Recent content by iionas

I will look over it again once I am home infront of my books. Thanks again for the help

Like I said this is a stab in the dark I don't want to come off as stupid...so I am thinking A = 2 x pie*10^2/4 = 157 then wed use f = F / A so wed have 75400/157 = 480.25 then wed use what you suggested..480.25 = 120 B B = Boltsthis will equal 4 .. We multiply by 2 because of two...

Let me give it a shot; No diagram just working out of a book with next to no explanation.

Im Going to post another question here...This is something once again with shear... I don't quite understand why this is. Determine the required number of 10mm mild steel bolts to hold two over lapping strips of metal against a total shear force of 75.4kN if the allowable stress in shear is...

20+20+15+15 x 8 = 70 70 * 8 = 560 560 * 360 = 201600 / 1000 = 201.6 = 202kNI Thank you for the help. I may post more in here.Im going to contribute as much as I can here. You've been more help than my tutor has. Thanks again.

There is some good youtube tutorials. I found very useful.. Just google it!

I have just reg'd here and I have been lurking forever but its time to give in so I can get some help. Ive been working through some examples in this book; and so far so good however I am on a question which does not make sense the way I am working it out seems to me to be correct however...