Simply Supported Beam Numerical 9: Draw Shear Force Diagram and Bending Moment Diagram by Shubham Kola
Subject - Strength of Materials
Chapter - Shear Force and Bending Moment Diagrams
Chapter - Shear Force and Bending Moment Diagrams
Transcript
00:07In this video
00:08we are going to learn
00:10how to draw shear force diagram
00:11and bending moment diagram
00:13for a simply supported beam AB
00:15as shown in figure
00:18so the statement is given as
00:21draw shear force
00:23and bending moment diagram
00:24for a simply supported beam
00:26AB
00:276 meter long
00:29is loaded
00:30as shown in figure
00:34so this is the
00:35simply supported beam AB
00:36of length 5 meter
00:39and carrying a uniformly distributed load
00:41of 3 kilo Newton per meter
00:43over a length of 2 meter
00:46so for this setup
00:48we have to draw shear force diagram
00:50and bending moment diagram
00:57so first of all
00:58I will draw free body diagram
01:00for this beam section
01:05so for solving this numerical problem
01:07first we have to convert
01:09this UDL
01:10into point load
01:14so to convert this
01:16I will multiply these UDL value
01:18with the length
01:20over which
01:21this UDL acts
01:24so the point load equal to
01:26UDL of 3 kilo Newton per meter
01:28multiply by distance DE
01:30that is 2 meter
01:33so here I got the converted point load
01:35of 6 kilo Newton
01:40now this converted point load
01:42will be acting on the midpoint
01:43of their UDL distance
01:47that is 6 kilo Newton load
01:49is acting on the midpoint
01:50of length DE
01:56now this type of problem
01:58we are going to solve in 3 steps
02:03In the first step
02:04we have to calculate
02:06the value of support reaction forces
02:08Ra and Rb
02:11so to calculate these values
02:13I will use
02:142 equations of equilibrium
02:17that is first equation is
02:19summation Fy
02:20equal to zero
02:22that means
02:23addition of all the forces
02:25in the vertical axis
02:27equal to zero
02:30and the second equation is
02:32summation of moment
02:34equal to zero
02:36that means
02:37addition of moments
02:39due to all the forces
02:40about any point
02:42must be 0
02:45so here in the first equation
02:47while I am doing the addition
02:48of all vertical forces
02:51I will consider
02:53upward forces as positive
02:55and downward forces as negative
03:00here Ra and Rb
03:01are the vertical reaction forces
03:03acting on the beam
03:05in the upward direction
03:07so as per the sign convention
03:09I will add these forces
03:10with positive sign
03:14and the point loads
03:15of 4 kilo Newton
03:172 kilo Newton
03:18are acting on the beam
03:20in downward direction
03:22so as per the sign convention
03:24I will add these forces
03:26with negative sign
03:29and the converted point load
03:31of 6 kilo Newton
03:33is acting on the beam
03:34in the downward direction
03:37so as per the sign convention
03:39I will add this force
03:40with negative sign
03:43therefore
03:44after calculating
03:46this will get the equation
03:48that is Ra plus Rb
03:50equal to 12 kilo Newton
03:53so I will give this as
03:55equation number 1
03:58now the next equation is
04:00summation of moment
04:02equal to zero
04:05so for calculating moment
04:07either we can take moment
04:09at point A
04:10or at point B
04:14and here my sign convention would be
04:17clockwise moment as positive
04:19and anticlockwise moment as negative
04:25so for taking moment at point A
04:27I will fixed beam at point A
04:31now here
04:324 kilo Newton force
04:34will be pushing this beam towards downward
04:36so it rotate the beam clockwise
04:38from fixed point
04:41and as per the sign convention
04:43for clockwise moment
04:45I will add this force
04:46with positive sign
04:50now as per decided
04:51moment is force into distance
04:53from fixed point
04:56so here plus 4 kilo Newton is the force
04:58into 1 meter is the distance
05:00from fixed point
05:04now here 6 kilo Newton converted point load
05:06will be pushing this beam toward downward
05:09so it rotate the beam clockwise
05:11from fixed point
05:14and as per the sign convention
