00:01
Let's start by calculating the reactions at support a and b.
00:05
So first we will calculate the reaction.
00:06
So let ra be the reaction force at support a.
00:14
So ra is the is at support a and rb is the reaction force at support.
00:24
Okay, they both are vertical force upwards.
00:27
Okay.
00:28
Now the sum of the vertical forces at equilibrium point is 0.
00:32
This is given to us.
00:33
So the sum is this is ra ra plus rb minus 5 is equals to 0.
00:40
Considering the concentrated load of 5 k ips.
00:45
Okay, here the unit is given to us k ips.
00:48
So when we will solve this the ra plus rb is equals to 5 and we are giving it equation 1.
00:56
Now the sum of movement about support a.
00:59
Sum of movement about support a is 0.
01:01
So to balance the bending moment caused by the concentrated load what we are doing here the rb multiply with 20 and minus 5 multiply by and the sum of movement.
01:17
So they are both movements.
01:19
So the sum is 0.
01:21
So from here the 20 rb is equals to 25.
01:27
So the value of rb is 1 .25 k ips.
01:34
Okay now from here we can easily find the value of ra okay, putting the written equation 1.
01:43
So writing down here from equation 1 ra will be 3 .75 k ips.
01:53
3 .75 k ips.
01:57
Okay, so we have calculated this they both now we will calculate the shear force v along the beam.
02:07
So now so we are solving it for shear force.
02:15
So now that we have the reactions we can calculate the shear force along the beam.
02:20
So we will analyze different sections of the beam.
02:24
So section 0 to 5 feet we are seeing first section 0 to 5 feet left end to the point of the concentrated load.
02:36
Okay, so at x is equals to 0 feet to x is equals to 5 feet.
02:42
There are no loads.
02:42
Okay, so there is no loads we can see so they so the shear force remain constant.
02:48
So from here v is equals to ra which is 3 .75 k ips and this is upward.
03:00
Okay, now section 5 to 10 feet.
03:05
This is section 0 to 5 feet now section 5 to 10 feet concentrated load to the right end of the uniform load.
03:15
Now at x is equals to 5 feet the concentrated load of 5 kps is acted downward and the shear force will decrease by 5 k ips from its previous value at the left end.
03:27
So v the shear force will be ra minus that load.
03:33
Okay, so load is 5 k ips.
03:35
So this is 3 .75 minus 5.
03:40
So we will get it minus 1 .25 and k ips or and it is downward.
03:49
Okay, because it is acting downward.
03:52
Now the section 10 to 15 feet 10 to 15 feet.
04:00
So right end of the uniform load at x is equals to 10 feet.
04:04
There is no concentrated load, but the uniform load acts throughout this segment.
04:08
So the uniform load is 2 k ips per foot and the length of the this segment is 5 feet.
04:15
So the total uniform load on this segment is so the total uniform load will be 20.
04:24
Okay 2 k ips per feet multiply with the 5 feet.
04:27
So we will get it 10 k ips.
04:30
Okay.
04:31
So since the uniform load acts downward the shear force will further decrease by 10 k ips from its previous value at the right end of the concentrated load.
04:41
So from here the shear force will be will minus is 1 .25 and minus the load.
04:48
So load is 10 k ips.
04:50
Okay.
04:50
So when we will calculate it 11 point minus 11 .25 and it will act downward.
04:57
Okay, and its unit is k ips clear.
05:01
Now we will see the last section which is 15 to 20 feet.
05:05
So i am removing this part and solving it here.
05:08
So it's for section 15 to 20 feet, which is right end of the uniform load to the right end of the beam.
05:16
Okay, so at x is equals to 15 feet there are no loads...