For determinacy check use this equation (3m+r)-(3j+c) Where m=member, r=Number of reaction, J=Number of joints C=Number of internal hinge, for Statically determinate (Ns)=(3m+r)-(3j+c) for externally determinancy (Ne)=r-3-c for internally determinancy (Ni)=Ns-Ni
M=3 (Member AB, BC and CD) R=5 ( Fixed Support= 3 + Pin Support=2) J=4 (Joint A, B, C and D), C=2 (Hinge A + Hinge B) NU = 3m+r = 3×3+5 = 14 EQ = 3j+c = 3×4+2 = 14 NU = EQ
Thank you Sir I was searching for the exact frames with support but different loading. I was trying to solve it using moment distribution method without first considering its determinacy. This post save my time in doing so.
GOOD JOB! Wonderful explanations except i still dont know Statically determinancy method. Is it with, | Unknowns = Equations x Members | , or | Reactions = Number of Equilibriums + Number of Supports | ... I know it is shown differently even in Hibbeler's book but it should be something... @structurefree
Bro Love your Videos.Found it to be quite helpful..am currently doing structural Priciples and this is what we are basically doing when studying portal frames(Frames) Cheers bro.. Keep Posting..!!!
with the hinge joint, what happen to the moment Ma... to maintain equilibrium of member AB?... ( i get that hinge doesnt transfer moment... but how would you draw the "blow" of structure and why do we not draw this moment??)
If the pin connection at point D is also fixed, meaning, both A and D are fixed supports, how am I going to draw its shear and moment diagram? Thank you.
@structurefree hi when we calculate the By and Cy in the member CD, you missed to add the total load in the equation which is 75kn isnt it? can you please check this
Can anyone explain why there is no horizontal force at Bx? Maybe I'm just misunderstanding hinges but I would assume the uniform load is spread over Ax and Bx.
Ugh. I did the left side member second.... Take moment about A trying to figure out what Bx is..... Idk why you can't do this gives the wrong answer for Bx 😩😔😔😔
@structurefree hi when we calculate the By and Cy in the member CD, you missed to add the total load in the equation which is 75kn isnt it? can you please check this
the solution is wrong. there should be transfer of moment at B and C too...u r transfering only the loads...what about the moment...even it should be transfered from A to B and B to C... And when u check dis sum in frame design app..the value of Ma comes out to be 144.5 KNm pls correct this
You explained every part of structural analysis in a detailed and simple way! the best Structural tutorial on UA-cam!
Thank you from Brazil!
I have watched a great deal of your videos through a Master's degree in structural engineering. They are very helpful!!!
For determinacy check use this equation (3m+r)-(3j+c)
Where m=member, r=Number of reaction, J=Number of joints C=Number of internal hinge,
for Statically determinate (Ns)=(3m+r)-(3j+c)
for externally determinancy (Ne)=r-3-c
for internally determinancy (Ni)=Ns-Ni
i know this is a long time ago but can anyone explain what m,r,j and c are on this particular frame, since im doing this is slovenian.
M=3 (Member AB, BC and CD)
R=5 ( Fixed Support= 3 + Pin Support=2)
J=4 (Joint A, B, C and D),
C=2 (Hinge A + Hinge B)
NU = 3m+r = 3×3+5 = 14
EQ = 3j+c = 3×4+2 = 14
NU = EQ
Thank you Sir I was searching for the exact frames with support but different loading. I was trying to solve it using moment distribution method without first considering its determinacy. This post save my time in doing so.
could u please visit my channel ^__^
Love watching your videos... they helped me through a lot during my degree courses.
Yes, that is what was done in the video at 18:05. Ma = 5KN/m * 10m * 10m/2.
Beat teacher ever
🙏
Internal moments are zero at the hinge. As long as equilibrium is maintained for each individual element, you are good to go.
Sir
How to calculate deflection for same member (frame)
Do you have any video on deriving the geometric stiffness matrix?
u r genius.....just made it reaaalllll simple
Great job
Yea I loved how he tried to call every single member member CD. He corrected himself on the last one but two out of three aint bad
* like what you done in Frame Analysis Example 2 (Part 1) about the joint
Zab Momochi
A could never be a roller, check the representation.. Great explanations!
Great example and explanations! Thank you.
MrFishhunter That's because we have Sum of Fx, Sum of Fy and Sum of M(moment) you might have got your answer since you asked 6 months ago, but still:)
GOOD JOB! Wonderful explanations except i still dont know Statically determinancy method. Is it with, | Unknowns = Equations x Members | , or | Reactions = Number of Equilibriums + Number of Supports | ... I know it is shown differently even in Hibbeler's book but it should be something... @structurefree
Bro Love your Videos.Found it to be quite helpful..am currently doing structural Priciples and this is what we are basically doing when studying portal frames(Frames)
Cheers bro..
