1WI)
1 + cos 4>
we) +2
wed~ -.4 Sin '+'
(co,2<1»
723
11.12 Step-by-Step Procedure for the Design of Circular Prestressed Concrete Tanks and Dome Roofs
and h = total depth of rim beam b = ring beam width W D = intensity of self-weight of shell per unit area (dead load) w L = intensity of live-load projection. 16. Compute the ring-edge beam cross section Pi
Ae
=
Ie
where Pi = initial prestressing force = PfY "9 = residual stress percentage Ie = allowable compressive stress in the concrete, not to exceed not more than 800-900 psi, in the edge beam.
0.21~,
but
17. Compute the area of the edge beam prestressing tendon
p Aps
=
.r'
lSI
where lsi is the allowable stress in the prestressing steel before losses, or Wcot
Aps
= 2. F 'T'iJpe
if accurate analysis is not performed. In the latter, W is the total dead and live load on the dome due to W D + W L andlpe is the effective prestress after losses. 18. Check the minimum dome thickness required to withstand buckling, i.e., . ~ Mm.hd=a\j~
where
a = radius of dome shell P u = ultimate uniformly distributed design unit pressure due to dead load and live load = (l.4D + 1.7L)/144
strength reduction factor for material variability = 0.7 buckling reduction factor for deviations from true spherical surface due to imperfections ~i = (aJrJ 2 , where ri :=; 1.4a ~e = buckling reduction factor for creep, material nonlinearity, and cracking = 0.44 + O.003Wv but not to exceed 0.53 Ec =initial modulus of concrete = 57,000~ psi. =
~i =
Figure 11.23 gives a step-by-step flowchart for a recommended sequence of operations to be performed in the design of circular prestressed concrete tanks and their shell roofs.
Chapter 11
724
Prestressed Concrete Circular Storage Tanks and Shell Roofs
(
START
l
,
Input: d, H, r, h, a, LiP, h', "1, P, fpu, 'PI" fpy,
WOI
'P.'
WL ,
f:, f~i' fr, fel fc~'
fp.
Assume wall thickness t and type of wall base joint. Compute F = 'Y( H - y) r for freely sliding base. Select membrane coefficient C from Tables 10.4-10.16 (3( 1 - 1'2) ]1/4 (3
=
(rt)1/2
Compute max. My at y above base My
= chH3 + pH2)
Mo. 0 0, il.O y and Oy
00
= +(2(3H -
il.O y Oy
=+
=F
rt 'Y )12(1 _1'2)
1)
6(\-1'2) (3 rt 2
+ 009((3y»)
[(3Mo 1/I((3y)
- il.O y
l Choose vertical prestress Pv' Compute concrete fiber stresses at critical base section when tank is empty and when full Pv MLc f=- ± -A I
Mpc
+ -I
where M L = liquid load vertical unit moment
= prestress vertical unit moment = 0.45f:
Mp
Max fe
,
Min t = 7 in. with vertical prestress Max. allow. residual axial fev = 200 psi Max. allow. tensile stress f, = 3.../i/,
Revise wall section details
CD
I I
No
/
\
t adequate?
'\
Yes
/
Compute factored moment Mu using load factors: Initial liquid pressure
1.3
Internal lateral pressure from dry material
1.7
Final prestress after losses
1.7
Strength reduction factor
0.9
Rqd Mn S available Mn Avail. Mn
I-
Mu
=-
0.9
~) + A,fy (d - ~)
= Ap,fp,(dp -
! Figure 11.23 dome roofs.
Flowchart for the design of circular prestressed tanks and their flat
11.12 Step-by-Step Procedure for the Design of Circular Prestressed Concrete Tanks and Dome Roofs
Revise wall section
1 I
No
l
/
Yes
Mu <0.9M.? " \.
l Compute slab base ring length L and thickness h 2CH 2 L2 = 1 + (tlh)3 (dt) 112
= wall thickness = base slab thickness d = tank interior diameter
where t
h
Compute percentage R
= _1_ 1+S
of moment to be
transferred to wall where S = 1. 1(hlt) y'(dttj or S' = ~ (hlt)..[(diii when only the outer rim of the slab ring is compressed by redial thrust at the rim, 2 2 where K = {d + d /l } ,/l = 0.2, d = (do - 2L) do - d
-T---T -
l Check if the elastic radial long-term deflection A f acceptable, where F,
= 1.7
{.!.L. } is tco E
= initial thrust, r = ~d, tco = thickness of wall = 57,OOO../f:
core at top or bottom of wall, Ec
l Anchor steel from base to wall up to minimum distance Y2 above base, where Y2
= Vrt;;; but not less than 3 ft. = 0.005t...
above top of base.
Min. vertical steel A,
l Max. allow crack width = 0.004 in. for liquid·retaining tanks
Wmax = 4.1 X 10-6 ec,Ep , VI; where I.
S t
b) = -8 (S2 - 1-
"'1 "'1 =diameter of wire in main direction 11"
S2 = spacing of wire in perpendicular direction tb = cover to center of reinforcement
ec,
= tensile surface strain in the concrete X, fpi Ep ,
where X, fp fpi
fp/fpi
= actual stress in steel = initial prestress
! Figure 11.23
Continued
725
726
Chapter 11
Prestressed Concrete Circular Storage Tanks and Shell Roofs
,
1
1
l Design roof shell dome:
ri~ h' ~ ~. Assume ring beam
section b X h = Ac. Select shell thickness t and check for min. t required to resist buckling from step 15. Edge ring beam prestressing force: P=
tbh
(N. -IJN~)
d
(N~COS4»
+"2
where 1 = -wod . - [-- -
tangential N. meridional N b
2 Sin 4>
~
1 + cos 4>
wLd (cos 2c/» cos 4> ] - ~ Sin
WL }
= -8 {Wo --- + 1 + cos 4> 2
= beam width,
h
= beam depth,
Wo
= dead
= live load
load, w L
J Compute rqd. Ac = P;ltc' where P; = P/;Y, ;Y = residual stress percentage'l tc = allowable concrete compressive stress ~ 0.20f; ~ 800 to 900 psi
No
Revise ring beam Ac
l
Assumed Ac :2: rqd. Ac?
Yes
J Compute edge ring beam prestress reinforcement A p,
=
A P'
= P;lf,; or
W cot 4> if accurate analysis is not performed. 21rfpe
,
W = total dead and live load (wo + WL) on the dome = effective prestress after losses
fpe
Check min. dome thickness t to withstand buckling,
.
Min. hd
=8
J!f£ -13-4>13; cEc
8 = radius of dome shell = 1.20 + 1.6L, 4> = 0.65,13;::0 0.50, = 0.44 + 0.OO3W L ~ 0.53,
where Pu
13c
Ec =57,OOO../f:
( Figure 11.23
l END
)
Continued
11.13 Design of Circular Prestressed Concrete Water-Retaining Tank and Its Domed Roof
727
11.13 DESIGN OF CIRCULAR PRESTRESSED CONCRETE WATER-RETAINING TANK AND ITS DOMED ROOF Example 11.3 Determine the maximum horizontal ring forces and vertical moments, and design the wall prestressing reinforcement, for a circular prestressed concrete tank whose diameter d = 125 ft (38.1 m) and which retains a water height H = 25 ft (7.62 m) for the following conditions of wall base support: (a) hinged, (b) fully fixed, (c) semisliding, and (d) partially fixed. Also, design the prestressed concrete ring edge beam for the domed roof shell assuming that the shell rise-span ratio h'ld = ~. Use a flat shell roof having shell angle
,.
(A ,! ,.
M. O.5N.
+ A p1! ps) ;;:!: -, - + ~ + cot {
v~f
"'~
aJ(V. - IPv
)' + (0.452A-'. 7~ P o)'
O.5V, - Vp
0 ,+"
,
(12.44)
12.7 AASHTO-lRFD Flexural-Strength Design Specifications vs. ACI Code Provisions
773
where Po = Perimeter of the shear now path N" = Applied axia l force , taken as posi ti ve if compressive
12.7 AASHTO-LRFD FLEXURAL-STRENGTH DESIGN SPECIFICATIONS VS. ACI CODE PROVISIONS
There are fundamen tal differences in the approaches of the AA SHTO-LRFO f1ex uralstrength design specifications and the ACI-3 l8 code provisions. The LRFD approach is based on strain limit values as discussed in Section 12.3 and con tro ll ed by the ratio of the neutral axis depth , c, to the effective depth . d, (Ref. 12.14-12. 16). This approach is variably termed as a unified approach since it is applicable to reinfo rced, prestressed. and partially prestressed concrete ultimate limit state design. The ACJ-3 18 code strength provisions have been applied for detennining the ulti mate design strength Ips in severa l examples in o ther sections of the book such as sections 4.9 and 4.10. They have to be applied in the design of full y and partially prestressed concrete members in build ing structu res. The current AAS HTO standa rd specifications (Re f. 12.2) for proportioning prestressed concrete members in flex ure generally follow the ACI code provisions. The LRFD alternat ive, which is a rational design approach, requires applying the strain limits unified procedure. It is useful to give a comparison summary showing the differences between the expressions specifi ed in these two approaches as shown in Table 12.9 adapted from Ref. 12.15.
Photo U.3 Segmental bridge launching. (Courtesy Mon ica Schuhese_ MidAtIan tic Precast Concrete Association)
Chapter 12
774 Table 12.9
LRFD and Standard AASHTO Design of Concrete Bridges
LRFD and ACI Provisions for Ultimate Strength Flexural Design
AASHTO·LRFD
ACICode Notation d to nonprestressed reinforcement
ds to nonprestressed reinforcement
d p to prestressed reinforcement
Aps/psdp + As/yds d = ----"--'--'---e Aps/ps + As/y
d' to compression steel
b
As Pp, p, p' ...... p = bd
dp
t:;.s
_ Ips _ ApsIps wp - Pp I'c - bdp I'c
de ds
1--
s
--
A~ Maximum Flexural Reinforcement Reinforced Concrete and Prestressed Concrete
All Cases - RC, PC, PPC Rectangular or T Section:
Maximum redistribution factor = 1000 E t
20 (1 - 2.36
percent, where
Et
= 0.003 (~ - 1).
;J
in %
provided:
provided: ~
de
for rectangular sections and:
::; 0 28 but never to exceed 0.42 . and
)0
for T-section behavior An alternative ACI Code check requires that ddt for the section is either tension controlled or in the transition zone of Fig. 4.45 -L-------1--~ -'-
T
775
12.8 Step-by-Step Design Procedure (LRFD) Table 12.9
Continued
ACI Code
AASHTO-LRFD Minimum Reinforcement
Reinforced Concrete 200 3.5vf:: P~Pmin~-~--fy fy
All Cases:
(For T-sections, P is based on web only) Prestressed and Partially Prestressed Concrete:
Particular result for reinforced concrete: 0.03f:
P~Pmin=h
(For T-sections, P is based on web only) Stress in Bonded Prestressing Steel at Ultimate Resistance in Bending Prestressed and Partially Prestressed ConcreteBonded Tendons
PCandPPC Bonded Tendons:
'Yp { fpu fps = fpu [ 1 - 131 Pp f:
fps
d p
, }]
+ d (w - w )
=
fpu(1 - k
;J
k = 2 (1.04 _ fpy) fpu
where:
b
fpu [ Pp -
te
d'
<
+ -d (w - w),] dp
~
0.17
0.15 dp
'Yp = 0.28 for fpy ~ 0.90 fpu (Low Lax) 0.40 for fpy ~ 0.85 fpu [normal] 0.55 for fpy ~ 0.80 fpu (bars)
131 = 0.85 for f; 0.65 for f; 0.85 - 0.05
-
::; 4 ksi ~
'------'-
8 ksi
(f; - 4)
for 4 ::; f;
::; 8 ksi
12.8 STEP-BY-STEP DESIGN PROCEDURE (LRFD)
The following is a summary of a recommended sequence of design steps: 1. Determine whether or not partial prestressing is to be chosen. 2. Select the bending moments and shear forces from Table 12.2(a)&(b), Section 12.7. 3. Follow the step sequence for flexural design of the member outlined in steps 2 through 10 of Section 4.13 in Chapter 4 and the flowchart of Figure 12.10 when using the LRFD method for flexure. Generally, d v = (de - a!2).
Chapter 12
776
LRFD and Standard AASHTO Design of Concrete Bridges
4. Determine the factored shear force Vu due to all applied loads at the critical section located at a distance d v or 0.5 d v cote from the face of the support, whichever is larger, where de
=
effective depth as shown in Table 12.8(a) and (b)
=
dp if no mild steel is used.
5. Compute the tendon shear component Vp" The factored shear stress is:
v= The nominal available shear stress vc = vlh. 6. Compute the quantity
v/f~
and assume a value of
e = 25°
prestressed beams is
e. A good initial assumption for
7. Compute the strain in the tensile reinforcement in order to enter Table 12.8 to ob-
tain a trial value of e and
~
from Equations 12.28 (a), (b) or (c) that applies, namely,
Section contains at least the minimum transverse reinforcement:
(~u + 0.5 Nu + 0.5(Vu ex
v
=
Vp) cote - Apsfpo)
2 (Es As + Ep Aps)
~
0.001
Section contains less than the minimum transverse reinforcement: (
~ u + 0.5 Nu + 0.5(Vu -
ex =
Yp) cot e - A ps/po ) ~
v
EsAs+EpAps
0.002
If the value of ex from Equations 12.28 (a) and (b) is negative, compute the longitu-
dinal tensile strain from the following expression and enter it in Table 12.8(a) or Table 12.8(b) where applicable.
(~ + 0.5 Nu + 0.5 (VU -
Yp) cot e - Apsfpo)
ex = Ac = area of the concrete in the flexural tension side of the member.
8. Enter LRFD Figure 12.9 again, with the value of v / f~ and strain Ex if the strut angle
e
is not close to the one assumed in the first trial, in order to obtain an ~usted value of ~. Otherwise, compute Vc from Equation 12.23, namely Vc = ~ V f~ bvdd (lb) or Vc = 0.0316~ bvd v (kip) using the ~ value obtained from the chart in Table 12.8(a). If Table 12.8(b) applies, enter the vaue of sxe and ex to determine the trial e and ~ values.
vFc
9. Compute Vs for the web reinforcement after the value of Vc has been determined.
Find the corresponding shear reinforcement spacing from:
777
12.8 Step-by-Step Design Procedure (LRFD)
10. In regions of high shear stresses, ensure that the amount and development of the
longitudinal reinforcement As and Aps should satisfy the following expression:
It is recommended that this check be made at the face of the bearing which lies
within the transfer length of the strands where the effective prestressing force is not fully developed. 11. When torsion exists combined with shear and flexure, the following steps need to
be followed: nominal torsion Tn
2AoAJy cot e
= -----
s
Obtain strain in tensile reinforcement from Equations 12.28 adjusted for torsion from Equation 12.38. where fpo = 0.70 fpu Nominal shear resistance:
where d v = (dp
-
all)
Shear reinforcement: Av
Vn - 0.0316~~bvdv
s
+ Vp
fyrdv cote
Forces are in kips and the stresses in ksi. For using lb and psi units, remove factor 0.0316. Torsion reinforcement:
Total web closed ties reinforcement:
778
Chapter 12
LRFD and Standard AASHTO Design of Concrete Bridges
Shear stress v for obtaining angle 8: (a) Box sections:
(b) Other sections
For avoiding yield of the longitudinal tensile reinforcement:
12. Check the horizontal interface shear:
where
where
(fy is in ksi) Take the nominal shear resistance as the lesser of
or
c = cohesion factor f.1 = friction factor Acr = concrete interface area = bJv Avt = area of shear reinforcement crossing the shear plane within area Acv = strength reduction factor = 0.90. Limit Avt to cases in which vuhl is greater than 100 psi. Figure 12.10 gives flowchart (Ref. 12.15) for steps to be followed in determining the nominal moment strength, namely, the nominal bending resistance for bonded and unbonded tendons. 13. Maximum allowable spacing of web shear reinforcement:
779
12.S Step-by-Step Design Procedure (LRFD)
Given materials and cross-sectional properties
Determine fps from strain compatibiblity fps == fpe + 15 ksi
No
fps == fpe +;:: 103 MPa
No If c::; 3 d~ Assume: f' y == 0 and repeat computation of c
Yes
~~--+i
Rectangular section:
bw == b ; hf == 0
de == (Apsfpsd ps + Asfyd s)/ (Apsfps + Asfy)
Compute: a == P1 c and c/d e
Over-reinforced section is not recommended, unless it is shown by test or analysis that performance will not be impaired
• For rectangular section behavior: b w == b • If c ::; 3d's, assume f'y == O. • To insure minimum reinforcement check that:
Figure 12.10 Nominal moment strength of prestressed section with bonded and unbonded reinforcement (Ref. 12.15).
780
Chapter 12
(a) If Vu < 0.125 fd, (b) Ifv u >0.125fd,
s = 0.8 d v s = 0.4 d v
LRFD and Standard AASHTO Design of Concrete Bridges
::s24 in. ::s12 in.
Note that the design yield strength of non-prestressed transverse reinforcement should not exceed 60.0 ksi. The design yield strength of prestressed transverse reinforcement should be taken as the effective prestress after allowing for all prestress losses plus 60.0 ksi, but not greater than fpy" For Dowel reinforcement spacing: 0.05 b v If Av 2:: -1-- where b v = width of interface and spacing as presented in item 13. yt
12.9 LRFD DESIGN OF BULB-TEE BRIDGE DECK Example 12.1 Design for flexure an interior beam of a 120 ft (36.6 m) simply supported AASHTO-PCI bulb-tee composite bridge deck with no skews (adapted from Ref. 12.11). The superstructure is composed of six pretensioned beams at 9'-0" (2.74 m) on centers as shown in Figure 12.11. The bridge has an 8-in. (203-mm) situ-cast concrete deck with the top ~-in to be considered as wearing surface. The design live load is the HL-93 AASHTO-LRFD fatigue loading. Assume the bridge is to be located in a low seismicity zone.
