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CHAPTER 2
Bridge Engineering Concept
2.1 INTRODUCTION
In the early development of highways there were very few of structures built
for crossing the roads. Basically, the roads were built to follow the contour of the
ground. In the design of the road, no visibility in vertical and horizontal direction was
considered. When the roads encounter the water, the bridge was constructed. At the
end of the bridges, the level of the road was raised. Since there were no heavy
equipment to carry the components of the bridges, the bridges were light and The
bridges were narrow so that they can be handled by man-power.
The early steel bridge was designed for short span (about 10 m). The type of
the bridges was truss bridge. The truss members were made of steel angles and plates
and they are connected with rivet. The rived angles and the plates were used to support
the floor beams. Often time timber stringers were used for floors.
Since the advancement of construction techniques and the availability of
heavy equipment and also the need of wider and stronger bridges to carry heavy loads,
the bridge engineers started to develop different type of bridges. From the short span
bridges until the long span of bridges. The steel, concrete and prestressed concrete
bridges are the most popular types of bridges. In this note, we will discuss the loads
and the steel bridges.
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2.2 BRIDGE COMPOSITION AND TERMS
Broadly speaking, we can divide the bridges into three parts. The first part isthe super-structure. This part directly supports the loads that come from the
automobiles, trains, pedestrian and the weight of the structure included the pavement.
The second part is the sub-structure. The sub-structure supports the supper structure
and transfers the loads to the ground. The sub-structure can be abutments, piers,
columns and the foundation. The third part is the bearings. The super-structure and the
sub-structure are connected with the bearings. The bearing is a small part of the bridge
structure. However, it may cause a big problem if the bearing does not function
properly particularly after a certain period of time. Schematically, the longitudinal
section of the bridge is shown in Figure 2.1.
Several components of super-structure are as follows:
• Bridge floor . The bridge floor can be made of concrete floor, metal deck
floor or timer floor. This floor directly supports the wheel load of the
automobiles, own weight of the floor slab and the finishing such as,
pavement.
•
The beam system. The beam system can be made of cross beams and
longitudinal beams. The function of the beam system is to transmit the
loads from the floor to the main structure system. The steel plate girder,
box girder, truss structure can considered as the main girder.
•
Cross beams and cross-frames. The cross-beams are constructed to prevent the bending in horizontal direction due to horizontal loads such as
wind or earthquake loads.
•
Support bearings. The support bearings are installed at the contact points
from the super-structure to the sub-structure and used to transmit the loads
from the super-structure to the sub-structure.
Several components of the sub-structure are as follows:
•
Abutments or piers. The abutments and piers are used to transmit to the
loads from the supper structure to the foundations. For multi-span bridges,
the abutments are constructed at the two ends of the bridges and the piers
are constructed between the two ends. The abutments are also used to
support the lateral loads from the soil.
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Clearance
spaneffective
spanstructuresub -
pilesdriven
bearing
length
spaneffective
AbutmentPier
structuresuper
Caison
pilesBored
Figure 2.1: The general cross section
• Foundation. The foundations are used to transmit the loads from the
abutments or piers to the ground. Several types of foundation are
commonly used such as, driven piles, bored piles, caisson etc. The choice
of the type of foundation depends on the soil investigation and the
topography of the area.
Several types of bearings are as follows:
•
Fixed bearings. These bearings are used to transmit the horizontal loads
from the super-structure such as wind loads, earthquake loads, impact
loads etc.
• Movable bearings. The movable bearings are used to accommodate the
possible deformation of the structure such as temperature, shrinkage,
creep, the movement of the foundation etc.
2.3 CLASSIFICATION OF BRIDGES
In the early the development of highways there are a few of crossings are
required to connect between two separate locations. However, since the availability of
heavy equipments and high strength materials several types of bridges were designed
to face the challenge to the need for wider and longer bridges. These days not only the
design but also the construction will play an important role in the decision to choose
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the type of bridges that suitable to a certain condition. Based on several different
criterias, we can make classification of bridges.
2.3.1 CLASSIFICATION OF BRIDGES BASED ON THE MATERIAL
Based on the material used in the super structure the bridges can be classified
as follows:
•
Wooden bridges. The wooden bridge is constructed for short span and
light loads.
•
Stone bridges. The stone-bridge is also constructed for short span and
usually it is constructed with consideration the aesthetic.
•
Steel bridges. The steel bridges are popular in the remote area where thematerial is difficult. It is suitable for short to medium span. The
advantageous of steel bridge for short span are easy to construct and the
quality can be controlled easily.
• Concrete bridges. The concrete bridges are the most popular for
interchanges. The advantageous of concrete bridges are cheaper
particularly for short span and the shape is more flexible than the steel
bridges. However, the dead load of concrete bridges is heavier than the
steel bridges.
• Composite steel bridges. In the composite steel bridge, the concrete slab is
an integral part of the superstructure beams. This type of bridge is popularsince the availability of high strength shear connectors to resist the
horizontal shear between the concrete deck and the steel.
• Aluminum bridges. These type bridges are not popular and very seldom
used in the design.
2.3.2 CLASSIFICATION OF BRIDGES BASED ON THE SHAPE
The bridges based on the shape can be classified as follows:
• Girder bridges. Broadly, these bridges can be classified into concrete
girder bridges and steel girder bridges (plate girder bridges). Most of the
girder bridges are short span bridges. The advantageous of plate girder
bridges and prestressed I girder bridges that they can be fabricated in the
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shops. This type of bridges is commonly used in the highway
intersections.
•
Truss bridges. When the longer span is required, the truss bridges become
the option. The truss bridges more economically for the medium span
however, it is difficult to give a good shape to truss bridges. In Indonesia
particularly, the truss bridges are used to cross small rivers.
• Arch bridges. This type of bridge is occasionally used in the construction.
It has a good shape but the construction is more difficult than the truss or
girder bridges.
• Rigid frame bridges
• Cable stayed bridges
• Suspension bridges
The basic concept of the bridge design is to lay the support system (main girders)
parallel to the road alignment and support the vertical loads. The girder, truss and rigid
frame bridges are popular in Indonesia. One cable stayed bridges constructed in Batam
which is part of Barelang project.
2.4 SURVEYS AND STUDIES
There are many parameters involved before we make the decision of
constructing a bridge. Several surveys and studies have to be conducted to support our
decision. Broadly speaking there are three parameters have to be studied before thedecision is made. The first parameter is called the feasibility study. In this parameter
we evaluate the advantageous and disadvantageous of constructing the bridges. In this
study we will determine whether this bridge is worthwhile to be constructed unless we
have no option. The second parameter, is the design. In the design, we will study the
types of bridges that can support the feasibility studies. The third parameter is the
construction. This is one of the most important aspects in the design of a bridge. For
instance, we design a bridge across a deep valley where there is no possibility of
constructing the middle support, the method of construction becomes a dominant
parameter in making the decision. Eventhough in this studies several different
disciplines are involved, but they have to work together as a team and the results
related each other. When the decision has been made that we are planing to construct
the bridge the next step is to conduct a survey. In order to give a guidance in the
survey, several surveys that can be considered are tabulated in Table (2.1)
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Table 2.1 Types of surveys
TYPE OF SURVEY CONTENT OF SURVEY PURPOSE OF SURVEY
1. Topographical
survey
Drawing up topographical map Selection of bridge position,
length and spacing.
2. Geological survey Collection of geological and soil
history data.
Drawing up geological map and
physical properties
Same as the above.
Selection of position and
Structural plan of substructure
work.
3.
Intersecting roadsetc.
Condition and future plans forwidth, altitude, clearance limit,
structure crossing, longitudinal
grade etc. of intersecting roads
and railways.
Underground installations.
Selection of bridge length,span spacing, clearance,
construction method.
4. Rivers survey Rivers cross section.
Volume and speed of flow.
High and low water level and
river gradient.
Navigating vessels.
Selection of span spacing.
Clearance, pier forms,
foundation, sunken depth of
pier.
Determination of overflow
and prevention section.
Determination of construction
method
Determination of impact load.
5. Survey of sea,
lakes and
marches.
Tidal level, wave height, tide.
Navigating vessels.
Selection of estimated high-
water level.
Selection of water pressure,
construction period and
method.Determination of span
spacing, height under girder,
impact load.
