝛹


𝑑𝑤





Let us consider Lagrange interpolation of 𝑤 and 𝜓 in the form

where 𝜓𝑗1

, 𝜓𝑗𝟐

are the Lagrange interpolation functions of

degree m-1 and n-1 respectively. In general, m and n are independent of each other, although m=n is most common. However, when m=n=2 (i.e, linear interpolation of both w and𝜓 is used), the derivative of 𝑤 is

which is elementwise-constant. The rotation 𝜓, being linear, is not consistent with that predicted by 𝑤(𝑥). For thin beams, the transverse shear deformation is negligible, and we must have𝜓 = −𝑑𝑤/𝑑𝑥, which requires

𝛿1𝑒𝑥𝑏 − 𝑥

ℎ𝑒+ 𝑆2

𝑒𝑥 − 𝑥𝑎

ℎ𝑒= −

𝑤2𝑒 − 𝑤1

𝑒

ℎ𝑒

𝑑𝑤

𝑑𝑥

𝑒

=𝑤2

𝑒 − 𝑤1𝑒

ℎ𝑒

𝑤 =

𝑗=1

𝑚

𝑤𝑗 𝜓𝑗1, 𝛹 =

𝑗=1

𝑛

𝑆𝑗 𝜓𝑗2



or, equivalently (by equating like coefficients on both sides)

which in turn requires

This implies that 𝜓(𝑥) is a constant, i.e., 𝑆1𝑒= 𝑆2

𝑒= 𝑆𝑒

𝛹(𝑥) = 𝑆1𝑒𝑥𝑏 − 𝑥

ℎ𝑒+ 𝑆2


ℎ𝑒= 𝑆𝑒

𝑆1𝑒 = 𝑆2

𝑒 = −𝑤2

𝑒 − 𝑤1𝑒

ℎ𝑒

𝑆1𝑒𝑥𝑏 − 𝑆2

𝑒𝑥𝑎 = − 𝑤2𝑒 − 𝑤1

𝑒 , 𝑆2𝑒 − 𝑆1

𝑒 = 0



However, a constant state of 𝜓(𝑥) is not admissible because the bending energy of the element.

would be zero. This numerical problem is known in the finite element literature as shear locking. To overcome shear locking, two alternative procedures have been used in the literature

Consistent interpolation. Use an approximation of 𝑤 and 𝜓such that 𝑑𝑤/𝑑𝑥 and 𝑤 are polynomials of the same degree (i.e, m=n+1)

Reduced integration. Use equal interpolation (i.e, m=n) for 𝑤 and 𝜓 and evaluate the bending stiffness coefficients using a numerical integration rule consistent with the actual interpolation of 𝜓. However, the stiffness coefficients associated with the shear energy

𝑥𝑎

𝑥𝑏 𝐺𝐴𝐾𝑠

2

𝑑𝑤

𝑑𝑥+ 𝛹

2

𝑑𝑥

𝑥𝑎

𝑥𝑏 𝐸𝐼

2

𝑑𝜓

𝑑𝑥

2

𝑑𝑥



must be evaluated using a numerical integration rule that treats𝜓 as if it is the same order polynomial as 𝑑𝑤/𝑑𝑥. Thus, if w and 𝜓 are approximated with linear polynomials, 𝑑𝑤/𝑑𝑥 is a constant and 𝜓 is linear. In evaluating the stiffness terms coming from the shear energy, we must use one-point integration, as dictated by 𝑑𝑤/𝑑𝑥 and not 𝜓. Note that one-point integration in this case is sufficient to evaluate the bending energy exactly but not the shear energy because it is quadratic in 𝜓 . Thus, it amounts to underintegrating the term. This is known as the reduced integration technique

