Recent content by didaw

Thank you

No, I will have a look into it a bit later though. The closest thing I have been taught is about UDL point loads, conversion of moments, equilibrium conditions, moments of force but I have a 20 page document covering all of them and a few others so only a few things are in a good amount of detail

Looking at my notes it looks like it doesn't provide information on how to work it out. However it has one example of it just not anything to determine how it was worked out

The force is 410 on the hinge since that is the required pressure to open the hinge?

The only think I can think of is to used point loads but I don't know how to work that out with only one force and no provided UDL. Is that the right method to use?

so with using a side view like the question says i will only have to do a load at the top and bottom? What method would i need to use to find out what the loads are? or since its a 3rd of the way up would i need to 1226.25/3=408.75 for the bottom and 1226.25-408.75=817.5 for the top? or do you...

im not to sure, would the force come from the bottom of the tank? the force would equal the amount of pressure the water is pushing on it and that would be different for every point of the side

so would it be a force the other side of a of 1226.25 to make it equal? i will do next time i didn't realize that it was 1/3 when i started this equation but now i do, thanks

i don't suppose you can help me out on the next 2 questions? the same data applies and the picture. A hinged water storage is shown below. Its top side is denoted as C, the narrow side as B and the long side as A. Assume water density is 1000 kg/m3 and gravity is 9.81 m/s2 Side A is hinged at...

thank you, i didn't realize that if i used the full numbers it would get me to the right answer

so both attempt's are right because i rounded some numbers to 3 dp if i used the full numbers i would get 0.16 recurring

is this what i should have done IG=bd^3/12 = 1 x 0.5^3 / 12 = 0.0104 H= IG / Ah + h = 0.0104 / (1 x 0.5) x 0.25 + 0.25 = 0.3332m so the force would be 0.3332m from the top and 0.1668 from the bottom

the bottom will have the most force on it since their is more water mass above it but i have the force 0.1672 from the bottom of the box surely that sounds about right? unless i have used the wrong formula to work out moment two ' Ix = bh^3 / 3 ' should i have used Ixc=bh^3/12

ive just posted it