International Journal of Scientific & Engineering Research, Volume 6, Issue 1, January-2015 93

ISSN 2229-5518

Modal Analysis of Small Frames Using

High Order Timoshenko Beams

Chadi Azoury, Assad Kallassy, Pierre Rahme

we model the corners where the horizontal and vertical bar meet with a special matrix. The results of our new model (3-noded Timoshenko beam for the horizontal and vertical bars and a special element for the corners based on the Q4 elements) are very satisfying when performing the modal analysis.

Notations

—————————— ——————————

E | Young’s modulus of elasticity | v | Poisson coefficient |

ρ | density of the material | h | height of the beam cross-section |

b I M | width of the beam cross section moment of inertia of the beam cross section mass matrix of the element | A K dof | area of the beam cross section stiffness matrix of the element degree-of-freedom |

𝐿 ℎ/2

𝑏𝜇 𝐿

ℎ/2

𝑈 = 𝑏 ∫ ∫ 𝜀𝑇 𝐸 𝜀 𝑑𝑑 𝑑𝑑 +

∫ ∫ 𝛾𝑇 𝐺 𝛾 𝑑𝑑 𝑑𝑑 (1)

he Euler-Bernoulli beam element is the most used element

2 0 −ℎ/2

2 0 −ℎ/2

for performing the modal analysis of beams and frames. This type of beam element gives an exact solution for the modal analysis problem given that we have long and slender beam. Whether the beam is clamped, pinned, or free, from any side, meshing it with EB elements will produce excellent results. However, when the beam becomes more and more short, i.e., when the ratio of the width of the beam to its length is > 0.1, the EB beam elements are no longer valid for a modal analysis. We must use Timoshenko beam elements for these cases. The 2-noded Timoshenko beam element is very much used in most software and analyses [1]. In the next paragraph, we will explain how the stiffness and mass matrix of such a beam element are calculated using linear simple shape functions. Following the same procedure, but using high-order shape functions, say quadratic ones, we will formulate the stiffness and mass matrix for this new 3-noded element that

we will call “Timo3” element.

Where, *L *is the length of the element, *b *and *h *are the width and the height of the beam respectively. µ is the correction factor for shear energy; generally taken 5/6 for beams with standard rectangular cross sections and 9/10 for circular section beams [3]. Many formulations of the Timoshenko beam exist, [4], [5], [6], and [7].

The degrees of freedom of this element are:

v1 : transverse displacement of the beam at the left node

θ1 : rotation of the beam section at the left node

v2 : transverse displacement of the beam at the right node

θ2 : rotation of the beam section at the right node

In this model, *v *and θ are independent variables, thus they can be interpolated independently. By using isoparametric linear shape functions for both variables *v *and θ:

1 1

𝑁1 = 2 (1 − 𝜉) 𝑁2 = 2 (1 + 𝜉)

𝑣(𝜉) = [𝑁 (𝜉) 𝑁 (𝜉)][𝑣

𝑣 ]𝑇

In a Timoshenko beam theory, plane sections remain plane after deformation but not necessary perpendicular to the neutral axis. The plane section rotates by an amount, *θ*, equal

1 2 1 2

𝜃(𝜉) = [𝑁1 (𝜉) 𝑁2(𝜉)][𝜃1 𝜃2 ]𝑇

The bending strain is [8]:

to the rotation of the neutral axis, *μ*, minus the shear strain γ*.*

𝑑𝜃

𝑑𝜃 𝑑𝜉

𝑑𝑁1

𝑑𝑁2

𝑑𝜉

The strain energy for an element of length *L *is [2]:

𝜅 = = = � 𝜃1 + 𝜃2 �

𝑑𝑑 𝑑𝜉 𝑑𝑑 𝑑𝜉 𝑑𝜉 𝑑𝑑

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The transverse shear strain is:

0 0 0 0

𝛾 =

𝑑𝑣

𝑑𝑑

− 𝜃 = 𝑑𝑁1

𝑑𝜉

𝑣1 +

𝑑𝑁2

𝑑𝜉

𝑣2�

𝑑𝜉

𝑑𝑑

− [𝑁1 𝜃1 + 𝑁2 𝜃2]

Kb =

0 EI/L 0 -EI/L

0 0 0 0

0 -EI/L 0 EI/L

With

We get

𝑑𝜉 = 𝐿,

𝑑𝑥 2

𝑑𝑁1

𝑑𝜉

= − 1

2

𝑑𝑁2 = 1

𝑑𝜉 2

µGA/L µGA/2 -µGA/L µGA/2

𝜅 = 𝐵𝑏 [𝑣1 𝜃1 𝑣2 𝜃2 ]𝑇

µGA/2 µGAL/4 -µGA/2 µGAL/4

Where 𝐵 = �0

𝐿

element.

