18.085 QUIZ 2 SOLUTIONS
Problem 1. (a) There are 6 bars and 4 non-fixed nodes, hence 4x2=8 unknowns. A is 6 by 8.
(b)-(c) We should expect 8-6=2 mechanisms. The first mechanism is when everything shifts horT
izontally, given by u1 = [1 0 1 0 1 0 1 0] . The second mechanism is when nodes 2, 3 are pushed
towards each other (and nodes 1, 4 move upwards and downwards respectively). This is given by
u2 = [0 1 1 0 − 1 0 0 − 1]T . The matrix A is given by



1 
A= √ 
2


1
−1
0
0
0
0
1 −1
1 0
0 −1
0 0
0 0
0 0
−1
0
1
0
√
2
0
0 0
1 −1
0 0
1 1
0 √0
0
2
0
0
1
−1
0
0
0
0
−1
−1
0
0








where the rows of A correspond to the following (chosen) order of the bars: 12, 23, 24, 34, left vertical
bar, right vertical bar. It is straightforward to check that Au1 = Au2 = 0.
(d) Adding two bars is needed to achieve stability. One way of doing this is by connecting node 4
to both fixed ends.
Problem 2. The boundary conditions are of type fixed-free, hence as trial functions we use two
hat functions φ1 , φ2 and one half-hat function φ3 (with supports [0, 32 ], [ 13 , 1], [ 32 , 1] respectively). The
test functions can be chosen to be the same as the trial functions, since they all obey v(0) = 0. Note
that all the slopes of φi are 3,-3 or 0.
(a) We have
" ´ 2/3 0 0 ´ 2/3 0 0 #
φ φ
φ φ
1 (−3) · (−3) (−3) · 3
3 −3
1/3 1 1
1/3 1 2
Ke = ´ 2/3 0 0 ´ 2/3 0 0
=
.
=
(−3) · 3
3·3
−3 3
3
φ1 φ2
φ2 φ 2
1/3
1/3
´1
´1 0 0
0
0
(b) We have K11 = 0 φ1 φ1 = 32 · 9 = 6, K12 = 0 φ1 φ2 =
common sense, we can easily find all other entries of K:
T
T
u1 = [1 0 1 0 1 0 1 0] u1 = [1 0 1 0 1 0 1 0]


6 −3 0
K =  −3 6 −3 
0 −3 3
1
3 (−3)
· 3 = −3. By symmetry and
The (discretized) right hand side F is given by
 ´1
 
 1 
δ(x − 21 )φ1 (x)dx
φ1 ( 21 )
0
2
´


F =  01 δ(x − 21 )φ2 (x)dx  =  φ2 ( 12 )   12  .
´1
0
φ3 ( 12 )
δ(x − 21 )φ3 (x)dx
0
The Galerkin formulation KU = F solves the weak problem with homogeneous boundary conditions,
that is, u(0) = 0, u0 (1) = 0. In our problem we have u(0) = 1, hence u − 1 is a homogenous solution,
which translates into


1
K(U −  1 ) = F
1
1
18.085 QUIZ 2 SOLUTIONS
i.e.

6
 −3
0
−3
6
−3
2

 7 
0
2
−3  U =  12 
3
0
Problem 3. (a) For a point load at x = 2/3, the displacement u(x) satisfies the 4th order equation
(assuming constant density c(x) = 1)
2
u0000 = δ(x − )
3
Zero displacements at x = 0, 13 translate into u(0) = 0, u( 13 ) = 0, while the free end at x = 1 means
no momentum: M (1) = 0, M 0 (1) = 0, i.e. u00 (1) = u000 (1) = 0.
(b) The solution will have the form of a cubic spline about x = 32 (hence its third derivative is a
step function which jumps at x = 23 ):
(
A + Bx + Cx2 + Dx3
x < 32
u(x) =
A + Bx + Cx2 + Dx3 + 16 (x − 32 )3 x ≥ 23
We have
(
(
2C + 6Dx
x < 2/3
6D
x < 2/3
00
000
u =
u =
2C + 6Dx + x − 23 x ≥ 2/3
6D + 1 x ≥ 2/3
5
hence u00 (1) = u000 (1) = 0 imply D = − 61 , C = 13 . Now u(0) = 0 ⇒ A = 0 and u( 13 ) = 0 ⇒ B = − 54
.
Hence the solution is
(
5
− 54
x + 31 x2 − 16 x3
x < 23
.
u(x) =
5
x + 31 x2 − 16 x3 + 16 (x − 23 )3 x ≥ 32
− 54