Sir, for simply supported using formula 3EI/L... we get other valves of final moments and for simple supported using normal formula 2EI/2... Getting different values why?
The formula which you have asked for I myself have consulted to my teacher when i was in college. And they suggest me to use the formula which i have used in this video. No doubt the formula which you are showing is correct but it's actually not used when you have a look in some higher books.
Clockwise means the movement which is similar to the movement of ⏰ like _ 1, 2, 3 and so on. While anti clock wise is movement opposite to the direction of movement of ⏰. Eg. 12, 11,10,9 and so on.
Clockwise +ve and anticlock wise -ve if we are taking moment about RHS and ( clockwise will be -ve and anticlock will be +ve if we are taking moment about LHS.)
Yes you are correct. I am extremely sorry for this. It has happened because i have mistakenly written 25/24(EI.ib) =15 as 24/25 (EI. ib) =15. Thanks alot for letting me know. By the way i have another video in hindi on same topic. And i am sure you are going to find it very helpful. Have a nice day.
Here is your answer. We need to use that equation when our hinged support is at the end of the beam. If we use the same equation then we will have one variable more than the number of equation.
And to prevent the solution getting lengthy we use that equation. And it is easy to recognize whether the support is hinged or fixed just by observation.
Here is my another video on slope deflection method here i have provided all the reason of using that equation , recognizing the fixed and hinged support as well as the problem that we face if we use general equation of slope and deflection. If yuo still face any kind of difficulty then let me know. Fell free to ask. ua-cam.com/video/qQAqGbK19UA/v-deo.html
Sir, for simply supported using formula 3EI/L... we get other valves of final moments and for simple supported using normal formula 2EI/2... Getting different values why?
Here in this numerical joint C is hinged support. And always remember that we apply the condition "equilibrium of joint" only at intermediate supports. Here in the figure you can see that only joint B is intermediate and the joint A and D are lying at ends. And if there were given one more intermediate joint(D assumed) after joint C then we would have applied one more consideration taking the equilibrium of joint C. I hope i am able to clear your doubts.
It's because our support C is hinged and lies at end. Do follow the step shown in the video whenever this case arises. It reduces one unknown and the solution becomes easy.
@@Civillearningonline sir even though the point B is hinged we shouldn't use that equation...we should use that equation only for the 2 ndor 3rd span right??
Yes dear. While calculating fixed end moment we use WL/8. And at the time of drawing BMD we need to use wab/l for point load and wl^2/8 for udl. Hope you understood.
Kani's method: It is suitable for multi storey building. As there is no need of deriving equation and solving it. So it's time saving. While, Slope deflection: It is used to analyse both statically determinate and indeterminate structure. Considering deformation due to bending only.
Dear paruvu, I have created a seperate playlist on my channel for slope Deflection Method you should check out some other examples also. Then I am sure you will be able to clear all your doubts.
Hi ged support are represented by arrow headed upward or triangular shape. Go through some problems which will help you to distinguish different types of supports.
Hello dear, First of all thanks for your comment. Here is your answer, It has happened because of overhanging portion at support C. Your question is we take moment 18 at C. You will find your answer at 8:59 - 9:30. Do watch that step you will find your answer. Have a nice day....
Because humar end c hinded he and support c k right pe overhanging portion bhi he so jab overhanging portion right side rahe ga then hum us k karan jo moment aaye ga usko -ve lenge and agar humara overhanging left side hota to hum moment +ve lete. I hope aap ko samajh aaya hoga. Best of luck for your exam. Description me aap ko 2nd example ka link mil jaye ga. Dekhlo wo bhi. Once again best of luck.
While solve equation we do not need to enter Deta in casio. There is another way of solving the equation in calculator. 1.Press mode 2. Press 5 Now Select the type of equation you want to solve. Now enter the constants.(i.e.numbers) Hope you got your answer.
