DSM - Torsion Example PDF

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DSM - Torsion Example...

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DSM – TORSION

SOLUTION k1

1

k2

2

Note:

J

d 4 32

Calculate torsional stiffness for each element:





ROD:

k1 

G1J 1 80 103    2.54  l 32 1000

TUBE 1:

k2 

G2 J 2 80 103    44  34  l 32  1000









=

0.31 kN.mm/rad

=

1.37 kN.mm/rad

Stiffness matrix:

T1   k1  k 2   k1  k 2   1  T     k  k  k  k      2  1 2 1 2  2 

k1  k 2  = 1.68 kN.mm/rad Then let:

θ1 = 0,

θ2 = 20° = 0.349 rads

Thus we have:

T1   1.68  1.68  0  T       2   1.68 1.68  0.349 For Row 1:

T1 = - 1.68 x 0.349

For Row 2:

T2 = + 1.68 x 0.349 = +0.58686 kN.mm = 586.8 N.mm (as required)

= -0.58686 kN.mm = -586.8 N.mm

Hence,

 max 

Tr 586.8  2  32  = 68.3 N/mm2 J   44  34



...