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@@JF59122 Yes, absolutely correct. He has written wrong in the equation but has written the correct magnitute of the answer. Please check my calculation
4m (5m for UDL from the end - 1m for Rb from the end) is correct as per his explanation. Sketch a free body diagram and convert the udl into point load, you'll understand it!!
3:34 and just ahed of ….. It’s still confusing how you can calculate distance … plz show how you calculate bcz it doesn’t seems right bcz if we take 1800 N at center at the beam which is 5mm then with respect to B Load must be at 3.5 not 4. So clarifying it properly not assume or take 4 in calculations and i am still confusing with the sign of 1800 N force in B.M calculations. 🙏🏻 anyway good effort but elaborate properly so we can easil understand.
4:21 Here the calculation will be:- (-200×1)+(-Rc×7)+(-180×10×(0.5+3.5+1))=0 or, -200-7Rc-9000=0 or, -7Rc=9200 or, Rc= -1314.28N Please check if there is any mistake!!!
On the calculation of shear force at point C when you have taken the section toward the right of C it will be +2000 in place of +1000 at the calculation time you have taken 1000 two time it should be 2000 at one place as it is the support reaction at point C you have taken it 1000 also
because in udl the load is different on both side of section. e.g at point B 1st load of A = 200 n the udl load of 180*1 (1 is the distance btw a and b) and after section B we have another load of RB
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Although ur comment is 6 months old but I would like to correct you It is in clockwise direction if taken point b as reference and clock wise direction is taken positive in find moment ...
HERE IS YOUR CORRECT SOLUTION ; here is some confusion going on , the teacher just follo the book and in the book its not simplified so let me calculate for moment at B .....-RcX7 + 180X9X9/2 - 200 - 180X1x1/2 = 1000 its a very good question , the 180 udl load is ating on right and left side of the section thats why i took +180x9x9/2 on right section where as on left there is a span of 1 mtr so again - 180x1x1/2 this is how you can get it
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Sir definition aur is solving me main confuse Hoon ki section ke right ya section ke left calculation karna hai aap dono kyun Kar Rahe hain.please answer
may be = lets again draw a diagram, an u will see the total lenght is 10. now divide 10/2=5 here 5 is in ur centre but wait we have to take the centre of Rb and Rc and the distance between rb and rc is 7 so if we divide 7/2 it comes 3.5 on each side. Now the trick pART IS we have to take the centre not whole beam but between rb and rb so we know half of section = 5 and when we add 3.5 + 2 = 5.5 on RHS and 4.5 on LHS so shift .5 to LHS then it comes 1+4=5 and on RHS 3+2 = 5 it will come like this . i know u r not going to understand this but u can mail me i also made a diagraam.
may be = lets again draw a diagram, an u will see the total lenght is 10. now divide 10/2=5 here 5 is in ur centre but wait we have to take the centre of Rb and Rc and the distance between rb and rc is 7 so if we divide 7/2 it comes 3.5 on each side. Now the trick pART IS we have to take the centre not whole beam but between rb and rb so we know half of section = 5 and when we add 3.5 + 2 = 5.5 on RHS and 4.5 on LHS so shift .5 to LHS then it comes 1+4=5 and on RHS 3+2 = 5 it will come like this . i know u r not going to understand this but u can mail me i also made a diagraam.
HERE IS YOUR CORRECT SOLUTION ; here is some confusion going on , the teacher just follo the book and in the book its not simplified so let me calculate for moment at B .....-RcX7 + 180X9X9/2 - 200 - 180X1x1/2 = 1000 its a very good question , the 180 udl load is ating on right and left side of the section thats why i took +180x9x9/2 on right section where as on left there is a span of 1 mtr so again - 180x1x1/2
In Point load calculation (Previous videos) your have drawn centre line on the right side for bending moment but in this video left side for point "B" Totally confusing, To match with the answer you are switching left and right ??