05:16for clockwise moment
05:18I will add this force
05:19with positive sign
05:22now as per decided
05:24moment is force into distance
05:26from fixed point
05:28so here 6 kilo Newton is the force
05:31into 3 meter is the distance
05:33from fixed point
05:36now here 2 kilo Newton force
05:38will be pushing this beam toward downward
05:41so it rotate the beam clockwise
05:43from fixed point
05:46and as per the sign convention
05:48for clockwise moment
05:50I will add this force
05:51with positive sign
05:55now as per decided
05:56moment is always force into distance
05:59from fixed point
06:01so here plus 2 kilo Newton is the force
06:04into 5 meter is the distance
06:06from fixed point
06:09now here reaction force Rb
06:11will be pushing this beam toward upward
06:14so it rotate the beam anticlockwise
06:16from fixed point
06:18so as per the sign convention
06:20for anticlockwise moment
06:22I will add this force
06:24with negative sign
06:26and moment is force into distance
06:28from the fixed point
06:30so I will multiply this force
06:32with the distance
06:33that is 6 meter
06:36so these are the moments
06:39so therefore
06:40by calculating
06:41this will get the value
06:43of reaction force Rb
06:45as 5.34 kilo Newton
06:49now I will put this value of Rb
06:51in equation number 1
06:54and after calculating
06:56this will get the value
06:57of reaction force Ra
06:59as 6.66 kilo Newton
07:04so now with the help
07:04of these calculated values of Ra and Rb
07:07I will further calculate
07:09the values of shear forces
07:11at all the points of beam
07:14so the next step is
07:16calculations of shear forces
07:20and for the shear force calculations
07:22our sign convention is
07:25upward forces are considered as positive
07:28downward forces are considered as negative
07:32and here you should note that
07:34while calculating the shear force
07:35at a particular point load
07:38you can calculate the shear force values
07:40for left side
07:42and right side
07:43of that particular point load
07:47but while calculating the shear force
07:49at uniformly distributed load
07:51you should calculate the shear force values
07:54at start point
07:55And end point
07:57of uniformly distributed load
08:00that is shear force at point D
08:02and shear force at point D
08:04we need to calculate
08:07and here at point A
08:09and point B
08:11there are support reaction forces
08:13Ra and Rb
08:14which are the point loads
08:17and since these are the point loads
08:20hence as per the rule
08:21for point load
08:23I will calculate the shear force values
08:25at left side and right side
08:27of point A and point B
08:30now at point C
08:32there is one point load
08:34hence as per the rule
08:36for point load
08:37I will calculate the shear force values
08:40at left side and right side
08:42of point C
08:44now at point F
08:46there is one point load
08:48hence as per the rule
08:49I will calculate the shear force values
08:52at left side and right side
08:54of point F
08:57so I will start the shear force calculation
09:00form left hand side of the beam
09:03therefore first to calculate shear force
09:05at point A
09:06to its left
09:08I will take the section
09:10to the left of point A
09:12that is SF at a to the left
09:14equal to
09:17so as you can see
09:18there is no force is acting
09:20at the left side of point A
09:23therefore SF at A to the left
09:25equal to zero
09:28so to draw the shear force diagram
09:31I will first draw
09:32a horizontal reference line
09:34of zero kilo Newton shear force
09:38so here I will mark
09:39 this point of zero kilo Newton shear force
09:42on the reference line
09:44that is shear force at point A
09:46to the left
09:47is zero kilo Newton
09:51now if I go to the section
09:53to the right side of point A
09:56that is SF at A to the right
09:57equal to
10:00then there is reaction force Ra
10:02that we had calculated as 6.66 kilo Newton
10:05which is acting on the beam
10:07in the upward direction
10:09so as per the sign convention
10:11I will consider upward forces as positive
10:14so here the shear force is
10:16plus 6.66 kilo Newton
