Simply Supported Beam Numerical 5: Draw Shear Force Diagram [SFD] and Bending Moment Diagram [BMD]
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 the shear force diagram
00:11and bending moment diagram
00:13for Simply Supported beam
00:15as shown in figure
00:18So the statement is given as
00:21Draw shear force
00:22and bending moment diagram
00:24for a simply supported beam AB
00:267 meter long
00:28is loaded
00:29as shown in figure
00:32So this is the Simply Supported beam AB
00:35of length 7 meter
00:37and carrying a uniformly distributed loads
00:40of 2 kilo Newton per meter
00:42over a length of 3 meter
00:44and another UDL of 1 kilo Newton per meter
00:47over a length of 2 meter
00:50and there are 2 point loads
00:52of 3 kilo Newton
00:54and 5 kilo Newton
00:55are acting on the beam
00:56as shown in figure
00:59So for this setup
01:00we have to draw
01:02the shear force diagram
01:03and bending moment diagram
01:09So first of all
01:10I will draw the free body diagram
01:12for this beam section
01:18So for solving this numerical problem
01:20first we have to convert
01:21these uniformly distributed loads
01:23into point loads
01:28so to convert this
01:29I will multiply these UDL values
01:32with the length
01:33over which
01:34these UDL Acts
01:37So here firstly I will convert
01:39this AC and CD portions of UDL
01:41into point loads
01:44So the first point load
01:45equal to
01:47UDL of 2 kilo Newton per meter
01:50multiply by distance AC
01:52that is 3 meter
01:54And the second point load
01:56equal to
01:57UDL of 1 kilo Newton per meter
02:00multiply by distance CD
02:02that is 2 meter
02:05So here I will get
02:06first converted point load of 6 kilo Newton
02:09and second converted point load
02:11of 2 kilo Newton
02:16Where first converted point load
02:17is acting on the midpoint
02:19of UDL distance AC
02:21and the second converted point load
02:23is acting on the midpoint
02:25of UDL distance CD
02:30Now this type of problem
02:32we are going to solve in 3 steps
02:36in the first step
02:38we have to calculate
02:39the values of support reaction forces
02:42Ra and Rb
02:44So to calculate these values
02:47I will use 2 equations of equilibrium
02:51where first equation is
02:53summation Fy equal to 0
02:56That means
02:57Addition of all the forces
02:59in the vertical axis
03:01equal to 0
03:04and the second equation is
03:05summation of moment
03:07equal to 0
03:09that means
03:11Addition of moments
03:12due to all forces
03:14about any point
03:16must be zero
03:19So here In the first equation
03:21while I am doing the Addition
03:22of all the vertical forces
03:25I will consider
03:26upward forces as positive
03:29and downward forces as negative
03:33Here Ra and Rb
03:34are the vertical reaction forces
03:37acting on the beam
03:38in the upward direction
03:41so as per the sign convention
03:42I will add these forces
03:44with positive sign
03:47And the point loads
03:49of 3 kilo Newton
03:50and 5 kilo Newton
03:52are acting on the beam
03:53in the downward direction
03:56so as per the sign convention
03:58I will add these forces
03:59with negative sign
04:02And the converted point loads
04:04of 6 kilo Newton
04:05and 2 kilo Newton
04:07are acting on the beam
04:08in the downward direction
04:10so as per the sign convention
04:12I will add these forces
04:14with negative sign
04:17Therefore
04:18after calculating
04:20this will get the equation
04:22that is Ra plus Rb
04:23equal to 16 kilo Newton
04:27so I will give this as
04:29equation number 1
04:32Now the next equation is
04:33summation of moment
04:35equal to 0
04:38so for calculating moment
04:40either we can take moment
04:42at point A
04:43or at point B
04:46and here my sign convention would be
04:48Clockwise moment as positive
04:50and anticlockwise moment as negative
04:55So for taking moment at point A
04:57I will fixed beam at point A
05:01Now here
05:02converted point load of 6 kilo Newton
05:05will be pushing this beam toward downward
05:07so it rotate the beam clockwise
05:10from fixed point
05:12and as per the sign convention
05:13for clockwise moment
05:15I will add this force
05:16with positive sign
05:20Now as per decided
05:21moment is force into distance
05:23from fixed point
05:26so here plus 6 kilo Newton
05:27is the force
05:29into 1.5 meter
05:31is the distance
05:32from fixed point
05:35Now here 3 kilo Newton force
05:37will be pushing this beam toward downward
05:40so it rotate the beam clockwise
05:42from fixed point
05:44so for clockwise moment