𝑥𝑎

𝑥𝑏 𝐺𝐴𝐾𝑠

2

𝑑𝑤

𝑑𝑥+ 𝛹

2

𝑑𝑥



For illustrative purposes, we take a detailed look at the expression

where 𝑥=𝑥𝑎+1/2ℎ𝑒 is the midpoint of the element and he is its length. Substituting

into this expression (with m=n=2) and equating to zero for thin beams, we obtain

𝐺𝐴𝐾𝑠

2

𝑥𝑎

𝑥𝑏

𝑑𝑤

𝑑𝑥+ 𝛹

2

𝑑𝑥 =𝐺𝐴𝐾𝑠

2

𝑑𝑤

𝑑𝑥+ 𝛹

2

|𝑥=𝑥𝑜+ ℎ𝑐 2 ℎ𝑒

𝑤 =

𝑗=1

𝑚


𝑗=1

𝑛

𝑆𝑗 𝜓𝑗2

𝐺𝐴𝐾𝑠ℎ𝑒

2

𝑤2𝑒 − 𝑤1

𝑒

ℎ𝑒+ 𝑆1

𝑒𝑥𝑏 − 𝑥

ℎ𝑒+ 𝑆2


ℎ𝑒

2

|𝑥=𝑥𝑎+ ℎ𝑒 2

=𝐺𝐴𝐾𝑠ℎ𝑒

2

𝑤2𝑒 − 𝑤1

𝑒

ℎ𝑒+

𝑆1𝑒 + 𝑆2

𝑒

2

2

= 0



(1) is a weaker requirement than (2)， i.e.， if (2) holds, then (1) also holds but (1) does not imply (2)

Note that (2) must hold only for problems for which the transverse shear energy is negligible

（1）

（2）

In summary, we use either consistent interpolation (m=n+1) or equal interpolat with reduced integration in the evaluation of the transverse shear stiffness coefficients in the Timoshenko beam element

𝐺𝐴𝐾𝑠ℎ𝑒

2

𝑤2𝑒 − 𝑤1

𝑒

ℎ𝑒+

𝑆1𝑒 + 𝑆2

𝑒

2

2

= 0

𝑆1𝑒 = 𝑆2

𝑒 = −𝑤2

𝑒 − 𝑤1𝑒

ℎ𝑒



0 =

𝑗=1

𝑚

𝐾𝑖𝑗11 𝑤𝑗 +

𝑗=1

𝑛

𝐾𝑖𝑗12 𝑠𝑗 − 𝐹𝑖

1 (𝑖 = 1,2, … ,𝑚)

0 =

𝑗=1

𝑚


𝑗=1

𝑛


2 (𝑖 = 1,2, … , 𝑛)

𝑤 =

𝑗=1

𝑚


𝑗=1

𝑛

𝑆𝑗 𝜓𝑗2

0 =

𝑥𝑜

𝑥𝑏

𝐺𝐴𝐾𝑠

𝑑𝑣1

𝑑𝑥𝛹 +

𝑑𝑤



0 =

𝑥𝑎

𝑥𝑏

𝐸𝐼𝑑𝑣2

𝑑𝑥

𝑑𝛹


𝑑𝑤





0 =

𝑗=1

𝑚


𝑗=1

𝑛


1 (𝑖 = 1,2, … ,𝑚)

0 =

𝑗=1

𝑚


𝑗=1

𝑛


2 (𝑖 = 1,2, … , 𝑛)

𝐾𝑖𝑗11 =

𝑥𝑎

𝑥𝑏

𝐺𝐴𝐾𝑠

𝑑𝜓𝑖1

𝑑𝑥

𝑑𝜓𝑗1

𝑑𝑥+ 𝑐𝑓𝜓𝑖

1𝜓𝑗

1𝑑𝑥

𝐾𝑖𝑗12 =

𝑥𝑎

𝑥𝑏

𝐺 𝐴𝐾𝑠

𝑑𝜓𝑖1

𝑑𝑥𝜓𝑗

2𝑑𝑥 = 𝐾𝑗𝑖

21 i.e. , 𝐾21 = 𝐾12 𝑇

𝐾𝑖𝑗22 =

𝑥𝑎

𝑥𝑏

𝐸𝐼𝑑𝜓𝑖

2

𝑑𝑥

𝑑𝜓𝑗2

𝑑𝑥+ 𝐺𝐴𝐾𝑠𝜓𝑖

2𝜓𝑗

2𝑑𝑥

𝐹𝑖1 =

𝑥𝑎

𝑥𝑏

𝑞 𝜓𝑖1𝑑𝑥 + 𝑄2𝑖−1, 𝐹𝑖

2 = 𝑄2𝑖



In the interest of clarity, the element label e on the quantities is omitted. Equations can be written in matrix form as

