Engineering maths boffins

MattHirst · World Chat Champion Joined: 25 Jun 2006 Karma :

Right, got an assignment due in about 3 weeks, part of it is on Simply Distributed Loads and Uniformly Distributed Loads.

I know we have a few engineers on here, and a few maths boffins so i'm hoping someone will be able to do what i can't Laughing

Apologies if it's in the wrong section.

Firstly, SDL's. I know these are fairly easy to do and i don't have a problem with doing these generally but just to re-cap. I've added an attatchment of an example of how they are set out in our assignments. The idea of them is to illustrate a supported beam (imagine a bridge with a support at either end).

So to work out Ra -

We know Ra - 8.5m

So from there 8.5Ra = (2x5) + (5x4) + (6.5x5)

Therefore 8.5Ra = 62.5 - 62.5/8.5 = 7.353KN exerted on Ra in an upwards motion.

Can't be bothered typing out the shizzle for Rb but we know there is a 15KN downward force.

So 15 - 7.353 = 7.647KN, so Rb has an upward force of 7.647KN.

Right, onto UDL's. These are the ones i can't grasp. I've drawn another shoddy using the same figures as the other one just to make things easier.

I know my UDL is 2.7 x L x (L x 0.5)

I know Ra is calculated the same as the SDL, but you place the result of the UDL in the centre of Ra and Rb (or is it the centre of the beam?) and use this as a downward force when working out Ra. But the problem i have is calculating Rb while taking into account the overhang.

Hope someone can help me as i'm totally baffled haha.

You will be karma'd accordingly.

Thanks
Matt
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It's all fun and games until someone gets hurt, then it's bloody hilarious!!!

Suitor_Stu · Could Be A Chat Bot Joined: 19 Jun 2006 Karma :

Answer is 42.

Laughing

MattHirst · World Chat Champion Joined: 25 Jun 2006 Karma :

You can GTFO with your 42 aswell Laughing

____________________
It's all fun and games until someone gets hurt, then it's bloody hilarious!!!

instigator · Super Spammer Joined: 19 Oct 2004 Karma :

Not sure if I'm reading this correctly but a UDL is usually illustrated as semi-circles along the length on which it is distributed. I don't see this on your diagram, otherwise I am slightly puzzled.

A UDL should be expressed as a point load in the centre of the UDL if memorysrves me correctly, so 2.7kN/m over a 3m section would be 3 x 2.7 = 8.1kN 1.5m from your reference point.

I may be wrong as I haven't done it on yonks but I presume you're trying to work out your bending moments.

MattHirst · World Chat Champion Joined: 25 Jun 2006 Karma :

Yeah the UDL should be shown as 'semi-circles' accross the whole length of the beam, to show that it is evenly distributed. I just couldn't be arsed to draw them Laughing

I, for some reason, was led to believe that the UDL was 2.7 * L * (L*0.5) where L = Length.

Not sure why, sure my lecturer taught us that way..meh i dunno Laughing

____________________
It's all fun and games until someone gets hurt, then it's bloody hilarious!!!

instigator · Super Spammer Joined: 19 Oct 2004 Karma :

I don't have my books on this any more which is a shame as Structural mechanics was one of my favourite classes and this is one of the first calculations you learn. I did this back at Higher (a-level) Embarassed

This may help - https://www.mechguru.com/?p=299 I'm fookin nackered otherwise I'd quite like to re-learn it.

Or this - www.mathfiles.com/view.php/2420.doc

What are you studying anyway? Mechanical or structural?