2010

Prepared by:Mohammed Al Telbani 12010/0599

Instructor:Dr. Mohammed Arafa

Islamic University – Gaza

Faculty of Engineering

Civil Engineering Department

Design and Rehabilitation of Structures

Design of Special Concrete Structures ENGC 6330

ACI CODE & EUROCODE APPROACHES FOR FLEXURAL

CRACKS CONTROL IN WATER TANKS

1. Introduction.

2. Factors Affecting the Width of Flexural Cracks.

3. Distribution of Flexural Reinforcement in One-way Slabs

and Walls.

• ACI 350-01 Requirements.

• ACI 350-06 Requirements.

• ACI 224-01 Requirements.

• Euro Code Requirements.

4. Distribution of Flexural Reinforcement in Two-way Slabs

and Plates

• ACI 224-01 Requirements

When designing a reinforced concrete structure, three limit

states must be considered:

• Ultimate Limit State.

• Serviceability Limit State.

• Special Limit State.

Excessive Crack Width

Cracks in concrete structures can indicate major structural

problems and detract from the appearance of monolithic

construction. There are many specific causes of cracking.

Cracking can be the result of one or a combination of

factors. Some examples include:

• Drying Shr inkage. This occurs as water used in the mix design

evaporates.

• Thermal Contraction/Expansion. Due to temperature changes.

• Subgrade Settlement (or Expansion). Resulting from poor soil

conditions or changes in soil moisture content.

• Differential Bear ing Capacity. Harder soils under part of the

foundation can cause stresses as the building “settles in.”

• Applied Stresses. Forces such as building load, earth load, or

hydrostatic pressure which is the cause of flexural cracks.

2. FACTORS AFFECTING THE WIDTH OF FLEXURAL CRACKS

The crack width of a flexural crack depends on the following

quantities:

1. The reinforcing steel stress is the most important variable;

2. The thickness of the concrete cover is an important variable but

not the only geometric consideration;

3. The area of concrete surrounding each reinforcing bar is also an

important geometric variable;

4. The bar diameter is not a major variable. Several bars at

moderate spacing are much more effective in controlling

cracking than one or two larger bars of equivalent area.

3. DISTRIBUTION OF FLEXURAL REINFORCEMENT IN

ONE-WAY SLABS AND WALLS

This section prescribes rules for distribution of flexural reinforcement to

control flexural cracking in one-way slabs, and one-way walls (reinforced

to resist flexural stresses in only one direction) that are not compression

controlled sections.

We will explain the following ACI codes approaches for control of

flexural cracks:

1. ACI 350-01, “Code Requirements for Environmental Engineering

Concrete Structures and Commentary”.

2. ACI 350-06, “Code Requirements for Environmental Engineering

Concrete Structures and Commentary”.

3. ACI 224R-01, “Control of Cracking in Concrete Structures”.

Cross sections of maximum positive and negative moment shall be so

proportioned that the quantity z given by:

3 Adfz cs= ACI 350R-01 (10-5)

does not exceed:

1. 115 kip/in normal environmental exposure.

2. 95 kip/in severe environmental exposure .

where,• z = quantity limiting distribution of flexural reinforcement, kip/in.

• fs = Calculated flexural stress in reinforcement at service load, (ksi).

fs shall be computed as the service moment divided by the product

of steel area and internal moment arm. In place of such

computations, it is permitted to take fs as 45% of specified yield

strength fy.

• dc = distance between the extreme tension fiber to the centroid of

the closet bar, in.

• A = effective tension area of concrete, which is the area of concrete

symmetric with reinforcing steel divided by number of bars, in2.

See Figure (1).

Figure (1): Stress Diagram and Effective tension area of concrete.

For liquid retention:

• Normal environmental exposure is defined as exposure

to liquids with a pH greater than 5, or exposure to sulfate

solutions of 1000 ppm or less.

• Severe environmental exposures are conditions in

which the limits defining normal environmental exposure

are exceeded.

