DOUBLY REINFORCED RECTANGULAR BEAMS Ok… ahmmm... We have to classify the problem whether an analysis or a design problem. This is again a section subject to a positive moment wherein you have tension at the bottom and compression at the top. So you have your tension bar AS and your compression bar A’S, your distances d and d’and ofcourse the width b although not indicated here. For analysis, the actual situation is we divide this into two couples (M1 and M2) where the first couple is composed of the force CC which is the concrete compression resultant treated as a rectangle with a corresponding tension force that will have to be equal to Cc from the equilibrium equation that the summation of forces must be equal to zero. In USD, it is easy to determine the location of the compression resultant because the stress distribution is rectangularwith a uniform stress of 0.85fC’. This Cc is in the center of the rectangle, so if the height of the rectangle is “a” CC will be acting at a distance 2 a from the top, therefore this will give you a level arm of2 a d− . So the force multiplied by this level arm gives you the Nominal Moment M1. You have to this; you have a couple made of a compression steel CS with a corresponding tension force TS2. The level arm will be just the difference of “d” and d’. That level arm multiplied by the force CS will give you the second couple M2. As usual you have to use the strain diagram. Always, the strain of concrete is 0.003 (ultimate stress design). Meaning this is the limit design and you want to know the capacity of the compression bar if it will fail. And this will happen wh en concrete crushes because steel just like “tira-tira” will just keep on elongating even if it had already yielded. When the strain here becomes 0.003, then you have the Ultimate stage. The analysis working equations are again taken from the basic equation from the summation of forces be equal to zero where CS + CC = TS where TS is made up of TS1 and TS2 corre spondi ng to CS and CC. Again, take note from this drawing that TS1= AS1fy and TS2=AS2fy. In USD, what makes it simpler is that the stress in the tension bar is equal to fy so it always have to be underreinforced as prescribed by the code to ensure ductile failure. Therefore, laging mag-yi-yield ang tension bars. REQUIREMENT NG CODE YAN!!!! SO STRESS ñ an is always fy. To continue, TS=ASfy and AS of course is made up of AS1 and AS2 because the stresses here are equal. CC is 0.85fC’ab, CS=A’S fS. From similar triangles ' 003 . 0 dc c SC− ∈ = , so you will have 003 . 0 * ' c dc SC− = ∈ . If the strain of the steel is less than the yield strain, it will follow the Hooke’s Law. But if the strain of the steel is greater than the yield stra in, again as I have said, you are in the flat porti on; your stress will be equal to fy. You can skip the
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Ok… ahmmm... We have to classify the problem whether an analysis or a design problem. This is againa section subject to a positive moment wherein you have tension at the bottom and compression at the
top. So you have your tension bar AS and your compression bar A’S, your distances d and d’and of
course the width b although not indicated here. For analysis, the actual situation is we divide this into
two couples (M1 and M2) where the first couple is composed of the force CC which is the concretecompression resultant treated as a rectangle with a corresponding tension force that will have to be equal
to Cc from the equilibrium equation that the summation of forces must be equal to zero. In USD, it is
easy to determine the location of the compression resultant because the stress distribution is rectangularwith a uniform stress of 0.85f C’. This Cc is in the center of the rectangle, so if the height of the rectangle
is “a” CC will be acting at a distance2
a from the top, therefore this will give you a level arm of 2
ad −
. So the force multiplied by this level arm gives you the Nominal Moment M 1. You have to this; youhave a couple made of a compression steel C S with a corresponding tension force TS2. The level arm will
be just the difference of “d” and d’. That level arm multiplied by the force C S will give you the second
