Complete Question
The complete question is shown on the first uploaded image
Answer:
The cross-sectional area is [tex]A =1.6815 m^2[/tex]
Explanation:
The free body diagram of the link is shown on the second uploaded image
From the question we are told that
Ultimate normal stress in the link [tex]AB = 450MPa[/tex]
Factor safety [tex]=3.59[/tex]
From our free diagram we can see that the moment about B is 0 Mathematically
[tex]\sum M_b =0[/tex]
But [tex]\sum M_b = -(20*0.4)-\frac{8(1.2)^2}{20} + D_y (0.8)[/tex]
Hence [tex]-(20*0.4)-\frac{8(1.2)^2}{20} + D_y (0.8) =0[/tex]
Making [tex]D_y[/tex] the subject
[tex]D_y = 17.2kN[/tex]
At equilibrium summation of all force is 0 mathematically
This means
[tex]\sum F_y =0[/tex]
i.e [tex]F_{BA} sin 35^o +D_y - 8(1.2) -20 =0[/tex]
[tex]F_{BA} = \frac{8(1.2)+20-17.2}{sin35^o}[/tex]
[tex]F_{BA} =21.62kN[/tex]
The factor of safety is mathematically
Factor of safety [tex]= \frac{\sigma _u}{\sigma _{ all}}[/tex]
Where [tex]\sigma_u[/tex] is the normal stress
[tex]\sigma_{all}[/tex] is the allowable stress this mathematically given as
[tex]\sigma_{all} = \frac{F_{AB}}{A}[/tex]
[tex]3.5 = \frac{21.62*10^3}{A}[/tex]
[tex]Factor\ of \ safety =\frac{450*10^6}{[\frac{21*10^3}{A} ]}[/tex]
Making A the subject
[tex]A = \frac{3.5*21*10^3}{450*10^6}[/tex]
[tex]= 1.6815*10^{-4} m^2[/tex]
