MecMovies 4.0 : M5.10: End-to-end Rods with Misfit

scene 10 of 10

The internal force in each of the axial members has been determined:

$F_{1} = 6.49351 kips = F_{2}$

The normal stress in each axial member can now be computed.

$σ_{1} = \frac{F_{1}}{A_{1}} = \frac{6.49351 kips}{0.60 {in.}^{2}} = 10.82 ksi (T)$

and

$σ_{2} = \frac{F_{2}}{A_{2}} = \frac{6.49351 kips}{0.20 {in.}^{2}} = 32.5 ksi (T)$

Finished