HERE IS YOUR CORRECT SOLUTION ; here is some confusion going on , the teacher just follo the book and in the book its not simplified so let me calculate for moment at B .....-RcX7 + 180X9X9/2 - 200 - 180X1x1/2 = 1000 its a very good question , the 180 udl load is ating on right and left side of the section thats why i took +180x9x9/2 on right section where as on left there is a span of 1 mtr so again - 180x1x1/2
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180(10 :which is the sum of the distances going to the load) (4 :because the half of the total (10) is 5 but it only stops at B so the 1mw hich is the distance from A to B is not included so 180(10)(5-1)
I think so he is not wrong because he is taking the whole beam and if you take the whole beam then it's 10 and half of 10 is 5 and the moment is at B then 5-1 =4 i.e.,180×10×4
I do matrix formulation for any beam problems , my method is based on graphical,numerical,and analytic analysis.what would you do if I add 10 or more support connection to your problem?not software are allowed to use.
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The matrix formulation you are talking about is used in Finite Element Analysis..in this video i have solved the problem by using the concepts of SOM(Strength of Materials)...yess i can solve it by matrix formulation also.
Sir ek simply beam hai uski length 8 miter pr Uniformly Distribute Load (UDL) 2 KN/M laga hai aur us pr point Load bhi lga hai 6 KN sir isko solve kaise ho ga
UDL that is converted into Point/Concentrated Load is 5m away Point A. And Point B is 1m away from Point A. So, Distance between the Point Load and Point B = (The distance from Point Load to Point A) - (Distance between Points A and B)
It's 180 into 10 into 10(full lenght) divided by 2 coz udl is taken as a point load at the centre, subtracted by 1 because its on the left side of the B which has to be subtracted. Hope it helps im struggling myself here
Shilpika Parida first take the entire udl as point load acting at the centre, then, subtract the overhanging lenght on the left of B since we are taking moment about B. That's the only possible way he could've done it,either that or he's wrong, coz udl is present over the entire lenght and hence it has to be taken as a whole. You can't do 180*7*7÷2 that would be bending moment.
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How to get notes for sfd and BMD sir please tell me fast tommorow our exam🙏
Wrong
Bad concepts
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-200x1-1800x5+7RC=0 is the correct eqn for moment at B which comes out to be RC=1314.28N
Wait isn't RC7 a hogging force so it should be negative right
@@JF59122 Yes, absolutely correct. He has written wrong in the equation but has written the correct magnitute of the answer. Please check my calculation
4m (5m for UDL from the end - 1m for Rb from the end) is correct as per his explanation. Sketch a free body diagram and convert the udl into point load, you'll understand it!!
Final year ho gya ?
Why minus 1m for Rb?
It's 7m from b to c..... So bending moment at that B will be -Rc×7 +1800×3.5 - 200 = ....... Right?
thanks a lot for this comment dude,i was struggling with that part
These are really the best lectures I have seen
-200-180*1*1/2+180*9*4.5-7Rc=0
-200-90+7290-Rc=0
Rc=1000
Why 7290 is plus? Not minus ?
even he is sleepy as i am.. XD 4:11 But these vids are really helpful and step by step at correct pace! Cheers man!
3:34 and just ahed of ….. It’s still confusing how you can calculate distance … plz show how you calculate bcz it doesn’t seems right bcz if we take 1800 N at center at the beam which is 5mm then with respect to B Load must be at 3.5 not 4. So clarifying it properly not assume or take 4 in calculations and i am still confusing with the sign of 1800 N force in B.M calculations. 🙏🏻 anyway good effort but elaborate properly so we can easil understand.
4:21
Here the calculation will be:-
(-200×1)+(-Rc×7)+(-180×10×(0.5+3.5+1))=0
or, -200-7Rc-9000=0
or, -7Rc=9200
or, Rc= -1314.28N
Please check if there is any mistake!!!
On the calculation of shear force at point C when you have taken the section toward the right of C it will be +2000 in place of +1000 at the calculation time you have taken 1000 two time it should be 2000 at one place as it is the support reaction at point C you have taken it 1000 also
Its kind request can u give description about sfd nd bmd ofcantilever beam..and few problems related to it...
