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 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
00:41load of 3 kN per m over a length of 2 m.
00:47So for this setup, we have to draw shear force diagram and bending movement diagram.
00:57So first of all, I will draw free body diagram for this beam section.
01:05So for solving this numerical problem, first we have to convert this UDL into point load.
01:15So to convert this, I will multiply this UDL value with the length over which this
01:22UDL acts.
01:24So the point load equal to UDL of 3 kN per m multiplied by distance DE that is 2 m.
01:33So here I have got the converted point load of 6 kN.
01:40Now this converted point load will be acting on the midpoint of their UDL distance.
01:47That is 6 kN load is acting on the midpoint of length DE.
01:57Now this type of problem we are going to solve in three steps.
02:03In the first step, we have to calculate the values of support reaction forces Ra and Rb.
02:11So to calculate these values, I will use two equations of equilibrium.
02:18That is first equation is, summation of Fy equal to zero.
02:23That means addition of all the forces in the vertical axis equal to zero.
02:30And the second equation is, summation of moment equal to zero.
02:36That means addition of moments due to all the forces about any point must be zero.
02:46So here in the first equation, while I am doing the addition of all vertical forces,
02:51I will consider upward forces as positive and downward forces as negative.
03:00Here Ra and Rb are the vertical reaction forces acting on the beam in upward direction.
03:07So as per the sign convention, I will add these forces with positive sign.
03:14And the point loads of 4 kN and 2 kN are acting on the beam in downward direction.
03:23So as per the sign convention, I will add these forces with negative sign.
03:30And the converted point load of 6 kN is acting on the beam in downward direction.
03:37So as per the sign convention, I will add these forces with negative sign.
03:43Therefore, after calculating, this will get the equation, that is Ra plus Rb equal to
03:5112 kN.
03:54So I will give this as equation number one.
03:59Now the next equation is, summation of moment equal to zero.
04:05So for calculating moment, either we can take moment at point A or at point B.
04:15And here my sign convention would be, clockwise movement as positive and anticlockwise movement
04:21as negative.
04:25So for taking moment at point A, I will fix the beam at point A.
04:31Now here, 4 kN force will be pushing this beam towards downward.
04:37So it rotates the beam clockwise from fixed point.
04:42And as per the sign convention, for clockwise movement, I will add this force with positive
04:47sign.
04:50Now as per decided, moment is force into distance from fixed point.
04:56So here plus 4 kN is the force into 1 m is the distance from fixed point.
05:04Now here 6 kN converted point load will be pushing this beam towards downward.
05:10So it rotates the beam clockwise from fixed point.
05:14And as per the sign convention, for clockwise movement, I will add this force with positive
05:20sign.
05:23Now as per decided, moment is force into distance from fixed point.
05:29So here 6 kN is the force into 3 m is the distance from fixed point.
05:37Now here 2 kN force will be pushing this beam towards downward.
05:42So it rotates the beam clockwise from fixed point.
05:46And as per the sign convention, for clockwise movement, I will add this force with positive
05:52sign.
05:55Now as per decided, moment is always force into distance from fixed point.
06:02So here plus 2 kN is the force into 5 m is the distance from fixed point.
06:09Now here reaction force Rb will be pushing this beam towards upward.
06:14So it rotates the beam anticlockwise from fixed point.
06:19So as per the sign convention, for anticlockwise movement, I will add this force with negative
06:24sign.
06:26And moment is force into distance from fixed point.
06:30So I will multiply this force with the distance that is 6 m.
06:37So these are the moments.
06:39So therefore, by calculating, this will get the value of reaction force Rb as 5.34 kN.
06:50Now I will put this value of Rb in equation number 1.
06:55And after calculating, this will get the value of reaction force Ra as 6.66 kN.
07:04So now with the help of these calculated values of Ra and Rb, I will further calculate
07:09the values of shear forces at all the points of beam.
07:15So the next step is calculations of shear forces.
07:20And for the shear force calculations, our sign convention is, upward forces are considered
07:26as positive and downward forces are considered as negative.
07:32And here you should note that, while calculating the shear force at a particular point load,
07:38you can calculate shear force values for left side and right side of that particular point
07:45load.
07:47But while calculating the shear force at uniformly distributed load, you should calculate shear
07:53force values at start point and end point of uniformly distributed load.
08:00That is shear force at point D and shear force at point E, we need to calculate.
