00:01
Excuse my diagram, but if you're getting this out of a text, you've got access to the better diagram.
00:06
And we are asked to determine the largest permissible load, w, for a beam.
00:10
Knowing the normal allowable stress is positive 80 megapascals in tension and negative 130 megapascals in compression.
00:19
Okay, so the relationships at the supports will be, and that will equal the sum of my y divided by 2.
00:33
So my rb and that will be my work times 0 .9 divided by 2 that'll be 0 .45 work okay my sheer forces so a 0 .0 v of b equal a minus 0 .2 w so my v of b equals a minus 0 .2 w so my v of b equals 0 .2 w negative and my d positive will equal.
01:50
So that will equal 0, negative 0 .2 plus 0 .45w.
02:04
That'll be w equals 0 .25w.
02:10
And then i've got c plus and c minus and d plus and d minus.
02:20
The c plus is do vc minus first.
02:25
That will equal vb plus and a 0 .5 w that'll equal 0 .25w minus 0 .5 w equals 0 .25w negative and my c plus will equal and that will equal negative negative 0 .252 .5 w plus 0 .45w equal 0 .2w and d will equal 0 .0.
03:22
Okay, got those.
03:24
Now let's do our bending moments.
03:26
I'm sick of that.
03:34
And for a, it'll be 0.
03:39
For b, it'll be a plus the derivative of b to a, a b dx.
03:46
So this will equal 0 plus 1 1ā2 times negative 0 .2 w, and that will equal negative 0 .02w.
04:13
Then let's do e, that'll be mb plus e to b, and that will be equal to negative 0 .022 .2.
04:36
Minus 1 1 1ā2 times 0 .25 times 0 .25 w and that will equal 0 .01125 w.
04:55
The centroid of section will equal and that will equal 1 ,200 plus 70 plus times 70 plus 1200 times 30 divided by 2 ,400 and that will equal 50 millimeters.
05:48
The moment of inertia of section will equal so that will be 40 40 times 10 to the third plus 1200 times 10 to the 2 to the 2nd plus 360 times 10 to the third plus 1 ,200 times 20 to the second...