0 1 � is the bending matrix of the

𝐿

Ks =

-µGA/L -µGA/2 µGA/L -µGA/2

µGA/2 µGAL/4 -µGA/2 µGAL/4

And 𝛾 = 𝐵𝑠 [𝑣1 𝜃1 𝑣2 𝜃2 ]𝑇

Where 𝐵 = �

𝐿

element?

𝜉−1

2

1 −𝜉−1

𝐿 2

� is the shear strain matrix of the

2 0 1 0

0 0 0 0

The virtual displacement is dv = N.[𝑑𝑣1 𝑑𝜃1 𝑑𝑣2 𝑑𝜃2]𝑇

and the virtual strains are:

𝑑𝜅 = 𝐵𝑏 [𝑑𝑣1 𝑑𝜃1 𝑑𝑣2 𝑑𝜃2 ]𝑇

𝑑𝛾 = 𝐵𝑠 [𝑑𝑣1 𝑑𝜃1 𝑑𝑣2 𝑑𝜃2 ]𝑇

M = ρAL/6

1 0 2 0

0 0 0 0

The bending moment is

M = Db .Bb .[𝑣1 𝜃1 𝑣2 𝜃2 ]𝑇 where Db = E I

And the shear force is

V = Ds .Bs .[𝑣1 𝜃1 𝑣2 𝜃2]𝑇 where Ds = µ G A

Let us consider the Timoshenko beam element with 3 nodes shown in Figure 1.

y

The bending stiffness matrix for the element is computed from:

𝐾𝑏 = � 𝐵𝑇 𝐷 𝐵 𝑑𝑑

Ω

θ 1-γ

θ 2-γ

θ 3-γ

The shear stiffness matrix:

𝑇

The deflection

𝐾𝑠 = � 𝐵𝑠 𝐷𝑠 𝐵𝑠 𝑑𝑑

Ω

And the consistent mass matrix is computed from:

𝑀 = � 𝜌𝜌𝑁𝑇 𝑁𝑑𝑑

v1

1 L/2

Initial

v2 v3

2 L/2 3

Ω

Using natural coordinates,

𝐾 = ∫1 𝐵𝑇 𝐷 𝐵 𝐿 𝑑𝜉

𝑏 −1 𝑏 𝑏 𝑏 2

Figure 1. 3-noded Timoshenko beam element

If we use the same procedure as in the previous paragraph, taking 3 nodes per elements, the new Lagrange quadratic shape functions will be:

1 𝑇 𝐿

𝐾𝑠 = ∫−1 𝐵𝑠 𝐷𝑠 𝐵𝑠

1

𝑀 = � 𝜌𝜌𝑁𝑇 𝑁

𝑑𝜉

2

𝐿

𝑑𝜉

𝑁1

1

= (−𝜉 + 𝜉2 ) 𝑁2

2

= (1 − 𝜉2 ) 𝑁3

1

= (𝜉 + 𝜉2 )

2

−1 2

In order to avoid shear locking, *K*s is obtained using the

reduced integration technique (one order less than required) [9]. Upon integrating, we get:

𝑣(𝜉) = 𝑁1 (𝜉)𝑣1 + 𝑁2 (𝜉)𝑣2 + 𝑁3 (𝜉)𝑣3

𝜃(𝜉) = 𝑁1 (𝜉)𝜃1 + 𝑁2 (𝜉)𝜃2 + 𝑁3 (𝜉)𝜃3

We get

𝜅 = 𝐵𝑏 [𝑣1 𝜃1 𝑣2 𝜃2 𝑣3 𝜃3]𝑇

Where 𝐵𝑏 = �0

2 �𝜉 −

𝐿

1� 0

2

−4𝜉 0

𝐿

2 �𝜉 +

𝐿

1� � is the bending

2

matrix of the element.

