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Calculation of a beam’s moment capacity:
The triangular beam indicated is made of 4ksi concrete and carries three
#9 rebars of fy=60ksi.
Note that geometry offers itself for calculations regarding the Compression, the
leverarm, and the depth of a.
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Calculation of a beam’s moment capacity:
The triangular beam indicated is made of f`c=4ksi concrete and
carries three (3) #9 rebars of fy=60ksi.
Note that geometry offers itself for calculations regarding the Compression,
the leverarm, and the depth of α.
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We begin by the fundamental assumption that fs=fy (we verify this at
a later step) and we calculate the area of steel As:
3 #9 rebars give us a total area of 3 sq. in.
T=As(fy) = (3in^2)(60ksi) = 180 kips
fs=Calculated stress of reinforcement at service loads, fy=specified yield strength of non prestressed reinforcement.
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We compute the compression block so that we have C=T:
Given C=T = 180 kips
The specific geometry of the beam allowed us to generate
the formula for C. That can be reversed to solve for a:
…which gives us a value of 10.29 in.
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Verifying our assumption that fs=fy:
As mentioned earlier,
ε
cu will carry the value of 0.003
c=α/β1…thus c=10.29in/0.85…..= 12.11in
Using the similar triangles method as seen above…
thus
ε
s is 0.00394 >
ε
y which is f`c/fy that is 60ksi/29000ksi yielding a
value of 0.00207, proving that fs=fy
ε
s=strain in steel,
ε
y= yield strain.
fs=Calculated stress of reinforcement at service loads, fy=specified yield strength of non prestressed reinforcement.
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Computing the Mn:
We can derive from the geometry that jd=d(2a/3)
Thus it would be safe to claim that
..which yields the result of 285.4 k`
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That wonderful property of structural materials to bend, crack and yet not
break, is one of the possible characteristics of RC.
When flexural forces surpass the limit My, steel reinforcement continues to
elongate. Resistance increases slightly, related to the increase of distance
between C and T. That distance increases as the depth of the concrete
stress block decreases until the concrete fractures. Although the stress of
the steel remains constant, the strain at the point of failure is several times
greater than the steel yield strain
ε
y
, ..approximately ε
y=fy/Es ≈.002
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On a section that fractures when the strain of steel is 0.006, if the As
was multiplied by a certain factor (let’s say doubled), then the Whitney
block would take a similar magnification (double in this case). The strain
at the tension could only be 0.003. The stretching of the steel at the
range between yield and beam failure would only be 0.001 instead of
0.004 as it was for the section with half as much tension reinforcement.
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There are three methods
of flexural failure of
concrete members:
Tension
Compression
Balanced
Source: http://www.shef.ac.uk/content/1/c6/04/71/91/fig32_3_concrete_crushing.jpg, Sept.20/09
Source: http://www.tfhrc.gov/structur/pubs/06115/images/fig29.jpg, Sept.20/09
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The section in the left carries four rebars to counterbalance the
compression force. The middle section indicates a Tbeam with a
larger cross sectional area in compression, and it is set to
equilibrium through the application of more tensile rebars. Inversely,
the notched section on the right carries less tensile reinforcement.
Note that values of â1c are independent of the section’s shape.
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The capacity reduction
“φ” factor also reflects
the relative ductility of
the cross section at
failure expressed in
terms of the strain εt
as presented by ACI.
To determine the
value of εt the
proportional triangle
method can be applied
as seen in the
diagram.
εt=0.003[(d/c)1]
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The capacity reduction
“φ” factor also reflects
the relative ductility of
the cross section at
failure.
For tension controlled,
ductile, flexural failure at
εt>0.005, (i.e 2.5 times
larger than εy for Grade
60), φ= 0.90
Failure at εt≤fy/Es is
compression controlled,
nonductile, and φ=0.65
For further details and info, please refer to in class reading: pp 40
41.
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ACI recognizes that the
magnitude of moments at
critical locations of a flexural
member, estimated through
elastic analysis, can not be
totally precise. Therefore,
designers are allowed to
“redistribute” moment values
(from support regions to span)
provided that:
At failure, the extreme fiber
tension strain εt exceeds
0.0075
The relative amount of
moment value that will be
redistributed is 1000(εt)
percent<20% of the moment
at the end of the beam.
Source: R.W. Furlong: Basic Decisions for Designing Reinforced Concrete Structures, Morgan Printing, Austin , TX, Sept. 2003
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ACI recognizes that the
magnitude of moments at
critical locations of a flexural
member, estimated through
elastic analysis, can not be
totally precise. Therefore,
designers are allowed to
“redistribute” moment values
(from support regions to span)
provided that:
At failure, the extreme fiber
tension strain εt exceeds
0.0075
The relative amount of
moment value that will be
redistributed is 1000(εt)
percent<20% of the moment
at the end of the beam.
Source: R.W. Furlong: Basic Decisions for Designing Reinforced Concrete Structures, Morgan Printing, Austin , TX, Sept. 2003
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Source: R.W. Furlong: Basic Decisions for Designing Reinforced Concrete Structures, Morgan Printing, Austin , TX, Sept. 2003
Equilibrium of forces at limit
load are maintained by insuring
that the avg of required end
moments plus the max +ve
moment is never reduced.
Midspan moments are
increased when end moments
are reduced, or else end
moments are increased if
midspan moment is reduced.
Practically, designing for
redistributed moments transfers
more reinforcement on midspan
and less at supports.
Moment redistributed should
not exceed 10(εt) multiplied by
the moment at support.