Keep Posting..!!!
Thank you. Could you please organize your calculation on the screen to get a screen shot for future reference to have a look.
Thank You. Your video is very helpful.
very nice example
A is a fixed connection.
Amazing and Simple videos
Ay should not equal Dy. The loading and boundary conditions are not symmetric.
Do we have to assume the direction of reaction forces whatever we do by our own and just balance the other?
very nice!
Thanks a lot. A stupid question; I got that we have 5 reactions but how did you reach that we have 3 equilibrium equations?
just as confused
The 3 Eq equations as in the Fx, Fy & Moment equations. We don't have any other equations that could help us.
@structurefree why aint we using a point load for member AB when solving the sum of forces in the horizontal direction
@ 6:10 "for each drawing here, I have per drawing" ?
y didn't u do 5*15 in member BC?
Can you make a video for Force Method for Frames? Thanks.
structurefree ! nice. Are you an engeer ? or ?
thank u so much bro for dis perfect explnation and good work.
just cant find composite structures.
thank you.
when calculating the vertical forces at member Ma, why did you not say (5kn)(10m)(5)?
Is there any difference in pin and hinge joint?
a question, if say the determinancy for this case was 0, as in 3 unkowns and 3 formulas, can you solve this without separating the frame into pieces?
with the hinge joint, what happen to the moment Ma... to maintain equilibrium of member AB?... ( i get that hinge doesnt transfer moment... but how would you draw the "blow" of structure and why do we not draw this moment??)
Ma = 5kn/m * 10 m * 5m.
What text book do you recommend?
Brilliant
I would like to see the same example with fixed supports in points B and C.
This is awesome....!Thanks
If the pin connection at point D is also fixed, meaning, both A and D are fixed supports, how am I going to draw its shear and moment diagram? Thank you.
if the frame is rigid then its statically indeterminate, if it is pin joined then it is statically determinate. right?
@structurefree hi when we calculate the By and Cy in the member CD, you missed to add the total load in the equation which is 75kn isnt it? can you please check this
if it is not a hinge, cant we make free-body-diagram?
can you add a video about frames and machines (structural analysis) that is slightly more difficult. Meaning with angles and all.
It sucks majorly that none of the examples are just a simple frame and point load. I just want to see how it's done, and this doesn't do it.
How would I draw this on Space Gass? Well how would i put the hinge in? i.e what is the fixity code?
Can anyone explain why there is no horizontal force at Bx? Maybe I'm just misunderstanding hinges but I would assume the uniform load is spread over Ax and Bx.
The number equations on the top and right part of the structure why is it 3?
really helpful keep it up.
isnt your support A a roller? how can it take force in X direction
i love you mate, legit!
structurefree Do you have a video of calculation reactions (the reaction of a redundant element) of a frame that is structurally indeterminate?
Thanks a lot
could u please visit my channel ^__^
Ma why teken 10 m
you have 50 k.n you should be take half 10 not 10
yes or no ?
in third step for horizontal member calculation, its not member CD its member BC...
thanks sir..
Lol you'd make a cool professor
isnt it number of parts and not equations?
Just since we have 3 beams :S?
Ugh. I did the left side member second.... Take moment about A trying to figure out what Bx is..... Idk why you can't do this gives the wrong answer for Bx 😩😔😔😔
Member CD,why cx=0
Thank you
Thnx alot
i want from you more exampel about frimes
i have exam after 4 days and my doctoer so bad
he give me hard exam and bad graeds
XD
why ?
By is +37.5 ryt???
m sorry my bad. got it.
+realistic chap word! That's how we ball.
WOWWWWWWWWWWWWWW
@structurefree hi when we calculate the By and Cy in the member CD, you missed to add the total load in the equation which is 75kn isnt it? can you please check this
I don't understand at all ...
the shape semetric so Ya equl Yd
the solution is wrong.
there should be transfer of moment at B and C too...u r transfering only the loads...what about the moment...even it should be transfered from A to B and B to C...
And when u check dis sum in frame design app..the value of Ma comes out to be 144.5
KNm
pls correct this
at B and C there is hinge support, hinges don't have moments.
Lol pins and hinges dont carry moments
sanjay kumawat ,,,ya u write.....he not even explaining the matrix method.....
Very bad
could u please visit my channel ^__^
Can I have your email bro?