- 3 months ago
Category
π
LearningTranscript
00:00In this question, we will give this beam, and in this beam, we will give this fixed support, and in this beam, we will give this roller support.
00:08Now, in this DEF section, we will have a uniform distributed load of 20 km per meter active way.
00:17In point C, we will have a clockwise moment of 60 km active way, and in point B, we will have a point load of 200 km per meter active way.
00:28In this case, we will give this beam to the beam, the axial force diagram, shear force diagram, bending moment diagram.
00:37In this case, we will have a first step to calculate the support reactions.
00:43In this case, we will have a fixed support, which is fixed support, which is fixed support.
00:49In this case, we will have a reaction, REY, that is vertical reaction.
00:55Basically, we will have a fixed support, which is horizontal, vertical, and moment.
01:02So, we will have a reaction to write RAY, RAX, and moment at AIMA.
01:09So, all together, we will have a 4 unknown support reactions to calculate.
01:14Our 4 unknown reactions, we will have a 3 equation of static equilibrium.
01:24So, our first step is...
01:34The first step is taking moment about point D, which is right side equals to 0, and we are considering anti-clockwise moment as positive.
01:48So, our first step is taking moment about point D, which is right side equals to 0, and we are considering anti-clockwise moment as positive.
01:53So, our first step is taking moment about point D, which is right side equals to 0, and we are considering anti-clockwise moment as positive.
02:03So, our first step is to take moment of force, which is right side equals to 0, and we are considering anti-clockwise moment as positive.
02:12That is, REY into 4.
02:16This step is taking moment of UDL, which is clockwise moment as positive.
02:20So, the first step is to take moment of UDL, which is 20 into D total distance, 8 into 4.
02:27So, at this step, from UDL, which is a 0, it is equal to 0.
02:33So, at this step is our first step of A, we have to calculate it by the moment.
02:41The value of R, we are considering, that we are considering the direction of R, we are acting.
02:47So, we have to calculate reaction value.
02:50Now, in this step, we calculate Rx value.
02:53because there is a horizontal reaction here
02:56so rax calculate the answer
02:58so our next step is summation
03:01force along x direction equals to 0
03:04we are considering force along right side as positive
03:06we are considering force along right side as positive
03:08we are considering force on the right side
03:12this is the cost component of the cost component
03:15that is 200 cost 30 degree
03:18minus 200 cost 30 degree
03:23and we are considering force
03:25which is the moment
03:27which is completely vertical direction
03:29so now this equals to 0
03:31and with this equation
03:33rax value
03:35rax value here
03:37173.2 kilonewton
03:41so rax we are now positive
03:43that means we assume
03:45rax is active
03:47now calculate
03:49ray or moment
03:51so
03:53so
03:55sum of vertical forces
03:57equal to 0
03:59that is
04:01sum of force along y direction
04:03equal to 0
04:05considering upward force as positive
04:07that means
04:09we are going to calculate
04:11ray
04:13200
04:15sine component
04:17which is acting along downward
04:19that is
04:21minus 200
04:23sine
04:25into 8
04:27and
04:29right
04:31and
04:33right
04:35which is acting along upward
04:37and its value is 160
04:39that is
04:41plus
04:43160 equals to 0
04:45that is
04:47100 kilonewton
04:49value positive
04:51assume
04:52that is
04:53upward direction
04:54that is
04:55upward direction
04:56that is
04:57upward direction
04:58so
05:00we can
05:02either
05:03moment about point
05:04equal to 0
05:05or moment about point
05:06equal to 0
05:07so
05:08moment about point
05:10equal to 0
05:11that is
05:12sum of moment about point
05:14equal to 0
05:15we are considering
05:16anticlockwise moment as positive
05:18but
05:19some moment about point
05:20equal to 0
05:21so
05:22moment about point
05:23equal to 0
05:24so
05:25ma
05:26so
05:27anticlockwise
05:28positive
05:29ma
05:30equal to clockwise direction
05:31negative
05:32that is
05:33minus
05:34ma
05:35this
05:36ray
05:37into
05:392
05:402
05:414
05:426
05:43plus
05:444
05:4510
05:46100
05:47into
05:4810
05:49now
05:50this
05:51moment
05:52is
05:53200
05:54this
05:55component
05:56is
05:57200
05:58sin
05:59component
06:00that is
06:01200
06:02sin
06:0330
06:04moment
