lecture-05 serviceability requirements & development of

1

Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures

Department of Civil Engineering, University of Engineering and Technology Peshawar

Lecture-05

Serviceability Requirements &

Development of Reinforcement

By: Prof Dr. Qaisar Ali

Civil Engineering Department

UET Peshawar

[email protected]

www.drqaisarali.com

1



Topics Addressed

Section 1: Deflections

Deflection in RC One-way Slabs and Beams, Deflection in RC Two-way

Slabs, Examples

Section 2: Cracking in RC Members

Crack Formation, Equations for Maximum Crack Width, Reasons for

Crack Width Control, Crack Control Reinforcement in Deep Members,

ACI Provisions for Crack Control, Example

Section 3: Development & Splices of Reinforcement

Bond Strength, Bond failure, ACI provisions for Development Length,

Splices of Reinforcement

2

mailto:[email protected]

2



Section 1: Deflections

3



Topics Addressed

Introduction to deflections

Deflection in RC One-way Slabs and Beams

Deflection in RC Two-way Slabs

Example 1

Example 2

4

3



Introduction to Deflections

Deflection Effects

It is important to maintain control of deflections so that members

designed mainly for strength at prescribed overloads will also

perform well in normal service.

Excessive deflections can lead to cracking of supported walls and

partitions, ill-fitting doors and windows, poor roof drainage,

misalignment of sensitive machinery and equipment, and visually

offensive sag etc.

5



Introduction to Deflections

Deflection Types

Immediate (short-term)

Due to applied loads on the member.

Long-term

Due to creep and shrinkage of concrete.

6

4



7

w

w

Mid-span deflection

1 – Ends of beam crack

2 – Cracking at midspan

3 – Full service load3 – Due to shrinkage and creep

Yielding of reinforcement

45

Introduction to DeflectionsDeflection history of

fixed ended beam

subjected to uniformly

distributed load



Deflection in RC One-way Slabs and

Beams

Immediate (short-term) Deflection

Immediate deflection at a given load in a structure is calculated

using equations of elastic deflection.

Δ = f(loads, spans, supports)/EI

EI is the flexural rigidity.

f(loads, spans, supports) is a function of particular load, span and support

arrangement.

8

5




Beams


Elastic equation in a general form for deflection of cantilever and

simple and continuous beams subjected to uniform loading can

be written as:

Δi = K(5/48)Mal2/EcIe

Where,

Ma = the mid-span moment (when K is so defined) for simple and continuous

beams. For cantilever beams, Ma will be the support moment.

l = Span length.

9




Beams


Determination of K: Theoretical values of deflection coefficient

“K” for uniformly distributed loading w:

10

Reference: PCA Notes.

6




Beams


Alternatively, deflections can be directly computed for different

conditions of loading and end conditions as below:

11




Beams


Determination of l: In elastic deflection formulae, l is the span

length and is least of:

ln + h

c/c distance between supports.

12

Support

c/c distance c/c distance

ln ln

Slab

h

7




Beams


Determination of Ec (modulus of elasticity of concrete)

Unless stiffness values are obtained by a more comprehensive

analysis, immediate deflection shall be computed with the modulus

of elasticity (Ec) of section as given in ACI 19.2.2.1.

Ec strongly influences the behavior of reinforced concrete members

under short-term loads, particularly showing more variation with

concrete quality, concrete age, stress level, and rate or duration of

load.

13

Ec = wc1.5 33 fc

Note: The formula is used for values of wc between 90 and 160 lb/ft3.

(in psi)



14


Beams


Effective Moment of Inertia

Mo

me

nt,

Ma

Deflection DDe

Mcr

Ig

Dt

M

Dcr

Icr

Ie

Graph shows the portion of

load deflection curve up to

stage 3 as discussed earlier

“Idealized short term deflection of RC beam”

8




Beams


Determination of Ie: In elastic deflection equation, the effective

moment of inertia Ie is calculated as (ACI 24.2.3.5a):

Where, Ma = maximum service load moment at the stage for which

deflections are being considered.

Mcr = Cracking moment = fr Ig /yt (where fr = 7.5fc)

Ig = Gross moment of inertia

Icr = Moment of inertia of cracked section

15




Beams


Determination of Ie:

Un-cracked section moment of Inertia (Ig) [about centroid]:

For an un-cracked concrete section, full section will be effective

so that:

Ie = Ig = bh3/12

16

h

b

9




Beams



Un-cracked section moment of Inertia (Ig) [about base]:

Ig = Ic.a = bh3/12 is the moment of inertia of rectangular section

about the centroid.