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TYPE OF SURVEY CONTENT OF SURVEY PURPOSE OF SURVEY
6. Soil survey. Boring.
Standard penetration test.
Soil test.
Test pit.
Plate bearing test.
Pressiometer test
Measurement of ground water
level.
Selection bearing strata for
use in substructure design.
Determination of allowable
bearing capacity.
Determination of unit weight
of soil, internal friction angle
and viscosity.
Determination of quantity of
settlement due to consoli-
dation.
Selection of constructionmethod.
7. Seismological
survey
Records of earthquakes and
earthquake damages.
Measurement of ground
microtremor.
Confirmation of bedrock.
Determination of design seis-
mic design coefficient.
8. Meteorological
survey
Survey of meteorological
observation records (wind
velocity, temperature, snow and
climate)
Determination of temperature
variation, wind load and snow
load.
Selection of protective
covering, materials, construc-
tion period and method.
9. Survey of
additions to
bridge.
Underground installation at
bridge position.
Plans for construction of
waterpipes, sewers, electricity
and telegraph wires.
Determination of measure-
ment and weight of addition.
10.
Corrosion. Extent of corrosion of existing
structures.
Organic matter, PH.
Exposure test for paints.
Selection of material to be
used.
Determination of rust preven-
tion methods.
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TYPE OF SURVEY CONTENT OF SURVEY PURPOSE OF SURVEY
11.
Materials survey. Aggregate and water (quality and
quantity) for concrete.
Selection and calculation of
materials.
12. Construction. Situation of material transporta-
tion routes.
Determination of maximum
transportation load.
Road plans for construction
works.
(JICA, 1977). The purpose of this survey is to give more detail information for
selection the type of the bridge. However, not all surveys have to be conducted but itdepends on the scale of the bridges.
2.5 SELECTION THE BRIDGE TYPE (SUPER-STRUCTURE)
Besides the information from the survey, several considerations in selecting
the type of the supper-structure of the bridge can be listed as follows:
a) The span length. The large portion of the forces in the design of long span
bridges comes from the weight of the structure itself. Usually, the weight
of the concrete structure is much heavier than the steel structure.
b)
The purpose of constructing the bridge. For instance, the type of the
bridge over the road (intersection) might be totally different with the type
of the bridge over the river or deep valley eventhough the span of the
bridge is the same.
c) The soil condition. When it is difficult to find a good bearing strata, we
might consider the light bridge.
d) The layout and the alignment of the bridge. When the layout makes an
angle (skewed bridge) or the curve bridge, special attention should be
given in the design.
e)
The availability of the material. In some areas crush stone is not available
or it is difficult to deliver good quality of concrete, the steel structure may
become a good option.
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2.6 SELECTION SUB-STRUCTURE
The sub-structure can be divided into two parts. The first part is the structureto the resist directly the loads from the upper structure such as abutments or piers.
Several considerations in the selection of this part:
• The location of the substructure. Mostly, the reinforced concrete structure
is used to support the super-structure. Unless for some reasons such as
concreting is not possible, the steel structure will be used.
• Within the rivers, the flow of the water during the normal seasons and the
flood. During flood, the direction of the flow of water might be different
from the normal flow. For normal flow of water the oval might be a good
shape. However, where there is a potential flood and the direction of the
flow of water is uncertain, the circular might be a good shape.
• The connections between the piers and foundation. It depends on the
connection of the super-structure and the first part of the substructure. The
stability of the piers has to be considered.
The second part of the sub-structure is foundation. The foundation is part of the
structure below the ground. The function of the foundation mainly is to transmit the
loads from the substructure to the foundation. Basically, the considerations in making
the selection of the foundation are based on the bearing stratum. The types of
foundation can be listed as follows:
• The bearing stratum within 5 m from the ground. When the bearing
stratum is near the surface the direct foundation can be considered.
• The bearing stratum between 5 – 20 m from the ground, the concrete piles
can be used.
• The bearing stratum over 30 m, the steel piles or the bored piles can be
used.
2.7 COMPOSITE GIRDER BRIDGE
The composite girder bridges can be constructed between prestressed concrete
girder and reinforced concrete slab or between steel girder and reinforced concrete
slab. When the composite construction between steel beam and concrete slab is
considered, creep factor should be considered. In designing the composite beams, there
are four types of loadings have to be considered.
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a) The first type is called long-term loading. Basically the dead load is considered as
long-term loading. When the dead load is working on the steel section, the creep
of the concrete is not considered. However, when the dead load is working on the
composite section such as pavement, curbs and railings the creep of the concrete
has to be considered.
b) The second type is called short-term loading. The live load is considered as short-
term loading. There is no need to consider creep of slab due to live load.
c)
The third type is impact loading. The span contributes to this load and basically
the impact load is considered for live load.
d) The third type is called temporary loading. Earthquake load and wind load are the
temporary loads.
It is important to note that the composite girder is designed based on the assumptionthat the slab and the girder act monolithically. There is no sliding between the steel
and concrete. In the actual construction, there is a stress in concrete after concrete is
setting. This phenomenon has to be considered in the design. Finally, the temperature
difference between the steel and the concrete also contributes to the additional stresses
in the composite section. The stress computation due to the above loads will be
discussed in the examples.
2.7.1 LOADS
The composite girders are designed to resist the loads imposed on the girders
and transfer those loads to the substructure. Based on the sequence of construction, the
loads imposed on the composite girders can be categorized into two categories. The
first category is the loads during construction. Mostly, these loads are acting on the
steel section where the composite action has not been developed. The second category
is the load after construction. These loads are acting after the composite action has
been developed. Based on the duration, the loads can be divided into two categories.
The first category is called permanent loads. The gravity loads are considered as
permanent loads. The second category is called temporary loads. The wind and
earthquake loads are considered as temporary loads. When the creep is considered, the
loads can be divided into two types. The first type is called long-term loading. The
dead loads are considered as long-term loading. The second type is short time loading.
The live loads can be considered as short-time loading. The detail of this load will be
discussed in the examples. Besides these loads, there are many other loads that may
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occur on the structure as shown in Kapsel (Soegiarso, 1999). In this class we will
study the composite girders due to dead and live loads.
a) Dead Loads
Several examples of the dead load in composite girders are as follows:
• The dead load due to the concrete floor or steel floor.
• The dead load due to the weight of the structure such as girders, bracings,
truss structure, gusset plates etc.
• The loads due to the wearing surface such as, pavement, handrails and
curbs.
• Other loads that may added due to the circumstances such as the electric
poles, catwalk etc.
• The loads during construction such as construction equipment.
b)
Live loads
• The loads due to moving vehicles (automobiles, carts, trains). The
magnitude and the distribution of these loads should conform to the
specifications for highway bridges.
• The loads due to pedestrian.
c) Impact coefficient
Since the live loads are the fast moving vehicles, a certain impact allowance due to
live loads has to be considered. The structure elements that are affected to this impact
allowance will discussed in the subsequent section.
2.8 AMERICAN ASSOCIATION OF STATE HIGHWAY AND TRANSPORTATION
OFFICIALS (AASHTO)
2.8.1 Live loads
In designing the bridge components loads imposed on the structure have to be
regulated in the codes. There are many codes that are available and each code isusually used in a particular country. In order to give a reference how the loads are
imposed, in this section we will discuss the AASHTO loads.
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Basically in the design of the bridge, there are two types of loadings have to be
considered. The first type is called truck loading and the second type is the lane
loading. Which ever governs between these two loadings will dictate the design of the
bridge. Based on the intensity of the loads, these two types of loadings (truck loading
and lane loading) can be further divided into two types. The first type is called H-
loadings and the second type is called HS-loadings. The HS loadings are heavier than
the H loadings. Based on the truck axle loads, H loadings are divided into two classes
i.e., H15-44 and H20-44. The affixing 44 indicates the year when the code was
instituted which is in 1944. Similarly, HS loadings are divided into two classes i.e., the
HS20-44 and HS15-44. Hence, there are four standard classes of loading: H15, H20,
HS15 and HS20. The difference between H15 and H20 or HS 15 and HS 20 is the
magnitude of the loads. The loading of H20 is 75 percent of HS20. The magnitude and
the intensity of the lane loads are shown in Figure 2.2. The magnitude of truck loads
for H20-44, H15-44 are shown in Figure 2.3 and the magnitude of truck loads for
HS20-44 and HS15-44 are shown in Figure 2.4.
