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 Betti-Bohmian
00:13diagram for a simply supported beam as shown in figure.
00:20So the statement is given as, Draw shear force and Betti-Bohmian diagram
00:26for a simply supported beam AB 6 m long is loaded as shown in figure.
00:36So this is the simply supported beam AB of length 6 m and carrying a UDL of 4 kN per
00:44m over a length of 1.5 m and another UDL of 2 kN per m over a length of 3 m.
00:57And there is one point load of 5 kN is acting on the beam at point E.
01:05So for this setup, we have to draw shear force diagram and Betti-Bohmian diagram.
01:13So first of all, I will draw free body diagram for this beam section.
01:22So for solving this numerical problem, first we have to convert these UDL values into point
01:28loads.
01:29So to convert this, I will multiply these UDL values with the length over which these
01:39UDL acts.
01:44So here I have got the converted point loads of 6 kN, 3 kN and 3 kN.
01:54Now these converted point loads will be acting on the midpoints of their UDL distance.
02:02That is 6 kN load is acting on the midpoint of distance AC and 3 kN load is acting on
02:10the midpoint of distance DE.
02:14And another 3 kN load is acting on the midpoint of distance EB.
02:23Now this type of problem we are going to solve in three steps.
02:30In the first step, we have to calculate the values of support reaction forces Ra and Rb.
02:39So to calculate these values, I will use two equations of equilibrium.
02:46That is first equation is, summation Fy is equal to zero.
02:52That means addition of all the forces in the vertical axis is equal to zero.
03:01And the second equation is, summation of moment is equal to zero.
03:07That means addition of moments due to all the forces about any point must be zero.
03:17So here in the first equation, while I am doing the addition of all vertical forces,
03:24I will consider upward forces as positive and downward forces as negative.
03:33Here Ra and Rb are the vertical reaction forces acting on the beam in upward direction.
03:40So as per the second equation, I will add these forces with positive sign.
03:48And the point load of 5 kN acting on the beam in downward direction.
03:54So as per the second equation, I will add these forces with negative sign.
04:01And the converted point loads of 6 kN, 3 kN and 3 kN are acting on the beam in downward
04:10direction.
04:11So as per the second equation, I will add these forces with negative sign.
04:18Therefore, after calculating, this will get the equation, that is Ra plus Rb is equal
04:27to 17 kN.
04:30So I will give this as equation number 1.
04:35Now the next equation is, summation of moment equal to zero.
04:43So for calculating moment, either we can take moment at point A or at point B.
04:52And here my second equation would be, clockwise movement as positive and anticlockwise movement
04:57as negative.
05:02So for taking moment at point A, I will fix the beam at point A.
05:08Now here, converted point load of 6 kN will be pushing this beam towards downward.
05:15So it rotates the beam clockwise from fixed point.
05:20And for clockwise movement, I will add this force with positive sign.
05:27And the moment is forcing to distance from fixed point.
05:32So we will multiply this force with the distance, that is 0.75 m.
05:41Now here converted point load of 3 kN will be pushing this beam towards downward.
05:48So it rotates the beam clockwise from fixed point.
05:52And for clockwise movement, I will add this force with positive sign.
06:00And the moment is forcing to distance from fixed point.
06:04So I will multiply this force with the distance.
06:11Now here 5 kN force will be pushing this beam towards downward.
06:16So it rotates the beam clockwise from fixed point.
06:21So for clockwise movement, I will add this force with positive sign.
06:28And the moment is forcing to distance from fixed point.
06:33So I will multiply this force with the distance, that is 4.5 m.
06:41Now here converted point load of 3 kN will be pushing this beam towards downward.
06:48So it rotates the beam clockwise from fixed point.
06:52So for clockwise movement, I will add this force with positive sign.
07:00And the moment is forcing to distance from fixed point.
07:04So I will multiply this force with the distance, that is 5.25 m.
07:13Now here reaction force Rb will be pushing this beam towards upward.
07:18So it rotates the beam anticlockwise from fixed point.
07:23So as per the sign convention, for anticlockwise movement, I will add this force with negative
07:29sign.
07:32And as per decided, moment is forcing to distance from fixed point.
07:38So here minus Rb is the force into 6 m is the distance.
07:45So these are the moments.
