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:

$δ = \frac{FL}{AE}$

And the deformation of aluminum rod $(3)$ is:

$δ_{3} = \frac{F_{3} L_{3}}{A_{3} E_{3}} = \frac{(80, 000 N) (1, 800 mm)}{(400 {mm}^{2}) (70, 000 N / {mm}^{2})} = 5.14 mm$