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And 𝛾 = 𝐵𝑠 [𝑣1 𝜃1 𝑣2 𝜃2 𝑣3 𝜃3]𝑇

Where

Table 1. Modal analysis of the cantilevered beam for different modelling

𝐵 = 2 �𝜉 − 1

𝐿 2

1 (𝜉 − 𝜉2 )

2

−4𝜉

𝐿

𝜉2 − 1

2 �𝜉 + 1

𝐿 2

1 (−𝜉 − 𝜉2 )�

2

is the shear strain matrix of the element.

Again, matrix *K*b is obtained using the exact integration whereas *K*s is obtained using the reduced integration

technique (one order less than required). Upon integrating, we

Kb

=µGA/(36L)

84 18L -96 24L 12 -6L

18L 4L2 -24L 4L2 6L -2L2

-96 -24L 192 0 -96 24L

24L 4L2 0 16 -24L 4L2

12 6L -96 -24L 84 -18L

-6L -2L2 24L 4L2 -18L 4L2

As we can see, since our beam is short (length = 5 m, width = 1 m), the Timo3 beam element shows better performance even in small number of mesh elements (10).

Let’s test the Timo3 beam element on a small L-frame clamped at its bottom as shown in the next figure:

M = ρAL/30 | 4 | 0 | 2 | 0 | -1 | 0 |

M = ρAL/30 | 0 | 0 | 0 | 0 | 0 | 0 |

M = ρAL/30 | 2 0 | 0 0 | 16 0 | 0 0 | 2 0 | 0 0 |

M = ρAL/30 | -1 | 0 | 2 | 0 | 4 | 0 |

0 | 0 | 0 | 0 | 0 | 0 |

Let’s perform the modal analysis of a clamped-free beam.

h=1m

Figure 3. Small L-frame test

All beams have rectangular sections (2x2 m2).

Poisson coefficient v = 0.3; Density ρ = 7800 Kg/m3; Elastic modulus *E *= 200 GPa.

Table 2 shows the results of the modal analysis of the L-frame

L = 5m

E = 200 GPa, v = 0.3

b=1m

using 2 types of meshing: the first type is the volumetric meshing with H8 elements and the second type is the surface meshing (plane stress condition) with Q4 elements.

Figure 2. Cantilever beam

The results of the modal analysis of this clamped-free beam are listed in Table 1.

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Table 2. Modal analysis of the small L-frame

Let’s model our frame using regular EB beam elements or Timo3 beam elements and compare the results of the modal analysis with the reference model the Q4 model.

Table 3 lists the results of the first six mode shapes with their corresponding natural frequencies for different modeling.

Table 3. Modal analysis of the L-frame using beam elements modeling

Freq

Mode

No

Mode

shape

(Hz)

Q4

EB Error Timo3 Error

1 54 60 11% 56 4%

2 146 151 3% 111 24% 3 294 344 17% 324 10% 4 454 639 41% 418 8%

5 584 853 46% 613 5%

Since the modal analysis of the Q4 model is close enough to the results of the H8 model, we will drop the volumetric H8 meshing and focus our attention on the 2D meshing with Q4 elements. In the next paragraphs, our reference model will be the Q4 model.

Figure 4. Modeling the test L-frame using Timo3 elements

6 656 1089 66% 728 11%

Average 31% 10%

As we can see from the previous table, the Timo3 model gives closer results than the EB model when compared to the results from the reference model the Q4 model. Still, the average error of the first six mode shapes of the Timo3 is 10.3% in comparison to the reference model the Q4 model. That shows that we need to make some further adjustments to our linear model in order to decrease that error. In the next paragraph, we will explain the method used that will reduce the error.

The bars of our test L-frame are modeled using Timo3 beam elements, but the connection that is large in our small L-frame must be given special consideration. We propose a special element for that connection based on the utilization of Q4 surface elements. There are many ways in the finite element method to connect two different element types like in [10], [11], and [12].

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Special element

(64Q4condensed)

Timo3 elements

1 0 0 0 0 0

0 1 -3*a 0 0 0

1 0 0 0 0 0

0 1 -2*a 0 0 0

1 0 0 0 0 0

0 1 -1*a 0 0 0

1 0 0 0 0 0

0 1 0 0 0 0

1 0 0 0 0 0

Figure 5. Special treatment for the connection

(at the corner)

The interface between the new “64Q4condensed” element and the Timo3 elements (on the vertical and the horizontal bars) is considered a rigid link. The corner element of

2 m width and 2 m height is meshed using 64 Q4 elements as shown in the next Figure.