07:47So therefore, by calculating, this will get the reaction force Rb as 9 kN.
07:56Now I will put this value of Rb in equation number 1.
08:02And after calculating, this will get the value of reaction force Ra as 8 kN.
08:10So now with the help of these calculated values of Ra and Rb, I will further calculate the
08:16values of shear forces at all the points of beam.
08:21So the next step is calculations of shear forces.
08:27And for shear force calculation, our sign convention is, upward forces are considered
08:32as positive and downward forces are considered as negative.
08:41And here you should note that, while calculating the shear force at a particular point load,
08:48you can calculate shear force values for left side and right side of that particular point
08:55load.
08:58But while calculating the shear force at uniformly distributed load, you should calculate shear
09:04force values at start point and end point of uniformly distributed load.
09:13That means, for the first UDL, I will calculate the shear force values at point A and at point
09:21C and for second UDL, I will calculate the shear force values at point D and at point
09:30B. But in this problem, at point A, there is
09:37reaction force Ra, which is a point load.
09:42And since it is a point load, hence as per the rule, for point load, I will calculate
09:48the shear force values at left side and right side of that point load.
09:54That is, shear force at point A to its left and shear force at point A to its right, we
10:01need to calculate.
10:05And also at point B, there is reaction force Rb, which is a point load.
10:11And since it is a point load, hence I will calculate the shear force values at left side
10:16and right side of point B. And at point E, there is a point load of
10:255 kN, hence as per the rule, for point load, I will calculate the shear force values for
10:33left side and right side of point E. So, we will start the shear force calculation
10:42from left-hand side of the beam.
10:44Therefore, first to calculate the shear force at point A to its left, I will take the section
10:52to the left of point A. That is, Sf at A to the left is equal to.
10:59So, as you can see, there is no any force acting at the left of point A.
11:04Therefore, Sf at A to the left is equal to zero.
11:10So, to draw the shear force diagram, I will first draw a horizontal reference line of
11:17zero kN shear force.
11:21So, here I will mark this point of zero kN shear force on the reference line.
11:28That is, Sf at point A to its left is zero kN.
11:35Now, if I go to the section to the right of point A, that is, Sf at A to the right is
11:42equal to.
11:45Then there is reaction force Ra, that we had calculated as 8 kN, which is acting on the
11:52beam in upward direction.
11:55So, as per the sign convention, I will consider upward forces as positive.
12:00So, here the shear force is plus 8 kN.
12:05Therefore, Sf at A to the right is equal to plus 8 kN.
12:12Here as the shear force value is positive, so I will mark this point of shear force above
12:19the reference line of zero kN shear force.
12:25And I will connect these two points with a vertical line.
12:31Now, the point C is the endpoint of VDL.
12:34So, I am taking section to point C. That is, Sf at point C is equal to.
12:43And here I will carry forward previous value of shear force up to point A to its right,
12:50which is 8 kN.
12:53And to the left side of point C, there is VDL of 4 kN per meter, which we had already
13:00converted into point load of 6 kN, which is acting on the beam in downward direction.
13:09So, as per the sign convention, I will consider downward forces as negative.
13:15Hence, I will add this point load with negative sign.
13:20Therefore, after calculating, this will get the shear force value as 2 kN.
13:28That is, Sf at point C is equal to 2 kN.
13:34Here as the shear force value is positive, hence I will mark this point of shear force
13:41above the reference line of zero kN shear force.
13:48And here the type of load is VDL over the length of 1.5 m.
13:52Hence, to draw the shear force diagram, I will indicate VDL with inclined line.
13:59So, I will connect these two points with inclined line.
14:06Now, the point D is the starting point of VDL.
14:10Hence, I am taking section to point D and I will calculate shear force at point D.
14:17That is, Sf at point D is equal to
14:22Now, there is no load on the beam between the right side of point C and point D.
14:29Therefore, shear force remains constant.
14:32That is, shear force at point D is equal to 2 kN.
14:38Here as there is no variation in shear force values, so I will make the horizontal line
14:44with shear force value as 2 kN.
14:49Now, at point E, there is a point load.
14:54Hence, I will calculate the shear force values at left side and right side of point E.
