Cantilever Beam Problem 3

Shubham Kola

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00:00In this video, we are going to learn how to draw shear force and bending movement diagram

00:13for a cantilever beam as shown in figure. So, the statement is given as, Draw shear

00:20force and bending movement diagram for a cantilever beam AB, 1.5m long, is loaded with uniformly

00:30distributed load of 2kN per meter and point load of 3kN as shown in figure.

00:38So this is the cantilever beam AB of length 1.5m and carrying a UDL of 2kN per meter and

00:48point load of 3kN. So, for this setup, we have to draw the shear force diagram and bending

00:55movement diagram. So, first of all, I will draw the free body

01:00diagram for this beam section. Here, first we have to convert this UDL into

01:07point load. So, to convert this, I will multiply this UDL value that is 2kN per meter with the

01:17length over which the UDL act, that is 1.25 meter. So, here I will get the point load of

01:242.5kN. Now, this converted point load is act on the

01:30midpoint of length over which the UDL act. Now, this type of problem, we are going to

01:37solve in three steps. In the first step, we have to calculate the

01:43value of support reaction force, R A. So, to calculate this value, I will use the condition

01:50of equilibrium, that is summation of Fy equal to 0. That means, addition of all forces in

01:58the vertical axis will be 0. While doing the addition of all vertical forces,

02:05I will consider upward forces as positive and downward forces as negative.

02:12Here, R A is the vertical reaction force. So, I will add this force with plus sign and

02:18converted point load acting on the beam in the downward direction. So, I will add this force

02:243kN with minus sign. So, from this, I will get the value of reaction

02:39force R A equal to 5.5kN. So, now with the help of this calculated value

02:46of R A, I will further calculate the values of shear forces at all the points of beam.

02:54So, the next step is calculations of shear forces. So, here the sign convention is, upward forces

03:01are considered as positive and downward forces are considered as negative.

03:08And here, I will start the shear force calculation from left-hand side of the beam. That is, Sf

03:13at A to the left equal to. Here, there is no force acting at the left side of point A. Therefore,

03:22Sf at A to the left equal to 0. So, to draw a shear force diagram, I will

03:29first draw a horizontal reference line of 0kN shear force. And here, I will mark 0kN value

03:36on the reference line. Now, if I go to the section to the right

03:42of point A, then there is reaction force R A that we had calculated as 5.5kN which is

03:49acting on the beam in the upward direction. So, as per the sign convention, I will consider

03:54upward forces as positive. So, here the shear force is plus 5.5kN.

04:02Here as the shear force value is positive, here I will draw a vertical line of 5.5kN shear force.

04:09Here, one thing you should remember. While calculating the shear force at a particular

04:15point load, you can calculate shear force values for left side and right side of that particular

04:22point load. But, while calculating the shear force at uniformly distributed load, i.e. UDL,

04:30shear force at point C and shear force at point B, we need to calculate.

04:43The shear force at point C, i.e. SF at point C. Now, there is no load on the beam between

04:50right side of point A and point C. Therefore, shear force remains constant.

04:57That is, shear force at point C equal to 5.5kN. Here as there is no variation in the shear force

05:05value, so I will make the horizontal line with shear force value as 5.5kN.

05:11Now, I am taking section to the left of point D and I will calculate the shear force at point

05:19D to its left, i.e. SF at point D to its left. And here I will carry forward previous value

05:26of shear force up to point C which is 5.5kN. Here, as we are calculating the shear force upto

05:35point D to its left. Hence, I will only convert this much part of UDL into point load acting

05:43in the downward direction. Hence, I will add minus 2kN per meter into distance 1 meter.

05:52So this will get the shear force as 3.5kN. Here as the shear force value is positive and

05:59here the type of load is UDL. Hence, to draw a shear force diagram, I will indicate UDL with

06:05an inclined line. So, I will connect these two points with

06:09inclined line. Now, if I go to the section to the right

06:14of point D, i.e. SF at point D to its right equal to, sooner I will carry forward previous

06:22value of shear force upto point D to its left, i.e. 3.5kN. And when we move towards the right

06:30of point T, then there is point load. So, as per the sign convention, downward force

06:36has negative. So, I will add this force with minus 3kN. So, this will get the shear force

06:43plus 0.5kN. Here as the shear force value is positive, here I will mark the point of 0.5kN

06:51shear force above the reference line of 0kN shear force. And I will join these two points with

06:58a vertical line. Now, the point B is the end point of UDL.

07:05So, I am taking section to the point B, i.e. SF at point B. And here I will carry forward

07:12previous value of shear force upto point D to its right, which is 0.5kN. And to the right

07:20side of point B, there is UDL of 2kN per meter. Now, in this case, I will only convert this much

07:28part of UDL into point load, which is acting on the beam in the downward direction. And

07:35the downward force, I will consider as negative. Here is minus 2kN per meter into distance 0.25m.

