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