1. The problem statement, all variables and given/known data Rigid beam ABC is supported as shown in the figure. The pin connections at B, C, and D are each double shear connections, and the ultimate shear strength of the pin material is 640 MPa. Tie rod (1) has a yield strength of 350 MPa. A uniformly distributed load of w = 11 kN/m is applied to the beam as shown. A factor of safety of 3.5 is required for all components. Assume a = 575 mm, b = 725 mm, c = 250 mm, and d = 525 mm. Determine (a) the minimum diameter required for tie rod (1). (b) the minimum diameter required for the double shear pins at B and D. (c) the minimum diameter required for the double shear pin at C. http://edugen.wileyplus.com/edugen/courses/crs6838/art/qb/qu/ch0/13425408508040_6521478752676759.jpg 2. Relevant equations σ = F/A τ = F/A σallow = σfailure/FS τallow = τfailure/FS 3. The attempt at a solution First I got the resultant force of w → 14.3 kN/m. I started by finding the x and y components of F1(force on the rod), which were F1cos35.9 and F1sin35.9, respectively. Then I summed the moments about C which gave me ΣM = (-14.3 kN/m)(13m) + (0.810F1)(0.25m) = (0.5864F1)(0.725) ⇒ F1 = 14.8 kN Using σ ≤ σallow, I plugged in my values → 14809.4 N/π(d/2)2 ≤ 350 N/mm2/3.5 ⇒ drod = 188.6 mm 14809.4 N/2π(d/2)2 ≤ 640 N/mm2/3.5 ⇒dB,D = 51.6 mm Then I summed the x and y forces to get the force at C, which was C = 13.2 kN 13242.3 N/2π(d/2)2 ≤ 640 N/mm2/3.5 ⇒ d = 46.1 mm All of my d values were wrong. I have recalculated and changed the signs on my forces and moments just to see if I did that wrong but I still can't get the right answer. Am I going about the problem the wrong way? I'm just not sure where I'm going wrong and I've been working on this and a similar problem for a couple days now to no avail.