For continuous span, in computing the maximum negative bending moment,
besides the concentrated load shown in Figure 2.2, an additional concentrated load is
provided. These two loads are placed as such to produce the maximum negative
bending moment.
In AASHTO specifications, the truck loading is assumed to occupy a width of
10 feet ( ≈ 3.0 m) and these loads shall be placed in a 10-foot wide design traffic lanesas shown in Figure 2.2.
2.8.2 Impact coefficient
In considering the impact allowance, the bridge structure is classified into two
groups i.e., group A and group B. Only the structure in group A, the forces obtained
from the live loads have to be multiplied with the impact coefficients. The impact
allowances shall not be applied to the structure in group B. The following are the
structures categorized in groups A and B.
Structure in group A (AASHTO 3.8.1.1)
(1)
Superstructure, including steel or concrete supporting columns, steel towers, legs
of rigid frames, and generally those portions of the structure that extend down to
the main foundation.
(2) The portion above the ground line of concrete or steel piles that are rigidly
connected to the superstructure as in rigid frame or continuous structure.
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P
load ed Concentrat
shear for lbs000,26
momentfor lbs000,18 *
laneload of footlinear lbs640load Uniform
Loading4420HS
Loading4420H
−
−)a
P
load ed Concentrat
shear for lbs500,19
momentfor lbs500,13 *
laneload of footlinear lbs640load Uniform
Loading4415HS
Loading4415H
−
−) b
* For the loading of continuous span involving lane loading refer to Article 3.11.3
which provides for an additional concentrated load.
Figure 2.2: The lane loading
Structure in Group B (AASHTO, 3.8.1.2)
(1) Abutments, retaining walls, piers, piles, except Group A(2)
(2) Foundation pressures and footings
(3)
Timber structures.
(4) Sidewalk loads.
(5) Culverts and structures having 3 feet or more cover.
3.8.2 Impact Formula
3.8.2.1 The amount of the impact allowance or increment is expressed as a fraction of
the live load stress, and shall be determined by the formula:
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125L
50I
+=
where I is impact fraction (maximum 30 percent) and L is length in feet of the portion
of the span that is loaded to produce the maximum stress in the member.
2.9 DESIGN EXAMPLE CONCRETE BRIDGES
In these design examples, we focus our discussion on the concrete part of the
bridges. The first is the design of the concrete slab. And the second is the design of the
slab bridges. The continuous slab bridge is adopted in the second example where the
influence lines have to be drawn in obtaining the maximum moments at particular
cross section.
2.9.1 Bridge slab
In order to be familiar to those loads, let consider the following example. A
multi-stringer composite steel bridge deck has the cross section shown in Figure 2.3.
The slab is continuos over the stringers and half width of flanges is assumed to be 0.15
m. The yield stress of the reinforcement is 24 MPa and the concrete strength is 30
MPa. The deck is cast reinforced concrete slab. The reinforcement is perpendicular to
the traffic. Design the bridge is based on Standard Specifications for Highway Bridges
(AASHTO, 1983) and the specified loading of HS20-44. In the AASHTO
specifications, all formulas are based on US customary units. The future units in theUSA will be metric units.
m40.2 m40.2 m40.2
m90.0 m90.0m20.7
surfacewearingm/kg150 2 m25.0
h
'm/kg50
Figure 2.3: The cross section
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Solution:
a)
Compute the bending moment of the slab.
In order to compute the maximum bending moment in the slab, we have to identify
direction of the reinforcement with respect to the traffic. In AASHTO specification it
is regulated in the following formulas:
The maximum bending moment based on AASHTO specifications 3.24.3.1.
Case A- Main reinforcement perpendicular to traffic (spans 2 to 24 feet inclusive)
HS 20 loading:
=
+
20P32
2s Moment in foot-pounds per foot-width of slab.
HS 15 loading:
=
+15P
32
2s Moment in foot-pounds per foot-width of slab.
In slabs continuous over three or more supports, a continuity of 0.80 shall be applied
to the above formulas for both positive and negative moment. In accurate results can
be obtained from accurate analysis such as using finite element method.
b) Compute impact coefficient
( )30.0
125L
50I ≤+=
In order to use the formulas, the units have to be converted.
c) The thickness of slab
30
10sd
+= (Table 8.9.2)
where s is the span length in feet according to section 8.8 (clear span plus the depth of
the member).
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m40.2
d
)L(spanlearc c
Figure 2.4: Effective span of concrete slab
d Ls c +=
Let assume the span s = 7.5 feet.
"96.6'58.030
105.730
10sd ≈=+=+≥ . Let consider the thickness of the slab is 7.5”
d) The moment due to the dead load:
Own weight = 150.0x12
5.7 = 0.094 Ksf
Weight of overlay (assumed) = 0.030 Ksf
Total DL = 0.124 Ksf
ft/Kft698.08
5.7x124.0x80.0
MM
2
DLDL ===
−+
e) Moment due to lane load
Type of loading HS20-44, the wheel load 20P = 16 Kips (Article 3.24.3)
The main reinforcement is perpendicular to the traffic (Article 3.24.3.1)
( )ft/Kft75.4ft/lbft750,4000,16x
32
25.7P
32
2sM 20 ==
+=
+=
The slab is continuous, the bending moment is multiplied by 0.8 for both positive nad
negative moment (Article 3.24.3.1)
ft/Kft8.375.4x80.0MM LLLL === −+ (Non factored)
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f) Impact coefficient and moment (Article 3.8.2.1)
Impact: 377.01255.7
505.12s
50I =+=+=. The maximum impact fraction is 0.3.
Impact moment:
ft/Kft14.180.3x30.0MM II === −+
Factored moment: AASHTO 3.22.1
( )[ ]ILD.MM LD +β+βγ== −+ (Article 3.22, Eq. (3.10))
( )[ ]
( )[ ] ft/Kft63.1114.18.367.1698.03.1
MM67.1M3.1MM ILD
=++=
++== −+
Effective depth (d):
The effective depth can be computed as follows (Figure 2.5):
)clearance(c
hd
2. d steel diaonDistributi
1d entreinforcemMain
)clearance(c
hd
2. d steel diaonDistributi
1d entreinforcemMain
a) Positive moment b) Negative moment
Figure 2.5: Effective depth for positive moment
Positive bending moment
clearanced 5.0hd 1 −−=
Let assume the clearance is 1 in. and 1d is"75.0 , then the effective depth becomes,
( ) "125.6175.05.05.7d """ =−−=
Negative bending moment:
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For negative bending moment we may consider the wearing and particularly when
snow is considered the clearance must be thicker than the area where snow is not
considered. This is due to the salt that can penetrate at the negative moment area. If we
consider the bridge on the snowy area then the effective depth becomes,
wearingclearanced d 5.0hd 21 −−−−=
Let assume the clearance and the wearing is "5.2 , 1d is"75.0 and 2d is "5.0 , then
the effective depth becomes,
( ) "125.45.25.075.05.05.7d """" =−−−=
g) Design reinforcement
In order to make self assessment to the code, let review the following
equations. From Figure 2.6 we obtain the following:
h) To compute bρ
ys b'c f ATand baf 85.0C == (2.1)
ba
'cf 85.0 b
d 2/ad b−
bC
003.0c =ε
T
C
yε cε
sA
Figure 2.6: Stress block diagram
From equilibrium we obtain, bf 85.0
f Aa
'c
ys
b = (2.2)
ba is the balance depth equivalent rectangular stress block.
