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SuggestedAnswerCh6

SuggestedAnswerCh6
SuggestedAnswerCh6

6.6 REVIEW QUESTIONS

1. Explain why the superposition technique can be used to formulate the frame elements

simply using the formulations of the truss and beam elements. On what conditions this superimposition technique will fail. Suggested answer: The frame element has properties of both the truss element and the beam element carrying loads in the axial directions as well as the transverse directions. As such, the frame elements will have axial translations (like the truss), transverse translation and rotation in the x -y plane about the z -axis (like the beam) as its degrees of freedom. Since small deformation is assumed, so that the axial translational deformation does not affect the transverse deformation, and the transverse deformation does not affect the axial deformation, the superposition technique can be applied to formulate the equations for the frame elements.

The superimposition technique will fail if the axial deformation of the element is actually coupled with the transverse deformation, which occurs when the deformations are very large. In such a large deformation, a transverse deformation can actually cause axial deformation and vice versa and the superposition technique will not work.

2. In the transformation from the local coordinate system to the global coordinate

system in a planar frame, does the rotation degree of freedom undergo any transformation? Why? (Hint): Review Section 6.2.2

3. Work out the displacements of the planar frame structure shown in Figure 6-12. All

the members are of the same material (E = 69.0 GPa, =0.33) and with circular cross sections. The areas of the cross-sections are 0.01 m 2. Compare the results with those obtained for the review question 6 in Chapter 4.

1 m

Figure 6-12 3 member planar truss structure

Suggested answer:

1 m

Analysis of the problem

It is clear that the structure is symmetric. Hence, we can use only a half of the structure as shown in the above figure. The cross-sectional area of bar 1 is reduced by half becoming 0.005 m 2. The dimensions of all the members are given in Table 3.1.

Table 3.1 Dimensions and properties of truss members

Element number Cross-sectional area, A e m 2 Length l e m Young’s modulus E N/m 2 2nd moment of

area, I z m 4 1 0.005 0.5 69 x 109 1.989 x 10-6

2 0.01

1.118 69 x 109 7.958 x 10-6

Note for circular cross-sections: 42

44

z r A

I ππ==

Step 1: Obtaining the direction cosines of the elements

Knowing the coordinates of the nodes

in the global coordinate system, the first step would be to take into account the orientation of the elements with respect to the global coordinate system. This can be done by computing the direction cosines using Eqs. (6.10) and (6.11). The coordinates of

all the nodes and the direction cosines of l x and l y are

shown in Table 3.2. 500N

2 D D 5

D 6 D 7

Table 3.2 Global coordinates of nodes and direction cosines of elements

Element

number Global node corresponding to Coordinates in global coordinate system Direction cosines using Eq. (4.21) local node 1 (i) local

node 2 (j)

X i , Y i X j , Y j l x m x l y m y

1 1

2 0, -0.5 0, 0 0 1 -1 0 2 2

3 0, 0

0.5, 1.0 0.447 0.894 -0.894 0.447 Due to the symmetry, nodes 1 and 2 can only move vertically. As the external force is also vertical, the displacement of node 1 relative to node 2 can be obtain straight away as ()()71290.55007.24610 m 69100.005

lP d EA --===??? (3-1)

and the stress in bar 1 is simply

5-22500 1.010 Nm 0.005

P A σ===? (3-2) These data can be used to check FEM solution.

Step 2: Calculation of element matrices in global coordinate system

After obtaining the direction cosines, the element matrices in the global coordinate system can be obtained. Note that the problem here is static. Hence, the element mass matrices need not be computed. What is required is thus the stiffness matrix. Recall that the element stiffness matrix in the local coordinate system is a 6×6 matrix since the total degrees of freedom is six for each element. Note that there will be no change in the dimension of the matrix after transformation. The stiffness matrix for each element in the local coordinate system can be obtained by Eq. (6.4):

(1)8-26.900 6.9000.1320.032900.1320.03290.010900.03290.005510Nm 6.900.0.1320.03290.0109e sy -????-????-=???????-??????

k (3-3)

(2)8-26.1700 6.17000.04710.026300.04710.02630.019600.02630.009810Nm 6.1700.0.04710.02630.0196e sy -????-????-=???????-??????

k (3-4) Using Eq.(6.13), the stiffness matrix in the global coordinate system for each element is as follows:

18-20.13200-0.0329-0.13200-0.03296.900-6.900.01090.032900.005510Nm 0.132000.0329. 6.900.0109e sy ????????=?????????????