The finite element model with the coefficients 𝐾𝑖𝑗 is the most general displacement finite element model of the Timoshenko beam theory. It can be used to obtain a number of special finite element models, as discussed next

𝐾11 𝐾12

𝐾21 𝐾22

𝑤

𝑠=

𝐹′

𝐹2

0 =

𝑗=1

𝑚


𝑗=1

𝑛


1 (𝑖 = 1,2, … ,𝑚)

0 =

𝑗=1

𝑚


𝑗=1

𝑛


2 (𝑖 = 1,2, … , 𝑛)



Consistent Interpolation Elements

The first consistent interpolation element (CIE) that we will consider is the one in which quadratic interpolation is used for w and linear interpolation for 𝜓 so that 𝑑𝑤/𝑑𝑥 and 𝜓 are of the same order polynomial (and hence, no shear locking occurs)

That is, we select 𝜓𝒊𝟏 to be quadratic polynomials and 𝜓𝒊

𝟐 to be linear polynomials.

For this choice of interpolation, 𝐾11 is 3x3, 𝐾1𝟐 is 3x2, and 𝐾𝟐𝟐

is 2x2.

The explicit forms of the matrices, when 𝐸𝐼 and 𝐺𝐴𝐾𝑠 are constant



when 𝑐𝒇 = 0, the finite element equations for this choice of interpolation are given by

𝐾11 =𝐺𝑒𝐴𝑒𝐾𝑠

3ℎ𝑒

7 −8 1−8 16 −81 −8 7

+𝑐𝑓𝑒ℎ𝑒

30

4 2 −12 16 2−1 2 4


6

−5 −14 −41 5

= 𝐾21 𝑇

𝐾22 =𝐸𝐼

ℎ𝑒

1 −1−1 1

+𝐺𝑒𝐴𝑒𝐾𝑠ℎ𝑒

62 11 2



(𝑄1𝑒, 𝑄𝟐

𝑒, 𝑄𝟑𝑒, 𝑄𝟒

𝑒) are the generalized forces, 𝑤c𝑒 and 𝑄𝑐

𝑒 are the deflection and applied external load, respectively, at the center node of the quadratic element and

This element is designated as CIE-1

𝐺𝐴𝐾𝑠

6ℎ𝑒

14 −16 2 −5ℎ𝑒 −ℎ𝑒

−16 32 −16 4ℎ𝑒 −4ℎ𝑒

2 −16 14 ℎ𝑒 5ℎ𝑒

−5ℎ𝑒 4ℎ𝑒 ℎ𝑒 2ℎ𝑒2

𝑒ℎ𝑒2𝛩𝑒

−ℎ𝑒 −4ℎ𝑒 5ℎ𝑒 ℎ𝑒2𝛩𝑒 2ℎ𝑒

2 𝑒

𝑤1𝑒

𝑤𝑐𝑒

𝑤2𝑒

𝑆1𝑒

𝑆2𝑒

=

𝑞1𝑒

𝑞𝑐𝑒

𝑞2𝑒

00

+

𝑄1𝑒

𝑄𝑐𝑒

𝑄3𝑒

𝑄2𝑒

𝑄4𝑒

𝛬𝑒 =𝐸𝑒𝐼𝑒

𝐺𝑒𝐴𝑒𝐾𝑠ℎ𝑒2 , 𝜇𝑒 = 12𝛬𝑒 , 𝛩𝑒 = 1 − 6𝛬𝑒 , 𝑒 = 1 + 3𝛬𝑒

𝑞𝑖𝑒 =

𝑥𝑎

𝑥𝑏

𝜓𝑖1

𝑞𝑑𝑥, 𝑖 = 1,2, 𝑐 , 𝜓𝑖1

= quadratic



Note that node c, which is the center node of the element, is not connected to other elements, and the only degree of freedom there is the transverse deflection