For determination of moment of inertia about base of the

section, transfer formula may be used:

Ibase = Ic.a + Ae2 = bh3/12 + Ae2

Where, A = bh ; e = h/2 (for given case)

Ibase = bh3/3

17

h

b

e



N.A


Beams



Cracked section Moment of Inertia (Icr) [about neutral axis]:

Similarly, moment of inertia of cracked section about neutral axis

can be determined through transfer formulae (I = bh3/3 + Ae2)

For the section shown:

Icr = [bw (kd)3 /3] + nAs(d – kd)2

kd in above equation is unknown.

n = Es / Ec

18

h

bw

nAs

kd

de

10




Beams



Cracked section Moment of Inertia (Icr):

Determination of kd:

Taking moments of area about the neutral axis:

kd bw kd/2 = nAs (d – kd)

kd2bw/nAs = d – kd

Let B = bw/nAs, then

kd2B = (d – kd)

Simplification through quadratic formula gives,

kd = {√(1 + 2Bd) – 1}/B

19

h

bw

nAs

kd

dN.A




Beams



Cracked section Moment of Inertia (Icr): Alternatively, gross and

cracked moment of inertia of rectangular and flanged sections can

be calculated from the given tables:

20

11




Beams




21




Beams




22

Note: Effective width of T-section as per ACI 6.3.2.1

12




Beams




23

Note: Effective width of T-section as per ACI 6.3.2.1




Beams


Application of Ie for simple and continuous members

We have seen that Ie depends on Ma , which will be different at different

locations . Therefore according to (ACI R 24.2.3.7); for

Simply supported Beams:

Ie = Ie,midspan

Beams with one end continuous:

Ie, average = 0.85Ie,midspan + 0.15 (Ie,continuous end)

Beams with both ends continuous:

Ie, average = 0.70Ie,midspan + 0.15 (Ie1 + Ie2)

Where, Ie1 and Ie2 refer to Ie at the respective beam ends. (ACI R

24.2.3.7)

24

13




Beams

Long-term Deflections

Shrinkage and creep due to sustained loads cause additional long-

term deflections over and above those which occur when loads are

first placed on the structure.

Such deflections are influenced by:

Temperature,

Humidity,

Curing conditions,

Age at the time of loading,

Quantity of compression reinforcement, and

Magnitude of the sustained load.

25




Beams

Long-term Deflections (ACI 24.2.4)

Additional Long-term deflection resulting from the combined

effect of creep and shrinkage is determined by multiplying the

immediate deflection caused by the sustained load with the factor

lD .

Δ (cp +sh) = λΔ(Δi)sus

26

lD =x

1 + 50r

r = As/bd

Time x

3 months 1.0

6 months 1.2

12 months 1.4

5 years 2.0

x is a time dependent factor for sustained load.

14




Beams

Long-Term Deflections

It is important to note here that long term deflections are function

of immediate deflections due to sustained load only i.e., Δ (cp +sh) =

λΔ(Δi)sus

Sustained loads are loads that are permanently applied on the

structure e.g., dead loads, superimposed dead loads and live

loads kept on the structure for long period.

27




Beams

Deflection Control according to ACI code:

Two methods are given in the code for controlling deflections.

Deflections may be controlled DIRECTLY by limiting computed

deflections [see Table 24.2.2] or INDIRECTLY by means of

minimum thickness [Table 7.3.1.1] for one-way systems.

28

15




Beams


In DIRECT approach, the deflections are said to be within limits if

the combined effect of immediate and long term deflections does

not exceed the limits specified in ACI table 24.2.2.

29




Beams


In the INDIRECT approach, the deflections for beams and one-way slabs

are controlled indirectly by the minimum requirements given in ACI table

9.3.1.1 and 7.3.1.1.

However, this method is applicable only to the cases of loadings and

spans commonly experienced in buildings and cannot be used for

unusually large values of loading and span.

30

16





The calculation of deflections for two-way slabs is complicated

even if linear elastic behavior can be assumed.

For immediate deflections 2D structural analysis is required in

which the values of Ec and Ie for one-way slabs and beams

discussed earlier may be used.

31




Long-Term Deflections

The additional long-term deflection for two-way construction is

required to be computed using the multipliers given in ACI

24.2.4.1.

32

17




Deflection Control according to ACI code

Deflections may be controlled directly by limiting computed deflections

[see Table 24.2.2] or indirectly by means of minimum thickness, table

8.3.1.1 and table 8.3.1.2 for two-way systems

33



Example 1

Analyze the simply supported reinforced concrete beam for short-

term deflections and long-term deflections at ages 3 months and 5

years.

Data:

fc′ = 3000 psi (normal weight concrete); fy = 40,000 psi

Es = 29,000,000 psi

Superimposed dead load (not including beam weight) = 120 lb/ft

Live load = 300 lb/ft (50% sustained) ; Span = 25 ft

34

18



Example 1

Data:

As = 3 #7 = 1.80 in2

ρ = As/bd = 0.0077

As′ = 3 #4 = 0.60 in2 (As′ not required for strength)

ρ′ = As′ /bd = 0.0026

35



Example 1

Solution:

Step 01: Minimum beam thickness, for members not supporting or

attached to partitions or other construction likely to be damaged by

large deflections:

hmin = (l/16) {multiply by 0.8 for fy = 40000 psi}

hmin = (25 12/16) 0.8 = 15″

22″ is the given depth, therefore O.K.