05:46I will add this force
05:47with positive sign
05:50And moment is force into distance
05:52from fixed point
05:54so here plus 3 kilo Newton
05:56is the force
05:57into 3 meter is the distance
05:59from fixed point
06:03Now here converted point load
06:04of 2 kilo Newton
06:06will be pushing this beam toward downward
06:09so it rotate the beam clockwise
06:11from fixed point
06:13so for clockwise moment
06:15I will add this force
06:17with positive sign
06:20And the moment is force into distance
06:22from fixed point
06:24so here plus 2 kilo Newton
06:26is the converted point load
06:28into 4 meter is the distance
06:30from fixed point
06:33Now here 5 kilo Newton force
06:35will be pushing this beam toward downward
06:38so it rotate the beam clockwise
06:40from fixed point
06:42so for clockwise moment
06:44I will add this force
06:46with positive sign
06:48And moment is force into distance
06:50from fixed point
06:53so here plus 5 kilo Newton
06:55is the force
06:56into 5 meter is the distance
06:58from fixed point
07:02Now here reaction force Rb
07:03will be pushing this beam toward upward
07:06so it rotate the beam anticlockwise
07:08from fixed point
07:10so as per the sign convention
07:12for anticlockwise moment
07:14I will add this force
07:16with negative sign
07:18and moment is force into distance
07:21from the fixed point
07:23so here minus Rb is the force
07:25into 7 meter is the distance
07:27from fixed point
07:30so these are the moments
07:33so therefore
07:34by calculating
07:36this will get the value
07:37of reaction force Rb
07:39as 7.285 kilo Newton
07:44Now I will put this value of Rb
07:46in equation number 1
07:48and after calculating
07:50this will get the value of
07:51reaction force Ra
07:53as 8.715 kilo Newton
07:58So now with the help of
07:59these calculated values of Ra and Rb
08:01I will further calculate
08:03the values of shear forces
08:05at all the points of beam
08:08so the next step is
08:09calculations of shear forces
08:13And for the shear force calculations
08:15our sign convention is
08:17upward forces are considered as positive
08:20and downward forces
08:21are considered as negative
08:26And here you should note that
08:28While calculating the shear force
08:30at a particular point load
08:32You can calculate the shear force values
08:35for left side
08:36and right side
08:38of that particular point load
08:41But while calculating the shear force
08:43at uniformly distributed load
08:45You should calculate
08:46shear force values
08:48at start point
08:49and end point
08:51of uniformly distributed load
08:54That is shear force at point A
08:56and shear force at point D
08:59we need to calculate
09:01But in this problem
09:03at point A
09:05and at point D
09:06there are point loads
09:09hence as per the rule
09:10for point loads
09:12I will calculate the shear force values
09:15at left side
09:16and right side
09:18of that point loads
09:20That is shear force
09:22at left side
09:23and right side
09:24of point A and point D
09:26we need to calculate
09:30and also at point C
09:32there is a point load
09:34hence as per the rule
09:35for point load
09:37I will calculate the shear force values
09:40at left side and right side
09:41of point C
09:44That is shear force at point C
09:46to its left
09:47and shear force at point C
09:49to its right
09:50we need to calculate
09:54and here
09:55at point B
09:57there is support reaction force Rb
09:59which is a point load
10:01and since it is a point load
10:03hence as per the rule
10:05for point load
10:07I will calculate the shear force values
10:09at left side
10:10and right side
10:12of point B
10:16So I will start the shear force calculation
10:18form left hand side of the beam
10:22Therefore
10:23first to calculate shear force
10:25at point A
10:26to its left
10:28I will take the section
10:29to the left of point A
10:31That is SF at A to the left
10:33equal to
10:36So as you can see
10:38there is no any force is acting
10:40at the left side of point A
10:42Therefore SF at A to the left
10:44equal to 0
10:47So to draw the shear force diagram
10:50I will first draw
10:51a horizontal reference line
10:53of 0 kilo Newton shear force
10:56So here I’ll mark
10:57this point of 0 kilo Newton shear force
11:00on the reference line
11:02that is shear force at point A
11:03to the left
11:04is zero kilo Newton
11:08Now if I go to the section
11:10to the right side of point A
11:12that is SF at A to the right
11:15equal to
11:17Then there is reaction force Ra
