Any structural or civil engineers here?

[andy_con]

i understand elasticity i just dont undertand the working behind it.

or the out come measurments? or the ^10 etc we dont do that in class so....

ive got lots of notes from class on bending moments so fingers crossed ill be ok. but ill post here anyway to see if im right.

cheers for helping me out guys

[andy_con]

ah gee i so badly hate college why did i say i would do this course..........

got find R 1 and R2



now i cant remember if i ignore the first 3m and start with 20kn??

i wana say this

3 × 20 + 6 × 50 + 10 × 40 - 13

[danielbriggs]

OK, if you ignore the distances and look purely at the vertical components, then:
R1+R2=20+50+40
R1+R2=110kN

Next take moments about R1:
(Sum of clockwise moments equals sum of anticlockwise moments)
(20x3)+(50x6)+(40x10)=(R2x13)
Which reduces to: 13R2=760
R2=760/13
R2=58.46 kN

Do the same for moments about R2:
(40x3)+(50x7)+(20x10)=(R1x13)
Which reduces to: 13R1=670
R1=670/13
R1=51.54 kN

Now to check you have them correct, use the first equation we established: (R1+R2=110)
51.54+58.46=110 kN
So we know we have the right values!

Dan

[FourDee]

Dan, don't do all the work for him or he won't learn anything

[Frixxion]

Well, ΔL is negative, so it would seem logical that the sign of the stress should also be. A negative elasticity modulus doesn't make sense. So not only do you have to get the units right, but also the signs. I suspect mechanical engineers are a bit sloppy with this though.

But what do I know... my area is chemical engineering.

Yeah you're right I remember now . Compressive stress is negative and tensile stress is positive. But that proves your point, we are a bit sloppy :P. Also ME's don't really care about a 10% or so deviation, at least that's what my teachers always do

[andy_con]

OK, if you ignore the distances and look purely at the vertical components, then:
R1+R2=20+50+40
R1+R2=110kN

Next take moments about R1:
(Sum of clockwise moments equals sum of anticlockwise moments)
(20x3)+(50x6)+(40x10)=(R2x13)
Which reduces to: 13R2=760
R2=760/13
R2=58.46 kN

Do the same for moments about R2:
(40x3)+(50x7)+(20x10)=(R1x13)
Which reduces to: 13R1=670
R1=670/13
R1=51.54 kN

Now to check you have them correct, use the first equation we established: (R1+R2=110)
51.54+58.46=110 kN
So we know we have the right values!

Dan

many thanks, so i was actually right with 3 × 20 + 6 × 50 + 10 × 40 - 13

so maybe i have learnt something from class LOL

believe it or not i actually fully understand all this part

[andy_con]

im so close yet so far, just finished doing the shear diagram and bending momnt diagram.

im now onto bows notation! anyone good at bows notation?

[FourDee]

you just have to integrate

[andy_con]

????

[danielbriggs]

????

I agree...
Bow's notation is a way of drawing forces on a diagram.