08:08And here at point A and point B, there are support reaction forces Ra and Rb, which are
08:15the point loads.
08:18And since these are the point loads, hence as per the rule, for point load, I will calculate
08:24the shear force values at left side and right side of point A and point B.
08:31Now at point C, there is one point load.
08:35Hence as per the rule, for point load, I will calculate the shear force values at left side
08:41and right side of point C. Now at point F, there is one point load.
08:48Hence as per the rule, I will calculate the shear force values at left side and right
08:53side of point F. So, I will start the shear force calculation
09:00from left-hand side of the beam.
09:02Therefore, first to calculate shear force at point A to its left, I will take the section
09:10to the left of point A, that is Sf at A to the left equal to.
09:17So as you can see, there is no forces acting at the left side of point A, therefore Sf
09:24at A to the left equal to zero.
09:29So to draw the shear force diagram, I will first draw a horizontal reference line of
09:35zero kilo Newton shear force.
09:38So here I will mark this point of zero kilo Newton shear force on the reference line,
09:45that is shear force at point A to its left is zero kilo Newton.
09:52Now if I go to the section to the right side of point A, that is Sf at A to the right equal
09:58to, then there is reaction force Re that we had calculated as 6.66 kilo Newton which is
10:06acting on the beam in upward direction.
10:10So as per the sign convention, I will consider upward forces as positive.
10:15So here the shear force is plus 6.66 kilo Newton, therefore Sf at A to the right equal
10:22to plus 6.66 kilo Newton.
10:27Here as the shear force value is positive, so I will mark this point above the reference
10:32line of zero kilo Newton shear force.
10:36And I will connect these two points with a vertical line.
10:43Now at point C, there is one point load, so I will calculate shear force values at left
10:49side and right side of point C. Therefore first to calculate shear force at
10:56point C to its left, I will take the section to the left of point C, that is Sf at C to
11:04the left equal to, and here I will carry forward previous value of shear force up to point
11:12A to its right which is 6.66 kilo Newton.
11:18And when we go to the left side of point C, then there is no load acting on the beam at
11:25left side of point C. Therefore Sf at C to the left equal to plus
11:306.66 kilo Newton.
11:34Here as you can see there is no variation in shear force values at right side of point
11:40A and left side of point C. Hence I will make the horizontal line which
11:45shear force value adds plus 6.66 kilo Newton.
11:51Now next to calculate shear force at point C to its right, I will take the section to
11:58the right of point C, that is Sf at point C to its right equal to, so here I will carry
12:07forward previous value of shear force up to point C to its left which is 6.66 kilo Newton.
12:17And when we go to the right side of point C, then there is one point load of 4 kilo
12:23Newton acting on the beam in downward direction.
12:28So as per the sign convention, I will consider downward forces as negative.
12:34So I will add this point load with negative sign.
12:39So therefore by calculating, this will get the shear force value at point C to its right
12:46as 2.66 kilo Newton.
12:50That is Sf at point C to its right equal to 2.66 kilo Newton.
12:57Here as the shear force value is positive, so I will mark this point above the reference
13:03line of zero kilo Newton shear force.
13:07And I will connect these two points with a vertical line.
13:13Now the point D is the starting point of UDL, hence I am taking section to point D and I
13:19will calculate the shear force at point D, that is Sf at point D equal to.
13:28Now there is no load on the beam between the right side of point C and point D, therefore
13:35shear force remains constant.
13:38That is Sf at point D equal to 2.66 kilo Newton.
13:44Here as there is no variation in shear force values, so I will make the horizontal line
13:50with shear force value as 2.66 kilo Newton.
13:56Now the point E is the end point of UDL, so I am taking section to point E, that is Sf
14:03at point E equal to.
14:08And here I will carry forward, previous value of shear force up to point D, which is 2.66
14:14kilo Newton and to the left side of point D, there is UDL of 3 kilo Newton per meter,
14:21which we had already converted into point load of 6 kilo Newton, which is acting on
14:26the beam in downward direction.
14:30But as per the circumvention, I will consider downward forces as negative, hence I will
14:35add this point load with negative sign.
14:40Therefore after calculating, this will get the shear force value as minus 3.34 kilo Newton.
14:49That is Sf at point D equal to minus 3.34 kilo Newton.
14:55Here as the shear force value is negative, hence I will mark this point of shear force
15:01below the reference line of 0 kilo Newton shear force.
