4.5 REVIEW QUESTIONS
1.
What are the characteristics of the joints in a truss structure and what are the effects of
this on the deformation and stress properties in a truss element?
Suggested answer: The joints of a truss structure are hinges (not welded together) that
transmits only forces but not the moments. Therefore, a truss member is stressed only
in the axial direction of the truss member, and only the axial deformation is the
primary interest of analysis. This allows us to set the local coordinate along the axial
axis, and only one DOF at a node is needed for the formulation based on the local
coordinate.
2.
How many DOFs does a 2-nodal, planar truss element have in its local coordinate
system, and in the global coordinate system? Why is there a difference in DOFs in
these two coordinate systems?
Suggested answer: Local: DOF=2; global: DOF=4. As the local coordinate is taken
along the axial direction of the truss member that is stressed only in one (axial)
direction and only the axial deformation is concerned, there is hence only one DOF at
a node. The global coordinate system is used for all members of the truss structure,
and will not be possible at all the times for the coordinate axes to coincide with the
axial directions of all the truss members. Therefore, freedom of a node in a planner
space needs two displacement components to represent, because the node can move in
two directions. The relation between the one local DOF and the two global DOF is
given by the transformation matrix T defined in Eq.(4.33).
3.
How many DOFs does a 2-nodal, space truss element have in its local coordinate
system, and in the global coordinate system? Why is there such a difference?
(Hint): Similar argument as those given for equation 2.
4.
Write down the expression for the element stiffness matrix, ke, with Young’s modulus,
E, length, le, and cross-sectional area, A  0.02x  0.01 . (Note: non-uniform crosssectional area).
Suggested answer:
Assume
A  A1 x  A0
Using Equation (4.15), we have
 1
 l  
le
e
T
 E  1
B cBdV   A 
0
 1   le
 l 
 e 
ke  
Ve

le
0
5.
1
 dx
le 
Al
( 1 e  A0 ) E  1 1
E  1 1
( A1 x  A0 ) 2 
dx  2
 1 1 
le  1 1 
le


Write down the expression for the element mass matrix, me, with the same properties
as that in question 4 above.
Suggested answer:
Assume
A  A1 x  A0
Using Equation (4.16), we have
le
N N
m e    NT NdV    A  1 1
0
 N 2 N1
Ve
N1 N 2 
dx
N 2 N 2 
le
 N N N1 N 2 
   ( A1 x  A0 )  1 1
 dx
0
 N 2 N1 N 2 N 2 
x
x x

(1  ) 2 (1  ) 

le
le le
le
 dx  A0  le  2 1 
   A1 x 
0

6 1 2 
x2 
 sym.

le2



x2
(
x

2


l
le
e
  A1  
0

sym.


 1 2 1
( 2  3  4 )
2
  A1le 
 sym.


6.
 A1le2 1 1

12 1 3
x3
x 2 x3 
)
(
 )
le2
le le2 
dx 

x3

le2

1 1 
(  )
A l 2
3 4
 0 e 
1 
6 1
4 
A0  le
6
A0  le
6
2 1
1 2


1
2 
2 1
1 2 


Work out the displacements of the truss structure shown in Figure 4-9. All the truss
members are of the same material (E = 69.0 GPa) and with the same cross-sectional
area of 0.01 m2.
0.5 m
0.5 m
1m
0.5 m
1000N
Figure 4-9 3 member planar truss structure
0.5 m
D6
0.5 m
D5
Y
3

1m
2
D4
2
1
D3
0.5 m
D2
X
1
500N
1
D1
1000N
Suggested answer:
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 2 is reduced by half becoming
0.005 m2. The dimensions of all the members are given in Table 6.1.
Table 6.1 Dimensions and properties of truss members
Element
number
1
2
Cross-sectional area,
Ae m2
0.005
0.01
Young’s modulus
E N/m2
69 x 109
69 x 109
Length
le m
0.5
1.118
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 Eq. (4.21). Since this
problem being a planar problem, there is no need to compute nij. The coordinates of all the
nodes and the direction cosines of lij and mij are shown in Table 6.2.
Table 6.2 Global coordinates of nodes and direction cosines of elements
Global node
corresponding to
Element
local
local
number
node 1 node 2
(i)
(j)
1
1
2
2
2
3
Coordinates in global
coordinate system
Direction cosines
using Eq. (4.21)
Xi, Yi
Xj, Yj
lij
mij
0, -0.5
0, 
0, 0
0.5, 1.0
0
0.4472
1
0.8945
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
 0.5 500   7.246 107 m
lP
(6-1)
d12 

