Simply Supported Beam Numerical 6: Draw Shear Force and Bending Moment Diagram
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
00:12and bending moment diagram
00:13for a 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
00:265 meter long
00:28is loaded
00:29as shown in figure
00:33so this is the simply supported beam AB
00:35of length 5 meter
00:38and carrying a uniformly distributed loads
00:4020 kilo Newton per meter
00:42over a length of 5 meter
00:45and there is 1 point load
00:47of 45 kilo Newton
00:49is acting on the beam
00:50at point C
00:54so for this setup
00:55we have to draw
00:56the shear force diagram
00:58and bending moment diagram
01:03so first of all
01:04I will draw the free body diagram
01:05for this beam section
01:10so for solving this numerical problem
01:12first we have to convert
01:14this uniformly distributed loads
01:16into point load
01:20so to convert this
01:21I will multiply this UDL value
01:24with the length
01:25over which their UDL act
01:29so here I will get
01:30the converted point load of 100 kilo Newton
01:35now this converted point load
01:36is acting on the mid point
01:38of the length
01:39over which this UDL acts
01:45now this type of problem
01:46we are going to solve in 3 steps
01:51in the first step
01:53we have to calculate
01:54the values of support reaction forces
01:57Ra and Rb
02:00so to calculate these values
02:02I will use
02:032 equations of equilibrium
02:07that is first equation is
02:08summation Fy equal to 0
02:12that means
02:13addition of all the forces
02:15in the vertical axis
02:16equal to 0
02:19and the second equation is
02:21summation of moment
02:23equal to 0
02:25that means
02:26addition of moments
02:28due to all forces
02:29about any point
02:31must be zero
02:35so here in the first equation
02:37while I am doing the addition
02:38of all the vertical forces
02:41I will consider
02:42upward forces as positive
02:44and downward forces as negative
02:49here Ra and Rb
02:50are the vertical reaction forces
02:52acting on the beam
02:54in the upward direction
02:56so as per the sign convention
02:58I will add these forces
02:59with positive sign
03:03and the point load
03:04of 45 kilo Newton
03:06is acting on the beam
03:07in the downward direction
03:09so as per the sign convention
03:11I will add this force
03:13with negative sign
03:15and the converted point load
03:17of 100 kilo Newton
03:19is acting on the beam
03:20in the downward direction
03:22so as per the sign convention
03:24I will add this force
03:26with negative sign
03:29therefore
03:30after calculating
03:31this will get the equation
03:33that is Ra plus Rb
03:35equal to 145 kilo Newton
03:39so I will give this as
03:40equation number 1
03:44now the next equation is
03:46summation of moment
03:47equal to 0
03:51so for calculating the moment
03:53either we can take moment
03:55at point A
03:56or at point B
03:59and here my sign convention would be
04:02clockwise moment as positive
04:04and anticlockwise moment as negative
04:09so for taking moment at point A
04:11I will fixed beam at point A
04:15now here 45 kilo Newton force
04:17will be pushing this beam towards downward
04:20so it rotate the beam clockwise
04:22from fixed point
04:24and as per the sign convention
04:26for clockwise moment
04:28I will add this force with positive sign
04:33now as per decided
04:35moment is force into distance
04:37from fixed point
04:39so here plus 45 kilo Newton
04:41is the force
04:42into 2 meter is the distance
04:44from fixed point
04:48and here converted point load
04:50of 100 kilo Newton
04:52will be pushing this beam toward downward
04:55so it rotate the beam clockwise
04:57from fixed point
04:59and for clockwise moment
05:01I will add this force
05:03with positive sign
05:05and the moment is force into distance
05:08from fixed point
05:10so here plus 100 kilo Newton
05:12is the converted point load
05:14into 2.5 meter is the distance
05:16from fixed point
05:20now here reaction force Rb
05:22will be pushing this beam toward upward
05:25so it rotate the beam anticlockwise
05:27from fixed point
05:29so as per the sign convention
05:31for anticlockwise moment
05:33I will add this force
05:34with negative sign
05:37and as per decided
05:39the moment is force is distance
05:40from the fixed point
05:43so here minus Rb is the force