Thus, there are different number of degrees of freedom at different nodes of the element, and this therefore complicates the assembly of elements and its implementation on a computer



Hence, we eliminate the node c dependence in the system of element equations by condensing out 𝑤𝑐

𝑒. The second equation can be used to express wec in terms of 𝑤1

𝑒, 𝑤𝟐𝑒, 𝑆1

𝑒, 𝑆2𝑒, 𝑞𝑐

𝑒 and 𝑄𝑐𝑒

Substituting for wec into the remaining equations, eliminate wec and rearranging the equations, we obtain

𝑤𝑐𝑒 =

6ℎ𝑒

32𝐺𝑒𝐴𝑒𝐾𝑠𝑞𝑐𝑒 + 𝑄𝑐

𝑒 +𝑤1

𝑒 + 𝑤2𝑒

2+ ℎ𝑒

𝑆2𝑒 − 𝑆1

𝑒

8

𝐺𝐴𝐾𝑠

6ℎ𝑒

14 −16 2 −5ℎ𝑒 −ℎ𝑒

−16 32 −16 4ℎ𝑒 −4ℎ𝑒

2 −16 14 ℎ𝑒 5ℎ𝑒




2 𝑒

𝑤1𝑒

𝑤𝑐𝑒

𝑤2𝑒

𝑆1𝑒

𝑆2𝑒

=

𝑞1𝑒

𝑞𝑐𝑒

𝑞2𝑒

00

+

𝑄1𝑒

𝑄𝑐𝑒

𝑄3𝑒

𝑄2𝑒

𝑄4𝑒



2𝐸𝑒𝐼𝑒

𝜇0ℎ𝑒3

6 −3ℎ𝑒 −6 −3ℎ𝑒

−3ℎ𝑒 ℎ𝑒2 1.5 + 6𝛬𝑒 3ℎ𝑒 ℎ𝑒

2 1.5 − 6𝛬𝑒

−6 3ℎ𝑒 6 3ℎ𝑒

−3ℎ𝑒 ℎ𝑒2 1.5 − 6𝛬𝑒 3ℎ𝑒 ℎ𝑒

2 1.5 − 6𝛬𝑒

𝑤1𝑒

𝑆1𝑒

𝑤2𝑒

𝑆2𝑒

=

𝑞1𝑒 +

1

2 𝑞𝑐𝑒

−1

8 𝑞𝑐𝑒ℎ𝑒

𝑞2𝑒 +

1

2 𝑞𝑐𝑒

1


+

𝑄1𝑒

𝑄2𝑒

𝑄3𝑒

𝑄4𝑒

𝑞𝑐𝑒 = 𝑞𝑐

𝑒 + 𝑄𝑐𝑒


𝑒 + 𝑄𝑐𝑒

𝐺𝐴𝐾𝑠

6ℎ𝑒

14 −16 2 −5ℎ𝑒 −ℎ𝑒

−16 32 −16 4ℎ𝑒 −4ℎ𝑒

2 −16 14 ℎ𝑒 5ℎ𝑒




2 𝑒

𝑤1𝑒

𝑤𝑐𝑒

𝑤2𝑒

𝑆1𝑒

𝑆2𝑒

=

𝑞1𝑒

𝑞𝑐𝑒

𝑞2𝑒

00

+

𝑄1𝑒

𝑄𝑐𝑒

𝑄3𝑒

𝑄2𝑒

𝑄4𝑒

𝑤𝑐𝑒 =

6ℎ𝑒

32𝐺𝑒𝐴𝑒𝐾𝑠𝑞𝑐𝑒 + 𝑄𝑐

𝑒 +𝑤1

𝑒 + 𝑤2𝑒

2+ ℎ𝑒

𝑆2𝑒 − 𝑆1

𝑒

8



For simplicity, but without loss of generality, we will assume

that 𝑄𝑐𝑒 = 𝟎 (i.e, no external point force is placed at the center

of the element) so that 𝑞𝑐𝑒 = 𝑞𝑐

𝑒. Note that the load vector is equivalent to that of the Euler-Bernoulli beam element. Thus, the CIE-I element, for all analysis steps, one must keep in mind that 𝑤 and 𝑞𝒊

