MecMovies 4.0 : M5.4: Deformations in a Three-rod Assembly

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The figure shows the free body diagram of the rigid bar A C B and the supporting rods. The free body diagram cuts through rods 1 and 2. The force along rods 1 and 2 are F subscript 1 which equals 24 kilo newtons and F subscript 2 which equals 56 kilo newtons both of which are directed vertically upwards. The force at the end point D of the support rod C D is F subscript 3 which equals 80 kilo newtons and is directed vertically downwards.

The relationship between the internal force in an axial member and its deformation is given by:

$δ = \frac{FL}{AE}$

Similarly, the deformation of aluminum rod $(2)$ can be calculated as:

$δ_{2} = \frac{F_{2} L_{2}}{A_{2} E_{2}} = \frac{(56, 000 N) (1, 800 mm)}{(400 {mm}^{2}) (70, 000 N / {mm}^{2})} = 3.60 mm$