Given: Maximum allowable stresses: Deck
I; = 4000 psi, normal weight
Bulb-tee
I; = 6500 psi
Ie
= 0.60
I; = 2400 psi
I;i = 5500 psi Ie = 0.60 I; = 3900 psi, Service III Ie = 0.45 I; = 2925 psi, Service I lei = 0.60 I; = 3480 psi t, = 6 Vi'c = 484 psi Ipu = 270,000 psi Ipy = 0.90 Ipu = 243,000 psi
Figure 12.11
Section Properties
Ae = 767 in. 2 h =72 in.
Ie
=
545,894 in.4
cb = 36.60 in. ct = 35.40 in.
Sb
=
st =
14,915 in? 15,421 in. 3
Bulb-tee Bridge Deck Cross Section in Example 12.1 (Ref. 12.11)
781
12.9 lAFD Design of Bulb-Tee Bridge Deck
,
fpl :: 0.75 f,.. :: 202.500 psi
r ""
/, :: /,1:: 60,000 psi Ep :: 28.5 )( 1(1 psi E6 :: 29.0)( Icr psi.
I~
-
A<
545,894
.
~
= - - - "" 712m· 767
.
Wo = 799plf.
Solution: I. TransjoNnro Deck slab controlling width
Compute the transfonn ed flan ge width:
Ea "" 33w l .5
Vi; =
33 x (1.5)1.5 Y4000 = 3830 ksi
E..- attransfer = 33( 1.5)1.5 Y5500 = 4500 ksi
E" at service:: 33( J.5tl Y6500 "" 4890 lui Effective flange width is the lesser of 120 x 12 . (i) (l) span :: 4 - 360 m. (ii) 12 "r+ greater of web thick ness or I-beam top flllnge width. b:: 12 x 7.5 + 0.5 x 42
:: III in. (iii) average spacing between beams = 9 x 12 = 108 in.
hence. controlling flange width = 108 in.
. E., 3830 Modular rall O n. = - - - ."" 0.78 E< 490 Tmnsformed width b", = n,b =0.78 x 108 = 84 in.
Photo 12.4 Chesapeake and Delaware Canal cable-stayed bridge on S. R. 1. Length 4650 fl. with two 12-fl. dee p precast girders carrying II 11 2-fl. wide roadway; Iypical spans are 150 ft. with precast box piers (Counesy Figg Engineering Group. Tallahassee. Florida)
782
Chapter 12
LRFD and Standard AASHTO Design of Concrete Bridges
2. Properties of composite section Disregard as insignificant the contribution of the deck concrete haunch to l' e which is needed because of the precast element camber. A' c = 1397 in. 2 h =80 in.
Icc
= 1,095,290 in.4
cbe = 54.6 in. to the bottom fibers cte = 72 - 54.6 = 17.4 in. - precast ctse = 80 - 54.6 = 25.4 in. - deck top _ 1,095,290 _ . 3 54.6 - 20,060 Ill.
Sbc -
S~ =
1,095,290 _ . 3 17.4 - 62,950 Ill.
ts _ 1,095,290 _ . 3 Se - 25.4 x 0.78 - 55,284 Ill.
3. Bending moments and shear forces Slab: W SDl
=
Barrier weight: W SD2 = 2 in. future-wearing surface:
WSD3
8 12 x 9
X
150
=
900 lb/ft
2 barriers (300 lb/ft) = 100lb/ft 6 beams
= 22 X 48 ft X 150 = 200lb/ft 1 6 beams
Live load (truck load) in LRFD would be based on HL-93 truck fatigue loading. Clear width from figure 12.12 = 48 ft (14.6 cm) Number of lanes
48
= 12 = 4 lanes
(a) Distribution factor for moment For two or more lanes loaded (Ref. 12.3), the distribution factor for bending moment (Table 12.3b)
_
(~)O.6 (~)O.2 (~)O.l L L' 12C;L
DFM - 0.075 + 95 . provided that beam spacing: 3.5 :::; S :::; 16 deck slab: 4.5 :::; Ts :::; 12 span: 20 :::; L :::; 240 no. of beams: Nb > 4 eg
n
Actual S = 9.0 ft Actual Ts = 7.5 in. Actual L = 120 ft Actual Nb=6
O.K. O.K. O.K. O.K.
=
distance between the center of gravity of the beam and the slab
=
7~5 + 0.5 + 35.4 =
=
Ee Esc
=
4890 3830
=
39.65 in.
1.28
Kg = n(Ie + Aee~) =
1.28 [545,894
+ 767 (39.65?l
=
2,242,191 in. 4
783
12.9 LRFD Design of Bulb-Tee Bridge Deck
hence, DFM = 0.075
+ (..2...)0.6 (~)0.2 [ 2,242,191 9.5
=
120
]0.1
12(7.5? (120)
0.732 lanes/beam
For one design lane loaded, from Table 12.3b,
DFM
=
(
(-Kg-)0.1 S)0.4 (S)0.3 -
0.06 + 14
= 0.06
9
+ ( 14
12~L
L
)0.4 ( 9 )0.3 [ 120
2242191 12(7.5);(120)
]0.1
= 0.499 lanes/beam;
consequently, the case of two or more lanes loaded controls so that the DFM = 0.732 lanes per beam. Fatigue Moments: The moment is taken for a single design truck having the same axle weight as in all other limit states, but with a constant spacing of 30 ft between the 32-kip axles. A multiple lane factor of 1.2 for fatigue is used to reduce the controlling DFM factor. From table 12.2a, the load factor is 0.75 and the impact factor (1M) for fatigue =15%. Hence, the fatigue truckload bending moment becomes:
Mf = (bending moment per lane) (DFMj1.2)(1 + 1M) · moment per 1ane ) (0.499)(1 + 0.15) or Mf = (ben dmg 1.2 (0.415)(1 + 0.15)
=
(bending moment perlane)
=
(0.478) (bending moment per lane)
(b) Distribution factor for shear
From Table 12.3(a), For two or more lanes loaded
DFV provided that: beam spacing:3.5 ~ S ~ 16 deck slab: 4.5 ~ Ts ~ 12 span: 20 ~ L ~ 240 10,000 ~ Kg ~ 7,000,000 hence,
DFV
=
0.2 +
(~) - (~y
Actual S = 9.0 ft Actual Ts = 7.5 in. Actual L = 120 ft Actual Kg = 2,242,191 in.4
O.K. O.K. O.K. O.K.
= 0.2 + (:2)~;6Y = 0.887 lanes/beam.
For one design lane loaded (Table 12.3a)
DFV = 0.36
+
C:.o) = 0.36 (;;~o) = 0.720 lanes/beam;
consequently, the case of two or more lanes loaded controls and DFV lanes per beam.
4. Load Combinations Total factored load, Q = TJ"'i.'Yiqi
= 0.887
LRFD and Standard AASHTO Design of Concrete Bridges
Chapter 12
784
where 11 'Yj qj
= a factor relating to ductility, redundancy and operational importance. = load factors = special loads;
use 11 = 1.0 for all practical purposes in this example. Investigate all the load combinations from table 12.2( a) and (b). The cases that control are as follows: (a) Service I for compressive stresses in the prestressed concrete components: Q = 1.0 (DC + DW) + 1.0 (LL + 1M) (b) Service III for tensile stresses in the prestressed concrete components: Q = 1.0 (DC + DW) + 0.8 (LL + 1M) (c) Strength I for ultimate strength: Maximum
Q = 1.25 DC + 1.50 DW + 1.75 (LL + 1M)
Minimum
Q
=
0.90 DC + 0.65 DW + 1.75 (LL + 1M)
(d) Fatigue for checking stress range in the strands
Q = 0.75 (LL + 1M) (The fatigue Q is a special load combination for checking the tensile stress range in the strands due to live load and dynamic allowance.)
5. Unfactored shear forces and bending moments (a) Truck Loads
Truck load shear force: V LT = (shearforce per lane)(DFV)(1
+ 1M)
= (shear force per lane)(0.887)(1 + 0.33) = 1.180 (shear force per lane) kips.
Truck load bending moment: MLT
= (moment per lane)(DFM)(1 + 1M) = (moment per lane)(0.732)(1 + 0.33) = 0.974 (moment per lane) ft-kips.
L T = Truck live load (b) Lane Loads
For lane loads, no dynamic allowance is applied, hence, V LL = (shear force per lane )(DFV)
= (shear force per lane)(0.887) kips MLL = (moment per lane)(DFM)
= (moment per lane)(0.732) ft-kips. The lane loads from Figure 12.4, the load on this bridge is as follows in Figure 12.13.
6. Computation of moments and shears (a) Lane live load (DFV = 0.887, DFM = 0.732) (i) Support section:
shear at the left support (x = 0) from equation 12.6(a) and Figure 12.12: V LL
=
0.64 2 2L (L - x) (DFV) 2
~6:20 (12of (0.887) = 34.1 kips
785
12.9 LRFD Design of Bulb-Tee Bridge Deck 0.64 kip/Mane
111111111111111111111111111111111111111111111111111111111111111111111111
1
right reaction
'eft ,eacUon
x
(120 - x) > x
I:
120' Figure 12.12
Truck load per lane
From equation 12.6(b), and DFM = 0.732 MLL =
0.64(x)(L - x) 2 (DFM)
(ii) Section at 24 ft from support: As an example, find V LL and M LL at x V LL
_
-
MLL =
=
=
. 0 ft-klp
24 ft from the left support.
0.64 ( )2 )_ . 2 X 120 120 - 24 (0.887 - 21.8 kips 0.64(24)(120 - 24) 2 (0.732)
=
. 539.7 ft-klp
(b) Truck live load (DFV = 1.180, DFM = 0.974)
Here, the impact factor 1M DFMvalues.
=
33% has to be included, hence, larger DFV and
(i) Support sections:
From Table 12.4,
V LT =
=
72[(L - x) - 9.33J L (DFV)
72[(120 - 0.0) - 9.33J 120 (1.180) = 78.1 kips
From Table 12.5, MLT
=
72(x)[(L - x) - 9.33J L (DFM)
= 0 ft-kip for the support moment.
(ii) Section at 24 ft from support:
V LT
=
72[(120 - 24) - 9.33J 120 (1.180)
MLT
=
72(24)[(120 - 24) - 9.33J . 120 (0.974) = 1215.0 ft-kIp
=
. 61.4 kips
(c) Fatigue moment at 24 ft (DFF= 0.478)
From Table 12.7, Mf =
72(x)[(L - x) - 18.22J L (DFF)
Chapter 12
786
LRFD and Standard AASHTO Design of Concrete Bridges
1
1 j
From before, DFF = 0.478 hence, Mf =
72(24)[(120 - 24) - 18.22J 120 (0.478) = 535.8 ft-kip
(d) Shears and moments due to dead loads: The loads to be considered are beam weight (WD) plus deck slab and haunches (WSDl ), and future wearing surface (WSD3)' The beam is simply supported, hence, the shear and moment at any cross section along the span are: Vx = W D (0.5L - x)
Mx = O.5WD x(L - x) As an example, consider a section at 24 ft from the left support and compute the shear and moment due to self-weight WD = 0.799 Kip/ft: Vx = 0.799(0.5 X 120 - 24) = 28.8 kips
Mx = 0.5 X 0.799 X 24(120 - 24) = 920.4 ft-kip. Tables 12.10 and 12.11 (Ref. 12.11) list the forces and moments required for the design of the interior beam elements. It should be noted that long-hand computations to develop such a table are time consuming. Computer programs developed by several state DOTs are available, some on the internet, such as the Washington State DOT Program.
7. Design of the Bulb-tee prestressed interior beam (1) Selection of Prestressing Strands
For Service-III load combination, bottom fiber stress fb is: fb
=
MD + Ms
Mb + Mws + 0.8(MLT + M LL ) +---------
Sb
where
MD Ms Mb Mws MLT MLL
Sbe
= unfactored self-weight moment, ft-kip
= unfactored moment due to slab and haunch weight, ft-kip = unfactored barrier moment, ft-kip
= unfactored future wearing surface moment, ft-kip
= unfactored truck load moment, ft-kip = unfactored lane load moment, ft-kip
From before, Sb = 14,915 in. 3 Sbe = 20,090 in?
From Tables 12.10 and 12.11, Midspan stresses at bottom fibers at service: fbe = =
(180 + 360) + 0.8(1830.3 + 843.2) (1438.2 + 1659.6) 14915 (12) + 20090 (12) , , 2.50
+ 1.60 == 4.10 ksi (T).
The 4.10 ksi (T) will be neutralized by prestressing the beam. Maximum allowable tensile stress:
It = 6.0-v:t psi = 6Y6500 = 484 psi = 0.484 ksi Required prestress compressive stress at the bottom fibers: feb = (4.1 - 0.48) = 3.62 ksi
787
12.9 LRFD Design of Bulb-Tee Bridge Deck Table 12.10
LRFD Service Shear and Moment Due to Dead Load Beam Weight
(Slab + Haunch) Weight W SD1
Wo
Distance
X
XlL
ft
Wearing Surface
WOS2
WSD3
Moment
Moment
Section
Barrier Weight Moment
Moment
Shear
Mg
Shear
Ms
Shear
Mb
Shear
Mws
kips
ft-kips
kips
ft-kips
kips
ft-kips
kips
ft-kips
0
0.0
47.9
0.0
55.3
0.0
6.0
0.0
12.0
0.0
6.00'
0.05
43.1
274.3
49.8
315.3
5.4
34.2
10.8
68.4
12
0.1
38.4
517.8
44.3
597.5
4.8
64.8
9.6
129.6
24
0.2
28.8
920.4
33.2
1,062.1
3.6
115.2
7.2
230.4
36
0.3
19.2
1,208.1
22.1
1,394.1
2.4
151.2
4.8
302.4
48+
0.4
9.6
1,380.7
11.1
1,593.2
1.2
172.8
2.4
345.6
60
0.5
0.0
1,438.2
0.0
1,659.6
0.0
180.0
0.0
360.0
'Critical section for shear
+ Harp point
Assume that the distance from the centroid of the prestressing reinforcement and the section bottom fibers = 0.05h
= (0.05)(72) = 3.6 in; use 4.0 in., hence ec = 36.6 - 4.0 = 32.6 in. As presented in the examples in Chapter 4, Pe fbp due to prestress = Ac Pe
or fbp = 767 +
Table 12.11
P X e
+ -e - -c Sb
P e X 32.6 . 14,915 = 3.62 ks!
LRFD Service Shear and Moment Due to Truck and Lane Loads
Truck Load with Impact WLT+I Distance
Section
X
XlL
ft
Lane Load
Fatigue Truck with Impact
W LL
W,
Shear VLT
Moment
Shear VLL
Moment
MLT
MLL
Moment M,
kips
ft-kips
kips
ft-kips
ft-kips
0 6.00'
0.0
78.1
0.0
34.1
0.0
0.0
0.05
73.8
367.8
30.6
160.2
165.0
12
0.1
69.6
691.6
27.5
303.6
309.2
24
0.2
61.4
1,215.0
21.8
539.7
535.8
36
0.3
52.7
1,570.2
16.6
708.3
692.7
48+
0.4
44.2
1,778.9
12.2
809.5
776.2
60
0.5
35.7
1,830.2
8.5
843.3
776.9
'Critical section for shear
+ Harp point
788
Chapter 12
LRFD and Standard AASHTO Design of Concrete Bridges
to give prestressing force Pe = 1037 kips Assume total prestress loss = 25% 1037 . Pi = 1 _ 0.25 = 1383 kips assume using ~ in.-dia 7-wire 270-K low-relaxation strands (Aps = 0.153 in.2) . ReqUlred number of strands
1383 X 202.5
= 0.153
= 44.6 strands.
After two trials and adjustments, 48 strands with the configuration shown in Figure 12.13 are tried. Less than 48 strands result in tensile stresses at the bottom fibers at service which exceed the maximum allowable It = 484 psi. Twelve strands are harped at 0.4 L. Accordingly, 36 strands remain straight at the beam (see Figure 12.13). From data, Cb = 36.60 in. and ct = 72 - 36.60 = 35.40 in. ee
=
[2 X 70 + 2 X 68 + 2 X 66 + 2 X 64 + 2 X 62 + 2 X 60 + 4 X 8
Cb -
+8
X
6 + 12 X 4 + 12 X 2J/48
= 36.60 - 19.42 = 17.28 in. ec =
Cb -
[2 X 12 + 12 X 4 + 8 X 6 + 8 X 4 + 2 X 10 + 2 X 12 + 2 X 14
+2
X
16 + 2 X 18 + 2 X 20J/48
= 36.6 - 6.92 = 29.68 in. Given!pi = 0.75 !pu = 202,500 psi, Pi
=
(48)(0.153)(202.5)
=
1488 kips.
After running a detailed step-by-step analysis of prestress losses as in chapter 3, Section 3.9, the total prestress loss was determined to be 26.4%.
!pe = 202.5(1
- 0.264)
= 149.0 ksi
hence, Pe = 1488(1 - 0.264) = 1095.0 kips
No. Distance from bottom (in.) Strands
2 2 2 2 2 2
1
72"
70 68 66 64 62 60
-6" No. Distance from Strands bottom (in.)
6"
I
~ ~ 1 spacj, @2"
II-r-
4 8 12 12
8 6 4 2
2 2 2 2 2 2 4 8 12 12
Harped strands
I
20 18 16 14 12 10 8 6 4 2
~ ~ 1 spac;.11-r-
At ends (ee = 17.28 in.)
Figure 12.13
No. Distance from Strands bottom (in.)
@ 2"
At midspan (ee = 29.68 in.)