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b1 b Ca β= (2.3)
where from ACI/318R, Article 10.2.7.3 and AASHTO, Article 8.16.2.7,
≥=β
≥
−−=β
≤=β
psii8000f for 65.0
ps4000f for 1000
4000f 05.085.0
psi4000f for 85.0
'c1
'c
'c
1
'c1
(2.4)
Let assume psi000,000,29E s =
000,29/f E/f yysyy =ε→=ε (2.5)
From Figure 2.6 we obtain,
d
C b
yc
c =ε+ε
ε (2.6)
The maximum usable strain at extreme concrete compression fiber shall be assume
equal to 0.003 (ACI/318R-95, Article10.2.4)
Substitute Eq. (2.5) into Eq. (2.6) yields,
( )d f 700,8700,8
Cf 700,8
700,8
d
C
y b
y
b
+=→+= (2.7)
( )d f 700,8
700,8a
y
1 b +β= (2.8)
From Figure (2.26), the equation for the balance failure is as follows:
y bys b'c bdf f A baf 85.0 ρ== (2.9)
( ) by
'c
b ad f
f 85.0=ρ (2.10)
Substitute Eqs. (2.8) into Eq. (2.10) we obtain,
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( )
+
β
=ρ
y
1
y
'c
b
f 000,87
000,87
f
f 85.0 (2.11)
The maximum and minimum reinforcement:
−≤ρ=ρ
−ρ=ρ
)1.5.10Article,108R 318/ACI( f
200 and
f
f 3
)3.3.10Article,108R 318/ACI(75.0
y
min
y
'c
min
bmax
(2.12)
i) To compute
Let consider Figure 2.6, where ba is replaced by a.
( )2/ad Cor TM n −= (2.13)
( )
−=
bf 85.02
f Ad f AM
'c
ys
ysn (2.14)
ρ=→=ρ bd A bd
As
s (2.15)
From Eqs. (2.14) and (2.15) yield,
( )
ρ−ρ=
bf 85.02
bdf d bdf M '
c
yyn (2.16)
( ) bf 85.02
f d bf bd M
'c
22y
22
y2
n
ρ−ρ= (2.17)
0 bd
Mf
f 7.1
f
2
ny'
c
22y =+ρ−ρ
(2.18)
φ=→=φ ununM
MMM (2.19)
Substitute Eq. (2.20) into Eq. (2.18) yields,
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0 bd
Mf
f 7.1
f
2
uy'
c
22y =
φ+ρ−
ρ (2.20)
Let2
uu
bd
MR φ= (2.21)
'c
2y
u'c
2y2
yy
f 7.1
f 2
R f 7.1
f 4f f
−±
=ρ
−−==ρ
'c
u
y
'c
f 85.0R 211
f f 85.0 (2.22)
j) Reinforcement due to moment positive:
"125.6d and ft/'K 63.11M ==+
From Eq. (2.21),
( )( )
( )( )( )344.0
125.6129.0
1263.11R
2u ==
Substitute uR into Eq. (2.22) yields,
( ) ( )( )
00602.05.485.0
344.0211
000,60
500,485.0=
−−==ρ
From Eq. (2.11) we obtain,
( )( )03018.0
000,60000,87
000,87
000,60
500,4825.085.0 b =+=ρ
( ) 0226.003018.075.0max ==ρ
OK 00602.0 max →ρ〈=ρ
( )( ) 2s in442.0125.61200602.0A ==+
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Moment negative:
"
125.4d and ft/'K 63.11M ==+
From Eq. (2.21),
( )( )
( )( )( )759.0
125.4129.0
1263.11R
2u ==
Substitute uR into Eq. (2.22) yields,
( ) ( )( ) max
0142.05.485.0
759.0211
000,60
500,485.0ρ〈=
−−==ρ
( )( ) 2s in705.0125.4120142.0A ==−
k) Reinforcement due to moment negative
The reinforcement should be able to carry 1.2 cr M (AASHTO, Article 8.17.1)
Where cr M is the cracking moment.
t
gr
cr y
If M = (AASHTO, Article 8.13.3)
where r f is the modulus of rupture , gI is the moment of inertia of the concrete
section about the centroidal axis, neglecting the reinforcement,t
y is the distance
from the centroidal axis to the extreme fiber. (AASHTO, Article 8.15.2.1)
psi503500,45.7f r == (AASHTO, Article 8.15.2.1.1)
( )"K 58.56
5.75.0
12/)5.7x12(503M
3
cr ==
( ) "K 67.958.562.1M2.1 cr ==
Check for bottom steel ( 2s in442.0A =+ )
( ) ( )( )( )"
'c
ys 578.0125.485.0
60442.0 bf 85.0
f Aa ===
( )2/ad f AM ysn −φ=φ
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( )( )( ) cr '
n MK 29.1392/578.0125.660442.09.0M 〉=−=φ
Check for top steel ( 2s in705.0A =− )
( )( )
( )( )"
'c
ys922.0
125.485.0
60705.0
bf 85.0
f Aa ===
( )2/ad f AM ysn −φ=φ
( )( )( ) cr '
n MK 49.1392/922.0125.460705.09.0M 〉=−=φ
l) Check Cracking
For psi000,40f y 〉
Ad f z cs= (ACI/318R, Article 10.6.4)
where,
armlinternaisd ,dA
MMMf
s
ILDs αα
++= , cd is the distance from the centroid of the
rebars to the outer surface of the slab and A is the area of concrete in tension zone.
Positive bending moment:
( ) "375.1175.05.0d ""
c =+= Let assume the bar spacing is "8 , the total area of concrete surrounding the bar is
A = 1.375 x 2 x 8 = 2in0.22
Let assume 87.0d =α
( )( )( )
72.284461.0125.687.0
1214.18.3698.0f s =
++= Ksi
( ) ( )( ) Ksi14549.890.22375.172.28z 3 〈==
Negative bending moment:
( ) "375.35.25.075.05.0d """ =++= Let assume the bar spacing is "7 , the total area of concrete surrounding the bar is
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A = 3.375 x 2 x 7.0 = 2in25.47
( )( )( )
74.26705.0125.487.0
1214.18.3698.0f s =++= Ksi
( ) ( )( ) Ksi1450.14525.47375.374.26z 3 〉==
2.9.2 Continuous bridge
This example is a continuous concrete slab bridge with three spans as shown
in Figure 2.7. The first span length is 24 feet, the second span (middle span) is 32 feet
and the third span is 24 feet. The concrete is the normal weight concrete with the
ultimate compressive strength psi500,4f 'c = . The live loads are based on HS20-44. Inthis example we compute the following:
'24 '24'32
h
Figure 2.7: Longitudinal section
a)
Draw the maximum and minimum bending moment diagrams and maximumand minimum shear diagrams.
b) Design the slabs due to the positive and negative bending moments.
c) Check the cracks requirements and the fatigue stress limits.
Solution:
The computation is based on one-way slab with the main reinforcement is parallel to
the traffic.
1. The minimum thickness of the slab:
542.030
10sh ≥
+= (AASHTO, Table 8.9.2)
where s is the clear span plus the depth of the member (AASHTO, Article 8.8).
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Let assume the clear span is the 36 feet.
542.04.130
1032h ' ≥=+= . Let h = "17
2. Loads
2.1 Dead loads:
Own weight of the slab: 0.15 x 1.40 x 1 = 0.21 K/ft
Weight of asphalt is assumed = 0.03 K/ft
Weight of the curb is assumed = 0.02 K/ft
Total dead load = 0.26 K/ft
2.2 Live loads:2.2.1 Truck loads
The wheel loads of HS20-44 are shown in Figure 2.8. These wheel loads are
distributed over the slab. The width of slab over which the wheel load is
distributed (E). When the main reinforcement is parallel to the traffic the width E
is computed based on AASHTO, Article 3.24.3.20
E = (4 + 0.06S) feet7≤ where S is the span length
( )
( )
≤=+=→=
≤=+=→=
feet792.53206.04E32S
feet744.52406.04E24S
''
''
'14 '' 3014 −
K 4 K 16 K 16
E
load Wheel
Figure 2.8: The wheel loads of HS20-44
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2.2.2 Lane loads
Qfor lbs000,26
Mfor lbs000,18ft/lbs640
The width of slab over which the wheel load is distributed (2E)
The main reinforcement parallel to the traffic (AASHTO, Article 3.24.3.20
E = (4 + 0.06S) feet7≤
2.3 Impact loads
≤=+=→=
≤=+=→=
3.031.012532
50I32S
3.033.012524
50I24S
'
'
2.4 Influence lines
The influence line is based on the configuration of the supports as shown in Figure 2.9.
This configuration is totally set in a paper with unanimous reference. This is only used
for study. Later on in this notes a computer program is developed to compute the
maximum will be developed.