10:19therefore SF at a to the right
10:22equal to plus 6.66 kilo Newton
10:26here as the shear force value is positive
10:29so I will mark this point
10:31above the reference line
10:32of zero kilo Newton shear force
10:36and I will connect
10:37these two points
10:38with a vertical line
10:42now at point C
10:44there is one point load
10:46so I will calculate the shear force values
10:48at left side and right side
10:50of point C
10:53therefore first to calculate shear force
10:55at point C
10:57to its left
10:58I will take the section
11:00to the left of point C
11:02that is SF at C to the left
11:05equal to
11:07and here I’ll carry forward
11:09previous value of shear force
11:11up to point A
11:12to its right
11:14which is 6.66 kilo Newton
11:17and when we go to the left side
11:19of point C
11:21then there is no load
11:23acting on the beam
11:24at left side of point C
11:27therefore SF at C to the left
11:29equal to plus 6.66 kilo Newton
11:34here as you can see
11:35there is no variation in shear force values
11:38at right side of point A
11:40and left side of point C
11:43hence I will make the horizontal line
11:44with shear force value as
11:47plus 6.66 kilo Newton
11:51now next to calculate
11:53shear force at point C
11:54to its right
11:56I will take the section
11:58to the right of point C
12:00that is SF at point C
12:02to its right
12:03equal to
12:06so here I’ll carry forward
12:08previous value of shear force
12:10up to point C
12:11to its left
12:13which is 6.66 kilo Newton
12:17and when we go to the right side
12:19of point C
12:20then there is one point load
12:22of 4 kilo Newton
12:24acting on the beam
12:25in downward direction
12:28so as per the sign convention
12:30I will consider downward force as negative
12:33so I will add this point load
12:35with negative sign
12:38so therefore
12:39by calculating
12:41this will get the shear force value
12:43at point C
12:44to its right
12:46as 2.66 kilo Newton
12:49that is SF at point C
12:51to its right
12:53equal to 2.66 kilo Newton
12:57here as the shear force value is positive
13:00so I will mark this point
13:02above the reference line
13:03of zero kilo Newton shear force
13:06and I will connect
13:08these two points
13:09with a vertical line
13:12now the point D
13:14is the starting point of UDL
13:16hence I am taking section to the point D
13:19and I will calculate
13:20the shear force at point D
13:23that is SF at point D
13:24equal to
13:28now there is no load
13:29on the beam
13:30between the right side of point C
13:33and point D
13:34therefore shear force remains constant
13:37that is SF at point D
13:39equal to 2.66 kilo Newton
13:44here as there is no variation
13:46in shear force values
13:48so I will make the horizontal line
13:49with shear force value as
13:512.66 kilo Newton
13:56now the point E
13:57is the end point of UDL
13:59so I am taking section to point E
14:02that is SF at point E
14:04equal to
14:07and here I’ll carry forward
14:09previous value of shear force
14:11up to point D
14:12which is 2.66 kilo Newton
14:16and to the left side of point E
14:18there is UDL
14:19of 3 kilo Newton per meter
14:21which we had
14:22already converted into point load
14:24of 6 kilo Newton
14:26which is acting on the beam
14:27in downward direction
14:30and as per the sign convention
14:31I will consider downward force as negative
14:34hence I will add this point load
14:36with negative sign
14:40therefore
14:41after calculating
14:42this will get the shear force value
14:45as minus 3.34 kilo Newton
14:48that is SF at point E
14:50equal to minus 3.34 kilo Newton
14:55here as the shear force value is negative
14:57hence I will mark
14:59this point of shear force
15:01below the reference line
15:02of zero kilo Newton shear force
15:06and here the type of load is UDL
15:08over the length of 2 meter
15:10hence to draw the shear force diagram
15:13I will indicate UDL
15:14with an inclined line
15:17so I will connect
15:18these two points
15:20with an inclined line
15:24now at point F
15:25there is one point load
15:27so I will calculate the shear force values
15:30at left side and right side