Its very bad explanation
Explanation
-200*1-180*1*1/2+180*7*3.5-Rc*7+180*2*8=0
Then Rc=1000N
This is correct process✅
uu r correct.... what the hell the teacher was doing ... making confusions
You are being confused a lil
He just converted udl into point load that is acting at centre 5unit from A now if u go for calculation it won't be wrong
Why u took 180*2*8 i think it should be 180*2*1 or 180*9*9/2
Calculation is right 100%..thankyu sir
ahmed khan are you sure??? Everyone are saying his calculations are wrong
sir, please make a video on sfd and bmd for fixed beams with calculations. sir your explaination is awesome and easy to understand
Thappu
Is it simply supported or overhang
why did he took two sections for each point?
because in udl the load is different on both side of section. e.g at point B 1st load of A = 200 n the udl load of 180*1 (1 is the distance btw a and b) and after section B we have another load of RB
@@ankushjaswal1992 thanks bro 😊
Sir u r great thx .god be with you all the time thanks alot
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Sir when we get single value or when we get double value ?
The diagram is overhanging, but in question it's written as simply supported!!
Really a very nice explanation sir.... helps a lot.... thank you sir
You are doing a great job sir😘 yoi saved my lyf 😍
Excellent effort and lucid presentation. Thanks
Thank you for your words!
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Shudn't that (1800*4) be preceded with "-" (minus) as u hv shown it is an anti clock-wise load?
Although ur comment is 6 months old but I would like to correct you It is in clockwise direction if taken point b as reference and clock wise direction is taken positive in find moment ...
@@craftart1365 isn't the moment on left side of B should be +ve , or sign convention for moment and bending moment are different?
@@debashisdasjnvp it should be - since the convention sign for bm starting from the left as in the example above is anticlockwise
Its in clockwise
HERE IS YOUR CORRECT SOLUTION ;
here is some confusion going on , the teacher just follo the book and in the book its not simplified so let me calculate for
moment at B .....-RcX7 + 180X9X9/2 - 200 - 180X1x1/2 = 1000
its a very good question , the 180 udl load is ating on right and left side of the section thats why i took +180x9x9/2 on right section
where as on left there is a span of 1 mtr so again - 180x1x1/2
this is how you can get it
Bhut achha se aapne padhya sir
Sb samjh aa gya .
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Thank you for your love!
@@Ekeeda do yu know Kenya sir?
@@Antonio-ji9jn yes my buddy we know ☺️
Sir...Why do we get two values of SF at B??
Sf at point B
180 nm = 1800 n
Please correct it...!
Nice sir ❤️👍
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Thnkyu sir....thnkyu very much....it helped a lot
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you made a wrong calculation for moment at B
also at SF calculation for point C, where he multiplied 180 with 7 instead it should have been multiplied by 8
Yer u r right bro he made a mistake at point B
@@lakshaygangwal8860 because he already calculated the load over b by multiply 180*1, so next time we only have to go for 180*7 at c
@@divyanshjangid9773 because he already calculated the load over b by multiply 180*1, so next time we only have to go for 180*7 at c
Distance between c and b is 7 so it's absolutely right go through the video once
Kindly explain 1800×4?
How is it possible?
I'm not able to understand this question..shear force calculation is so confusing .
its easy calculation
supposed to multiply with 4.5m from point B which 4.5m is the centre where the UDL is acting for the +moment
Abdul Asyraf Yusuf bro are you sure because according to concept you r correct 4.5 me also but which concept is taught in this 4
Thank you very much sir aap bht aacha samjate Hain
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U are just awesome sit
Sir definition aur is solving me main confuse Hoon ki section ke right ya section ke left calculation karna hai aap dono kyun Kar Rahe hain.please answer
2:43 how the moment is zero?
Can anyone explain me this?
Nidra po velli full wrong in moments finding
Telugu viewer spotted
@@pranaykumar333 😁
Why are we getting two shear forces for B and C points?
It's very easy types solved
cant get better than this
Moments at point 'B'
1800*4
How it's possible
It's possible
@@RadhaKrishna-oc4yy
How explain it
@@RadhaKrishna-oc4yy he made a mistake
we see he is yelling.... so made mistake that is what i guess
may be = lets again draw a diagram, an u will see the total lenght is 10. now divide 10/2=5 here 5 is in ur centre but wait we have to take the centre of Rb and Rc and the distance between rb and rc is 7 so if we divide 7/2 it comes 3.5 on each side. Now the trick pART IS we have to take the centre not whole beam but between rb and rb so we know half of section = 5 and when we add 3.5 + 2 = 5.5 on RHS and 4.5 on LHS so shift .5 to LHS then it comes 1+4=5 and on RHS 3+2 = 5 it will come like this . i know u r not going to understand this but u can mail me i also made a diagraam.
