00:01
In this question you have not submitted the diagram that what kind of loading is it is it is so i have taken it from the web so the loading is like this this beam is fixed on both the sides and load informally distributed load is acting having magnitude w zero zero newton per meter let us say and this is length l alright so the loading is symmetric so reaction at both lengths would be equal so that means w l by and wl by 2 also in case of fixed beam the end movements are wl square by 12 wl square by right so how do i calculate it these moments that is a matter of another question i can solve it but over here i will tell you only how to draw the bending moment diagram so now if i take the section from a x distance like this if this is vx and this is mx like this, so the reaction was wl by 2 and the bend movement was wl squared by 12.
01:17
This is udl acting.
01:20
So for cr force, that means vx, sigma f vertical must be equal to 0.
01:31
So vx must be equal to wl by 2 minus wl into x.
01:39
This is vx at any point x away from the a end.
01:43
And similarly for mx, sigma m about this where we cut it must be 0.
01:53
So this would be mx would be equal to minus wl squared by 12 because they are both in same nature minus wl squared by 12 plus wl by 2 into x minus wx square by.
02:13
This would be the just wait a second.
02:18
And yeah, moment about this point must be zero.
02:26
This is clockwise, this load gives clockwise and this gives anti -clockwise.
02:30
So like this, it will become the parmuga.
02:32
So if i try to draw those equations, i will get something like this.
02:37
For seer force diagram, this was wl by 2 minus w into x.
02:44
So like this, wl by 2 negative positive.
02:52
This is sear force diagram...