06:05and
06:06moment
06:07will be
06:08anticlockwise
06:09that is
06:10plus
06:11200
06:12sin
06:1330
06:14degree
06:15that is
06:17minus
06:1860
06:19so
06:20the
06:21point
06:22about
06:23moment
06:24and
06:25the
06:26point
06:27is
06:28right
06:29so
06:30this
06:31portion
06:32d
06:33e
06:34portion
06:35is
06:36symmetric
06:38or
06:40point
06:41e
06:42is
06:44equal to
06:45this
06:46moment
06:47is
06:48equal to
06:49this
06:50moment
06:51is
06:52equal to
06:53this
06:54portion
06:55is
06:5620
06:57into
06:584
06:59into
07:004
07:01by
07:022
07:03and
07:04this
07:05is
07:06minus
07:0720
07:08into
07:094
07:10into
07:114
07:12into
07:134
07:14by
07:152
07:16and
07:17this
07:18is
07:19equal to
07:200
07:21the
07:23remaining
07:24portion
07:25we
07:26solve
07:27the
07:28value
07:29calculate
07:30ma
07:31minus
07:32260
07:33kilonewton
07:34meter
07:35minus
07:36260
07:37kilonewton
07:38meter
07:39clockwise
07:40so
07:41we assume
07:42opposite direction
07:43moment
07:44act
07:45that is
07:46260
07:47kilonewton
07:48meter
07:49and it is not acting along clockwise it is acting along anti clockwise direction
07:53so
07:54so
07:55we calculate the support reaction
07:57now we calculate the forces
07:59so
08:00first we calculate the axial force
08:02so
08:03axial force
08:04we calculate the axial force
08:05so
08:06axial force
08:07we calculate the axial force
08:08we
08:09now
08:10the reaction force
08:11is
08:12given beam
08:13and
08:14make x free
08:15so
08:17so
08:18so
08:19this
08:20is
08:21the
08:22reaction force
08:23here
08:24is
08:25the reaction force
08:26is
08:27the value of
08:28160
08:29kilonewton
08:30upward
08:31direction
08:32so
08:33vertical
08:35reaction force
08:36is
08:37100
08:38horizontal
08:39reaction force
08:40is
08:41173.2
08:42and
08:43moment
08:44moment
08:45moment
08:46about
08:47point
08:48a
08:49i
08:50is
08:51260
08:52so
08:53this
08:54is
08:55the
08:56reaction force
08:57is
08:58260
08:59so
09:00this
09:01is
09:0220
09:03kN
09:04per
09:05meter
09:06and
09:07this
09:08load
09:09is
09:10moment
09:11at
09:12point
09:13c
09:1460
09:15km
09:16clockwise
09:17direction
09:1860
09:19kilonewton
09:20moment
09:21point
09:22b
09:23point
09:24load
09:2530
09:26degree
09:27inclination
09:28so
09:29point
09:30load
09:31resolve
09:32two
09:33factors
09:34components
09:35so
09:36200
09:37kilonewton
09:38two
09:39two
09:40hundred
09:41sine
09:42thirty
09:43two
09:44hundred
09:45sine
09:46thirty
09:47value
09:48by
09:49100
09:50cos
09:51component
09:52x
09:53x
09:54x
09:55x
09:56x
09:57x
09:58x
09:59x
10:00x
10:01x
10:02x
10:03x
10:04x
10:05x
10:06x
10:07x
10:08x
10:09x
10:10x
10:11x
10:12x
10:13x
10:14x
10:15x
10:16x
10:17x
10:18x
10:19x
10:20x
10:21x
10:22x
10:23x
10:24x
10:25x
10:26x
10:27x
10:28x
10:29x
10:30x
10:31x
10:32x
10:33x
10:34x
10:35x
10:36x
10:37x
10:38I am sure force to go to point A
10:47so we are left to go right
10:50so we are left to go right
10:54left to go right to go up positive and down negative
10:59so point A
11:01point A we are left to go left
11:03so left to go right to go left
11:05so we are left to go left
11:08so point A we are not going to start the structure
11:11that is it will be zero
11:13now the shear force at point A
11:15when the shear force is left to go right
11:18the shear force is left to go right to the reaction force
11:21that is 100 kN which is acting along upward
11:23so the positive shear force is 100 kN
11:28now we are going to point B
11:30shear force at point B
11:33so the shear force is left to go right to the shear force
11:36so B and A is not force
11:39so the shear force is left to go right to the shear force
11:41so B is not force
11:43B is the vertical shear force
11:45100 kN
11:46B is the right to go right
11:49we are left to go left to go right to the 100 kN
11:52so shear force at B
11:54so B is the right to go right to the 100 kN
11:57so the shear force is left to go right to the 100 kN
12:00which is acting along downward direction
12:02so minus 100 kN
12:04that is shear force at B
12:07value is 0
12:09now point C
12:09that is shear force at point C
12:12left to go left to the left
12:13C to go left to the first
12:14that is pahε
12:32so scheer force at point C
12:35Now, shear force at point D-right.
12:37D-right, D-back has a point load.
12:42This is UDL, which has just started here.
12:45So, D-right, shear force is 0.
12:48Now, point E.
12:50Here, shear force at point E-left.
12:52Point E-left.
12:53E-left means, E-left means,
12:55we can use UDL for shear force here.