15:32of point F
15:35therefore first to calculate
15:37shear force at point F
15:38to its left
15:40I will take the section
15:42to the left of point F
15:44that is SF at F to the left
15:46equal to
15:48and here I’ll carry forward
15:50previous value of shear force
15:52up to point E
15:54which is minus 3.34 kilo Newton
15:58and when we go to the left side
16:00of point F
16:02then there is no load
16:03acting on the beam
16:05at left side of point F
16:07therefore SF at F to the left
16:10equal to minus 3.34 kilo Newton
16:15here as you can see
16:16there is no variation
16:18in shear force values
16:20at point E
16:21and left side of point F
16:24hence I will make the horizontal line
16:26with shear force value as
16:28minus 3.34 kilo Newton
16:32now next to calculate
16:34shear force at point F
16:36to its right
16:37I will take the section
16:39to the right of point F
16:41that is SF at point F
16:43to its right
16:44equal to
16:47so here I’ll carry forward
16:48previous value of shear force
16:51up to point F
16:52to its left
16:53which is minus 3.34 kilo Newton
16:57and when we go to the right side of point F
17:00then there is one point load
17:01of 2 kilo Newton
17:03acting on the beam
17:04in downward direction
17:07so as per the sign convention
17:09I will consider downward forces as negative
17:13so here I will add
17:14this point load
17:15with negative sign
17:19so by calculating
17:20this will get the shear force value
17:22as minus 5.34 kilo Newton
17:26that is SF at point F
17:27to its right
17:29equal to minus 5.34 kilo Newton
17:33here as the shear force value is negative
17:36hence I will mark
17:37this point of shear force
17:39below the reference line
17:41of zero kilo Newton shear force
17:44and I will connect
17:46these two points
17:47with a vertical line
17:50now at point B
17:52there is reaction force Rb
17:54acting on the beam
17:55in the upward direction
17:57and since it is point load
17:59hence as per the rule
18:01for point load
18:02I will calculate the shear force values
18:05at left side and right side
18:07of that point load
18:10therefore first to calculate
18:11shear force at point B
18:13to its left
18:15I will take the section
18:16to the left of point B
18:19that is SF at B to the left
18:21equal to
18:23and here I’ll carry forward
18:25previous value of shear force
18:27up to point F
18:28to its right
18:30which is minus 5.34 kilo Newton
18:34and when we go to the left of point B
18:37then there is no load
18:38acting on the beam
18:40at left side of point B
18:42therefore SF at B to the left
18:44equal to minus 5.34 kilo Newton
18:48here as you can see
18:50there is no variation
18:51in shear force values
18:53between the right side of point F
18:55and left side of point B
18:58hence I will make the horizontal line
19:00with shear force value as
19:02minus 5.34 kilo Newton
19:06now next to calculate
19:08shear force at point B
19:10to its right
19:12I will take the section
19:14to the right of point B
19:16that is SF at point B
19:17to its right
19:19equal to
19:22so here I’ll carry forward
19:23previous value of shear force
19:25up to point B
19:27to its left
19:28which is minus 5.34 kilo Newton
19:32and when we go to the right side
19:34of point B
19:36then there is reaction force Rb
19:38of 5.34 kilo Newton
19:40which is acting on the beam
19:42in upward direction
19:44so as per the sign convention
19:46I will consider upward force as positive
19:50so I will add this upward force
19:51with positive sign
19:55so here minus 5.34
19:57plus 5.34
19:59gives me the value of shear force
20:01as zero kilo Newton
20:04therefore SF at point B
20:06to its right
20:07equal to zero kilo Newton
20:10so I’ll mark
20:11this point of zero kilo Newton shear force
20:14on the reference line
20:16that is SF at point B
20:18to its right
20:19equal to zero kilo Newton
20:22and I’ll connect
20:24these two points
20:25with a vertical line
20:29and here in shear force diagram
20:31whatever the portion drawn
20:33above the reference line
20:34I will show this portion
20:36by positive sign
20:38and the portion which is drawn