Bulb-Tee Prestressing Strand Pattern
789
12.9 LRFD Design of Bulb-Tee Bridge Deck
(2) Check of Concrete Unfactored Stresses (a) Stresses at Transfer Initial fpi = 0.7 fpu = 0.7 x 270 = 202.5 ksi. Common practice assumes that initial relaxation losses at prestressing amount to 9 to 10%. Use 10% reduction infpi· Pi
=
0.90 X 1488
=
1339 kips
Hence, Pi = (0.9)(202.5)(0.153x48) = 1338 kips (i) Support Section From Chapter 4, Equation 4.1 (a), f t = _ Pi Ae = -
(1 _
eeCt) _ Mo
1338 ( 17.28 X 35.4) 767 1 712 - 0
fb = - Pi -(1 Ae = _
st
r2
=
-0.25 ksi (C), no tension, O.K.
+eeCb) - +Mo Sb
r2
1339 (1 + 17.28 X 36.3) + 767 712 0
= 3.29 ksi (C) < allowable fe = 3.48 ksi O.K. (ii) Midspan Section
ft
= _
1338 (1 _ 29.68 X 36.60) _ 1438 X 12 767 712 15,421
= 0.917 - 1.119 = -0.202 ksi (C), no tension allowed, hence, O.K.
Ii
= -
b
1339 (1 + 29.68 X 36.6) + 1438 X 12 767 712 14,915
= -4.513 + 1.157 = -3.356 ksi (C)
<
than allowable f~i = 5.50 ksi, O.K.
(b) Stresses at Service: (i) Midspan Section:
From chapter 4, Equations 4.3( a) and 4.3(b): f
t
fb
= -
Pe ( eeCt) Ae 1 -
MT 7 - Sf
$
fe
$
It
P e ( 1 + -eeCb) + -MT Ae r Seb
= --
Since the loads are placed at different stages of construction, for Service-I precast sections, ft
= _
= _
=
Pe ~
(1 _
eeCt) _ Mo + Ms _ Mws + Mb
?
~
?
1095 (1 _ 29.68 X 35.40) _ (1438 + 1660)12 _ (360 + 180)12 767 712 15,421 62,950
0.679 - 2.411 - 0.103
=
-1.835 ksi (C)
< Service allowable fe
= 2925 psi
O.K.
1095 ( 29.68 X 36.60) (1438 + 1660)12 (360 + 180)12 fb = - 767 1 + 712 + 14,915 + 20,060
= -3.605 + 2.493 + 0.323 = -0.789 ksi (C) O.K.
790
Chapter 12
LRFD and Standard AASHTO Design of Concrete Bridges
(3) Including stresses due to the transient lane and truck loads
A factor of 0.80 is used for Type III loading (Table 12.2a)
t'
= -1.835 -
0.8 (1830 + 843)12 62,950
= -1.835 - 0.408 = -2.243 ksi (C) < Service-Illfe = 3900 psi O.K. fb
=
-0.789 +
0.8 (1830 + 843)12 20,060
= -0.789 + 1.279 = 0.490 ksi (T)
= Allowable It
= 0.484, O.K.
(4) Concrete stresses at top deckfibers (i) Under permanent Service I loads Mws + Mb
ste
f; =
(360 + 180)12 55 ,284
=
= -0.117ksi(c) < allowablefe = 2.4ksi OK (ii) Under permanent and transient lane and truck loads, Service I:
f; =
Mws + Mb
ste
= -0.117 -
MLT + MLL
ste
+
(1830 + 843)12 55 ,284
= -0.697 ksi (C)
<
allowable fe = 2.4 ksi,
OK
(5) Concrete Stresses at beam bottom fibers, Service III (See step 3) fb
Pe ( 1 Ae
= --
(Mws + M B ) + 0.8(MLT + M LL ) + -eeCb) + MD + Ms + ---------r2
Sb
+ 29.68
= _ 1095 (
Sbe
X 36.60) 712
+ (1438 + 1660)12
767
1
[(360
+ 180) + 0.8(1830 + 843)J12
+
14,915
20,060
= - 3.605 + 2.492 + 1.603 = 0.490 ksi (T) ::; allowable ft = 0.484 ksi, O.K. (6) Fatigue stresses LRFD specifies that in regions of compressive stress due to permanent loads and prestress, fatigue is only considered if the compressive stress is less than twice the maximum tensile live load stress resulting from the fatigue: Thus, for permanent loads only, the term (MLT + MLL)/S~ is taken out to give:
fb
Pe ( 1 Ae
= --
(Mws + M B ) + -eeCb) + MD + Ms + -'----'-------'r2
= -3.605 + 2.492 +
Sb
Sbe
(360 + 180)12 20,060 = -0.790 ksi (c) O.K.
From table 12.11, fatigue moment M f = 777 ft-kips, Tensile fatigue stress at the bottom fibers,
fb = 0.75Mf = 0.75 X 777 X 12 = 0.348 ksi (T) Sbe 20,060 Since twice 0.348 = 0.696 < 0.790 ksi (which is a compressive stress), a fatigue check is unnecessary.
1
1
791
12.9 LRFD Design of Bulb-Tee Bridge Deck
From the forgoing computations, the flexural design is O.K. at the initial and service load conditions. To be complete and also determine the reserve strength available for overload conditions, the limit state at failure design is necessary as in the following section. The total design has to include shear, torsion, if any, and serviceability as in Example 12.2.
8. Ultimate strength (Limit state offailure) (a) Nominal flexural resistance moment From Tables 12.2 (a) and (b) total factored moment for Strength I Load: Mu = 1.25DC + 1.5DW + 1.75(LL + 1M) From Table 12.10 Mu = 1.25(1438 .
+ 1660) + 1.5(360 + 180) + 1.75(1830 + 843)
9316 ft-kip
9316 . = - - = 9316 ft-kip 1.0
Mu
ReqUIred M, = -
=
c\>
Average stress in the prestressing reinforcement whenlpe ~ 0.5 Ipu' from
Ips = Ipu (1 - k
equation 12.9:
;J
where k
= 2 (1.04 -
For the depth of the compressive block, use slab
i~).
I: = 4.0 ksi.
k = 0.28 for low-relaxation steel dp = (h-cover to c.g.s.) = [(72 + 8) - 6.92] = 73.08 in. b = effective compression flange width = 9' - 0" = 108 in. Aps = 48 x 0.153 = 7.344 in. 2
Ipu = 270 ksi From equation 12.10, c=
Aps/pu
+ Asly -
0.85/~J3lb
A~/;
Ipu
+ kAps d
p
7.344 + 0 - 0 0.85 =
X
6.20 in.
4.0
X
<
ts
0.85 =
X
108 + 0.28
X
270 ) 7.344 ( 73.08
7.5 in.
a = J31C = 0.85 X 6.20 = 5.27 in.,
hence, neutral axis is within the flange and the section is considered rectangular. Average design reinforcement strength Ips: 6.20) . Ips = 270 ( 1 - 0.28 73.08 = 263.6 kSI nominal flexural resistance M, = Aps/ps (d -
Mn = M, = 7.344
X
263.6(73.08 -
~)
5.~7)C12)
= 11,364 ft-kips > required Mn = 9316 ft-kip
O.K.
..£. :S 0.42 for ductile behavior discussed in section 12.3.1 and table 12.1 (a). de
C 6.20 Actual de = 73.08 = 0.085
< 0.42 O.K.
792
Chapter 12
lRFD and Standard AASHTO Design of Concrete Bridges
(b) Minimum reinforcemellt
As discussed in section 12.3.4, the minimum reinforcement has to be the lesser of 1.2 M" or 1.33 M~ required by the applicable load combinations. Sec also flow-
chart Figure 12.1\ and Tables 12.10 and 12.1 1.
h
=
7.Sv'!;
=
7.5Y65OO - 605 psi = O.6ksi
Iff '" compressive stress due to effective
prestress only al the bonom fibers as de-
fined in section 12.3.4.
P, ( 1 + "" - A(
7"C,) =
-3.606 ksi from before.
Non-composite M dN '" M n + M s = 1438 + 1660 = 3098 (I-kip SI><- '" 20.060 in .~
Sb = 14,915 in,· From equation 12.1 3,
I)
M<. .. (f, + 1..)5,,- Md""(~: -
(20.060 _ )
= (0.6 + 3.6) 14.915 _ 3098 12 14.915
I
= 5220 - 1069 = 4151 fl-kip
1.2 M", = 1.2 X 4151 = 4981 fl-kip 1.33 M~ - 1.33 X 9316 = 12,390 (I-kip> 4981 ft-kip
Photo 12.5 Hanging Lake Viaduct, Glenwood Ca nyon, Colorado, lola I lenglh 1297 ft .• consisting of 34 spans, primarily 200-(1. lenglhs (Courtesy Figg Engineering Group, Tallahassee, Florida)
12.10 LRFD Shear and Deflection Design
793
hence, the lesser of the two moments controls, namely, 1.2 Mer = 4981. Mn or Mr = 11,364 > 4981 O.K.
9. Pretensioned Anchorage Zone The zone reinforcement is designed using the force in the strands just prior to release transfer. The LRFD specifications require that the bursting resistance, P r should not be less than 4.0% of the force in the strands, Fpi' before release, namely: P r=l.As ;:: 0.04Fpi Fpi = 48 x 0.153 x 202.5 = 1488 kips P r = 0.04 x 1488 = 59.5 kips Use a stress, Is, in the anchorage reinforcement not exceeding 20 ksi. Required area = 59.5/20 = 2.98 in. 2 Try No.5 closed ties; As = 2 x 0.31 = 0.62 in. 2 Number of ties = 2.98/0.62 = 4.8 Distance within which anchorage reinforcement has to be provided from beam end = h/5 = 72/5 = 14.4 in. Use No.5 closed ties at 3 in. center-to-center, with the first tie starting at 2 in. from the beam end.
Conclusion: Accept the design of the bulb-tee bridge for flexure. For the design to be complete, design for shear, interface shear transfer and deflection/camber checks have to be performed as in Example 12.2.
12.10 LRFD SHEAR AND DEFLECTION DESIGN Example 12.2
Design the web shear reinforcement for the bulb-tee beam in Example 12.1 at the critical section near the supports and the interface shear transfer reinforcement at the interface plane between the precast section and the deck situ-cast concrete. Also, verify if the span deflection is within the allowable limits. Solution:
1. Web Shear Design (a) Strain at centroid level of reinforcement =
0.90
ee
17.28 in.
=
Provide web steel when Vu > 0.5<1> (Ve + V p ) Critical section is the greater of 0.5 d v cotO or d v de = de = h - ee = 80.0 - 17.28 = 62.72 in. dv =
(d -"2a) = 62.72 e
5.27
~2-
. = 60.08m.
2:
0.9de = 0.9 X 62.72 = 56.45 in.
2:
O.72h
= 0.72
X
80
= 57.6 in.
d v = 60.08 in. controls as the largest of the three values
Assume 0 = 23° for a first trial 0.5d v cotO = 0.5 X cot 23°
= 70.77 > 60.08 in., use d v = 60.08 in the Ex equation
794
Chapter 12
LRFD and Standard AASHTO Design of Concrete Bridges
As the support bearing width is not yet determined, assume it conservatively = O. Consequently, the critical section for shear is 74.35 in. == 6.2 ft from the support, being larger than the dimension d v = 60.08 in. as stipulated by the LRFD AASHTO requirement; hence distance 74.35 in. controls for the critical shear section. x
L
6.2
= 120 == O.05L from the support face.
From equation 12.23,
In order to determine the value of 13 several computations have to be performed. Reinforcement strain Ex from equation 12.28 is initially using either Equations 12.28(a) or (b). Try Equation 12.28(a), Mu
d Ex
=
+ 0.5Nu + 0.5VuCot8 - Apsfpo
v
::;
2[ EsAs + EpsApsJ
0.002
At plane 0.05L, from Table 12.10 Mu
= 1.25(275 + 315 + 34) + 1.5 (68) + 1.75(368 + 160) = 1802 ft-kip
Corresponding shear:
+ 50 + 5) + 1.50(11) + 1.75(74 + 31) = 323 kips Nu = applied normal force at 0.05L plane = 0 Vu = 1.25(43
fpo
=
jacking stress = 0.70 fpw
=
fpe
fceEps
.
+ - - . It can however, be conservatively taken Ee
as the effective prestress fpe fee = concrete compressive stress at the centriod of the composite section due to both prestressing and the bending moments resisted by the precast section acting alone.
Distance from the c.g.c. of the composite section to the c.g.c. of the precast section. CI = Cbe -
At the critical section eel Section modulus
Cb =
54.6 - 36.6
=
18.0 in.
18.9 in.
=
Ie
545,894 18.0
Sl = -
= --- =
Cl
. 4 30,327 m.
ee = 17.28 in. f2 = 712 in. 2 ];
= _
Pe Ae
ee
(1 _eeCl) _ (MD + MSD ) f2
Sl
= _ 1095 ( _ 18.9 X 18.0) _ (275 + 315)(12) 767 1 712 30,327 = -0.746 - 0.233 = -0.979 ksi (c) 1; = pe
48
1095 = 149.0 ksi X 0.153
Ee = 4890 ksi
1I
795
12.10 LRFD Shear and Deflection Design
Vu = 323 kips
Mu
= 1802 kip-ft
Aps
= 48 x 0.153 = 7.344
Vu = 323 kips, Vp = 24.5 kips from Fig. 12.13 and the computations from the slope of the harped tendon to follow.
From Equation 12.29 /po
=
149.0 + 0.979 x 28,500/4890
or fpo
=
0.70 fpu
=
0.7
X
270
=
=
155.0 ksi
189.0 ksi
Use fpe = 155 ksi in the computations. Assume that the section contains at least the minimum Av to be verified later, using Equation 12.28 (a) for a first trial and no longitudinal mild steel, calculate the strain ex as follows:
(~ + 0.5 Nu + 0.5(Vu - ~) cot e ex =
Apsfpo)
2(Es As + Ep Aps)
1802 X 12 60.08 + 0 + 0.5(323 - 24.5)cot 23° - 7.344 2(0 + 28,500
X
X
155
7.344)
-426.8 . = 418608 = negatlve value. , hence, Eq. 12.28(c) would have to be used:
(~ + 0.5 Nu + O.5(Vu ex =
Vp)cote - Apsfpo)
2(Ee Ae + Es As + Ep Aps)
Ae = area of the concrete section on the flexural tension side of the member as shown in Figure 12.9. Hence, from Figure 12.15 of the flanged section used in this example, 1 Ae = 26 X 6 + "2 (26 - 6) X 4.5 + 6 X 4.5 + 6 X 25.5 = 156 + 45 + 27 + 153 = 381 in. 2 Ee = 4890 ksi from before when the concrete is beyond the 28 day strength
Hence, 1602 X 12 60.08 + 0 + 0.5(323 - 24.5) cot 23° - 7.344 ex =
2(4890 -426.8 4145 X 103
X
381 + 0 + 28,500
0 10
X
= -.
X
X
155
7.344
1 -3' I' 0 m. m.
Use loo0ex = -0.10 in Table 12.8(a). (b) Web shear strength, V", from 6 - 13 analysis.
Vu = 323 kips
From equation 12.15(b) Shear stress v =
(Vu - V p) bd '
v v
fpe from before = 149.0 ksi
where = 0.90 for shear
·, 796
Chapter 12
LRFD and Standard AASHTO Design of Concrete Bridges
Plane of crack
36 strands
12 strands
48'-6"
12'-0"
Beam Length = 121 '-0"
Figure 12.14
Beam tendon geometry
Figure 12.14 shows the inclination angle, 1jI, of the 12 harped strands, (65 - 15)
= 48.5 x 12 = 0.086
sin IjI Harped tendon force
= 12 x Vp
=
0.153
X
155.0 = 284.6 kips
284.6 sin a
=
284.6 X 0.086
=
24.5 kips
. d 323 - 0.9 X 24.5 093 k . R eqUlre v = 0.9 x 6 x 60.08 = . SI
. v 0.93 RatIo t; = 6.5 = 0.143 Entering Table 12.8 (a) for the values of 1000sx = -0.10 and (vJt;)
= 0.143, we get
a = 23.3°, which is close to the value of 23° assumed, and 13 = 2.80. Hence, Ve
=
0.031613
v7'c bvdv
= 0.0316 x 2.80V6.5
X
6
X
60.08 = 81.3 kips
(c) Selection of web reinforcement
From equation 12.26, check whether web reinforcement is needed, namely, if Vu > 0.5<1> (Ve + V p) 0.5<1> (Ve
+ Vp) = 0.5
X
+ 24.5)
0.9(81.3
=
47.6 kips,
<
Vu
=
323 kips
Use web steel. Required V,
=
Vu - Ve - Vp
=
(323) - 81.3 - 24.5 0.9
=
253.0 kips
From Equation 12.25, Available V, Trying No.4 stirrups, Av = 2
X
AJytdv cota = ---'-------
s
0.20 = 0.40 in. 2 cot a
_ 0.4 x 60.0 hence, 253 .0 -
X
= cot 23° = 2.356
60.08 s
X
(i) Maximum allowable web stirrup spacing: If vu < 0.125 s = 0.8 d v :;; 24 in. If vu > 0.125 s = 0.4 d v :;; 12 in.
t;
t;
2.475
,
giving s = 13.4 in.
797
12.10 LRFD Shear and Deflection Design
Limit Vu = 0.125 f~ = 0.125 X 6.5 = 0.81 ksi Actual Vu = 0.93 ksi > limit Vu = 0.81, hence use s = 0.40 d v = 0.40 = 24 but not to exceed 12 in. clc. Therefore, space the transverse web reinforcement at 12 in. c/c.
X
60.08
(ii) Minimum area of transverse reinforcement:
Avlft = 0.0316
• r;:; bvS V f~-I yt
_ . (6
- 0.0316 V ~ 6.5
X 12) ~
. 2/ft -_ 0.10 Ill.
