Recent content by takio

Thank you Deserfox. So did you consider F1 location as pivot? Would you please help me with this once again? What would happen if there are two hollowed structure with one shaft? Would that be.. -P*a=F2*2B=F3*(2B+C)=F4*(4B+C)

Hello Can anyone please help me with this? The shaft is located through the two holes on the rectangular hollowed structure. When applying the load P, How can I calculate the force 1 and 2? Thank you

Hello rb1957 I have calculated bolt group as below.

Hello Thank you for your comment. I have calculated the frame, using the moment and direct load. I hope this makes sense. Thank you

Hello rb1957 Thank you so much for your comments ! I am sorry that I didn't specify the other length. I edited it. and, I am s Ideal truss - Pin-connected : no moments. - Internal force in members : only compression or tension, no shear - Assumptions : frictionless joints, nodes are only at...

Hello The two A-frames stands side by side and linked by the top cross beam and middle cross beam. and the middle bar has a rod sticking out at the center. Force is applied at a distance D from the middle bar. I wanted to calculate the internal forces in the beams AC, CB, and AB. I think I...

Thank you very much all. I refreshed my strength of materials, and I started getting hang of bolt groups. I used https://mechanicalc.com/reference/bolt-pattern-force-distribution This website and calculated some problems, by following the equations and explanations.

Hello I have a question on eccentric load cases as below. Would anyone could help me with this? [link Loading of the connection by a force inclined to the contact surface not running through the connection gravity centre.]https://www.mitcalc.com/doc/boltcon/help/en/boltcon.htm[/url] [link...

Hello all Thank you so much for helping me out. I think I got hang of the load calculations! I appreciate all the comments !

Hello all I ran FEA to see SF on the carbon steel member. I used Carbon steel yield limit = 46000 PSI The carbon steel surface area that bolt contacts to : 0.3in^2 Applied preload = 10440 lbf (109 lb-in) Stress = 34800 psi SF = 1.3 Thank you for the information on HSLA steel

Hello, Thank you for your comment again. Yes that's what I meant,, it's carbon steel material.

Hello When applying recommended torque from clamp manufacturer. (They said the clamp safety factor is 5, when using 109 ft lbs for 0.375 steel member.) I get safety factor of 1.5~2 in members at bolt joint surface. but my requirement for the member safety factor is 3. In this case, is this...

Hello Warose For me, I got 502 lbf. (250 lbs force cantilevered 21.5" to the c.g. of the (4) bolt group) I just calculated with the same method above, but using 250 lbf Thank you

Hello I have recalculated as below. I hope this is right. 1) V=500 LBF 2)Total moment at centroid Ma=250 LBF * 1.75 in Mb=250 LBF * (18.5+2.75) in Mt = 250 * (18.5+2.75+1.75) lbf in = 5750 lbf in 3) r = 3.02 in (bolt distance from centroid) 4) Primary shear load per bolt F' = 500 lbf / 4 =...

Hello Thank you for your comment again! I just quickly calculated. 1)Moment caused by the load = 250*1 2)Assuming all the bolts are equally spaced in a circle. All the bolts are 3 inches away from the centroid. 3)Load that is tangent to the circle = 250 lbf in / 3 in = 166.67 lbf 4) X and Y...