15:01So, first to calculate shear force at point E to its left, I will take the section to
15:11the left of point D. That is, Sf at point D to its left is equal to
15:18And here I will carry forward previous value of shear force up to point D, which is 2 kN.
15:27And to the left side of point D, there is VDL of 2 kN per meter, which we have already
15:36converted to point load of 3 kN, which is acting on the beam in downward direction.
15:43And as per the sign convention, I will consider downward forces as negative.
15:49Hence, I will add this point load with negative sign.
15:55Therefore, after calculating, this will get the shear force value as minus 1 kN.
16:04That is, Sf at point D to its left is equal to minus 1 kN.
16:11Here as the shear force value is negative, hence I will mark this point of shear force
16:17below the reference line of 0 kN shear force.
16:23And here the type of load is UDL over the length of 1.5 m.
16:29Hence, to draw the shear force diagram, I will indicate UDL with inclined line.
16:36So, I will connect these two points with inclined line.
16:43Now, if I go to the section to the right of point E, that is, Sf at point E to its right
16:50is equal to So, here I will carry forward previous value
16:57of shear force up to point T to its left, which is minus 1 kN.
17:04And when we move towards the right side of point T, then there is one point load of 5
17:11kN, which is acting on the beam in downward direction.
17:16So, as per the sign convention, I will consider downward forces as negative.
17:23So, I will add this point load with negative sign.
17:27Therefore, after calculating, this will get the shear force as minus 6 kN.
17:36That is, Sf at point T to its right is equal to minus 6 kN.
17:43Here as the shear force value is negative, hence I will mark this point of shear force
17:50below the reference line of 0 kN shear force.
17:56And I will join these two points with a vertical line.
18:01Now, at point B, there is reaction force Rb acting on the beam in upward direction.
18:10And since it is a point load, hence as per the rule, for point load, I will calculate
18:16the shear force values at left side and right side of that point load.
18:21Therefore, first to calculate shear force at point B to its left, I will take the section
18:29to the left of point B, that is Sf at B to the left equal to, so here I will carry forward
18:38previous value of shear force up to point T to its right, which is minus 6 kN.
18:47And to the left side of point B, there is UDL of 2 kN per meter, which we had already
18:54converted into point load of 3 kN, which is acting on the beam in downward direction.
19:03And as per the sign convention, I will consider downward forces as negative, hence I will
19:09add this point load with negative sign.
19:13Therefore, after calculating, this will get the shear force value as minus 9 kN.
19:22That is Sf at point B to its left equal to minus 9 kN.
19:29Here as the shear force value is negative, hence I will mark this point of shear force
19:35below the reference line of 0 kN shear force.
19:41And here the type of load is UDL over the length of 1.5 meter, hence to draw the shear
19:47force diagram, I will indicate UDL with inclined line.
19:54So I will connect these two points with inclined line.
20:01Now next to calculate, shear force at point B to its right, I will take the section to
20:08the right of point B, that is Sf at point B to its right equal to, so here I will carry
20:17forward previous value of shear force up to point B to its left, which is minus 9 kN.
20:27And when we go to the right side of point B, then there is reaction force Rb of 9 kN,
20:35which is acting on the beam in upward direction.
20:39So as per the sign convention, I will consider upward forces as positive.
20:45So I will add this upward force with positive sign.
20:51So here minus 9 kN plus 9 kN gives me the value of shear force at 0 kN.
21:00Therefore Sf at point B to its right equal to 0 kN.
21:07So I will mark this point on the reference line and I will connect these two points with
21:14a vertical line.
21:18And here in shear force diagram, whatever the portion drawn above the reference line,
21:24I will show this portion by positive sign.
21:28And the portion drawn below the reference line, I will show this portion by negative
21:33sign.
21:36So here I have completed the shear force diagram.
21:43Now the next step is calculations of bending movement.
21:49So bending movement at a section of beam is calculated as the algebraic sum of the movement
21:57of all the forces acting on one side of section.
22:03So to calculate bending movement, we can start either from left end of beam or from right
22:10end of beam.
22:13Here I will start from left end of beam.
22:17So whenever you are calculating the bending movement, you should remember these conditions.
22:25So here for simply supported beam, the condition is, at the ends of simply supported beam,
22:32the bending movement will be zero.
22:35So bending movement at point A and bending movement at point B will be zero.