Use No.4 stirrups at 12 in. center-to-center with the spacing to be increased along the span. (iii) Maximum shear resistance:
To ensure that the concrete in the web does not crush prior to yielding of the stirrups, :S 0.25f~bvdv
(Vn - Vp )
(Vn - Vp ) = Ve + Vs = 75.0 + 253.0 = 335.8 kips 0.25f~bvdv
= 0.25
X
6.5
6
X
X
60.08
= 585.8 kips> 335.8, O.K.
2. Interface shear transfer (a) Dowel reinforcement design
Assume that the critical section for shear transfer is the same as the vertical shear at plane 0.05 L from the support face. From load combination Strength I: Vu = 1.25(5.4) + 1.5(10.8) + 1.75(73.8 + 30.6) = 205.5 kips
d v = 74.35 in.
205.5
276 k· /. Ip Ill.
V uh = 74.35 =.
. V uh 2.76 . . ReqUired Vn = ~ = 0.9 = 3.07 kip/ill.
From Equation 12.32, available Vn = cAev + IL[Avffy + Pel For concrete placed clean, hardened concrete with interface contact not intentionally roughened, c
= 0.075 ksi
IL
= 0.6
b v = contact width between slab and precast flange top = 42 in. A~
. d
Ill.
epth
_
- 42.0
X
_.
1.0 - 42.0 Ill.
2
hence, 3.07 = 0.075 X 42.0 +O.6(A vf X 60 + 0) to give AVf = 0.0 in 2/in., < Av =0.40 in. 2 at 12 in. c/c vertical stirrups. On this basis, no special additional dowel reinforcement is needed. LRFD, however, also requires that if the width b v exceeds 36 in., a minimum of four bars are required as dowel reinforcement. Thus, use also two No.3 dowels at 12 in. c/c in addition to the No.4 vertical stirrups at 12 in. c/c, to give total A vf = 0.62 in.2/ft. (b) Maximum and minimum dowel reinforcement f~ =
Actual provided Vn = 0.075
4.0 ksi for the deck concrete X
42 + 0.6 (0~~2
X
60) = 5.01 kips/in.
798
Chapter 12
LRFD and Standard AASHTO Design of Concrete Bridges
From Equations 12.36 (a) and (b), the maximum allowable: 0.2f;A cv
= 0.2 x 4.0 x 42.0 = 33.6 kip/in. O.K.
In both cases, more than provided Vm
3. LRFD Minimum Longitudinal Reinforcement The longitudinal reinforcement at each beam section along the span has to satisfy equation 12.29: Asfs + Apsfps
2:
dMu A.. v~
+ 0.5 -Nu + (Vu A:"""
0.5V, - Vp ) cote
o/v
From Tables 12.8 and 12.9 at x = 0 from support, Vu = 1.25(47.9 + 55.3 + 6.0) + 1.50(12.0) + 1.75(78.1 + 33.9) = 350 kip Vs based on only the No.4 stirrups = 260.4 kips Mu = 0 Nu = 0 cot e hence,
= cot 22° = 2.475
-Mu- + 0.5 -Nu + (Vu - dv
0.5V, - Vp ) cot I)
= 0 + 0 + (~~~ - 0.5
X
253.0 - 24.5) (2.356)
= 560.5 kips
Number of straight strands at the support =36 Number of draped strands at the support = 12 From the assumed crack plane intersection with the strands in Figure 12.14, the distance of the intersection from the support = 6 + 4.22 cot 23° = 15.9 in. where the strand stops at 6 in. from face of the support. Transfer length = 60 x strand diameter = 30 in. The available prestress of the 36 straight strands at the support face is a portion of the effective prestress, fpe. Hence, use fpe
=
149.0
X
15.9 30.0
=
. 79.0 kSl
For the top draped strands, the crack in Figure 12.15 intersects the strands at a distance == 140 in. from the support face (compute from geometry of the dimensions in Figure 12.14). Consequently the effective prestress can be approximated at fpc = 149.0 ksi. Asfs + Apsfps = 0 + 36
X
0.153(79.0) + 12 x 0.153(149.0)
= 435.1 + 273.6 = 708.7 kips
>
560.5 kips, hence, no additional longitudinal reinforcement is needed.
4. Deflection and camber (i) Immediate deflection due to permanent loads
Compute the camber and deflection of the beam as detailed in the discussions and numerous examples of chapter 7. From the deflection table in Figure 7.6, Midspan 8
=
here, a = =
8
PL2 8 [ec + (ee - eJ.±3 a:] EJg L L
2 PL [~+ (ee - ec)] EJg 8 24 2
799
12.10 LRFD Shear and Deflection Design
From Example 12.1:
Icc = 1,095,290 in. 4
Pi = 1488 kips ee = 29.68 in. ee = 17.7 in. Eei
ws+h
= 0.922 kip/ft
Wbarrier
= 0.300 kip/ft
Ae = 767 in. 2
= 4620 ksi
14,915 in?
Eee = 4890 ksi
Sb
w D = 0.779 kip/ft
S t = 15,421 in?
Ie
o= I
=
= 545,897 in. 4
1488(120 X 12? [29.68 + (17.7 - 29.68)] 4620 X 545,897 8 24
= 1.22 (3.71 - 0.50) = 3.92 in. t (camber) WD
per inch = 0.799/12 = 0.065 kip/in.
5wL4 384Ee;Ig
OD=--= Ws+h
5(0.065)(120 X 12)4 384 X 4620 X 545,894
1.44in.-!-
= 0.922 kip/ft = 0.077 kip/in. 5(0.077)(120
X
12)4
.
OD = 384 X 4888 X 545,894 = 1.61 m. -!Wbarrier =
0.300 kip/ft
=
5(0.025)(120
0.025 kip/in. X
12)4
.
OD = 384 X 4888 X 1,095,290 = 0.26 m.-!-
(ii) Immediate deflection due to transient loads Live load deflection limit = Ll800. LRFD specifications require that all the bridge deck beams be assumed to deflect equally under applied live load and impact. They also stipulate that the long-term deflection may be taken as four times the immediate deflection. This stipulation is too general and the designer is well-advised to use other more refined methods. The larger is the span the more is the needed accuracy. It should be emphasized that computed deflection values can differ from actual deflections by as much as 30 to 40% depending on the concrete modulus and stress-strain relationship assumed and the degree of accuracy of the method used in the computation. The following deflection computation methods from chapter 7 can give reasonable step-by-step values during the loading history
• PCI multipliers method (Sec. 7.7.1) • Incremental time-step method (Sec. 7.7.2) • Approximate time-step method (Sec. 7.7.3) From Figure 12.11 in Example 12.1, the number of bridge beams = 6 and the number of lanes = 4 DFD
=
distribution factor for deflection
= number of lanes divided by number of beams
4
= 6 = 0.667 lanes/beam It is more conservative to use moment distribution factor DFM = 0.732
800
Chapler 12
Table 12.12
LRFD and Standard AASHTO Design of Concrete Brid9.es
Long-Term Camber and Deflection Noncomposite PCI
Transfer (in .)
Multipliers
Composite PCI Multipliers
3.921 -1.44t NC12.48i
1.80 1.85
2.20 2.40
8.62i -3.47!
-t.6J !
1.85 1.85
2.40 2.30
Net 5. 151 -3.84t --O.60t
"
Prestress
w, W S-h
IVbo,..ier
'u
'"
Final S
--0.26> --OAI t --O.78t 2.481
511'''''
(in,)
-OA) J,
--O.78t +0.48.
Design lane load. W= 0.64 DFM =:
0.64 kip/ ft (0.732) "" 0.468 kipj ft/ bcam
= 0.039 kip/ in/ beam
5
8LL
=
384EJ
5(0.039)(120 X 12)'
_
.
t
384 X 4888 x 1.095.290 - 0.4110.
The transient truck load and impact deflection is dctcnnincd from innuence lines of wheel position for maximum moment. For a 120-ft span. the 72 kip
Photo 12.6
West Kowloon Expressway V iaduct, Hong Kong, 1997. 4.2-Km dunl
three-lane causeway connecting Western Harbor Crossi ng to new airport (Collrtesy Institution of Civil Engineers. London)
801
12.11 Standard AASHTO Flexural Design of Prestressed Bridge Deck Beams
resultant of the axial loads falls at 2.33 ft from the midspan. The deflection at midspan = 0.8 in. ,J.. BLT =
0.8(IM)(DFM)
=
0.8(1.33)(0.732) = 0.78 in. -J,
Using the PCI multipliers from Table 7.1, a summary of the long-term cambers and deflections are given in Table 12.11. .
L
Allowable deflectIon B = 800
=
> actual
120 X 12 . 800 = 1.80 m. (down) =
0.49 in.
O.K.
Adopt the bridge deck design of the interior beam in Example 12.1 and 12.2.
12.11 STANDARD AASHTO FLEXURAL DESIGN OF PRESTRESSED BRIDGE DECK BEAMS (LFD) ExampleU.3 Design for flexure, an interior beam of the bridge deck in Example 12.1 (adopted from Ref. 12.11) using the standard AASHTO Design Specifications for HS-20 lane and truck loads. Use the same data and allowable stresses of the materials as in the indicated example except where they differ from the LRFD allowable stresses. Solution: 1. Transformed deck slab controlling width From example 12.2 Step 1, Ees =
3830 psi
Eei =
4620 psi at transfer
Eee =
4890 psi at service
Average spacing between beams = 108 in. Transformed flange width bm = 84 in. 2. Properties of Section Non-Composite Ae = 767 in.
2
Composite Ae = 1402 in. 2
h = 72 in.
h
Ie = 545,894 in. 4
=
80 in.
Icc = 1,095,290 in.4
36.60 in.
Cbe
= 54.6 in.
Ct
= 35.40 in.
C te
= 17.4 in.
?
2
Cb =
= 1051 in.
Sb = 14.915 in. 3
st =
15,421 in?
Ctse
=
25.4 in.
?
=
712 in. 2
Sbe
=
20,060 in. 3
S~ = 62,950 in. 3 S~s =
55,284 in?
self-weight
WD
= 7991b/ft
slab
WSDl = 9001b/ft
3. Bending moment and shear forces
0.5
1/
haunch
=
22 lb/ft
Barrier weight
WSD2 = 100lb/ft
2-in. future wearing surface
WSD3 =
2001b/ft
802
Chapter 12
LRFD and Standard AASHTO Design of Concrete Bridg.es
Live load (truck load) in AASHTO standard specifications would be based on HS-20 trucks. Number of lanes = 48/12 = 4 lanes (a) Distribution/actor/or moment
Live load in the standard specifications is either the standard truck or lane loading corresponding to HS-20. In LRFD, both have to be used in the design. From Section 12.2.2, the live load distribution factor for moment for a precast beam is DFm = S/5.5 = 9.0/5.5 = 1.636 wheels per beam, where S = average spacing between beams in feet.
"21 DFm =
0.818 lanes per beam
the live load impact factor 1= L
1
0 125
~ 30%
or,
50 2 120 + 150 = O. 04 In LRFD, This factor has a maximum 33% value hence, VLL+I
= (shear force per lane) (DFM) (1 =
+ I)
(shear force per lane) (0.818) (1 + 0.204) kips
= (shear force per lane) (0.985) kips MLL+I
= (moment per lane) (DFM) (1 + I) = (moment per lane) (0.818) (1 + 0.204) kips = (moment per lane) (0.985) kips.
Load contributions from Equation 12.5 and Table 12.1 show that load combination Group I controls. Group I service load design = 1.00 D + 1.00 (L + 1) Group I factored load design = 1.3 [1.00D + 1.67(L + I)] (b) Shear and bending moments Vx = (w) (0.5 L - X)
Mx
=
0.5 (w) (X) (L - X)
As an example, the following are computations for the shear and moment at midspan, at the support and at the critical shear section: At midspan, Vx = 0
MD = 0.5(0.799) (60) (60) = 1438 ft-kip At support, V = 0.799(60) = 47.9 kip At critical shear section, The cgs of the prestressing steel is e = 17.1 in. near the support section (see subsequent computations) d = 80 - 17.1 = 62.9 in.
> 0.80 h =
64 in.
Use d = 64.0 in. Critical shear section at h/2 = 80/2 = 40 in. = 3.33 ft V3.33 M3.33
= 0.799 [(0.5) (120) - 3.33] = 45.3 kips = 0.5(0.799) (3.33) (120 - 3.33) = 155.4 ft-kip
803
12.11 Standard AASHTO Flexural Design of Prestressed Bridge Deck Beams Table 12.12
Standard AASHTO (LFD) Service Shear and Moment Due to Dead Load
Beam Weight WD
(Slab + Haunch) Weight W SD1
Distance Section Shear Moment Shear Moment X XlL Vg Mg Vs Ms
ft 0 3.33* 12 24 36+ 48 60
0.0 0.028 0.1 0.2 0.3 0.4 0.5
Barrier Weight
Wearing Surface WSD3
WS02
Live Load Plus Impact
Shear Moment Shear Moment Shear Moment Vb
Mb
Vws
Mws
VLL+I
MLL+I
kips
ft-kips
kips
ft-kips
kips
ft-kips
kips
ft-kips
kips
ft-kips
47.9 45.3 38.4 28.8 19.2 9.6 0.0
0.0 155.4 517.8 920.4 1,208.1 1,380.7 1,438.2
55.3 52.2 44.3 33.2 22.1 11.1 0.0
0.0 179.3 597.5 1,062.1 1,394.1 1,593.2 1,659.6
6.0 5.7 4.8 3.6 2.4 1.2 0.0
0.0 19.4 64.8 115.2 151.2 172.8 180.0
12.0 11.3 9.6 7.2 4.8 2.4 0.0
0.0 38.9 129.6 230.4 302.4 345.6 360.0
65.4 63.6 58.3 51.2 44.1 37.0 29.9
0.0 211.5 699.7 1,229.1 1,588.4 1,799.6 1,851.6
*Critical section for shear +Harp point
The values for shear and moment for all permanent and transient loads are tabulated in Table 12.12 (Ref. 12.11). Compare the tabulated values with those computed by the LRFD method in Table 12.9 and 12.10.
4. Design of Bulb-Tee Prestressed Interior Beam 1. Selection of prestressing strands Due to applied gravity loads, the unfactored stress at bottom fibers: fb
=
MD
+ MSDl + ----::..::..::...--='-=------==--=MSD2 + MSD3 + M LL + 1 Sb
Sbe
(1438 + 1660)12 (180 + 360 + 1852)12 . = 14,915 + 20,060 = 3.923 kSI
Vi":
Allowable tensile stress It = 6 = 6 V 6500 = 484 psi = 0.484 ksi Required precompressive stress at the bottom fibers after losses = (fb fbp = 3.923 - 0.484 = 3.439 ksi Assume that the tendon c.g.s. is at a distance Yb = 4 in. from the bottom fibers.
ee = Ce
-
Yb
It)
= 36.60 - 4.00 = 32.60 in.
Pe P e X 32.6 P e Peee fbp due to prestress = Ae + S; or 3.439 = 767 + 14,915 Pe = 986 kips
Assume total prestress loss = 25% Pi
=
986 1 _ 0.25
=
15k' 31 IpS
Assume using l/2-in. diameter 7-wire 270-K low-relaxation strands (Aps 0.153 in.2), . ReqUlred number of strands
=
1315 02 5 0.153 X 2 .
=
42.44 strands
Try 44 strands. After trials and adjustments, assume that the strands pattern is as shown in Figure 12.16 with 10 strands harped at 48 ft from the support. From data, cb = 36.60 in. and c, = 72 - 36.60 = 35.40 in.
804
LRFD and Standard AASHTO Design of Concrete Brid~es
Chapter 12
ee =
Cb -
[2
x 70 + 2 x 68 + 2 x 66 + 2 x 64 + 2 x 62 + 2 x 8 + 8 x 6
+ 12 x 4 + 12 x 2]/44
= 36.60 - 18.09 =
e3.33
18.51 in.
= 17.1 in. at the critical shear section.
ec =
Cb -
+ 12
+2
[2
X
16
X
4
+ 12
X
X
14
+2
X
12
+2
X
10
+4
X
8
+8
X
6
2]/48
= 36.6 - 5.81 = 30.79 in. Computing the total losses in prestress by the detailed method and the examples of Chapter 3, the total loss of prestress was 24.9%. fpe before losses
= 0.75(270) = 202.5 ksi
hence, adjusted fpe = 202.5(1 - 0.249) = 152.1 ksi Pe
= 48(0.153) (15.2) = 1024 kips
Common practice assumes that a prestress relaxation and other losses at prestressing amount to 9 to 10% Use 9% here to get fpi = 202.5(1 - 0.09) = 184.3 ksi
Pi = 44(0.153) (184.3) = 1240 kips From Example 12.2, fei
= 0.6 f~ = 0.6(5500) = 3300 psi (c)
f,i
= 7.5~ = 7.5Y55oo = 556 psi (T)
If the computed tensile stress at transfer exceeds 200 psi or 3 ~ = 220 psi, whichever is small, bonded reinforcement has to be provided to resist the total tensile force in the concrete, computed on the basis of uncracked section.