K (3-5) 28-21.2705 2.4468-0.0235-1.2705-2.4468-0.02354.94070.0118-2.4468-4.94070.01180.01960.0235-0.01180.098010Nm 1.2705 2.44680.0235. 4.9407-0.01180.0196e sy ????????=?????????????

K (3-6)

Step 3: Assembly of global FE matrices

The next step after getting the element matrices will be to assemble the element matrices into a global finite element matrix. Since the total global degrees of freedom in the structure are nine, the global stiffness matrix will be a 9×9 matrix. The assembly is done by adding up the contributions for each node by the elements that share the node. DOFs 4,5 and 6 are shared by the 2nd local node of element 1 and the 1st local node of element 2. Therefore, the corresponding terms of these DOFs must be added together. In summary, the global stiffness matrix is

1.2700.13200.03290.13200.03290000 6.900 6.900000.032900.01090.032900.00550000.132

00.03290.132000.03290 6.9005 2.44680.0235 1.2705 2.44680.02352 6.90.4468 4.94070.0118 2.4468 4.94070.01180.0++----+++-------=---K 0.02350.01180.01960.02350.01180.09801.2705 2.44680.0235 1.2705 2.44680.02352.4468 4.94070.0118 2.44684329

00.00550.032900.010*********.94070.0118-0.02350.01180.09800.0235-0.01180.01960????????????-++-------8-210Nm ???????????????????

(blue: element 1; red: element 2)

(3-7)

Step 4: Applying boundary conditions

The global matrix assembled initially is usually not PD. It becomes PD after applying boundary conditions, which can leads to a deduction in dimension (condensation). In this case,

D 1 = D 3 = D 4 = D 6 = D 7 = D 8 = D 9 = 0 (3-8) This implies that the corresponding rows and columns will actually give no effect to the solving of the matrix equation. Hence, we can simply remove the corresponding rows and columns.

0.13200.03290.13200.03290000 6.900 6.900000.032900.01090.032900.00550000.132

00.0329 1.4025 2.44680.0094 1.2705 2.44680.02350 6.90 2.446811.84070.0118 2.4468 4.94070.01180.0329

00.00550.00940.01180---------=----K 8-2

10Nm .03050.02350.01180.0980000 1.2705 2.44680.0235 1.2705 2.44680.0235000 2.4468 4.94070.0118 2.4468 4.94070.0118000-0.02350.01180.09800.0235-0.01180.0196?????????????????-????--??----???????? (3-9)

The condensed global matrix becomes a 2x 2 matrix as follows:

8-26.9 6.9 10 Nm 6.911.8407-??=???-??K (6-10)

It can easily confirm that this condensed stiffness matrix is SPD. The constrained global FE equation is

KD = F (6-11) where

[]25T D D =D (6-12) and the force vector F is given as

5000-??=????

F (6-13) Note that the only force applied is at node 1 in the downward direction of D 2. Equation (6-11) is actually equivalent to two simultaneous equations (of SPD) involving the three unknowns D 2 and D 5 as shown below.

[][]8258256.9 6.9105006.911.8407100

D D D D -?=--+?= (6-14)

Step 5: Solving the FE matrix equation

The final step would be to solve the FE equation, Eqs. in (6-14), to obtain the solution for D 4 and D 6. This can be done manually, since this only involves two unknowns. To this end, we obtain first

D 2 =1.716 D 5

(6-15) and then

D 2 =-1.0121×10-6 m

(6-16)

D 5 =-1.7367×10-6 m

We can also calculate

712257.24610 m d D D --=-=? (6-17) which is the same as that in Eq. (3-1) and therefore confirms our solution.

In comparison with the review question 6 in Chapter 4 whereby truss elements are used instead, it can be seen that exactly the same solution is obtained. The rotation and transverse DOFs of the frames used above does not play a part in the structure due to the symmetry of the structure and the external loading.

What if the structure is not symmetrical? And what if the loading is not symmetrical, even if the structure is symmetrical?

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