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`

Step-by-step 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)-(8(1.2)^2)/(20) + D_y (0.8)`

Hence
`-(20*0.4)-(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) = (8(1.2)+20-17.2)/(sin35^o)`

`F_(BA) =21.62kN`

The factor of safety is mathematically

Factor of safety
`= (\sigma _u)/(\sigma _( all))`

Where
`\sigma_u` is the normal stress

`\sigma_(all)` is the allowable stress this mathematically given as

`\sigma_(all) = (F_(AB))/(A)`

`3.5 = (21.62*10^3)/(A)`

`Factor\ of \ safety =(450*10^6)/([(21*10^3)/(A) ])`

Making A the subject

`A = (3.5*21*10^3)/(450*10^6)`

`= 1.6815*10^(-4) m^2`