2. Check of concrete unfactored stresses The standard AASHTO allowable stresses are as follows, Case I for all load combinations:
No. Distance from Strands bottom (in.) 2 70
Harped strands
No. Distance from Strands bottom (in.) 2 16 2 14 2 12 2 10 4 8 8 6 12 4 12 2
2 2 2 2
Figure 12.15
64 62
No. Distance from Strands bottom (in.) 2 8 8 6 12 4 12 2
!:.l121 spac.11L @2" At midspan (e c = 30.79")
68 66
At ends (ee = 18.51") Bulb-tee prestressing strand pattern
805
12.11 Standard AASHTO Flexural Design of Prestressed Bridge Deck Beams
Precast beam fe = 0.60 f~ = 0.6(6500) = 3900 psi Deck slab fe = 0.60 f~ = 0.6(4000) = 2400 psi Case (II) for effective pretension force + permanent dead loads: Precastbeamfe =
0.40f~
= 0.4(6000) = 2400 psi
Deck slab fe = 0.40 f~ = 0.4(4000) = 1600 psi
~ (pre tensioning force
Case (III) for live load +
+ dead load)
Precast beamfe = 0.40 f~ = 0.4(6500) = 2600 psi Deck slab fe = 0.40 f~ = 0.4(4000) = 1600 psi Allowable tensionft = 6~ = 6Y6500 = 484 psi Allowable ft at transfer = 3
v7'c = 242 psi
a. Stresses at Transfer Initial fpi = 0.7 fpu = 202.5 ksi. Common practice assumes that initial relaxation losses at prestressing amount to 9 to 10%. Use 9% reduction infpi. hence, Pi = (0.9)(202.5)(0.153 x 48) = 1338 kips i. Support Section From Chapter 4, Equation 4.1 (a),
t'
fb
= -
~J 1 - e~t) - ~~
= -
1~6~0 (1
=
-
18.517~235.40) -
0 = -0.129 ksi (c),
O.K.
Pi ( eecb) MD -::4 1+7 + Sr e
= _
1240 (1 + 18.51 X 36.30) + 0 767 712
= -3.14 ksi (C) == fei = 3.3 ksi O.K. (ii) Midspan Section.
f
= -
1240 (1 _ 30.79 X 35.40) _ 1438 X 12 767 712 15,421
= 0.858 - 1.119 = -0.261 ksi (C), no tension allowed, hence, O.K.
11 = - 1240 (1 + 30.79 767
b
X
712
36.60) + 1438 X 12 14,915
= -4.175 + 1.157 = -3.018 ksi (C) < 3.300 ksi allowed O.K. b. Stresses at Service load: (i) Midspan Section precast section fiber stresses:
concrete stress at top fibers at midspan due to all loads:
t' =
- Pe Ae
ee
=
(1 _ r
30.79 in.
eeCt) _ MD + M DS1 _ M DS2 + M DS3 _ M LL + 1
st
S~
S~
ee = 18.51 in.
From the moment values at midspan tabulated in table 12.10 for load combinations:
806
Chapler 12
LRFD and Standard AASHTO Design of Concrete Bridges
Photo 12.7 Natchez Parkway Arches, Nashville Tennessee. America's firs t segmental arch bridge; principal arch span is 582-ft. long and has a vertical clearance of 137 ft (Collrtesy Figs Engineering Group, Tallahassee. Florida) Case(I):
_ 1024 (I _ 30.79 x 35.40) _ ( 1438 + 1660)12 _ (360 + 180)12 767 712 15.421 62.950
=
f
( 1852)12 62.950 = 0.708 - 2.411 - 0.103 - 0.353 = -2.159ksi (C)
Case (II):
f = 0.708 - 2.41 I - 0.103
= -
1.806 ksi (C)
Case ( /II):
f
= 0.5(0.708 - 2.411 - 0.103) = - 1.256 ksi (C)
All compressive stress arc less than the a llowablcf~= 3900 psi O.K. (ii) Midspa/l sectiOll bottom fiber slresses
P, ( M v + Mas! MOSl + MvSJ L.L"/ ) i b=-l +erc -b+ + +MA< r' Sb S~ SIK
f~
=
_ 1024 (I + 30.79 X 36.60) + ( 1438 + 1660)12 + (360 + 180)12 767 712 14,9 15 20.060 ( 1852)12
+ 20,060
"" - 3.437 + 2.493 + 0.323 + 1.10 = 0.486 ksi (T) ill:
allowablcf, = 0.484 ksi
O .K.
807
12.11 Standard AASHTO Flexural Design of Prestressed Bridge Deck Beams
(iii) Midspan slab top-fiber stresses, assuming concrete strength same as of the pre-
cast beam Case (I): ft=
+ M DS1
MD
Sb
ft=
M LL + 1 Sbe
---
(180 + 360) 1852(12) 55,284 - 55,284
=
-0.117 - 0.402
= -0.519 ksi (C) < allowable fe = 2400 ksi O.K. Case (II): ft = -0.117 ksi (C)
O.K.
Case (III):
f = 0.5( -0.117)
= -0.461 ksi (C)
- 0.402
O.K.
E c = 33 W1.5vt::t.'e
=
v'4oOo (slab)
= 0.78 6500 (beam) Hence reduce these stresses by the 0.78 multiplier in order to get the true stresses at the top slab fibers. Modular ratio
<
~
V
3. Ultimate Strength (Limit State at Failure)
a. Normal flexural resistance moment Mu = 1.3 [MD + MSDl + MSDZ + MSD3 + 1.67 (MLL+l)]
= 1.3(1438 + 1660 + 180 + 360 + 1.67
X
1852) = 8750 ft-kip
From equation 12.9(c) 'Y fpu] fps = fpu [ 1 - 131 p f;
For the depth of the compressive block use the slab f;
= 4.0 ksi, 131 = 0.85
'Y = 0.28 for low-relaxation strands
b = flange wdith = 108 in.
ee = 30.79 in.,
= 36.60,
Cb
hence, Yb = 36.60 - 30.79 = 5.81 in. dp = distance from the top of the deck to the centroid
of the prestressing strands. = beam depth (h)
+ haunch + slab thickness (hf
+ (0.5 + 7.5) - 5.81 Aps = 44(0.153) = 6.732 in.2 = 72
Aps p
6.732 X 74.19
= bdp = 108
-
Yb)
= 74.19 in.
= 0.00084
Consequently, fps
=
0.28 270] 270 [ 1 - 0.85 0.000844:0 Apsfps
6.732 X 265.0 X 4.0 X 108
a = 0.85f;b = 0.85
=
. 265.0 kSI .
.
= 4.86 Ill. < 7.5 Ill.
808
Chapter 12
LRFD and Standard AASHTO Design of Concrete Bridges
hence design as rectangular section. Available M u = 4>Mn = Aps/ps
(d - ~) 4.~6)]
= 1.0 [ 6.732 (265.0) (74.19 _
= 128,018 in.-kip > required M u = 8750 ft-kip, O.K. c=
a/131 = 4.86/0.85 = 5.72 in.
ddt = 5.72174.19 = 0.24 < 0.42 limit in Sec. 12.3.1; hence, the section is tension controlled and 4> = 1.0 is verified. b. Maximum Reinforcement
fps p f;
:S
0.36131
p ;;
=
0.00084
:S
0.36
X
0.85 = 0.306
X
~~~ =
0.0557
<
0.306
O.K.
c. Minimum Reinforcement The total amount of pre tensioned and post-tensioned reinforcement should be adequate to develop an ultimate moment such that 4> Mn
2:
1.2 Mcr
From equation 12.13,
s: - 1 ]
Mcr = (fr + fce)Sb - M dnc [ SbC
f,. = 7.5Vfc = 7.5v'6500 = 605 psi = 0.605 ksi Concrete stress due to prestressing only, after all losses is fb = 3.437 ksi (C)
M dnc = non-composite dead load moment due to beam self-weight and slab weight =
1438
+ 1660
=
3098 ft-kip
1.2 Mcr = 1.2 [11 (0.605 + 3.437)20,060 - 3098 2
G~:~~~
-
1)] = 6896 ft-kip
It should be noted that contrary to the LRFD specifications, the standard specifica-
tion stipulates that this requirement has to be satisfied only at the critical section. d. Pretensioned anchorage zone
Before initial losses, fpi
=
Ap,(0.75 fpu)
= 6.732(0.75 4 % fpi
= 0.04
X
X
1363
270) = 1363 kips
= 54.5 kips
Allowable Is = 20 ksi Hence, required Av
= 5;~5 = 2.73 in. 2
Try No.5 vertical stirrups in the rectangular anchorage zone region Av = 2 X 0.305 = 0.61 in. 2
. 2.73 No. of shrrups = -6o. 1 Precast Section dp
=
=
4.5
(h - Yb) = 72 - 5.81 = 66.19 in.
12.12 Standard AASHTO Shear-Reinforcement Design of Bridge Deck Beams
809
( 2"a) = 66.19 - -2- = 63.76 in.
d v = dp
4.86
-
Distance within which anchorage reinforcement has to be provided
= ; =
63~76
= 15.94 in.
Use five # 5 closed U-stirrups at 3 in. center-to-center at each beam end.
12.12 STANDARD AASHTO SHEAR·REINFORCEMENT DESIGN OF BRIDGE DECK BEAMS Example 12.4 Design for shear, an interior beam of the bridge deck in Example 12.3 using the standard AASHTO design specifications for HS-20 Lane and Truck loads. Use the same data and allowable stresses of the materials as in the indicated example. Use the refined flexural and web shear approach for determining the nominal strength of the plain concrete in the web. Also design the interface shear transfer reinforcement and check the deflection and camber of the beam. Solution:
1. Shear Reinforcement The AASHTO standard specification follows the ACI-318 code for shear and torsion which are detailed in Chapter 5, Sections 5.5 and 5.6 as well Section 5.18 for torsion. Vs ::s
<1>
(Ve + Vs), where <1> = 0.90 vs.
<1>
= 0.85 in ACI.
Other strength reduction factors, <1>, also differ from the ACI factors. The computations have to be based on a factored shear value at a distance 1/2 h from the face of the support. The nominal shear strength, Ve , of the plain concrete in the web has to be the lesser of the flexural shear, V ci , and the web shear, Yew'
a. Flexural shear, Vc; From Equation 5.11, V ci = 0.6>.. ~ bwdp + Vd +
~ (Mer) 2:: 1.7>" ~ bwd
p
Mmax
where b w = 6 in. From table 12.12 for standard AASHTO loads in Example 12.3 Vd = total unfactored dead load at the critical section
= 45.3 + 52.2 + 5.7 + 11.3 = 114.5 kips VLL+l
(unfactored) = 63.6 kips V u = factored shear force at the critical section
+ 1.67VLL + 1) = 1.3(114.5 + 1.67 X 63.6) = 286.9 kips Md = 155.4 + 179.3 + 19.4 + 38.9 = 393.0 kips = 1.3(Vd
M LL + 1 = 211.5 kips
Mu = 1.3 (Md + 1.67 M LL + 1)
= 1.3(393.0 + 1.67
X
211.5)
= 970.1 ft-kips
Vi = factored shear force at the section due to externally applied loads occurring simultaneously with Mmax.
= (VU - V D ) = 286.9 - 114.5 = 172.4 kips. This is on the conservative side since the factored V u is reduced
810
Chapter 12
LRFD and Standard AASHTO Design of Concrete Bridges
by the unfactored Yd. Mmax
= (Mu + Md ) = 970.1 - 393.0 = 577.1 ft-kip
From equation 5.12,
Mer
Sbe (6~ + fee - fd). Note that the factor "6" in the term 6~
=
is conservative and maybe unjustified since the modulus of rupture,!" is taken as 7.5~ and tests indicate even higher values.
fee = compressive stress due to prestress after losses at the extreme fibers of the section where tensile stress is caused by externally applied load. e
P (1 + 7 eelCb)' were h . at t h · · I sectIOn . = - Ae eel = 195 . Ill. e cnttca
= _ 1024 (1 + 19.5 767
fd
X 36.60) 712
= -2.673 ksi (C)
stress due to unfactored dead load at the extreme fibers of the section where tensile stress is caused by externally applied load
=
. _ [(155.4 + 179.3)(12) + (19.4 + 38.9)12] _ 20,060 - 0.304 kSl 14,915 Hence, Mer
=
20,060 (6Y6500 ----u1000 + 2.673 -
0.304
)
.
= 4776 ft-kip
Yb = distance of cgs of the prestressing strands at the critical section from the bottom fibers = 17.12 in. dpe
= he - Yb = 80 - 17.12 = 62.88 in.
0.8 he = 64 in., controls
used dp = 64 in. Hence,
Vci =
0.6Y6500(6.0 X 64) 172.4 X 4776 . 1000 + 114.5 577.1 = 1,559.8 kips
Minimum Vci = 1.7~ bwd
=
0.6Y6500(6.0 1000
X
64)
= 52.6 kips « 1,559.8 kips O.K.
b. Web shear Vew From Equation 5.15,
Vew
=
(3.5>.. ~ + 0.3fe)b wdp + V p
where fe is termed as Fpe in AASHTO notation. It is the concrete stress at the centroid of the section resisting all externally applied load. -
f, = e
Pe ( ee1(Cbe - Cb)) MD(cbe - Cb) 1+ --=--'---"-'----"-Ae r Ie
--
From section properties, cbe = 54.6 in. and cb = 36.6 in. _ = _ 1024 (
fe
767 =
1
_ 19.5(54.60 - 36.60)) + 334.7 X 12(54.60 - 36.60) 712 545,894
-0.676 + 0.132
=
-0.545 ksi (C)
12.12 Standard AASHTO Shear-Reinforcement Design of Bridge Deck Beams
Vp
=
811
vertical component of prestressing force.
From Figure 12.15, the tangent of angle tV subtended by the 10 harped tendons
65 - 15 X 12
== 48.5
.
== sm tV == 0.086
Vp = ApJpe sin tV = (10
0.153) 149.0 X 0.086 = 19.61 kips
X
Hence,
Vew = ( =
3.5 X 1.0Y6500 1000
171.15
+ 19.61
=
+ 0.3
X
) 0.545 6.0 X 64
+ 19.61
190.8 kips, controls since it is less than Vci
c. Selection of web steel
Ve = 190.8 kips VU<(Ve+Vs)
or,
Vs=(:u- Vc)
. 286.9 . ReqUired Vs = 0.90 - 190.8 = 128.0 kips , ~
, r;:;
Maximum allowable Vs = 8 v f~ bwdv = 9 v 6500
>
6.0 X 64 . 1000 = 247.7 kips
128.0 kips O.K. (the section depth adequate for shear).
Avlydv V =s s Required Av = [,Vd = 6 ~28.~ = 0.033 in.2jft = 0.0028 in. 2jin. s y v O. X 4.0 s
. . Av _ 50b w Mmlmum ----; -
T - (5060,0006.0) -_ 0.005 m.. 2'jm., controls. _
X
Using No.4 two-legged U-stirrups in the rectangular end section, Av = 2 X 0.20 = 0.40 in. 2
Av- = --5 0.40 = 80 In. . · s = --. Spacmg umtAv 0.00 Maximum allowable spacing s = 0.75 he or 24 in.
= 0.75(72 + 7.5 + 0.5) = 60 in. or 24 in. Use No.4 U-stirrups at 12 in. center-to-center in the rectangular end block section over a width =h = 80 in. Beyond the end of the anchorage block, stirrups would no longer be needed. However, it is useful to use minimum vertical mesh reinforcement in the web along the span. The 12-in. spacing is necessitated by the interface horizontal shear requirement.
2. Intetface shear reinforcement Determine the interface shear force for the critical section at ! he from the support.
a. Contact sutface roughened or minimum ties used V u = 286.9 kips V nh
<
V nh
=
nominal horizontal shear strength
2:
V U = 286.9 = 318 8 k' 0.9 . IpS
812
Chapter 12
LAFD and Standard AASHTO Design of Concrete Bridges
Allowable V"" == b.,dpc where
b< = width of cross-section at the COnlaet surface being investigated fo r ho rizontal shear .. 42.0 in.
dpc = 62.88 in. from before Available VoM =
80(42.0 x 62.88) 1000
- 2 11.3 kips
<
3 18.9 ki ps
he nce. dowel reinforcement is needed.
b. Minimum ties provided and contact surface roughened
V.....
=
35Ob,dpt =
350(42.0 X 62.88) . . 1000 .. 924.3 kips > 3 18.8 kIps O. K.
Use V,.jo=3 18.8kips From equation 5.33,
. . MII111llUm A ,'h ""
50bj.,~
h
50b.,s -T per dowel
Assume 12-in. dowel spacing. Minimum A,,~ per dowel
(50X42 x 12) '=
60,000
='
0.42 in.2/ft
Photo 12.8 Pier support for Sioney Trail Dow Ri ver segmental Bridge. Calgary, Alberta, span 1562 fl , deck width 69 n. the deck rises 89 to 118 ft above the river valley (Collrtesy J ames Skeet- Reid Crowthcr Enginecring. Calgary)
813
12.13 Shear and Torsion Reinforcement Design of a Box-Girder Bridge
Available dowels form shear reinforcement: = No.4 V-stirrups at 12 in. c./c. = 0.40 in.2/ft
O.K.
Beyond the rectangular end block zone of 80 in., add in the web additional #4 dowels at 12 in. c.lc. to compliment the single #4 bars available in the 6-in. thick web. Maximum allowable spacing
=
4 bw
=
4 X6
=
24 in.
O.K.
Note that the vertical web shear reinforcement is utilized here to provide for the required dowel reinforcement.
3. Deflection Computations The deflection computations are similar to those given in Example 12.2 except that in the standard AASHTO specifications, fatigue live load for deflection is disregarded. from Table 12.11, Example 12.2, the final long-term deflection becomes: 1)
= -8.62 + 3.46 + 3.86 + 0.60 + 0.41 = -0.29 in. (camber)
<
L
800
=
120 X 12 800
=
. 1.80 Ill.
O.K.
Adopt the bridge-deck design of the interior prestressed beam in Examples 12.3 and 12.4.
12.13 SHEAR AND TORSION REINFORCEMENT DESIGN OF A BOX-GIRDER BRIDGE Example 12.5
A single span composite two-lane box girder bridge has a span of 90'-0" (27.5m) The deck is composed of seven AASHTO BIII-48 box beams at 4'-0" on centers to form a 28'-0" bridge deck with a traffic pathway width = 25'-0" as shown in Figure 12.16. Each beam is subjected to a factored shear V u = 140 kips at the critical support section, a corresponding moment at that section = 320 ft-kip and a torsional moment T u = 165 ft-kip. Design the shear and torsional reinforcement for this bridge section, using the LRFD expressions, given:
n=
5.0 ksi
f;i = 4.0 ksi fpu = 270.0 ksi fpy
=
0.90 fpu
fpi
<
0.75 fpu = 202.5 ksi
=
243.0 ksi
Total prestress loss
fy = 60.0 ksi Eps
=
28,500 ksi
Es = 29,000 ksi Ac = 813 in. 2
h
=
39 in.
