1(1)
260 AL
Assignment:3
150 AL (3)
For the truss structure shown:
(2)ST
Find displacements of joints 2 and 33000.4 kN
Find stress, strain, & internal forces
2
in each member.
AAL = 200 mm2 , AST = 100 mm2
All other dimensions are in mm.
Solution
Let the following node pairs form the elements:
ElementNode Pair
(1)1-3
(2)2-1
(3)2-3
E (AL) = 69kN/ mm2, E (ST) = 207kN/mm2
A(1) = A(2) = 200mm2, A(3) =100mm2
Find the stiffness matrix for each element
Element (1)u1yu3y
(1)
L(1 ) = 260 mm,u1xu3x
E(1) = 69kN/mm2,1260 mm3
A(1) = 200mm2
= 0c2 = 1
c = cos = 1,
s = sin = 0, s2 = 0
cs = 0
EA/L = 69 kN/mm2 x 200 mm2 x 1/ (260mm) = 53. 1 kN/mm
c2cs- c2-cs
[Kg](1) = (AE/L) xcss2-cs- s2
-c2-csc2cs
-cs-s2css2
[Kg](1) = (53.1) x10-10
0000
-1010
0000
Element 2u1y
= 900
c = cos 900 = 0, c2 = 01u1x
s = sin 900 = cos 00 = 1, s2 = 1
cs = 0(2)
EA/L = 69 x 200 x (1/150) = 92 kN/mm
2u2x
u2xu2yu1xu1yu2y
0000u2x
010-1u2y
[kg](2) = 0000u1x
0-101u1y
Element 3u3y
= 300
c = cos 300 = 0.866, c2 = 0.753u3x
s = cos 600 = .5, s2 = 0.25
cs = 0.433u2y(3)
EA/L = 207 x 100 x (1/300) = 69 kN/mm300 mm
2 = 300
u2xu2yu3xu3yu2x
u2x.75.433-.75-.433
u2y-.433.25-.433-.25
[kg](3) = u3x-.75-.433.75.433
u3y-.433-.25.433.25
Assembling the stiffness matrices
Since there are 6 deflections (or DOF), u1 through u6, the
matrix is 6 x 6. Now, we will place the individual matrix element
from the element stiffness matrices into the global matrix
according to their position of row and column members.
Element [1]
u1xu1yu2xu2yu3xu3y
u1x53.1-53.1
u1y
u2x
u2y
u3x-53.153.1
u3y
The blank spaces in the matrix have a zero value.
Element [2]
u1xu1yu2xu2yu3xu3y
u1x92-92u1y
u2x
u2y-9292
u3x u3y
Element [3]
u1xu1yu2xu2yu3xu3y
u1x
u1y
u2x51.729.9-51.7-29.9
u2y29.917.2-29.9-17.2
u3x-51.7-29.951.729.9
u3y-29.9-17.229.917.2
Assembling all the terms for elements [1], [2] and [3], we get
the complete matrix equation of the
structure.u1xu1yu2xu2yu3xu3y
53.1000-53.10u1xF1
0920-9200u1yF1
0051.729.9-51.7-29.9u2x= F1
0-9229.9109.2-29.9-17.2u2yF1
-53.10-51.7-29.9104.829.9u3xF1
00-29.9-17.229.917.2u3yF1
Boundary conditions
Node 1 is fixed in both x and y directions, whereas, node 2 is
fixed in x-direction only and free to move in the y-direction.
Thus,
u1x= u1y = u2x = 0.
Therefore, all the columns and rows containing these elements
should be set to zero. The reduced matrix is:
109.2 29.9 17.2 0
104.829.9 u2y0
29.9 u3x
17.229.917. 2
u3y 0.4
Writing the matrix equation into algebraic linear equations, we
get,
109.9u2y- 29.9 u3x - 17.2u3y =0
-29.9u2y+ 104 u3x + 29.9u3y =0
-17.2u2y + 29.9u3x + 17.2u3y =-0.4
solving, we getu2y = -0.0043
u3x = 0.0131
u3y = -0.0502
Stress, Strain and deflections
Element (1)
Note that u1x, u1y, u2x, etc. are not coordinates, they are
actual displacements.
L = u3x = 0.0131
= L/L = 0.0131/260 = 5.02 x 10-5 mm/mm = E = 69 x 5.02 x 10-5 =
0.00347 kN/mm2
Reaction R = A = 0.00347 kN
Element (2)
L = u2y = 0.0043
= L/L = 0.0043/150 = 2.87 x 10-5 mm/mm = E = 69 x 2.87 x 10-5 =
1.9803 kN/mm2
Reaction R = A = (1.9803 x 10-3) (200) = 0.396 kN
Element (3)Since element (3) is at an angle 300, the change in
the length is found by adding the displacement components of nodes
2 and 3 along the element (at 300). Thus, L = u3x cos 300 + u3y sin
300 u2y cos600
0.0131 cos300 -0.0502 sin300 + 0.0043 cos600
-0.0116
= L/L = -0.0116/300 = -3.87 x 10-5 = 3.87 x 10-5 mm/mm = E = 207
x -3.87.87 x 10-5 = -.0080 kN/mm2
Reaction R = A = (-0.0087) (100) = 0.-0.800 kN