20:39below the reference line
20:41I will show this portion
20:42by negative sign
20:45so here I have
20:46completed the shear force diagram
20:52now the next step is
20:53calculations of bending moment
20:57so the bending moment
20:58at a section of beam
21:00is calculated
21:02as the algebraic sum
21:04of the moment of all the forces
21:06acting on one side of the section
21:10so to calculate bending moment
21:12we can start
21:14either from left end of beam
21:16or from right end of beam
21:19here I will start
21:20from left end of beam
21:24so whenever
21:24you are calculating the bending moment
21:26you should remember these conditions
21:29so here for simply supported beam
21:32the condition is
21:34at the ends of simply supported beam
21:37the bending moment will be zero
21:40so bending moment at point A
21:42and bending moment at point B
21:44will be zero
21:46that is BM suffix A
21:48equal to zero kilo Newton meter
21:50and BM point B
21:52equal to zero kilo Newton meter
21:57so to draw bending moment diagram
21:59firstly I will draw
22:01the reference line
22:02of bending moment zero kilo Newton meter
22:05so here I’ll mark
22:07these values
22:08with a point
22:09on the reference line
22:14so now we have to calculate
22:15bending moment at point C
22:19so here
22:20in case of simply supported beam
22:22while you are doing the calculations
22:24for bending moment
22:25at a particular point
22:28you should always add
22:30moment of all the forces
22:32present
22:33either from left end of beam
22:35or from right end of beam
22:38up to that particular point
22:40at which
22:41you are calculating the bending moment
22:44it means
22:46while I am calculating
22:47the bending moment at point C
22:50I will add moment
22:52of all the forces
22:54present
22:55either from left end of beam
22:57or from right end of beam
22:59up to point C
23:03and for bending moment calculation
23:04our sign convention is
23:07for sagging effect of beam
23:08the force is considered as positive
23:11for hogging effect of beam
23:13the force is considered as negative
23:18so first to calculate bending moment
23:20at point C
23:22that is BM suffix C
23:24equal to
23:27here at left hand side of point C
23:30there is reaction force Ra
23:31of 6.66 kilo Newton
23:34which is acting on the beam
23:36in the upward direction
23:39due to this force
23:40the beam shows sagging effect
23:43and for sagging effect of beam
23:45I will consider this force as positive
23:49so I will add this force
23:50with positive sign
23:54and as per decided
23:55moment is force into distance
23:58so I will multiply this force
24:00with the distance
24:02from point of action of force
24:04that is 1 meter
24:07therefore
24:08after calculating
24:10this will get the value
24:11of bending moment at point C
24:14equal to 6.66 kilo Newton meter
24:17that is BM suffix C
24:19equal to 6.66 kilo Newton meter
24:24so as it is positive value
24:25of bending moment
24:27hence I will mark this point
24:29above the reference line
24:31of bending moment zero kilo Newton meter
24:35now there is no any load
24:36present on the beam
24:38between point A
24:39and point C
24:41therefore
24:42to draw the bending moment diagram
24:44I will connect these two points
24:46with inclined line
24:51now next we have to calculate
24:53bending moment at point D
24:55that is BM suffix D
24:57equal to
25:01here at left hand side of point D
25:03there is reaction force Ra
25:05of 6.66 kilo Newton
25:08which is acting on the beam
25:09in upward direction
25:12due to this force
25:13the beam shows sagging effect
25:16and for sagging effect of beam
25:18I will consider this force as positive
25:23so I will add this force
25:24with positive sign
25:28and as we know
25:29moment is force into distance
25:32so I will multiply this force
25:34with the distance
25:35from point of action of force
25:37that is 2 meter
25:42and here
25:43at point C
25:45there is one force
25:46of 4 kilo Newton
25:48which is acting on the beam
25:50in downward direction
25:52due to this force
25:54the beam shows hogging effect
25:56and for hogging effect of beam