12:59And, this reaction is in E-right.
13:03Now, E-left means,
13:05first, D-right means, value 0 transfer.
13:08Then, this force means,
13:1020 into this distance is 4.
13:12That is,
13:14that means,
13:16that means, 20 into 4 minus 80.
13:20So, E-left means, shear force comes,
13:22minus 80 kilonewton square force.
13:25Now, shear force at E-right.
13:28First, E-left means, shear force as it is.
13:30Now, point E-right means,
13:32point E-right means,
13:33that means,
13:34this reaction force is upward direction.
13:36Now, this is UDL.
13:37And, this is upward,
13:39then, positive.
13:40That is,
13:41plus 160.
13:42And, E-right means,
13:43shear force value is 80 kilonewton.
13:46Now, we can do point F.
13:48So, point F first,
13:49shear force at point F-left.
13:51So, first,
13:52this force transfer to 80.
13:54Now, we can do the shear force.
13:56So, we can do the shear force here.
13:58So, we can do 20 into 4.
14:00And, this is negative.
14:02That is,
14:03minus 20 into 4.
14:05And, this is equal to 0.
14:06So, the shear force at F-right,
14:08first,
14:09we can do the force transfer.
14:10And, point F-right means,
14:11we can do the loading action.
14:13So, we can do 0.
14:15So, we can do 0.
14:17So, we can do the shear force calculation.
14:19just need to do the shear force.
14:20Now, we can calculate bending moment.
14:22We can calculate bending moment.
14:27Bending moment in the beginning,
14:29if there is a assumption that we can do left,
14:30that we can do right,
14:31so, left the right is more difficult,
14:33clockwise moment is positive,
14:34and anti-clockwise moment is negative.
14:36So, this assumption statistic is 0.
14:39Now, the first point is here.
14:42If point is already moment,
14:44when the same point is already moment,
14:46then the point is left and right,
14:48So, moment about point 1 left, when the structure starts with 0, now moment about point 1 right, this is 260 kNm will come, now this is 260 kNm will come, now this is 260 kNm, anti-clockwise direction, as I am saying left to the right, clockwise positive, so this is negative moment of 260 kNm, now we have point B map, so moment at point B, so now we have to transfer this moment as it is,
15:17if we load this moment, we have to include this moment, so this moment is 100 kNm, so this is 100 kNm, that is plus 100 kNm, that is clockwise moment, so this moment is about point B, value is minus 60 kNm,
15:42now point C ma already moment act by gocha, so c ko pa ni left and right, we have to consider the moment, so moment about point C ko left, C ko left ma 60 kNm, so first we have to transfer this 260 kNm as it is, that is minus 260,
15:58after this moment, here, the moment was 100pNm, that is plus 100pNm, again, this 100pNm, this point is 100pNm, and then this moment isena, that is minus 100pNm,
16:12so this moment about point C ko left ma faery, minus 60 kNm, now moment about C ko right, moment about C ko right,
16:22so this moment, if we have to turn C to the left ma moment, that is minus 60 kNm,
16:28This moment is clockwise moment, so we have to have plus left, plus 60 and this equals to 0.
16:34So, C to right moment is 0. Now, moment is about point D.
16:39Point D in question, we have to give internal hints.
16:42So, where internal hints is 0.
16:45So, if we have to have a solution, then we can do that in the right direction.
16:52So, now, moment is about point D.
16:55Moment is about point D.
16:57First, we have to transfer U260.
17:00That is, minus 260.
17:02This is U100 kLt.
17:05Point D is about 100 into 6.
17:08That is, 600 equal clockwise moment.
17:11That is, plus 100 into 6.
17:14Again, U100 is about anti-clockwise moment.
17:21That is, minus 100 into 1.
17:24That is, 4.
17:25This distance is 4.
17:27Then, this 60 kLt is about clockwise moment.
17:32So, plus 60 is about 60.
17:34So, this moment is about point D.
17:36We have to have zero.
17:37That is, our solution is going in the right direction.
17:40Now, calculate E to F point moment.
17:43So, moment at point E.
17:45So, now, moment about point E calculate.
17:48So, moment about point E calculate.
17:50So, moment about point E calculate.
17:51So, first I transfer it to 260.
17:53So, again minus 260.
17:55Then, minus 260.
17:57That is, this 100.
17:59This 100 is about point E.
18:01That is, the moment is about point E.
18:03Plus 100 into 10.
18:07Then, minus 100 into 8.
18:10That is, this 100 is about point E.
18:13That is, the moment is about point E.
18:15So, this 60 kLt.
18:16We have to write here.
18:18Clockwise 60.
18:19That is, plus 60.
18:21So, this moment is the moment.
18:24This moment is the moment.
18:26So, this moment is the moment.
18:28This moment is the value of 20 into 4 into 4 by 2.