𝑒 are determined by quadratic interpolations functions

2𝐸𝑒𝐼𝑒

𝜇0ℎ𝑒3

6 −3ℎ𝑒 −6 −3ℎ𝑒


2 1.5 − 6𝛬𝑒

−6 3ℎ𝑒 6 3ℎ𝑒


2 1.5 − 6𝛬𝑒

𝑤1𝑒

𝑆1𝑒

𝑤2𝑒

𝑆2𝑒

=

𝑞1𝑒 +

1

2 𝑞𝑐𝑒

−1


𝑞2𝑒 +

1

2 𝑞𝑐𝑒

1


+

𝑄1𝑒

𝑄2𝑒

𝑄3𝑒

𝑄4𝑒


𝑒 + 𝑄𝑐𝑒



The second consistent interpolation element, denoted CIE-2, is based on Lagrange cubic interpolation of 𝑤(𝑥) and quadratic interpolation of 𝛹(𝑥)

This element leads to 7X7 element stiffness matrix with seven degrees of freedom (𝑤1

𝑒, 𝑤𝟐𝑒, 𝑤𝟑

𝑒, 𝑤𝟒𝑒, 𝑆1

𝑒, 𝑆𝟐𝑒, 𝑆𝟑

𝑒) per element. Elimination of the internal nodal degrees of freedom will result in a 4x4 matrix

Alternatively, the same element can be derived directly by assuming Hermite cubic interpolation of 𝑤(𝑥) and a dependent quadratic interpolation of 𝛹(𝑥)

This approach leads to an element that yields, as a special case, the Euler-Bernoulli beam element



The exact solution for the homogeneous case (e,g 𝑐𝑓=0) is

This suggests that one may use cubic approximation of 𝑤(𝑥) and an interdependent quadratic approximation of 𝛹(𝑥).The resulting finite element model is termed an interdependent interpolation element（IIE）

Since the solutions are valid also for a typical finite element (replace 𝑐𝑖with 𝑐𝑖

𝑒), we proceed to express cei in terms of the nodal values of 𝑤 and 𝛹

The resulting four relations among（𝑤1𝑒, 𝑤2

𝑒, 𝑆1𝑒, 𝑆2

𝑒) and (𝑐1𝑒,

𝑐2𝑒, 𝑐3

𝑒, 𝑐4𝑒), when inverted and back, give

𝑤 𝑥 = −1

𝐸𝐼𝑐1

𝑥3

6+ 𝑐2

𝑥2

2+ 𝑐3𝑥 + 𝑐4 +

1

𝐺𝐴𝐾𝑠𝑐1𝑥

𝐸𝐼𝑌 𝑥 = 𝑐1𝑥2

2+ 𝑐2𝑥 + 𝑐3



The approximation functions:

Here 𝑥 is the nondimensional local coordinate

𝜙1𝑒 =

1

𝜇𝑒𝜇𝑒 − 12𝛬𝑒 𝑥 − 3 − 2 𝑥 𝑥2

𝜙2𝑒 = −

ℎ𝑒

𝜇𝑒1 − 𝑥 2 𝑥 + 6𝛬𝑒 1 − 𝑥 𝑥

𝜙3𝑒 =

1

𝜇𝑒3 − 2 𝑥 𝑥2 + 12𝛬𝑒 𝑥

𝜙4𝑒 =

ℎ𝑒

𝜇𝑒1 − 𝑥 𝑥2 + 6𝛬𝑒 1 − 𝑥 𝑥

𝑤 𝑥 ≈ 𝑤ℎ𝑒 𝑥 =

𝑗=1

𝑚

𝜙𝑗𝑒 𝛥𝑗

𝑒 , 𝛹 𝑥 ≈ 𝛹ℎ𝑒 𝑥 =

𝑗=1

𝑛

𝜑𝑗𝑒 𝛥𝑗

𝑒

𝛥1𝑒 = 𝑤1

𝑒 , 𝛥2𝑒 = 𝑆1

𝑒 , 𝛥3𝑒 = 𝑤2

𝑒 , 𝛥4𝑒 = 𝑆2

𝑒

𝜑1𝑒 =

6

ℎ𝑒𝜇𝑒1 − 𝑥 𝑥

𝜑2𝑒 =

1

𝜇𝑒𝜇𝑒 − 4 𝑥 + 3 𝑥2 − 12𝛬𝑒 𝑥

𝜑3𝑒 = −

6

ℎ𝑒𝜇𝑒1 − 𝑥 𝑥

𝜑4𝑒 =

1

𝜇𝑒3 𝑥2 − 2 𝑥 + 12𝛬𝑒 𝑥

𝑥 =𝑥 − 𝑥𝑎

ℎ𝑒, 𝜇𝑒 = 1 + 12𝛬𝑒 , 𝛬𝑒 =

𝐸𝑒𝐼𝑒

𝐺𝑒𝐴𝑒𝐾𝑠ℎ𝑒2



Substitution into the total potential energy functional and differentiating it with respect to 𝛥𝑖

𝑒, yields the finite element model

𝑄𝑗𝑒 have the same meaning as before

𝐾𝑖𝑗𝑒 =

𝑥𝑎

𝑥𝑏

𝐸𝑒𝐼𝑒𝑑𝜑𝑖

𝑒

𝑑𝑥

𝑑𝜑𝑗𝑒

𝑑𝑥+ 𝐺𝑒𝐴𝑒𝐾𝑠 𝜑𝑖

𝑒 +𝑑 𝜙𝑖

𝑒

𝑑𝑥𝜑𝑗

𝑒 +𝑑 𝜙𝑗

𝑒

𝑑𝑥𝑑𝑥

𝑞𝑖𝑒 =

𝑥𝑎

𝑥𝑏

𝜙𝑖𝑒𝑞 𝑥 𝑑𝑥

𝐾𝑒 𝛥𝑒 = 𝑞𝑒 + 𝑄𝑒

𝑤 𝑥 ≈ 𝑤ℎ𝑒 𝑥 =

𝑗=1

𝑚

𝜙𝑗𝑒 𝛥𝑗

𝑒 , 𝛹 𝑥 ≈ 𝛹ℎ𝑒 𝑥 =

𝑗=1

𝑛

𝜑𝑗𝑒 𝛥𝑗

𝑒

𝛥1𝑒 = 𝑤1

𝑒 , 𝛥2𝑒 = 𝑆1

𝑒 , 𝛥3𝑒 = 𝑤2

𝑒 , 𝛥4𝑒 = 𝑆2

𝑒



For uniformly distributed load, 𝑞(𝑥) = 𝑞0, the load vector yields

which is exactly the same as that in the Euler-Bernoulli beam element (EBE). For nonuniform loads the load vectors of the two elements are different

𝑞𝑒 =𝑞0ℎ𝑒

12

6−ℎ𝑒

12ℎ𝑒

2𝐸𝑒𝐼𝑒

𝜇0ℎ𝑒3

6 −3ℎ𝑒 −6 −3ℎ𝑒

−3ℎ𝑒 2ℎ𝑒2

𝑒3ℎ𝑒 ℎ𝑒

2𝛩𝑒

−6 3ℎ𝑒 6 3ℎ𝑒

−3ℎ𝑒 ℎ𝑒2𝛩𝑒 3ℎ𝑒 2ℎ𝑒

2 𝑒

𝑤1𝑒

𝑆1𝑒

𝑤2𝑒

𝑆2𝑒

=

𝑞1𝑒

𝑞2𝑒

𝑞3𝑒

𝑞4𝑒

+

𝑄1𝑒

𝑄2𝑒

𝑄3𝑒

𝑄4𝑒


𝐺𝑒𝐴𝑒𝐾𝑠ℎ𝑒2 , 𝜇𝑒 = 1 + 12𝛬𝑒 , 𝛩𝑒 = 1 − 6𝛬𝑒 ,

𝑒= 1 + 3𝛬𝑒

𝐾𝑒 𝛥𝑒 = 𝑞𝑒 + 𝑄𝑒



Reduced Integration Element

When equal interpolation of 𝑤(𝑥) and 𝛹(𝑥) is used (m=n), all submatrices are of the same order (nXn), where n is the number of terms in the polynomial (or n-1 is the degree of interpolation)