25:58I will consider this force as negative
26:02so I will add
26:03this force with negative sign
26:06and I will multiply this force
26:09with the distance
26:10from point of action of force
26:12that is 1 meter
26:16therefore
26:17after calculating
26:18this will get the value
26:20of bending moment at point D
26:22equal to
26:249.32 kilo Newton meter
26:27that is BM suffix D
26:29equal to 9.32 kilo Newton meter
26:33so as it is positive value
26:35of bending moment
26:37hence I will mark this point
26:38above the reference line
26:40of bending moment zero kilo Newton meter
26:45now there is no any load
26:46present on the beam
26:48between point C
26:50and point D
26:52therefore
26:52to draw the bending moment diagram
26:54I will connect these two points
26:56with inclined line
27:02now here you should note that
27:04in shear force diagram
27:05when the shear values
27:08changes from positive to negative
27:10then at that point
27:12the bending moment value is maximum
27:15so I will represent this point
27:17with point M
27:19so at point M
27:21we need to find out
27:22maximum bending moment value
27:26but here we don’t know
27:27the location of point M
27:29that is we don’t know the distance X
27:32but we can find out this distance X
27:35by using equation
27:36of similarity of triangles
27:40because
27:41here we known the distance DE
27:43which is 2 meter
27:45and here the values of shear forces
27:47we are considered as heights of triangles
27:51therefore
27:52for the first triangle
27:54the height is 2.66
27:57and base is X
27:59and for second triangle
28:01the height is 3.34
28:03and base is 2 minus x
28:07therefore
28:08from similarity of triangles
28:10the base of first triangle
28:12that is X
28:13divided by
28:15height of first triangle
28:17which is 2.66
28:20is equal to
28:22base of second triangle
28:24that is 2 minus x
28:26divided by
28:27height of second triangle
28:29which is 3.34
28:33now here the unknown term is X
28:35therefore
28:36after calculating
28:38this will get the distance X
28:40equal to
28:410.886 meter
28:44so here I got the location of point M
28:48so from distance X
28:49next I will calculate
28:51the maximum bending moment at point M
28:56so next to calculate
28:57bending moment at point M
29:00that is BM suffix M
29:01equal to
29:06here at left hand side of point M
29:08there is reaction force Ra
29:10of 6.66 kilo Newton
29:13which is acting on the beam
29:15in the upward direction
29:18due to this force
29:19the beam shows sagging effect
29:22and for sagging effect of beam
29:24I will consider this force as positive
29:27so I will add this force with positive sign
29:31and as we know
29:33moment is force into distance
29:36so I will multiply this force
29:38with the distance
29:40from point of action of force
29:42that is 2.886 meter
29:47and here
29:48at point C
29:50there is one force
29:51of 4 kilo Newton
29:53which is acting on the beam
29:55in downward direction
29:57due to this force
29:59the beam shows hogging effect
30:02and for hogging effect of beam
30:04I will consider this force as negative
30:06so I will add this force
30:08with negative sign
30:11and I will multiply this force
30:12with the distance
30:14from point of action of force
30:16that is 1.886 meter
30:21and here
30:23from point D
30:24to point M
30:26there is UDL
30:27of 3 kilo Newton per meter
30:29over the distance of 0.886 meter
30:33so I will first convert
30:34this portion of UDL
30:36into point load
30:38that is 3 kilo Newton per meter
30:41is the UDL value
30:42multiply with the distance
30:44over which
30:45the UDL acts
30:47that is 0.886 meter
30:51now this converted point load
30:53is acting at the midpoint
30:55of this much portion of UDL
30:57in downward direction
30:59due to this converted point load
31:02the beam shows hogging effect
31:04and for hogging effect of beam
31:06I will consider this force
31:08as negative
31:10so I will add this converted point load
31:12with negative sign
31:14so I will multiply this force
31:17with the distance
31:18from point of action of force
31:20that is half of 0.886 meter
