1. The problem statement, all variables and given/known data In the steel structure shown, a 6‐mm‐diameter pin is used at C and 10‐mm‐diameter pins are used at B and D. The ultimate shearing stress is 150 MPa at all connections, and the ultimate normal stress is 400 MPa in link BD. Knowing that a factor of safety of 3.0 is desired, determine the largest load P that can be applied at A. Note that link BD is not reinforced around the pin holes. 2. Relevant equations F.S = σult/σallow σ = F/A 3. The attempt at a solution I understand the parts concernging the ultimate shearing stress, my confusion is with the normal stress. When I draw a FBD of beam AB, I assign the force at B, By, to point upwards and the force at C, Cy, to point downwards. ƩMc = -By(.120) + P(0.280) By = 2.33P My question is, does this this result mean that member BD is in compression or tension? Having drawn a FBD of AC I understand that the forces are acting on beam AC, thus while By points in the +y-direction, if I were to draw a FBD of BD, the force, By, would point in the opposite direction thereby indicating that member BD is in tension. Hence when I use the normal stress equation σ = F/A where F is the internal force and A is the cross sectional area, why is the diameter of the pin removed form the width of member BD?