18:31That is, minus 20 into 4 into 4 by 2.
18:36This moment at point E is the value of minus 160 kNm.
18:42And, moment about point F calculate.
18:45So, moment about point F.
18:47So, this moment is the value of minus 260.
18:51Here, minus 260 plus 100 into 14 minus 100 into 12 plus 60.
19:03So, this moment is the value of UDL.
19:0520 into 8 into 8 by 2.
19:08That is, minus 20 into 8 into 8 by 2.
19:13So, this moment is the value of 160.
19:15So, this moment is the value of 160 into 4.
19:19That is, plus 160 into 4.
19:21That is, plus 160 into 4.
19:23So, this moment is the value of 0.
19:25So, now we have to plot this value.
19:27So, now we have to make xl force diagram.
19:30We have to make xl force.
19:31We have to make xl force value.
19:33xl force member AB is minus 173.2.
19:37So, xl force member AB is minus 173.2.
19:39So, this is the value of minus i.
19:41So, we have to make xl force.
19:42So, we have to make xl force.
19:44So, here we have to make xl force.
19:46So, this portion of xl force has to make xl force.
19:49And, this is the value of 173.2.
19:53Now I will make a shear force diagram
19:55So the shear force diagram
19:57Shear force at A left is 0
19:59A right is 100
20:01So positive
20:03I have to show you
20:05Ordinate value is 100
20:07And B left
20:09B left is 100
20:11So this 100 is B left
20:13And B right is 0
20:15So 0
20:17So 0
20:19So shear force at C left
20:21C right is 0
20:23D left is 0
20:25So here is 0
20:27Now here is shear force
20:29Minus 80
20:33So here is shear force
20:35Here is 0
20:37And E left is minus 80
20:39So negative
20:41As you can see
20:43This is positive
20:45So here is straight line
20:47And value is minus 80
20:49So here is the right
20:51Shear force is plus 80
20:53That means here is the shear force
20:55And plus 80
20:57Shear force at point F
21:01F left is 0
21:03So point F is 0
21:05So here is the shear force diagram
21:07And this is the shear force diagram
21:09For the given beam
21:11Bending moment diagram
21:13So bending moment
21:15Moment about point
21:17Moment about point
21:19Moment about point
21:20A right is minus 260
21:21And bending moment diagram
21:23Negative aound the right
21:25Positive aound the right
21:27That is minus 260
21:29This negative value is 260
21:31This moment at point A right
21:33So point B
21:35So point B
21:37Moment of moment is minus 60
21:39So point B moment
21:41So point B moment
21:4360 is 60
21:45Let's try straight line
21:47Moment at point C
21:49C left is minus 60
21:51And C right is 0
21:53That is minus 60 constant
21:55C left and C right is 0
21:57But this is still negative
21:59Now here is less the slack
22:00Moment at point E
22:01μ ννλ borders
22:03So point B
22:05If only the bulb rate
22:06It is 0
22:07To draw thecard
22:09Thenζ
ζ³
22:10This is so
22:12ing will be 0
22:13γ γ
22:15Moment at point E
22:17So point E
22:19That means
22:19fundamental value
22:21So point F
22:22and point B
22:24Moment at point F
22:25is 0
22:26But
22:26Bending moment
22:27This bending moment is under uniform distributed load
22:32Uniform distributed load is a parabolic nature
22:37So we have to draw a curve
22:40If we have to draw a curve, we have to draw a curve
22:43So we have to draw a calculation
22:46Now we join a straight line
22:49If we join a mid value, we can find a mid value
22:52The mid value is 0 plus 160 by 2
22:54So this value is 80
22:56Similarly, 160 plus 0 by 2, this value is 80
23:00Now bending moment at DE mid
23:03And the bending moment at EF mid, we calculate
23:06So this figure is reference to the moment at DE mid
23:14So the moment at DE mid, we calculate this figure
23:20Similarly, we calculate the value of minus 40
23:24Similarly, moment at EF mid, we calculate the value of minus 40
23:32So we have to calculate the moment at EF mid
23:38So we calculate the moment at EF mid
23:40So we calculate the moment at F point
23:42So we calculate the value of minus 40
23:44So we calculate the value of minus 40
23:46So we calculate the value of minus 40
23:48So we calculate the value of minus 40
23:50So we calculate the value of minus 40
23:54So we calculate the value of minus 40
23:56So we calculate the value of minus 40
23:58So we calculate the value of minus 40
24:00So we calculate the value of minus 40
24:02So we calculate the value of minus 40
24:04So we calculate the value of minus 40
24:06So we calculate the value of minus 40
24:08So we calculate the value of minus 40
24:10So we calculate the value of minus 40
24:12So we calculate the value of minus 40
24:14So now we have the bending moment diagram, now we will erase the line, now this is the
24:23required bending moment diagram for the given beam.