Ig
=
168,367 in.4
Cb
= 19.29 in.
Ct
19.71 in.
=
Sb
=
8,721 in?
st =
8,542 in. 3
Ed
=
3840 ksi
Ec
=
4290ksi
=
22 %
814
Chapter 12
LRFD and Standard AASHTO Design of Concrete Bridges
25'-0"
,
(2) 7/8 in.dia. - 150 ksi bars through 2 in dia hole in each diaphragm
/1 ;.
\ I
B~
B;;um;nous we.,;ng surtace
B
'/':
~r1 ~
'/':
8V1
Er1
8V1
~
/.
~
'/
7 Beams @ 4'-0" = 28'- 0"
c
·1
(a)
3/4"
-.----L--t -,--11-_ _ _----, 51)/2"
.
.
~Shear key
3" (Typ)
1
2@36"
39" 28"
jl 5
1'2" /;
debonded strands at the support section ••
•••
• iii • • • iii • • iii • • iii • • iii • • iii • • • iii·
6@4" 23 @ 211
38"~5"-
t -5"/'"
(b)
Figure 12.16 Two-Lane Box-Girder Bridge. (a) Roadway cross section, (b) Cross section of a component beam unit at midspan and at end sections. The end section has seven strands de-bonded.
Solution: 1. Effective shear depth, d. The strands are horizontal, Vp = 0 From Figure 12.17,
At midspan, Aps = 31 !-in. dia., 7-wire, low-relaxation 270-K strands
At support, Aps
=
4.437 in. 2 at the bottom fibers.
=
24 stands, since 7 are de-bonded
=
3.366 in. 2
Midspan c.g.s. of strands from bottom: _ 23 Ybe -
X
2+6
X
31
4+ 2
Midspan dp = h - he = 39.0 - 4.58 = 34.42 in.
X
36 _ 4 58 . - . Ill.
12.13 Shear and Torsion Reinforcement Design of a Box-Girder Bridge
815
Support cgs of strands from bottom: Ybe
(23 - 7)2 + 6 X 4 + 2 X 36 31 _ 7
=
.
= 5.33 m.
support dp = h - he = 39.0 - 5.33 = 33.67 in. Aps/psdp + As/yd
----=----=----=----'--
where, de =
Apsfps + As/y
d v = effective shear depth = distance between resultants of tensile and compressive forces, but not less than 0.90de or 0.72 h.
To determine the neutral axis depth, c, and the equivalent rectangular block depth, a, use the midspan section for the computations as the section of maximum moment (Mu= 0 at support). Assume the neutral axis within the 5! in. "flange." c=
Apsfpu + As/y - A;f; 0.85f;131b + kAps
=
4.437(270) + 0 + 0
(~:)
0.85
X
5.0
X
0.80
X
48 + 0.28
X
4.437
C~~~2)
6.95 in.
a = 131c = 0.80
6.95 = 5.56 in.
X
>
5.5 in., hence treat as a flanged
section with width b = web width b w bw
=2
c=
= 10 in.
5
X
Apsfpu + Asfy - A;f; - 0.85f;131(b - bw)hf 0.85f;131bw + kAps
4.437
=
270 + 0 - 0 - 0.85
X
0.85
(~:)
X
11.14 in.
5.0
X
0.80
> 5.5 in.
X
X
0.80
10 + 0.28
X
X
5.0(48 - 10)5.5
270 ) 4.437 ( 34.42
O.K.
a = 131c = 0.80 X 11.14 = 8.91 in.
c)p = 270 (1 - 0.28 {11.14}) 34.4 = 245.4 kSI.
(
fps = fpu 1 - k d
8.91 . = 29.22 m. ( 2"a) = 33.67 - -2-
d v = dp
0.9 de = 0.9
-
X
O.72h = 0.72
33.67
X
39
= 30.30 in. (controls) = 28.08 in.
2. Angle of inclination 6 of the diagonal compression struts Critical section near the support is the larger of 0.5d v cot 6 or d v from the face of the support. 6 is obtained from Table 12.8 using the values of v/f~ and Ex' From equation 12.43(a), v=
Vu -
TpPh
+- A;h
= 0.9 from Table 12.1(a)
AOh = (48 - 2
X
1.5 for clear cover - 2
X
0.25 for stirrups)
X
816
Chapter 12
LRFD and Standard AASHTO Design of Concrete Bridges
(39 - 2 x 1.5 - 2 x 0.25)
= 44.5 x 35.5 = 1580 in?
Ph = 2(44.5 + 35.5) = 160 in. bv = 2
v
=
X
5 = 10 in.
140 - 0 165 X 12 X 160 + ------:--0.9 X 10 X 30.3 0.9(1580)2
= 0.515 + 0.141 = 0.656 ksi ~
f;
= 0.656 =
5.0
o.
131
For torsion adjustment, change the numerator in Eqs. 12.28(a), (b), and (c) to the following from Eq. 12.43(b):
dMu
+ O.5Nu + 0.5
u
v.. cot 6
Use fpo =.fpe = fp;(l- 0.22) = 202.5 X 0.78 = 157.9 ksi Nu = 0, Vp
= 0, Vu = 140 kips, Tu = 165 ft-kip, Mu = 320 ft-kip.
Ao =. 0.85A oh = 0.85 X 1580 = 1343 in. 2
Try 6 = 23.5 for a first trial; cot 23S = 2.30 0
Trying Equation 12.28(a) for less than minimum longitudinal tensile reinforcement case, 320 X 12 30.3 + 0 + 0.5 cot 23S
=
E x
(140)2 +
(160 X 165 x 12)2 2 x 1343 - (3.366 x 157.9)
2 (0 + 28,500 x 3.366)
= -1.01 X 10- 3 in.jin.
< 0.002 in.jin.
but negative.
Hence, the longitudinal tensile strain, ex, has to be computed adjusting Equation 12.28(c) for the torsional moment, Tu , from Equation 12.43(b), namely,
where Ac = area of concrete in the flexural tension side of the member (Fig. 12.9) and
box girder vertical walls (Fig. 12.18) Ao = 0.85 AOh = 0.85 x 1580 in. 2 = 1343 in. 2 b w = 2 X 5 in. Ac = 10 (33.67 - 11.14) = 225.3 in. 2
c
= 11.14 in.,
dp = 33.67 in.,
= 10 in.
320 X 12 + 0 + 0.5 X 2.30 (140)2 + (160 in. X 165 ~ 12) - 3.366 in. 2 X 157.9 ksi 30.3 2 X 1343 in. ~=--------2(-4-29-0-k-s-i-X-22-5-.3-1-·n~.2-+-0-+--2-8,-5-00-ks-i-X-3.-3-66-in~.2~)-------
=
126.73 + 0.5 X 2.30 X 183.06 - 531.49 = -0.086 X 10- 3 2 X 1.062,468
Enter Table 12.8 (a) with (1000 ex = 0.086) and ( 2.787, say, 2.79
~=
0.131) to get 6 = 23.3 0 and 13 =
817
12.13 Shear and Torsion Reinforcement Design of a Box-Girder Bridge
Photo 12.9 Launching the segmental bridge segments for Stoney Trail Bow River segmental Bridge. Calgary, Alberta, span 1562 fl. dec k width 69 ft . the deck rises 89 10118 ft above the rive r valley (Courtesy James Skeel- Reid Crowther Engineering. Calgary)
3. Daign of trans~'use closed stirrups Vr :; 0.031613 = 0.0316
VI; b.,il . using ksi uni ts
vs:o (10 x 30.3)
x 2.79
= 59.7 kips
<
140 kips
hence web shear reinforcement is necessary. From equation 12.4O(a)
A. ~ V. - (O.0361~ Vt;b.d. + V,l f~td. cot
s
140 _ (0.0316
a
x 2.79
V5.O x
10 x 30.3 + 0)
~~O~.9____~~~~~~______ 60.0 x 30.3 x 2.3
'" 0.023 in ,z/in'/two legs. From equation 12.40(b), 165
x 12
At T~ 0.9 0006' 2· I --; = 2Aof,t cot9 - 2 x 1343 x 60 x 2.3:D· m./m.oneleg A .. A . At ( 006 . ,. -, = - + 2 ~ = 0.023 + 2 O. ) = 0.035 m. / m./two legs
,
,
Trying NO.4 closed stirrups with each of the two legs of a stirrup in each ve rtical wall.
.
spacings:; Maximum allowable s '" 12 in.
2 x 0.20 0.035
.
= 11.4 m.
818
Chapler 12 lAFD and Standard AASHTO Design of Concrete Bridges Usc No.4 closed stirrups at 10 in. center to center thro ughout thc sprln. Note thal lhe spacing of the transverse reinfo rcement can be increased along thc spun if the shear and torsion envelopes warran t it.
4. Longitudinal re;nforr:~mrn' check From Equation 12.40.
)' (0.4'''")'
M , + 0.5./ N + COlO J(~ (AI, + A,JI") ~ ----.!!... -'L + O.SY, - V"
$ ,.
¢l1( , ' t ' r
+
2w
..
Ap'i p< :0 3.366 X 245.5 ;;; 826 kips A " - 0.85 A M;;; 0.85 x 1580 '" 1343 in,l
P.. - 0.85 p~ "" 0.85 x 160 = 136 in. V
s
~
AJA·
-- = S
0.40 x 60 X 30.3 . = 108 kIps 6.75 .
Hence. 320 x 12 + 0 + 2.30 ' (140 + 0.5 0.9 x 30.2 0.9
V
x
102)' + (0.450.9x 165 X 12 X 136)' X 2 X 1343
= 140.8 + 23(212.6) = 630 kips
< 826 kips.
Hence no additional longitudinal reinforcement is nceded. Adopt thc NO. 4 vertical lies at 10 in. on centers in each of the two beam box walls. Each vertical transverse tie, if not in a single piece. has 10 be fully developed to Slltisfy the development length requirements of the specifications.
Photo 12. 10 Slate Route 509 Elevated Single-Point Urban Interchange. Tacoma, Washingto n: a sit u-<:ast post-tensioned box girder bridge fea turing tight radius curved exterior webs; the footprint of the interchange is approximately two football field s in size (Designed by BERGE RIA BAM Engi neers. Federal Way. Washing. to n, courtesy Robert Mast. Senior Principal)
12.14 LAFD Major Design Expressions In SI Format
Photo l2.11 Valley Ave. Bridge. Fife. Washington: 4-span bridge consisting of two slrucluraltypes, situ-cast post-tensioned concrete box girder with precast prestressed end girders (Designed by IlERGER/A IlAM Engineers. Federal Way. Washington, courtesy Robert Mast. Senior Principal)
12.14 LRFD MAJOR DESIGN EXPRESSIONS IN SI FORMAT
Eq. 12.9(0)'
For non-bonded tendons.
819
820
Chapter 12
<
LRFD and Standard AASHTO Design of Concrete Bridges
0.94 !py where Ll
=
span and tendon length
L2
=
stresses in MPa.
Eq.12.12(a): c - ::; 0.42 de
Moment distribution factor Pd
=
20 (1 - 2.36
;J
Eq. 12.24(a) Vc
=
0.083[3
Vi'c bvdv
where band d y (mm),f~ (MPa) Eq. 12.29, when torsion is present,
s Avt = 0.35b v J; , where b v, s (mm). Eq.12.38: Vu =
2
Vu
where Vu (Newton), Tu (N-mm), Ao (mm
+ (0.9PhT u )2 2A
o
2 ).
Eq. 12.40(b): Tn where s (mm). 2A of,y cot e
SELECTED REFERENCES 12.1 ASCE, "Minimum Design Loads for Buildings and Other Structures," ANSI-ASCE 7-95 Standard, American Society of Civil Engineers, Reston, VA, 1995, pp. 214. 12.2 AASHTO, "Standard Specifications for Highway Bridges," 18th Ed., Supplements, American Association of State Highway and Transportation Officials, Washington, D.C., 2006-2009. 12.3 AASHTO, "LRFD Bridge Design Specifications," American Association of State Highway and Transportation Officials, Washington, D.C., 2004. 12.4 ACI, "Building Code Requirements for Structural Concrete (ACI 318-08) and Commentary (ACI 318R-08), American Concrete Institute, Farmington Hills, MI. 12.5 Nawy, E.G., Reinforced Concrete-A Fundamental Approach, 6th Ed., Prentice Hall, Upper Saddle River, NJ., 2009, 936 pp. 12.6 Collins, M. P., and Mitchell, D., Prestressed Concrete Structures, Prentice Hall, Upper Saddle River, NJ, 1991. 12.7 Collins, M. P., and Mitchell, D, "Shear and Torsion Design of Prestressed and Non-Prestressed Concrete Beams," PCI Journal, Precast/Prestressed Concrete Institute, Chicago, 1980, pp. 12-100.
821
Problems for Solution
12.8 Hsu, T. T. c., Torsion in Reinforced Concrete, Van Nostrand Reinhold, New York, 1983. 12.9 Hsu, T. T. c., "Torsion in Structural Concrete-Uniformly Prestressed Members Without Web Reinforcement," PCI Journal, V. 13, Precast/Prestressed Concrete Institute, Chicago, 1968, pp.34-44. 12.10 Hsu, T. T. c., Unified Theory of Reinforced Concrete, CRC Press, Boca Raton, FL, 1993. 12.11 PCI, Bridge Design Manual, Precast/Prestressed Concrete Institute, Chicago, 1999. 12.12 PT!, Post-Tensioning Manual, 5th Ed., Post-Tensioning Institute, Phoenix, AZ, 1991. 12.13 Kudlapur, S. T., and Nawy, E. G., "Early Age Shear Friction Behavior of High Strength Concrete Layered Systems at Subfreezing Temperatures," Proceedings, Symposium on Designing Concrete Structures for Serviceability and Safety, ACI SP-133.9, E. G. Nawy and A. Scanlon, ed., American Concrete Institute, Farmington Hills, MI, 1992, pp. 159-185. 12.14 Naaman, A. E., "Unified Design Recommendations for Reinforced, Prestressed and Partially Prestressed Concrete Bending and Compression Members," ACI Structural Journal, American Concrete Institute, Farmington Hills, MI, March-April 1992, pp. 200-210. 12.15 Naaman, A. E., "Unified Bending Strength Design of Concrete Members: AASHTO-LRFD Code," Journal of Structural Engineering, American Society of Civil Engineers, Reston, VA, June 1995,pp.964-970. 12.16 Mast, R. F., "Unified Design Provisions for Reinforced and Prestressed Concrete Flexural and Compression Members," ACI Structural Journal, American Concrete Institute, Farmington Hills, MI, April 1992, pp. 185-199. 12.17 Badie, S. S., Baishya, M. c., and Tadros, M. K., "NUDECK-An Efficient and Economical Precast Prestressed Bridge Deck System," PCI Journal, Vol. 43 No.5, Precast/Prestressed Concrete Institute, Chicago, September-October 1998, pp. 56-71. 12.18 Ma, Z., Huo, x., and Tadros, M. K., "Restraint Moment in Precast/Prestressed Concrete Continuous Bridges," PCI Journal, Vol. 43, No.6, Precast/Prestressed Concrete Institute, Chicago, November-December 1998, pp. 40-57. 12.19 Nawy, E. G., editor-in-chief, Concrete Construction Engineering Handbook, 2nd Ed., CRC Press, Boca Raton, FL, 2008, pp. 1-1560. 12.20 Nawy, E. G., Fundamentals of High Performance Concrete, 2nd Ed., John Wiley & Sons, New York, 2001, 466 p. 12.21 Hsu, T. T. c., Zhu, R. H. and Lee, J. Y., "A Critique on the Modified Compression Field Theory," Presented at the 78 th Annual Meeting of the Transportation Research Board, Washington, D.C., January 10-14, 23 pp. 12.22 Nawy, E. G., "Concrete-The Sustainable Infrastructure Material for the 21 st. Century," Keynote Address Paper, No. E-C103, Transportation Research Board, National Research Council, Washington, D.C., September 2006, pp. 1-23. 12.23 Nawy, E. G., "Concrete-The Sustainable Infrastructure Material for the 21 st. Century," Keynote Address Paper, Proceedings, The First International Conference on Recent Advances in Concrete Technology, Crystal City, Washington, D.C., September 19, 2007, pp. 1-24.
PROBLEMS FOR SOLUTION 12.1 Design for flexure a 100 ft (30.5m) simply supported AASHTO-PCI bulb-tee composite bridge deck with no skews using the LRFD AASHTO specifications. The superstructure is composed of six pre tensioned beams at 9'-0" (2.74 m) on centers. The bridge has an 8 in. (203 mm) situ-cast concrete deck with the top one half inch to be considered as wearing surface. The design live load is the HL-93 AASHTO-LRFD fatigue loading. Assume the bridge is to be located in a low seismicity zone. Given, the following maximum allowable stresses: Deck
Bulb-tee
f~
= 4000 psi, normal weight
fe
=
f~
= 6500 psi
0.60 f~
=
2400 psi
822
Chapter 12 f~i =
LRFD and Standard AASHTO Design of Concrete Bridges
5500 psi
fe = 0.60 f~ = 3900 psi, Service III fe
= 0.45 f~ = 2925 psi, Service I
fei = 0.60 f~ = 3480 psi
f, = 6~ = 484 psi fpu
=
270,000 psi
= 0.90 fpu = 243,000 psi /Pi = 0.75 fpu = 202,500 psi
fpy
fy = 60,000 psi
28.5
X
106 psi
Es = 29.0
X
106 psi.
