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 a shear force diagram and bending
00:12movement diagram for a simply supported beam as shown in the figure.
00:19So the statement is given as, Draw a shear force and bending movement diagram
00:24for a simply supported beam AB 5 m long is loaded as shown in the figure.
00:34So this is the simply supported beam AB of length 5 m and carrying a uniformly distributed
00:40load of 20 kN per m over a length of 5 m and there is one point load of 45 kN is acting
00:49on the beam at point C.
00:54So for this setup, we have to draw the shear force diagram and bending movement diagram.
01:03So first of all, I will draw the free body diagram for this beam section.
01:10So for solving this numerical problem, first we have to convert this uniformly distributed
01:15loads into point loads.
01:20So to convert this, I will multiply this UDL value with the length over which the UDL acts.
01:29So here I will get the converted point load of 100 kN.
01:35Now this converted point load is acting on the midpoint of the length over which this
01:40UDL acts.
01:45Now this type of problem we are going to solve in three steps.
01:52In the first step, we have to calculate the values of support reaction forces RA and RB.
02:00So to calculate these values, I will use two equations of equilibrium.
02:07That is, first equation is, summation fi equal to zero.
02:12That means, addition of all the forces in the vertical axis equal to zero.
02:20And the second equation is, summation of moment equal to zero.
02:25That means, addition of moments due to all the forces about any point must be zero.
02:35So here in the first equation, while I am doing the addition of all the vertical forces,
02:41I will consider upward forces as positive and downward forces as negative.
02:49Here RA and RB are the vertical reaction forces acting on the beam in the upward direction.
02:56So as per the sign convention, I will add these forces with positive sign.
03:03And the point load of 45 kN is acting on the beam in the downward direction.
03:10So as per the sign convention, I will add these forces with negative sign.
03:16And the converted point load of 100 kN is acting on the beam in the downward direction.
03:23So as per the sign convention, I will add these forces with negative sign.
03:29Therefore, after calculating, this will get the equation.
03:34That is, RA plus RB equal to 145 kN.
03:39So I will give this as equation number one.
03:44Now the next equation is, summation of moment equal to zero.
03:51So for calculating moment, either we can take moment at point A or at point B.
04:00And here my sign convention would be, clockwise moment as positive and anticlockwise moment
04:05as negative.
04:09So for taking moment at point A, I will fix beam at point A.
04:15Now here 45 kN force will be pushing this beam towards downward.
04:21So it rotates the beam clockwise from fixed point.
04:25And as per the sign convention, for clockwise moment, I will add this force with positive
04:31sign.
04:34Now as per decided, moment is force into distance from fixed point.
04:40So here plus 45 kN is the force into 2 m is the distance from fixed point.
04:49And here converted point load of 100 kN will be pushing this beam towards downward.
04:55So it rotates the beam clockwise from fixed point.
05:00And for clockwise moment, I will add this force with positive sign.
05:06And the moment is force into distance from fixed point.
05:10So here plus 100 kN is the converted point load into 2.5 m is the distance from fixed
05:17point.
05:21Now here reaction force Rb will be pushing this beam towards upward.
05:25So it rotates the beam anticlockwise from fixed point.
05:30So as per the sign convention, for anticlockwise moment, I will add this force with negative
05:35sign.
05:38And as per decided, moment is force into distance from fixed point.
05:43So here minus Rb is the force into 5 m is the distance from fixed point.
05:52So these are the moments.
05:54So therefore, by calculating, this will get the reaction force Rb as 68 kN.
06:04Now I will put this value of Rb in equation number 1.
06:09And after calculating, this will get the value of reaction force Ra as 77 kN.
06:17So now with the help of these calculated values of Ra and Rb, I will further calculate the
06:23values of shear forces at all the points of beam.
06:28So the next step is calculations of shear forces.
06:33And for shear force calculations, our sign convention is, upward forces are considered
06:39as positive and downward forces are considered as negative.
06:46And here you should note that, while calculating the shear force at a particular point load,
06:52you can calculate shear force values for left side and right side of that particular point
06:58load.
06:59But while calculating the shear force at a uniformly distributed load, you should calculate
07:06shear force values at start point and end point of uniformly distributed load.
07:14That is shear force at point C and shear force at point D, we need to calculate.
07:21But here at point A and at point B, there are support reaction forces Ra and Rb, which
07:29are the point loads.
07:31And since these are the point loads, hence as per the rule, for point load, I will calculate
07:37the shear force values at left side and right side of point A and point B.