EA 69 109  0.005
and the stress in bar 1 is simply
P
500

 1.0  105 Nm -2
A 0.005
These data can be used to check FEM solution.
1 
(6-2)
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 2×2 matrix since the total degrees of freedom is two
for each element. However, in the transformation to the global coordinate system, the degrees
of freedom for each element becomes four, therefore, the stiffness matrix in the global
coordinate system is a 4×4 matrix. The stiffness matrices can be computed using Eq. (4.35)
and is shown below.
0 0 0 0
9 
 0.005 69  10 
1 0 1
e1
K 

0 0
0.5


1
 sy.
(6-3)
0 0
0 
0

6.9 0 6.9 

 108 Nm-2

0
0 


6.9 
 sy.
0.2 0.4 0.2 0.4 
9 
 0.01 69  10 
0.8 0.4 0.8
K e2 

0.2 0.4 
1.118


0.8 
 sy.
(6-4)
1.234 2.469 1.234 2.469 

4.938 2.469 4.938


 108 Nm-2

1.234
2.469 


4.938 
 sy.




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
six, the global stiffness matrix will be a 6×6 matrix. The assembly is done by adding up the
contributions for each node by the elements that share the node.
Element 1 contribute to DOFs of D3 to D6. In summary, we have the final global stiffness
matrix.
D1
D2
D3
D4
D5
D6






0
0
0
0
0   D1
0

6.9
0
6.9
0
0   D2


0  1.234 0  2.469 1.234 2.469   D3
K  108  
6.9  4.938 2.469 4.938   D4


1.234
2.469   D5
 sy.
4.938   D

6
(6-5)
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, D1, D2, D3 and D5 are constrained and thus
D1 = D3 = D5 = D6 = 0
(6-6)
This implies that the first to third and fifth 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.
D1
D2
D3
D4
D5
D6






0
0
0
0
0 
0

6.9
0
6.9
0
0 


1.234 2.469 1.234 2.469 
8

K  10 
11.838 2.469 4.938 


1.234
2.469 
 sy.
4.938 

 D1
 D2
 D3
 D4
 D5
 D6
(6-7)
The condensed global matrix becomes a 2×2 matrix given as follows.
6.9 
 6.9
K
 108 Nm-2

 6.9 11.838
(6-8)
It can easily confirm that this condensed stiffness matrix is SPD. The constrained global FE
equation is
KD = F
where
(6-9)
DT   D2 D4 
(6-10)
and the force vector F is given as
500
F
(6-11)

 0 
Note that the only force applied is at node 1 in the downward direction of D2. Equation (6-9)
is actually equivalent to two simultaneous equations (of SPD) involving the two unknowns D2
and D4 as shown below.
6.9 D2  6.9 D4   108  500
(6-12)
 6.9 D2  11.838D4   108  0
Step 5: Solving the FE matrix equation
The final step would be to solve the FE equation, Eqs. in (6-12), to obtain the solution for D2
and D4. This can be done manually, since this only involves two unknowns. To this end, we
obtain first
(6-13)
D2 = D4
and then
D2 =1.7367×10-6 m
(6-14)
D4 =1.0121×10-6 m
We can also calculate
d1 2  D2  D4  7.246 107 m
(6-15)
which is the same as in Eq. (6.1), which confirms our solution.
To obtain the stresses in the elements, Eq. (4.37) is used as follows:
0



6 
 1 1  0 1 0 0   1.7372×10 
 1  EBTDe  69 109 



0
 0.5 0.5  0 0 0 1  

(6-16)
  1.0126 106 
1.7372 
6
5
 34.5 109  1 1 
 10  1.0 10 Pa
1.0126 
1 
 1
 2  EBTDe  69  109 
1.118 1.118 
0




0
0 
0
0.4472 0.8945


(6-17)



0
0.4472 0.8945 
0
 0

1.0126  106 
 0 
6
4
 61.717  109  1 1 
  10  5.59  10 P
0.9057


which is the same as in Eq. (6.1), which confirms our solution once again.
Note that the FEM produces exact solution for this case.