36

19



Example 1

Solution:

Step 02: Moments

wd = 0.120 + 0.150 12 22/144 = 0.395 kips/ft

Md = wdl2/8 = 0.395 252/8 = 30.9 ft-kips

Ml = wll2/8 = 0.300 252/8 = 23.4 ft-kips

Md+l = 54.3 ft-kips

Msus = Md + 0.5Ml = 30.9 + 0.5 23.4 = 42.6 ft-kips

37



Example 1

Solution:

Step 03: Modulus of rupture, modulus of elasticity, modular ratio

fr = 7.5 √ (fc′) = 7.5 √ (3000) = 411 psi

Ec = wc1.533 √ (fc′) = 1501.5 33 √ (3000) = 3.32 106 psi

n = Es/Ec = 29 106/ 3.32 106 = 8.7

38

20



Example 1

Solution:

Step 04: Gross and cracked section moments of inertia

Ig = bh3/12 = 12 223/12 = 10650 in4

B = b/nAs = 12/{8.7 1.80} = 0.766″

r = (n – 1)As′/nAs = {(8.7 – 1)0.60/(8.71.80) = 0.295

kd = [√ {2dB(1 + rd′/d) + (1+r2)} – (1 + r)]/B = 5.77″

Icr = {b(kd)3/3} + nAs (d – kd)

2 + (n – 1) As′ (kd - d′)2 = 3770 in4

Ig/Icr = 2.8

39



Example 1

Step 05: Effective moment of inertia

Mcr = frIg/yt = [411 10650/11]/12000 = 33.2 ft-kips

a. Under dead load only:

Mcr/ Md = 33.2/30.9 > 1, hence (Ie)d = Ig = 10650 in4

b. Under sustained load:

(Mcr/Msus)3 = (33.2/42.6)3 = 0.473

(Ie)sus = (Mcr/Ma)3Ig + {1 - (Mcr/Ma)

3}Icr ≤ Ig (where Ma= Msus)

= 0.47310650 + (1 – 0.473) 3770 = 7025 in4

c. Under dead + live load:

(Mcr/Md+l)3 = (33.2/54.3)3 = 0.229

(Ie)d+l = (Mcr/Ma)3Ig + {1 – (Mcr/Ma)

3}Icr ≤ Ig (where Ma= Md+l)

= 0.22910650 + (1 – 0.229) 3770 = 5345 in4

40

21



Example 1

Step 06: Initial or short term deflection

(Δi)d = K(5/48)Mdl2/Ec(Ie)d = (1)(5/48)(30.9)(25)2(12)3/(3320)(10650) = 0.098″

K = 1 for simply supported spans

(Δi)sus = K(5/48)Msusl2/Ec(Ie)sus = (1)(5/48)(42.6)(25)2(12)3/(3320)(7025) = 0.205″

(Δi)d+l = K(5/48)Md+ll2/Ec(Ie)d+l = (1)(5/48)(54.3)(25)2(12)3/(3320)(5345) = 0.344″

(Δi)l = (Δi)d+l – (Δi)d = 0.344 – 0.098 = 0.246″

41



Example 1

Step 06: Initial or short term deflection

Allowable deflection: For flat roofs not supporting and not attached to

nonstructural elements likely to be damaged by large deflections:

(Δi)l ≤ l/180

l/180 = 300/180 = 1.67″

(Δi)l =0.246″ < l/180, OK

Allowable deflection: Floors not supporting and not attached to

nonstructural elements likely to be damaged by large deflections:

(Δi)l ≤ l/360

l/360 = 300/360 = 0.83″

(Δi)l = 0.246″ < l/360, OK

42

22



Example 1

Step 07: Long-term deflections at ages 3 months and 5 yrs

Allowable deflection: From Table 24.2.2, the most stringent requirement has

been placed on “Roof and floor construction supporting or attached to

nonstructural elements likely to be damaged by large deflections (sum of the

long-term deflection due to all sustained loads and the immediate deflection

due to any additional live load)= l/480

l/480 = 300/480 = 0.63 inch

43

Duration ξ λ = ξ/(1 + 50ρ′) Δisus Δ (cp +sh)=λΔisus Δil

5 years 2.0 1.77 0.205″ 0.363 0.246″

3 months 1.0 0.89 0.205″ 0.182 0.246″



Example 1

Step 07: Long-term deflections at ages 3 months and 5 yrs.

Which computed deflection should be considered for this case ?