0 1 122 3 4 5 6 7 8 9 10 11 13 14 15 16 17 18 19 20
L 1.333L L
Figure 2.9: The segments of the bridge
The ratio of middle span length to the first and the third span length is 1.333. The bridge is divided into 20 segments and a unit load is placed from point 1 to 19
successively. Plots of influence diagrams are shown in Figures 2.10-2.12. Based on the
influence line tables, the ordinate in Figures 2.10-2.12 have to be multiplied by L/99
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where L is the length of the span. Basically, there are two types of loading have to be
considered. The first type is the truck loading either HS15 or HS20. Only one truck
load is considered for the whole bridge. Therefore, for long span bridge, most probably
the truck loading may not govern. The second type is lane load. In producing the
maximum bending moment, for continuous span, the concentrated loads may be
applied to different spans (AASHTO specifications Article 3.11.3). For maximum
positive moment, only one concentrated load shall be used with uniform loads that
produce the maximum moment.
The bending moments and shear forces at particular section can be obtained by
taking the algebraic sum of the influence line ordinates and multiplied by the length of
each segment. It is important to note that for bending moment only the influence line
ordinates have to be multiplied by L/99. In order to illustrate the computation of
bending moment due to lane load, let consider the influence line of section 1 as shownin Figure 2.9. The algebraic sum of the influence line ordinates between sections 1
through 6 is as follows:
55.3647.163.333.650.972.12T =++++=
( )( )( )96/24d TArea = where d is the distance between points.
( )( )( ) 65.3299/24455.36Area ==
The maximum bending moments and shear forces at each section are as follows:
Table 2.2: Ordinates for influence diagram for bending moments
Moment Unit Load at point
at point 1 2 3 4 5 6 7 8 9
1 12.72 9.50 6.35 3.63 1.47 0.00 -1.18 -1.98 -2.07
2 9.44 19.00 12.70 7.27 2.93 0.00 -2.37 -3.97 -4.13
3 5.90 12.00 19.05 10.90 4.40 0.00 -3.55 -5.95 -6.20
4 2.37 5.00 8.90 14.50 5.87 0.00 -4.73 -7.93 -8.27
5 -1.17 -2.00 -1.25 1.70 7.08 0.00 -5.92 -9.92 -10.30
6 -4.70 -9.00 -11.40 -11.20 -7.70 0.00 -7.10 -11.90 -12.40
7 -3.90 -7.55 -9.56 -9.39 -6.45 0.00 7.84 1.17 -1.738 -3.10 -6.10 -7.73 -7.58 -5.20 0.00 6.28 14.20 8.95
9 -2.30 -4.65 -5.89 -5.76 -3.95 0.00 4.71 10.80 19.60
10 -1.65 -3.20 -4.05 -3.95 -2.70 0.00 3.15 7.35 13.80
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Table 2.2- continued
Moment Unit Load at point
at point 10 11 12 13 14 15 16 17 18 19
1 -1.93 -1.58 -1.07 -0.52 0.00 0.38 0.55 0.55 0.43 0.23
2 -3.87 -3.17 -2.13 -1.03 0.00 0.77 1.10 1.10 0.87 0.47
3 -5.80 -4.75 -3.20 -1.55 0.00 1.15 1.65 1.65 1.30 0.70
4 -7.73 -6.33 -4.27 -2.07 0.00 1.53 2.20 2.20 1.73 0.93
5 -9.67 -7.92 -5.33 -2.58 0.00 1.92 2.75 2.75 2.17 1.17
6 -11.60 -9.50 -6.40 -3.10 0.00 2.30 3.30 3.30 2.60 1.40
7 -3.35 -3.67 -2.96 -1.54 0.00 1.05 1.49 1.46 1.15 0.64
8 4.90 2.15 0.47 0.03 0.00 -0.20 -0.33 -0.38 -0.30 -0.13
9 13.20 6.97 3.92 1.59 0.00 -1.45 -2.14 -2.21 -1.75 -0.89
10 21.40 13.80 7.35 3.15 0.00 -2.70 -3.95 -4.05 -3.20 -1.65
The bending moment at each point is obtained by multiplying the ordinate of the
particular point in Table 2.2 by L/99, where L is the length based on Figure 2.9
Table 2.3: Ordinates for influence diagram for shears
Moment Unit Load at point
at point 1 2 3 4 5 6 7 8 91 0.786 0.576 0.385 0.220 0.089 0.00 -0.072 -0.111 -0.215
2 -0.214 0.576 0.385 0.220 0.089 0.00 -0.072 -0.111 -0.215
3 -0.214 -0.424 0.385 0.220 0.089 0.00 -0.072 -0.111 -0.125
4 -0.214 -0.424 -0.615 0.220 0.089 0.00 -0.072 -0.111 -0.125
5 -0.214 -0.424 -0.615 -0.780 0.089 0.00 -0.072 -0.111 -0.125
6L -0.214 -0.424 -0.615 -0.780 -0.911 0.00 -0.072 -0.111 -0.125
7 0.046 0.088 0.111 0.108 0.076 0.00 0.905 0.785 0.647
8 0.046 0.088 0.111 0.108 0.076 0.00 -0.095 0.785 0.647
9 0.046 0.088 0.111 0.108 0.076 0.00 -0.095 -0.215 0.647
10 0.046 0.088 0.111 0.108 0.076 0.00 -0.095 -0.215 -0.352
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Table 2.3- continued
Shear Unit Load at point
at point 10 11 12 13 14 15 16 17 18 19
1 -0.117 -0.096 -0.065 -0.031 0.00 0.023 0.033 0.033 0.026 0.014
2 -0.117 -0.096 -0.065 -0.031 0.00 0.023 0.033 0.033 0.026 0.014
3 -0.117 -0.096 -0.065 -0.031 0.00 0.023 0.033 0.033 0.026 0.014
4 -0.117 -0.096 -0.065 -0.031 0.00 0.023 0.033 0.033 0.026 0.014
5 -0.117 -0.096 -0.065 -0.031 0.00 0.023 0.033 0.033 0.026 0.014
6L -0.117 -0.096 -0.065 -0.031 0.00 0.023 0.033 0.033 0.026 -0.046
7 0.500 0.353 0.215 0.095 0.00 -0.076 -0.108 -0.111 -0.088 -0.046
8 0.500 0.353 0.215 0.095 0.00 -0.076 -0.108 -0.111 -0.088 -0.046
9 0.500 0.353 0.215 0.095 0.00 -0.076 -0.108 -0.111 -0.088 -0.046
10 0.500 0.353 0.215 0.095 0.00 -0.076 -0.108 -0.111 -0.088 -0.046
In the concrete design for bridges, the shear force may not be crucial in the
design. But in steel beam, the shear or reaction at the support is carried by the web. In
addition, in the design of shear connectors to resist fatigue, the range of shear has to be
computed. The range of shear is the difference between minimum and maximum shear
envelopes due to live load. The shear force at each point is obtained by multiplying the
ordinate of the particular point in Table 2.3 by the concentrated load. By computing
the maximum and minimum shears at particular section, the range of shear can be
obtained.
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0 1 122 3 4 5 6 7 8 9 10 11 13 14 15 16 17 18 19 20
- 0 .
2 1
4
0 .
7 8 0
0 .
0 8 9
- 0 .
0 7
2
0 .
0 2 3
0 .
0 3 3
0 .
0 3 3
0 .
0 2 6
0 .
0 1 4
- 0 .
4 2
4
- 0 .
6 1
5
4sectionfordiagramInfluence
0 .
2 2
0
- 0 .
1 1
1
- 0 .
1 2
5
- 0 .
1 1
7
- 0 .
0 9
6
- 0 .
0 6
5
- 0 .
0 3
1
- 0 .
2 1 4
0 .
7 8 0
0 .
0 8 9
- 0 .
0 7 2
0 .
0 2 3
0 .
0 3 3
0 .
0 3 3
0 .
0 2 6
0 .
0 1 4
- 0 .
4 2 4
- 0 .
6 1 5 5sectionfordiagramInfluence
- 0 .
9 1 1
- 0 .
1 1 1
- 0 .
1 2 5
- 0 .
1 1 7
- 0 .
0 9 6
- 0 .
0 6 5
- 0 .
0 3 1
- 0 .
2 1 4
- 0 .
7 8 0
- 0 .
0 7 2
0 .
0 2 3
0 .
0 3 3
0 .
