Pavement Design - Rigid Pavement Design.pdf

Pavement Design - Rigid Pavement Design
Design of rigid pavement

1. Steps for the design of CC pavement thickness as per the IRC guidelines
The design parameters and design approach mentioned above are to be followed.
The wheel load stresses due to dual wheel load assembly of single axle and tandem
axles applied near the edge of the pavement are determined using the set of charts
given in the
IRC Gui<l:elines. Some of these stress charts are given in Fig. 7 .21 to 7 .29.
Suggested steps for the thickness design
of rigid pavement are given below. ,
(a) The various design parameters such as design axle load (98
th
percentile load),
temperature differential
of the locality, K-value of the supporting layers and
the relevant properties of cement concrete are noted
1
(b) The spacing between the longitudinal joints and transverse contraction joints
are decided
( c) A trial thiclmess
of pavement is ~ssumed keeping in view the magnitude of
design axle load and th~ locality where the pavement is to be constructed ..
( d) Using the stress chart the edge load stress due to the design axle load and the
stress ratio are determined; if the stress ratio is a little lower than 0.44, the
trial thickness is tentatively accepted;
if not another suitable trial thickness is
assumed and the calculations are repeated until the requirement is fulfilled
''"

2. 478
DESIGN OF HIGHWAY PAVEMENT
( e) Number of repetition of different axle. loads _ of ~agni~de hi~~er than the
design axle load expected
dming the design penod 1s computed/estimated
( f) Stres~es due to the different single arid
tandem axle_ loads of n:agnitudes higher
than the design axle load
on the pavement of tentatively finalised thickness
are
determined using the edge
load stress charts
(g) The stress ratios, the expected number
of repetitions of the respective loads
during the design life
and the cumulative damage due to fatigue-life consumed
are tabulated. Fatiaue life consumed
or cumulative fatigue damage is total of
(C/N) ratios
or :EC/N (see Column 7 of Table 7. 7)
(h)
If the cumulative fatigue damage exceeds 1.0, hig~er
tri_al thickness is assumed
and the above exercise
of determination of cumulative fatigue ,damage
is
repeated until this value is slightly less than 1.0
( i) The warping stress Ste
at the · edge of the slab of selected thickness . is
determined, consid~ring the highest temperature differ-ential
at the locality
(j) Let the edge load stress due to the highest magnitude
of load Pm expected
during the design life
be = Sem and the total flexural stress
due to warping and
the highest load
be = (Ste + St:m), If the total stress is less than the flexural
strength, the design thickness is aGcepted.
If this exceeds the flexural str'ength
the trial is repeated
by assuming a still higher
pavement thickness until this
requirement is also fulfilled
(k) The final pavement thickness is checked for
comer load stress Sc.
({) After designing the pavement thickness requirement,
the next requirement is
design of (a) dowel bar system at the expansion
joints and (b) tie bars along
_the longitudinal joints.
The design details of dowel b~rs and tie bars are given
in subsequent paragraphs .
The loads based on the axle
load distribution studies may be multiplied by a safety
factor
of 1.2, to take care of further possible overloading. ' - ·
Table 7.7 Stress ratios and fatigue analysis
~
Axle load Axle
Edge load
Stress
Expected No.
· Fatigue life
stress, Se
Fatigue
of repetitions
type load, t 2 ratio, Se/Sr life, N during desig~
consumed,
kg/cm C/N
life, C
. 0) (2) (3) (4) (5) (6) (7)
Single
Pi Se1 Se1/Sr N1 C1 C1/N1
P2 Se2 Se2/Sr N2
C2 C2/N2
P3 St3 Se3/Sr N3
C3 C3/N3
-- -- ---- -~ -- ----
Tandem
Ps Ses Ses/Sr
Ns Cs Cs/Ns
-- -- ---- -- ------ -
Total
.tCFN .
Example 7 .20
From the analysis of axle load distribution studies
it is found that the 9g'h