The code says that this should include “Sum of the long-term

deflection due to all sustained loads and the immediate deflection

due to any additional live load”

As the partition will be installed after the immediate deflection due to

dead load has occurred, total deflection which would affect the partition

would include

1. Only long term dead load deflection (not including dead load immediate

deflection)

2. Both Short and long term sustained live load portion

3. Additional live load immediate deflection. This is immediate deflection due to

total live load minus sustained live load

44

23



Example 1


Therefore total deflection for this case would be equal to

λ Δd + Δi sus live + λ Δi sus live+ Δi additional live

(λ Δd + λ Δi sus live ) + ( Δi sus live + Δi additional live )

λ (Δd + Δi sus live ) + Δil total

(Δd + Δi sus live ) is total sustained load and Δil total would be

denoted by Δil

Finally we have

λ (Δisus) + Δil

45



Example 1


The deflection [ λ Δisus + Δil ] is given as follows:

[ λ (Δisus) + Δil ] should be ≤ l/480 ; where l/480 = 300/480 = 0.63″

[ λ (Δisus) + Δil ] = 0.61 < 0.63 ; OK

46

Duration λΔisus Δil λ Δisus + Δil

5 years 0.363 0.246″ 0.61″

3 months 0.182 0.246″ 0.428″

24



Example 2

Find the deflection under full service load for the beam of the hall

shown below.

Data:

fc′ = 3000 psi (normal weight concrete); fy = 40,000 psi

Dead load (wd) = 2.26 kip/ft; Live load (wl) = 0.40 kip/ft

Full service load (ws) = wd + wl = 2.66 kip/ft

Note: No live load is sustained.

47



Example 2

Solution:

Summary of Strength Design of Beam:

According to ACI 9.3.1 {table 9.3.1.1} = hmin = l/16 = 36.9″

Taken h = 5′ = 60″ ; d = h – 3 = 57″ ; bw = 18″ (assumed)

Also, l = 61.5′

Flexural Design = 12 #8 bars

48

25



Example 2

Solution:

Immediate deflection (under full service load):

Δi = K(5/48)Mal2/EcIe

K = 1 (for simply supported beam)

l = 61.5′ = 738″

Md = wdl2/8 =12 (2.26 61.52/8) = 12821 in-kip

Ml = wll2/8 =12 (0.40 61.52/8) = 2269.35 in-kip

Md+l = wd+ll2/8 =12 (2.66 61.52/8) = 15091 in-kip

Ec = 57 √fc′ = 57 √ (3000) = 3122 ksi

49



Example 2

Solution:


For calculation of Ie, various parameters are required:

Note: Effective width as per ACI 6.3.2.1

50

26



Example 2

Solution:


Effective width of T-beam (ACI 6.3.2.1)

This effective width of T-section is different from that used in design

for torsion and direct design method.

51

T - Beam

1 bw + 16h

2 bw + sw

3 bw + ℓn/4

swsw

Least of the above values is selected



Example 2

Solution:


n = Es/Ec = 29000/3122 = 9.289

C = bw/(nAs) = 18/ (9.289 12 0.79) = 0.204

b = 114″ (ACI 6.3.2.1)

f = hf(b – bw)/ (nAs) = 6.54

yt = h – ½ [ {b – bw}hf2 + bwh2]/ [{b – bw}hf + bwh] = 39.39″

Ig = (b–bw)hf3/12+bwh3/12+(b–bw)hf(h–hf/2–yt)

2+bwh(yt–h/2)2 = 599578 in4

Mcr = 7.5 √fc′Ig/yt = 6252 in-kip < Md (therefore section is cracked for dead load)

kd = [ √{C(2d + hff) + (1+f)2} – (1 + f)]/C = 9.05

Icr = (b–bw)hf3/12+bw(kd)3/12+(b – bw)hf(kd – hf/2)2 + nAs(d – kd)2 = 229722 in4

52

27



Example 2

Solution:


For Dead load (Ma = Md = 12821 in-kip):

Ie(d) = (6252/12821)3 599578 + [1–(6252/12821)3] 229722= 272608 in4

For Dead + Live loads (Ma = Md+l = 15091 in-kip):

Ie(d+l) = (6252/15091)3 599578 + [1–(6252/15091)3] 229722= 256030 in4

53



Example 2

Solution:


Deflection due to dead load:

Δi(d)=K(5/48)Mdl2/EcIed = 1(5/48)12821(61.5×12)2/(3122272608)= 0.85″

Deflection due to dead + live loads:

Δi(d+l)=K(5/48)Md+ll2/EcIe(d+l)=1(5/48)15091(61.5×12)2/(3122256030)= 1.07″

Deflection due to live load:

Δil =Δi(d+l) − Δi(d) = 1.07 − 0.85 = 0.22″

Note: In this case, Δi(d) is Δi(sus) , the deflection due to sustained load.