05:45into 5 meter is the distance
05:47from fixed point
05:51so these are the moments
05:54so therefore by calculating
05:56this will get the reaction force Rb
05:59as 68 kilo Newton
06:03now I will put this value of Rb
06:05in equation number 1
06:09and after calculating
06:10this will get
06:11the value of reaction force Ra
06:12as 77 kilo Newton
06:17so now with the help of
06:18these calculated values of Ra and Rb
06:21I will further calculate
06:22the values of shear forces
06:24at all the points of beam
06:28so the next step is
06:29calculations of shear forces
06:33and for shear force calculations
06:35our sign convention is
06:37upward forces are considered as positive
06:40and downward forces
06:41are considered as negative
06:46and here you should note that
06:47while calculating the shear force
06:50at a particular point load
06:52you can calculate the shear force values
06:54for left side
06:56and right side
06:57of that particular point load
07:00but while calculating the shear force
07:02at uniformly distributed load
07:05you should calculate the shear force values
07:07at start point
07:09and end point
07:10of uniformly distributed load
07:14that is shear force at point C
07:15and shear force at point D
07:17we need to calculate
07:20but here
07:22at point A
07:23and point B
07:25there are support reaction forces
07:27Ra and Rb
07:28which are the point loads
07:31and since these are the point loads
07:33hence as per the rule
07:35for point load
07:36I will calculate the shear force values
07:39at left side and right side
07:41of point A and point B
07:45Now at point C
07:46there is a point load
07:48hence as per the rule
07:49for point load
07:51I will calculate the shear force values
07:53at left side and right side
07:55of point C
07:58so I will start the shear force calculation
08:01form left hand side of the beam
08:04therefore
08:05first to calculate shear force
08:07at point A to its left
08:10I will take the section
08:11to the left of point A
08:13that is SF at A to the left
08:15equal to
08:18so as you can see
08:20there is no any force
08:21is acting at the left side of point A
08:24therefore SF at A to the left
08:26equal to 0
08:31so to draw the shear force diagram
08:33I will first draw
08:35a horizontal reference line
08:36of 0 kilo Newton shear force
08:39so here I’ll mark
08:41this point of 0 kilo Newton shear force
08:43on the reference line
08:45that is shear force at point A to its left
08:48is zero kilo Newton
08:53now if I go to the section
08:55to the right side of point A
08:57that is SF at A to the right
08:59equal to
09:02then there is reaction force Ra
09:03that we had calculated
09:05as 77 kilo Newton
09:07which is acting on the beam
09:09in the upward direction
09:12so as per the sign convention
09:13I will consider upward forces as positive
09:17so here the shear force is
09:19plus 77 kilo Newton
09:22therefore SF at A to the right
09:24equal to
09:25plus 77 kilo Newton
09:30here as the shear force value is positive
09:32so I will mark this point
09:34above the reference line
09:35of 0 kilo Newton shear force
09:39and I will connect
09:40these two points
09:41with a vertical line
09:46now at point C
09:48there is a point load
09:49hence I will calculate
09:51the shear force values
09:52at left side and right side
09:54of point C
09:58so first to calculate
09:59shear force at point C to its left
10:03I will take the section
10:04to the left of point C
10:06that is SF at point C to its left
10:09equal to
10:13and here I’ll carry forward
10:14previous value of shear force
10:17up to point A to its right
10:19which is 77 kilo Newton
10:23and to the left side of point C
10:25there is UDL
10:26of 20 kilo Newton per meter
10:30so I will convert this AC portion of UDL
10:32into point load
10:34that is UDL of 20 kilo Newton per meter
10:37multiply by distance 2 meter
10:40now this converted point load
10:42is acting on the beam
10:44in the downward direction
10:46and as per the sign convention
10:48I will consider downward force as negative
10:51hence I will add this point load
10:53with negative sign
10:57therefore
10:58after calculating
11:00this will get the shear force value
11:02equal to 37 kilo Newton
11:06here as the shear force value is positive
11:09hence I'll mark
11:10this point of shear force
11:12above the reference line
11:13of zero kilo Newton shear force
