Subject - Strength of Materials
Chapter - Shear Force and Bending Moment Diagrams
Chapter - Shear Force and Bending Moment Diagrams
Category
📚
LearningTranscript
00:00In this video, we are going to learn, how to draw shear force diagram and bending moment
00:13diagram for a simply supported beam AB as shown in figure.
00:19So the statement is given as, Draw shear force and bending moment diagram
00:25for a simply supported beam AB, 5 m long is loaded as shown in figure.
00:34So this is the simply supported beam AB of length 5 m and carrying a uniformly distributed
00:41load of 10 kN per m over a length of 2 m.
00:48So for this setup, we have to draw shear force diagram and bending moment diagram.
00:56So first of all, I will draw the free body diagram for this beam section.
01:04So for solving this numerical problem, first we have to convert this uniformly distributed
01:10load into point load.
01:14So to convert this, I will multiply this UDL value with the length over which this UDL
01:22acts.
01:24So here I will get the converted point load of 20 kN.
01:31Now this converted point load will be acting on the midpoint of their UDL distance.
01:40Now this type of problem, we are going to solve in three steps.
01:46In the first step, we have to calculate the values of support reaction forces Ra and Rb.
01:55So to calculate these values, I will use two equations of equilibrium.
02:01That is first equation is, summation Fy is equal to zero.
02:07That means addition of all the forces in the vertical axis is equal to zero.
02:15And the second equation is, summation of moment is equal to zero.
02:21That means addition of moments due to all the forces about any point must be zero.
02:31So here in the first equation, while I am doing the addition of all vertical forces,
02:37I will consider upward forces as positive and downward forces as negative.
02:46Here Ra and Rb are the vertical reaction forces acting on the beam in the upward direction.
02:54So as per the sign convention, I will add these forces with positive sign.
03:01And the converted point load of 20 kN is acting on the beam in downward direction.
03:08So as per the sign convention, I will add these forces with negative sign.
03:15Therefore, after calculating, this will get the equation, that is Ra plus Rb is equal
03:22to 20 kN.
03:25So I will give this as equation number one.
03:29Now the next equation is, summation of moment is equal to zero.
03:35So for calculating moment, either we can take moment at point A or at point B.
03:43And here my sign convention would be, clockwise movement as positive and anticlockwise movement
03:49as negative.
03:52So for taking moment at point A, I will fix beam at point A.
03:57Now here 20 kN converted point load will be pushing this beam towards downward.
04:04So it rotates the beam clockwise from fixed point.
04:09So as per the sign convention, for clockwise movement, I will add this force with positive
04:14sign.
04:18Now as per decided, moment is force into distance from fixed point.
04:24So here plus 20 kN is the force into 2 m is the distance from fixed point.
04:32Now here reaction force Rb will be pushing this beam towards upward.
04:36So it rotates the beam anticlockwise from fixed point.
04:41So as per the sign convention, for anticlockwise movement, I will add this force with negative
04:46sign.
04:49And as per decided, moment is force into distance from fixed point.
04:55So here minus Rb is the force into 5 m is the distance from fixed point.
05:02So these are the moments.
05:04So therefore, by calculating, this will get the value of reaction force Rb as 8 kN.
05:12Now I will put this value of Rb in equation number 1.
05:17And after calculating, this will get the value of reaction force Ra as 12 kN.
05:25So now with the help of these calculated values of Ra and Rb, I will further calculate the
05:30values of shear forces at all the points of beam.
05:35So the next step is, calculations of shear forces.
05:40And for shear force calculations, our sign convention is, upward forces are considered
05:46as positive and downward forces are considered as negative.
05:54And here you should note that, while calculating the shear force at a particular point load,
06:00you can calculate shear force values for left side and right side of that particular point
06:06load.
06:08But while calculating the shear force at uniformly distributed load, you should calculate shear
06:14force values at start point and end point of uniformly distributed load.
06:21That is shear force at point C and shear force at point D, we need to calculate.
06:29And here at point A and point B, there are support reaction forces Ra and Rb, which are
06:37the point loads.
06:39And since these are the point loads, hence as per the rule, for point load, I will calculate
06:45the shear force values at left side and right side of point A and point B.
06:50Therefore, first to calculate shear force at point A to its left, I will take the section
07:00to the left of point A, that is Sf at A to the left equal to.
07:07So as you can see, there is no any force is acting at the left side of point A. Therefore,
07:13Sf at A to the left equal to zero.
07:18So to draw the shear force diagram, I will first draw a horizontal reference line of
07:24zero kilo Newton shear force.
07:28So here I will mark this point of zero kilo Newton shear force on the reference line.
07:34That is shear force at point A to its left is zero kilo Newton.
07:41Now if I go to the section to the right side of point A, that is Sf at A to the right equal
07:47to, then there is reaction force Ra, that we had calculated as 12 kilo Newton, which
07:54is acting on the beam in upward direction.
07:59So as per the sign convention, I will consider upward forces as positive.
08:04So here the shear force is plus 12 kilo Newton.