31:25therefore
31:26after calculating
31:28this will get the value
31:30of bending moment at point M
31:32equal to
31:3310.50 kilo Newton meter
31:37that is BM suffix M
31:39equal to 10.50 kilo Newton meter
31:44so as it is positive value
31:46of bending moment
31:48hence I will mark this point
31:50above the reference line
31:52of bending moment zero kilo Newton meter
31:56and here
31:57between point D and point M
32:00there is UDL
32:01therefore to draw bending moment diagram
32:04I will indicated UDL
32:06with a parabolic curve
32:08hence I will join
32:09these two points
32:11with a parabolic curve
32:17now next we have to calculate
32:19bending moment at point E
32:21that is BM suffix E
32:23equal to
32:26here at right hand side of point E
32:28there is reaction force Rb
32:30of 5.34 kilo Newton
32:33which is acting on the beam
32:34in upward direction
32:37due to this force
32:39the beam shows sagging effect
32:41and for sagging effect of beam
32:44I will consider this force as positive
32:47so I will add this force
32:48with positive sign
32:52and as we know
32:53moment is force into distance
32:56so I will multiply this force
32:58with the distance
32:59from point of action of force
33:01that is 2 meter
33:05and here
33:06at point F
33:07there is one force
33:09of 2 kilo Newton
33:11which is acting on the beam
33:12in the downward direction
33:14due to this the force
33:16the beam shows hogging effect
33:18and for hogging effect of beam
33:20I will consider this force as negative
33:23so I will add this force
33:25with negative sign
33:27and I will multiply this force
33:29with the distance
33:31from point of action of force
33:33that is 1 meter
33:37therefore
33:38after calculating
33:39this will get the value
33:41of bending moment at point E
33:43equal to 8.68 kilo Newton meter
33:47that is BM suffix E
33:49equal to 8.68 kilo Newton meter
33:55so as it is positive value
33:56of bending moment
33:58hence I will mark this point
34:00above the reference line
34:02of bending moment zero kilo Newton meter
34:06and here
34:07between point M
34:08to point E
34:10there is UDL
34:12therefore to draw bending moment diagram
34:15I will indicated
34:17with a parabolic curve
34:19hence I will join
34:21these two points
34:22with a parabolic curve
34:26now next we have to calculate
34:28bending moment at point F
34:30that is BM suffix F
34:32equal to
34:36here at right hand side of point F
34:38there is reaction force Rb
34:40of 5.34 kilo Newton
34:42which is acting on the beam
34:44in the upward direction
34:47due to this force
34:48the beam shows sagging effect
34:51and for sagging effect of beam
34:53I will consider this force
34:55as positive
34:57so I will add this force
34:58with positive sign
35:01and as we know
35:03moment is force into distance
35:06so I will multiply this force
35:08with the distance
35:09from point of action of force
35:11that is 1 meter
35:15therefore
35:16after calculating
35:18this will get the bending moment value
35:20at point F
35:21equal to 5.34 kilo Newton meter
35:26that is BM suffix F
35:27equal to 5.34 kilo Newton meter
35:33so as it is positive value
35:34of bending moment
35:36hence I will mark this point
35:38above the reference line
35:40of bending moment zero kilo Newton meter
35:44and there is no any load
35:45present on the beam
35:47between the point E
35:49and point F
35:51therefore to draw bending moment diagram
35:54I will connect these two points
35:56with inclined line
36:00and also
36:01there is no any load
36:03present on the beam
36:05between the point F
36:06and point B
36:08therefore
36:08to draw the bending moment diagram
36:10I will connect these two points
36:13with inclined line
36:17now since I can see
36:19this bending moment diagram is drawn
36:21above the reference line
36:22of bending moment zero kilo Newton meter
36:26so whatever the portion I have drawn
36:28all the values
36:28of bending moment are positive
36:30hence I'll show this portion
36:32by positive sign
36:36so here I've completed
36:37the shear force diagram
36:38and bending moment diagram
36:40for this simply supported beam