Hello dear, First of all thanks to you for your comment. You might have noticed that we have got (EI/ib) in all three equations. And we have calculated the value of (El/ib) using the portion AB of the beam. Also, I have not used Mba + Mbc = O here in this numerical. And that's why we have got that answer. And believe me it won't create any kind of problem. And one more suggestion to you to solve problems related to slope deflection method. Try to use both the equation i.e. equation for fixed end and hinged support whenever you have such questions. Because incase if u are using the same equation through out the the Numerical gets trouble some i myself have faced that. This was my first video on this topic. I will soon be adding more examples to clear ur all doubts so stay connected by subscribing @civillearningonline. And i will show the application of Mba + Mbc =O in some other example soon. Thanks alot for ur comment. Have a nice day.
im from srilanka . can you make moment distribution method little hard question in english. before 15th . i will share your video with my friends. thank you .
If there is no simple support at c or if there is fixed support at c then use the usual slope deflection formula and and there will be no need of providing free moment.
@@srujanatenneti4827 if you understand Hindi then you can watch my another video on same method and i am quite sure the these videos are going to be very helpful to you. Do check out the description section. Best of luck for your exam.
Hello dear, Aap ka question he ki Mcb/2 ko 18/2 q liya gaya he? Ans: Humara jo support c he waha pe 2 moments aa raha he ek to point load jo span BC pe he and dursra overhanging portion CD. To aisa case me kya hota he ki hum total moment at C ko lenge ge. And is k liye C pe lag raha sara moments ko hum add karenge. And humne to fixed end moment at CB +30 calculate kiya he. And overhanging portion ka load 6KN , negative moment generat karta he at C and jab hum add karenge (+30-12) at C pe to wo humra total moment at C hota he . So Mcb/ 2 = +18/2 le te he.
Maine isi liye such type question kiya tah ta ki koi confusion na ho and yaad rakhna aap jab bhi overhanging portion aaye to us k karan jo moment generat hoga us ko hum include karenge.
Agar samajh me aagaya to reply ko like kardo aap ta ki muje pata chaljaye ki aap ko samjh aaya. And aap ko ek aur video mile ga same method pe hindi me aap us ko bhi dekh sakte ho.
Moment Distribution Method with sinking support:
ua-cam.com/video/OXLrPodRUF0/v-deo.html
It was crystal clear and explained very well... Very well executed the sum explanation.. We want this type of teacher in our country
ua-cam.com/channels/gSu05skHs7WxhTd3doO1eg.html
Mistakes are the sign of something good..
You are doing such a amazing work sir... Respect 💐 and keep it up.stay blessed.
Bro I think you had made mistake in the step of 2/3+3/8 which is 25/24. Thus the value of EI(ib) is 14.4
You are right bro.
Thanks bro I was confused because of this idiot
Exactly
He took deflection at c zero
3/5 is the right value
Wonderful✨😍
Welldone
Summation of moment at point B ought to be equals to zero (MBA+MBC=0)
Yes. Right
Bro handwriting is superb 👍
Superb
Wt about step 3 final that is 24/25 didn't get how 25 will come ?
I think he made mistake when we do the lcm it becomes 25/24 because (2/3+3/8)
Bro formula is wl/4 not wab/L
Yes you are correct.
And i have used both the formulas at the end the result is same.
So don't hesitate just prepare well.
Thanks.
Nice
Thanks alot
Sir shouldn't it be 25/24 EI ib
In last line of 3rd step
bhai everything is perfect but agar question mein find the slope bole toh final answer kya likhna hai?
Sir, for simply supported using formula 3EI/L... we get other valves of final moments and for simple supported using normal formula 2EI/2... Getting different values why?
The formula which you have asked for I myself have consulted to my teacher when i was in college. And they suggest me to use the formula which i have used in this video. No doubt the formula which you are showing is correct but it's actually not used when you have a look in some higher books.