54

28



Example 2

Solution:

Long term deflection (under full service load):

Δ (cp +sh) = λΔ(Δi)sus

Allowable deflection (case 1): With no false ceiling attachments, the case for

“Roof and floor construction supporting or attached to nonstructural elements not

likely to be damaged by large deflections” applies: λΔisus+ Δil ≤ l/240

l/240 = 61.5 ×12/240 = 3.07″

λΔisus+ Δil = 1.92″ ; As 1.92″ < 3.07″, OK

lD =x

1 + 50r r = As/bd

Duration ξ λ Δisus λΔisus Δil λΔisus+ Δil

5 years 2 2 0.85″ 1.7″ 0.22 1.92″

3 months 1 1 0.85″ 0.85″ 0.22 1.07″

55



Example 2

Solution:


Allowable deflection (case 2): With false ceiling attachments, the case for

“Roof and floor construction supporting or attached to nonstructural elements

likely to be damaged by large deflections” applies: λΔisus+ Δil ≤ l/480

l/480 = 61.5 ×12/480 = 1.54″

λΔisus+ Δil = 1.92″ ;

As 1.92″ > 1.54″ , N.G.

(The deflection should not have exceeded 1.54 inch required by ACI code)

56

29



Example 2

Solution:


Now for the beam in question, table below displays the live load deflections

due to various depths of beam (ranging from minimum depth as per ACI

code requirement up to 60″).

From the above table, it is concluded that minimum depth requirements of

ACI does not govern for unusually large values of loading and/or span.

Depth (inches) λΔisus+ Δil (inches) ACI Limitation Remarks

36.9 (hmin of ACI) 5.42 3.07 Not Governing

40 4.60 3.07 Not Governing

50 2.91 3.07 OK

60 1.92 3.07 OK

57



Section 2: Cracking in RC Members

58

30



Topics Addressed

Crack Formation

Reasons for Crack Width Control

Parameters Affecting Crack Width

ACI Code Provisions for Crack Control

Maximum Spacing Requirements

Crack Control Reinforcement in Deep Flexural Members

Example 1

Equations for Maximum Crack Width

Example 2

59



Crack Formation

All RC beams crack, generally starting at loads well below service

level, and possibly even prior to loading due to restrained shrinkage.

In a well designed beam, flexural cracks are fine, so-called hairline

cracks, almost invisible to a casual observer, and they permit little if

any corrosion to the reinforcement.

60

31



Crack Formation

As loads are gradually increased above the cracking load, both

the number and width of cracks increase, and at service load

level a maximum width of crack of about 0.016 inch (0.40 mm) is

typical.

If loads are further increased, crack widths increase further,

although the number of cracks do not increase substantially.

61



Crack Formation

In ACI code prior to 1995, the limitation on crack width for interior

and exterior exposure was 0.016 and 0.013 inch respectively.

Research in later years has shown that corrosion is not clearly

correlated with surface crack widths in the range normally found

with reinforcement stresses at service load levels. For this

reason, the former distinction between interior and exterior

exposure has been eliminated in later codes.

62

32



Crack Formation

Therefore, the limiting value of crack width both for interior and

exterior exposures is now taken as 0.016 inch.

63



Crack Formation

Because of complexity of the problem, present methods for

predicting crack widths are based primarily on test observations.

Most equations that have been developed predict the probable

maximum crack width, which usually means that about 90 % of

the crack widths in the member are below the calculated value.

64

33



Reason for Crack Width Control

Appearance is important for concrete exposed to view such as

wall panels.

Corrosion is important for concrete exposed to aggressive

environments.

Water tightness may be required for marine/sanitary

structures.

65




Concrete cover

Experiments have shown that both crack spacing and crack width

are related to the concrete cover distance (dc), measured from

center of the bar to the face of concrete. Increasing the concrete

cover increases the spacing of cracks and also increases crack

width.

66

w1 (> w)

34




Distribution of reinforcement in tension zone of beam

Generally, to control cracking, it is better to use a larger number of

smaller diameter bars to provide the required As than to use the

minimum number of larger bars, and the bars should be well

distributed over the tensile zone of the concrete.

67

Smaller diameter bars distributed

over tension zone

This portion

may crack




Stress in Reinforcement (crack width fsn)

Where fs is the steel stress and “n” is an exponent that varies in the

range of 1.0 to 1.4. For steel stress in the range of practical interest,

say from 20 to 36 ksi, n may be taken equal to 1.0.

Steel stress may be computed based on elastic cracked section

analysis. Alternatively, fs may be taken equal to 0.60fy (ACI 24.3.2.1)

68

35




Bar Deformations

Beams with smooth round bars will display a relatively small number

of wide cracks in service, while beams with bars having proper

surface deformations will show a larger number of very fine, almost

invisible cracks.

69



ACI Code provision for crack control

Crack width is controlled in the ACI Code: (ACI 24.3.2)

s (inches) = (540/fs) – 2.5cc OR 12(36/fs) (whichever is less)

The center-to-center spacing between the bars in a concrete

section shall not exceed “s” as given by the above equation.