couple M2. As usual you have to use the strain diagram. Always, the strain of concrete is 0.003 (ultimatestress design). Meaning this is the limit design and you want to know the capacity of the compression
bar if it will fail. And this will happen when concrete crushes because steel just like “tira-tira” will just
keep on elongating even if it had already yielded. When the strain here becomes 0.003, then you have
the Ultimate stage. The analysis working equations are again taken from the basic equation from thesummation of forces be equal to zero where CS + CC = TS where TS is made up of TS1 and TS2
corresponding to CS and CC. Again, take note from this drawing that TS1= AS1f y and TS2=AS2f y. In USD
what makes it simpler is that the stress in the tension bar is equal to f y so it always have to be underreinforced as prescribed by the code to ensure ductile failure. Therefore, laging mag-yi-yield ang tension
bars. REQUIREMENT NG CODE YAN!!!! SO STRESS ñan is always f y. To continue, TS=ASf y and AS
of course is made up of AS1 and AS2 because the stresses here are equal. CC is 0.85f C’ab, CS=A’S f S. From
similar triangles'
003.0
d cc
SC
−∈
= , so you will have 003.0*'
c
d cSC
−=∈ . If the strain of the steel is less
than the yield strain, it will follow the Hooke’s Law. But if the strain of the steel is greater than the yield
strain, again as I have said, you are in the flat portion; your stress will be equal to f y. You can skip the
Monday 5:00 – 9:00 / Thursday 6:00 – 9:00 Engr. Alberto S. Cañete - Professor
Ultimate Stress Design (USD)
Lecture Notes
2k6 – 2k7
step of comparing the strain. Instead, what you can do is simply compute this formula for the stress of
the compression steel and then check whether it is greater than the f y or not. If it is greater than f y, thennag-yield siya. Sa excel, min [E*∈ , f y]. Again, according to the procedure that we followed sa WSD, CC
+ CS = 0.85f c’ [ab – A’s] + A’S f sc{ ito ang meaning niya 0.85 fc’ times the concrete area (a*b) minus
the area occupied by the compression bar}. Rearranging, you now have 0.85f c’ (ab) + A’S (f sc-0.85f c’)With this you are now dealing with the solid rectangle and theoretically a solid circle. Therefore the
analysis for this is a de-kahon na procedure, given everything including the section dimension. Take
note, ang first choice ninyo is singly, dahil pareho lang ng design at pareho lng ng bayad. So only when
you are forced na magdodoubly kayo unless, dagdagan ang bayad mo at magba-value engineering kathen you can exert an extra effort. In reality, ang beam ninyo ay may top bars and bottom bars according
to the code, minimum ang 2 sa taas at 2 sa baba otherwise wala kang pagtatalian ng stirrups mo.
)c'f 85.0s(f A'c(ab)0.85f'
f sA's]A'-c[ab0.85f'CC
zone,concretein the barsncompressio by thedisplacedearea/volumconcreteheconsider tto:note
fyf
Ef if
E
fystrain)(yieldAlso,
0.003*cd'-c
:ianglessimilar tr from
fsA'Ccab,0.85f'C fy;AT
:Where
TCC0F
ANALYSIS
sc
scsc
scysc
scsscysc
sy
sc
sscss
scs
−+=
+=+
=→∈∈
∈=→∈∈
=∈
=∈
===
=+→=∑
>
<
Ok... Your first step is to solve for max ρ . You don’t have to compute the min ρ anymore. For this to be a
candidate for doubly, kailangan marami ang tension bars. This is an analysis problem, given ang A S, if
actual ρ < max
ρ then you can analyze it as singly because the tension reinforcements are not that much,
kayang-kaya siyang i-match nung concrete (CC) ∴ hindi na kailangan ng tulong ng A’S para lang
balansehin yung AS. Pag sinuswerte kau, then analyze it as singly. But when we say na iaanalyze sa
singly, it doesn’t mean na hindi na i-che-check ang min ρ , kailangan lagpas xa sa min ρ , otherwise you
still have brittle failure. BUT!!!!!!!!! PAG MINAMALAS-MALAS KAYO and your actual ρ > max
ρ
then you have to analyze it as doubly… ü If it is doubly, you have to compute for the ' ρ which is
equal to bd
AS '
and then check the difference between this and the'
ρ which you have to compare with
max ρ (paki-open ang code please page 4-35 article 410.4.3 last sentence).
“Classmates, yung iba ay self explanatory na lang. Binasa na lang ni sir ungformula at ung iba ay previous transcription natin. Mapapamahal kasi kungmadami ang ipiprint, nakakawindang lang naman basahin.” ÜANALYSIS:
1.) Determine the maximum safe live load WL that the beam below can carry. Use f’c = 32 and Grade 60 bars. Assume that the total dead load WD = 25 kN/m.m-kN73.26910*