Bro u told that upwards direction is - it's downwards means +
It's sign convention for find shear force.
👍👍😊
You played many gemics in this video not yet clear
You can check for some other videos. I have created a seperate playlist on this method with extra examples.
Bro explain clockwise and anticlockwise direction
Clockwise means the movement which is similar to the movement of ⏰ like _ 1, 2, 3 and so on.
While anti clock wise is movement opposite to the direction of movement of ⏰. Eg. 12, 11,10,9 and so on.
@@Civillearningonline I mean in the problem Clockwise means +or -
Clockwise +ve and anticlock wise -ve if we are taking moment about RHS and ( clockwise will be -ve and anticlock will be +ve if we are taking moment about LHS.)
Sir..pls confrm the value of EIib..it's not 125/8 ..correct value is 72/5
Then only the final moments value will be same..
Yes you are correct.
I am extremely sorry for this.
It has happened because i have mistakenly written 25/24(EI.ib) =15 as 24/25 (EI. ib) =15.
Thanks alot for letting me know.
By the way i have another video in hindi on same topic.
And i am sure you are going to find it very helpful.
Have a nice day.
Slope deflection method in hindi:
ua-cam.com/video/qQAqGbK19UA/v-deo.html
Ramro sanga van na ehh
Playlist hera. Aru pani video xan. That will help you a lot.
Sir can we get still more problems
There more such problems i have discussed you can check out the playlist or description.
Sir in slope deflection eqn how at Mbc taking hinge
Plz reply anyone why can't we do same procedure as we did
Here is your answer.
We need to use that equation when our hinged support is at the end of the beam.
If we use the same equation then we will have one variable more than the number of equation.
And to prevent the solution getting lengthy we use that equation.
And it is easy to recognize whether the support is hinged or fixed just by observation.
Here is my another video on slope deflection method here i have provided all the reason of using that equation , recognizing the fixed and hinged support as well as the problem that we face if we use general equation of slope and deflection.
If yuo still face any kind of difficulty then let me know.
Fell free to ask.
ua-cam.com/video/qQAqGbK19UA/v-deo.html
Sir, for simply supported using formula 3EI/L... we get other valves of final moments and for simple supported using normal formula 2EI/2... Getting different values why?
👍👍
What about joint "C"
Here in this numerical joint C is hinged support.
And always remember that we apply the condition "equilibrium of joint" only at intermediate supports.
Here in the figure you can see that only joint B is intermediate and the joint A and D are lying at ends.
And if there were given one more intermediate joint(D assumed) after joint C then we would have applied one more consideration taking the equilibrium of joint C.
I hope i am able to clear your doubts.
@@Civillearningonline Is joint d is continuous...if not then we taken as joint na..
@@Civillearningonline send me your Email ID I have lot of doubts, in this topic, I will share that to you........ Plz 🙏🙏
civillearningon@gmail.com
Sir why was the expression for Mbc different that the other expressions
It's because our support C is hinged and lies at end.
Do follow the step shown in the video whenever this case arises. It reduces one unknown and the solution becomes easy.
Sir how do you identified point B is roller and point c is hinged
If the support is represented as arrow then it's hinged and if it has a circle then it's roller support.
@@Civillearningonline sir even though the point B is hinged we shouldn't use that equation...we should use that equation only for the 2 ndor 3rd span right??
How did you find that the end C is hinged
It's clear from the diagram.
As you can see that the hinged support is represented by arrow.
@@Civillearningonline then end B is also hinged am i right
Yes
Bro, u told that for midspan, moment is Wl/8 in the beginning, but 12:35 here u put Wl/8. Why?
Yes dear. While calculating fixed end moment we use WL/8.
And at the time of drawing BMD we need to use wab/l for point load and wl^2/8 for udl. Hope you understood.
@@Civillearningonline yes sir. Thanks
For point load..... WL²/8 ryt..... Or just WL/8...
WL/8
Sir was it only one hinge support at c?
