SS Beam carrying combination of UDL and Point Load | Lecture - 24

This Lecture includes How to draw Shear Force and Bending Moment Diagram of a Simply Supported Beam carrying a combination of Uniformly Distributed Load and Point Load.
1:04 Bending Moment Sign Conventions
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NEXT LECTURE: Simply Supported Beam carrying Uniformly Varying Load.
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[@UnacademyGATECivil_ES]

1:58 I'm sorry I've taken 5 by mistake, it's 6m. So overall moment of this UDL will be [ - 3 x 4 x (2+3+1)] = -36 kNm.

[@krishnakrish8429]

Civilocity ok sir thanks

[@Thereincarnatedgoon]

When can fm or dss start sir? Any plans ahead?

[@Thereincarnatedgoon]

Civilocity sir plzz x ki value confirm kr dijiye, how is it 1.19

[@UnacademyGATECivil_ES]

I'm planning for FM after SOM. It also depends on the demand of students so let's see. You can select your subject in the POLLS that appear mostly at the end of the videos. It will be helpful for both of us.

[@UnacademyGATECivil_ES]

Please check your other comment.

[@veerashettyhadapad4653]

Sir the value of G is 5.86m is there not 5.19m...

[@rayyansheikh503]

Here is How to find the value of x where shear force is zero which he got as 1.19 Let the shear force be zero at point G , now SF at G = 10.583 - 5 -3x =0 Implies x = 1.86 so it should be 1.86 and not 1.19 correct me if I am wrong

[@somthinghappen6987]

Sir how to calculate 5.19 at distace point A to G plz tell me fast sir i confused that

[@subhajitdas660]

SFD banake similar triangle wala formula lagado.. 😀

[@pranavpisal5422]

how did you get 5.19

[@kavitakhabri1274]

distance of point G from end A is 5.86, not 5.19..

[@rubymaurya3788]

Kavita Khabri mera bhi yahi aa rha h

[@Alimola563]

Kaisey kiya

[@sunnyprabhat4667]

6?

[@shreyamishra7990]

Yes even I'm getting the same value

[@SatishkumarME-]

RA 12-5×9-3×4×6-10×1=0 Ra=10.5kN

[@legendcreatz9783]

Han bhai yhi mera question hai ki sir ne 3*4*5 kaise kiya jabki vha 3*4*6 aata ....If iam not wrong!!!!!

[@GAURAVSINGH-ni1cv]

Ra = 9.58

[@haridarshan6388]

Sir 6:07 mai 1.19 kaise aya?...

[@manjunath-uy8vl]

At 2:13 (4*3*6)

[@saumyaarya8481]

Sir 5.19 bilkul smjh nhi aa ra ki kse aaya? Plz explain this with solution in the reply of this comment

[@AvinashKumar-bb4nc]

dear sir please upload more videos with different loading conditions & with different types of beam

[@UnacademyGATECivil_ES]

+Avinash Kumar yes, for sure!!

[@025hritikjaiswal6]

Distance from A to G is 5.86 not 5.19..

[@prakashkhilery4115]

How to convert UDL to point load

[@michelleanncenteno6355]

Thank you so much, such a great help

[@mahiprajapati7574]

Bmd me parabolic hogya wo kaise pta chalega

[@Plants_o_philic]

nro you cleared my basic concepts those i was like totally forgotten . and i was like clapping after watching this video. your way of explaining is awesome..... keep it up

[@ankushbhatti9583]

Sir tell me how u calculate the length which 5.91 it's so confusing Sir plzz tell

[@yogeshchouhan9394]

Sir symmetrical ∆ ki property or sf se length AG ki length find out krna konse vdo m btaya h pls tell me

[@imransayyad5039]

How do we get 5.19

[@user-wp5gw7td5r]

I actually didnt get distance value of G.

[@knowledgeablefact9047]

Bhai calculation tho shi se krna sikh le

[@alokmaurya3668]

How can be find 4 X 3 X 5

[@rashminavaneethan6112]

the distance of point G from point A is 5.86m

[@rashminavaneethan6112]

Also BM @ E = 4.38 kNm not 16.38

[@richieerich_]

Dude u have done some serious mistakes over here...when u're calculating BM....but gr8 efforts :)

[@mayukhborana3447]

sir put more vedios.

[@mohammadasifali7062]

Sir ji jb moment find kar rahe hai to direct multiply kyou kar diye samjh me nhi aa ya kaise moment nikale hai..... Please sir tell me... Please sir 🙏🙏🙏

[@1994prazwal]

please explain about distance from point G in BMD

[@nupursharma6432]

let me also know

[@amulyaadimulam4640]

sir ...can u tell me how did u calculate ....bm @ E ....why did u take 5 in udl calculation

[@Idrees.wani.]

Where can i get the whole set of lectures from start to end.... m getting random lectures...

[@krishnakrish8429]

Your lectures are very useful Way of teaching is very good thanks alot sir

[@UnacademyGATECivil_ES]

+ramanaboyina vk thank you so much :)

[@asthatewari2830]

now u r going a little fast. Please discuss some more problems on this. Where will I get practice problems on this topic?

