# 3 Hinged Arch Type 1 – Structural Analysis 1

Do subscribe to Ekeeda Channel and press bell icon to get updates about latest engineering HSC and IIT JEE main and advanced videos Hello friends today we are studying a new subject named as structural analysis one subtopic three Hinged arches in Hinged arches there are three types of problems comes in exam first one is supported same level second one is supported different level and third one is circular arch today we are seeing supported same level let’s see 3 Hinged arches supported at same level in supported same level first we have to see what these arches bodies look like arches arches is look like this the name indicates it’s a three hinges arches number one hints this is number two inch and this is number three inch and this one is your L stands for spend and this is your H is nothing but rise of the arches and this is your tension shine this one is your compression sign 10 check means two arrows are kissing each other is nothing but tension or two arrows are away from each other is nothing but compression in this chapter in this chapter only comes compression because of three inch arches and in this in this problem so you have to come three types of load in exam let’s see what are the three types of load this is your V this is your H a this is our V B so HB this is nothing but ground one low DC wheel other low DC UDL and the third one is point load the first load is point load second one is UDL and third one is u VL this types of loads comes in exam and you have been type one you have to remember one formulas is named as y is equal to 4 H upon L square into L X minus X square this is your main formula of this chapter you have to remember this formula y is equal to 4 H upon L square into L X minus X square and in this chapter H is always in word why H is always converge as you have seen this H a and HB are always in word why this is always in what because load are coming from upward in this for example this is your arch load is coming from upward your arch is trying to spread like this like this you have to spread for you to resist this you have to apply H that same direction that’s why in this chapter H is always inward now we are starting one problem type 1 this set of problem comes in example for question number 1 5 marks a three-inch arches of span 40 meter and lies 8 meter carries a concentrated load 200 kilonewton and 150 kilonewton at a distance of 8 meter and 16 meter from the left hand and and a ubian of 50 kilonewton per meter on the right half of the span find the horizontal thirst now I am explaining you what is horizontal thrust first we have to draw the diagram this is your solution this is your diagram ca3 in charge of spent 40 meter and dry state meter this is your span 40 meter and dry skate meter concentrated load of 200 kilonewton and 150 kilonewton at a distance of 8 meter and 60 meter from the left end this is your 200 kilonewton from 8 meter 150 from 16 meter from left side and a UDL of 50 greater on the right half of this fan and a 50 kilonewton is on the right half of the span now you have to calculate horizontal crust is nothing but we have to calculate this VA value this hv value and this H a and this H B you have to calculate H this problem how thre a set of Miyagi theory Victoria is familiar now you see first type we now first you have to calculate support reaction calculation in supporting action calculation we are calculating first salvation moment then second one summation FY then third one summation F X first we have to calculate the moment taking moment at he is equal to zero that sign conversion is clockwise positive clockwise positive in clockwise positive as you can see in beams is also like a is also a hint things first we have to see we have to take one rod and fix it from a as you are taking moment at a and take downward of 200 kilonewton and try to bend like this downward so 200 into the distance 200 into 200 into the distance is 6 meter 6 meter as you all know what is moment moment is nothing but force into that perpendicular distance and the second one is 150 into 16 this is also clockwise is like this 115 plus 150 into it’s a proper little idea Chun toward ready to 6 meter plus 150 into 150 into 16 as you all know this is your 150 into the distance force into the perpendicular distance now scum UDL as you all know how to converts point load into UDA sorry UDL into a point load this is 50 this 50 into plus 50 into 20 is nothing but it’s your not it’s your it’s your load not distance and your distance will be 60 divided by 2 plus this 20 positions will be into 30 this is your distance – – – this you’re fixed like this and your evb will be unknown – maybe into the distance 40 minus VB into 40 is equal to zero while solving this we get the value of VB is equal to eight 50 kilonewton after that you have to apply summation FY is equal to zero upward positive see your VB will be 850 kilonewton and upward positive nothing but V ay minus plus 120 so if V plus minus 20 200 minus 150 minus 15 to 20 plus 850 is equal to 0 we minus 200 minus 150 minus 50 into 20 plus 850 is equal to zero while solving this you get the value of VA will be 500 kilo Newton your GA will be 500 kilo Newtons now you have to solve third part third part is nothing but third part is nothing but horizontal thing but H taking bending moment at C is equal to zero this is a sign conversion of bending moment as you all know there are huge difference between moment and bending moment this is nothing but moment bending moment and rafter that we are saying this is moment and this is bending what on what is bending moment the fears we are taking bending moment from centre either we have to take the section from left or right so we are taking the section from left now we are taking a bending moment right now we are taking bending moment at C there is huge difference between moment and more bending moment notice moment moment is nothing but force into perpendicular distance bending moment is also nothing but force into perpendicular distance but here bending moment we are taking bending moment from via for example we are taking bending mode from a we have to consider part either left or right for example we are taking bending moment at C we have to consider either part C A or Part C B we are considering considering consider Part C a consider Part C a like this and you have to follow the sign conversion this from the left upward positive downward negative before you decide conversion for bending moment sagging or hogging this is sagging this is hogging now we have to take this bending mode see see for example this you’re taking bending movement from see your B may like to bend like this for example this like this the 500 will be this distance then your 500 Plus this anchor version follow 500 into the distances the distance is nothing but 2500 into 20 now now this one this is coming like this this follow the 200 is follows this sign conversion this is nothing but minus minus 200 this distance force into perpendicular distance 200 into this distance then 200 into 12 then minus 150 into this fourth 150 into 4 minus now now this H is also coming now H is nothing but for your purchase for ending a distance will be hate but this will be – or this will be – because if you open your ass your H will be like this village will be like this this is nothing but this and conversion follows this one this is nothing – H into we have perpendicular distance will be 8 is equal to 0 while solving this you get the value of H is nothing but it’s 75 kilo Newton as you all know in this chapter both the HR same H a is equal to HV is equal to H 75 kilo Newton H is always invert the H value will be always in but this edge will be 875 kilo Newton this edge will be at 75 kilo Newton and this is your final answer of horizontal thirst h is equal to 8 75 kilo newton thank you so much last week I saw Rebecca Tamil and this is your final answer thank you friends for watching this video I hope under Stephen understand this topic name I separated same level to subscribe this EQ channel and share this video with your friends thank you