70

b

s cc is the clear cover in inches from the nearest surface in tension

to the surface of the flexural tension reinforcement.

fs is the bar stress in ksi under service condition.

36




The stress fs is calculated by dividing the service load moment by

the product of area of reinforcement and the internal moment arm.

71

b

As

d

Asfs

a

n.a

For equilibrium,

Ms = Asfs(d – a/2)

fs = Mn/As(d – a/2)

Where,

Ms = service load

moment




Alternatively, the ACI code permits fs to be taken as 60 percent of

the specified yield strength fy (because seldom will be reinforcing

bars stressed greater than 60 % of fy at service loads).

The condition “fs = 0.60fy“ is for full service load condition. For

loading less than that, fs shall be actually calculated.

72

37



Maximum Spacing Requirement

Maximum spacing requirement corresponding to concrete

cover as per ACI equation {s = (540/fs) – 2.5cc}:

73

For various values of

concrete cover and fs of

24, 36 and 45 ksi, the

maximum center-to-

center spacing between

the reinforcing bars

closest to the surface of

a tension member to

control crack width can

be plotted as shown.

4.5″

7.5″

15″



Crack Control Reinforcement in Deep

Flexural Members

74

For deep flexural members with effective depth d exceeding 36

inches, additional longitudinal skin reinforcement for crack control

must be distributed along the side faces over the full depth of the

flexural tension zone.

ssk ≤ d/6, 12″ or 1000Ab/(d−30)

Ask = As/2

38



Example 1

Figure shows the main flexural reinforcement at mid span for a T

girder in a high rise building that carries a service load moment of

7760 in-kips. The clear cover on the side and bottom of the beam

stem is 2 ¼ inches. Determine if the beam meets the crack control

criteria in the ACI Code.

75



Example 1

Solution:

Since the depth of the web is less than 36 inches, skin

reinforcement is not needed.

To check the bar spacing criteria, the steel stress can be

estimated closely by taking the internal lever arm equal to

distance d – hf/2.

fs = Ms/ {As(d – hf/2)}

= 7760/ {7.9 × (32.25 – 6/2)} = 33.6 ksi

76

39



Example 1

Solution:

Alternately, the ACI code permits using fs = 0.60fy = 0.60 × 60 = 36 ksi.

Therefore, according to ACI 24.3.2, the maximum center-to-center spacing

between reinforcing bars required to control crack width is:

s = (540/fs) – 2.5cc = {540/ (33.6)} – 2.5 × 2.25 = 10.4 inches OR

s = 12 × (36/fs) = 12 × {36/ (0.60 × 60)} = 12″ (10.4 inches governs)

Therefore, required spacing is 10.4″. The provided spacing shall be less than or

equal to 10.4″. From given figure, c/c sprovided = 5.375″. Therefore the crack

control criteria of ACI code is met.

77




As alternate to ACI crack width equation, crack width can be

calculated using following equations. These equations have

been used in deriving the ACI Code equation on crack control.

Gregely and Lutz equation:

w = 0.076βfs(dcA)1/3

Frosch equation:

w = 2000(fs/Es)β√[dc2 + (s/2)2]

78

40




Where,

w = maximum width of crack, thousandth inch

fs =steel stress at loads for which crack width is to be

determined, ksi

Es = modulus of elasticity of steel, ksi

dc = thickness of concrete cover measured from tension face to

center of bar closest to that face, inch

s = maximum bar spacing, inch.

79




β = ratio of distance from tension face to neutral axis to distance

from steel

centroid to neutral axis = e/f

80

f e

41




A = concrete area surrounding one bar which is equal to total

effective tension area of concrete surrounding total reinforcement

divided by number of bars, in2.

A = 2dcg b/n (where n = number of bars)

81



Example 2

Check the maximum crack width of the beam shown below. The

beam is subjected to service load moment of 7760 in kips. The

clear cover on the side and bottom of the beam stem is 2 ¼ inches.

The material strengths are fc′ = 4 ksi and fy = 60 ksi. Use:

Gregely & Lutz equation,

Frosch equation,

82

36″ 32 ¼″

3 ¾″

27″

10 #8

6″

42



Example 2

83

a=3.68″

27″

6″

f = 28.57″e = 32.32″

Solution:

Gregely & Lutz equation:

w = 0.076βfs(dcA)1/3

Determine the location of neutral

axis for the determination of β.

a = Asfs/ (0.85 fc′ b) = (10 0.8)

0.6 × 60/ (0.85 4 27) = 3.13″

c = a/β1 = 3.13/ 0.85 = 3.68″

β = e/f = 32.32/ 28.57 = 1.13



Example 2

Solution:

Gregely & Lutz equation:

w = 0.076βfs(dcA)1/3

Determine fs.