0 3 3
0 .
0 2 6
0 .
0 1 4
- 0 .
4 2 4
- 0 .
6 1 5
- 0 .
9 1 1
- 0 .
1 1 1
- 0 .
1 2 5
- 0 .
1 1 7
- 0 .
0 9 6
- 0 .
0 6 5
- 0 .
0 3 1
- 1 .
0 0
0 . 9
0 5
- 0 . 0
7 6
7sectionfordiagramInfluence
- 0 .
0 9 5
0 .
7 8 5
0 .
6 4 7
0 .
5 0 0
0 .
3 5 3
0 .
2 1 5
0 .
0 9 5
- 0 . 1
0 8
- 0 . 1
1 1
- 0 . 0
8 8
- 0 . 0
4 6
- 0 .
2 1 5
- 0 .
0 7 6
8sectionfordiagramInfluence
- 0 .
0 9 5
0 .
7 8 5
0 .
6 4 7
0 .
5 0 0
0 .
3 5 3
0 .
2 1 5
0 .
0 9 5
- 0 .
1 0 8
- 0 .
1 1 1
- 0 .
0 8 8
- 0 .
0 4 6
- 0 .
3 5 3
- 0 .
0 7 6
9sectionfordiagramInfluence
- 0 .
0 9 5
- 0 .
2 1 5
0 .
6 4 7
0 .
5 0 0
0 .
3 5 3
0 .
2 1 5
0 .
0 9 5
- 0 .
1 0 8
- 0 .
1 1 1
- 0 .
0 8 8
- 0 .
0 4 6
0.80
6.57
2.47
0.52
9.95
0.18
2.47
0.52
0.52
2.47
13.77
1.72
0.19
12.18
1.72
1.72
8.80
0.81
1.72
1.72
1.96
5.96
1.72
0 .
0 7 6
0 .
1 0 8
0 .
1 1 1
0 .
0 8 8
0 .
0 4 6
0 .
0 7 6
0 .
1 0 8
0 .
1 1 1
0 .
0 8 8
0 .
0 4 6
0 .
0 7 6
0 .
1 0 8
0 .
1 1 1
0 .
0 8 8
0 .
0 4 6
0 .
9 0 5
- 0 .
0 7 6
0 .
7 8 5
0 .
6 4 7
0 .
5 0 0
0 .
3 5 3
0 .
2 1 5
0 .
0 9 5
- 0 .
1 0 8
- 0 .
1 1 1
- 0 .
0 8 8
- 0 .
0 4 6
1.7215.99
1.72 0 .
0 7 6
0 .
1 0 8
0 .
1 1 1
0 .
0 8 8
0 .
0 4 6
6R sectionfordiagram Influence
1 .
0 0
6Lsectionfordiagram Influence
Figure 2.12: Influence diagram for shear
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0 1 122 3 4 5 6 7 8 9 10 11 13 14 15 16 17 18 19 20
- 0 .
3 5 3
- 0 .
0 7 6
10sectionfordiagramInfluence
- 0 .
0 9 5
- 0 .
2 1 5
0 .
5 0 0
0 .
3 5 3
0 .
2 1 5
0 .
0 9 5
- 0 .
1 0 8
- 0 .
1 1 1
- 0 .
0 8 8
- 0 .
0 4 6
- 0 .
5 0 0
1.72
3.65
3.65
1.72
0 .
0 7 6
0 .
1 0 8
0 .
1 1 1
0 .
0 8 8
0 .
0 4 6
Figure 2.12: Influence diagram for shear (-continued)
2.5 Maximum bending moments due to truck loading
The points in the following are refereed to Figure 2.9
• Point 1 of Figure 2.9
( ) ft/K 89.1055.272.1244.5
16
99
24M '=
+=+
( ) ( ) ft/K 10.238.044.5
4
99
2433.198.1
92.5
16
99
24M '−=
+
+−=−
• Point 2
( ) ft/K 59.1446.10.1944.516
9924M '=
+=+
( ) ( ) ft/K 20.477.044.5
4
99
2497.365.2
92.5
16
99
24M '−=
+
+−=−
• Point 3
( ) ft/K 58.13005.1944.5
16
99
24M '=
+=+
( ) ( ) ft/K 30.615.144.5
4
99
2498.395.5
92.5
16
99
24M '−=
+
+−=−
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• Point 4
( ) ft/K 18.11185.15.1444.516
99
24
M'
=
+=
+
( ) ( ) ft/K 40.853.144.5
4
99
243.593.7
92.5
16
99
24M '−=
+
+−=−
• Point 5
( ) ft/K 05.508.744.5
16
99
24M '=
=+
( ) ( ) ft/K 20.1163.692.992.516
0.244.5
4
99
24
M'
−=
++−=
−
• Point 6
( ) ft/K 07.30.23.244.5
16
99
24M '=
+=+
( ) ( ) ( ) ft/K 23.1795.792.5
49.11
92.5
164.11
44.5
16
99
24M '−=
++−=−
• Point 7
( ) ( ) ( ) ft/K 72.515.144.5
453.0
44.5
1684.7
92.5
16
99
24M '=
++=+
( ) ( ) ft/K 12.951.392.5
1656.9
44.5
16
99
24M '−=
+−=−
•
Point 8
( ) ( ) ft/K 13.1020.044.5
4
99
2431.12.14
92.5
16
99
24M '=
−
+=+
( ) ft/K 77.639.610.344.5
16
99
24M '−=
+−=−
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( ) ( )
( ) ( ) ft/K 17.786.35
92.52
64.02.14
92.52
18
99
24M '=+
=+
( ) ( )
( ) ( ) ft/K 87.430.181.28
44.52
64.073.7
44.52
18
99
24M '−=
++−=−
• Point 9
( ) ( )
( ) ( ) ft/K 41.1095.58
92.52
64.06.19
92.52
18
99
24M '=+
=+
( ) ( )
( ) ( ) ft/K 13.418.887.21
44.52
64.089.5
44.52
18
99
24M '−=
++−=−
•
Point 10
( ) ( )
( ) ( ) ft/K 56.1188.67
92.52
64.04.21
92.52
18
99
24M '=+
=+
( ) ( )
( ) ( ) ft/K 40.308.1508.15
44.52
64.005.4
44.52
18
99
24M '−=+−
−=−
2.7 Shear forces due to truck loading
• Point 0
( ) K 83.3303.00.144.5
16Q =+=+
( ) K 41.0031.0118.092.5
16Q −=−−=−
•
Point 1
( ) K 76.2155.0786.044.5
16Q =+=+
( ) ( ) K 95.0214.044.516
118.092.5
16
Q −=−+−=−
• Point 2
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( ) K 84.105.0576.044.5
16Q =+=+
( ) ( ) K 58.1125.092.5
16424.0
44.5
16Q −=−+−=−
• Point 3
( ) K 13.1385.044.5
16Q ==+
( ) ( ) K 15.2125.092.5
16615.0
44.5
16Q −=−+−=−
•
Point 4
( ) K 65.022.044.5
16Q ==+
( ) ( ) ( ) K 66.2048.092.5
4125.0
92.5
1678.0
44.5
16Q −=−++−=−
• Point 5
( ) K 26.0089.044.5
16Q ==+
( ) ( ) K 71.3118.092.5
4324.0911.0
44.5
16Q −=−+−−=−
• Point 6L
( ) K 12.0014.0028.044.5
16Q =+=+
( ) K 11.452.00.192.5
16Q −=−−=−
•
Point 6R
( ) ( ) K 33.41.044.5
4574.000.1
92.5
16Q =++=+
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( ) K 42.0066.0076.044.5
16Q −=−−=−
• Point 7
( ) ( ) K 68.3110.044.5
4427.0905.0
92.5
16Q =++=+
( ) K 42.0066.0076.044.5
16Q −=−−=−
• Point 8
( ) ( )
K 83.244.5
076.04284.0785.0
92.5
16Q =−+=+
( ) ( ) ( ) K 89.0046.044.5
4092.0
44.5
16215.0
92.5
16Q −=−+−+−=−
• Point 9
( ) ( ) K 20.2038.044.5
4155.0647.0
92.5
16Q =++=+
( ) ( ) K 28.1110.044.5
16353.0
92.5
16Q −=−−=−
• Point 10
( ) ( ) ( ) K 71.1111.044.5
16047.0
92.5
4500.0
92.5
16Q =++=+
K 71.1Q =−
2.8 Shear forces due to lane loading
• Point 1
( ) ( )[ ] K 30.2786.0x2652.065.664.0
44.52
1Q =++=+
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( ) ( ) ( )[ ]
( ) ( ) ( )[ ]214.02643.064.0
92.52
1125.02647.264.0
44.52
1Q −+−+−+−=−
K 94.0Q −=−
•
Point 2
