7:01 you did say bx^3 but you wrote x^2😬...had to replay a few times just to make sure I heard it right if not the equation is wrong🤷🏻♀️I was expecting you’d add an ‘animation’ around the time you’re discussing it to inform the viewers the typo like in your other videos but when nothing appeared I had to verify first and replaying the video helps👍
At 4:13 I mention that we don't define the geometry of the transformed area, just the area itself. You'll see this same simplification in textbooks. If I recall correctly, we can ignore the other term because the actual geometry of the real steel is so small its contribution to the overall value is negligible.
Hey Sophia. I don't have a reinforced T-Beam, but I do have an example with a regular T-Beam here: www.engineer4free.com/4/pure-bending-in-the-elastic-range-example-2-t-beam bithbvidrisbfokkow essentially the same principles
@@Engineer4Free thank you for replying. Maybe you can answer a question for me? This is for a reinforced t beam using the transformer method. When calculating inertia, my book uses bh^3/12+Ady^2+ bh^3/3 + nAst(d-x)^2. Where as the one video I have seen work this problem uses two parallel theorems and then adds the nAst(d-x). My question is why does the book use the bh^3/12 & bh^3/3?
is the whole compressed area above the neutral line of concrete under plastic deformation? if yes, is there any concrete area under elastic compressed deformation, and why its contribution does not count?
Hey first of all there is a typo in the video regarding that expression. at 7:00 I say "cubed" but write "squared." I should have written "cubed" but my hand was off doing it's own thing. That first term should be written as I said: (1/3)bx^3. With that noted, the moment of inertia for a rectangle about an axis that passes through its centroid is (1/12)bh^3. The moment of inertia for a rectangle about an axis that is aligned with its bottom edge is (1/3)bh^3. The latter is the case here, because we are taking moment of inertia about the neutral axis, which is an axis aligned with the bottom edge of the top rectangle. You can find these two expressions in a moment of inertia table in any mechanics of materials textbook. If you want to know where the 1/3 comes from, you can derive the (1/3)bh^3 equation by using the parallel axis theorem from (1/12)bh^3+Ad^2 for the top rectangle. If A=bh and d=h/2 then (1/12)bh^3 + (bh)(h/2)^2 = (1/12)bh^3 + (bh)(h^2/2^2) = (1/12)bh^3 + (bh^3)/(1/4) = (1/12)bh^3 + (bh^3)(3/12) = (bh^3)(4/12) = (bh^3)(1/3) = (1/3)bh^3. Cheers hope that helps.
@@Engineer4Free So using (1/3)bh^3 essentially includes the dimensions from the top rectangle too? If for example, I use (1/12)bh^3 +A(x/2)^2 for the top rectangle and add that with A*(d-x) ^2 [I ignored the I for the bottom transformed steel as the height is approximately 0). Will i get that same answer as yours?
Can any one give suggestions how to solve - A rainforced column is 300mm dia and has 4 steel bar each of 12 mm dia embedded in it. If the allowable stress in steel and concrete are 65.0MN/m^2 and 4.0MN/m^2 respectively. Calculate the safe axial load which the coloumn can carry. Es =15Ec
Your explanation is really good, got it the first time
7:01 you did say bx^3 but you wrote x^2😬...had to replay a few times just to make sure I heard it right if not the equation is wrong🤷🏻♀️I was expecting you’d add an ‘animation’ around the time you’re discussing it to inform the viewers the typo like in your other videos but when nothing appeared I had to verify first and replaying the video helps👍
Very helpful, thank you.
Thanks for the feedback! Make sure you check out engineer4free.com/mechanics-of-materials if you haven't already :)
It clearly describes how reinforced problem is solved but for the moment of inertia why you don't use 1/12bh^3
Because that is for the gross moment of inertia, (1/3)bh^3+nAs(d-x)^2 is for the cracked moment of inertia.
im confused on why doesnt you calculate the inertia of the steel (disc shaped) but only use nAd^2? Thanks for all the help
At 4:13 I mention that we don't define the geometry of the transformed area, just the area itself. You'll see this same simplification in textbooks. If I recall correctly, we can ignore the other term because the actual geometry of the real steel is so small its contribution to the overall value is negligible.
Can you please do a video for pure bending of Reinforced concrete but for a T-Beam please
Hey Sophia. I don't have a reinforced T-Beam, but I do have an example with a regular T-Beam here: www.engineer4free.com/4/pure-bending-in-the-elastic-range-example-2-t-beam bithbvidrisbfokkow essentially the same principles
@@Engineer4Free thank you for replying. Maybe you can answer a question for me? This is for a reinforced t beam using the transformer method. When calculating inertia, my book uses bh^3/12+Ady^2+ bh^3/3 + nAst(d-x)^2.
Where as the one video I have seen work this problem uses two parallel theorems and then adds the nAst(d-x). My question is why does the book use the bh^3/12 & bh^3/3?
is the whole compressed area above the neutral line of concrete under plastic deformation? if yes, is there any concrete area under elastic compressed deformation, and why its contribution does not count?
Isn't the moment of inertia =1/12bx^3 + nAs(d-x)^2 for the rectangle?
Hey first of all there is a typo in the video regarding that expression. at 7:00 I say "cubed" but write "squared." I should have written "cubed" but my hand was off doing it's own thing. That first term should be written as I said: (1/3)bx^3. With that noted, the moment of inertia for a rectangle about an axis that passes through its centroid is (1/12)bh^3. The moment of inertia for a rectangle about an axis that is aligned with its bottom edge is (1/3)bh^3. The latter is the case here, because we are taking moment of inertia about the neutral axis, which is an axis aligned with the bottom edge of the top rectangle. You can find these two expressions in a moment of inertia table in any mechanics of materials textbook. If you want to know where the 1/3 comes from, you can derive the (1/3)bh^3 equation by using the parallel axis theorem from (1/12)bh^3+Ad^2 for the top rectangle. If A=bh and d=h/2 then (1/12)bh^3 + (bh)(h/2)^2 = (1/12)bh^3 + (bh)(h^2/2^2) = (1/12)bh^3 + (bh^3)/(1/4) = (1/12)bh^3 + (bh^3)(3/12) = (bh^3)(4/12) = (bh^3)(1/3) = (1/3)bh^3. Cheers hope that helps.
Yes
@@Engineer4Free So using (1/3)bh^3 essentially includes the dimensions from the top rectangle too?
If for example, I use (1/12)bh^3 +A(x/2)^2 for the top rectangle and add that with A*(d-x) ^2 [I ignored the I for the bottom transformed steel as the height is approximately 0). Will i get that same answer as yours?
Is there any software where I can automatically get dimension of x and stress and strain diagram in compression zone of a RC beam?
what program do you use for your drawings?
I think I did this one with Sketchbook. I’ve got a full list of hardware and software that I use at engineer4free.com/tools ✌️
Can you do in wolfram mathematica animation how stress and strain diagram of compression zone in RC beam varies in accordance to the value of As?
Can any one give suggestions how to solve - A rainforced column is 300mm dia and has 4 steel bar each of 12 mm dia embedded in it. If the allowable stress in steel and concrete are 65.0MN/m^2 and 4.0MN/m^2 respectively. Calculate the safe axial load which the coloumn can carry. Es =15Ec
You're too fast
The method to calculate x, the position of neutral axis is wrong.
It's absolutely correct.. stop confusing people
You have to say why. Just say that doesn't make any sense.