Sir
In this question, when we take moment about B, then why it is -1800×4?
Why isn't it -180×5? As moment will be on the centre..
may be = lets again draw a diagram, an u will see the total lenght is 10. now divide 10/2=5 here 5 is in ur centre but wait we have to take the centre of Rb and Rc and the distance between rb and rc is 7 so if we divide 7/2 it comes 3.5 on each side. Now the trick pART IS we have to take the centre not whole beam but between rb and rb so we know half of section = 5 and when we add 3.5 + 2 = 5.5 on RHS and 4.5 on LHS so shift .5 to LHS then it comes 1+4=5 and on RHS 3+2 = 5 it will come like this . i know u r not going to understand this but u can mail me i also made a diagraam.
so 4 is correct
HERE IS YOUR CORRECT SOLUTION ;
here is some confusion going on , the teacher just follo the book and in the book its not simplified so let me calculate for
moment at B .....-RcX7 + 180X9X9/2 - 200 - 180X1x1/2 = 1000
its a very good question , the 180 udl load is ating on right and left side of the section thats why i took +180x9x9/2 on right section
where as on left there is a span of 1 mtr so again - 180x1x1/2
he is just writing everything down without proper explanation
Everything is confusing
Yes bro u r absolutely right
Burst explaination👾😵
3:51 plz explain....4+1=4...how??
In Point load calculation (Previous videos) your have drawn centre line on the right side for bending moment but in this video left side for point "B" Totally confusing, To match with the answer you are switching left and right ??
HERE IS YOUR CORRECT SOLUTION ;
here is some confusion going on , the teacher just follo the book and in the book its not simplified so let me calculate for
moment at B .....-RcX7 + 180X9X9/2 - 200 - 180X1x1/2 = 1000
its a very good question , the 180 udl load is ating on right and left side of the section thats why i took +180x9x9/2 on right section
where as on left there is a span of 1 mtr so again - 180x1x1/2
Taking moment about a at bigining, called 4 -1 ie 4 mtr?how is it??pl rectify pl
Why sf at B two time calculate at right and left
How to watch the private videos, your videos are very simple
Well presented Sir.
Awesome...
Why did you use only 3.44 for the distributed load and not 4.44 for the very last moment?
Sir i have a doubt when you calculate the shear force at B point
-7*Rc+(1800*5)-(200*1)
why we calculated two share forces at point b and c?🤔
The equation for moment about point B has been wrongly done sir. Please rectify.
Sir sfd or bmd calculation krte time left to right ya right to left jane se graph ka positive or negative region change ho jta h kya?
What is the concept of multiplaying Centre of gravity in Shear force i am too confuse plz give me some clues
how 4 comes.you said 1+4=4
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Isko kon subscribe karta hain
Fucking asshole he is.
He writes much but explains less.
Sholdnt it be 1800 * 5 because the UDL is over the whole beam ?
I also thinking about this
1800 * 4 is the correct one bro
Fun AniMeme yes correct
From b
Why we have to take to calculations for pt B and pt C left and write side..
How did you get the value of 1800x4 in reaction of beam... explain it
Dear Sir,
The UDL which is converted into point load is 1800N ,but you have consider its as 180N ,Kindly reply with conclusion
180 * 10 i.e, load at center of it
@@Samreen-hn4ql yes it should be 1800
1800*4 at B is wrong rite?
sir why you take negativ sing for U D L to calculate support reaction??????????????
Few mistakes in calculation. I guess. For UDL. Its load×distance×distance/2. But he did in another way... while calculating Moment at B
no he is correct bcz he has to calculate moment at B and that 1cm is from point A. So 5-1=4 hence 4cm is distance
That is for uvl
Why did we take +ve value for the 1800N force acting downwards?
because when we sit at b, 1800N will rotate the beam in clockwise direction, clockwise is taken positive.
how do you calculate the value of rc
9:36 there is a mistake
There is -1000 not +1000
No
Sir ek hi point pe shear force ki do value kase ho skti ha
Best lecture.... Thank you so much sir
Don't just calculate it in your head and put down the numbers. It's supposed to be a you tube video elaborate man!
