Question

Link AB is to be made of a steel for which the ultimate normal stress is 450 MPa. Determine the cross-sectional area for AB for which the factor of safety will be 3.50. Assume that the link will be adequately reinforced around the pins at A and B.

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

The cross-sectional area is $A =1.6815 m^2$

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 $AB = 450MPa$

       Factor safety $=3.59$

From our free diagram we can see that the moment about B is 0 Mathematically

                    $\sum M_b =0$

       But     $\sum M_b = -(20*0.4)-\frac{8(1.2)^2}{20} + D_y (0.8)$

 Hence   $-(20*0.4)-\frac{8(1.2)^2}{20} + D_y (0.8) =0$

Making $D_y$ the  subject

                     $D_y = 17.2kN$

 At equilibrium summation of all force is 0 mathematically

          This means

                                $\sum F_y =0$

i.e        $F_{BA} sin 35^o +D_y - 8(1.2) -20 =0$

            $F_{BA} = \frac{8(1.2)+20-17.2}{sin35^o}$

             $F_{BA} =21.62kN$

The factor of safety is mathematically

                      Factor of safety $= \frac{\sigma _u}{\sigma _{ all}}$

Where $\sigma_u$ is the normal stress

           $\sigma_{all}$ is the allowable stress this mathematically given as

                      $\sigma_{all} = \frac{F_{AB}}{A}$

                     $3.5 = \frac{21.62*10^3}{A}$

       

      $Factor\ of \ safety =\frac{450*10^6}{[\frac{21*10^3}{A} ]}$  

Making A the subject

                 $A = \frac{3.5*21*10^3}{450*10^6}$

                     $= 1.6815*10^{-4} m^2$