11:18and here the type of load is UDL
11:20over the length of 2 meter
11:22hence to draw the shear force diagram
11:24I will indicate UDL
11:25with an inclined line
11:28so I will connect these two points
11:30with an inclined line
11:35now if I go to the section
11:36to the right of point C
11:38that is SF at point C to its right
11:41equal to
11:44so here I’ll carry forward
11:46previous value of shear force
11:47up to point C to its left
11:50that is 37 kilo Newton
11:54and when we move towards
11:56the right side of point C
11:58then there is one point load
11:59of 45 kilo Newton
12:02which is acting on the beam
12:03in downward direction
12:06so as per the sign convention
12:08I will consider downward forces as negative
12:11so I will add
12:12this point load
12:13with negative sign
12:16therefore
12:17after calculating
12:19this will get shear force value
12:21as minus 8 kilo Newton
12:24that is SF at point C to its right
12:27equal to minus 8 kilo Newton
12:30here as the shear force value is negative
12:33hence I will mark
12:34this point of shear force
12:36below the reference line
12:37of zero kilo Newton shear force
12:41and I will join
12:43these two points
12:44with a vertical line
12:48now at point B
12:49there is reaction force Rb
12:51acting on the beam
12:53in the upward direction
12:54and since it is point load
12:56hence as per the rule
12:58for point load
12:59I will calculate the shear force values
13:02at left side
13:03and right side
13:04of that point load
13:08therefore first to calculate
13:10shear force at point B
13:11to its left
13:13I will take the section
13:15to the left of point B
13:17that is SF at B to the left
13:19equal to
13:21and here I’ll carry forward
13:23previous value of shear force
13:25up to point C to its right
13:28which is minus 8 kilo Newton
13:31and to the left side of point B
13:33there is UDL
13:35of 20 kilo Newton per meter
13:38so I will convert this CB portion of UDL
13:40into point load
13:42that is UDL of 20 kilo Newton per meter
13:45multiply by distance 3 meter
13:49now this converted point load
13:50is acting on the beam
13:52in downward direction
13:54and as per the sign convention
13:56I will consider downward force as negative
13:59hence I will add this point load
14:01with negative sign
14:05therefore
14:06after calculating
14:08this will get the shear force value
14:09as minus 68 kilo Newton
14:12that is SF at point B to its left
14:15equal to minus 68 kilo Newton
14:19here as the shear force value is negative
14:22hence I'll mark this point
14:24below the reference line
14:25of zero kilo Newton shear force
14:28and here the type of load is UDL
14:30over the length of 3 meter
14:32hence to draw the shear force diagram
14:34I will indicate UDL
14:36with an inclined line
14:38so I’ll connect these two points
14:40with inclined line
14:45now next to calculate
14:46shear force at point B to its right
14:50I will take the section
14:51to the right of point B
14:53that is SF at point B to its right
14:57equal to
15:00so here I’ll carry forward
15:02previous value of shear force
15:04up to point B to its left
15:06which is minus 68 kilo Newton
15:10and when we go to the right side
15:12of point B
15:14then there is reaction force Rb
15:16of 68 kilo Newton
15:18which is acting on the beam
15:19in the upward direction
15:22so as per the sign convention
15:24I will consider upward forces as positive
15:27so here I will add this point load
15:29with positive sign
15:32so here minus 68
15:34plus 68
15:35gives me the value of shear force
15:37as zero kilo Newton
15:40therefore SF at point B
15:42to its right
15:43equal to zero kilo Newton
15:46so I’ll mark this point
15:48of zero kilo Newton shear force
15:50on the reference line
15:52and I’ll connect these two points
15:54with a vertical line
15:58and here in shear force diagram
16:00whatever the portion drawn
16:02above the reference line
16:04I will show this portion
16:05by positive sign
16:08and the portion which is drawn
16:09below the reference line
16:11I will show this portion
16:13by negative sign
16:16so here I have completed
16:17the shear force diagram
16:21now the next step is
16:23calculations of bending moments
16:26so bending moment
16:27at a section of beam
16:29is calculated
16:31as the algebraic sum
16:32of the moment of all the forces
