Recent content by Matthew Heywood

Ah okay. So you'd use ∂ = 0, FAB = FBC + 300kN = FCD = FDE + 600kN and E = 200GPa as it is steel?

So ∂ = 4.5mm = FAB(150mm)/125mm2EAB + (FBC + 300)(150mm)/125mm2EBC + FCD(150mm)/200mm2ECD + (FDE + 600)(150mm)/200mm2EDE ?

Using ΔL = FL/AE ?

Homework Statement stat indet.jpg Homework Equations FAB = FBC etc... The Attempt at a Solution FAB = FBC + 300kN = FCD = FDE + 600kN Sorry for another post. I'm really not sure where to go with this one...

so A = 20kN/200Mpa = 1x10-4 then r = √(1x10-4/π) = 5.64mm so d = 11.3mm Thanks :D

And then use σ = F/A?

Oh okay. So would it be 20kN?

Ah okay. So the for the force on each bolt: T = Fm ⇒ F = 1kNm/50mm = 20kN So 10kN on each bolt?

Homework Statement jan 2012 q3.jpg I've solved everything up to part d). I generally struggle with questions on bolts and I'm really not sure how to tackle this problem. Homework Equations τ = Tr/J, τ = 200MPa, T=1kNm The Attempt at a Solution I worked out the separation of the bolts (I...

Ahh okay. Well thank you for all of your help and patience :D

and then when x = 0, y = -w(0)/24EI + wL3(0)/6EI - wL4/8EI = -wL4/8EI Finally got closer :P although still not entirely right. I've got the negative of the answer on the sheet. Sorry for being a pain here :P

Ohhh hang on. So dy/dx = 0 when x = L ∴ c1 = wL3/6EI ∴ y = -wx4/24EI + wL3x/6EI + c2 and y = 0 when x = L ∴ c2 = -wL4/8EI Is this looking better? :P

I don't really know what else I can do. I get that the slope and deflection are 0 at the fixed point, but I don't know what else I can use.

So -wx3/6EI + c1 = 0 ⇒ c1 = wx3/6EI ∴ -wx4/24EI + c1x + c2 = -wx4/24EI + wx4/6EI + c2 = 0 ⇒ c2 = -wx4/8EI Am I right so far?

So -wx3/6EI + c1 = -wx4/24EI + c1x + c2 = 0 I'm still struggling to know where to go next.