Eps
=
12.2 Design the web shear reinforcement for the bulb-tee beam in Problem 12.1 at the critical section near the supports and the interface shear transfer reinforcement at the interface plane between the precast section and the deck situ-cast concrete. Also, verify if the span deflection is within the allowable limits. 12.3 A single-span two-lane unskewed AASHTO Type BIII-48 bridge has an overall span of 96 ft and the cross-section shown in Figure P12.1 (Adapted from the PCI Manual - Ref. 12.11). The total deck width is 28 ft and the clear roadway is 25 ft wide. The deck has a 3-in. bituminous wearing surface. Design for flexure and shear an interior box element using the AASHTO LRFD specifications in the design. Given: Effective span f~
=
95 ft.
= 5000 psi, normal weight
fe = 0.60 f~ = 3000 psi, Service III fe
= 0.45 f~ = 2250 psi, Service I
fei = 0.60 f~ = 3000 psi ft = 6~ = 424 psi
3/4"
-II-
--._....L.._
• 2
5112"
-1
3" (Typ)
@
~Shearkey
39" 28"
j
1
--V" • /;2 em
5
36"
Debonded strands at the support sect;an •• e •
t -5,,1..
•••
em· ·m· -8- em· ·m· . . .. 38"--1
Figure P12.1
6@4"
23 @ 2"
5"-
Box Beam Geometry
823
Problems for Solution
!pu = 270,000 psi !py = 0.90 !pu = 243,000 psi !pi = 0.75 !pu = 202,500 psi !y = 60,000 psi 28.5
X
106 psi
Es = 29.0
X
106 psi.
Eps
=
Section Properties: Ae = 813 in.2
h = 39 in.
Ie = 168,367 in. 4 Cb =
19.29 in.
C,
=
19.71 in.
Sb
=
8728 in. 3
S'
=
8542 in. 3
WD = 847Ib/ft.
12.4 Solve Problem 12.3 using the AASHTO Standard specifications for both flexure, shear and
deflection.
SEISMIC DESIGN OF PRESTRESSED CONCRETE STRUCTURES
13.1 INTRODUCTION: MECHANISM OF EARTHQUAKES
The carth crust is composed of several layers of hard "tectonic" plates, called lirho,~pheres,
which Ooal on the softer, underpinning. nuid medium ca lled mamle. These plates or rock masses, when fractured, form faili/ lilies. The adjoining plates or rock masses arc prevented by the interacting frictional forces from movi ng pasl one anot her most of the time. However, when this frictional ultimate resistance is reached because of the cOnlinuous motion of the underlying fluid , any two plales ca n impact on onc another, genera ting seismic waves that can cause large horizontal and vertical ground motions. These grou nd motions translate into inertia forces in structures. The length and width of a fault are interrelated to the magnitude of the earthquake. The fault is the cause rather than the result of the eart hquake. A fault can cause an earthquake due to the rollowing reasons (Ref. 13.5):
l. Cumulative st rain in the rault over a long period or time reaches the rupture level. 2. Slip of the tectonic plates at the fault zones causes a rebound, as in Fig. 13.I(a).
Northridge, California, 1994 earthquake structural failure. (COllrtesy, Dr. Murat Saatcioglu.) 824
825
13.1 Introduction: Mechanism of Earthquakes
Photo 13.1 311 S. Wacker Street, Chicago, 12,000 concrete (Courtesy Portland Ceme nt Association.)
3. Sudden push and pull forces at the fau lt lead to reverse momen t couples. as in Fig. 13. 1(b). The moment caused by these couples as a measure of earthquake size can be termed the seismic momems. The magnitude is equal to rock rigidity x fault area x amount of slip. The range of slip velocity in such faults as the San Andreas Fault in California is 30 to 100 mm per year. On this basis, a slippage or horizontal motion of 3 m at such fau lts in one single earthquake is expected to occur at intervals of 30 to 100 years.
'.1
Ibl
Figure 13.1 Mechanism of earthquakes: (a) slip of tectonic plates; (b) reverse moment couples.
826
Chapter 13
Seismic Design 01 Prestressed Concrete Structures
"holo 13.2 Bridge girder collapse in the San Francisco 1989 earthquake. (Courtesy Portland Cement Association.)
Earthquakes may be characterized by three calcgorics: low, moderate, and high intensity. The intensity is governed by ground motion accelera tions. represented by response spectra and coefficients derived from such spectra. A struclUre is expected to respond essentially e lastically 10 low-intensity earthquak es. In such a case, the stresses are expected 10 remain within the elastic range, with a slight possibility of developing limited inelasticity with no appreciable structural or no n-structural damage. Structural response is expected to be inelastic under high-intensity earthquakes having an intensity of 5 or higher on the Richter scale and in regions close to the epicen ter. For the design of st ructures in seismic zones, two me thods are presented in the IBC 2009 code, the spectral response method and the eq uivalent la teral force method. The latter has certai n limitations that wilt be discussed later. A detailed discussion of the subject of earthquakes is beyond the scope of th is book since the prima ry aim of this chapter is the proportioning of seismic resistant componen ts of concrete st ructures. However, some of the basic underlying characteristics are important to cover. They are intended to help define the magnitude of the lateral seism ic base shear forces that determine the geomet ry and form of the earthquake resisting components of a structure. namely, the lateral force resisti ng system (LFRS). Such a system has two components: horizontal and vertica l. The horizontal elements are the compone nts that resist the seismic forces . They can be diaphragms, coupling beams. and shear walls. The vertical component comprises the walls and vertical frames of the structure.
13.1.1 Earthquake Ground Motion Characteristics Ground motion. caused by seismic tremors, involves acceleration. velocity, and displacement. These are in the majority amplified, thereby producing forces and displacements. which can exceed those wh ich the structure is able to sustain (Rer. 13. 13). The maximum value of the ground motion magn itude. namely, the peak ground velocity, peak ground
827
13.1 Introduction: Mechanism of Earthquakes
acceleration and peak ground displacement become the principal parameters in the seismic design of structures. Additional factors also affect the response of a structure. They include frequency, amplitude of motion, shaking duration, and site soil characteristics. These can all be represented by a response spectrum which idealizes a structure into a dampened, single degree of freedom system (SDF) oscillating at various periods and frequencies. The maximum vibration magnitude reached during any time duration after the base ground motion is its spectral value. 13.1.2 Fundamental Period of Vibration
The basic natural period T of a simple one-degree-of-freedom system is the time required to complete one whole cycle during dynamic loading. In other words, it is the time required for a phase angle wt to travel from 0 to 21T, where w is the angular frequency of the system. Hence wt = 21T, leading to the expression T
=
2:
=
21T(7t2
(13.1)
where m = mass of system k = spring constant and damping is not considered Most reinforced concrete structures are multidegrees-of-freedom systems, as in Fig. 13.2. In this case the structural mass can be assumed to be concentrated in the vertical spring element at the floor level, resulting in multiple modes with frequencies (periods) for each mode. The compound natural period T is then evaluated with due consideration given to the distribution of mass and stiffness. Codes require that T be established using the structural properties and deformation characteristics of the resisting elements in a properly substantiated analysis using expressions such as those given by the International Building Code-IBC 2009 (Ref. 13.2), or The Uniform Building Code (Ref. 13.3) integrated into the IBC provisions. Since a structure is composed of a series of single degrees of freedom components subjected to the same base motion, a series of maximum values related to the SDF system's fundamental periods, T, would ensue. These, in turn, form a spectral curve for that base ground motion. By knowing the base motion, the SDF fundamental period and the percent critical damping, one can obtain from the applicable curve the maximum acceleration, velocity, and displacement relative to the base (Ref. 13.14). Evidently, computer
-vm2 hi
h
1 Figure 13.2
m,
JI Base /':
Base shear V
Modeling multistory structures.
828
Chapler 13
Seismic Design 01 Prestressed Concrete Structures
Photo 13.3 Northridge. California. 1994 earthquake struclurnl fai lure. (Courtesy Dr. Murat Saatcioglu.)
use is needed 10 obtain a complete spectral response of the multi-degree state of a structure. It should be recognized that a structure is designed to resisl enrthquake motion such that it is able to sustain and survive the earthquake through large inelastic deformations and energy dissipation through crack ing and limited locn! material failure. but without loss of stabi lity. It would be highly uneconomical to design the lateral force resisting system 10 the earthquake force s such that the structure deform s only elastically as a result of these forces. The codes have this as a basic philosophy particularly for major earthquakes in which some structural damage can resu lt. 13.1.3 Design Philosophy The International Building Code (lBC 2009. Ref 13.2) o n seism ic design consolidates the three major ex.isling regional codes into one major documen t that is now universally adopted covering the work of the foll owing regional codes: I. Building O[ficials Code Administration InlCrnational (BOCA)
2. International Conference of Building Officials (ICBO): Uniform Building Code (UBC) 3. Southern BuiJding Code Congress Intc rnalional (S BCCI) Underlying its seismology design provisions arc:
13.2 Spectral Response Method
829
1. Recommended design levels related to effective peak accelerations that can resist minor earthquakes without damage, moderate earthquakes without structural damage, and major earthquakes in which some structural damage can result. 2. Minimum design criteria for all types of buildings, low and high rise, with and without shear walls. 3. Spectral response values for various ground motion intensities, mainly within the elastic range. 4. Provide design criteria for lateral ground motion, unidirectional and bi-directional, addressing them one at a time. 5. Limit the story drift and displacement magnitudes of the building structures within acceptable ranges, through control of stiffness of components and shear walls, diaphragms, and coupling beams.
13.2 SPECTRAL RESPONSE METHOD 13.2.1 Spectral Response Acceleration Maps
As discussed in Ref. 13.15, prior to the Northridge and Kobe earthquakes, the Uniform Building Code (UBC) provisions performed satisfactorily in the United States in past earthquakes. The failures in these two cases were determined to be due to "related configurations of the structural systems, inadequate connection detailing, incompatibility of deformations and design or construction deficiencies. They were not due to deficiency in strength (Structural Engineers Association of California, 1995). The UBC provisions incorporated in the International Building Code (1EC) are based on consideration of the site conditions of the structure and the application of maximum considered earthquake ground motion maps for site class B, prepared by the United States Geological Survey (USGS). The equivalent maximum considered earthquake ground motion values for the ceiling were determined to be 1.50 g for the short period and 0.60 g for the long period (Ref. 13.15). The high seismicity regions, where the maximum considered earthquake ground motion values are greater than 0.75 g for the 1.0 sec, peak acceleration additional requirements are imposed on irregular structures exceeding five stories in height and a period T in excess of 0.5 sec, such as increasing the ground motion spectral acceleration values by 50 percent. The USGS large-scale maps for the 1.0 sec and the 0.2 sec levels of spectral response acceleration, site-B class, and 5 percent critical damping are condensed and abridged in Figs. 13.3(a) and (b) for general guidance. They show the relative values of the peak spectral response accelerations at the two ground motion levels of 0.2 and 1.0 sec. Values have to be extrapolated linearly from the USGS large-scale maps for use in the seismic design of structures.
13.2.2 Seismic Design Parameters
Both the spectral response method and the equivalent lateral force method are based on the same code principles and formulations presented in this chapter. Sites are classified into six categories A, B, C, D, E, and F as shown in Table 13.1 on site properties. Ground motion accelerations and the maximum considered earthquake spectral response acceleration are considered at 1.0 sec period (Sl) and at short periods (Ss) such as 0.2 sec obtained from seismic contour maps discussed in Section 13.2.1.
co o (.,)
,60
~\
1\\
\'III,
301
~
\ I' 1·1
r.- ____ - r r /
3.0
.--- ........... ""
""'-
Figure 13.3(a) Maximum considered earthquake ground motion for the United States, 0.2 sec. Spectral response acceleration Ss as a percent of gravity, site-class B with 5 percent critical damping.
/
I I I I I / /
I I
I I I I \ \ I I f
I f I / ./
/
___ .1:.Q----./
/
./
./
./
Figure 13.3(b) Maximum considered earthquake ground motion for the United States, 1.0 sec. Spectral response acceleration 8 1 as a percent of gravity, site-class B with 5 percent critical damping.
co
....
(0)
Chapter 13
832 Table 13.1
Seismic Design of Prestressed Concrete Structures
Site Class Definitions and Classifications Average Properties in Top 100Ft (30 m), As in Section 1615.1.5
Site Class
Soil Profile Name
Soil Shear Wave Velocity, VS1 (IUS)
Standard Penetration Resistance, N
Soil Unconfined Shear Strength 5 u (PSF)
A
Hard rock
Vs > 5,000
not applicable
not applicable
B
rock
2,500:0: Vs :0: 5,000
not applicable
not applicable
C
Very dense soil and soft Rock
1,200:0: Vs:O: 2,500
N>50
Su> 2,000
D
Stiff soil profile
600 :0: Vs :0: 2,500
15 :O:N:O: 50
1,000:0: Su:O: 2,000
E
Soft soil profile
Vs < 600
N< 15
Su
E
Any profile with more than 10 ft of soil having the following characteristics: -plasticity index PI > 20; -moisture content w > 40% and -unconfined shear strength Su < 500 psf
F
Any profile containing soil having one or more of the following characteristics: 1. Soils vulnerable to potential failure or collapse under seismic loading such as liquefiable soils, quick and highly sensitive clays, collapsible weakly cemented soils. 2. Peats and/or highly organic clays (H> 10 ft of peat and/or highly organic clay where H = thickness of soil) 3. Very high plasticity clays (H > 25 ft with plasticity index PI > 75) 4. Very thick soft/medium stiff clays (H > 120 ft)
For Sl: 1 ftlsec = 304.8 mm/sec; 1 psf = 0.0479 kP; 1 ft.
= 305 mm.
Steps and expressions for classifying and determining their values are detailed in Ref. 13.2 SMS
= FaSs
(13.2a)
SMI
=
FvS 1
(13.2b)
where, Fa = Site coefficient from Table 13.2a Fv = Site coefficient from Table 13.2b Ss = Mapped spectral acceleration for short periods (See Ref. 13.2 for map contour values) SI = Mapped spectral acceleration for l.O-sec periods (See Ref. 13.2 for map contour values) Table 13.2(a) Values of Site Coefficient Fa as a Function of Site Class and Mapped Spectral Response Acceleration at Short Periods (Ss) Mapped Spectral Response Acceleration at Short Periods Site Class
A B C D E F
5s
:S
0.25
0.8 1.0 1.2 1.6 2.5 Note b
5 s =0.50
5 s = 0.75
5 s = 1.00
0.8 1.0 1.2 1.4 1.7 Note b
0.8 1.0 1.1 1.2 1.2 Note b
0.8 1.0 1.0 1.1 0.9 Noteb
5s
::::: 1.25
0.8 1.0 1.0 1.0 0.9 Note b
833
13.2 Spectral Response Method
Table 13.2(b) Values of Site Coefficient Fvas a Function of Site Class and Mapped Spectral Response Acceleration at 1.0 Sec Periods (81) Mapped Spectral Response Acceleration at 1.0-Sec Periods Site Class A B C D
E F
~
0.1
51 =0.2
51 =0.3
51 =0.4
5 1 2'::0.5
0.8 1.0 1.7 2.4 3.5 Note b
0.8 1.0 1.6 2.0 3.2 Note b
0.8 1.0 1.5 1.8 2.8 Noteb
0.8 1.0 1.4
0.8 1.0 1.3 1.5 2.4 Noteb
51
1.6 2.4 Note b
NOTES: a-Straight line interpolation for intermediate values are to be made. b--Site geotechnical investigation and dynamic site response analyses are to be performed
The design spectral response accelerations at short periods (Ss) and at 1 sec (SI) are to be adjusted for site class effect (SMS) at short periods and (SMl) for 1 sec based on Table 13.1 in conjunction with Tables 13.2( a) and 13.2(b) for site coefficients. The maximum considered earthquake spectral response for short and one second periods are respectively defined by the following expressions where for 5 percent damped design, the spectral response acceleration becomes: (13.3a) (13.3b) 13.2.3 Earthquake Design Load Classifications and Seismic Categories
Structures in seismic areas have to be classified in separate categories than those that are sUbjected to low or negligible seismic loads. Regardless of the period of vibration of the structure they are classified as in tables 13.3 (a) and 13.3 (b) for short period response acceleration and I-second response acceleration respectively. The IBC 2009 assigns four seismic occupancy categories, I, II, III, and IV. They are defined by the magnitude of the seismic response acceleration, namely, the short period response SDS of 0.2 seconds and the I-second period response SDl' Tables 13.3 (a) and 13.3 (b) list the design categories A, B, C, and D corresponding to the SDS and SDllevels. The Code stipulates that Occupancy Category I, II, or III structures located where the map spectral response acceleration parameter at I-second period, SD1' is greater than or equal to 0.75, that structure should be assigned Seismic Design Category E. Occupancy Table 13.3(a) Seismic Design Category Based on Short Period Response Accelerations Occupancy Category Value of Sos
Sos < 0.167g 0.167g ~ Sos < 0.33g 0.33 g ~ Sos < 0.50g 0.50 g ~ Sos
lor II
III
IV
A
A
B
B
C D
C D
A C D D
1 834
Chapter 13
Seismic Design of Prestressed Concrete Structures
Table 13.3(b) Seismic Design Category Based on 1 Second Period Response Accelerations Occupancy Category Value of 501
lor II
III
IV
SDl < 0.067g 0.067g:5 SDl < 0.133g
A
A
A
B
B
C
0.133 g:5 SDl < 0.20g 0.20 g:5 SDl
C D
C D
D D
Category IV structures located where the mapped spectral response acceleration parameter at I-second period Sm, is greater than or equal to 0.75, the structure should be assigned to Seismic design Category F. All other structures should be assigned to a seismic design category based on their occupancy category and the design spectral response coefficients SDS and Sm as defined in Equations 13.3 (a) and (b) or the site-specific procedures of ASCE 7. But each structure should be assigned to more severe seismic category in accordance with Tables 13.3 (a) or 13.3 (b) irrespective of the fundamental period of vibration, T. To amplify, there are exceptions that allow for the Seismic Design Category to be determined from Table 13.3 (a) alone. In order to use this exception, Sl must be less than 0.75 and all of the following requirements have to be met (Ref. 13.13). 1. The approximate fundamental period of vibration, Tm as determined by Equation 13.13 in each of the two orthogonal directions is less than 0.8 Ts, where Ts = SmISDs' 2. The fundamental period of the structure that is used to calculate the story drift in the two orthogonal directions is less than Ts. 3. The seismic response coefficient, Cs , is determined by Equation 13.9. 4. The diaphragms are rigid, or where diaphragms are considered flexible, the spacing between vertical elements of the lateral force-resisting system does not exceed forty feet. The FEMA 302 Parts 1 and 2 defined seismic regions as follows in Ref. 13.15: Region 1-Regions of Negligible Seismicity with Very Low Probability of Collapse of the Structure (No Spectral Values)
Region definition: Regions for which Ss < 0.25 g and Sj < 0.10 g. Design values: No spectral ground motion values required. Use a minimum lateral force level of 1 percent of the dead load for seismic design Category A. Region 2-Regions of Low and Moderate to High Seismicity (Probabilistic Map Values) Region definition: Regions for which 0.25 g < Ss < 1.5 g and 0.25 g < Sl < 0.60 g. Maximum considered earthquake map values: Use Ss and Sl map values. Transition between Regions 2 and 3-Use values of Ss = 1.5 g and S1 = 0.60 g.