3. RIGID PAVEMENT DESIGN METHODS
479
ercentile load
0
~
sin~le _axles is 12 _t and that on tandem axles is 20 t._ The number of
P ·er loaded v_e 4hicles e
stl
rnated dunng the design life of 30 years are (a) on single
0'° · · f 4 '
xtes: 32 x l O repetitions
O 14
t load and 8 x 10 repetitions of 16 t load and (b) on
:;ndent axles: I
8
x
104
repetitio~ of 24 I load and 3 x 10
4 repetitions of 28 t load.
'[be contact_ p~essure is
7
-
5
kg/cm • The spacing between the longitudinal joints and
contraction Jomts are 3.5 m and 4.2 m respectively.
The maximum tem~era~e differential on concrete pavements of thickness 20, 25
and 30 cm in the location 1s found t? be 1 ~-6, 17 .4 and 18.0 °c. The supporting layer
consists of GSB a
nd
DLC layers with ~timated modulus K value of 30 kg/cm 3• The
flexural strength of concrete is 45 kg/cm .
Design the ~ickness of CC pavement as per IRC Guidelines, using the edge load
stress charts (Figs. 7.21 to 7.29) and Bardbury's warping stress coefficient chart (Fig.
7.
20). Assume other data suitably.
Solution
Design load on single axle = 12 t
Design load on tandem axles = 20 t
Assume a trial thickness, h
1= 24 cm
Edge load stress:
From the stress chart (Fig. 7.23}, corresponding to h = 24 and K = 30 kg/cm 3, edge
load stress dtte to single axle load of 12 t, Se= 17 kg/cm
2
.
From stress chart (Fig. 7.27}, corresponding to h = 24 and K = 30 kg/cm3, edge
2
load stress due to total load of 20 ton tandem axles, Se= 12 kg/cm
Fatigue analysis: .
The stress ratio due to design load on single axle= l 7 /45 = 0.38, less than 0.44 and
so the design load is safe under possible fatigue failure.
Stress due to load of 14 ton single axle from stress chart (Fig. 7.24) = 18.5 kg/cm2
Stress due to load of 16 ton single axle from stress chart (Fig. 7.25) =21.5 kg/cm
2
Stress due to load of24 ton tandem axles from stress chart (Fig. 7.28) = 13.0 kg/cm
2
Stress due to load of 28 ton t~dem axles from stress chart (Fig. 7.29) = 15.0 kg/cm
2
The fatigue analysis is to be carried out considering t~e magnitude of !oad and
number of repetitions of over loaded vehicles exceeding design axle load on smgle and
tandem axles. The
11
details are given below in tabular form.
Axle Edge load
Stress Fatigue
Expected
no. of Fatigue life
load
Axle
stress,
Se repetitions during consumed,
tyPe
load, t
2 ratio, Se/Sr life, N
design life, C
CIN - kg/cm
_Single
14 18.5 0.41 'Infinite 32 X 10
4
0
,o
8 X 10
4
0.033 -
16 21.5 0.48 2.4 X lQ
Tandem
24 13.0 0.29 Infinite 18 X 10
4
0 -
3 X 10
4
28 15.0 0.33 Infinite 0 --
Total
0.033

