Subject - Strength of Materials
Chapter - Shear Force and Bending Moment Diagrams
Chapter - Shear Force and Bending Moment Diagrams
Category
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LearningTranscript
00:00In this video, we are going to learn, how to draw shear force diagram and bending moment
00:12diagram for a simply supported beam as shown in figure.
00:19So the statement is given as, Draw shear force and bending moment diagram
00:24for a simply supported beam AB 7 m long is loaded as shown in figure.
00:33So this is the simply supported beam AB of length 7 m and carrying a uniformly distributed
00:39loads of 2 kN per m over a length of 3 m and another UDL of 1 kN per m over a length of
00:472 m and there are 2 point loads of 3 kN and 5 kN are acting on the beam as shown in figure.
00:59So for this setup, we have to draw the shear force diagram and bending moment diagram.
01:10So first of all, I will draw the free body diagram for this beam section.
01:18So for solving this numerical problem, first we have to convert these uniformly distributed
01:23loads into point loads.
01:28So to convert this, I will multiply these UDL values with the length over which these
01:35UDL acts.
01:37So here firstly, I will convert these AC and CD portions of UDL into point loads.
01:44So the first point load equal to UDL of 2 kN per m multiplied by distance AC that is
01:533 m and the second point load equal to UDL of 1 kN per m multiplied by distance CD that
02:02is 2 m.
02:05So here I will get the first converted point load of 6 kN and second converted point load
02:11of 2 kN.
02:16Where first converted point load is acting on the midpoint of UDL distance AC and the
02:22second converted point load is acting on the midpoint of UDL distance CD.
02:31Now this type of problem we are going to solve in three steps.
02:37In the first step, we have to calculate the values of support reaction forces Ra and Rp.
02:45So to calculate these values, I will use two equations of equilibrium.
02:52Here first equation is, summation fi equal to zero.
02:57That means addition of all the forces in the vertical axis equal to zero.
03:04And the second equation is, summation of movement equal to zero.
03:10That means addition of movements due to all the forces about any point must be zero.
03:19So here in the first equation, while I am doing the addition of all the vertical forces,
03:25I will consider upward forces as positive and downward forces as negative.
03:33Here Ra and Rp are the vertical reaction forces acting on the beam in the upward direction.
03:41So as per the sign convention, I will add these forces with positive sign.
03:48And the point loads of 3kN and 5kN are acting on the beam in the downward direction.
03:56So as per the sign convention, I will add these forces with negative sign.
04:03And the converted point loads of 6kN and 2kN are acting on the beam in the downward direction.
04:11So as per the sign convention, I will add these forces with negative sign.
04:17Therefore, after calculating, this will get the equation, that is Ra plus Rb equal to 16kN.
04:28So I will give this as equation number one.
04:32Now the next equation is, summation of movement equal to zero.
04:38So for calculating movement, either we can take movement at point A or at point B.
04:46And here my sign convention would be, clockwise movement as positive and anticlockwise movement
04:51as negative.
04:56So for taking movement at point A, I will fix beam at point A.
05:02Now here, converted point load of 6kN will be pushing this beam towards downward.
05:08So it rotates the beam clockwise from fixed point.
05:12And as per the sign convention, for clockwise movement, I will add this force with positive sign.
05:20Now as per decided, movement is force into distance from fixed point.
05:26So here plus 6kN is the force into 1.5m is the distance from fixed point.
05:36Now here 3kN force will be pushing this beam towards downward.
05:40So it rotates the beam clockwise from fixed point.
05:44So for clockwise movement, I will add this force with positive sign.
05:50And movement is force into distance from fixed point.
05:54So here plus 3kN is the force into 3m is the distance from fixed point.
06:03Now here converted point load of 2kN will be pushing this beam towards downward.
06:09So it rotates the beam clockwise from fixed point.
06:14So for clockwise movement, I will add this force with positive sign.
06:20And the movement is force into distance from fixed point.
06:25So here plus 2kN is the converted point load into 4m is the distance from fixed point.
06:33Now here 5kN force will be pushing this beam towards downward.
06:38So it rotates the beam clockwise from fixed point.
06:43So for clockwise movement, I will add this force with positive sign.
06:49And movement is force into distance from fixed point.
06:53So here plus 5kN is the force into 5m is the distance from fixed point.
07:02Now here reaction force Rb will be pushing this beam towards upward.