16:35acting on one side of the section
16:38so to calculate bending moments
16:41we can start
16:42either from left end of beam
16:44or from right end of beam
16:47here I will start
16:49from left end of beam
16:52so whenever you are calculating
16:54the bending moments
16:55you should remember
16:57these conditions
16:59so here for simply supported beam
17:01the condition is
17:03at the ends of simply supported beam
17:05the bending moment will be zero
17:09so bending moment at point A
17:10and bending moment at point B
17:12will be 0
17:14that is BM suffix A
17:16equal to 0 kilo Newton meter
17:18and BM suffix B
17:20equal to 0 kilo Newton meter
17:24so to draw bending moment diagram
17:26firstly I will draw
17:28the reference line of bending moment
17:300 kilo Newton meter
17:32so here I’ll mark these values
17:34with a points on the reference line
17:39so now we have to calculate
17:40bending moment at point C
17:44so here
17:45in case of simply supported beam
17:47while you are doing the calculations
17:49for bending moment
17:50at particular point
17:53you should always add
17:55moment of all the forces
17:57present
17:58either from left end of beam
18:00or from right end of beam
18:02up to that particular point
18:04at which
18:05you are calculating the bending moment
18:08it means
18:09while I am calculating
18:11the bending moment at point C
18:13I will add moment
18:15of all the forces
18:17present
18:18either from left end of beam
18:20or from right end of beam
18:22up to point C
18:25and for bending moment calculations
18:27our sign convention is
18:30for sagging effect of beam
18:32the force is considered as positive
18:35for hogging effect of beam
18:38the force is considered as negative
18:42so to calculate
18:44bending moment at point C
18:46that is BM suffix C
18:48equal to
18:51here at left hand side of point C
18:53there is reaction force Ra
18:55of 77 kilo Newton
18:57which is acting on the beam
18:59in the upward direction
19:02due to this force
19:03the beam shows sagging effect
19:06and for sagging effect of beam
19:08I will consider this force as positive
19:13so I will add
19:14this force with positive sign
19:17and as per decided
19:18the moment is force into distance
19:21so I will multiply this force
19:23with the distance
19:24from point of action of force
19:26that is 2 meter
19:30and here
19:31from point A
19:32to point C
19:34there is UDL
19:35of 20 kilo Newton per meter
19:38so firstly I will convert
19:39this AC portion of UDL
19:41into point load
19:44that is UDL of 20 kilo Newton per meter
19:46multiply by distance 2 meter
19:50now this converted point load
19:52is acting on the beam
19:54in the downward direction
19:56due to this converted point load
19:59the beam shows hogging effect
20:01and for hogging effect of beam
20:03I will consider this converted point load
20:05with negative sign
20:08so I will add
20:09this converted point load
20:10with negative sign
20:13and I will multiply this force
20:15with the distance
20:16from point of action of force
20:19that is 1 meter
20:22therefore
20:23after calculating
20:25this will get the value
20:26of bending moment at point C
20:28equal to
20:30114 kilo Newton meter
20:34that is BM suffix C
20:35equal to 114 kilo Newton meter
20:41so as it is positive value
20:43of bending moment
20:44hence I'll mark
20:45this point of bending moment
20:47above the reference line
20:48of 0 kilo Newton meter bending moment
20:53and here
20:54between point A and point C
20:57there is UDL
20:58therefore to draw bending moment diagram
21:01I will indicated UDL
21:02with a parabolic curve
21:05hence I will join
21:06these two points
21:08with a parabolic curve
21:12and also
21:13between point C and point B
21:16there is UDL
21:18therefore to draw bending moment diagram
21:20I will indicated UDL
21:22with a parabolic curve
21:24hence I will join
21:25these two points
21:27with a parabolic curve
21:30now since I can see
21:32this bending moment diagram is drawn
21:34above the reference line
21:36of bending moment 0 kilo Newton meter
21:39so whatever the portion I have drawn
21:41all the values of bending moment
21:42are positive
21:43hence I will show this portion
21:45by positive sign
21:48so here I have completed
21:50the shear force diagram
21:51and bending moment diagram
21:53for this simply supported beam