Cantilever Beam Problem 5

Shubham Kola

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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.

10:54護

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