4. 480
DESIGN OF 1-llGHW A Y PAVEMENT
Since total fatigue life consumed or cumulative_ fati~e damag~ due ~o axle loads
exceeding design load is
only 0.033 or 3.3 % of f~ttgue hfe ~e d_eSlgn t~ckness of 2
4
cm is safe due to repeated application of loads durmg the design hfe.
Warping stress:
Temperature differential, t for 24 cm pavement thickness (by interpolation) = 17 .24 °c
Length of slab, Lx = 4.2 m = 420 cm, thickness, h = 24 cm, K
3
= 30 kg/cm
Assuming E
= 3 x 10
5
kg/cm
2
,
Poisson's ratio,µ= 0.15
Radius
of relative stiffness, I
= [ Eh3 ]
114
12K(l-µ
2
)
= 58.6 cm (substituting the ~alues)
Corresponding to Lx =
420 cm, Lxf/ = 420/58.6 = 7.17; from
warping stress coefficient
chart (Fig. 7 .20),
Cx = 0.98
Corresponding to
LJI = 350/58.6 = 5.97, Cy= 0.9
Considering higher
of the two values, adopt warping stress coefficient
of 0.98.
Therefore
maximum warping stress at pavement edge,
Ste =
Total stress
Cx Eet = (0.98 x 3 x 10
5
x 1 x 10-
5
x 17.24)/2
2
2
= 25.34 kg/cm
Highest load stress at pavement edge·= 21.5
kg/cm
2
Total (load
stress+ warping stress) at edge = 46.84 kg/cm
2
As the total stress exceeds the flexural strength
of 45 kg/cm
2
,
the design is to be
revised, assuming a higher pavement thickness
in order to fulfil this criterion.
In trial no. 2, assume pavement thickness h =
30 cm
Radius of relative stiffness, (substituting the values) l = [ Eh
3
]
114
= 69.27 cm
· 12K(l-µ
2
)
Highest edge load stress due to single axle load of 16 t for h == 30 cm (from Fig. 7.25)
= 15.3 kg/cm
2
Temperature differential, t for pavement thickness of 30 cm (by interpolation== 18.0 °C
Lxf/ = 420/69. 27 = 6.06, Cx = 0.90
Maximum warping stress at edge = (0.90 x 3 x 10
5
x 1 x 10-s x 18.0)/2 = 24.3 kg/cm
2
Total (load stress + tarping stress) = 15.3 + 24.3 = 39.6 kg/cm2, less than flexural
strength of 45 kg/cm , hence safe.

5. RIGID PAVEMENT DESIGN METHODS
481
T
he factor of safety available for the maximum stress d t th hi h
· -
45139 6 -1 1 ue
O
e g est load and
warping stress - · -• 4, may be accepted
Check for comer load stress, Sc = !~ [ 1-( a -;2 r
2
]
Substituting design load P = 16 / 2 = 8.0 t = 8,000 kg h = 30 / = 6
9 27
3
== J(8000/ 7 .5,r) = 18.43 cm ' cm, · cm,
Substituting, the
values in the equation Sc= 18 4 kg/cm
2
and so th d · · "
. , • e es1gn 1s sa1e.
Adopt design thickness of CC pavement = 30 cm
7.9.4 De~ign of Dowel Bars at Load Transfer Joints
Objectiv~s of dowel bars
As mentioned in Art. 7 .6.3, expansion joints and construction joints are formed as
thro~gh
joints acr~ss ~~ full depth of the slab. A small gap of about 20 mm is
provided
at expans10~ JOmts to allow for expansion of long CC pavement slabs during
s~er seaso~. This gap or joint width helps to relieve the compressive stresses
dunng expansion
and also helps to prevent buckling of the slab near the joint Steel
dowel bars
are embedded at mid depth during construction as specified in the design in
order
to strengthen these weak locations and to provide desired load transfer to the
adjoining
slab across the joint. Though it is not considered very essential to provide
expansion
joints in CC pavements at regular intervals, they should invariably be
provided wh~re the pavement is designed to abut structures like b~dges. /
Functioning
Figure 7.30 (a) illustrates that there is no load transfer across the expansion joint
without dowel bars as each slab deflects independently under the axle load. The
functioning of an expansion joint with dowel bar is illustrated in Fig. 7.30 (b). When
wheel load is placed at the edge of the slab adjoining the expansion joint, a part of the
deflection
and load are transferred across to the adjoining slab with the help of a group
of dowel bars. The load stress diagram in dowel bar is shown in Fig. 7.30 ( c ).
Rounded
steel bars of diameter 25 to 32 mm and length about 500 mm are placed at
intervals of 250 to 300 mm as per the design. About 50 mm less t1ian half lengtli of
. the rounded steel dowel bars are embedded during concreting along one slab, so as to
develop bond with the concrete. The other half length of the dowel bars (plus 50 mm)
are covered with a suitable plastic sheathing to prevent development of bond between
the dowel bars and the concrete of the adjoining slab. This de-bonded half length of
the dowel bars can slide into the adjoining slab. The construction details are given in
Art. 8.5.5 of Chapter 8, 'Highway Construction'.
Stresses in dowel bars
The maximum bearing stress developed in the. dowel bar. and the allowable bearing
stress in cement concrete are to be considered dunng the design.
Maximum bearing stress
The bearing stress between concrete and dowel ba~ depends ~inly on diameter of
the dowel bar and spacing between them. The maximum beanng stress between the
concrete and the dowel bar is given by the equation:

6. 482 DESIGN OF HIGHWAY PAVEMENT
3
Sbm = Mc Pt (2 + Pz)/4(3-EsI
(Eq. 7.32)
where,
Sbm = maximum bearing stress between concrete and dowel bar, kgfcm
2
Mc = modulus of dowel-concrete interaction or the dowel sµppoit, kg/cm
2
/cm ==
41,500 kg/cm
3
P
1
= maximum load transferred by a dowel bar or the load transferred by the first
dowel
bar from the edge, kg
b = diameter ofrounqed dowel bar, cm (2.5 to 3.2 cm)
z = joint width, cm
Es = modulus of elasticity of the steel dowel, kg/cm
2
= 2 x 10
6
kg/cm
2
_
I = moment of inertia of the dowe 1 bar, cm
4
= 7r b
4
I 64 cm
4
p = relative stiffness of the dowel bar embedded in concrete, cm and is given
by: .
J3 = (Mc b/4 EI)°·
25
cm (Eq. 7.331
roERO
I DEFLECTION
--------
-DEFLECTION
dx1
(a) WITHOUT DOWEL BAR
. I
(b) WITH LOAD TRANSFER
DOWEL BAR
(c) LOAD DIAGRAM OF DOWEL BAR
Fig. 7 .30 Functioning of dowel bar
Allowable bearing stress
b
. . tlt of The eanng stress 1m concrete depends upon the ultimate compressive streng
concrete and the diameter of the dowel bar. The allowable bearing stress in concrete,
Fb is determined using the expression given by the American Concrete Institute:
- Fb = Fcs (10.16-b)/9.525 (Eq. 7.3
4
)

7. RIGID PAVEMENT DESIGN METHODS 483
where,
F
b = allowable bearing stress in concrete, kg/cm
2
Fcs = ultimate compressiv~ strength of concrete, kg/cm
2
(for M-40 concrete, j
Fcs = 400 kg/cm
2
)
b = diameter of the qowel bar, cm
Design principle
The diameter of the dowel bar and spacing between them are designed such that a
group of dowel bars can transfer 40 % of the design axle load ( which was considered
for the design of the CC pavement), across the joint to the adjoining slab. The most
unportant factors governing the design of a dowel bar are: (a) the maximum bearing
stress be~een the concre~e and the dowel bar, Sbm vide Eq. 7.32 and (b) allowable
bearing stress in concrete, Fb, vide Eq. 7.34.
Usually the first dowel
bar is placed at a distance of 15 cm from the pavement edge.
The critical loading position for the dowel bar is when the design load is placed directly
above the first dowel bar. Let the maximum load carried by ~e first dowel bar when the
wheel load is directly above the bar= Pt kg. A few of dowel bars that are next to the first'
dowel bar will also share the load transfer, but will carry lower magnitudes of load as the
distance from the first dowel bar increases. The load transfer py the dowel group near the
pavement edge assuming linear variation is illustrated in Fig. 7.31.
The maximum distance
up to which the dowel group can carry part of load has be~
theoretically calculated to be 1.8 times the radius of relative stiflhess, i.e., ( 1.8 [) as shown
in Fig. 7.31. However subsequent performance studies have shown that
due to repeated
load applications and relative movements of the dowel bars with respect the adjoining slabs
( during expansion and contraction
of CC slab), some loosening take~ place in the dowel
system and hence the effective load transfer is only up to ( 1.0
[). As per IRC Guidelines
the distance up to the dowel group action is to be considered in the design is 1.0 /.
A
@ ® ®
I ~-•I• ... ,. •----t
~150 mm'""'~-----
1
-81
------1 ►-I
~1.81-3 s)
1.8 I
EFFECTIVE DOWELS DUE TO LOAD AT EDGE A
Fig. 7.31 Dowel group action at the edge of CC pavement
8t
eps for the design of dowel bars as per IRC guidelines
(1) Properties of CC pavement: (a) design thickness of CC pavement = h cm
(b) design wheel load for dowel bar design, P kg= half of design axle load