15:06And here the type of load is UDL over the length of 2 meter, hence to draw the shear
15:12force diagram, I will indicate UDL with inclined line.
15:17So I will connect these two points with inclined line.
15:24Now at point F, there is one point load, so I will calculate the shear force values
15:30at left side and right side of point F. Therefore, first to calculate shear force at point F
15:38to its left, I will take the section to the left of point F, that is Sf at F to the left
15:46equal to and here I will carry forward previous value of shear force up to point T, which
15:54is minus 3.34 kilo Newton.
15:58And when we go to the left side of point F, then there is no load acting on the beam at
16:05left side of point F. Therefore, Sf at F to the left equal to minus 3.34 kilo Newton.
16:15Here as you can see, there is no variation in shear force values at point T and left
16:22side of point F, hence I will make the horizontal line with shear force value as minus 3.34
16:29kilo Newton.
16:31Now next to calculate shear force at point F to its right, I will take the section to
16:40the right of point F, that is Sf at point F to its right equal to, so here I will carry
16:48forward previous value of shear force up to point F to its left, which is minus 3.34
16:55kilo Newton.
16:58And when we go to the right side of point F, then there is one point load of 2 kilo
17:02Newton acting on the beam in downward direction.
17:08So as per the sign convention, I will consider downward forces as negative.
17:13So here I will add this point load with negative sign.
17:19So by calculating, this will get the shear force value as minus 5.34 kilo Newton.
17:26That is Sf at point F to its right equal to minus 5.34 kilo Newton.
17:34Here as the shear force value is negative, hence I will mark this point of shear force
17:39below the reference line of zero kilo Newton shear force.
17:45And I will connect these two points with a vertical line.
17:50Now at point B, there is reaction force Rb acting on the beam in the upward direction
17:58and since it is a point load, hence as per the rule for point load, I will calculate
18:03the shear force values at left side and right side of that point load.
18:09Therefore, first to calculate shear force at point B to its left, I will take the section
18:17to the left of point B, that is Sf at B to the left equal to and here I will carry forward
18:25previous value of shear force up to point F to its right which is minus 5.34 kilo Newton.
18:35And when we go to the left of point B, then there is no load acting on the beam at left
18:40side of point B. Therefore, Sf at B to the left equal to minus 5.34 kilo Newton.
18:49Here as you can see, there is no variation in shear force values between the right side
18:54of point F and left side of point B. Hence, I will make the horizontal line with
19:00shear force value as minus 5.34 kilo Newton.
19:07Now next to calculate shear force at point B to its right, I will take the section to
19:14the right of point B, that is Sf at point B to its right equal to, so here I will carry
19:23forward previous value of shear force up to point B to its left which is minus 5.34
19:30kilo Newton.
19:32And when we go to the right side of point B, then there is reaction force Rb of 5.34
19:39kilo Newton which is acting on the beam in upward direction.
19:45So as per the sign convention, I will consider upward forces as positive.
19:50So I will add this upward force with positive sign.
19:55So here minus 5.34 plus 5.34 gives me the value of shear force as zero kilo Newton.
20:03Therefore, Sf at point B to its right equal to zero kilo Newton.
20:11So I will mark this point of zero kilo Newton shear force on the reference line.
20:16That is Sf at point B to its right equal to zero kilo Newton.
20:23And I will connect these two points with a vertical line.
20:29And here in shear force diagram, whatever the portion drawn above the reference line,
20:35I will show this portion by positive sign.
20:39And the portion drawn below the reference line, I will show this portion by negative
20:43sign.
20:46So here I have completed the shear force diagram.
20:52Now the next step is calculations of bending movement.
20:57So bending movement at a section of beam is calculated as the algebraic sum of the movement
21:05of all the forces acting on one side of section.
21:10So to calculate bending movement, we can start either from left end of beam or from
21:16right end of beam.
21:19Here I will start from left end of beam.
21:24So whenever you are calculating the bending movement, you should remember these conditions.
21:30So here for simply supported beam, the condition is, at the ends of simply supported beam,
21:37the bending movements will be zero.
21:40So bending movement at point A and bending movement at point B will be zero.
21:46That is B on surface A equal to zero kilo Newton meter and B on surface B equal to zero
21:53kilo Newton meter.
21:57So to draw the bending movement diagram, firstly I will draw the reference line of bending
22:03movement zero kilo Newton meter.