08:08Therefore, Sf at A to the right equal to plus 12 kilo Newton.
08:16Here as the shear force value is positive, so I will mark this point of shear force above
08:22the reference line of zero kilo Newton shear force.
08:27And I will connect these two points with a vertical line.
08:34Now the point C is the starting point of UDL.
08:38Hence I am taking section to point C and I will calculate shear force value at point
08:43C, that is Sf at point C equal to, now there is no load between the right side of point
08:52A and point C.
08:55Therefore, shear force remains constant.
08:59That is Sf at point C equal to plus 12 kilo Newton.
09:06Here as there is no variation in shear force values, so I will make the horizontal line
09:11with shear force value as 12 kilo Newton.
09:17Now the point D is the end point of UDL.
09:22So I am taking section to point D, that is Sf at point D equal to, and here I will carry
09:30forward previous value of shear force up to point C, which is 12 kilo Newton.
09:37And to the left side of point D, there is uniformly distributed load of 10 kilo Newton
09:42per meter, which we had converted into point load of 20 kilo Newton, which is acting on
09:48the beam in downward direction.
09:52So as per the sign convention, I will consider downward forces as negative.
09:57Hence I will add this point load with negative sign.
10:01Therefore, after calculating, this will get the shear force value as minus 8 kilo Newton.
10:10That is Sf at point D equal to minus 8 kilo Newton.
10:16Here as the shear force value is negative, hence I will mark this point of shear force
10:22below the reference line of zero kilo Newton shear force.
10:28And here the type of load is UDL over the length of 2 meter.
10:32Hence to draw the shear force diagram, I will indicate UDL with inclined line.
10:39So I will connect these two points with inclined line.
10:46Now at point B, there is reaction force RB acting on the beam in upward direction and
10:53since it is a point load, hence as per the rule, for point load, I will calculate the
10:59shear force values at left side and right side of that point load.
11:04Therefore, first to calculate shear force at point B to its left, I will take the section
11:12to the left of point B, that is Sf at B to the left equal to, and here I will carry forward
11:21previous value of shear force up to point D, which is minus 8 kilo Newton.
11:28And when we go to the left of point B, then there is no load acting on the beam at left
11:33side of point B. Therefore, Sf at B to the left equal to minus 8 kilo Newton.
11:42Here as you can see, there is no variation in shear force values at point D and at point
11:49B to its left.
11:52Hence I will make the horizontal line with shear force value as minus 8 kilo Newton.
11:59Now next to calculate shear force at point B to its right, I will take the section to
12:07the right of point B, that is Sf at B to its right equal to, so here I will carry forward
12:17previous value of shear force up to point B to its left, which is minus 8 kilo Newton.
12:25And when we go to the right side of point B, then there is reaction force Rb of 8 kilo
12:31Newton, which is acting on the beam in upward direction.
12:36So as per the sign convention, I will consider upward forces as positive.
12:41So here I will add this upward force with positive sign.
12:47So here minus 8 kilo Newton plus 8 kilo Newton gives me the value of shear force as zero
12:53kilo Newton.
12:55Therefore, Sf at point B to its right equal to zero kilo Newton.
13:02So I will mark this point of zero kilo Newton shear force on the reference line, that is
13:09Sf at point B to its right equal to zero kilo Newton.
13:15And I will connect these two points with vertical line.
13:21And here in shear force diagram, whatever the portion drawn above the reference line,
13:27I will show this portion by positive sign.
13:31And the portion drawn below the reference line, I will show this portion by negative sign.
13:38So here I have completed the shear force diagram.
13:43Now the next step is calculations of bending moment.
13:48So the bending moment at a section of beam is calculated as the algebraic sum of the
13:55movement of all the forces acting on one side of section.
14:01So to calculate bending moment, we can start either from left end of beam or from right
14:07end of beam.
14:08Here I will start from left end of beam.
14:13So whenever you are calculating the bending moments, you should remember these conditions.
14:21So here for simply supported beam, the condition is, at the ends of simply supported beam,
14:27the bending moment will be zero.
14:30So bending moment at point A and bending moment at point B will be zero.
14:36That is B on surface A equal to zero kilo Newton meter and B on surface B equal to zero
14:43kilo Newton meter.
14:46So to draw the bending moment diagram, firstly I will draw the reference line of bending
14:51moment zero kilo Newton meter.
14:55So here I will mark these values on the reference line.
15:01So now we have to calculate bending moment at point C.
15:06So here in case of simply supported beam, while you are doing the calculation for bending
15:12moment at a particular point, you should always add movement of all the forces present either
15:20from left end of beam or from right end of beam up to that particular point at which
15:27you are calculating the bending moment.
15:31It means, while I am calculating the bending moment at point C, I will add movement of
15:39all the forces present either from left end of beam or from right end of beam up to point
15:46C.
15:49And for bending moment calculation, our sign convention is, for sagging effect of beam,
15:56the force is considered as positive and for hogging effect of beam, the force is considered
16:01as negative.