8. 484
(2)
(3)
(4)
(5)
(6)
(7)
DESIGN OF HIGHWAY PAVEMENT
on single rear axle <;_onsidered for pavement thickness design = half of.9&'h
percentile load on~ingle axle ( c) elastic modulus of concrete, E = 3 x 105
kg/cm
2
(
d) subgrade modulus = K kg/cm
3
(
e )' radius of relative stiffness of
the CC pavement=/ cm and (0 ultimate compressive strength of concrete
2 2 '
I:cs kg/cm (for M-40 concrete Fcs = 400 kg/cm·)
Total Joad to be sustained
by the dowel group= 0.4P
MaXJ.mum load sustained by the first dowel bar near the edge= Pt kg (to be
' '
determined)
Trial diameter
of dowel bar= b cm (25 to 32 cm) and trial spacing between
dowel bars = s
cm (25 to 30 cm) are assumed
Properties
of dowel bars: (a) Elastic modulus of rounded steel bars, Es= 2
6 2 . . Mk/ 3(
x 10 kg/cm , (b) modulus of dowel-concrete mteraction, c g cm may
be taken as 41,500 kg/cm
3
,(c) joint widt.p., z cm and (d) Moment of inertia
4 4 4
of dowel bar, I cm = ( 7rb /64) cm
' . '
·using Eq. 7.34 and substituting the above data, the allowable bearing stress
in concrete is calculated,
Fb = Fcs (10.16-b) / 9.525
Total load transferred
by the dowel group is calculated m terms of
maximum load carried, Pt' by the edge dowel= Pt { 1 + (/'- s)/l + (/ - 2s)/l +
(/-3s)/l + - - -} = Pt y
(8) Using known value
of design load, P (vide step 1-b), calculate the value of
r~ == (0.4P/y) kg
(9) -Determine the value of relative stiffness of the dowel bar embedded in
· ;concrete, using Eq. 7.33,
p = (Mcb/4 El)
0
·
25
cm
'
( 10) Using Eq 7.32, and substituting the relevant values check for maximum
bearing stress betw~en concrete and the dowel
bar sustaining load Pt, = Sbm
. = McPt (2 + Pz)/4 f3 Esl
( 11)
If the calculated value of Sbm is less than the allowable bearing stress_ in
concrete, Fb calculated in step ( 6) above, the design is safe. Otherwise,. the
trial design is repeated by changing the spacing, s between the dowel bars ,1
or the diameter, b or both the values until the value of Sbm is less than Fb
Example 7. 21
The design thickness of a CC pavement is 26 cm considering a design axle load ( 98
1h
percentile load)
of 12,000 Jc:g on single axle and M-40 concrete with characteristic
compressive strength
of 400 kg/cm
2
.
The radius of relative stiff~1ess i:s found to be 62.2
' ' 6
cm. (data from Example 7.20). If the elastic modulus of dowel bar steel is 2 x 10
kg/cm
2
,
modulus of dowel-concrete interaction is 41,500 kg/ cm
3
and joint width is 1.8
cm, design the dowel bars for 40 % load transfer considering edge loading.
Solution
Given data:
Radius
of relative stiffness of pavement,/ = 62.2 cm, Fcs = 400 kg/cm
2