22:06So here I will mark these values with a point on the reference line.
22:14So now we have to calculate bending movement at point C.
22:20So here in case of simply supported beam, while you are doing the calculations for bending
22:25movement at a particular point, you should always add movement of all the forces present
22:34either from left end of beam or from right end of beam up to that particular point at
22:40which you are calculating the bending movement.
22:45It means while I am calculating the bending movement at point C, I will add movement of
22:52all the forces present either from left end of beam or from right end of beam up to point
23:00C.
23:03And for bending movement calculation, our sign convention is for sagging effect of beam,
23:09the force is considered as positive and for hugging effect of beam, the force is considered
23:14as negative.
23:19So first to calculate bending movement at point C, that is B on surface C equal to.
23:28Here at left hand side of point C, there is reaction force R A of 6.66 kilo Newton,
23:35which is acting on the beam in upward direction.
23:39Due to this force, the beam shows sagging effect.
23:43And for sagging effect of beam, I will consider this force as positive.
23:49So I will add this force with positive sign.
23:54And as per decided, movement is force into distance, so I will multiply this force with
24:01the distance from point of action of force.
24:05That is 1 meter.
24:07Therefore, after calculating, this will get the value of bending movement at point C equal
24:14to 6.66 kilo Newton meter.
24:18That is B on surface C equal to 6.66 kilo Newton meter.
24:24So as it is positive value of bending movement, hence I will mark this point above the reference
24:30line of bending movement 0 kilo Newton meter.
24:35Now there is no any load present on the beam between point A and point C.
24:40Therefore, to draw the bending movement diagram, I will connect these two points with inclined
24:47line.
24:48Now next we have to calculate bending movement at point D.
24:56That is B on surface D equal to.
25:01Here at left-hand side of point D, there is reaction force array of 6.66 kilo Newton,
25:08which is acting on the beam in upward direction.
25:12Due to this force, the beam shows sagging effect.
25:17And for sagging effect of beam, I will consider this force as positive.
25:23So I will add this force with positive sign.
25:28And as we know, movement is force into distance.
25:32So I will multiply this force with the distance from point of action of force.
25:38That is 2 meter.
25:42And here at point C, there is one force of 4 kilo Newton, which is acting on the beam
25:50in downward direction.
25:53Due to this force, the beam shows hogging effect.
25:57And for hogging effect of beam, I will consider this force as negative.
26:02So I will add this force with negative sign.
26:07And I will multiply this force with the distance from point of action of force.
26:12That is 1 meter.
26:15Therefore, after calculating, this will get the value of bending movement at point D equal
26:23to 9.32 kilo Newton meter.
26:27That is B on surface D equal to 9.32 kilo Newton meter.
26:34So as it is positive value of bending movement, hence I will mark this point above the reference
26:39line of bending movement at 0 kilo Newton meter.
26:45Now there is no any load present on the beam between point C and point D.
26:52Therefore, to draw the bending movement diagram, I will connect these two points with inclined
26:57line.
27:02Now here you should note that, in shear force diagram, when the shear force values changes
27:08from positive to negative, then at that point, the bending movement value is maximum.
27:15So I will represent this point with point M.
27:20So at point M, we need to find out maximum bending movement value.
27:26But here we don't know the location of point M.
27:29That is we don't know the distance x.
27:33But we can find out this distance x by using the equation of similarity of triangles.
27:40Because here we know the distance Te, which is 2 meter.
27:46And here the values of shear forces, we are considering as heights of triangles.
27:51Therefore, for the first triangle, the height is 2.66 and base is x.
28:00And for second triangle, the height is 3.34 and base is 2-x.
28:07Therefore, from similarity of triangles, the base of first triangle, that is x, divided
28:14by height of first triangle, which is 2.66, is equal to base of second triangle, that
28:24is 2-x, divided by height of second triangle, which is 3.34.
28:33Now here the unknown term is x, therefore, after calculating, this will get the distance
28:40x, equal to 0.886 meter.
28:45So here I got the location of point M. So from distance x, next I will calculate
28:51the maximum bending movement at point M. So next to calculate bending movement at point
28:59M, that is BM surface M, equal to, here at left-hand side of point M, there is reaction
29:09force array of 6.66 kN, which is acting on the beam in upward direction.
29:18Due to this force, the beam shows sagging effect.
29:22And for sagging effect of beam, I will consider this force as positive.