fs = 0.6fy = 0.6 60 = 36 ksi

dcg = 2.75 + 1 = 3.75″

A = 2dcgb/n = 7.527/10 = 20.25 in2

w = 0.0761.1336(2.7520.25)1/3

= 11.80 thousandth inch OR 0.012″ (0.30 mm)

84

27″

2dcg = 7.5″dc = 2.75″

43



Example 2

Solution:

Frosch equation:

w = 2000(fs/Es)β√[dc2 + (s/2)2]

c/c spacing between bars (s) = 5.375″

Es = 29000 ksi

w = 2000(36/29000)1.13√[(2.75)2 +(5.375/2)2]

= 10.78 thousandth inch OR 0.011″ (0.28 mm)

85

27″

s = 5.375″



Section 3: Development & Splices of

Reinforcement

86

44



Topics Addressed

Definition of Development Length

Bond failure

ACI provisions for Development of Tension Reinforcement

ACI provisions for Development of Standard Hook in Tension

Dimensions & Bends for Standard Hooks

Various Scenarios where ldh must be satisfied

Splices of Deformed Bars

87



Definition of Development Length

Length of embedded reinforcement required to develop the design

strength of the reinforcement at critical section.

If the actual length (l) is equal to or greater than the development

length (ld), no premature bond failure occurs.

88

45



Bond Failure

There are two types of Bond Failure

1. Direct pullout of reinforcement: Direct pullout of reinforcement

occurs in members subjected to direct tension.

2. Splitting of concrete: In members subjected to tensile flexural

stresses, the reinforcement causes splitting of concrete as shown.

89



ACI Provision for Development of

Tension Reinforcement

Basic Equation (ACI 25.4.2.3)

For deformed bars or deformed wire, ld shall be:

In which the confinement term (cb + Ktr)/db shall not be taken greater than

2.5.

90

ℓd =3

40

𝑓𝑦

λ 𝑓𝑐′

ΨzΨ𝑒Ψ𝑠

𝑐𝑏 + 𝐾𝑡𝑟𝑑𝑏

𝑑𝑏 (25.4.2.3a)

46



Basic Equation (ACI 25.4.2.3)

• For the calculation of ℓd, modification

factors shall be in accordance with

Table 25.4.2.4.

91







Simplified Equations (ACI 25.4.2.2)

Section 25.4.2.2 recognizes that many current practical construction

cases utilize spacing and cover values along with confining

reinforcement, such as stirrups or ties, that result in a value of (cb +

Ktr)/db of at least 1.5.

For other cases, (cb + Ktr)/db shall be taken equal to 1.0

Based on these values of 1.5 and 1.0, equation (25.4.2.3a) can be

simplified as given on next slide.

92

47





Simplified Equations (ACI 25.4.2.2)

ld determined shall not be less than 12″.

93





94

ld for grades 40 and 60 (αβλ = 1 and fc′ = 3000 psi )

(Clear spacing of bars being developed or spliced not less than db, clear

cover not less than db, and stirrups or ties throughout ld not less than the

code minimum OR Clear spacing of bars being developed or spliced not

less than 2db and clear cover not less than db.)

Bar No Grade 40 Grade 60

#3 12 16

#4 15 22

#5 18 27

#6 22 33

#7 32 48

#8 37 55

#9 41 62

ld for grades 40 and 60

(αβλ = 1 and fc′ = 3000 psi )

(Other cases)

Bar No Grade 40 Grade 60

#3 16 25

#4 22 33

#5 27 41

#6 33 49

#7 48 72

#8 55 82

#9 62 92

48





95

Table: Generalized formulae for development lengths for usual cases of

reinforcement.

αβλ = 1

fc′ = 3000 psi

ld for No. 6 and smaller

bars

(inches)

ld for No. 7 and larger

bars

(inches)

Grade 40 Grade 60 Grade 40 Grade 60

Clear spacing of bars being developed or spliced

not less than db, clear cover not less than db,

and stirrups or ties throughout ld not less than

the code minimum

OR

Clear spacing of bars being developed or spliced

not less than 2db and clear cover not less than

db.

32db 45db 36db 55db

Other Cases 45db 66db 55db 82db





Further reduction in development length (ACI R25.4.2.3)

The development length can be reduced by increasing the term (c +

Ktr)/db in the denominator of equation 25.4.2.3a

“c” is the smaller of either the distance from the center of the bar to the nearest

concrete surface or one-half the center-to-center spacing of the bars being

developed.

Bars with minimum clear cover not less than 2db and minimum clear

spacing not less than 4db and without any confining reinforcement would

have (c + Ktr)/db value of 2.5.

Therefore ld can be further reduced by substituting the value of 2.5

instead of 1.5 in equation 25.4.2.3a

96

49




Standard hook in Tension (ACI 25.4.2)

Development length ldh, in inches, for deformed bars in tension

terminating in a standard hook shall be the greater of (a) through (c).

