Reinforced Concrete Design

Lecture no. 2 - Flexure

Flexure in Beams and Slabs

• Beams and slabs are subjected primarily to flexure (bending) and shear.

• At any section within the beam, the internal resisting moment is necessary to equilibrate the bending moments caused by external loads.

Fig. 1. One-way flexure (MacGregor 1997, Fig. 4-1)

Continuous one-way slab

Fig. 2. Internal forces in a beam (MacGregor 1997, Fig. 4-3)

Basic Assumptions in Flexure Theory

1) Plane section remains plane.2) The strain in the reinforcement is equal

to the strain in the concrete at the same level (perfect bond).

3) The stresses in the concrete and reinforcement can be computed from the strains using the stress-strain curves.

Basic Assumptions in Flexure Theory (cont’d)

4) The tensile strength of concrete is neglected.

5) Concrete is assumed to fail when the compressive strain reaches a limiting value, for example, a value of 0.003.

Plane Section Remains Plane

Fig. 3. Assumed linear strain distribution (Notes 1990, Fig. 6-5)

Fig. 4. Cracking of reinforced concrete beam (MacGregor 1992, p. 79)

BMD

SFD

+

Elastic Stresses, Cracked Section

E.N.A.c

d

nAs

b

fcc/3

d - c/3 M

fsT

Fig. 6. Elastic stresses and strains in cracked section (at service loads)

kd

jd= d-kd/3

kd/3

Static Test on Under-reinforced Beam

Failure of Under-reinforced Beam

Concrete fails at strain = 0.003

Analysis for Ultimate Moment Capacity of Beam Section

1) Stress and strain compatibility: stress-strain relationships are used.

2) Equilibrium: internal moments must balance the bending moment due to applied load.

To compute the moment capacity of the beam, two requirements must be satisfied:

Tension, Compression, and Balanced Failures

Flexural failures may occur in three different ways:

1. Tension failure. Reinforcement yields before concrete strain reaches its limiting value. (Under-reinforced)

2. Compression failure. Concrete strain reaches its limiting value before steel yields. (Over-reinforced)

3. Balanced failure. Concrete reaches its limiting value and steel yields at the time of failure.

Failure mode depends on the reinforcement ratio, sAbd

ρ =

Tension, Compression, and Balanced Failures (cont’d)

,s > ,y ,s = ,y ,s = ,y

0 0 0,u = 0.003 ,u = 0.003 ,u = 0.003 ,u = 0.003

,s = ,y

Balanced sectionStrength controlled bytension in reinforcement

Strength controlled bycompression in concrete

Pure

(underreinforced) (overreinforced)compression

Fig. 7. Strain distribution in concrete beam (Notes 1990, p. 6-21)

Balanced Failure

Equivalent Rectangular Stress Block in Concrete

• ACI permits the use of equivalent rectangular concrete stress distribution for ultimate strength calculations.

Equivalent Rectangular Stress Block in Concrete (cont’d)

• Uniform concrete stress: • Depth of stress block:

'0.85 cf1a cβ= Distance from the fiber of maximum

strain to the neutral axis

Strain Equivalent rectangular stress block

T = As fssε

Actual Stress Distribution

Fig. 9. Equivalent rectangular stress block (MacGregor 1997, Fig. 4-17)

Equivalent Rectangular Stress Block in Concrete (cont’d)

The factor is a function of compressive strength of concrete as follows:

1β

'

''

1

'

0.85 for 280 ksc

2800.85 0.05 for 280 560 ksc70

0.65 for 560 ksc

c

cc

c

f

f f

f

β

⎧ ≤⎪

⎛ ⎞−⎪= − ≤ ≤⎨ ⎜ ⎟⎝ ⎠⎪

⎪ ≥⎩

Balanced Failure b

d

cb

Asb = Dbbd

,s = ,y = fy/Es

,u = 0.003

ab = $1cb

0.85 fc’

ab/2

Cb = 0.85 fc’bab

Tb = Asbfy

N.A.

1. From similar triangles,

0.003 0.003b

y

c dε

=+

1 10.003

0.003b by

a c dβ βε

= =+

(1)

Balanced Failure (cont’d)

The balanced reinforcement ratio is then:

2. Force equilibrium,

'0.85sb y cA f f ba=

Substituting Eq. (1) into (2) gives:

(2)

'1 0.0030.85

0.003c

sby y

fA bdf

βε

⎛ ⎞= ⎜ ⎟⎜ ⎟+⎝ ⎠

'1 0.0030.85

0.003sb c

by y

A fbd f

βρε

⎛ ⎞= = ⎜ ⎟⎜ ⎟+⎝ ⎠

(3)

(4)

T C=

Maximum Reinforcement in Design

To ensure a ductile behavior, the maximum reinforcement ratio is given by:

Note: ACI defines a section as being tension-controlled if the net tensile strain in the layer of steel farthest from the compression face of the beam equals or exceeds 0.005 in tension.

max 0.75 bρ ρ= (5)

P.N.A.

c

b

0.85 f’c

C

AsT = Asfy

0.85 f’c

a/2

C = 0.85 f’cba

d - a/2

T = Asf y

c a

b

P.N.A.

As

As

d h

b

,c

,s

c

f’c

C

T = Asfy

P.N.A.

a = $1c

0.85 f’c

a/2C

d - a/2

T

Fig. 10. Stresses and strains at nominal flexural strength (Nawy 1996, p. 93)

Under-reinforced Section (Tension Failure)

Exceed yield strain of steel !s yε ε>

'0.85s s cA f f ba=

T C=From force equilibrium ,

'0.85s y

c

A fa

f b=

Because ,s yf f=

Therefore, the nominal (theoretical) flexural capacity is:

(7)

(8)

(9b)Or

2n s yaM A f d⎛ ⎞= −⎜ ⎟

⎝ ⎠

'0.852n caM f ba d⎛ ⎞= −⎜ ⎟

⎝ ⎠

(9a)

Finally, the actual flexural capacity of the section is:

nMφ

0.9 for tension-controlled section0.7 for compression-controlled section

(10)