22:41That is, B on surface A is equal to zero kNm and B on surface B is equal to zero kNm.
22:52So to draw the bending movement diagram, firstly I will draw the reference line of bending
22:57movement zero kNm.
23:01So here I will mark these values with the points on the reference line.
23:09So now we have to calculate bending movement at point C.
23:15So here, in case of simply supported beam, while you are doing the calculations for bending
23:21movement at a particular point, you should always add movement of all the forces present
23:30either from left end of beam or from right end of beam up to that particular point at
23:36which you are calculating the bending movement.
23:41It means, while I am calculating the bending movement at point C, I will add movement of
23:48all the forces present either from left end of beam or from right end of beam up to point
23:56C. And for bending movement calculation, our
24:02sign convention is, for sagging effect of beam, the force is considered as positive
24:10and for hugging effect of beam, the force is considered as negative.
24:17So first to calculate, bending movement at point C, i.e. beyond surface C, equal to,
24:27here at left-hand side of point C, there is reaction force Ra of 8 kN, which is acting
24:34on the beam in upward direction.
24:38Due to this force, the beam shows sagging effect and for sagging effect of beam, I will
24:45consider this force as positive.
24:49So I will add this force with positive sign.
24:54And as we know, movement is force into distance, so I will multiply this force with the distance
25:02from point of action of force, i.e. 1.5 m.
25:08And here, from point A to point C, there is UDL of 4 kN per meter, that we had converted
25:17into point load of 6 kN, which is acting on the beam in downward direction.
25:25Due to this load, the beam shows hugging effect and for hugging effect of beam, I will consider
25:32this force as negative.
25:35So I will add this converted point load with negative sign and I will multiply this force
25:43with the distance from point of action of force, i.e. 0.75 m.
25:51Therefore, after calculating, this will get the value of bending movement at point C,
25:59equal to 7.5 kNm.
26:03That is, beam surface C equal to 7.5 kNm.
26:10So as it is positive value of bending movement, here I will mark this point of bending movement
26:17above the reference line of bending movement at 0 kNm.
26:24And here, between point A and point C, there is UDL, therefore to draw the bending movement
26:31diagram, I will indicate UDL with a parabolic curve.
26:37Hence I will join these two points with a parabolic curve.
26:45Now next we have to calculate, bending movement at point D, i.e. beam surface D equal to.
26:55Here at left-hand side of point D, there is reaction force RA of 8 kN, which is acting
27:02on the beam in upward direction.
27:06Due to this force, the beam shows sagging effect and for sagging effect of beam, I will
27:14consider this force as positive.
27:17So I will add this force with positive sign.
27:22And as per decided, movement is forcing to distance, so I will multiply this force with
27:28the distance from point of action of force, i.e. 3 m.
27:35And here, from point A to point C, there is UDL of 4 kNm, that we have converted into
27:44point load of 6 kN, which is acting on the beam in downward direction.
27:51Due to this load, the beam shows hogging effect and for hogging effect of beam, I will consider
27:59this force as negative.
28:02So I will add this converted point load with negative sign.
28:07And I will multiply this force with the distance from point of action of force, i.e. 2.25 m.
28:17Therefore, after calculating, this will get the value of bending movement at point D equal
28:25to 10.5 kNm.
28:29That is, beam surface D equal to 10.5 kNm.
28:36So as it is positive value of bending movement, hence I will mark this point above the reference
28:43line of bending movement 0 kNm.
28:49Now, there is no any load present on the beam between point C and point D.
28:56Therefore, to draw the bending movement diagram, I will connect these two points with inclined line.
29:07Now, here you should note that, in shear force diagram, when the shear force value changes
29:13from positive to negative, then at that point, the bending movement value is maximum.
29:21So I will represent this point with point M. So at point M, we need to find out maximum
29:30bending movement value.
29:34But here we don't know the location of point M, i.e. we don't know the distance x.
29:41But we can find out this distance x, by using the equation of similarity of triangles.
29:49Because here we know the distance DE, which is 1.5 m.
29:55And here the values of shear forces, we are considering as sides of triangles.
30:01Therefore, for the first triangle, the height is 2 and base is x.
30:09And for second triangle, the height is 1 and base is 1.5-x.