07:06So it rotates the beam anticlockwise from fixed point.
07:11So as per the sign convention, for anticlockwise movement, I will add this force with negative
07:17sign.
07:19And movement is force into distance from fixed point.
07:23So here minus Rb is the force into 7m is the distance from fixed point.
07:31So these are the movements.
07:33So therefore, by calculating, this will get the value of reaction force Rb as 7.285 kN.
07:44Now I will put this value of Rb in equation number 1.
07:49And after calculating, this will get the value of reaction force Ra as 8.715 kN.
07:58So now with the help of these calculated values of Ra and Rb, I will further calculate
08:03the values of shear forces at all the points of beam.
08:08So the next step is calculations of shear forces.
08:14And for shear force calculations, our sign convention is upward forces are considered
08:19as positive and downward forces are considered as negative.
08:26And here you should note that, while calculating the shear force at a particular point load,
08:33you can calculate shear force values for left side and right side of that particular point
08:39load.
08:41But while calculating the shear force at uniformly distributed load, you should calculate shear
08:46force values at start point and end point of uniformly distributed load.
08:55That is shear force at point A and shear force at point D, we need to calculate.
09:02But in this problem, at point A and at point D, there are point loads.
09:08Hence, as per the rule, for point loads, I will calculate the shear force values at left
09:16side and right side of that point load.
09:21That is shear force at left side and right side of point A and point D, we need to calculate.
09:30And also at point C, there is a point load.
09:34Hence, as per the rule, for point load, I will calculate the shear force values at left
09:40side and right side of point C. That is shear force at point C to its left
09:47and shear force at point C to its right, we need to calculate.
09:55And here, at point B, there is support reaction force Rb, which is a point load.
10:02And since it is a point load, hence as per the rule, for point load, I will calculate
10:07the shear force values at left side and right side of point B.
10:16So I will start the shear force calculation from left-hand side of the beam.
10:21Therefore, first to calculate the shear force at point A to its left, I will take the section
10:29to the left of point A, that is Sf at A to the left equal to.
10:36So as you can see, there is no forces acting at the left side of point A, therefore Sf
10:43at A to the left equal to zero.
10:48So to draw the shear force diagram, I will first draw a horizontal reference line of
10:53zero kilo newton shear force.
10:56So here I will mark this point of zero kilo newton shear force on the reference line.
11:02That is shear force at point A to its left is zero kilo newton.
11:09Now if I go to the section to the right side of point A, that is Sf at A to the right equal
11:15to, then there is reaction force Ra, that we had calculated as 8.715 kilo newton, which
11:24is acting on the beam in the upward direction.
11:27So as per the sign convention, I will consider upward force as positive.
11:32So here the shear force is plus 8.715 kilo newton.
11:38Therefore Sf at A to the right equal to plus 8.715 kilo newton.
11:46Here as the shear force value is positive, so I will mark this point above the reference
11:50line of zero kilo newton shear force.
11:55And I will connect these two points with the vertical line.
12:03Now at point C, there is a point load, hence I will calculate the shear force values at
12:09left side and right side of point C. So first to calculate shear force at point
12:17C to its left, I will take the section to the left of point C, that is Sf at point C
12:25to its left equal to.
12:29And here I will carry forward previous value of shear force up to point A to its right,
12:35which is plus 8.715 kilo newton.
12:40And to the left side of point C, there is uniformly distributed load of 2 kilo newton
12:44per meter, which we had already converted into point load of 6 kilo newton, which is
12:50acting on the beam in downward direction.
12:54And as per the sign convention, I will consider downward force as negative, hence I will add
13:00this point load of 6 kilo newton with negative sign.
13:05Therefore, after calculating, this will get the shear force value as 2.715 kilo newton.
13:14That is Sf at point C to its left equal to 2.715 kilo newton.
13:22And as the shear force value is positive, hence I will mark this point above the reference
13:27line of 0 kilo newton shear force.
13:31And here the type of load is UDL or the length of 3 meter.
13:36Hence to draw the shear force diagram, I will indicate UDL with an inclined line.
13:42So I will connect these two points with inclined line.
13:46Now if I go to the section to the right of point C, that is Sf at point C to its right
13:56equal to, so here I will carry forward, previous value of shear force up to point C to its
14:04left, that is plus 2.715 kilo newton.