( ) ( ) ( )[ ] K 64.1576.02652.093.364.0
44.52
1Q =++=+
( ) ( ) ( )[ ]
( ) ( ) ( )[ ]125.02647.264.0
92.52
1424.02670.164.0
44.52
1Q −+−=−+−=−
K 52.1Q −=−
• Point 3
( ) ( )[ ] K 04.1385.02652.001.264.044.5
1Q =++=+
( ) ( ) ( )[ ]
( ) ( ) ( )[ ]125.02671.264.0
92.52
1615.02678.364.0
44.52
1Q −+−++−=−
K 11.2Q −=−
•
Point 4
( ) ( )[ ] K 60.022.02652.080.064.044.5
1Q =++=+
( ) ( ) ( )[ ]
( ) ( ) ( )[ ] K 66.2125.02647.264.0
92.52
178.02657.664.0
44.52
1Q −=−+−++−=−
• Point 5
( ) ( ) ( )[ ] K 25.0089.02652.018.064.0
44.52
1Q =++=+
( ) ( ) ( )[ ]
( ) ( ) ( )[ ] K 45.3125.02647.264.0
92.52
1911.02695.964.0
44.52
1Q −=−+−++−=−
• Point 6
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( )
( ) ( )[ ] K 11.0033.02652.064.044.52
1Q =+=+
( ) ( ) ( )[ ]
( ) ( ) ( )[ ] K 61.3125.02647.264.0
92.52
100.12677.1364.0
44.52
1Q −=−+−++−=−
•
Point 7
( ) ( )[ ]
( ) ( ) ( )[ ] K .752905.02618.1264.0
92.52
172.164.0
44.52
1Q =++=+
( ) ( ) ( )[ ]
( ) ( ) ( )[ ] K 59.0111.02672.164.0
44.52
1095.02619.064.0
92.52
1Q −=−+−+−−=−
• Point 8
( ) ( )[ ]
( ) ( ) ( )[ ] K 30.2785.0268.864.0
92.52
172.164.0
44.52
1Q =++=+
( ) ( ) ( )[ ]
( ) ( ) ( )[ ] K 88.0215.02681.064.0
92.52
1111.02672.164.0
44.52
1Q −=−−+−−=−
• Point 9
( ) ( )[ ]
( ) ( ) ( )[ ] K 84.1647.02696.564.0
92.52
172.164.0
44.52
1Q =++=+
( ) ( ) ( )[ ]
( ) ( ) ( )[ ] K 25.1353.02696.164.0
92.52
1111.02672.164.0
44.52
1Q −=−+−++−=−
• Point 10
( ) ( )[ ]
( ) ( ) ( )[ ] K 40.150.02665.364.0
92.52
172.164.0
44.52
1Q =++=+
( )
( ) ( )[ ]
( )
( ) ( )[ ] K 66.150.02665.364.092.52
1111.02672.164.0
44.52
1Q −=−+−++−=−
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ft/'K 40.22MIL =
−
+
The ultimate moment based on AASHTO 3.10 and group IA is as follows:
( )[ ]ILDu MM2.2M3.1M ++=
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P o
i n t
I L
A
r e a
P
o s i
t i v e
I L
A r e a
N e g a
t i v e
) f t /
K ( M '
D
) f t /
K ( M
'
m a x
L L
+
) f t /
K ( M
'
m a x
L L
−
) f t /
K ( M
'
L L
I m p a c t
−
) f t /
K ( M
'
L L
I m p a c t
+
)
f
t
/ K ( M '
I
L + +
) f t /
K ( M '
I
L − +
) f t /
K ( M
'
I
L
D +
+
+
) f t /
K ( M
'
I
L
D −
+
+
) f t /
K ( r a n g e
M '
1 2 3 4 5 6 7 8 9 1 0
3 4
. 7 3
5 3
. 9 6
5 6
. 9 2
4 3
. 8 6
1 8
. 5 3
1 2
. 5 1
8 . 3
1
3 5
. 8 6
5 8
. 9 5
6 7
. 8 8
1 0
. 0 2
2 0
. 0 2
3 0
. 0 6
4 0
. 0 8
5 4
. 0 6
5 3
. 9 8
3 0
. 1 1
3 0
. 0 5
3 0
. 1 6
1 0 2
. 7 9
6 . 4
2
8 . 8
2
6 . 9
9
0 . 9
8
- 9 . 2
4
- 2 3
. 4 8
- 1 1
. 8 7
1 . 5
0
7 . 5
1
9 . 8
1
1 0
. 8 9
1 4
. 5 9
1 3
. 5 8
1 1
. 1 8
5 . 0
5
3 . 0
7
5 . 7
2
1 0
. 1 3
1 4
. 2 9
1 5
. 3 1
- 2 . 1
0
- 4 . 2
0
- 6 . 3
0
- 8 . 4
0
- 1 1
. 2 0
- 1 7
. 2 3
- 9 . 1
2
- 6 . 7
7
- 5 . 3
0
- 4 . 6
6
3 . 2
7
4 . 3
8
4 . 0
7
3 . 3
5
1 . 5
2
0 . 9
2
3 . 0
4
1 . 7
2
4 . 2
9
4 . 5
9
- 0 . 6
3
- 1 . 2
6
- 1 . 8
9
- 2 . 5
2
- 3 . 3
6
- 5 . 1
7
- 2 . 7
4
- 2 . 0
3
- 1 . 5
9
- 1 . 4
0
1 4 . 1 6
1 8 . 9 7
1 7 . 6 5
1 4 . 5 3
6 . 5 7
3 . 9 9
7 . 4 4
1 3 . 7 7
1 8 . 5 8
1 9 . 9 0
- 2 . 7
3
- 5 . 4
6
- 8 . 1
9
- 1 0
. 9 2
- 1 4
. 5 6
- 2 2
. 4 0
- 8 . 8
0
- 1 1
. 8 6
- 6 . 8
9
- 6 . 0
6
2 0
. 5 8
2 7
. 7 9
2 4
. 6 4
1 5
. 5 1
- 2 . 6
7
- 1 9
. 4 9
- 4 5
. 8 8
- 2 3
. 7 3
- 7 . 3
0
- 4 . 4
3
1 5
. 2 7
2 6
. 0 9
2 9
. 7 1
3 . 6
9
3 . 3
6
- 1 . 2
0
- 9 . 9
4
- 2 3
. 8 0
0 . 6
2
3 . 7
5
1 6 . 2
9
2 4 . 4
3
2 5
. 8 4
2 5 . 4
5
2 1 . 1
3
2 9 . 3
9
1 9
. 3 0
2 2 . 5
7
2 5 . 4
7
2 5 . 9
6
l o a d
L i v e
a n
d
L o a
d
D e a d
t o
d u e
M o m e n
t
:
2 . 4
T a
b l e
(
)
2 6
. 0 x
A
A
M D L
−
+
−
=
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Table 2.6: The maximum shear forces due to Dead Load and Truck loading
Point +A −A )K (QD
)K (
QL+ )K (
QL− )K (
Q IL++ )K (
Q IL−+ )K (
Q ILD+ ++ )K (
Q ILD− ++
0 10.74 -2.47 2.15 3.83 -0.41 4.98 -0.53 7.13 1.62
1 7.17 -2.90 1.11 2.76 -0.95 3.58 -1.24 4.69 -0.13
2 4.45 -4.17 0.07 1.84 -1.58 2.39 -2.05 2.46 -1.98
3 2.53 -6.49 -1.03 1.13 -2.15 1.47 -2.80 0.44 -3.83
4 1.32 -9.04 -2.00 0.65 -2.66 0.85 -3.46 1.15 -5.46
5 0.70 -12.42 -3.05 0.26 -3.71 0.34 -4.82 -2.71 -7.87
6L 0.52 -16.24 -4.09 0.12 -4.11 0.16 -5.34 -3.93 -9.436R 17.71 -1.72 4.16 4.33 -0.42 5.23 -0.55 9.39 3.74
7 13.9 -1.91 3.12 3.68 -0.42 4.78 -0.55 7.90 2.57
8 10.52 -2.53 2.08 2.83 -0.89 3.67 -1.16 5.75 0.92
9 7.68 -3.68 1.04 2.20 -1.28 2.86 -1.66 3.90 -0.62
10 5.37 -5.37 0.00 1.71 -1.71 2.22 -2.22 2.22 -2.22
Table 2.7: The maximum shear forces due to Dead load and Lane loading
Point +A −A )K (
QD )K (
QL+
)K (
QL−
)K (
Q ILD+
++ )K (
Q ILD−
++
1 7.17 -2.90 1.10 2.30 -0.94 4.09 -0.12
2 4.45 -4.17 0.07 1.64 -1.52 2.20 -1.91
3 2.53 -6.49 -1.03 1.04 -2.11 0.32 -3.77
4 1.32 -9.04 -2.00 0.60 -2.66 -1.22 -5.46
5 0.70 -12.42 -3.05 0.25 -3.45 -2.73 -7.54
6 0.52 -16.24 -4.09 0.11 -3.61 -3.95 -8.78
7 13.9 -1.91 3.12 2.75 -0.59 6.70 2.358 10.52 -2.53 2.08 2.30 -0.88 5.07 0.94
9 7.68 -3.68 1.04 1.84 -1.25 3.43 -0.59
10 5.37 -5.37 0.00 1.40 -1.66 1.82 -2.16
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1 102 3 4 5 6 7 8 90