2 m

0 1 1*a 0 0 0

1 0 0 0 0 0

0 1 2*a 0 0 0

1 0 0 0 0 0

0 1 3*a 0 0 0

0 0 0 1 0 4*b

0 1 4*a 0 0 0

64 (Q4) elements

Rigid link

u 5

0 0 0 1 0 3*b

2 m0 0 0 0 1 0

u 6 | u 4 | 0 | 0 | 0 | 1 | 0 | 2*b | ||||

0 | 0 | 0 | 0 | 1 | 0 | ||||||

u 2 | 0 | 0 | 0 | 1 | 0 | 1*b | |||||

u 3 | Rigid link | 0 | 0 | 0 | 0 | 1 | 0 | ||||

u 1 | 0 | 0 | 0 | 1 | 0 | 0 | |||||

0 | 0 | 0 | 0 | 1 | 0 |

Figure 6. Corner modeled with 64 Q4 elements and rigid links at the interface

The stiffness matrix ke (8×8) and the mass matrix Me (8×8) of each Q4 element are known [13]. We assemble the 64 Q4 elements to find the global stiffness and mass matrix of the corner k(162×162) and M(162×162). Using static condensation (Guyan reduction technique), we can reduce the global stiffness matrices to kc (34×34) and Mc (34×34) after elimination of the 128 non-interface dof. The rigid link can only translate in

2 directions and rotate in one direction thus 3 dof are needed

at each interface. Therefore, our special element will have 6

dof, 3 at each node.

The reduced stiffness matrix kr (6×6) is given by: kr = LTkc L, Where L(34×6) is the matrix giving the values of the dof

relative to both interfaces given by:

L=[

0 0 0 1 0 -1*b

0 0 0 0 1 0

0 0 0 1 0 -2*b

0 0 0 0 1 0

0 0 0 1 0 -3*b

0 0 0 0 1 0

0 0 0 1 0 -4*b

0 0 0 0 1 0 ];

Where *a *and *b *are the width and column of the corner divided by 8 respectively.

Using partition matrix notation,

krr = kk(1:34,1:34); krc = kk(1:34,35:162);

kcr = kk(35:162,1:34); kcc = kk(35:162,35:162);

kkcond = krr – krc . (kcc )-1.kcr ;

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Therefore, we can find that k = LT.kkcond .L. Where k(6×6) was found using Matlab®.

The same procedure can be used to find the mass matrix M, keeping in mind that it is a function of the elementary mass matrix of each Q4 element and the stiffness matrix kcc [14].

Mrr = MM(1:34,1:34); Mrc = MM(1:34,35:162); Mcr = MM(35:162,1:34); Mcc = MM(35:162,35:162); MMcond = Mrr – Mrc .(kcc)-1.krc T- krc .(kcc)-1.Mrc T+ krc .(kcc )-

1.Mcc .(k cc )-1.krc T;

M=LT.MMcond .L

For our L-frame example, *a *= 2/8, *b *= 2/8, *E*=2e11, *v *= 0.3, *t *= 2, the matrix k for the corner element (64Q4condensed) is

k = [

5.0708e+11 1.6119e+11 -5.9359e+10 -5.0708e+11 -

1.6119e+11 -2.8653e+11

1.6119e+11 5.0708e+11 2.8653e+11 -1.6119e+11 -

5.0708e+11 5.9359e+10

-5.9359e+10 2.8653e+11 4.062e+11 5.9359e+10 -2.8653e+11

-6.0307e+10

-5.0708e+11 -1.6119e+11 5.9359e+10 5.0708e+11

1.6119e+11 2.8653e+11

-1.6119e+11 -5.0708e+11 -2.8653e+11 1.6119e+11

5.0708e+11 -5.9359e+10

-2.8653e+11 5.9359e+10 -6.0307e+10 2.8653e+11 -

5.9359e+10 4.062e+11 ]

m = [

10588 -2319.7 -1081 7093.5 1524.1 288.55

-2319.7 37625 -8288.2 3115.4 7093.5 -7805.7

-1081 -8288.2 7164.2 -7805.7 288.55 3108

7093.5 3115.4 -7805.7 37625 -2319.7 -8288.2

1524.1 7093.5 288.55 -2319.7 10588 -1081

288.55 -7805.7 3108 -8288.2 -1081 7164.2 ] Now that we have found the stiffness matrix k(6,6) and the

mass matrix M(6,6) of the corner element; the

“64Q4condensed” element with Timo3 elements for the horizontal and vertical bars, we can model our test L-frame and perform the modal analysis.

