MecMovies 4.0 : M5.6: Coaxial Tube and Core

scene 6 of 13

The figure shows the free body diagram of the tube near the free end B. The load on top of the tube is P which equals 30 kilo newtons directed vertically downwards. The reaction force on member 2 is F subscript 2 directed vertically downwards. The reaction force at the base of the tube is F subscript 1 directed vertically downwards.

Force-deformation relationships will be written for each of the axial components. The internal force in an axial member will be assumed to be tension.

For tube $(1)$ , the force-deformation relationship can be written as:

$δ_{1} = \frac{F_{1} L_{1}}{A_{1} E_{1}}$

and for core $(2)$ :

$δ_{2} = \frac{F_{2} L_{2}}{A_{2} E_{2}}$