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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.