thanks for the reply, but I am pretty sure that's not right, if you put in 1.22m you get Q=10.47 as required, but if you put in 1.891 you get Q=19.16 don't you?
Homework Statement
A 4m wide channel has a bed slope of 1 in 1500 and a Manning's roughness coef n=0.01. A broad crested measuring weir in the form of a streamlined hump 0.4m high spans across the full width of the channel. the discharge is 10.5m^3/s. Determine
i)the uniform depth of flow...
Yes I did examples like that.
The one I had problems with was having just one metal bar fixed by rigid walls and therefore no expansion was allowed what so ever
they will use the length they see, so the extended length surely?
my question is more that when we load the member to prevent it expanding before it is heated it isn't actually subjected to any stress, it only experiences stress when we heat it and I'm finding it hard to get my head around...
Why if we heat something so it expands and then squash it back to its original size is different from loading it so that it does not expand in the first place.
Yeah sorry about that, my original problem in a textbook asked for stress but the problem I had in the solutions was with strain.
Thanks a lot though, great help!
Can you also recommend anything for me to read on why there is a difference?
Thanks again
Say you have a member of original length L and after a change in temperature T it changes in length to L'=L(1+αT). If we apply a force so that the length is returned back to L, the change in length is LαT and the strain is ε=LαT/L=αT.
My question is why is the denominator in the strain L and...