18.085 HOMEWORK 6 SOLUTION
HAOFEI WEI
Problem (2.7.3) .
Solution.
(a). By inspection, we can see that there are two rigid motion and two mechanisms which are in A ’s nullspace. They are: i. Horizontal translation: u T
1 ii. Vertical translation: u T
2
=
=
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0 iii. Clockwise rotation about node 3: iv. Collapse mechanism: u T
4
(b). We can see that A
T w = f u T
3
= 1 0 sin pi
4
= 1 0 1 0 0 0 0 0
− cos
π
4
0 will have a solution when we can find the external
0 0 − 1 forces uniquely determine the internal forces of the square. This happens when we have two forces pulling or pushing on two oppsing ends of a bar.
Thus, our four f ’s are: i.
f T
1 ii.
f T
2 iii.
f T
3 iv.
f
T
4
=
=
=
=
1
0
0
0
0
1
0
0
−
0
0
0
1
0
1
0
0
0
0
1
0
−
0
0
0
1
0
0
0
0
−
−
0
0
1
1
(c). Finally, it is trivial to check that the products u T i f j
= 0 for all i, j .
Problem (2.7.4) .
Solution.
Recall from the definition of A that each row of A has number of elements equal to twice the number of free (not fixed) nodes. Each row of A describes the stretching of a single bar in the truss, so the number of rows is equal to the number of bars. We have here 4 unfixed nodes - 1, 2, 3, and 4 - and 8 bars, thus A will have 8 rows and 8 columns .
If node 1 moves a unit to the right, bar 1 will stretch horizontally by cos(45
◦
) and bar 3 will compress by the same amount, thus giving us the first column of A , which describes the horizontal stretching of the bars due to node 1’s motion, to be h
√
2
0 − √
2
0 0 0 0 0 i
T
.
Finally, we know that A
T w = 0 has a nontrivial solution because A is a singular matrix. We can see that A is singular because it has a nontrivial solution to Au = 0, namely a rotation of the truss about node 5. This inducates that the rank of A is less than 8, which means that A T also has rank less than 8. Since A T has 8 columns, this indicates that the columns must be linearly dependent, and so A T w = 0 has a nontrivial solution.
Problem (2.7.6) .
Solution.
First, we see that truss F has 4 bars and 3 free nodes, and thus 6 forces.
This gives us 6 − 4 = 2 solutions to Au = 0. Since translation and rotation are both impossible due to the two fixed nodes, we see that there must be 2 mechanisms.
1

2 HAOFEI WEI
To help find the mechanisms, we will write down A :
A =
 0

−


√
2
0
0
−
1
√
2
0
0
0
√
2
− √
2
0
0
√
2
√
2
0
0
0
√
2
0
0 
0
− √
2


1

The truss can collapse with the two side bars (1 and 3) either falling to the same side or the opposite sides. The first mechanism corresponds to u
1
= 1 0 1 0 1 0
T
, with all three nodes falling to the right. The second corresponds to u
2
= with nodes 1 and 3 falling inwards and node 2 going up to compensate.
1 0 0 1 − 1 0
To make the truss stable, we must prevent both mechanisms of collapse. We could connect nodes 1 and 3 to stop mechanism 2 and either the left fixed node to 3 or the right fixed node to 1 to stop mechanism 1. We could also connect the left fixed node to 3 and the right fixed node to 1, stopping these two nodes from moving horizontally and thus preventing both mechanisms. By adding in these extra bars, A is now a square, non-singular matrix since Au = 0 will only have a trivial solution. Thus, A T w = f will only have one solution, w = ( A T )
− 1 f .
T
,
Problem (2.7.10) .
Solution.
The drawing should look like the following:
 cos( θ ) 
Using the standard way of defining each row of and A
0
A , we have A T
0
=


 sin( θ )
− cos( θ )