07:42Now at point C, there is a point load.
07:48Hence as per the rule, for point load, I will calculate the shear force values at left side
07:54and right side of point C. So we will start the shear force calculation.
08:01From left-hand side of the beam, therefore, first to calculate the shear force at point
08:08A to its left, I will take the section to the left of point A, that is Sf at A to the
08:15left equal to, so as you can see, there is no any force is acting at the left side of
08:23point A. Therefore, Sf at A to the left equal to zero.
08:31So to draw the shear force diagram, I will first draw a horizontal reference line of
08:37zero kilo Newton shear force.
08:40So here I will mark this point of zero kilo Newton shear force on the reference line.
08:46That is shear force at point A to its left is zero kilo Newton.
08:53Now if I go to the section to the right side of point A, that is Sf at A to the right equal
08:59to, then there is reaction force Ra, that we had calculated as 77 kilo Newton, which
09:07is acting on the beam in the upward direction.
09:12So as per the sign convention, I will consider upward forces as positive.
09:17So here the shear force is plus 77 kilo Newton.
09:22Therefore, Sf at A to the right equal to plus 77 kilo Newton.
09:30Here as the shear force value is positive, so I will mark this point above the reference
09:35line of zero kilo Newton shear force.
09:39And I will connect these two points with a vertical line.
09:47Now at point C, there is a point load.
09:50Hence, I will calculate the shear force values at left side and right side of point C.
09:58So first to calculate shear force at point C to its left, I will take the section to
10:04the left of point C, that is Sf at point C to its left equal to, and here I will carry
10:14forward previous value of shear force up to point A to its right, which is 77 kilo Newton.
10:23And to the left side of point C, there is UDL of 20 kilo Newton per meter.
10:30So I will convert this AC portion of UDL into point load.
10:35That is UDL of 20 kilo Newton per meter multiplied by distance 2 meter.
10:41Now this converted point load is acting on the pin in the downward direction.
10:47And as per the sign convention, I will consider downward force as negative, hence I will add
10:52this point load with negative sign.
10:57Therefore, after calculating, this will get the shear force value equal to 37 kilo Newton.
11:06Here as the shear force value is positive, hence I will mark this point of shear force
11:12above the reference line of zero kilo Newton shear force.
11:18And here the type of load is UDL over the length of 2 meter.
11:22Hence to draw the shear force diagram, I will indicate UDL with an inclined line.
11:28So I will connect these two points with an inclined line.
11:35Now if I go to the section to the right of point C, that is SF at point C to its right
11:42equal to.
11:43So here I will carry forward previous value of shear force up to point C to its left.
11:50That is 37 kilo Newton.
11:54And when we move towards the right side of point C, then there is one point load of 45
12:00kilo Newton, which is acting on the beam in downward direction.
12:06So as per the sign convention, I will consider downward force as negative.
12:12So I will add this point load with negative sign.
12:16Therefore, after calculating, this will get the shear force value as minus 8 kilo Newton.
12:24That is SF at point C to its right equal to minus 8 kilo Newton.
12:31Here as the shear force value is negative, hence I will mark this point of shear force
12:36below the reference line of zero kilo Newton shear force.
12:42And I will join these two points with a vertical line.
12:48Now at point B, there is reaction force RB acting on the beam in the upward direction.
12:55And since it is a point load, hence as per the rule for point load, I will calculate
13:00the shear force values at left side and right side of that point load.
13:07Therefore, first to calculate shear force at point B to its left, I will take the section
13:15to the left of point B. That is SF at B to the left equal to, and
13:22here I will carry forward, previous value of shear force up to point C to its right,
13:29which is minus 8 kilo Newton.
13:32And to the left side of point B, there is UDL of 20 kilo Newton per meter.
13:38So I will convert this CB portion of UDL into point load.
13:43That is UDL of 20 kilo Newton per meter multiplied by distance 3 meter.
13:49Now this converted point load is acting on the beam in downward direction.
13:55And as per the sign convention, I will consider downward force as negative, hence I will add
14:00this point load with negative sign.
14:05Therefore, after calculating, this will get the shear force value as minus 68 kilo Newton.
14:13That is SF at point B to its left equal to minus 68 kilo Newton.
14:20Here as the shear force value is negative, hence I will mark this point below the reference
14:24line of 0 kilo Newton shear force.
14:29And here the type of load is UDL over the length of 3 meter.
14:33Next to draw the shear force diagram, I will indicate UDL with inclined line.
14:39So I will connect these two points with inclined line.