07:43So, this is converted point load. So, by calculating, this will get the shear force

07:51at 0kN. So, I will mark this point of 0kN shear force. And to draw shear force, I will

08:00indicate UDL with an inclined line. So, I will connect these two points with an inclined

08:06line. And here in shear force diagram, whatever the portion drawn above the reference line,

08:13I will show this by plus sign. So, here I have completed the shear force diagram.

08:21Now the next step is calculations of bending movement. The bending movement at a section

08:28of beam is calculated as an algebraic sum of movement of all the forces acting on one

08:34side of the section. So, to calculate bending movement, we can start either from left end

08:42of beam or from right end of beam. Here I will start from the right side.

08:49So, whenever you are calculating the bending movements, you should remember these conditions.

08:56Here for cantilever beam, the condition is, at the free end, the bending movement will

09:01be 0. That is beam suffix b equal to 0kN. So, to draw bending movement, firstly I will draw

09:11the reference line of 0kN bending movement. So, I will mark this value with a point on the

09:18reference line. So, now we have to calculate the bending

09:23movement at point D. Here you should remember, in case of cantilever beam,

09:30while you are doing the calculations for bending movement at a particular point, you should always

09:36add, movement of all the forces present from the free end of cantilever beam up to that

09:42particular point at which you are calculating the bending movement.

09:47And for bending movement calculations, our sign convention is, for sagging effect of beam,

09:53the force is considered as positive, and for hogging effect of beam, the force is considered

09:59as negative. So, in UDL of 2 kN per meter, acting off the beam in the downward direction.

10:09Due to this, the beam shows hogging effect. And for hogging effect of beam, I will consider

10:15this force as negative. So, firstly, I will convert this much part of UDL into point load,

10:23that is minus 2 kN per meter, into distance 0.25 meter. And this point load is acting on

10:30the midpoint of UDL, that is half of 0.25 meter, that is 0.125 meter. And here I will multiply

10:39this converted point load with the distance from point of action of force, that is 0.125 meter.

10:48So, by calculating, this will get the value minus 0.0625 kN per meter. So, as it is negative

10:55value of bending movement, yes, I will mark this point below the reference line of 0 kN per

11:01bending movement. And to draw bending movement, I will indicate UDL with a parabolic curve.

11:09Yes, I will join these two points with a parabolic curve.

11:15Now, next we have to calculate, bending movement at point C, that is beam suffix C equal to.

11:23Here at the right-hand side of point C, there is UDL of 2 kN per meter acting on the beam

11:31in the downward direction. Due to this, the beam shows hogging effect. And for hogging effect

11:37of beam, I will consider this force as negative. So, firstly, I will convert this whole part

11:44of UDL into point load, that is minus 2 kN per meter into distance 1.25 meter. So, this

11:54is the point load acting on the midpoint of UDL, that is half of 1.25 meter, that is 0.625

12:00meter. And I will multiply this converted point load with the distance from point of action

12:07of force, that is 0.625 meter. And also there is one force. So, due to this force, the beam

12:15shows hogging effect. And for hogging effect of beam, I will consider this force as negative.

12:21So, here I will add this force of minus 3 kN and I will multiply this force with the distance

12:28from point of action of force, that is 1 meter. So, by calculating, this will get the value

12:35of bending movement at point C equal to minus 4.5625 kN per meter. So, as it is negative value

12:45of bending movement, hence I will mark the point of bending movement below the reference line

12:50of 0 kN per meter bending movement. And to draw bending movement, I will indicate UDL with

12:57a parabolic curve. Hence, I will join these two points with a parabolic curve.

13:02Now, next we have to calculate the bending movement at point A. Here, at right side of point A, there

13:11is UDL. So, this UDL, which I had converted into point load in the downward direction, hence the

13:18beam shows hogging effect. And for hogging effect, our sine conversion is negative, hence the

13:25amount of force is minus 2.5 kN into distance from point of action of force, i.e. 0.875 meter.

13:34Because this converted point load is acting at distance of 0.875 meter from point A, and also

13:42there is one force of 3 kN. Due to this force, the beam shows hogging effect. And for hogging effect of

13:50beam, I will consider this force as negative. And multiply this force with the distance from point of

13:58friction of force, i.e. 1.25 meter. So, this will get the value of bending movement at point A, minus 5.975 kN meter.

14:08So, as it is negative value of bending movement, hence I will mark the point below the reference

14:15line of 0 kN meter bending movement. And as you can see in cantilever beam, there is no force present

14:23between point A and point C. Therefore, to draw a bending movement diagram, I will connect these two

14:30points with an inclined line. Now, since I can see this bending movement diagram is drawn below the

14:38reference line of bending movement 0 kN meter, hence I will show this portion by minus sign.

14:45So, here I have completed the shear force diagram and bending movement diagram for this cantilever beam.

15:00So, thank you very much.

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