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Source: R.W. Furlong: Basic Decisions for Designing Reinforced Concrete Structures, Morgan Printing, Austin , TX, Sept. 2003
To facilitate the redistribution process
as stated earlier, a minimum
ε
t
value of 0.01 should be set.
Given an fy value of 60ksi, that
translates to a stretch
approximately 5 times that of the
strain experienced at fy.
When
ε
t is 0.01 (see lower
configuration in figure) the value
“c” can be deduced geometrically
to be 0.003d/(0.003+0.001)
which returns: c=0.231d
By substituting this value we can
deduce that
As(fy)=0.196(f’c)b(â1)d
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*Note: The min effective depth suggests that after As is determined, a cover per ACI guidelines should be determined.
To recap and bring to surface a few standardized
values:
A recommended ρ_Max will be applied when
the objective is to minimize the depth of a
beam. The εt will be 0.004 and φ will be 0.81.
As example you can use the following formula for minimum effective
depth.*
The ρ_Min is a threshold value that we shall
never cross.
The ρ_lim is the limit that keeps us still within an
εt of .005
The ρ_10 is an ideal condition that we should
always aim for in order to have a comfortable
condition for our designed element and where
the εt is 0.010
OR
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Select flexural bars for the section and the required moment and determine the ΦMn, the b1
ratio (ratio of depth of rectangular stress block, a, to the depth to neutral axis c) the distance c
and the strain on the extreme fiber of reinforcement εt.
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Assignments will be received at the beginning of class period.
Reading: Furlong Chapt. 4.1 & 4.2
Assignment 2 will be due one week
from today.
BEAM DEFLECTION FORMULAE
BEAM TYPE
SLOPE AT FREE END
DEFLECTION AT ANY SECTION IN TERMS OF x
MAXIMUM
DEFLECTION
1. Cantilever Beam – Concentrated load P at the free end
2
2
Pl
E
I
θ=
()
2
3
6
Px
yl
x
EI
=
−
3
max
3
Pl
E
I
δ=
2. Cantilever Beam – Concentrated load P at any point
2
2
Pa
E
I
θ=
()
2
3f
o
r
0
6
Px
ya
x
x
a
EI
=
−<
<
()
2
3f
o
r
6
Pa
yx
a
a
x
l
EI
=
−<
<
()
2
max
3
6
Pa
la
EI
δ
=−
3. Cantilever Beam – Uniform
ly distributed load ω (N/m)
3
6
l
E
I
ω
θ=
()
2
22
64
24x
yx
l
l
x
EI
ω
=+
−
4
max
8
l
E
I
ω
δ=
4. Cantilever Beam – Uniform
ly varying load: Maximum intensity ω
o
(N/m)
3
o
24
l
E
I
ω
θ=
()
2
32
2
3
o
10
10
5
120
x
yl
l
x
l
x
x
lEI
ω
=−
+
−
4
o
max
30
l
E
I
ω
δ=
5. Cantilever Beam – Couple moment M at the free end
M
l
E
I
θ=
2
2
M
x
y
E
I
=
2
max
2
M
l
E
I
δ=
BEAM DEFLECTION FORMULAS
BEAM TYPE
SLOPE AT ENDS
DEFLECTION AT ANY SECTION IN TERMS OF x
MAX
IMUM
AND CENTER
DEFLECTION
6. Beam Simply Supported at Ends – Concentrated load P at the center
2
12
16
Pl
E
I
θ=
θ
=
2
2
3
for
0
12
4
2
Px
l
l
yx
x
EI
⎛⎞
=
−<
<
⎜⎟
⎝⎠
3
max
48
Pl
E
I
δ=
7. Beam Simply Supported at Ends – Concentrated load P at any point
22
1
()
6
Pb
l
b
lEI
−
θ=
2
(2
)
6
Pab
l
b
lEI
−
θ=
()
222
for
0
6
Pbx
yl
x
b
x
a
lEI
=
−−
<
<
()
()
3
22
3
6
for
Pb
l
yx
a
l
b
x
x
lEI
b
ax
l
⎡
⎤
=−
+
−
−
⎢
⎥
⎣
⎦
<
<
(
)
32
22
max
93
Pb
l
b
lEI
−
δ=
at
()
22
3
xl
b
=−
()
2
2
at the center, if
34
48
Pb
lb
EI
δ=
−
ab
>
8. Beam Simply Supported at Ends – Uniformly distributed load ω (N/m)
3
12
24l
E
I
ω
θ=
θ
=
()
32
3
2
24
x
yl
l
x
x
EI
ω
=−
+
4
max
5
384l
E
I
ω
δ=
9. Beam Simply Supported at Ends – Couple momen
t M at the right end
1
6
M
l
E
I
θ=
2
3
M
l
E
I
θ=
2
2
1
6
M
lx
x
y
E
Il
⎛⎞
=−
⎜⎟
⎝⎠
2
max
93
M
l
EI
δ=
at
3
l
x
=
2
16
M
l
E
I
δ=
at the center
10. Beam Simply Supporte
d at Ends –
Uniformly varying load: Maxim
um intensity ω
o
(N/m)
3
o
1
7
360
l
E
I
ω
θ=
3
o
2
45
l
E
I
ω
θ=
()
42
2
4
o
71
0
3
360
x
yl
l
x
x
lEI
ω
=−
+
4
o
max
0.00652
l
E
I
ω
δ=
at
0.519
x
l
=
4
o
0.0065
1
l
E
I
ω
δ=
at the center
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