Yes
Then at B was it simply supported beam
Simply supported beam are actually the beam having hinged support.
You will find this when you will practice some some numericals. It need practice and once you practice it's gonna be fun. Believe me😁
What is diff between slope defln and kanis mthd
Kani's method:
It is suitable for multi storey building.
As there is no need of deriving equation and solving it. So it's time saving.
While,
Slope deflection:
It is used to analyse both statically determinate and indeterminate structure.
Considering deformation due to bending only.
Sir the value of Mcb is zero right then why haven't you substituted while drawing in diagram?
Support C is simply supported but span CD is overhanging so the moment due load on CD is transferred to support C.
@@Civillearningonline ok sir thank you
How to find support reaction in this problem can anyone tell me plz
For last cantilever span only
If we consider only last cantilever portion of this beam, then it's reaction will be -6KN.
How it will get 125? In step3
Sir iam not understand because your talking letters at below. So background are totally dark exact step are not visible to eye
Please check playlist you will find some other examples on this method. And i am sure they are going to help you alot.
Here you go.
ua-cam.com/play/PL513Y7_xBTnAprzGbIpMFzGm-ju1ZiCuS.html
What if the other end was also fixed
Check playlist you will find an example related to that concept.
Sir why you haven't find Mcb?
Moment at hinged support is zero.
@@Civillearningonline sir point B also hinged then why we haven't kept zero sir?
But support B is at intermediate position.
Dear paruvu,
I have created a seperate playlist on my channel for slope Deflection Method you should check out some other examples also. Then I am sure you will be able to clear all your doubts.
Sir, why is Mab negative
Anti clockwise moment
CtoD not there? Lot of confused how it will get 125/8 &
I hoe you understand Hindi check out the playlist you will find some other examples of this method and they are going to help you alot.
@@Civillearningonline no sir can't understand in hindhi only kannada and english but
How it's came not understand
I will reply you shortly.
how do we know that “c” is hinged
Hi ged support are represented by arrow headed upward or triangular shape.
Go through some problems which will help you to distinguish different types of supports.
B is also hinged then
@@tiamiyuhamed9333 B is hinged in this question. We need to check whether the support at the farther end is hinged or not. That's it.
In 14:32 how do u get 18
Hello dear,
First of all thanks for your comment.
Here is your answer,
It has happened because of overhanging portion at support C.
Your question is we take moment 18 at C. You will find your answer at 8:59 - 9:30.
Do watch that step you will find your answer.
Have a nice day....
@@Civillearningonline tq so much 👍
Mcd=-6*2 kese hua thora bataiye
Because humar end c hinded he and support c k right pe overhanging portion bhi he so jab overhanging portion right side rahe ga then hum us k karan jo moment aaye ga usko -ve lenge and agar humara overhanging left side hota to hum moment +ve lete.
I hope aap ko samajh aaya hoga.
Best of luck for your exam.
Description me aap ko 2nd example ka link mil jaye ga. Dekhlo wo bhi.
Once again best of luck.
Sir how to get Deta A and Deta B in Casio
I am unable to understand your question.
Please clearfy.
While solve equation we do not need to enter Deta in casio.
There is another way of solving the equation in calculator.
1.Press mode
2. Press 5
Now
Select the type of equation you want to solve.
Now enter the constants.(i.e.numbers)
Hope you got your answer.
How come it's 24/25 it must be 25/24 right?
Yes. You are right.
Sorry for it.
@@Civillearningonline it's ok
Mba +Mbc= 0 so how that 34.41 and 33.14 came. must be both same.
Hello dear,
First of all thanks to you for your comment.
You might have noticed that we have got (EI/ib) in all three equations.
And we have calculated the value of (El/ib) using the portion AB of the beam.
Also, I have not used Mba + Mbc = O here in this numerical.
And that's why we have got that answer. And believe me it won't create any kind of problem.
