Answer
$\sigma_{DC} = {-4.97 MPa}$
Work Step by Step
Find the value of the reactions by summing moment at A.
$\circlearrowleft^{+} \Sigma {M@_A = 0}$
${-200(9.81)(1150) + F_f (850) = 0}$
${F_f = 2654.471 N \uparrow}$
${\uparrow^{+} \Sigma F_v = 0}$
${-200(9.81) + 2654.471 + F_Av = 0}$
${F_{Av} = -692.471 N}$
$Find \thinspace F_{DC} \thinspace by \thinspace summing \thinspace moment \thinspace at \thinspace point \thinspace E$
$\circlearrowleft^{+} \Sigma {M@_E = 0}$
$200(9.81)(200) + 692.471(1350) - F_{DC} \frac{675}{682.367} (550) =0$
$F_{DC} = 2439.493 N \searrow$
$The \thinspace area \thinspace of \thinspace rod \thinspace DC \thinspace is \colon$
$A_{DC} = \frac{\pi}{4} (25^2) = 490.874 mm^2$
$Then \thinspace the \thinspace stress \thinspace in \thinspace the \thinspace rod \thinspace is\colon$
${\sigma_{DC} = \frac{2439.493 N}{490.874 mm^2} = 4.97 MPa}$
${Since \thinspace the \thinspace force \thinspace is \thinspace directed \thinspace to \thinspace the \thinspace rod, \thinspace the \thinspace force \thinspace is \thinspace compressive \thinspace hence\colon}$
$\sigma_{DC} = -4.97MPa$