Load Dead toduemomentBendinga)
42.682.8
99.6
98.0
24.9−
48.23−
87.11−
50.1−
51.7
81.9
1 102 3 4 5 6 7 8 90
58.20
82.827.15
69.3 36.320.1−
94.9−
80.23−
88.45−
73.23−
30.7−
62.0
75.3
79.27
64.24
51.15
67.2−
49.19−
43.4−
09.26
71.29
pactImLLDLtoduemomentBendingMin.&Max. b) ++
Figure 2.13: The maximum and minimum bending moments at each section
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1 102 3 4 5 6 7 8 90
Load Dead todue forcesSheara)
10.1
05.3−00.2−
12.3
16.4
08.2
04.1
0.0
1 102 3 4 5 6 7 8 90
13.0−
35.2
69.4
46.2
32.0
55.1−
93.3−
75.5
90.3
59.0−
16.2−83.3−
46.5−
71.2−
90.7
94.0
43.9−
pactImLLDLtoduemomentBendingMin.&Max. b) ++
07.0
03.1−
09.4−
22.2
87.7−
98.1−
62.1
13.7
74.3
39.9
Figure 2.14:Maximum&Minimum shear forces at each section
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2.12 Moment positive reinforcement:
The maximum positive moment occurs at section 10:
ft/'K 59.4M,ft/'K 31.15M ,ft/'K 8.9M ILLDL === +++
Based on AASHTO 3.22.1 Group 1 and Article 3.5, the following load combination is
applied:
( )[ ]ILLDLu MM67.1M3.1M ++=
( )[ ] ft/'K 94.5559.431.1567.18.93.1M u =++=
There is no centrifugal force in this problem.
The depth (d) = h – clearance - 2/d 1
Let assume the clearance is "1 and 1d is "75.0 .
"625.15"375.0"1"17d =−−=
Considering Eq. (2.21) we obtain,
( )( )
( )( )( )254.0
625.15129.0
1294.55
bd 9.0
MR
22
uu ===
The required ρ from Eq. (2.11),
( ) ( )( )
0044.05.485.0
254.0211
000,60
500,485.0=
−−==ρ
Check the maxρ from Eq. (2.11),
( )( )0311.0
000,60000,87
000,87
000,60
500,4825.085.0 b =+=ρ
( ) 0233.00311.075.0max ==ρ
OK 0044.0 max →ρ〈=ρ
( )( ) 2s in766.05.14120044.0A ==+
Use bar # 6, the spacing between bars is,
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"89.6"12x766.0
44.0=
The #[email protected]”, the total cross sectional area is 0.81 2in
2.13 The cracking control
For psi000,40f y 〉
Ad f z cs= (ACI/318R, Article 10.6.4)
where,
armlinternaisd ,
dA
MMMf
s
ILLDLs αα
++= and cd is the distance from the centroid of
the rebars to the outer surface of the slab and A is the area of concrete in tension zone.
( ) "25.115.05.0d ""c =+=
Let assume the bar spacing is "5.6 , the total area of concrete surrounding the bar is
A = 1.25 x 2 x 6.5 = 2in25.16
Let assume 87.0d =α
( )( )( ) 22.34766.0625.1587.0
1259.43.158.9
f s =
++
= Ksi
( ) ( )( ) Ksi14538.9625.16375.136.32z 3 〈==
2.14 Moment negative reinforcement
The maximum negative moment occurs at section 6:
ft/'K 17.5M,ft/'K 23.17M ,ft/'K 48.23M ILLDL −=−=−= +++
Similarly with positive moment, based on AASHTO 3.22.1 Group 1 and Article 3.5,
the following load combination is applied:
( )[ ]ILLDLu MM67.1M3.1M ++=
( )[ ] ft/'K 15.7917.523.1767.148.233.1M u −=−−+−=
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)clearance("5.2
"17"5.13
Distribution steel dia. 0.5"
Main reinforcement #8
The arrangement of the negative reinforcement is assumed based on the above figure,
Let assume the clearance and the wearing is "5.2 , 1d is "75.0 and 2d is "5.0
"625.13"5.0"375.0"5.20.17d =−−−= ( )( )
( )( )( )47.0
625.13129.0
1215.79
bd 9.0
MR
22
uu ===
The required ρ from Eq. (2.11),
( ) ( )( )
0084.05.485.0
47.0211
000,60
500,485.0=
−−==ρ
The maxρ from Eq. (2.11),
( )( ) 0311.0000,60000,87
000,87
000,60
500,4825.085.0 b =+=ρ
( ) 0233.00311.075.0max ==ρ
OK 0084.0 max →ρ〈=ρ
( )( ) 2s in37.1625.13120084.0A ==−
Use steel bar # 8, the spacing is in9.612x37.1
79.0=
The #[email protected]”, the total cross sectional area is 1.45 2in
2.15 The cracking control
In computing cd the wearing surface thickness is reduced.
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"375.2"5.0"375.0"5.1d c =++=
Let assume the bar spacing is "5.6 , the total area of concrete surrounding the bar is
A = 2.375 x 2 x 6.5 = 2in875.30
Let assume 87.0d =α
( )( )( )
93.2745.1625.1587.0
1217.523.1748.23f s =
++= Ksi
( ) ( )( ) Ksi14594.130875.30375.393.27z 3 〈==
2.16 Minimum steel requirement
The reinforcement should be able to carry 1.2 cr M (AASHTO, Article 8.17.1)
Where cr M is the cracking moment.
t
gr
cr y
If M = (AASHTO, Article 8.13.3)
where r f is the modulus of rupture , gI is the moment of inertia of the concrete
section about the centroidal axis, neglecting the reinforcement, ty is the distance
from the centroidal axis to the extreme fiber. (AASHTO, Article 8.15.2.1)
psi503500,45.7f r == (AASHTO, Article 8.15.2.1.1)
( )"K 58.56
5.75.0
12/)5.7x12(503M
3
cr ==
( ) "K 67.958.562.1M2.1 cr ==
Check for bottom steel ( 2s in442.0A =+ )
( )( )
( )( )"
'c
ys578.0
125.485.0
60442.0
bf 85.0
f Aa ===
( )2/ad f AM ysn −φ=φ
( )( )( ) cr '
n MK 29.1392/578.0125.660442.09.0M 〉=−=φ
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Check for top steel ( 2s in705.0A =− )
( )( )
( )( )"
'c
ys922.0
125.485.060705.0
bf 85.0f Aa ===
( )2/ad f AM ysn −φ=φ
( )( )( ) cr '
n MK 49.1392/922.0125.460705.09.0M 〉=−=φ