National Taiwan University
Department of Civil Engineering
Fall 2023
Homework 3
Advanced Structural Theory
Instructor: L.-J. LEU
9/28/2023
Due: 10/5/2023
1. Consider a column between two rigid floors as shown below, which is a typical column in a
so-called shear building. Note that the column is rigidly connected to the floors and the degrees
of freedom for each end of the column are the in-plane (x-axis and y-axis) displacements and
(z-axis) rotation. The strain energy of the column can be expressed as
1
1
1
U = k x (ub − ua )2 + k y (vb − va ) 2 + k (b − a ) 2 . Note the all the primed (‘) variables are
2
2
2
referred to the local (member) axes and they are not derivatives.
(a) What are k x , k y , and k ? Write the answer in terms of E, G, h, I x , I y , and J.
1
(b) The strain energy can be expressed as U = ΔT k Δ , where Δ = ua , va ,a , ub , vb ,b T .
2
What is k  ? Write the answer in terms of k x , k y , and k .
1
(c) The strain energy can also be written as U = ΔT kΔ in terms of the degrees of freedom at
2
the mass center, where Δ = U a ,Va , a ,Ub ,Vb , b T . According to the congruent
transformation, k = TT k  T . Determine T.
coordinates (x, y) in the XY plane.
Y
x
Vb
X
h
Y
a
The a-th floor
Va
Ua
v
b
b
u
2 (x,y) b
X
Ub
The b-th floor
Y
y
2
b
Note that node 1 and node 2 have the same
The b-th floor
Y
va
 a
u
1 (x,y) a
1
X
X
The a-th floor
2. Obtain the stiffness matrix of the second element by choosing global node 3 as local node 1 and
global node 2 as local node 2. Prove the obtained element stiffness matrix is equivalent to the
element stiffness matrix on page 20 of course notes.
3. Derive the 6-th column of K on page 21 of course notes by using the definition of stiffness.
Namely, assume that v3 =1 and other DOFs=0. You need to calculate the internal force of each
member corresponding to v3 =1 first and then evaluate the summation of these internal forces at
nodes to obtain the 6-th column of K.
4. Consider a truss structure with
 1 0 −1 
K= [K] = 20000  0 1 0  kN/m


 −1 0 1.36 
 90 
 
and P= {P} = 135 kN.
 0 
 
We can obtain the eigenvalues and eigenvectors (already normalized) of K as follows:
 0.64142 
0.76719
0.


 


1 = 43922 , φ1 = 
0.
 ; 2 = 20000 , φ 2 = 1. ; 3 = 3278.6 , φ 3 =  0. 
− 0.76719 
0.64142
0.




 
Note that the unit for  i is kN/m. Let the solution be expressed as Δ = c1φ1 + c 2 φ 2 + c3 φ 3 .
(a) Solve Δ directly from the coupled system, K Δ =P.
(b) Determine ci , i = 1 ~ 3 first and then Δ by solving the (uncoupled) generalized system that
was discussed in class.
5. Figure (a) shows an old railway truss bridge. It is known that the downward displacement at
point b is 60 mm when P=120 kN.
(a) The bridge is now stiffened by adding three truss members as shown in Fig. (b). The
downward displacement at point b now becomes 20 mm when P=120 kN is applied at point
b of this stiffened truss bridge. What is the value of EA/L?
(b) Instead of adding three members, the same stiffening effect can be obtained by using a rigid
member bd and a tendon adc with pretension T. What would be the value of T/L?
a
c a
b
300
P
300
b
EA, 0.5L
EA, L
c
EA, L
d
Fig. (b)
Fig. (a)
6. Consider the truss shown below; there are nine degrees of freedom as shown. Draw a 9  9
matrix first. Mark (*) for element 1 and mark (X) for element 2 in the 9  9 matrix when these
two elements are assembled into it.
2
4
1
8
3
7
2
1
5
6
9