14:45Now next to calculate shear force at point P to its right, I will take the section to
14:51the right of point P. That is SF at point P to its right equal to.
15:01So here I will carry forward previous value of shear force up to point P to its left which
15:07is minus 68 kilo Newton.
15:11And when we go to the right side of point P, then there is reaction force Rp of 68 kilo
15:17Newton which is acting on the beam in the upward direction.
15:22So as per the sign convention, I will consider upward forces as positive.
15:28So I will add this point load with positive sign.
15:33So here minus 68 plus 68 gives me the value of shear force as 0 kilo Newton.
15:40Therefore SF at point P to its right equal to 0 kilo Newton.
15:46So I will mark this point of 0 kilo Newton shear force on the reference line.
15:53And I will connect these two points with a vertical line.
15:59And here in shear force diagram, whatever the portion drawn above the reference line,
16:04I will show this portion by positive sign.
16:08And the portion which is drawn below the reference line, I will show this portion by negative
16:13sign.
16:17So here I have completed the shear force diagram.
16:22Now the next step is calculations of bending movements.
16:27So bending movement at a section of beam is calculated as the algebraic sum of the movement
16:33of all the forces acting on one side of section.
16:39So to calculate bending movements, we can start either from left end of beam or from
16:45right end of beam.
16:48So here I will start from left end of beam.
16:52So whenever you are calculating the bending movements, you should remember these conditions.
17:00So here for simply supported beam, the condition is, at the ends of simply supported beam,
17:06the bending movements will be zero.
17:09So bending movement at point A and bending movement at point B will be zero.
17:15That is B on surface A is equal to zero kNm and B on surface B is equal to zero kNm.
17:24So to draw the bending movement diagram, firstly I will draw the reference line of bending
17:30movement zero kNm.
17:33So here I will mark these values with points on the reference line.
17:39So now we have to calculate bending movement at point C.
17:44So here in case of simply supported beam, while you are doing the calculation for bending
17:50movement at a particular point, you should always add movement of all the forces present
17:58either from left end of beam or from right end of beam up to that particular point at
18:04which you are calculating the bending movement.
18:09It means, while I am calculating the bending movement at point C, I will add movement of
18:15all the forces present either from left end of beam or from right end of beam up to point
18:23C.
18:26And for bending movement calculation, our sign convention is, for sagging effect of
18:31beam, the force is considered as positive and for hogging effect of beam, the force
18:38is considered as negative.
18:43So to calculate bending movement at point C, that is B on surface C is equal to.
18:51Here at left hand side of point C, there is reaction force array of 77 kN, which is acting
18:58on the beam in the upward direction.
19:02Due to this force, the beam shows sagging effect.
19:06And for sagging effect of beam, I will consider this force as positive.
19:13So I will add this force with positive sign.
19:17And as per decided, the movement is forced into distance, so I will multiply this force
19:23with the distance from point of action of force.
19:27That is 2 m.
19:30And here from point A to point C, there is UDL of 20 kN per meter.
19:38So first I will convert this AC portion of UDL into point load.
19:44That is UDL of 20 kN per meter multiplied by distance 2 m.
19:51Now this converted point load is acting on the beam in downward direction.
19:57Due to this converted point load, the beam shows hogging effect.
20:01And for hogging effect of beam, I will consider this converted point load with negative sign.
20:08So I will add this converted point load with negative sign.
20:14And I will multiply this force with the distance from point of action of force.
20:19That is 1 m.
20:22Therefore, after calculating, this will get the value of bending moment at point C equal
20:29to 114 kN per meter.
20:34That is beam surface C equal to 114 kN per meter.
20:41So as it is positive value of bending moment, here I will mark this point of bending moment
20:47above the reference line of bending moment 0 kN per meter.
20:53And here between point A and point C, there is UDL.
20:58Therefore, to draw the bending moment diagram, I will indicate UDL with a parabolic curve.
21:06Hence I will join these two points with a parabolic curve.
21:12And also between point C and point B, there is UDL.
21:18Therefore, to draw the bending moment diagram, I will indicate UDL with a parabolic curve.
21:25Hence I will join these two points with a parabolic curve.
21:31Now since I can see this bending moment diagram is drawn above the reference line of bending
21:37moment 0 kN per meter.
21:39So out of the portion I have drawn, all the values of bending moment are positive.
21:44Hence I will show this portion by positive sign.
21:49So here I have completed the shear force diagram and bending moment diagram for this simply
21:54supported beam.