,

− sin( θ )) is the transpose of that. Using matrix multiplication, we can see that
 cos 2 ( θ )
A
T
0
CA
0
= C


 sin( θ ) cos( θ )
− cos
2
( θ )
− sin( θ ) cos( θ ) sin( θ ) cos( θ ) sin
2
( θ )
− sin( θ ) cos( θ )
− sin
2
( θ )
− cos 2 ( θ )
− sin( θ ) cos( θ ) cos
2
( θ ) sin( θ ) cos( θ )
− sin( θ ) cos( θ ) 
− sin
2
( θ ) sin( θ ) cos( sin
2
( θ )
θ )




18.085 HOMEWORK 6 SOLUTION 3 f
1
For A
T
0
= − f
3 y = f and f
2 to be solvable,
= − f
4
.
f must be a multiple of A
T
0
, and so we must have
Problem (3.1.3) .
Solution.
Doing as the question suggests and integrating both sides from x = 0 to x = 1, we get w (0)
−
−
Z
1
0 w dw dx = dx
(1) = 0 =
Z
1 f ( x ) dx
0
Z
1 f ( x ) dx
0
Thus, the condition is that R
1
0 forces on the bar summing to zero.
f ( x ) dx = 0, which corresponds to the external
Problem (3.1.10) .
Solution.
To use the finite elements method, we need to find K and F . Since
− u 00 = 2, we know c ( x ) = 1 and f ( x ) = 2. Furthermore, we are given φ i
( x ) is the hat function with
Thus, we have h = 1 / 4. We can assume that our test functions to be v i
= φ i
.
K ij
=
Z
1 dφ i dv i
0 dx dx
Using dφ i dx
= dv i dx

=


1 h
− 1 h
0
We find that K =
 8 − 4 0 

− 4 8 − 4

0 − 4 8
F i is given by
( i − 1) h < x < ih ih < x < ( i + 1) h elsewhere
F i
=
Z
1
0 f ( x ) φ i dx = 2
Z
1
0
φ i
( x ) dx = 2( h ) =
1
2
Using MatLab, we can solve the system KU = F to find
U = K \ F =
 8 − 4 0 

− 4 8 − 4

0 − 4 8
\
1
2
 1 

1

1
=
 3 
1
16

4

3
Thus, we find that U =
1
16
(3 φ
1
( x ) + 4 function centered at i/ 4 with h = 1 / 4.
φ
2
( x ) + 3 φ
3
( x )), where φ i
( x ) is the hat
Comparing this to the function u ( x ) = x − x 2 at the mesh points, we find

4 HAOFEI WEI
U (1 / 4) = 3 / 16 = (1 / 4) − (1 / 4)
2
U (1 / 2) = 1 / 4 = (1 / 2) − (1 / 2)
2
U (3 / 4) = 3 / 16 = (3 / 4) − (3 / 4)
2
And so, our solution matches the exact solution at all the mesh points, as expected.
Problem (3.1.12) .
Solution.
We are asked to find M ij centered at x = i/ 4.
= R
0
1
V i
V j dx , where V i is the hat function
For i = j , we know the integral is just M ii
= 2 R
0
1 / 4
(4 x ) 2 dx = 1 / 6, where we use the symmetry of the hat function and only integrate from the left zero intercept to the peak. Thus, M ii
=
1
6
.
For neighboring i and j , we will have overlap in the area between the two peaks with an integral M ij
= R
1 / 2
(2 − 4 x )( − 1 + 4 x ) dx =
1 / 4
R
1 / 4
(1 − 4 x )(4 x ) dx =
0
1
24
. Thus,
M ij
= 1
24 for | i − j | = 1.
For i and j differing by more than 1, the hat functions don’t overlap, and so
M ij
= 0 for | i − j | > 1.
From these three integrals, we can find our mass matrix to be
 4 1 0 
1
24

1 4 1

0 1 4
Problem (3.1.14) .
Solution.
Simply following the instructions and applying the given equation gives us
Z
2 h
φ
5
( x ) = h
=
= h
φ
5
( h ) +
6
4 h
φ
5
(
6
3 h
) +
2 h
φ
5
(2 h )
6 h
(0) +
6
2 h
4 h
(1) +
6 h
(0)
6
3
Thus, R
2 h h
φ
5
( x ) =
2 h
3
.