Below are the first 10 and last 10 pages of uncorrected machine-read text (when available) of this chapter, followed by the top 30 algorithmically extracted key phrases from the chapter as a whole.
Intended to provide our own search engines and external engines with highly rich, chapter-representative searchable text on the opening pages of each chapter.
Because it is UNCORRECTED material, please consider the following text as a useful but insufficient proxy for the authoritative book pages.
Do not use for reproduction, copying, pasting, or reading; exclusively for search engines.
OCR for page 5
5
CHAPTER 1
INTRODUCTION AND RESEARCH APPROACH
The goal of this project was to develop proposed simpli- The concrete contribution is controlled by the value of the
fied shear design provisions for the AASHTO LRFD Bridge coefficient as follows:.
Design Specifications that would overcome perceived diffi-
culties with using the current shear design provisions, which Vc = 0.0316 fcbv d v where f c
is in ksi units (Eq. 2)
are the provisions of the Sectional Design Model (A5.8.3).
This Sectional Design Model constitutes the general shear The coefficient of 0.0316 is 1 / 1000 and is used to con-
design requirements in the first three editions of the AASHTO vert the relationship for Vc from psi to ksi units.
LRFD Bridge Design Specifications (1, 7, and 17). A variable angle truss model is used to calculate the con-
Section 1.1 describes the problem that led to this project tribution of the shear reinforcement. See Equation 3 where
and begins with a summary of the LRFD Sectional Design the angle of the field of diagonal compression, , is used in
Model (A5.8.3), followed by a brief description of the calculating how many stirrups, [dvcot()/s], are included in
basis of this model, and a discussion of the differences the transverse tie of the idealized truss.
between the AASHTO LRFD and Standard Specifications
(AASHTO, 2002) shear design provisions. Section 1.2 Av f y d v cot()
summarizes the information that was available to develop Vs =
s (Eq. 3)
the proposed simplified provisions. This information con-
sists of an overview of what is known about the mecha- where dv 0.9d or 0.72h, whichever is greater. (Eq. 4)
nisms of shear resistance, a summary of code provisions,
and descriptions of available experimental test data and The values for and are obtained from Table 1 for mem-
analysis methods for shear. Section 1.3 defines project bers that contain at least the minimum required amount of
objectives, the approach used for meeting these objectives, shear reinforcement (See Equation 5) and from Table 2 for
and project tasks. members that contain less than that amount.
bv s
1.1 THE AASHTO LRFD SHEAR Av,min 0.0316 fc where f c and fy are in ksi units (Eq. 5)
DESIGN SPECIFICATIONS fy
1.1.1 Summary of the LRFD Sectional To obtain values for and from Table 1 (Av < Av,min), the
Design Model (S5.8.3) designer selects the row in which to enter the table from the
shear design stress ratio (v/f c) and the column by the longi-
The AASHTO LRFD Section Design Model for Shear
tudinal strain x at mid-depth, which may be taken as one-half
(A5.8.3) is a hand-based shear design procedure derived
of the strain in the longitudinal tension reinforcement, t.
from the Modified Compression Field Theory (MCFT).
This strain is equal to the force in the longitudinal tension
Prior approaches focused on expressions for shear strength
reinforcement divided by the axial stiffness of the tension
that were then modified for the effect of other forces. This
reinforcement. As shown in Equation 6 and illustrated in
is a comprehensive design approach for structural concrete
Figure 1, the effects of all demands on the longitudinal rein-
members in which the combined actions of axial load,
forcement are taken into account:
flexure, and prestressing are taken into account when com-
pleting the shear design of any section of any member.
t Mu / d v + 0.5 N u + 0.5(Vu - Vp ) cot() - Aps f po
In this approach, the nominal shear capacity is taken as a x = =
sum of a concrete component, a shear reinforcement com- 2 2( Es As + E p Aps )
ponent, and the vertical (or transverse) component of the (Eq. 6)
prestressing:
Equation 6 assumes that the member is cracked and, there-
Vn = Vc + Vs + Vp (Eq. 1) fore, only the axial stiffness of the reinforcement need be
OCR for page 6
6
TABLE 1 Values of and for members with at least minimum shear reinforcement
considered when evaluating t and x. If x is negative, then where ag is the maximum aggregate size in inches and taken
the member is uncracked and the axial stiffness of the equal to 0 when f c 10 ksi.
uncracked concrete needs to be considered per Equation 7. Table 2 shows that as sxe and x increase, the value of
decreases and increases. The result is that, as the member
becomes deeper and the value of the moment increases,
t Mu / d v + 0.5 N u + 0.5(Vu - Vp ) cot() - Aps f po
x = = the contributions of the concrete and shear reinforcement
2 2( Es As + E p Aps + Act Ec ) decrease.
(Eq. 7) The LRFD Sectional Design Model introduced a new
requirement into shear design provisions--the direct consid-
where Act is the area of the concrete beneath mid-depth. eration of shear in determining the required capacity of the
Alternatively, the designer can conservatively take x = 0 longitudinal reinforcement at any point along the length of
if Equation 6 yields a negative value. the member (see Equation 9).