[@gamefanclub3129]

Sir ek video section kaise lete hai uss par banayeye. Mujhe bohot problem ho rhi hai section lene mein

[@BhagatSingh-yh6zv]

I can't understand why (wl2)/8 come here

[@mmaseebkhan2676]

Please explain how the distance of point G calculated from point A ,i was not get this thing

[@rakeshprasadram9408]

Apne ak mistake kiya hai...cg se distance 5 nhi 6 hoga..tavi 10.5 aaiga

[@syedafizabatool2567]

Plz sir mje apni wo video bta da jime apne simply supported ma point load generally smjaya tha mry marks ka mamla. Mje marks apki is video k reference sa mly ga

[@vamshikrishna-zq7qp]

Am unable to understand moment about G I mean perpendicular distances

[@ervikaskumar6481]

Ra=11.58 and 15.42 please check

[@sandeepgurjar1758]

Sir 5.19 kese aaya usko explain kr dijiye

[@sandeepgiri8278]

symmtrical triangle ya shear force wala method kis video mai hai udl ka bmd banane ke liye

[@waqas_noman]

UVL and point load pa be ak example bna dan... please

[@nishantpatel1056]

Not able to clarify Sign convention ...in BM and SF

[@BhagatSingh-yh6zv]

5.19

[@BhagatSingh-yh6zv]

I can understand 3x4x5

[@ganeshshukla7701]

Sir mujeh moment point G par kaise nikala h wo samajh mai nhi aaya

[@lalitchoudhary1095]

Sir phle question m 12 kN load Rf se 6 m k distance p lag rhi h

[@z-viralpictures2650]

Awesome..sir you just great.....

[@BhagatSingh-yh6zv]

How to calgulate this

[@statusforyou0786]

Sir think cylinder par vadio banaye wo strength of metals(SOM) chapter ka ek part hai diploma 2nd semester ke syllabus me samil hai please make vadio this topic

[@UnacademyGATECivil_ES]

Sure!!

[@saumyaarya8481]

Sr how -36KNm come ??

[@ankushbhatti9583]

RA ki value Shi nhi h

[@ashutoshluhania9507]

In udl c to d how do you get 4x3x5

[@princykumari6175]

Itna fst....... kuch v smjh ni aaya

[@praveenupadhyay.]

dear how you taken during reaction 4*3*5 ( 5? how is taken , can you confirm me?

[@soumojyotimondal732]

That will be 2+3+1=6 It was a small mistake

[@sandeepgiri8278]

means G point ki value kaise find kare

[@souvikmandal5678]

Start SA please

[@ekaneminyang7750]

Thanks

[@sarifulsardar7027]

Moment ki kise Kiya sir

[@gayatribhatere4232]

Sir 5 kese aaya 1st question mai

[@amravatikarnikhilpoharkar9476]

Point G Ka nahi samjh Aya sir

[@ajeetsingh-bz4nn]

tumhe bahot jaldi rahti h

[@lalitchoudhary1095]

Sir distance G kaise nikale

[@selfenquiry456]

UVL load ka jo calculation hain usko le kar discuss kare on different types of beam ke upor toh it would be better for me

[@UnacademyGATECivil_ES]

+Koustav Mukherjee sure!

[@selfenquiry456]

+Civilocity thnx for it sir

[@krishnakrish8429]

In udl taken -3×4×5 how sir It is confusingPlease explain me

[@UnacademyGATECivil_ES]

1:58 So negative sign is because UDL is on the left side of point F which is producing a HOGGING Moment and that's why taken NEGATIVE. 3kN/m is the rate of loading. 3 is multiplied by 4 to calculate it's magnitude i.e. 12. So when we take moments of UDL at a point, we multiply the magnitude of the load with the distance between the load and the point which is measured from the CG(center of gravity) of the load to the point where moment of all the forces are being taken. In this case the distance will be, 2+3+1=6. I'm sorry I've taken 5 by mistake, it's 6m. So overall moment of this UDL will be [ - 3 x 4 x (2+3+1)] = -36 kNm.

[@biswajitmahato1472]

Civilocity sir by your questions there was 1 not 2 so... why are you taken 2

[@UnacademyGATECivil_ES]

1:04 Bending Moment Sign Conventions

[@adityajhabharadwaj9551]

Esme bm at point g pe distance kaise nikala ???

[@Thereincarnatedgoon]

Pls confirm the value of x.

[@UnacademyGATECivil_ES]

6:05 I'm assuming that you already know how to calculate the distance of point G from A either by symmetrical triangles or by putting the equation of Shear Force equal to zero. Now after calculating, AG = 5.19 also, AB = 3 and BC = 1 So, the distance CG will come out to be, CG = (AG - AB - BC) CG = 5.19 - 3 - 1 CG = 5.19 - 4 CG = 1.19 or x = 1.19 If you still have any doubt then please elaborate your question and give it a time stamp so that I can help you even better.

[@Thereincarnatedgoon]

Civilocity sir my x value is is coming as 1.86.

[@Thereincarnatedgoon]

Civilocity sir plz confirm 'x' ki value, i believe it will be 1.86

[@arijeetdalalchoudhury1177]

Civilocity please explain using similar triangles.. cause i am also getting 5.86

[@deepakrathva]

I am also getting 1.86