Answer
$$\sigma_{BD}=101.5\space MPa\space (Tension) $$
$$\sigma_{CE}=21.7 \space MPa \space (Compression) $$
Work Step by Step
1)Free Body Diagram (FBD) of element $\overline{ABC}$:
Equilibrium of Moments
$\sum M_{B} = 0$
$F_{CE}*0.4 - 20 * 0.25 = 0$
$\rightarrow F_{CE} = 12.5\space kN$
Equilibrium of Forces in the y-direction
$\sum F_{y} = 0$
$20 + 12.5 - F_{BD} = 0$
$\rightarrow F_{BD} = 32.5\space kN$
2)Internal Forces in Links BD and CE
Each of the links consists of two parallel elements so the internal force in each one is halved.
$\Diamond$ For links BD:
Since the link is in tension, the maximum tensile stress will occur at the pin opening. Therefore, the effective cross sectional area is reduced by the opening width:
$A_{BD} = (36-16)*8 = 160\space mm^2$
The max axial stress in BD is then:
$\sigma_{BD,max} = \frac{F_{BD}}{A_{BD}} = \frac{(32.5 \div 2)}{160} = 101.5 \space MPa \space(Tension)\space\space\leftarrow ANS1$
$\Diamond$ For links CE:
Since the link is in compression, the maximum compressive stress will occur at the pin opening. Therefore, the effective cross sectional area is:
$A_{CE} = (36)*8 = 288\space mm^2$
The max axial stress in CE is then:
$\sigma_{CE,max} = \frac{F_{CE}}{A_{CE}} = \frac{(12.5 \div 2)}{288} = 21.7 \space MPa \space(Compression)\space\space\leftarrow ANS2$
