MecMovies 4.0 : M5.3: Deformation of a Compound Axial Member

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Note: Assume a positive internal force (i.e., tension).

The deformation of segment $(1)$ can be computed from:

$δ_{1} = \frac{F_{1} L_{1}}{A_{1} E_{1}} = \frac{(50 kN) (1, 800 mm) (1, 000 N / kN)}{(314.16 {mm}^{2}) (100, 000 N / {mm}^{2})} = 2.86 mm$

where the cross-sectional area of segment $(1)$ is:

$A_{1} = \frac{π}{4} {(20 mm)}^{2} = 314.16 {mm}^{2}$