most useullll video

sir the distance from the point load 200 is 8 not 6

abi chuteya ,,,,, pora galat barta raha hai

Is nothing but

Is nothing but

Is nothing but

I dont subscribe your channel

I don't want to be offensive but agar kuch galat bol diya to retake karo na….

Poore team mein kisi ko achhe se After Effects nahin aata kya? Kyun paper ko baar baar badalna pad raha hai?

Ek simple edit, question ko top left mein le jao, aur jab refer kar rahe ho to tumhe bhi aur viewer ko bhi dono ek saath dikhega.

You did not have to explain the difference between moment and bending moment. Even if you decide to do it, please make sure you are clear and concise. Man mein kuch bhi aa gaya wo bol diya…

Well presented thank you !!!

Tension sign is wrong and compression sign

You are too fast, sometimes you take the wrong number. In your Sum of Moment you took 6m instead of 8m. therefore all the calcs need to be revised ! Kind regards Rahman

agr nhi aata toh mt pdha…..par galaat kyun sikha rha h?

Sir mujhe varying load me max BM lena bta dijiye…..

Jb hm max bending moment nikalenge to x distance mante h to jo intesity h varying load ki use x distance ke liye nikalenge kya or fir act krne ki distance v kya x ke liye nikalenge kya…..ya firrr jo poore arch le liye intensity h jo wo intensity use krenge

There is nothing but wht sir???

this sign of b.m is hagging and sagging

Poor teaching skills….he himself is confused

How do u derive that foemula…??

Very nice bro…keep it up. U r a good teacher

good brother but i need after that how to find +ve and _ve BMD and RS and NT

Why remaining videos are private 👎👎👎👎

Hello sir kuch questions hai kya unko solved krva sakte ho kya??

Sir calculation of arches and beam are same ,

Last m kya bole h sir

Arch is nothing But a big tension 🤣🤣🤣🤣

Pura sum galat hai bhai.

VA=1000

VB=350

VB×20 – HB×8

350×20 – HB×8

HB= 350×20/8

HB=875

Maa ki aank… Galti pe galti aur confidence itna.

Sir eddy's theorem Ka derivation upload kije pls

Distance for load 200 kN is not 6 m, its 8 m

Thanks u sir

moment calculations me glt kiya ho…

200kN×6❌❌

200kN×8✔✔

👌

Nice video sir very helpful thanks 🙏

Apne value galt le liya h Vb nikalte shame 200×8 h apne 200×6 le liya h correct it

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Ye kon bhonk ra h bich bich me…

(50×20)x30 kha se aaaya

Madr chot. Kon si college se prke Aya he lours.

I wasted

20mb to download this video🤣🤣🤣😂😂😂

Horizontal thrust = -125

Aap intna kiu fast parahate ha…kahi jana ha kaya….

Always check your videos after preparing it… 6m??

200×8 not the 200×6

50/2 + 20 = 45 not 30

Whole problem is wrong

incomplete solution i want fully problom

It's not 30 it will be 10

Moment calculation is wrong

phile tension compression ka phark samjo

Bro you taken 8 r 6 from left hand side

i have always been relying on this channel,after watching this video i think my confidence level has dropped a bit

ps: come up with some professional lectures @ekeeda

Moment calculation bhale hi galat kiya ho pr last me answer sahi likha hai..padhane se pehle thoda revision baad me editing sikh lo ….time ki maa behen kr di..

thx bro

Jésus loves you

Believe in him and repent

last me kya bolna hai ?? 😂😂

It is nothing but worst

sir in this problem explain with bending moment its my request

Abeyo waste fellow worst teaching of yours

Sir distance=30??how? 50÷2+20=45?

Shagging hogging lena h ya clockwise anticlockwise🤦🤦

There in nothing but 🤣🤣 sir aur confused hogyi m toh