14:10And when we move towards the right side of point C, then there is one point load of 3
14:15kilo newton which is acting on the beam in downward direction.
14:20So as per the sign convention, I will consider downward force as negative.
14:26So I will add this point load with negative sign.
14:30Therefore, after calculating, this will get the shear force value as minus 0.285 kilo
14:38newton.
14:40That is Sf at point C to its right equal to minus 0.285 kilo newton.
14:47Here as the shear force value is negative, hence I will mark this point below the reference
14:52line of 0 kilo newton shear force and I will connect these two points with a vertical line.
15:03Now at point D, there is a point load.
15:07Hence I will calculate the shear force values at left side and right side of point D.
15:14So first to calculate shear force at point D to its left, I will take the section to
15:20the left of point D, that is Sf at point D to its left equal to, and here I will carry
15:28forward, previous value of shear force up to point C to its right which is minus 0.285
15:36kilo newton.
15:39And to the left side of point D, there is uniformly distributed load that we had converted
15:44into point load of 2 kilo newton, which is acting on the beam in downward direction.
15:51And as per the sign convention, I will consider downward force as negative, hence I will add
15:58this point load with negative sign.
16:02Therefore, after calculating, this will get the shear force value as minus 2.285 kilo
16:09newton.
16:10That is Sf at point D to its left equal to minus 2.285 kilo newton.
16:19Here as the shear force value is negative, hence I will mark this point below the reference
16:23line of 0 kilo newton shear force.
16:28And here the type of load is UDL over the length of 2 meter.
16:32Hence to draw the shear force diagram, I will indicate UDL with inclined line.
16:38So I will connect these two points with an inclined line.
16:45Now if I go to the section to the right of point D, that is Sf at point D to its right
16:52equal to, so here I will carry forward, previous value
16:56of shear force up to point D to its left, that is minus 2.285 kilo newton.
17:05And when we move towards the right side of point D, then there is one point load of 5
17:11kilo newton, which is acting on the beam in downward direction.
17:17So as per the sign convention, I will consider downward force as negative, so I will add
17:22this point load with negative sign.
17:27Therefore, after calculating, this will get the shear force as minus 7.285 kilo newton.
17:35That is Sf at point D to its right equal to minus 7.285 kilo newton.
17:44Here as the shear force value is negative, hence I will mark this point of shear force
17:49below the reference line of 0 kilo newton shear force.
17:55And I will join these two points with a vertical line.
18:02Now at point B, there is reaction force Rb acting on the beam in the upward direction.
18:09And since it is a point load, I will calculate the shear force values at left side and right
18:15side of that point load.
18:19Therefore, first to calculate shear force at point B to its left, I will take the section
18:26to the left of point B, that is Sf at B to the left equal to, and here I will carry forward
18:34previous value of shear force up to point D to its right, which is minus 7.285 kilo
18:41newton.
18:43And when we move towards the left of point B, then there is no load is acting on the
18:49beam at left side of point B. Therefore, Sf at B to the left equal to minus
18:567.285 kilo newton.
19:01Here as you can see, there is no variation in shear force values at point D to its right
19:08and at point B to its left, hence I will make the horizontal line with shear force
19:14value as minus 7.285 kilo newton.
19:21Now next to calculate shear force at point B to its right, I will take the section to
19:28the right of point B, that is Sf at point B to its right equal to, so here I will carry
19:36forward previous value of shear force up to point B to its left, which is minus 7.285
19:43kilo newton.
19:46And when we move towards the right side of point B, then there is reaction force Rb of
19:527.285 kilo newton, which is acting on the beam in the upward direction.
19:58So as per the sign convention, I will consider upward force as positive.
20:03So here I will add this upward force with positive sign.
20:09So here minus 7.285 plus 7.285 gives me the value of shear force as zero kilo newton.
20:17Therefore, Sf at point B to its right equal to zero kilo newton.
20:23So I will mark this point of zero kilo newton shear force on the reference line.
20:29That is Sf at point B to its right equal to zero kilo newton.
20:36And I will connect these two points with the vertical line.
20:43And here in shear force diagram, whatever the portion drawn above the reference line,
20:48I will show this portion by positive sign.
20:52And the portion which is drawn below the reference line, I will show this portion by negative
20:57sign.
20:59So here I have completed the shear force diagram.
21:05Now the next step is calculations of bending moment.