Table 4 shows the results of the modal analysis of our test L- frame using 64Q4condensed element at the corner and Timo3 beam elements for the horizontal and vertical bars.

Table 4. Modal analysis of the L-frame using improved modeling

Mode No | Shape | Freq (Hz) Q4 | Timo3 Error | 64Q4condensed Error | |||

1 | 54 | 56 | 4% | 55 | 2% | ||

2 | 146 | 111 | 24% | 150 | 2% | ||

3 | 294 | 324 | 10% | 311 | 6% | ||

4 | 454 | 418 | 8% | 473 | 4% | ||

5 | 584 | 613 | 5% | 602 | 3% | ||

6 | 656 | 728 | 11% | 700 | 7% | ||

Average | 10% | 4% |

The above table shows the results of the modal analysis of the test L-frame. By using a 64Q4condensed element at the corners the error of the average of the first 6 natural frequencies of the test L-frame was reduced from 10.3% (using only Timo3 beam elements for the horizontal and vertical bars) to 4.1% (using Timo3 beam elements for the horizontal and vertical bars but with a 64Q4condensed element at the corner).

We know that the stiffness matrix k(6×6) and the mass matrix M(6×6) are both a function of: The width of the corner, the height of the corner, the elastic modulus *E *of the material at the corner, the Poisson coefficient *v *of the material at the corner, and the density ρ of the material at the corner.

If we let *sl *= *b*/*a *= the ratio of both the height and the width of the corner, the expression of the stiffness matrix will be a function of: *E*, *v*, *sl, *and *a *only.

We can easily show that the stiffness matrix k can be written as k = *E*. [ki ] where E is the Young’s modulus of the material at the corner. Let’s find the expression of ki .

If we vary *v *from 0.1 to 0.9 and *sl *from 1 to 10, and by using the modeling technique (the multiple nonlinear regression) we can find the value of each number of the stiffness matrix using

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the function surface fitting in Matlab. For example, the value of k11 is:

k11 (*v*,sl) = p00 + p10**v *+ p01*sl + p20**v*^2 + p11**v**sl + p02*sl^2

+ p30**v*^3 + p21**v*^2*sl + p12**v**sl^2 + p03*sl^3 + p40**v*^4 + p31**v*^3*sl + p22**v*^2*sl^2 + p13**v**sl^3 + p04*sl^4 + p50**v*^5 + p41**v*^4*sl + p32**v*^3*sl^2 + p23**v*^2*sl^3 + p14**v**sl^4 + p05*sl^5

Coefficients (with 95% confidence bounds):

p00 = -1.317 (-1.446, -1.189) p10 = 39.28 (38.14, 40.43)

p01 = 0.3133 (0.2174, 0.4091)

p20 = -210.7 (-215.6, -205.9) p11 = -3.461 (-3.781, -3.141)

p02 = 0.03668 (0.00061, 0.072)

p30 = 496.4 (486.3, 506.5) p21 = 15.54 (14.89, 16.19)

p12 = 0.01262 (-0.044, 0.069)

p03 = -0.00452 (-0.011, 0.0022) p40 = -529.8 (-540, -519.6)

p31 = -26.81 (-27.5, -26.12)

p22 = -0.01135 (-0.068, 0.045) p13 = -0.00116 (-0.0065,

0.0042) p04 = 0.0003 (-0.0003, 0.0009)

p50 = 209.7 (205.7, 213.7) p41 = 16.33 (16.02, 16.64)

p32 = 0.0568 (0.0299, 0.0837)

p23 = -0.0031 (-0.0055, -0.00076) p14 = 0.0001 (-7.3e-5, 0.0003)

p05 = -1.07e-5 (-3.3e-5, 1.1e-5)

