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00:00In this video, we are going to learn how to draw shear force and bedroom movement agrams
00:13for a cantilever beam subjected to uniformly varying load, as shown in figure.
00:19The uniformly varying load, or also known as gradually varying load, is a type of load
00:25which is spread over a beam in such a manner that the rate of loading varies uniformly from
00:32point to point along the length of the beam, as shown in figure.
00:37The uniformly varying load is also known as gradually varying load or triangular load.
00:44So in today's lecture, we will solve a numerical problem based on this type of loading.
00:51So the statement is given as, draw shear force and bedroom movement diagrams for a
00:55cantilever beam AB, 4m long, carries gradually varying load, 0 kN per meter at free end,
01:04to 3 kN per meter at the fixed end.
01:08So for this setup, we have to draw the shear force diagram and bending movement diagram.
01:15So first of all, I will draw the free body diagram for this beam section.
01:23So for solving this numerical problem, first we have to convert this gradually varying load
01:29into point load, whose value will be equal to the area of this triangle.
01:36So the point load is equal to the area of this triangle, that is 1 by 2 based into height.
01:45Where base of this triangle is 4m and height of this triangle is 3 kN per meter.
01:52So it will get the point load of 6 kN.
01:55Now this converted load will be acting at the centroid of this triangle.
02:02So the centroid of this triangle will be at a distance of 1 by 3rd of the total length of
02:08the beam from left end or 2 by 3rd of the total length of the beam from the right end.
02:15So this converted point load is acting at this point.
02:21Now this type of problem, we are going to solve in 3 steps.
02:25In the first step, we have to calculate the value of support reaction force, Ra.
02:32So to calculate this value, I will use the condition of equilibrium, i.e. summation of
02:37Fy equal to 0.
02:40That means addition of all forces in the vertical axis equal to 0.
02:47While doing the addition of all vertical forces, I will consider the upward forces as positive
02:54and downward forces as negative.
02:56Here Ra is the vertical reaction force acting on the beam in the upward direction.
03:02So I will add this force with positive sign.
03:06And the converted point load of 6 kN force is acting on the beam in the downward direction.
03:13So I will add this 6 kN load with negative sign.
03:17So from this, I will get the value of reaction force Ra equal to 6 kN.
03:26So now with the help of this calculated value of Ra, I will further calculate the values of
03:31shear forces at all the points of beam.
03:35So the next step is, Calculations of shear forces.
03:39So here the sign convention is, for shear force calculations, the upward forces are considered
03:45as positive and downward forces are considered as negative.
03:49Here you should remember, while calculating the shear force at a particular point load, you
03:57can calculate shear force values for left side and right side of that particular point load.
04:04But while calculating the shear force at uniformly varying load, you should calculate shear force
04:10values at start point and end point of uniformly varying load.
04:16That is shear force at point A and shear force at point B we need to calculate.
04:22But in this diagram, at point A, the reaction force is acting on the beam and since the reaction
04:29force is a point load, here we have to calculate the shear force values at left side and right
04:36side of point A.
04:38So, here I will start the shear force calculation from left side of point A, that is SF at A to
04:45the left.
04:46So, as you can see, there is no force acting on the left side of point A. Therefore, SF at
04:52A to the left is equal to 0.
04:56So, to draw a shear force diagram, I will first draw a horizontal reference line of 0 kN shear
05:04force.
05:05So, here I will mark the point of 0 kN on the reference line, that is shear force at point
05:12A to its left equal to 0 kN.
05:15Now, if I go to the section to the right of point A, that is SF at A to the right, then
05:22there is reaction force array that we had calculated as 6 kN which is acting on the beam in the upward
05:29direction.
05:30So, as per the sign convention, I will consider upward forces as positive.
05:35So, here the shear force is plus 6 kN.
05:38Here, as the shear force value is positive, so I will mark this 6 kN shear force above the
05:45reference line of 0 kN shear force and I will connect these two points with a vertical line.
05:52Now, point B is the end point of uniformly varying load.
05:57So, I am taking section to the point B, that is SF at point B and here I will carry forward
06:04previous value of shear force up to point A to its right, which is 6 kN.
06:10And to the left side of point B, there is uniformly varying load that we had converted into
06:16point load of 6 kN acting on the beam in the downward direction.
06:21So, as per the sign convention, I will consider this downward force as negative.
06:26So, I will add this point load as minus 6 kN.
06:30So, here plus 6 kN minus 6 kN gives me the value of shear force as 0 kN.
06:38So, I will mark this point of 0 kN shear force at point B and here if you can see, in the
06:46loading diagram, between point A and point B, there is uniformly varying load and to draw
06:53the shear force, I will indicate uniformly varying load with a parabolic curve.
06:57So, I will connect these two points with a parabolic curve.
07:03And here in shear force diagram, whatever the portion drawn above the reference line,
07:08I will show this portion with plus sign.
07:11So, here I have completed the shear force diagram.
07:16Now, the next step is calculations of bedding movement.
07:21So, bedding movement at a section of beam is calculated as the algebraic sum of the movement
07:28of all the forces acting on the one side of the section.
07:32So, to calculate bedding movements, we can start either from left end or from right end
07:39of beam.
07:40Here I will start from right side.
07:43So, whenever you are calculating the bending movements, you should remember these conditions.
07:49So, here for cantilever beam, the condition is, at the free end of cantilever beam, the
07:55bending movement will be 0.
07:57That is, beam suffix B equal to 0 kNm.
08:01So, to draw a bending movement diagram, firstly I will draw the reference line of 0 kNm bending
08:08movement.
08:09So, I will mark this value with a point on the reference line.
08:14So, now we have to calculate the bending movement at point A.
08:20So, here, in case of cantilever beam, while you are doing the calculations for bending movement,
08:26at a particular point, you should always add movement of all the forces present from the
08:33free end of cantilever beam up to that particular point at which you are calculating the bending
08:38movement.
08:39It means, while I am calculating the bending movement at point A, I will add movement of
08:46all the forces present from the free end of cantilever beam up to point A.
08:52And for bending movement calculations, our science convention is, for sagging effect of beam,
08:59the force is considered as positive.
09:01And for hugging effect of beam, the force is considered as negative.
09:06So, the bending movement at point A, that is, beam of surface A equal to, algebraic sum of
09:14movement of all the forces at right-hand side of point A.
09:18So, here at right-hand side of point A, there is uniformly varying load, that is triangular load,
09:25which we had converted into point load of 6 kNm, which is acting on the beam in the downward
09:31force.
09:32Due to this downward force, the beam shows hogging effect.
09:35And for hogging effect of beam, I will consider this force as negative.
09:40As the amount of force is minus 6 kNm into distance from point of action of force, that
09:47is, 1 by 3rd of length 4m, that is, 1 by 3 into 4m.
09:53Therefore, after calculating this, we will get value minus 8 kNm.
10:00So, as it is negative value of bending movement, yes, I will mark the point of bending movement
10:06below the reference line of 0 kNm bending movement.
10:11And I will connect these two points with a vertical line.
10:15And to draw a bending movement diagram, I will indicate uniformly varying load with a cubic curve.
10:21Yes, I will join these two points with a cubic curve.
10:26Now, since I can see this bending movement diagram is drawn below the reference line of 0 kNm bending movement,
10:33yes, I will show this portion by minus sign.
10:36So, here I have completed the shear force and bending movement diagram for this triangular beam.
10:51There are several species that have had therise, and that has happened in a single stroke.
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