00:01
Let's discuss this question.
00:02
So the box girder is subjected to a share of b that is equal to 15 k and determine the share flow at point b and the maximum share flow in the girders web ad.
00:14
So let's start.
00:15
Here we can draw the diagram first.
00:19
So for example this is the diagram here.
00:32
This is the rub sketch here we can say this will be the section and this is a and this is b.
00:46
The length of this is 250 millimeter and the distance between this both point is 15 millimeter.
00:59
So here we can say the distance between this both is also 150 millimeter and here we can say the distance between this both is 25 millimeter.
01:12
Here we can say y1 dash is equal to 0 .1325 millimeter.
01:20
Meter and here y 2 or we can say y dash 2 will be equal to 0 .1325 meter so now moment of inertia that is i will be equal to i that is equal to 1 divided by 12 multiplied by 0 .375 multiplied by 0 .28 multiplied by 0 .3 multiplied by 0 .25 cube.
02:04
So here we can say this is equal to 0 .2953575 multiplied by 10 raised to minus 3 meter 4.
02:16
So now here we can say qb is equal to by y -dash -b.
02:24
So this will be equal to 0.
02:30
1325 multiplied by 0 .375 multiplied by 0 .015 that is equal to 0 .7453125 multiplied by 10 x .2 minus 3 meter cube...