The calculated stress fs in reinforcement closest to a surface in tension at

service loads shall not exceed that given by Eq. ACI 350-06 (10-4) and

(10-5) and shall not exceed a maximum of 36,000 psi:

• In normal environmental exposure areas:

psi 20,000

224

3202

2

max, ≥

++

=b

sds

f

β

ACI 350-06 (10-4)

ACI 350-06 REQUIREMENTS

psi 17,000

224

2602

2

max, ≥

++

=b

sds

f

β

•In Severe environmental exposure areas:

ACI 350-06 (10-5)

ACI 350-06 REQUIREMENTS

Important Note

The numerical limitations of z (ACI 350-01) and fs,max (ACI 350-06)

for normal environmental exposure and severe environmental

exposure respectively correspond to limiting crack widths of 0.010 in

(0.254 mm) and 0.009 in (0.2286 mm).

The following approach (ACI 224-01) gives a value of crack width

which will be compared with a reasonable crack width values. See

table (1).

Equation ACI 350-01(10-5) is written in a form emphasizing reinforcing

details rather than crack width. It is based on the (Gergely-Lutz)

expression:

33 10076.0 −×= Adfw csβ

where:

• w = most probable maximum crack width, in.

• β = ratio of distance between neutral axis and extreme tension

fiber to distance between neutral axis and the centroid of the main

reinforcing steel.

ACI 224-01 REQUIREMENTS

cdch

−−

=β

Nominal limit value of the crack width specified for cases with expected

functional consequences of cracking are stipulated in Table (1).

w < w lim

Exposure ConditionCrack Width

in. mm

Dry or protective membrane 0.016 0.41

Humidity, moisture air, soil 0.012 0.30

Deicing chemicals 0.007 0.18

Seawater and sweater spray, wetting and drying 0.006 0.15

Water-retaining structures 0.004 0.10

Table (1): Guide to reasonable* crack widths, reinforced concrete under service loads.

ACI 224-01 REQUIREMENTS

The code stipulates that the design crack width be evaluated from the

following expression:

Eurocode Requirements

smrmk sw εβ=

where,

= design crack width;

= average stabilized crack spacing;

= mean strain under relevant combination of loads and allowing for

the effect such as tension stiffening or shrinkage; and

= coefficient relating the average crack width to the design value

= 1.7 for load-induced cracking and for restraint cracking in sections

with minimum dimension in excess of 800 mm (32 in.).

kw

rms

smε

β

The strain in the section is obtained from the following expression:

Eurocode Requirements

where,= stress in the tension reinforcement computed on the basis of a

cracked section, MPa;= stress in the tension reinforcement computed on the basis of a

cracked section under loading conditions that cause the first crack,MPa;= coefficient accounting for bar bond characteristics= 1.0 for deformed bars and 0.5 for plain bars;= coefficient accounting for load duration= 1.0 for single short-term loading and 0.5 for sustained or cyclic

loading; and= Modulus of elasticity of the reinforcement, MPa.

smε

])/(1[/ 221 ssrsssm E σσββσε −=

sσ

srσ

1β

2β

sE

The average stabilized mean crack spacing is evaluated from thefollowing expression:

Eurocode Requirements

where,db = bar diameter, mm;

= effective reinforcement ratio = As / Act ; the effective concretearea in tension Act is generally the concrete area surrounding thetension reinforcement of depth equal to 2.5 times the distance fromthe tensile face of the concrete section to the centroid of thereinforcement. For slabs where the depth of the tension zone maybe small, the height of the effective area should not be taken greaterthan [(c – db)/ 3], where c = clear cover to the reinforcement, mm;

k1 = 0.8 for deformed bars and 1.6 for plain bars; andk2 = 0.5 for bending and 1.0 for pure tension.

rms

mm ,/25.050 21 tbrm dkks ρ+=

tρ

4. DISTRIBUTION OF FLEXURAL REINFORCEMENT IN TWO-WAY SLABS AND PLATES

ACI 224-01 REQUIREMENTS

Analysis of data on cracking in two-way slabs and plates (Nawy and Blair

1971) has provided the following equation for predicting the maximum

crack width:

Ifkw sβ=

where the terms inside the radical are collectively termed the grid index:

−==πρ8

1

21

1

21

b

c

t

b

ddsssdI

where,

• k = fracture coefficient with a value k = 2.8 x 10-5 for uniformly

loaded restrained two-way action square slabs and plates. For

concentrated loads or reactions or when the ratio of short to long span

is less than 0.75 but larger than 0.5, a value of k = 2.1 x 10-5 is

applicable. For span aspect ratios less than 0.5, k = 1.6 x 10-5;

• β = 1.25 (chosen to simplify calculations, although it varies between

1.20 and 1.35);

• fs = actual average service-load stress level or 40% of the specified

yield strength fy, ksi;

4. DISTRIBUTION OF FLEXURAL REINFORCEMENT IN TWO-WAY SLABS AND PLATES

ACI 224-01 REQUIREMENTS

where,

• db1 = diameter of the reinforcement in Direction 1 closest to the

concrete outer fibers, in.;

• s1 = spacing of the reinforcement in Direction 1, in.;

• s2 = spacing of the reinforcement in perpendicular Direction 2, in.;

• ρt1 = active steel ratio, that is, the area of steel As per ft width/ [12db1 +

2c1], where c1 is clear concrete cover measured from the tensile face

of concrete to the nearest edge of the reinforcing bar in Direction 1;

and

• w = crack width at face of concrete caused by flexure, in.

4. DISTRIBUTION OF FLEXURAL REINFORCEMENT IN TWO-WAY SLABS AND PLATES

ACI 224-01 REQUIREMENTS

Example

We have a water tank with:

• Wall thickness = 50.0 cm = 19.685 in

• Service Moment Ms = 9.0 ton.m = 781.165 kip.in

• Take 100 cm of the wall = 39.37 in

• Concrete cover = 5.0 cm = 1.968 in

• Concrete Compressive Stress

• Reinf. Yield Stress

ksi 267.4/ 300 2' == cmkgfc

ksi 738.59/ 4200 2 == cmkgf y

Example

1- Compute the area of flexural reinforcement:

d = 50.0 – 5.0 – 1.4/2 = 44.3 cm

min2

5

0012.0)300()3.44(100

)0.9()10(61.2114200

)300(85.0 ρρ <=

−−=

mcmAs / 62.143.441000033.0 2=××=

Use 1 Φ 14 @ 10 cm

Example

2- Check for crack width:

33 10076.0 −×= Adfw csβcdch

−−

=β

d =h – cover – 0.5 db = 19.685 – 1.968 – 0.5(0.551)= 17.44 in

1βac =

bffA

ac

ys'85.0

=

836.070

)280300(05.085.070

)280(05.085.0'

1 =−

−=−

−= cfβ

Number of bars = 100/10 = 10 bars

22 in 384.2)2/551.0(10 =×= πsA

Example

in 997.037.39267.485.0

738.59384.2=

×××

=a

in 193.1836.0997.0

==c

138.1193.1440.17193.1685.19

=−−

=β

2kip/in 34.19

2997.044.17384.2

165.781

2

=

−×

=

−×

=adA

Mfs

s

Example

in 244.2)2/551.0(968.1 =+=cd

2in 665.17937.3244.222 =××== sdA cc

in 0057.010665.17244.2 34.19138.1076.0 33 =××××= −w

mm 145.04.250057.0 =×=w > 0.10 mm NOT O.K.

3- Repeat the previous steps using 1 Φ 14 @ 7.5 cm

mm 0999.0=w O.K.

REFERENCES

1. ACI Committee 224 (ACI 224R-01), "Control of Cracking in

Concrete Structures", American Concrete Institute, Detroit,

Michigan.

2. ACI Committee 350 (ACI 350-01), "Code Requirements for

Environmental Engineering Concrete Structures and Commentary",

American Concrete Institute, Detroit, Michigan.

3. ACI Committee 350 (ACI 350-06), "Code Requirements for

Environmental Engineering Concrete Structures and Commentary",

American Concrete Institute, Detroit, Michigan.

4. "Cracking in Concrete Walls", Concrete Foundations Association of

North America.