11:19that we had calculated
11:20as 8.715 kilo Newton
11:24Which is acting on the beam
11:25in the upward direction
11:27So as per the sign convention
11:29I will consider upward forces as positive
11:32So here the shear force is
11:34plus 8.715 kilo Newton
11:38Therefore SF at A to the right
11:40equal to plus 8.715 kilo Newton
11:45Here as the shear force value is positive
11:48So I’ll mark this point
11:49above the reference line
11:51of 0 kilo Newton shear force
11:54And I will connect
11:56these two points
11:57with a vertical line
12:02Now at point C
12:04there is a point load
12:06Hence I will calculate
12:07the shear force values
12:09at left side
12:10and right side
12:11of point C
12:14So first to calculate
12:15shear force at point C
12:17to its left
12:19I will take the section
12:21to the left of point C
12:23that is SF at point C
12:25to its left
12:26equal to
12:29And here I’ll carry forward
12:30previous value of shear force
12:32up to point A
12:33to its right
12:35which is plus 8.715 kilo Newton
12:39And to the left side of point C
12:41there is uniformly distributed load
12:43of 2 kilo Newton per meter
12:45Which we had
12:46already converted into point load
12:47of 6 kilo Newton
12:50which is acting on the beam
12:51in downward direction
12:54And as per the sign convention
12:56I will consider downward force as negative
12:59Hence I will add
13:00this point load of 6 kilo Newton
13:02with negative sign
13:05Therefore
13:06after calculating
13:08This will get the shear force value
13:10as 2.715 kilo Newton
13:13That is SF at point C
13:15to its left
13:16equal to 2.715 kilo Newton
13:21Here as the shear force value is positive
13:24Hence I will mark this point
13:26above the reference line
13:27of zero kilo Newton shear force
13:31And here the type of load is UDL
13:33over the length of 3 meter
13:35Hence to draw the shear force diagram
13:37I will Indicate UDL
13:39with an inclined line
13:41So I’ll connect
13:43these two points
13:44with an inclined line
13:49Now if I go to the section
13:50to the right of point C
13:53That is SF at point C
13:54to its right
13:55equal to
13:59So here I’ll carry forward
14:00previous value of shear force
14:02up to point C
14:03to its left
14:05that is plus 2.715 kilo Newton
14:09And when we move towards
14:11the right side of point C
14:13Then there is one point load
14:15of 3 kilo Newton
14:16Which is acting on the beam
14:18in downward direction
14:20So as per the sign convention
14:22I will consider downward forces
14:23as negative
14:25So I will add
14:26this point load
14:28with negative sign
14:31Therefore
14:32after calculating
14:33This will get shear force value
14:35as minus 0.285 kilo Newton
14:39That is SF at point C
14:41to its right
14:42equal to minus 0.285 kilo Newton
14:47Here as the shear force value is negative
14:49Hence I’ll mark this point
14:51below the reference line
14:52of zero kilo Newton shear force
14:55And I'll connect
14:57these two points
14:58with a vertical line
15:03Now at point D
15:04there is a point load
15:06Hence I will calculate the
15:07shear force values
15:09at left side
15:10and right side
15:11of point D
15:14So first to calculate
15:15shear force at point D
15:17to its left
15:18I will take the section
15:20to the left of point D
15:22that is SF at point D
15:24to its left
15:25equal to
15:27And here I’ll carry forward
15:29previous value of shear force
15:31up to point C
15:32to its right
15:34which is minus 0.285 kilo Newton
15:38And to the left side of point D
15:40there is uniformly distributed load
15:43that we had converted into point load
15:45of 2 kilo Newton
15:47which is acting on the beam
15:48in the downward direction
15:51And as per the sign convention
15:53I will consider downward force as negative
15:56Hence I will add
15:57this point load
15:59with negative sign
16:02Therefore
16:03after calculating
16:05This will get the shear force value
16:07as minus 2.285 kilo Newton
16:10That is SF at point D
16:12to its left
16:13equal to minus 2.285 kilo Newton
16:18Here as the shear force value is negative
16:21Hence I will mark this point
16:22below the reference line
16:24of zero kilo Newton shear force
16:27And here the type of load is UDL
16:29over the length of 2 meter
16:32Hence to draw the shear force diagram
16:34I will Indicate UDL
16:35with inclined line
16:38So I will connect
16:39these two points
16:40with an inclined line
16:45Now if I go to the section
16:46to the right of point D
16:48That is SF at point D