16:06So first to calculate bending moment at point C, that is B on surface C equal to, here at
16:14left hand side of point C, there is reaction force array of 12 kilo Newton, which is acting
16:20on the beam in the upward direction.
16:24Due to this force, the beam shows sagging effect.
16:28And for sagging effect of beam, I will consider this force as positive.
16:34So I will add this force with positive sign.
16:39And as per decided, moment is force into distance.
16:44So I will multiply this force with the distance from point of action of force.
16:49That is 1 meter.
16:52Therefore, after calculating, this will get the value of bending moment at point C equal
16:59to 12 kilo Newton meter.
17:02That is B on surface C equal to 12 kilo Newton meter.
17:07So as it is positive value of bending moment, hence I will mark this point of bending moment
17:14above the reference line of bending moment 0 kilo Newton meter.
17:20Now there is no any load present on the beam between the point A and point C.
17:25Therefore, to draw the bending moment diagram, I will connect these two points with inclined
17:32line.
17:37Now here you should note that in shear force diagram, when the shear force values changes
17:43from positive to negative, then at that point the bending moment value is maximum.
17:50So I will represent this point with point M.
17:54So at point M, we need to find out maximum bending moment value.
18:01But here we don't know the location of point M.
18:04That is why we don't know the distance x.
18:08But we can find out this distance x by using the equation of similarity of triangles.
18:15Because here we know the distance CD i.e. 2 meter and here the values of shear forces
18:23were considered as heights of triangles.
18:27Therefore, for the first triangle, the height is 12 and base is x.
18:36And for second triangle, the height is 8 and base is 2-x.
18:43Therefore, from similarity of triangles, the base of first triangle i.e. x divided by height
18:52of first triangle which is 12 is equal to base of second triangle i.e. 2-x divided by
19:03height of second triangle which is 8.
19:08So here the unknown term is x.
19:11Therefore, after calculating, this will get the distance x equal to 1.2 meter.
19:19So here I got the location of point M. So from distance x, next I will calculate the
19:27maximum bending moment value at point M. So next to calculate bending moment at point
19:35M, that is beam of surface M equal to, here at left-hand side of point M, there is reaction
19:44force RA of 12 kN which is acting on the beam in the upward direction.
19:52Due to this force, the beam shows sagging effect.
19:56And for sagging effect of beam, I will consider this force as positive.
20:02So I will add this force with positive sign.
20:06And as per decided, moment is forced into distance, so I will multiply this force with
20:13the distance from point of action of force i.e. 2.2 meter.
20:20And here from point C to point M, there is UDL of 10 kN per meter.
20:29So I will first convert this portion of UDL into point load.
20:34That is 10 kN per meter is the UDL.
20:38And I will multiply with the distance over which this UDL acts i.e. 1.2 meter.
20:45Now this converted point load is acting on the midpoint of this much portion of UDL in
20:51downward direction.
20:52Due to this converted point load, the beam shows hogging effect.
20:58And for hogging effect of beam, I will consider this force as negative.
21:04So I will add this converted point load with negative sign.
21:10And I will multiply this force with the distance from point of action of force i.e. half of
21:161.2 meter.
21:20Therefore, after calculating, this will get the value of pending moment at point M equal
21:28to 19.2 kN meter.
21:32That is beam surface M equal to 19.2 kN meter.
21:38So as it is positive value of pending moment, here I will mark this point of pending moment
21:45above the reference line of pending moment 0 kN meter.
21:51And here between the point C and point M, there is UDL.
21:56Therefore, to draw the pending moment diagram, I will indicate UDL with a parabolic curve.
22:02Hence, I will join these two points with a parabolic curve.
22:07Now next we have to calculate pending moment at point D i.e. beam surface D equal to.
22:18Here at right-hand side of point D, there is reaction force Rb of 8 kN which is acting
22:24on the beam in the upward direction.
22:28Due to this force, the beam shows sagging effect.
22:33And for sagging effect of beam, I will consider this force as positive.
22:38So I will add this force with positive sign.
22:43And as per decided, moment is force into distance, so I will multiply this force with the distance
22:51from point of action of force i.e. 2 m.
22:56Therefore, after calculating, this will get the value of pending moment at point D equal
23:04to 16 kN meter i.e. beam surface D equal to 16 kN meter.
23:13So as it is positive value, hence I will mark this point of pending moment above the reference
23:19line of pending moment 0 kN meter.
23:25And here between point M to point D, there is UDL.
23:30Therefore, to draw the pending moment diagram, I will indicate UDL with a parabolic curve.
23:38Hence I will join these two points with a parabolic curve.
23:44And there is no any load present on the beam between the point D and point B.
23:48Therefore, to draw the pending moment diagram, I will connect these two points with inclined
23:55line.
23:56Now since I can see, this pending moment diagram is drawn above the reference line of pending
24:04moment 0 kN meter.
24:07So whatever the portion I have drawn, all the values of pending moment are positive.
24:11So, hence I will show this portion by positive sign.
24:17So here I have completed the shear force diagram and pending moment diagram for this simply
24:22supported beam.