Chapter 6. Compression Reinforcement - Flexural Members

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¡ä

? u = 0.003

A¡¯s

h

d

A bs

d¡¯

d-c b

b

? s¡ä

cb

a b = ¦Â 1c b

Cs

Cc

T bs

h-c b

?s = ?y

strains

stresses

forces

From geometry we can find the strain in compression steel at failure as:

d¡ä

? s¡ä = 0.003 c ?

c

CIVL 4135

(6.1)

118

Compression Reinforcement

6.3. Nominal Resisting Moment When Compression Steel Yields

d¡¯

? u = 0.003

A¡¯s

h

c

?y

0.85f c¡ä

a

0.85f c¡ä

Cs

=

d

As

d-c

Cc

+

A s¡äf y

Ts

T s = (A s ? A s¡ä)f y

> ?y

b

a

A s¡äf 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¡ä) f y (d ?

¦Â 1c

) + A s¡ä f y (d ? d¡ä)

2

(6.2)

and from equilibrium:

(6.3)

0.85f c¡ä ab = (A s ? A s¡ä)f y

Solve for ¡°a¡±:

a=

CIVL 4135

A s ? A s¡ä

f

0.85f c¡ä 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¡ä

? u = 0.003

A¡¯s

c

? s¡ä = ? y

a

Cs

Cc

d

h

As lim

d-c

Ts

¡Ý ?y

b

From geometry:

?u

c

=

?¡ä s

c ? d¡ä

if compression steel yield

(6.5)

?¡ä s = ? y then:

?u

c

?y = c ? d ¡ä

¡ú

c =

?u

d¡ä

? ?y

?u

(6.6)

Equilibrium for case II:

? A¡ä s)f y = 0.85 ¡Á (¦Â 1c) b f ¡ä 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¡ä s)f y

s

= 0.85 ¡Á ¦Â 1 ¡Á b ¡Á f ¡ä c

bd

??

?u

?u

d¡ä

? ?y

A lim

f¡ä

A ¡äs

s

=

+ 0.85 ¡Á ¦Â 1 ¡Á c

bd

bd

fy

?

?u

?u

d¡ä

? ?y d

or

? lim = ? ¡ä s + 0.85 ¡Á ¦Â 1 ¡Á

? actual > ? lim

if

if

? ?

?

?

?

f ¡äc

87, 000

d¡ä

¡Á

fy

87, 000 ? f y d

then compression steel will yield

?

?

f¡ä

As ? A ¡äs

87, 000

d¡ä

¡Ý 0.85 ¡Á ¦Â 1 ¡Á c ¡Á

bd

fy

87, 000 ? f y d

CIVL 4135

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¡ä = 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 ¡ä c ¦Â 1 cb = 0.85 ¡Á (5 ksi) ¡Á (0.80) ¡Á c ¡Á (12) = 40.8c

C s = A s¡äf y = 3.8 ¡Á (60ksi) = 228 kips

T s = (7.62 in 2) ¡Á ( 60 ksi) = 457 kips

Equilibrium:

Cs + Cc = Ts

solve for c:

c =

457 ? 228

= 5.6 in

40.8

d¡¯

? u = 0.003

0.85f c¡ä

check assumption

?¡ä s

c

c ? d¡ä

c

5.6 ? 2.5

= 0.003

= 0.0017

5.6

?¡ä s = 0.003

?¡ä s = 0.0017 <

d

d-c

Ts

?y

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.

CIVL 4135

121

Compression Reinforcement

Assume f ¡¯s < fy

C s = A¡ä sf ¡ä s = A¡ä s ?¡ä s E s

c ? 2.5

c ? 2.5

= (3.8 in 2) ¡Á (0.003

) ¡Á (29, 000 ksi) = 330

c

c

Now for equilibrium:

40.8c + 330 ¡Á

Cs + Cc = Ts

c ? 2.5

= 457 kips

c

¡ú solve for c

c = 6.31 in

¡ú

check assumption

f ¡ä s = 0.003 ¡Á

6.31 ? 2.5

¡Á 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

¦Õ

0.75

0.65

We are in the tension-controlled section and satisfy

the ACI code requirements.

¦Õ = 0.75 + (? t ? 0.002)(50)

SPIRAL

OTHER

Compression

Transition

Controlled

CIVL 4135

¦Õ = 0.9

Tension

Controlled

? t = 0.002

? t = 0.005

c = 0.600

dt

c = 0.375

dt

122

Compression Reinforcement

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