Problem 1:Analysis of continuous beam using stiffness matrix method

Name of the Subject: Analysis of Indeterminate Structure
Subject Code: 18CV52
University: Visvesvaraya Technological University
Year of Study: III year (V Sem)
Tutor: Avinash Gornale
Affiliation: MIT Mysore
Lecture Session: M5S2
Topics Covered: Analysis of continuous beam using stiffness matrix method
Related Links
www.mitmysore.in
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[@BGMs08]

I was struggling too much to get this understood but I couldn't thank you sir for this valuable lecture

[@manojnagsharma888]

Mod 1;
1,SSP, Dsi=r-e+no of hinges
Dk=3j-R+no of hinges..
2,open truss Dsi=3m+r-3j
Dk=3j-R..
3,closed truss Dsi=m+r-2j
Dk=2j-R..
Fix-2, rol-1, ss-1, hinged-3
MOD 2;
Mohr slope theorem = the change in the slope between two points on a straight member under flexure is equal to the area of M/EI diagram between two points..
Slope tita=$M/EI × dx = Area of BMD/EI..
Deflection theorem = The moment about a point A of the M/EI diagram between points A and B will give the deflection of point A relative to the tangent at point B...
Y=$AtoB dy =$AtoB Mx bar/EI × dx...
Macaulay
1, support reaction
2, consider a sec x-x
BM w.r.t x-x
Mx=..,
Apply diff equa,
M=Mx=EI×d^2y/dx^2..
Integrate equa
EI×dy/dx= x becomes x^2/2+c1
Again integrate
EIy= x^2/2 becomes x^3/3+c1x+c2..,
3, apply Boundary conditions
@x=0, y=0 in equa and find c2=?
@x=span length, y=0 find c1=?
4,to find deflection apply
C1=, c2=, x=required
EIy=...., y@c=..... /EI..
5, max deflection
dy/dx=0.., 0=....+c1
Find x=.
From equa EIymax=...., ymax=..... /EI.
Mod 3; Castiglianos.,
1,vertical deflection @C
will=$int 0 to L, (P×x)^2/2E×_I ×dx +$0 to l (P×(first L))^2/2E×_I ×dx...
wi=P^2/2EI × x^3/__ ]0toL +P^2/2EI × (second L)^2/__ × x]0toL...,
wi=..
The work done=1÷2 ×P×∆=we.
We k t, we=wi..
1÷2 ×P×∆=_____/EI..
∆=? mm..
2,Horizontal deflection @C
Introduce a dummy point load at point C in horizontal direction.
The strain energy,
wi= _______dx + $0toL (P×1st L +Qx)^2/2E×_I ×dx..
The deflection is given by.,
∆=dowi/doQ=0+$0toL 2(P+Qx)/2E×_I ×x×dx
Split this into 2..
Put Q=0, we get
∆=2×P/__EI × x^2/2 ]0toL
∆=? mm..
Unit load 1, moment calculation at x-x
Mem, limit, origin, Mx, m1, m2, EI...
2, vertical deflec
∆cv=$0toL Mxm1/EI × dx..
3, horizontal deflec
∆ch= $0toL Mxm2/EI × dx..
MOD 4; Hinge,
1, supports €v=0, €Ma=0, va=?, VB=?, €Mc=0(LHS), Ha=?=Hb..
2, To calcu vertical ordinate Yd and slope tita..
Yd=4hx/L^2 ×(L-x)..
@x=given m, Yd=..,
tan tita=4h/L^2 ×(L-2x)
Tita=...,
3, To calcu BM@9m frm left
€Md=va×9-Ha×Yd..
BM@D=___KN-m...
4,To calcu NT and RS@9m frm left..
V=va-udl×distance..
H=..
NT=Hcos tita +Vsin tita.. in KN..
RS=H sin tita - V cos tita in KN..
5, To calcu max BM along span.. Max+ve BM
Sec in AC portion..
Mx=0(LHS)
Yx= find
DoMx/dox(Diffretiate) =
X=find.. M@x find Mmax=.... KN-m..
Maximum -ve BM
Sec in BC portion..
Same as follows..
6, Draw BMD..
CABLE ;
va, VB, ha=hb
Tmax=Ra=√H^2+X^2..
Lc=L+8h^2/3L..
Dia(sigma) = Tmax/pai/4 × d^2..
MOD 5;
A curve or graph that represents a function like reaction at a support, shear force at a section, bending moment at a section of a structure etc for various positions of unit load on the span of the structure in called ILD..
Used,
1, the effects of loads that occupy different position on a structure can be studied by means of influence lines..
2, to identify the positions of loads for maximum shear and BM at specified sec..
3, ILD is very useful in along with rolling loads..
Mod 1
DS: can equi,BMandSF at any sec is independent of material, same...of cross sec or MOIner, extra conditions like compatibility of displacement are nit required, ex: SSB, cantilever, 3 hinged..,
Linear; material has linear stress strain relationship and a small deformation in that case law of super position holds good..
Non linear; material does not have relationship, large deformation, change of geometry cannot be neglected..

[@somashekharsunnadakumpi2448]

Add first introduction section

[@mdmizanurmizanur5810]

so effective ❤️

[@dipendranathbhowmick6577]

Excellent teaching

[@upekakuruppu170]

Well explained. Thank you.

[@akshaysharma003]

Ra should be 248.89 pls correct

[@zahidbashir4406]

Well sir can you name the book where this method is done this way... Please

[@kannikagowda2905]

Sir plz put the introduction to stiffness and flexibility matrix method

[@sourabhwarke5258]

sir Mfba is -160 ..u have taken it +160 while calculating load matrix

[@vivekagnivamsha2304]

Thank you very much

[@147-rutujapatil4]

Thanks

[@darshxn_007]

Thank you sir 🖤

[@RajuDarling-yg5kh]

Sir Rb value is wrong
That is 231.13

[@a-02sagarjadhav53]

👍🏻👍🏻

[@hemanthkumarb7689]

Ra = 231.09

[@aubaidbeigh8112]

It confused me so much i calculated it 100 times i m also getting 231

[@carefuldrawstipsters7834]

@@aubaidbeigh8112 Rba

[@shivarajushivu7023]

Tq sir

[@ranjanyadav1591]

How u ll get notes

[@kodapas...3715]

Tqq sir

[@Abdiannnn]

Tqsm sir

[@thilakprasadbm8089]

Sir pls provide full notes

[@Catloverss95]

FEM video 01 - Introduction and bar element
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FEM video 02 - Bar element example
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FEM video 03 - beam element
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FEM video 04 - beam element example 01
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FEM video 05 - beam element example 02 with UDL
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FEM video 06 - truss element introduction
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FEM video 07 - Truss example FEM
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[@Mohammedyahiyakhan-pq4xv]

Sir explain in hindi