9. RIGID PAVEMENT DESIGN METHODS
485
Elastic modulus
of dowel bar steel, Es = 2 x 1'06 kg/cm2,
modulus of dowel -concrete interaction, u = 41 500 kg/ J d • • .dth
1 • "'"C , cm an Jomt w1 z = .8 cm
Design load (98t11 percentile axle load) on single axle ~ 12,000 kg
Therefore design wheel load for dowel bar design, p =
6000 kg
Total load to be sustained by dowel g.roup
= 0 ,4 p = 0 ,4 X 6000
= 2400 kg
Let the maximum load s~stained by the first dowel bar near the edge= P1 kg
Assume diameter of dowel bar b = 3 o cm and sp · b
in the first trial , . acmg etween dowel bars, s = 25 cm
Substituting the relevant values,
Moment
of inertia of dowel bar, I = ( 7rb
4
/64) cm
4 =
3
_
976
cm
4
Allowable bearing stress in conc_rete is calculated using Eq. 7 .34
ie., Fb = Fcs (10.16-b)/9,525 = 300.7 kg/cm
2
T~tal ~oad transferre~ by the dowel group for the assumed dowel diameter and
spacmg m terms of maximum load carried,
Pt = P1 {l +(/-s)//+(/-2s)//+(/-3s)//+---}
= Pt Y = Pt {l + (62.2-25)/62.2 + (62. 2-50)/62.2} = 1.794 P
1
= 2400
Pt = 2400/1.794 = 1337.8 kg
Relative stiffness
of the dowel bar embedded · in concrete, p = (Mcb/4 El) o.2
5
cm
(substituting the relevant values from above)= 0.
25
Check for maximum bearing stress between concrete and the dowel bar sustaining
load Pt(vide Eq. 7.32), Sbm = McPt (2 + Pz)/4 p
3
E
5I = (substituting the relevant
. 2
values from above)= 273.7 kg/cm .
As the maximum bearing stress between concrete and the dowel bar (273.7 kg/en/)
is less than the allowable bearing strength in concrete (300.7 kg/cm
2
)
the dowel bar
design is safe and therefore may be accepted. Rounded dowel bars of diameter 3.0 cm
and spacing 25 cm may. be provided at expansion joints.
(Note: In case it is necessary to reduce some quantity of steel, the trial may be repeated assuming
a higher spacing of 27 or 30 cm)
7.9.5 Design of Tie Bars at Longitudinal Joints
Objectives of tie bars
Tie bars are used across the longitudinal joints of cement concrete p_avements. Tie
bars are embedded at mid depth during concreting ensure th~ two adJacent sl~bs on
either side of the longitudinal joint to remain firmly to~et~er a~d to_ prev~nt opem~g up
oftheJ·o· t S F. 7 17 and 7.32. The ]ongitudinalJomts with tie bars act as lunges
m . ee ig. . Tl . b t
and help to relieve part of warping stresses in CC pavement. 1e tie ars are no
designed to act as load tran~_fer devices.

10. 486
DESIGN OF HIGHWAY PAVEMENT
Tie bars are thus designed to withstand tensile stresses, the
maximum tensile force
in tie bars being
equal to the force required to overcome frictional force between the
bottom
of the pavement slab and the base course. The force is estimated between the
longitudinal
joint under consideration and the adjacent joint or free edge.
I-
z
0
w
-,
w
C)
~ Cl
0 <(
0
w z w
w 0
w
w
::,
w
a:
I-
a:
u. <3 u.
z
0
~
bm bm
~v-
PLAN
~-TIE BARS OF LENGTH ½ _i_
=i--=======[=3,--E----_------ihmm
--------FRICTION FORCE
CROSS SECTION
Fig. 7.32 Functioning of tie bars
Area of cross section steel tie bars
-T
The weight per ~etre length of the pavement slab of unit weight W kg/m
3
,
width B
m and thickness h cm= B h W /100 kg.
If the friction coefficient between the bottom of the slab and the base course is f,
the frictional force developed per m length to pull the CC pavement slab =
(B h W t) /100 kg.
The total frictional force per m length is to be resisted by the steel tie bars of cross
section area, As
cm
2
..
If the permissible tensile stress in steel is S
5 kg/cm
2
,
the total
force in tie bars
per m length = A
5 S
5