And one more suggestion to you to solve problems related to slope deflection method.
Try to use both the equation i.e. equation for fixed end and hinged support whenever you have such questions.
Because incase if u are using the same equation through out the the Numerical gets trouble some i myself have faced that.
This was my first video on this topic. I will soon be adding more examples to clear ur all doubts so stay connected by subscribing @civillearningonline.
And i will show the application of Mba + Mbc =O in some other example soon.
Thanks alot for ur comment.
Have a nice day.
I hope this video is going to help u alot.
ua-cam.com/video/qQAqGbK19UA/v-deo.html
I have no idea whether you are familiar with Hindi language or not.
Let me know by replying this.
I am ready to help you out with all your confusion.
im from srilanka . can you make moment distribution method little hard question in english. before 15th . i will share your video with my friends. thank you .
Of course I will.
On each and every type soon.
Can you please explain step 3 I'm not understand how u get values EI (ib) 🙄
Simply solve the equation using calculator or manually by elimination method.
Yr sahi sahi likh ta kite thirty likhi jnda kite thirty nine
Bhai playlist check karlo ek video hindi me bhi he. Hope Ap ko achha lagega.
ua-cam.com/video/qQAqGbK19UA/v-deo.html
Slope deflection method in Hindi.
How to do it when there is no simple support at C?
Thank you sir
If there is no simple support at c or if there is fixed support at c then use the usual slope deflection formula and and there will be no need of providing free moment.
@@Civillearningonline ok thank you sir
@@srujanatenneti4827 if you understand Hindi then you can watch my another video on same method and i am quite sure the these videos are going to be very helpful to you. Do check out the description section. Best of luck for your exam.
@@Civillearningonline thank you sir. I do understand hindi.
talk loudly
Vedant Auti 🦋
Roll number 26
#binod 😁😁
@Pratik Aute pratik auto😂😂
@Pratik Aute it's not so hard. You just need some practice. Prepare well. Best of luck.
@Pratik Aute well done 👍👍
Shhlurrppp ..........
Mcb/2=15 OR AAP NE 18 LIYA HAI...18/2....DON'T UNDERSTAND .......PLZ REPLAY
Or aap n 18 likha hai diffrence
formula mai aap n mcb/2 hai ....mcb ki value 30 h.... 30/2 lena hai to 18/2 kiyu liya hai....
Hello dear,
Aap ka question he ki Mcb/2 ko 18/2 q liya gaya he?
Ans:
Humara jo support c he waha pe 2 moments aa raha he ek to point load jo span BC pe he and dursra overhanging portion CD.
To aisa case me kya hota he ki hum total moment at C ko lenge ge.
And is k liye C pe lag raha sara moments ko hum add karenge.
And humne to fixed end moment at CB +30 calculate kiya he.
And overhanging portion ka load 6KN , negative moment generat karta he at C and jab hum add karenge (+30-12) at C pe to wo humra total moment at C hota he .
So Mcb/ 2 = +18/2 le te he.
Maine isi liye such type question kiya tah ta ki koi confusion na ho and yaad rakhna aap jab bhi overhanging portion aaye to us k karan jo moment generat hoga us ko hum include karenge.
Agar samajh me aagaya to reply ko like kardo aap ta ki muje pata chaljaye ki aap ko samjh aaya. And aap ko ek aur video mile ga same method pe hindi me aap us ko bhi dekh sakte ho.
Dongare Rohit
#binod 😂😂
where is Mdc bro?
Span DC is overhanging portion and there is no support at D so there will no moment (Mdc).
English aya tho baat kar bhai ni tho hindi may hi bol plz🙏
Ye lo bhai hindi bhi he
ua-cam.com/video/qQAqGbK19UA/v-deo.html
@@Civillearningonline bura mat manna fir 😊
Koi baat nai yar,
Just prepare well.
Best of luck
Hindi bolna bhaiyyo
Ok