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Worked example...

Description

A(mm2)

L(mm)

E(kN/mm2)

K (kN/m)

Aluminium

400

280

70

100

Brass

200

100

100

200

Steel

70

100

200

140

K=

AE L

First step: determine element stiffness matrix Element 1:

¿ 1 2 k ¿ 2¿−k 1 ¿ k 1 ¿ 1 k1 1 Element 2:

¿ 2 3 k ¿ 3 ¿−k 2 ¿ k 2 ¿ 2 k2 2 Element 3:

¿ 3 4 k ¿ 4 ¿−k 3 ¿ k 3 ¿ 3 k3 3 This is a 4 d.o.f. system therefore global stiffness matrix is a matrix of order 4x4

¿ 1 2 3 4 1 k1 −k 1 ¿ 3 ¿ ¿−k 2 ¿ k 2+ k 3 ¿−k 3 ¿ 4 ¿ ¿ ¿−k 3 ¿ k 3 ¿ 2 −k 1 k 1 +k 2 −k 2

Global stiffness matrix

[

100 −100 0 0 0 K global = −100 300 −200 0 −200 340 −140 0 0 −140 140

Now have a set of simultaneous equations

]

[F] = [K][U] To solve need to apply boundary conditions U1=0, U4=0, F2=-50, F3=100

[ ][

][ ]

0 F1 100 −100 0 0 U2 0 −50 = −100 300 −200 0 −200 340 −140 U 3 100 0 0 −140 140 0 F4

Eliminate terms with displacement equal to zero and therefore reduce matrix

][ ]

[ ][

−50 300 −200 U 2 = 100 −200 340 U 3

Solve by Gaussian Elimination: Step 1: Produce augmented matrix (combine force and stiffness matrixes). This ensures that any manipulations to the rows we perform makes the equations consistent.

300 [−200

−200 −50 340 100

]

Want to get the bottom left hand corner reduced to zero therefore need to add

2 3

2nd row

[

300 2 x 300−200 3

[3000

−200 2 x (−200 )+ 340 3

−200 −50 206.67 66.67

−50 2 x (−50 ) 100 3

]

]

Equation now in appropriate form, bring back augmented matrix and solve equations

x first row to the

−50 [66.67 ]=[3000

][ ]

−200 U 2 206.67 U 3

66.67 =206.67 u 3 u3=0.323mm

−50=300 u2 −200u 3 −50=300 u2 −64.6

u2=0.048 mm

Now we know displacements we can calculate forces

F1=−100∗U 2

F1=−4.8 kN F 4=− 140∗U 3

F 4=−45.22 kN...