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 movement
00:12diagram for a simply supported beam AB as shown in figure.
00:19So the statement is given as, Draw shear force and bending movement diagram
00:25for a simply supported beam AB 6 m long is loaded as shown in figure.
00:34So this is the simply supported beam AB of length 6 m and carrying a uniformly distributed
00:40load of 2 kN per m over a length of 6 m.
00:47So for this setup, we have to draw the shear force diagram and bending movement diagram.
00:56So first of all, I will draw the free body diagram for this beam section.
01:03So for solving this numerical problem, first we have to convert this uniformly distributed
01:08load into point load.
01:12So to convert this, I will multiply.
01:16This UDL value with the length over which this UDL acts.
01:23So the point load equal to UDL of 2 kN per m multiplied by distance AB that is 6 m.
01:32So here I will get the converted point load of 12 kN.
01:39Now this converted point load is acting on the midpoint of the length over which this
01:44UDL acts.
01:47That is 12 kN load is acting on the midpoint of length AB.
01:55Now this type of problem we are going to solve in three steps.
02:02In the first step, we have to calculate the values of support reaction forces RA and RB.
02:10So to calculate these values, I will use two equations of equilibrium.
02:16Where first equation is, summation fi equal to zero.
02:21That means addition of forces in the vertical axis equal to zero.
02:28And the second equation is, summation of movement equal to zero.
02:33That means addition of movements due to all the forces about any point must be zero.
02:42So here in the first equation, while I am doing the addition of all vertical forces,
02:48I will consider upward forces as positive and downward forces as negative.
02:56Here RA and RB are the vertical reaction forces acting on the beam in the upward direction.
03:04So as per the sign convention, I will add these forces with positive sign.
03:10And the converted point load of 12 kN is acting on the beam in the downward direction.
03:16So as per the sign convention, I will add this force with negative sign.
03:22Therefore, after calculating, this will get the equation, that is RA plus RB equal to
03:2912 kN.
03:31So I will give this as equation number one.
03:34Now the next equation is, summation of movement equal to zero.
03:42So for calculating movement, either we can take movement at point A or at point B.
03:50And here my sign convention would be, clockwise movement as positive and anticlockwise movement
03:55as negative.
03:58So for taking movement at point A, I will fix beam at point A.
04:04Here converted point load of 12 kN will be pushing this beam towards downward.
04:10So it rotates the beam clockwise from fixed point.
04:14And for clockwise movement, I will add this force with positive sign.
04:20And the movement is forced into distance from fixed point.
04:24So here plus 12 kN is the converted point load into 3 m is the distance from fixed point.
04:33Now here reaction force RB will be pushing this beam towards upward.
04:38So it rotates the beam anticlockwise from fixed point.
04:42So as per the sign convention, for anticlockwise movement, I will add this force with negative
04:47sign.
04:50And as per decided, movement is forced into distance from fixed point.
04:55So here minus RB is the force into 6 m is the distance from fixed point.
05:03So these are the movements.
05:05So therefore, by calculating, this will get the value of reaction force RB as 6 kN.
05:14Now I will put this value of RB in equation number 1.
05:19And after calculating, this will get the reaction force RA as 6 kN.
05:27So now with the help of these calculated values of RA and RB, I will further calculate the
05:32values of shear forces at all the points of beam.
05:37So the next step is, calculations of shear forces.
05:42And for shear force calculations, our sign convention is, upward forces are considered
05:47as positive and downward forces are considered as negative.
05:55And here you should note that, while calculating the shear force at a particular point load,
06:01you can calculate shear force values for left side and right side of that particular
06:06point load.
06:09But while calculating the shear force at uniformly distributed load, you should calculate shear
06:15force values at start point and end point of uniformly distributed load.
06:22That is shear force at point C and shear force at point D, we need to calculate.
06:29But here at point A and at point B, there are support reaction forces RA and RB, which
06:37are the point loads.
06:39And since these are the point loads, hence as per the rule, for point load, I will calculate
06:44the shear force values at left side and right side of point A and point B.
06:51So I will start the shear force calculation from left-hand side of the beam.
06:55Therefore, first to calculate shear force at point A to its left, I will take the section
07:02to the left of point A. That is Sf at A to the left equal to.
07:10So as you can see, there is no forces acting at the left side of point A. Therefore, Sf
07:15at A to the left equal to zero.
07:20So to draw the shear force diagram, I will first draw a horizontal reference line of
07:25zero kN shear force.
07:29So here I will mark this point of zero kN shear force on the reference line.
07:34That is shear force at point A to its left is zero kN.
07:41Now if I go to the section to the right side of point A, that is Sf at A to the right equal
07:46to, then there is reaction force Ra that we had calculated as 6 kN, which is acting
07:53on the beam in the upward direction.
07:56So as per the sign convention, I will consider upward forces as positive.
08:01So here the shear force is plus 6 kN.
08:04Therefore, Sf at A to the right equal to plus 6 kN.
08:10Here as the shear force value is positive, so I will mark this point of plus 6 kN shear
08:15force above the reference line of zero kN shear force.
08:22And I will connect these two points with a vertical line.
08:29Now at point B, there is reaction force Rb acting on the beam in the upward direction.
08:35And since it is a point load, hence as per the rule, for point load, I will calculate
08:41the shear force values at left side and right side of that point load.
08:46Therefore, first to calculate shear force at point B to its left, I will take the section
08:54to the left of point B, that is Sf at B to the left equal to, so here I will carry forward
09:02previous value of shear force up to point A to its right, which is 6 kN.
09:10And to the left side of point B, there is UDL of 2 kN per meter, that we had converted
09:16into point load of 12 kN, which is acting on the beam in the downward direction.