97



For the calculation of ℓdh,

modification factors shall be in

accordance with Table 25.4.3.2.

Factors ψc and ψr shall be

permitted to be taken as 1.0. At

discontinuous ends of members,

25.4.3.3 shall apply.

98



50




Standard hook in Tension (ACI 12.5)

For bars being developed by a standard hook at discontinuous ends of

members with both side cover and top (or bottom) cover to hook less than 2-

1/2 in., (a) through (c) shall be satisfied:

(a) The hook shall be enclosed along ℓdh within ties or stirrups perpendicular to ℓdh

at s ≤ 3db

(b) The first tie or stirrup shall enclose the bent portion of the hook within 2db of the

outside of the bend

(c) ψr shall be taken as 1.0 in calculating ℓdh in accordance with 25.4.3.1(a)

where db is the nominal diameter of the hooked bar.

99





100

ldh (inches) for grades 40 and 60(αβλ = 1 and fc′ = 3000 psi )

ldh = (fy/50√fc′)db

Bar No. Grade 40 Grade 60

#3 6 8

#4 7 11

#5 9 14

#6 11 16

#7 13 19

#8 15 22

#9 16 25

51





101

ldh (inches) for grades 40 and 60 (generalized formulae)

Grade 40 Grade 60

16db 22db



Dimensions and bends for standard

hooks (ACI 25.3.2)

Minimum inside bend diameters for bars used as transverse reinforcement and

standard hooks for bars used to anchor stirrups, ties, hoops, and spirals shall

conform to Table 25.3.2. Standard hooks shall enclose longitudinal reinforcement.

102

52



Various scenarios where ldh

must be

satisfied

Beam Column Joint

103

Development of beam

reinforcement in column

shall be > ldh

Development of column

reinforcement in beam

shall be > ldh

ColumnBeam



Various scenarios where ldh

must be

satisfied

Development of column reinforcement in foundation

104

53




Introduction

Splice means “to join”.

In general, reinforcing bars are stocked by supplier in

lengths upto 60′. For this reason, and because it is often

more convenient to work with shorter bar lengths, it is

frequently necessary to splice bars.

Splices in the reinforcement at points of maximum stress

should be avoided.

Splices should be staggered.

105




Types

Bar splicing can be done in three ways:

Lap Splice

Mechanical Splice

Welded Splice

106

54




Lap Splice:

Splices for #11 bars and smaller are usually made simply lapping the

bars by a sufficient distance to transfer stress by bond from one bar

to the other.

The lapped bars are usually placed in contact and lightly wired so

that they stay in position as the concrete is placed.

According to ACI 12.14.2.3, bars spliced by noncontact lap splices in

flexural members shall not be spaced transversely farther apart than

one-fifth the required lap splice length, nor 6 inches.

107




Lap Splice:

According to ACI 25.5.2.1, minimum length of lap for tension lap

splices shall be as required for Class A or B splice, but not less

than 12 inches, where:

Class A splice ................................................... 1.0ld

Class B splice ................................................... 1.3ld

Where ld is as per in ACI 25.4 (discussed earlier).

108

55




Lap Splice:

Lap splices in general must be class B splices according to ACI

25.5.2.1, except that class A splice is allowed when the area of the

reinforcement provided is at least twice that required by analysis

over the entire length of the splice and when ½ or less of the total

reinforcement is spliced within the required lap length.

The effect of these requirements is to encourage designers to

locate splices away from regions of maximum stress to a location

where the actual steel area is at least twice that required by

analysis and to stagger splices

109




Lap Splice:

According to ACI 25.5.5.2, tension lap splices shall not be used for

bars larger than #11 (Because of lack of adequate experimental

data on lap splices of No. 14 and No. 18 bars).

110

56




Mechanical Splice:

In this method of splicing, the bars in direct

contact are mechanically connected through

sleeves or other similar devices.

According to ACI 25.5.7.1, a full mechanical

splice shall develop in tension or compression,

as required, at least 125 percent of specified

yield strength fy of the bar.

This ensures that the overloaded spliced bar

would fail by ductile yielding in the region away

from the splice, rather than at the splice where

brittle failure is likely.

111




Welded Splice:

Splicing may be accomplished by welding in which bars in direct

contact are welded so that the stresses are transferred by weld

rather than bond.

According to ACI 25.5.7.1, A full welded splice shall develop at

least 125 percent of the specified yield strength fy of the bar.

This is for the same reason as discussed for mechanical splices.

112

57




Lap splice location

The splicing should be avoided in the critical locations, such as at

the maximum bending moment locations and at the shear critical

locations.

113



References

Control of Deflection in RC Structures, Reported by ACI

Committee 435

ACI 318-14

Design of Concrete Structures by Nilson, Darwin and Dolan

[Chapter 5, 6].

114

58



The End

115

lecture-05 serviceability requirements & development of

Documents