16:50to its right
16:51equal to
16:54So here I’ll carry forward
16:56previous value of shear force
16:58up to point D
16:59to its left
17:00that is minus 2.285 kilo Newton
17:04And when we move towards
17:06the right side of point D
17:09Then there is one point load
17:10of 5 kilo Newton
17:12Which is acting on the beam
17:14in downward direction
17:16So as per the sign convention
17:18I will consider downward forces as negative
17:21So I will add this point load
17:23with negative sign
17:26Therefore
17:28after calculating
17:29This will get shear force
17:31as minus 7.285 kilo Newton
17:35That is SF at point D
17:37to its right
17:38equal to minus 7.285 kilo Newton
17:44Here as the shear force value is negative
17:46Hence I’ll mark
17:47this point of shear force
17:49below the reference line
17:51of zero kilo Newton shear force
17:55And I will join
17:56these two points
17:57with a vertical line
18:02Now at point B
18:03there is reaction force Rb
18:05acting on the beam
18:07in the upward direction
18:09and since it is point load
18:11I will calculate the shear force values
18:13at left side
18:15and right side
18:16of that point load
18:19Therefore first to calculate
18:20shear force at point B
18:22to its left
18:24I will take the section
18:25to the left of point B
18:28That is SF at B to the left
18:30equal to
18:32And here I’ll carry forward
18:34previous value of shear force
18:36up to point D
18:37to its right
18:38which is minus 7.285 kilo Newton
18:42And when we move towards
18:44the left of point B
18:46then there is no load is acting
18:48on the beam
18:49at left side of point B
18:53Therefore SF at B
18:54to the left
18:55equal to minus 7.285 kilo newton
19:01Here as you can see
19:02there is no variation
19:03in shear force values
19:05at point D
19:06to its right
19:08and point B
19:09to its left
19:11Hence I will make the horizontal line
19:13with shear force value as
19:15minus 7.285 kilo Newton
19:20Now next to calculate
19:22shear force at point B
19:24to its right
19:26I will take the section
19:27to the right of point B
19:30That is SF at point B
19:31to its right
19:32equal to
19:35So here I’ll carry forward
19:36previous value of shear force
19:38up to point B
19:40to its left
19:41which is minus 7.285 kilo newton
19:46And when we move towards
19:47the right side of point B
19:49Then there is reaction force Rb
19:51of 7.285 kilo Newton
19:54which is acting on the beam
19:56in the upward direction
19:58So as per the sign convention
20:00I will consider upward force as positive
20:03So here I will add
20:04this upward force
20:06with positive sign
20:09So here minus 7.285
20:11plus 7.285
20:12gives me the value of shear force
20:14as zero kilo Newton
20:17Therefore SF at point B
20:19to its right
20:20equal to zero kilo Newton
20:23So I will mark
20:24this point of zero kilo Newton shear force
20:27on the reference line
20:29That is SF at point B
20:31to its right
20:32equal to zero kilo Newton
20:35And I’ll connect these two points
20:37with a vertical line
20:42And here in shear force diagram
20:44whatever the portion drawn
20:46above the reference line
20:47I will show this portion
20:49by positive sign
20:51And the portion which is drawn
20:53below the reference line
20:55I will show this portion
20:56by negative sign
20:59So here I have completed
21:00the shear force diagram
21:05Now the next step is
21:06calculations of bending moments
21:09so the bending moment
21:11at a section of beam
21:12is calculated
21:14as the algebraic sum
21:16of the moment of all the forces
21:18acting on one side of section
21:22So to calculate bending moments
21:24we can start
21:26either from left end of beam
21:28or from right end of beam
21:31 Here I will start
21:32from left end of beam
21:35so whenever you are calculating
21:37the bending moments
21:39You should remember
21:40these conditions
21:43so here for simply supported beam
21:44the condition is
21:47at the ends of simply supported beam
21:49the bending moment will be zero
21:52So bending moment
21:54at point A
21:55and bending moment at point B
21:57will be 0
21:59that is BM suffix A
22:01equal to 0 kilo Newton meter
22:03and BM suffix B
22:05equal to 0 kilo Newton meter
22:09So to draw bending moment diagram
22:12firstly I will draw
22:13the reference line
22:14of bending moment 0 kilo Newton meter
22:18So here I’ll mark
22:19these values
22:20with a point on the reference line
22:25So now we have to calculate
22:26bending moment at point C
22:30So Here
