Chapter 6.
Compression Reinforcement - Flexural Members
If a beam cross section is limited because of architectural or other considerations, it may happen that the concrete cannot develop the compression force required to resist the give bending moment. In this case, reinforcing is added in the compression zone, resulting in a so-called doubly reinforced beam, i.e., one with compression as well as tension reinforcement. Compression reinforced
is also used to improve serviceability, improve long term deflections, and to provide support for stirrups throughout the beam.
6.1. Reading Assignment:
Text Section 5.7; ACI 318, Sections: 10.3.4, 10.3.3, and 7.11.1
6.2. Strength Calculations
0.85f c′
Á u = 0.003
A’s
h
d
A bs
d’
d-c b
b
Á s′
cb
a b = β 1c b
Cs
Cc
T bs
h-c b
Ás = Áy
strains
stresses
forces
From geometry we can find the strain in compression steel at failure as:
d′
Á s′ = 0.003 c −
c
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(6.1)
118
Compression Reinforcement
6.3. Nominal Resisting Moment When Compression Steel Yields
d’
Á u = 0.003
A’s
h
c
Áy
0.85f c′
a
0.85f c′
Cs
=
d
As
d-c
Cc
+
A s′f y
Ts
T s = (A s − A s′)f y
> Áy
b
a
A s′f y
Cc
Case I
Case II
Doubly Reinforced Rectangular Beam
Total resisting moment can be considered as sum of:
1. Moment from corresponding areas of tension and compression steel
2. The moment of some portion of the tension steel acting with concrete.
M n = (A s − A s′) f y (d −
β 1c
) + A s′ f y (d − d′)
2
(6.2)
and from equilibrium:
(6.3)
0.85f c′ ab = (A s − A s′)f y
Solve for “a”:
a=
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A s − A s′
f
0.85f c′ b y
(6.4)
119
Compression Reinforcement
6.4. Compression Steel below Yield Stress (strain compatibility check).
Whether or not the compression steel will have yielded at failure can be determined as follows:
d’
0.85f c′
Á u = 0.003
A’s
c
Á s′ = Á y
a
Cs
Cc
d
h
As lim
d-c
Ts
≥ Áy
b
From geometry:
Áu
c
=
Á′ s
c − d′
if compression steel yield
(6.5)
Á′ s = Á y then:
Áu
c
Áy = c − d ′
→
c =
Áu
d′
− Áy
Áu
(6.6)
Equilibrium for case II:
− A′ s)f y = 0.85 × (β 1c) b f ′ c
(A lim
s
(6.7)
Substitute for “c” from Eq. (6.6) and (6.7) and divide both sides by “bd” gives:
(A lim
− A′ s)f y
s
= 0.85 × β 1 × b × f ′ c
bd

Áu
Áu
d′
− Áy
A lim
f′
A ′s
s
=
+ 0.85 × β 1 × c
bd
bd
fy

Áu
Áu
d′
− Áy d
or
à lim = à ′ s + 0.85 × β 1 ×
à actual > à lim
if
if
 



f ′c
87, 000
d′
×
fy
87, 000 − f y d
then compression steel will yield


f′
As − A ′s
87, 000
d′
≥ 0.85 × β 1 × c ×
bd
fy
87, 000 − f y d
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1
bd
120
(6.8)
(6.9)
(6.10)
this is common for shallow
beams using high strength
steel
then compression steel
will yield
Compression Reinforcement
6.5. Example of analysis of a reinforced concrete section having compression reinforcement.
Determine the nominal moment, Mn , and the ultimate moment capacity, Mu , of the reinforced
concrete section shown below.
2.5”
A’s= 3.8 in2
22.2
f c′ = 5, 000 psi
f y = 60, 000 psi
As= 7.62 in2
12”
Solution
Mn can be calculated if we assume some conditions for compression steel.
Assume that compression steel yields:
C c = 0.85f ′ c β 1 cb = 0.85 × (5 ksi) × (0.80) × c × (12) = 40.8c
C s = A s′f y = 3.8 × (60ksi) = 228 kips
T s = (7.62 in 2) × ( 60 ksi) = 457 kips
Equilibrium:
Cs + Cc = Ts
solve for c:
c =
457 − 228
= 5.6 in
40.8
d’
Á u = 0.003
0.85f c′
check assumption
Á′ s
c
c − d′
c
5.6 − 2.5
= 0.003
= 0.0017
5.6
Á′ s = 0.003
Á′ s = 0.0017 <
d
d-c
Ts
Áy
fy
60
=
= 0.00207
Es
29, 000
wrong assumption
This means the compression steel does not yield. Therefore, our
initial assumption was wrong. We need to make a new assumption.
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121
Compression Reinforcement
Assume f ’s < fy
C s = A′ sf ′ s = A′ s Á′ s E s
c − 2.5
c − 2.5
= (3.8 in 2) × (0.003
) × (29, 000 ksi) = 330
c
c
Now for equilibrium:
40.8c + 330 ×
Cs + Cc = Ts
c − 2.5
= 457 kips
c
→ solve for c
c = 6.31 in
→
check assumption
f ′ s = 0.003 ×
6.31 − 2.5
× 29, 000 = 52.5 ksi < f y = 60 ksi
6.31
assumption o.k.
check ACI Code requirements for tension failure
c = 6.31 = 0.284 < 0.375
22.2
d
0.90
φ
0.75
0.65
We are in the tension-controlled section and satisfy
the ACI code requirements.
φ = 0.75 + (Á t − 0.002)(50)
SPIRAL
OTHER
Compression
Transition
Controlled
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φ = 0.9
Tension
Controlled
Á t = 0.002
Á t = 0.005
c = 0.600
dt
c = 0.375
dt
122
Compression Reinforcement
Calculate forces:
C c = 40.8 × (6.31 in) = 258 kips
C s = 3.8 × (52.5ksi) = 200 kips
258+200=458
T s = (7.62 in 2) × ( 60ksi) = 457 kips
Equilibrium
is
satisfied
Take moment about tension reinforcement to determine the nominal moment capacity of the section:

Mn = Cc d −
β 1c
2
 + C (d − d′)
s
Nominal moment capacity is:
M n = (258 kips) × (22.2 − 0.80 × 6.31) + 200(22.2 − 2.5)
2
= 5080 + 3940 = 9020 in − kips
Ultimate moment capacity is:
M u = φ M n = 0.9 × 9020 = 8118 in − k
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Compression Reinforcement
6.6. Example of analysis of a doubly reinforced concrete beam for flexure
Determine whether the compression steel yield at failure.
2.5”
2 No. 7
21”
f c′ = 5, 000 psi
f y = 60, 000 psi
A’s= 1.2 in2
4 No. 10
As= 5.08 in2
14”
Solution
à =
As
= 5.08
= 0.0173
14 × 21
bd
à − Ã′ = 0.0173 − 0.0041 = 0.0132
A′ s
1.2
Ã′ =
=
= 0.0041
14 × 21
bd
Check whether the compression steel has yielded, use Eq. (6.10):
?
0.0132 ≥
0.85 × β 1 ×


f ′c
87, 000
d′
×
fy
87, 000 − f y d


?
87, 000
2.5
0.0132 ≥ 0.85 × 0.80 × 5 ×
60
87, 000 − 60000 21
?
0.0132 ≥ 0.0217
Therefore, the compression steel does not yield.
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124
Compression Reinforcement
6.7. Example: Design of a member to satisfy a nominal moment capacity.
Assume we have the same size beam as Section 6.6. example and
wish to satisfy the same nominal conditions:
2.5 ”
f y = 60, 000 psi
Required
A’s = ? in 2
22.2”
f c′ = 5, 000 psi
M n = 9020 in − k
As = ? in 2
12”
Solution
For singly reinforced section:
use
c = 0.375
d
f′
à = 0.85β 1 c × c
d
fy
5 ksi
= 0.0213
60 ksi
Maximum As1 for singly reinforced section then is:
à = (0.85)(0.80)(0.375)
A s1 = Ã × b × d = (0.0213) × (12) × (22.2) = 5.66 in 2
fy
M n = Ãf y bd 2 1 − 0.59Ã
f c′




M n = (0.0213 in 2)(60 ksi)(12 in)(22.2 in) 2 1–0.59(0.0213) × 60 = 6409 in.kips
5
M u2 = φM n = 0.9 × 6409 = 5747 in.kips
Moment which must be resisted by additional compression and tension reinforcement
M u1 = M u1 − M u2
M u1 = 0.9 × 9020 – 5747 = 2365 in.kips
Assuming compression steel yields we will have:
M u1 = φA′ s f y (d − d′) = 0.9 × A′ s × (60) × (22.2 − 2.5) = 1063.8 × A′ s
2365 in–k = 1063.8 × A′ s
A′ s =
→
2365 = 2.23 in 2
1063.8
Therefore, the design steel area for tension and compression reinforcement will be:
A s = 5.66 + 2.23 = 7.89 in 2
A′ s = 2.23
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8-#9
3-#8
in 2
125
Compression Reinforcement
2.5 ”
A’s = ? in 2
22.2”
As = ? in 2
12”
Check whether the compression steel has yielded, use Eq. (6.10):


f′
As − A ′s
87, 000
d′
≥ 0.85 × β 1 × c ×
bd
fy
87, 000 − f y d


87, 000
8 − 2.37 ≥ 0.85 × 0.80 × 5 ×
2.5
22.2 × 12
22.2
60
87, 000 − 60000
0.0211 ≥ 0.206
Therefore the compression steel yields at failure
Check to make sure that the final design will fall under “tension-controlled”
a=
(A s − A s′)f y
0.85f c′b
a=
(8.00–2.37)60
= 6.62 in
0.85(5)(12)
c = a = 6.62 = 8.28 in
0.80
β1
c = 8.28 = 0.373 < 0.375
22.2
d
Tension controlled
see the following page for the rest of the solution done in a speadsheet.
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Compression Reinforcement