30:16Therefore, from similarity of triangles, the base of first triangle, i.e. x, divided by
30:26height of first triangle, i.e. 2, is equal to base of second triangle, i.e. 1.5-x divided
30:38by height of second triangle, which is 1.
30:44Now here the unknown term is x.
30:45Therefore, after calculating, this will get the distance x equal to 1 m.
30:56So here I got the location of point M. So from distance x, next I will calculate the
31:03maximum bending movement at point M.
31:08Using x to calculate, bending movement at point M, i.e. beam surface M, equal to, here
31:17at left-hand side of point M, there is reaction force RA of 8 kN, which is acting on the beam
31:26in upward direction.
31:28Due to this force, the beam shows sagging effect.
31:33And for sagging effect of beam, I will consider this force as positive.
31:39So I will add this force with positive sign.
31:44And as per decided, moment is force into distance.
31:49So I will multiply this force with the distance from point of action of force, i.e. 4 m.
31:58And here, from point A to point C, there is UDL of 4 kN per meter, that we had converted
32:06into point load of 6 kN, which is acting on the beam in downward direction.
32:15Due to this load, the beam shows hogging effect.
32:19And for hogging effect of beam, I will consider this force as negative.
32:25So I will add this converted point load with negative sign.
32:30And I will multiply this force with the distance from point of action of force, i.e. 3.25 m.
32:40And here, from point D to point M, there is UDL of 2 kN per meter, over the distance of
32:491 m.
32:51So I will first convert this portion of UDL into point load, i.e. 2 kN per meter is the
32:59UDL value multiplied with the distance over which this portion of UDL acts, i.e. 1 m.
33:09Now this converted point load is acting at the midpoint of this much portion of UDL in
33:16downward direction.
33:18Due to this converted point load, the beam shows hogging effect.
33:24And for hogging effect of beam, I will consider this force as negative.
33:29So I will add this converted point load with negative sign.
33:34And I will multiply this force with the distance from point of action of force, i.e. 1.5 m.
33:44Therefore, after calculating, this will get the value of pedibohmmet at point M equal
33:51to 11.5 kNm, i.e. beyond surface M equal to 11.5 kNm.
34:04So as it is positive value of pedibohmmet, here I will mark this point above the reference
34:09line of pedibohmmet 0 kNm.
34:15And here, between point D and point M, there is UDL.
34:21Therefore, to draw the pedibohmmet diagram, I will indicate UDL with a parabolic curve.
34:28Hence, I will join these two points with a parabolic curve.
34:37Now next we have to calculate pedibohmmet at point T, i.e. beyond surface E equal to.
34:46Here at right-hand side of point T, there is reaction force Rb of 9 kN, which is acting
34:53on the beam in upward direction.
34:58Due to this force, the beam shows sagging effect.
35:02And for sagging effect of beam, I will consider this force as positive.
35:09So I will add this force with positive sign.
35:14And as per decided, moment is force into distance.
35:18So I will multiply this force with the distance from point of action of force, i.e. 1.5 m.
35:29And here, from point E to point B, there is UDL of 2 kNm, that were converted into point
35:39load of 3 kN, which is acting on the beam in downward direction.
35:46Due to this load, the beam shows hogging effect.
35:50And for hogging effect of beam, I will consider this force as negative.
35:56So I will add this converted point load with negative sign.
36:02And I will multiply this force with the distance from point of action of force, i.e. 0.75 m.
36:12Therefore, after calculating, this will get the value of pedibohmmet at point E equal
36:19to 11.25 kNm.
36:24That is, beam surface E equal to 11.25 kNm.
36:32So as it is positive value of pedibohmmet, here I will mark this point of pedibohmmet
36:39above the reference line of pedibohmmet 0 kNm.
36:46And here, between point T to point B, there is UDL.
36:52Therefore, to draw the pedibohmmet diagram, I will indicate UDL with a parabolic curve.
36:58Hence, I will join these two points with a parabolic curve.
37:05Now, since I can see, this pedibohmmet diagram is drawn above the reference line of pedibohmmet
37:140 kNm, so whatever the portion I have drawn, all the values of pedibohmmet are positive.
37:21Hence, I will show this portion by positive sign.
37:27So here I have completed the shear force diagram and pedibohmmet diagram for this simply supported
37:32beam.
37:33.