Absolutely right brother… give a brief or elaborate that how you put that value ..👍🏻
SF at point B why we take 2 values???
how to write SF at point D = 0 ???
Nice! Make Some more videos Sir.
how 1800*4 comes there??😓😓 you didn't tell that
Sagar Negi I will 180×9×4.5
180(10 :which is the sum of the distances going to the load) (4 :because the half of the total (10) is 5 but it only stops at B so the 1mw hich is the distance from A to B is not included so 180(10)(5-1)
Jose Miguel Betanio great bro but one doubt we are taking from rc then why we put the value or C-D 2m
pls I can't understand how it's possible pls help me
Sagar Negi 1m plus 7m it's 8m.... And half of it is 4m.... I think so...
Thank u sir plz solve influence line diagram problems sir
1800 will be multiplied with 3.5 ...... your calculation is wrong...
I think so he is not wrong because he is taking the whole beam and if you take the whole beam then it's 10 and half of 10 is 5 and the moment is at B then 5-1 =4 i.e.,180×10×4
You are telling conversion from N/m to N right ??
But here 180x10 (Total length 10m)
@@shivashelva_ 180*10 =1800 N and multiply 4 because its in the centre
@@ankushjaswal1992 Thanks for your feedback
Good sir
How 1800×4
180 is a load multiply by total distance i.e 10 into 4 i.e the distance of force from B...u will get 180x10x4= 1800x4.....
Sir please upload more videos like this and especially on cantilever beams
can i have a overhanging beam and it has udl entire over its length and point loads in between
Hello
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Explanation is awesome btw No need of your picture in video bv its distracting
why we take two position for shear force i.e left side and right side while
in last question we only take left side for shear force
In step 2 waht if we take moment at poit c
Hey Sir,On the distance of the point load 1800 am i supposed to multiple it by 10 as the distance load or 5?
@Pavan Kumar y don't we take 3.5 there... As half of 7
@@Samreen-hn4ql UDL is on whole beam that's why we should take half of the whole beam
Thank you sir you teach very well
I do matrix formulation for any beam problems , my method is based on graphical,numerical,and analytic analysis.what would you do if I add 10 or more support connection to your problem?not software are allowed to use.
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The matrix formulation you are talking about is used in Finite Element Analysis..in this video i have solved the problem by using the concepts of SOM(Strength of Materials)...yess i can solve it by matrix formulation also.
Can i get ur nuumber plzz..
I need to know how it can be solved using matrix
Sir ek simply beam hai uski length 8 miter pr Uniformly Distribute Load (UDL) 2 KN/M
laga hai aur us pr point Load bhi lga hai 6 KN sir isko solve kaise ho ga
SFD & BMD
Sir how it comes 1800*4..?? Sir plz clear this confusion . Waiting for your rply
UDL that is converted into Point/Concentrated Load is 5m away Point A.
And Point B is 1m away from Point A.
So, Distance between the Point Load and Point B = (The distance from Point Load to Point A) - (Distance between Points A and B)
@@SoudagerAamer mujhe sumjhaao please
How to draw shear force diagram and bending moment calculation and bending moment diagram
be half of 7 is 3.5....but u took 4 .....why...?
Yes how sir
It's 180 into 10 into 10(full lenght) divided by 2 coz udl is taken as a point load at the centre, subtracted by 1 because its on the left side of the B which has to be subtracted. Hope it helps im struggling myself here
Plzzzzzz mujhe samjhao
180×10×4 eya 4 kahase ayaa.
Shilpika Parida first take the entire udl as point load acting at the centre, then, subtract the overhanging lenght on the left of B since we are taking moment about B. That's the only possible way he could've done it,either that or he's wrong, coz udl is present over the entire lenght and hence it has to be taken as a whole. You can't do 180*7*7÷2 that would be bending moment.
how come is it a simply supported beam??? Isn't it over hanging on both sides?????????
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Sir how to know that moment is clockwise or anticlockwise..?
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we cant take the moment of both side about reference point at once.