The element coefficient matrices 𝐾𝑖𝑗11, 𝐾𝑖𝑗

12 as well as the first

part of 𝐾𝑖𝑗22 must be evaluated exactly. The second part of 𝐾𝑖𝑗

11

is to be evaluated using reduced integration For the choice of linear interpolation functions, and for

elementwise constant values of 𝐺𝐴𝐾s and 𝐸𝐼, the matrices

have the following explicit values (when 𝑐𝑓𝑒 = 0)

𝐾11 𝐾12

𝐾21 𝐾22

𝑤

𝑠=

𝐹1

𝐹2



where one-point integration is used to evaluate the second part of 𝐾22. Note that 𝐾11, 𝐾12 and the first part of 𝐾22 can be evaluated exactly with one-point quadrature (i.e, numerical integration) when 𝐸𝐼 and 𝐺𝐴𝐾s, are constant because the integrands of these coefficients are constant. Hence, one-point integration for Kab satisfies all the requirements.

The resulting beam element is termed the reduced integration element (RIE)


ℎ𝑒

1 −1−1 1

, 𝐾12 =𝐺𝑒𝐴𝑒𝐾𝑠

2−1 −11 1

𝐾22 =𝐸𝑒𝐼𝑒ℎ𝑒

1 −1−1 1

+𝐺𝑒𝐴𝑒𝐾𝑠ℎ𝑒

ℎ𝑒

1 11 1



The element equations of the RIE are

or, in alternative form（to resemble those of the EBE）

𝜓 are the linear interpolation functions

𝐺𝑒𝐴𝑒𝐾𝑠

4ℎ𝑒

4 −2ℎ𝑒 −4 −2ℎ𝑒

−2ℎ𝑒 ℎ𝑒2 + 𝑎𝑒 2ℎ𝑒 ℎ𝑒

2 − 𝑎𝑒

−4 2ℎ𝑒 4 2ℎ𝑒

−2ℎ𝑒 ℎe2 − 𝑎𝑒 2ℎ𝑒 ℎ𝑒

2 + 𝑎𝑒

𝑤1𝑒

𝑆1𝑒

𝑤2𝑒

𝑆2𝑒

=

𝑄1𝑒

𝑄2𝑒

𝑄3𝑒

𝑄4𝑒

+

𝑞1𝑒

0𝑞2𝑒

0

𝑎𝑒 = 4𝐸𝑒 𝐼𝑒 𝐺𝑒 𝐴𝑒𝐾𝑠

2𝐸𝑒𝐼𝑒

𝜇0ℎ𝑒3

6 −3ℎ𝑒 −6 −3ℎ𝑒


2 1.5 − 6𝛬𝑒

−6 3ℎ𝑒 6 3ℎ𝑒


2 1.5 − 6𝛬𝑒

𝑤1𝑒

𝑆1𝑒

𝑤2𝑒

𝑆2𝑒

=

𝑄1𝑒

𝑄2𝑒

𝑄3𝑒

𝑄4𝑒

+

𝑞1𝑒

0𝑞2𝑒

0

𝑞𝑖𝑒 =

𝑥𝑎

𝑥𝑏

𝜓𝑖𝑒 𝑞𝑑𝑥, (𝑖 = 1,2)


𝐺𝑒𝐴𝑒𝐾𝑠ℎ𝑒2 , 𝜇0 = 12𝛬𝑒



Discussion:

It is interesting to note that the element stiffness matrix of the linear RIE is the same as that of the CIE-1 obtained using quadratic approximation of w and linear approximation of 𝜓

The only difference is the load representation. In CIE-1, the load vector is equivalent to that of the Euler-Bernoulli beam theory, whereas in RIE it is based on integration, which contributes only to the force degrees of freedom and not to the moment degrees of freedom

The quadratic interpolation of both 𝑤 and 𝜓 with full integration of the element coefficient matrices also suffers slightly from the shear-locking phenomenon. As the degree of approximation and/or the number of elements in the mesh is increased shear locking will disappear and reduced integration is not necessary

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