Lec 08 Trusses II
ฝัง
- เผยแพร่เมื่อ 1 ต.ค. 2024
- Analysis of statically determinate trusses, How to accommodate the weight of the individual member or snow in roof tops in analysis, Method of joints, Isolation of a joint, Representation of joint as a circle, Identification of tension or compression in the members and its representation in the sketch, Solution for a simple truss.
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Thnku sir....it sounds like teacher is building their childs 💜❤
Very passionate teaching! Really helps in building subject interest. Thank you, Sir.
Nice teaching good explanation reaches easily
Thank you sir for giving such a beautiful explanation 💐💐💐
Thank you sir and govt of india for the funding of projects like this
Thank you sir! You are truly passionate in what you do, & it shows! keep it up!
Thank you sir for giving such a Great explanation
Can any one help with this
case 1 i.e Join A
The sum of froces at Fy is as Fa + Faf sin 45 and Fx is as Fb + Faf cos 45 we solve by this equation
but when we solve with case 2 i.e at Join F
The sum Fy is as Faf cos 45 +Fb and Fx as a Ffe +Faf sin 45 we are solving with this equation
So why we use likewise please help with this
Amazing lecture
outstanding teaching sir
Good explanation
Sir,You are a real teacher
great lecture sir
Can anyone help with cordinate selection,
@time 35.18 in +y direction Fx=0
Eq should be
-Ffb+FafCos45=0 instead of Fab-Fafcos45=0
Although final answer be same,
We are taking forces w.r.t point F (equilibrium at point F). How will F(ab) will act on point F?
Your observation is correct. As per the positive coordinate directions, the equations you have mentioned would be correct.
However, as long as '-1' is multiplied to both sides of the equation, it does not affect the result and either of the two equations can be used.
Ffb is assumed to be away from joint F. When you resolve the force Faf, the vertical component acts opposite to the assumed force Ffb. So when adding the forces, you attach a negative symbol to indicate that the other force (Faf cos(theta)) is acting opposite to the assumed force.
At 12:14 how we got RAY+RDY=2+2=4KN and also RDY X 3a-2 X 2a-2 x a =0 ???
1) Taking net vertical forces = 0 (since truss are in equilibrium)
2) Taking moment at point A. Since it is assumed to be in equilibrium, net moment = 0
As @Arkit Kabir correctly pointed out, these are the equilibrium equations applied to the free body of the truss
R(Ay) + R(Dy) = 2 + 2 = 4kN.
the right hand side represents the sum of applied loads.
Since the truss is in equilibrium, one can take moments about any point and it should add up to zero.
For convenience and eliminating one of the unknown reaction, we are taking moment about point A.
Clockwise moments = negative
Anti-clockwise moments = positive
+R(Dy)⋅3a - 2⋅2a - 2⋅a +R(Ay)⋅0 = 0 [One unknown gets eliminated]
Very useful
amazing proffesor _/\_
What if we consider those forces on pin joint as internal stresses of the member .. it will make it easy to grasp
Or also we can say the pin is getting compression as per drawing because it's actually pushing the member that's why memeber putting equal and opposite force on it .. this also directs above statement correct
Why we are not showing compression at the joints because we are not showing what joint is exerting but what joint is taking
Sir, What if the top and and bottom chords are continuous over 2/3 joints and spliced at 12 m, will it remain a ideal truss.
practically this is done in steel trusses, so the bending is controlled .
Very good teaching(no exaggeration)got all the doubts clarified which were during offline class
Can u explain how -Fde sin45 +. Fbe sin45 + Ffe = 0 . Bit confusing because of the sign conventions. @39:40
@@sameermasood2289 Bro we covered it in 1st semester. Now I am in 4th semester. I can look at it and tell you something but it will only increase your confusion
AT 43:04 nature of force in member CE is written wrong in final result
kuladeviam Mae Ramesh sir
Rahul was here
ow te anle is 45 deress in question
Thank you sir.
Beautiful
Tnq you sir...
Love you sir
21:00
At 36:00
F(fe) should be 4
No bro
You are multiplying F(AF) by √2 when it should be divided by √2 instead.
Sin 45 = 1/√2. Hence, the answer provided in the lecture is correct and F(FE) = 2 kN
Excellent explanation.. Thank you sir.