29:28So I will add this force with positive sign.
29:32And as we know, moment is force into distance.
29:37So I will multiply this force with the distance from point of action of force, that is 2.886
29:44meter.
29:48And here at point C, there is one force of 4 kN, which is acting on the beam in downward
29:56direction.
29:57Due to this force, the beam shows hogging effect.
30:02And for hogging effect of beam, I will consider this force as negative.
30:07So I will add this force with negative sign.
30:11And I will multiply this force with the distance from point of action of force, that is 1.886
30:18meter.
30:22And here from point D to point M, there is UDL of 3 kN per meter over the distance of
30:31point 886 meter.
30:33So I will first convert this portion of UDL into point load.
30:39That is 3 kN per meter is the UDL value multiplied with the distance over which this UDL acts,
30:48that is 0.886 meter.
30:52Now this converted point load is acting at the midpoint of this much portion of UDL in
30:57downward direction.
31:00Due to this converted point load, the beam shows hogging effect.
31:04And for hogging effect of beam, I will consider this force as negative.
31:10So I will add this converted point load with negative sign.
31:15So I will multiply this force with the distance from point of action of force, that is half
31:21of 0.886 meter.
31:25Therefore, after calculating, this will get the value of bending movement at point M equal
31:33to 10.50 kNm.
31:38That is beam surface M equal to 10.50 kNm.
31:45So as it is positive value of bending movement, hence I will mark this point above the reference
31:51line of bending movement at 0 kNm.
31:56And here between point D and point M, there is UDL, therefore to draw the bending movement
32:03diagram, I will indicate UDL with a parabolic curve.
32:08Hence I will join these two points with a parabolic curve.
32:17Now next we have to calculate bending movement at point E, that is beam surface E equal to.
32:26Here at right-hand side of point E, there is reaction force Rb of 5.34 kN, which is
32:33acting on the beam in upward direction.
32:37Due to this force, the beam shows sagging effect.
32:42And for sagging effect of beam, I will consider this force as positive, so I will add this
32:48force with positive sign.
32:52And as we know, movement is force into distance, so I will multiply this force with the distance
32:59from point of action of force, that is 2 meter.
33:05And here at point F, there is one force of 2 kN, which is acting on the beam in downward
33:13direction.
33:14Due to this force, the beam shows hogging effect.
33:19And for hogging effect of beam, I will consider this force as negative, so I will add this
33:24force with negative sign.
33:28And I will multiply this force with the distance from point of action of force, that is 1 meter.
33:36Therefore, after calculating, this will get the value of pedding movement at point E equal
33:44to 8.68 kNm, that is beam surface E equal to 8.68 kNm.
33:55So as it is positive value of pedding movement, here I will mark this point above the reference
34:01line of pedding movement at 0 kNm.
34:06And here, between point M to point T, there is UDL, therefore to draw the pedding movement
34:14diagram, I will indicate UDL with a parabolic curve.
34:18Hence, I will join these two points with a parabolic curve.
34:27Now next we have to calculate pedding movement at point F, that is beam surface F equal to
34:33Here at right-hand side of point F, there is reaction force RB of 5.34 kN, which is
34:43acting on the beam in upward direction.
34:47Due to this force, the beam shows sagging effect.
34:51And for sagging effect of beam, I will consider this force as positive, so I will add this
34:58force with positive sign.
35:02And as we know, movement is forced into distance, so I will multiply this force with the distance
35:10from point of action of force, that is 1 meter.
35:15Therefore, after calculating, this will get the pedding movement value at point F equal
35:22to 5.34 kNm.
35:26That is beam surface F equal to 5.34 kNm.
35:33So as it is positive value of pedding movement, hence I will mark this point above the reference
35:39line of pedding movement 0 kNm and there is no any load present on the beam between
35:48the point T and point F. Therefore, to draw the pedding movement diagram, I will connect
35:54these two points with inclined line and also there is no any load present on the beam between
36:05point F and point B. Therefore, to draw the pedding movement diagram, I will connect these
36:12two points with inclined line.
36:18Now since I can see, this pedding movement diagram is drawn above the reference line
36:23of pedding movement 0 kNm, so whatever the portion I have drawn, all the values of pedding
36:29movement are positive.
36:30Now, hence I will show this portion by positive sign.
36:36So here I have completed the shear force diagram and pedding movement diagram for this simply
36:41supported beam.