21:10So the bending moment at a section of beam is calculated as the algebraic sum of the
21:16movement of all the forces acting on one side of section.
21:22So to calculate bending moments, we can start either from left end of beam or from right
21:29end of beam.
21:31Here I will start from left end of beam.
21:35So whenever you are calculating the bending moments, you should remember these conditions.
21:43So here for simply supported beam, the condition is, at the ends of simply supported beam,
21:49the bending moment will be zero.
21:53So bending moment at point A and bending moment at point B will be zero.
22:00That is B on surface A equal to zero kilo newton meter and B on surface B equal to zero
22:06kilo newton meter.
22:10So to draw the bending moment diagram, firstly I will draw the reference line of bending
22:15moment zero kilo newton meter.
22:18So here I will mark these values with a point on the reference line.
22:25So now we have to calculate bending moment at point C.
22:30So here in case of simply supported beam, while we are doing the calculations for bending
22:36moment at a particular point, we should always add moment of all the forces present either
22:44from left end of beam or from right end of beam up to that particular point at which
22:51we are calculating the bending moment.
22:55It means while I am calculating the bending moment at point C, I will add moment of all
23:01the forces present either from left end of beam or from right end of beam up to point
23:09C.
23:11And for bending moment calculation, our sign convention is, for sagging effect of beam,
23:17the force is considered as positive and for hugging effect of beam, the force is considered
23:23as negative.
23:26So first to calculate bending moment at point C, that is B on surface C equal to.
23:35Here at left hand side of point C, there is reaction force array of 8.715 kilo newton
23:42which is acting on the beam in the upward direction.
23:46Due to this force, the beam shows sagging effect.
23:50And for sagging effect of beam, I will consider this force as positive.
23:56So I will add this force with positive sign.
24:01And as per decided, moment is always force multiplied by distance.
24:06So I will multiply this force with the distance from point of action of force.
24:11That is 3 meter.
24:15And here from point A to point C, there is UDL of 2 kilo newton per meter.
24:23That we have converted into point load of 6 kilo newton which is acting on the beam
24:28in the downward direction.
24:31Due to this load, the beam shows hugging effect.
24:35And for hugging effect of beam, I will consider this force as negative.
24:42So I will add this converted point load with negative sign.
24:47And I will multiply this force with the distance from point of action of force.
24:52That is 1.5 meter.
24:55Therefore, after calculating, this will get the bending moment value at point C equal
25:02to 17.145 kilo newton meter.
25:08That is B of surface C equal to 17.145 kilo newton meter.
25:16So as it is positive value of bending moment, here I will mark this point above the reference
25:22line of bending moment 0 kilo newton meter.
25:27And here between point A and point C, there is uniformly distributed load.
25:33Hence to draw the bending moment diagram, I will indicate uniformly distributed load
25:39with a parabolic curve.
25:40Hence I will join these two points with a parabolic curve.
25:48Now next we have to calculate bending moment at point D.
25:52That is B of surface D equal to.
25:57Here at right-hand side of point D, there is reaction force Rp of 7.285 kilo newton,
26:05which is acting on the beam in the upward direction.
26:09Due to this force, the beam shows hugging effect.
26:13And for sagging effect of beam, I will consider this force as positive.
26:19So I will add this force with positive sign.
26:23And the moment is force into distance.
26:26So I will multiply this force with the distance from point of action of force.
26:31That is 2 meter.
26:34Therefore, after calculating, this will get the value of bending moment at point D equal
26:41to 14.575 kilo newton meter.
26:46That is B of surface D equal to 14.575 kilo newton meter.
26:53So as it is positive value of bending moment, hence I will mark this point of bending moment
27:00above the reference line of bending moment 0 kilo newton meter.
27:05Like here, between point C and point D, there is UDL.
27:12Hence to draw the bending moment diagram, I will indicate UDL with a parabolic curve.
27:18Hence I will join these two points with a parabolic curve.
27:24And there is no load present on the beam between point D and point B.
27:31Therefore to draw the bending moment diagram, I will connect these two points with an inclined
27:35line.
27:36Now since I can see, this bending moment diagram is drawn above the reference line of bending
27:44moment 0 kilo newton meter.
27:47So whatever the portion I have drawn, all the values of bending moment are positive.
27:51Hence I will show this portion by positive sign.
27:57So here I have completed the shear force diagram and bending moment diagram for this simply
28:02supported beam.