11. r1terefore,
RIGID PAVEMENT DESIGN METHODS
As Ss = BhWf/100
487
As = BhWf/100S
5 (Eq. 7.35)
In the Eq. 7.35, the allowable working stress in steel tie bars may be taken as, S
5 =
2 · · h 3
12
50 kg/cm , umt weig t of CC, W = 2400 kg/m and maximum value of friction
ffi
cient, f = 1.5.
coe
Assuming a suitable diameter, d of tie bar ( 12 or 16 mm), the cross section area per
•e
bar and the number of tie bars per m length to provide a total area of As and the
:~quired spacing can be determined.
Length of tie bar
The bond strength developed along the periphery of each tie bar up to the
embedded length on each side of the concrete slab has to be equal to or more tha.n the
tensile force developed in the tie bar. Therefore the total length, Lt of tie bar should be
at }east twice the embedded length.
Let the total length of tie bar = Lt cm, the bond stress developed = Sb kg/cm
2
,
diameter = d cm.
The
embeddec:l peripheral area of a tie bar on one slab = ( 11' d Lt/2)
Total bond force developed in each half
of the tie bar = (Sb 11' d Lt)/2
The tensile force developed in each tie bar of diameter b '. cm = (Ss 11' d2)/4
The tensile force developed in each tie bar may be equated to the bond force
developed on each embedded half length of the tie bar.
2
i.e., (Ss 11' d )/4 = (Sb 11' d. Lt)/2
Therefore, minimum length of tie bar,
Lt
= dSs/2 Sb
The recommended values are:
Allowable tensile stress in plain tie bar,
Ss
Allowable tensile stress in deformed tie bar, Ss
Allowable bond stress in plain bars
Allowable bond stress in deformed bars
2
= 1250 kg/cm
,
= 2000 kg/cm-
2
= 17.5 kg/cm
2
= 24.6 kg/cm
Recommended dimeQsions and spacing of tie bars/or longit11dinaljoints
(Eq. 7.36)
Th . d d d
. t r
ofti·e bars for CC pavements of thickness more than 25 e recommen e 1ame e
cm are 12 and 16 mm; the minimum length of deformed bars ~re 64 and 80 cm
respectivel . the maximum spacing between tie bars for slabs of thicknes~ 25 and 30
.
yh, f 60 t 100 cm However if rounded bars are used, higher length
cm are m t e range o o · .
of tie bars '-':'ith closer spacin~ are reqmred.
Example 7 .22 .
, h thickness of 26' cm and lane width of 3.5 m.
A cement concrete pavement as a . . . b
1
Design the tie bars along the longitudinal jomts usmg the data given e ow.

12. 488 DESIGN OF HIGHWAY PAVEMENT
Allowable working stress in steel tie bars, Ss -1250 kg/cm
2
Unit weight of CC, W -2400 kg/m
3
Maximum value of friction coefficient, f -1.2
Allowable tensile stress in deformed tie bar, S
5
- 2000 kg/cm
2
. ~ 2
Allowable bond stress in deformed bars, Sb -24.6 kg/cm
Solution
Given data: B = 3.5 m, h = 26 cm, W = 2400 kg/cm
2
,
f= 1.2, S
5 = 2000 kg/cm
2
Total area of steel tie bars perm length of longitudinal joint is given by Eq. 7.35
As= BhWf/I00S
5
/
Substituting the values, As · l .3 lcm
2
Use 12 mm diameter deformed bars as tie qars, of cross section area= 0.95 cm
2
1
I
No. of tie bars perm length = 1.31/0.95 = 1.38
I
Spacing of 12 mm diameter tie bars = 100/1.38 = 72.5 cm
I
2 ·-2
d = 1.2 cm, S
5 = 2000 kg/cm , Sb = 24.6 kg/cin
Using
Eq. 7 .3 6 Minimurp length of ti~' ~ar, Lt = d. Ss/2 Sb , ·
Substituting the values,
Lt = 48.8 cm
The~efore provide 12 mm diameter tie bars of length 55 to 60 cm at a spacing of 70
cm.