09:23And as per the sign convention, I will consider downward forces as negative, hence I will
09:28add this point load with negative sign.
09:31Therefore, after calculating, this will get the shear force value as minus 6 kN, that
09:39is Sf at point B to its left equal to minus 6 kN.
09:45Here as the shear force value is negative, hence I will mark this point of shear force
09:51below the reference line of 0 kN shear force.
09:56And here the type of load is UDL over the length of 6 m.
10:01Hence to draw the shear force diagram, I will indicate UDL with inclined line.
10:07So, I will connect these two points with an inclined line.
10:13Now next to calculate shear force at point B to its right, I will take section to the
10:20right of point B, that is Sf at point B to its right.
10:26So here I will carry forward previous value of shear force up to point B to its left,
10:32which is minus 6 kN.
10:35And when we go to the right side of point B, then there is reaction force Rb of 6 kN,
10:42which is acting on the beam in the upward direction.
10:46So as per the sign convention, I will consider upward forces as positive, so I will add this
10:51upward force with positive sign.
10:56So here minus 6 kN plus 6 kN gives me the value of shear force at 0 kN.
11:04Therefore, Sf at point B to its right equal to 0 kN.
11:12So I will mark this point on the reference line, that is shear force at point B to its
11:17right equal to 0 kN.
11:21And I will connect these two points with a vertical line.
11:26And here in shear force diagram, whatever the portion drawn above the reference line,
11:31I will show this portion by plus sign and the portion drawn below the reference line,
11:37I will show this portion by negative sign.
11:41So here I have completed the shear force diagram.
11:46Now the next step is calculations of bending movements.
11:52So the bending movement at a section of beam is calculated as the algebraic sum of the
11:58movement of all the forces acting on the one side of the section.
12:04So to calculate these movements, we can start either from left end of beam or from right
12:10end of beam.
12:13Here I will start from left end of beam.
12:17So whenever you are calculating the bending movements, you should remember these conditions.
12:24So here for simply supported beam, the condition is, at the ends of simply supported beam,
12:30the bending movement will be zero.
12:34So bending movement at point A and bending movement at point B will be zero.
12:41That is B on surface A equal to 0 kNm and B on surface B equal to 0 kNm.
12:49So to draw the bending movement diagram, firstly I will draw the reference line of
12:54bending movement 0 kNm.
12:58So here I will mark these values with points on the reference line.
13:04Now here you should note that, in shear force diagram, when the shear force changes from
13:09positive to negative, then at that point, the bending movement value is maximum.
13:16So I will represent this with point M. So at point M, we need to find out maximum
13:23bending movement value.
13:26But here we don't know the location of point M.
13:30That is we don't know the distance x.
13:33But we can find out this distance x by using the equation of similarity of triangles.
13:39Because here we know the distance AB, which is 6 m.
13:44And here the values of shear forces were considered as heights of triangles.
13:49Therefore, for the first triangle, the height is 6 kN and the base is distance x.
13:57And for second triangle, the height is 6 kN and base is distance 6-x.
14:04Therefore, from similarity of triangles, the base of first triangle, that is x, divided
14:12by height of first triangle, which is 6 kN, is equal to base of second triangle, that
14:19is 6-x, divided by height of second triangle, which is 6 kN.
14:27Now here the unknown term is x, therefore, after calculating, this will get the distance
14:34x equal to 3 m.
14:38Now here I will get the location of point M. So, from distance x, next I will calculate
14:45the maximum bending movement value at point M.
14:49That is B of surface M equal to.
14:53So here, in case of simply supported beam, while you are doing the calculations for bending
14:58movement at a particular point, you should always add movement of all the forces present
15:07either from left end of beam or from right end of beam up to that particular point at
15:13which you are calculating the bending movement.
15:17It means, while I am calculating the bending movement at point M, I will add movement of
15:23all the forces present either from left end of beam or from right end of beam up to point
15:30M.
15:33And for bending movement calculations, our sign convention is, for sagging effect of
15:38beam, the force is considered as positive.
15:42And for hugging effect of beam, the force is considered as negative.
15:49Here at left end side of point M, there is reaction force Ra of 6 kN, which is acting
15:55on the beam in the upward direction.
15:59Due to this force, the beam shows sagging effect.
16:02And for sagging effect of beam, I will consider this force as positive.
16:09So I will add this force with positive sign.
16:14And as per decided, movement is forced into distance.
16:18So I will multiply this force with the distance, that is 3 m.
16:26And here, from point A to point M, there is UDL of 2 kN per meter over the distance
16:34of 3 m.
16:35So, I will convert this AM portion of UDL into point load.
16:40That is 2 kN per meter is the UDL.
16:44Multiply with the distance over which this UDL acts.
16:50Now this converted point load is acting at the midpoint of this much part of UDL in downward
16:55direction.
16:57Due to this converted point load, the beam shows hogging effect.
17:02And for hogging effect of beam, I will consider this force as negative.
17:07So I will add this converted point load with negative sign.
17:12And I will multiply this force with the distance from point of action of force, that is 1.5
17:20m.
17:23And here, between point A and point M, there is UDL.
17:28Therefore, to draw the bending moment diagram, I will indicate UDL with a parabolic curve.
17:34Hence, I will join these two points with a parabolic curve.
17:41And also, between point M and point P, there is UDL.
17:46Therefore, to draw the bending moment diagram, I will indicate UDL with a parabolic curve.
17:51Hence, I will join these two points with a parabolic curve.
17:58Now since I can see this bending moment diagram is drawn above the reference line of bending
18:04moment 0 kN m, hence I will show this portion by positive sign.
18:11So here I have completed the shear force diagram and bending moment diagram for this simply
18:16supported beam.