22:31In case of simply supported beam
22:33while you are doing the calculations
22:35for bending moment
22:37at particular point
22:39you should always add
22:41moment of all the forces
22:43present
22:44either from left end of beam
22:46or from right end of beam
22:47Up to that particular point
22:49at which
22:50you are calculating the bending moment
22:54It means
22:55while I am calculating
22:57the bending moment at point C
22:59I will add moment
23:01of all the forces
23:03present
23:04either from left end of beam
23:06or from right end of beam
23:08up to point C
23:10And for bending moment calculations
23:13our sign convention is
23:15For sagging effect of beam
23:17the force is considered as positive
23:20and for hogging effect of beam
23:22the force is considered as negative
23:26So first to calculate
23:28bending moment at point C
23:30That is BM suffix C equal to
23:34Here at left hand side of point C
23:37There is reaction force Ra
23:39of 8.715 kilo Newton
23:42which is acting on the beam
23:43in the upward direction
23:46Due to this force
23:47the beam shows sagging effect
23:49And for sagging effect of beam
23:51I will consider this force
23:53as positive
23:56so I will add
23:57this force with positive sign
24:00and as per decided
24:02moment is always
24:03force multiply by distance
24:06so I will multiply this force
24:08with the distance
24:09from point of action of force
24:11that is 3 meter
24:14And here
24:15from point A
24:17to point C
24:18there is UDL
24:20of 2 kilo Newton per meter
24:22that we had converted into point load
24:24of 6 kilo Newton
24:27which is acting on the beam
24:28in the downward direction
24:31Due to this load
24:32the beam shows hogging effect
24:35And for hogging effect of beam
24:37I will consider this force as negative
24:41so I will add this converted point load
24:43with negative sign
24:46And I will multiply this force
24:48with the distance
24:50from point of action of force
24:52that is 1.5 meter
24:55Therefore
24:57after calculating
24:58this will get the bending moment value
25:00at point C
25:02equal to
25:0317.145 kilo Newton
25:08That is BM suffix C
25:09equal to
25:1117.145 kilo Newton meter
25:16So as it is positive value
25:18of bending moment
25:19Hence I will mark this point
25:21above the reference line
25:22of bending moment 0 kilo Newton meter
25:27And here
25:28between point A and point C
25:31there is uniformly distributed load
25:33Hence to draw the bending moment diagram
25:36I will indicated uniformly distributed load
25:38with a parabolic curve
25:41Hence I will join
25:42these two points
25:43with a parabolic curve
25:48Now next we have to calculate
25:49bending moment at point D
25:52That is BM suffix D
25:54equal to
25:57Here at Right hand side
25:58of point D
26:00There is reaction force Rb
26:02of 7.285 kilo Newton
26:04which is acting on the beam
26:06in the upward direction
26:08Due to this force
26:10the beam shows sagging effect
26:12And for sagging effect of beam
26:14I will consider this force
26:16as positive
26:18so I will add this force
26:20with positive sign
26:23and the moment is force into distance
26:26so I will multiply this force
26:28with the distance
26:29from point of action of force
26:31that is 2 meter
26:34Therefore
26:35after calculating
26:36this will get the value
26:38of bending moment at point D
26:40equal to
26:4214.575 kilo Newton meter
26:46That is BM suffix D
26:47equal to 14.575 kilo Newton meter
26:53So as it is positive value
26:55of bending moment
26:56Hence I'll mark
26:57this point of bending moment
26:59above the reference line
27:01of bending moment 0 kilo Newton meter
27:05And here
27:06between point C and point D
27:09there is UDL
27:11hence to draw the bending moment diagram
27:14I will indicated UDL
27:15with a parabolic curve
27:18Hence I will join
27:19these two points
27:20with a parabolic curve
27:24And there is no load
27:25present on the beam
27:27between point D and point B
27:30Therefore
27:31to draw the bending moment diagram
27:32I will connect these two points
27:34with an inclined line
27:38Now since I can see
27:39This bending moment diagram
27:41is drawn
27:42above the reference line
27:43of bending moment 0 kilo Newton meter
27:46So whatever the portion I have drawn
27:48all the values of bending moment
27:50are positive
27:52Hence I will show this portion
27:53by positive sign
27:57so here I have completed
27:58the shear force diagram
27:59and bending moment diagram
28:01for this Simply Supported beam