Table 1 shows that as the longitudinal strain becomes
larger, the values for decrease and the values for increase. Tmin 0.5 N u + 0.5Vu cot + Mu dv - Aps f ps (Eq. 9)
This means that as the moment and longitudinal strain
increase, both the magnitude of the concrete and shear rein- In the end regions of prestressed concrete members, the
forcement contributions to shear resistance decrease. development length of the strands at the location of the first
To obtain values for and when Av < Av,min, Table 2 is diagonal crack must be taken into consideration when satis-
used. As for members containing at least minimum shear fying the requirements of Equation 9.
reinforcement, the column by which the designer enters In the design of a member by the LRFD Sectional Design
Table 2 is based on the value of the longitudinal strain at mid- Model, the member can be considered to be divided into
depth, x. To determine the row, the spacing of the layers of design spans of length dvcot() as shown in Figure 3. Each
crack control reinforcement is used, sxe (see Equation 8 and design span can be designed for the shear force midway
Figure 2). along the length of the span. If the load is applied to the top
of the member, then a staggered shear design concept may be
1.38sx used in which each design span is designed for the lowest
sxe =
0.63 + ag (Eq. 8) value of shear occurring within the design span.
OCR for page 7
7
TABLE 2 Values of and for members with less than minimum shear reinforcement
The Sectional Design Model was developed for regions in designed by the Sectional Design Model for the shear force
which engineering beam theory applies and there is a uni- at dvcot()/2 from the support.
form flow of the diagonal compressive stresses. However, Figure 4 is a flowchart of the entire procedure for use of
the LRFD specifications also permit the end region of mem- the LRFD Sectional Design Model. To further illustrate this
bers (the distance between the support and dvcot()/2 from procedure, a brief example is given for the design of a sec-
the support) that are subject to a complex state of stress to be tion of the 72-inch-deep bulb-tee girder in Figure 5. (This
Figure 1. Effects of axial load, moment, shear, and prestressing on
longitudinal strain in non-prestressed member.
OCR for page 8
8
h/2
bw
sz = dv sz x
Act Flexural
h/2
tension
As Area 0.003bv sz
side
Figure 2. Evaluation of crack spacing parameter Sx.
example was extracted from a design of a 120-foot single- 2. Assume x as -0.10 × 10 -3 x -0.05 × 10 -3, then
span AASHTO-PCI bulb-tee beam bridge with no skew. The obtain
example briefly illustrates the shear design procedure in
= 22.8° and = 2.94 from Table 1 (S5.8.3.4-1).
LRFD specifications. The critical section is taken at 0.06L
from centerline of a support.) 3. Compute x
Vu = 316.2 kips, Mu = 2134.0 ft-kips, N u = 0 kips,
, Mu d v + 0.5 N u + 0.5(Vu - Vp ) cot - Aps f po
x =
Vp = 23.4 kips 2( Es As + E p Aps )
As = 0, Aps = 5.508 in 2 , f po = 189.0 ksi, fc = 6.5 ksi, = -1.091 × 10 -3 0.002
E p = 28500 ks
Given that x is negative, recalculate
bv = 6 in, dv = 73.14 in, = 0.9
Mu d v + 0.5 N u + 0.5(Vu - Vp ) cot - Aps f po
x =
2( Ec Ac + Es As + E p Aps )
1. Compute shear stress ratio vu fc
= -0.080 × 10 -3
Vu - Vp
vu = = 0.7473 ksi
bv d v
if x satisfies the assumed range, then = 22.8° and =
vu fc = 0.115 2.94 are O.K.
Design Section Design Section Design Section Design Section
0.5dvcot 0.5dvcot 0.5dvcot2 0.5dvcot2
dvcot dvcot2
dvcot dvcot2 Design Span
Design Span
Design Span Design Span
Vu Vu
Vr
Vr
Shea r
Sh ea r
Location Location
Figure 3. Design regions and shear demand using the sectional design model.
OCR for page 9
9
Start
Assume value of x
and take and from
corresponding cell of
Determine b v and dv Table 1.
Eq. 5.8.2.9 Can
longitudinal
Yes
Calculate x reinforcement resist End
Calculate Vp required tension?
Eq. 5.8.3.4.2-1
Eq.5.8.3.5
Calculate shear stress
ratio v/fc, Eq. 5.8.2.9-1 Is No
No calculated x less
than assumed
If the section is within value? Can you
the transfer length of use excess shear
any strands, calculate Yes capacity to reduce the No Provide additional
the average effective longitudinal steel longitudinal
value of fpo Is requirements in reinforcement
assumed x too Eq.5.8.3.5-1?
Yes
conservative?
If the section is within ( too high?)
the development length Yes
of any reinforcing bars, No
calculate the effective Choose values of
value of As Determine transverse and corresponding
reinforcement to to large x , Table 1
ensure V < Vn
Eq. 5.8.3.3
Figure 4. Flowchart for LRFD design procedure.
4. Determine shear reinforcement Av fy dv (cot + cot )sin
From Vs = , and set fy = 60.0
s
Vc = 0.0316 fcbv dv = 103.9 kips, ksi, then
Av Vs
Vu = = 0.021 in 2 /in
Vs = - Vc - Vp = 224.0 kips s fy dv cot
Av
Use #4 bar double legs @12 in., = 0.033 in 2 /i
s
/in > 0.021 in 2 /in
Av fy dv (cot + cot )sin
This provides Vs = = 344.6 kips
s
5. Compute maximum limit check: Vc + Vs 0.25 fc bv dv
Vc + Vs = 448.5 kips 0.25 fc bv dv = 713.1 kips, O.K.
6. Compute longitudinal reinforcement check at the end
of beam
Mu N V
As fy + Aps f ps + 0.5 u + ( u - 0.5 Vs - Vp ) cot
dv
Mu N V
Figure 5. Design example + 0.5 u + u - 0.5 Vs - Vp cot = 456.4 ki
ips
dv
implementing the LRFD sectional
design model. As fy + Aps f ps = 460.1 kips 456.4 kips, O.K.