Goodness of fit: SSE: 39.05 R-square: 0.9986 Adjusted R- square: 0.9986 RMSE: 0.07289

For the mass matrix:

M11 (*v*,*sl*) = p00 + p10**v *+ p01**sl *+ p20**v*^2 + p11**v***sl *+ p02**sl*^2 + p30**v*^3 + p21**v*^2**sl *+ p12**v***sl*^2 + p03**sl*^3 + p40**v*^4 + p31**v*^3**sl *+ p22**v*^2**sl*^2 + p13**v***sl*^3 + p04**sl*^4

+ p50**v*^5 + p41**v*^4**sl *+ p32**v*^3**sl*^2 + p23**v*^2**sl*^3 +

p14**v***sl*^4 + p05**sl*^5

Coefficients (with 95% confidence bounds):

p00 = 13.45 (13.41, 13.5) p10 = -5.285 (-5.661, -4.909)

p01 = -0.750 (-0.781, -0.718)

p20 = -20.78 (-22.36, -19.21) p11 = 0.551 (0.446, 0.656)

p02 = 0.4257 (0.4139, 0.4376)

p30 = 57.59 (54.3, 60.89) p21 = 0.1597 (-0.05333, 0.3728)

p12 = -0.0226 (-0.041, -0.0039)

p03 = -0.0478 (-0.0500, -0.0455) p40 = -57.38 (-60.72, -54.04)

p31 = -3.889 (-4.115, -3.663)

p22 = 0.3484 (0.3298, 0.367) p13 = -0.01205 (-0.0138, -

0.0102) p04 = 0.0034 (0.0032, 0.0037)

p50 = 21.83 (20.52, 23.15) p41 = 2.482 (2.381, 2.584)

p32 = -0.02902 (-0.0378, -0.02)

p23 = -0.016 (-0.0168, -0.0152) p14 = 0.00089 (0.00082,

0.00096) p05 = -0.00011 (-1e-4, -1e-4)

Goodness of fit: SSE: 4.199 R-square: 1 Adjusted

R-square: 1 RMSE: 0.0239

The R-square value from all the regression analyses of all terms *k*ij of the stiffness matrix k and the mass matrix Mij is at least 0.99. Thus, the model fits very well. Using this surface fitting in Matlab, we were able to find the analytical expression of all terms of the stiffness matrix k and mass matrix M of the

64Q4condensed element.

After finding the analytical expression for the stiffness matrix and the mass matrix for the corner element, and by modeling the horizontal and vertical bars with Timo3 elements, let’s run the modal analysis on another L-frame. For example:

3 m

4 m

3 m

2 m

Figure 7. New test L-frame

Running the modal analysis again on this new test L-frame, we get the results shown in Table 5.

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Table 5. Modal analysis of another test L-frame using improved modeling

Mode No | Shape | Freq (Hz) Q4 | Timo3 Error | 64Q4condensed Error | |||

1 | 20 | 20 | 1% | 20 | 2% | ||

2 | 66 | 51 | 23% | 68 | 2% | ||

3 | 174 | 184 | 6% | 183 | 5% | ||

4 | 281 | 258 | 8% | 297 | 6% | ||

5 | 371 | 378 | 2% | 377 | 2% | ||

6 | 429 | 455 | 6% | 453 | 6% | ||

Average | 8% | 4% |

The previous table shows again that our special modeling at the corner with 64Q4condensed elements gives closer results to the reference model (Q4 model) in modal analysis.

Replacing the 3D modeling of prismatic frames by traditional linear elements (Timoshenko or Euler-Bernoulli beams) leads

beams are thin and slender, the Euler-Bernoulli beam elements are well suited but they perform poorly in the case of small frames i.e. short beams. By using Timo3 beam elements for the bars and “64Q4condensed” elements at the corner, the model rapidly and easily converges to the exact solution even if we used a small number of elements. First, because our shape functions in the Timo3 beam element are quadratic. Second, the corners are modeled by a 64Q4condensed element that is based on the assembling and condensation of 64 Q4 elements.

Whenever you are running a modal analysis of frames, if the length of beams is short with respect to the width of the cross section of that beam, use the quadratic Timoshenko beam element and give special consideration for the connection at the corner. The best approach is to use Timo3 beam elements

for the horizontal and vertical bars and “64Q4condensed”

element at the corner.

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