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 33
33
Theta However x shall not be taken as less than -0.2 × 10-3.
f pc
cot = 1.0 + 3 1.8 For members having less than minimum shear reinforce-
fc ment, as required by Equation 32, the equivalent crack spac-
when Vcw < Vct (where stress is in ksi units) (Eq. 24) ing parameter, sxe, is calculated as:
which is equivalent to:
1.38sx
f pc sxe = (in. units) (Eq. 30)
cot = 1.0 + 0.095 1.8 0.63 + ag
fc
(where stress is in psi units) (Eq. 25) where ag is the maximum aggregate size (in.). Then, the fac-
tor accounting for the shear resistance of cracked concrete,
This expression was selected so that cot() was equal to 1.0 , can be computed from:
( = 45 degrees) when fpc = 0 (i.e. non-prestressed member).
The slope of the influence of fpc on provides a good 4.8 51
correlation with test data. = (in. units) (Eq. 31)
(1 + 1500 x )(39 + sxe )
The complete design procedure is shown in Figure 21.
The minimum area of shear reinforcement is:
3.2 CHANGE PROPOSAL 2: MODIFICATION OF
LRFD SECTIONAL DESIGN MODEL (S5.8.3) bv s
Av ,min = fc (in., psi) (Eq. 32)
fy
The shear design provisions in the 1994 Canadian Stan-
dards Association code for the Design of Concrete Structures It should be noted that minimum shear reinforcement is
(6) were essentially the same as the Sectional Design Model required when the factored shear force exceeds Vc, rather
in the first three editions of the LRFD Bridge Design Specifi- than Vc/2 as required by the ACI 318-02 code. Furthermore,
cations (1, 7, 17). In order to simplify the CSA shear design
the minimum amount of shear reinforcement is greater than
provisions, the 2004 code introduced equations for evaluating
the minimum amount required by ACI 318-02 and the
and that replaced the tables. Furthermore, a new equation
AASHTO Standard Specifications.
for x was introduced by assuming that was 30 degrees when
For members having at least minimum transverse rein-
evaluating the influence of shear on the longitudinal strain, x.
forcement, the angle of the diagonal compression field, , is
Change proposal 2 is the adoption of the CSA relationships
calculated as:
for , , and x. These provisions are herein referred to as the
CSA Method. This method is presented below. = 29 + 7000 x (Eq. 33)
Vn = Vc + Vs + Vp 0.25 fc bv d v + Vp (Eq. 26)
and the coefficient, , is obtained from Equation 31 with the
where: Vc = fcbv d v (in., psi) : concrete equivalent crack spacing parameter, sze, set to 12 inches.
contribution (Eq. 27) In the modified LRFD design provisions presented in
Appendix F, the contractor proposes that when the member is
Av f y d v (cot + cot )sin not continuous, or cast integrally with the support, the end
and Vs = : steel
s contribution (Eq. 28) region is designed by the strut-and-tie method in LRFD Arti-
cle 5.6.3 (7) when the design shear stress exceeds 0.18f c at the
As shown in Figure 1, the longitudinal strain, x, is com- first critical section from the support. This is to guard against
puted at mid-depth of the cross-section by: a diagonal compression failure that could occur due to fun-
neling of the diagonal compression above a simple support.
Mu d v + 0.5 N u + Vu - Vp - Aps f po The complete CSA design procedure is presented in
x = Figure 22.
2( Es As + E p Aps ) (Eq. 29a)
When x is negative, it is taken as either zero or recalcu- 3.3 DISCUSSION OF DESIGN EXAMPLES
lated by changing the denominator of Equation 29a such that
the equation becomes: To illustrate the use of these two proposed methods in dif-
ferent design situations, eight design examples were pre-
Mu d v + 0.5 N u + Vu - Vp - Aps f po pared. These examples were selected from existing PCI
x =
2( Es As + E p Aps + Ec Act ) (Eq. 29b) examples, suggestions from project panel members, and new
examples selected by the contractor. Each example begins
where Act is the area of concrete in tension. with the completed flexural design at a section with specified
OCR for page 34
34
Figure 21. Flowchart for use of proposed simplified provisions.
OCR for page 35
35
Figure 22. Flowchart for shear design in accordance with CSA.
