The normal stress is given by:
$\sigma = \frac{F}{A}$
where $\sigma$ is the normal stress, $F$ is the force, and $A$ is the cross-sectional area.
Given:
$\sigma = 42 MPa = 42 \times 10^6 Pa$
$A = 800 mm^2 = 800 \times 10^{-6} m^2$
Therefore, the force in member BD
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