835
13.2 Spectral Response Method
Region 3-Regions of High Seismicity Near Known Faults (Deterministic Values)
Regional definition: Regions for which 1.5 g < Ss and 0.60 g < SI' The structural analysis based on the worst load combinations should be the basis for determining the seismic forces E for combined gravity and seismic load effects when they are additive and the maximum seismic load effect Em' The value of E and Em are determined from the following expressions detailed in Ref. 13.2 for additive seismic force and dead load: (13.4a) E = pQE + 0.2 SDsD
E
=
noQE
+ 0.2 SDsD
(13.4b)
For counteracting seismic forces and dead load:
where,
E
=
pQE - 0.2 SDsD
(13.5a)
E
=
noQE - 0.2 SDsD
(13.5b)
E = combined effect of horizontal and vertical earthquake-induced forces
P = reliability factor based on system redundancy = 1.0 for categories A, B, andC QE = effect of horizontal seismic forces SDS = spectral response acceleration at short periods obtained from mc Sec. 1613.5.4. no = system over-strength factor given in Table 13.4 D = effect of dead load 13.2.4 Redundancy
A redundancy coefficient P has to be assigned to all structures based on the extent of structural redundancy inherent in the lateral force resisting system. For structures in seismic design categories A, B, and C, the value of the redundancy coefficient p is to be taken as 1.0. For structures in seismic design categories D, E, and F, the redundancy coefficient p has to be taken as the largest of the values PI computed at each story level "i" of the structure in accordance with the expression
20
PI =2 - - - - -
rmaxi~
(13.6a)
In SI Units, the expression becomes PI = 2 -
6.1 ----=:=
rmaxi~
(13.6b)
where
rmax i = ratio of the design story shear resisted by the most heavily loaded single element in the story to the total story shear for a given loading condition, Ai = Floor area in square feet (m2) of the diaphragm level immediately above the story
The value of p cannot be less than 1.0 and need not exceed 1.5. 13.2.5 General Procedure Response Spectrum
The design response can be idealized by the fundamental period-response acceleration relationship shown in Fig. 13.4 for three fundamental period levels.
Table 13.4
Design Coefficients and Factors for Basic Seismic-Foree-Resisting Systems (abridged from Ref. 13.4)
BASIC SEISMIC-FORCE-RESISTING SYSTEM
RESPONSE MODIFICATION COEFFICIENT R"
SYSTEM OVERSTRENGTH FACTOR 00
SYSTEM LIMITATIONS DEFLECTION AND BUILDING HEIGHT AMPLIFICATION FACTOR, CDb LIMITATIONS (FT) BY SEISMIC DESIGN CATEGORYc AS DETERMINED IN IBC SECTION 1616. A&B
C
Dd
Ee
F'
Bearing Wall System Special reinforced concrete shear walls Ordinary reinforced concrete shear walls Detailed plain concrete shear walls Ordinary plan concrete shear walls Ordinary precast shear wall
NL ]60 NL NP NL NP
160 100 NP NP NP NP
NP NP
NP NP
NP NP
NP ND
NL NL NP
NP NP
4~
NL
NL
NP NP NP 40
NP NP
1~
NL NL NL
NP 40
NP 40
5 4 2
2.5 2.5 2.5
5 4 2
1.5 3
2.5 2.5
1.5 3
NL NL NL NL NL
Building Frame System Ordinary reinforced concrete shear walls Detailed plain concrete shear walls Ordinary plain concrete shear walls Indeterminate precast shear wall
5 2 1~
5
2.5 2.5 2.5 2.5
4.5 2.5
Moment Resistant Frames Special reinforced concrete moment frames Intermediate reinforced concrete moment frames Ordinary reinforced concrete moment frames
8
3
5.5
NL
NL NL
NL
NL
5
3
4.5
NL
NL NP
NP
NP
3
3
2.5
NLh
NP
NP
NP
NP
NL NL
NL NL
NL NP
NL NP
NL 160 100 100 NL NP NP NP 160 NP NP NP
NP
Dual System with Special Moment Frames Special reinforced concrete shear wall Ordinary reinforced concrete shear wall Special reinforced masonry wall
7
6 5~
2.5 2.5 3
5~
5 5
NL
NP
Dual System with Intermediate Moment Frames Special reinforced concrete shear wall Ordinary reinforced concrete shear wall Ordinary reinforced masonry shear wall Shear Wall-Frame interactive system with ordinary reinforced concrete moment frames and ordinary reinforced concrete shear walls
6~
5.5 3
4~
2.5 2.5
5 4.5
3
2~
NL NL NL
2.5
4
NL
NP
NP
NP
For SI, 1 ft = 305 mm aResponse modification coefficient R, for use throughout bDefiection amplification factor, Cd CNL = not limited and NP = not permitted dlimited to buildings with a height of 240 ft or less. elimited to buildings with a height of 160 ft or less. fOrdinary moment frame is permitted to be used in lieu of Intermediate moment frame in Seismic Design Categories B. and C. gThe tabulated value of the overstrength factor, 00 may be reduced by subtracting! for structures with flexible diaphragms but shall not be taken as less than 2.0 for any structure. hOrdinary moment frames of reinforced concrete are not permitted as a part of the seismic-force-resisting system in Seismic Design Category B structures founded on Site-Class E or F soils
836
837
13.3 Equivalent Lateral Force Method C/)'"
c
o
~Q) Q5
u
Sos - 1 - - -....
:i.
1
Q)
1
(Jl
1
c
o
1
0..
1
(Jl
Q)
a::
Sa= SolT
1
SOt
1
1
1 1 1 1 1
1 1 1 1 1
- , - - - - - -1- - - - -
~ 0.4 Sos
t5 Q) 0.. C/)
TS
To
1.0
Period T Figure 13.4
Design response spectrum.
1. For periods in seconds less than or equal to To, the design spectral response acceleration Sa is determined from the following equation:
Sa
=
SDS T + 0.4 SDS To
0.6 -
(13.7a)
2. For periods greater than or equal to To, and less than or equal to Ts ' the design spectral response acceleration Sa' is taken equal to SD' 3. For periods greater than Ts' the design spectral response acceleration, Sa' is deter-
mined from the expression: (13.7b) where, SDS = the design spectral response acceleration at short periods SDl = the design spectral response acceleration at I-sec periods T = Fundamental period (in seconds) of the structure To = 0.2 SD/SDS T=SD/SDs
The sites have to be classified for determining the shear wave velocity and the maximum considered earthquake ground motion. Details are given in the IBe (Ref. 13.2) section 1615. 13.3 EQUIVALENT LATERAL FORCE METHOD 13.3.1 Horizontal Base Shear
In this method, a building is considered fixed at the base. The seismic base shear, V, in a given direction is determined from the expression(Ref. 13.2): V = Cs W
(13.8)
where, Cs = seismic response coefficient W = The effective seismic weight of the structure, including the total dead loads and other loads listed herein:
838
Chapter 13
Seismic Design of Prestressed Concrete Structures
1. In areas used for storage, a minimum of 25 percent of the reduced floor live load (floor live load in public garages and open parking structures need not be included). 2. Where an allowance for partition load is included in the floor load design, the actual partition weight or a minimum weight of 10 psf (500 Pa/m2) of floor area, whichever is greater. 3. Total operating weight of permanent equipment. 4. 20 percent of flat roof snow load where the flat roof snow load exceeds 30 psf. SDS
Cs = (RjI)
(13.9)
But Cs cannot exceed the value:
(13.10)
nor can it be taken less than: Cs = 0.044
SDS
(13.11)
where, SDS = Design spectral response acceleration at short period as determined in Section 13.2.2 R = Response modification factor from Table 13.4 I = Occupancy importance factor from Table 13.5 T = fundamental period of building (seconds) For buildings and structures in seismic design categories E or F and in buildings and structures for which the I-sec spectral response, SI is equal to or greater than 0.6 g, the value of the seismic coefficient Cs should not be taken less than: 0.5S1 Cs = Rj I
(13.12)
The fundamental period T in the direction under consideration has to be determined by analysis based on the structural and deformational characteristics of the resisting element. In lieu of an analysis, an approximate fundamental period Ta , in seconds, can be used from the following expression:
(13.13) where, CT = Building Period Coefficient • 0.035 for moment resisting frame systems of steel in which the frames resist 100 percent of the required seismic force and are not enclosed or adjoined by more rigid components that will prevent the frames from deflecting when subjected to seismic forces (the metric coefficient is 0.085) • 0.030 for moment resisting frame systems of reinforced concrete in which the frames resist 100 percent of the required seismic force and are not enclosed or adjoined by more rigid components that will prevent the frames from deflecting when subjected to seismic forces (the metric coefficient is 0.073) • 0.030 for eccentrically braced steel frames (the metric coefficient is 0.073) • 0.020 for all other building systems (the metric coefficient is 0.049) h n = the height (ft or m) above the base to the highest level of the building.
13.3 Equivalent Lateral Force Method Table 13.5(a) Occupancy Category
839
Occupancy of Buildings and other Structures for Floods, Wind, Snow, Earthquake and Ice Loads Nature of Occupancy
I
Buildings and other structures that represent a low hazard to human life in the event of failure, including, but not limited to: • Agricultural facilities • Certain temporary facilities • Minor storage facilities
II
All buildings and other structures except those listed in Occupancy Categories I, III, and IV
III
Buildings and other structures that represent a substantial hazard to human life in the event of failure, including, but not limited to: • Buildings or structures whose primary occupancy is public assembly with an occupant load greater than 300. • Buildings and other structures containing elementary school or day care facilities with an occupant load greater than 250. • Buildings and other structures containing adult education facilities, such as colleges and universities with an occupant load greater than 500. • Group 1-2 occupancies with an occupant load of 50 or more resident patients but not having surgery or emergency treatment facilities. • Group 1-3 occupancies. • Any other occupancy with an occupant load greater than 5000a . • Power generating stations, water treatment facilities for potable water, waste water treatment facilities and other public utilities facilities not included in occupancy category IV. • Buildings and other structures not included in occupancy category IV containing sufficient quantities of toxic or explosive substances to be dangerous to the public if released.
IV
Buildings and other structures designated as essential facilities, including, but not limited to: • Group 1-2 occupancies having surgery or emergency treatment facilities. • Fire, rescue, ambulance and police stations or emergency treatment facilities. • Designated earthquake, hurricane or other emergency shelters. • Designated emergency preparedness, communications, and operations centers and other facilities required for emergency response. • Power-generating stations and other public utility facilities required as emergency backup facilities for Occupancy Category IV structures. • Structures containing highly toxic materials as defined by Section 307 where the quantity of the material exceeds the maximum allowable quantities of Table 307.1 (2) of Ref. 13.2. • Aviation control towers, air traffic control centers and emergency aircraft hangers. • Building and other structures having critical national defense functions. • Water storage facilities and pump structures required to maintain water pressure for fire suppression.
'Reference 13.2
840
Chapter 13 Table 13.5(b)
Seismic Design of Prestressed Concrete Structures
Importance Factors*
Occupancy Category
lor II
1.0 1.25 1.5
III
IV
* Reference 13.2 In cases where moment resisting frames do not exceed 12 stories in height and having a minimum story height of 10 ft (3 m), an approximate period Ta in seconds in the following form can be used: Ta
=
0.1 N
(13.14)
where N
= number of stories
The computed fundamental period, T, cannot exceed the product of the coefficient, Cn' in Table 13.6 for the upper limit on the computed period times the approximate fundamental period, Ta.The base shear V is to be based on a fundamental period, T, in seconds, of 1.2 times the coefficient for the upper limit on the calculated value, Cw taken from Table 13.6 times the approximate fundamental period Ta' 13.3.2 Vertical Distribution of Forces
The lateral force Fx (kips or kN) induced at any level can be determined from the following expressions: (13.15a) (13.15b)
where Cvx = vertical distribution factor
V = total design lateral force or shear at the base of the building (kips or kN), Wi and Wx = the portion of the total gravity load of the building, W, located or assigned to level i or x hi and hx = the height (ft or m) from the base to level i or x k = a distribution exponent related to the building period as follows:
Table 13.6
Coefficient for Upper Limit On Computed Fundamental Period
Design Spectral Response Acceleration at 1-sec period, SOl
0.4 0.3
2':
Coefficient Cu
1.2
0.2
1.3 1.4
0.15 :s 0.1
1.5 1.7
841
13.3 Equivalent Lateral Force Method
• For buildings having a period of 0.5 sec or less, k = 1 • For buildings having a period of 2.5 sec or more, k = 2 • for buildings having a period between 0.5 and 2.5 seconds, k shall be 2 or shall be determined by linear interpolation between 1 and 2 13.3.3 Horizontal Distribution of Story Shear Vx
The seismic design story horizontal shear in any story, Vx (kips or kN) should be determined from the following expression: (13.16) where Fi = the portion of the seismic base shear, V (kips or kN) introduced at level i.
13.3.4 Rigid and Flexible Diaphragms
(a) Rigid diaphragms: The seismic design story shear, V x , has to be distributed to the various vertical elements of the system in the story under consideration. This distribution is to be based on the relative stiffness of the vertical resisting elements and the diaphragms. (b) Flexible Diaphragms: The seismic design story shear, V x , in this case has to be distributed to the various vertical elements based on the tributary area of the diaphragms to each line of resistance. The vertical elements of the lateral force resisting system can be considered to be in the same line of resistance, if the maximum out of plane offset between such elements in less than 5 percent of the building's dimension perpendicular to the direction of the lateral load. 13.3.5 Torsion If the diaphragms are not flexible, the design has to include the torsional moment M(
(Kip-ft or kN-m) resulting from the difference in location between the center of mass and the center of stiffness. Dynamic amplification of torsion for structures in seismic design category C, D, E or F has to be accounted for by multiplying the torsional moments by a torsional amplification factor presented in Ref. 13.2, Sec. 1613.5.3. 13.3.6 Story Drift and the P-Delta Effect
(a) Drift: The design story drift, Ll, is computed as the difference between the deflections of the center of mass at the top and bottom of the story being considered. If allowable stress design is used Ll is computed using earthquake forces without dividing by 1.4. The deflection of level X is to be determined from the following expression, Cdo xe Ox = - 1 -
(13.17)
where, Cd = Deflection amplification factor (Table 13.4) Ox = Deflections (in. or mm) determined by an elastic analysis of the seis-
mic forces resisting system. 1 = Occupancy importance factor (Table 13.5) The design story drift, Ll, has to be increased by an incremental factor relating to the P-delta effects. The redundancy coefficient, p, in the case of drift should be taken as 1.0.
842
Chapter 13
Seismic Design of Prestressed Concrete Structures
(b) P-delta effects: The P-delta effects can be disregarded if the stability coefficient, from the following expression is equal or less than 0.10,
e=
Pi::' "V,;hsxC d
e,
(13.18)
where, P x = The total unfactored vertical design load at and above Level x (kip or
.:l
Vx hsx Cd
kN); when computing the vertical design load for purposes of determining P-delta, the individual load factors need not exceed 1.0 = The design story drift (in. or mm) occurring simultaneously with Vx = The seismic shear force (kip of kN) acting between level x and x-I = The story height (ft or m) below level x = The deflection amplification factor in Table 13.4.
The stability coefficient, e, shall not exceed emax determined as follows:
emax =
0.5 C l3 ~ 0.25 d
where:
13
=
The radio of shear demand to shear capacity for the story between level x and x-I. Where the ratio 13 is not calculated, a value of 13 = 1.0 shall be used.
When the stability coefficient, e, is greater than 0.10 but less than or equal to emax , inter-story drifts and element forces shall be computed including P-delta effects. To obtain the story drift for including the P-delta effect, the design story drift shall be multiplied by 1.0/(1 - e). Table 13.7
Allowable Story Drift, d (in. or mm)a,b,C,d,e
Occupancy Group Building
lor II
III
IV
Structures, other than masonry shear wall structures, 4 stories or less with interior walls, partitions, ceilings and exterior wall systems that have been designed to accommodate the story drifts.
O.025hsxc
O.020hsx
O. 015h sx
Masonry cantilever shear wall structuresc
O.OlOhsx
O.010hsx
O.OlOhsx
Other masonry shear wall buildings
O. OO7h sx
O.OO7sx
O.OO7sx
All other buildings
O.020h sx
O. 015hsx
O.OlOhsx
a·hsx is the story height below Level x. seismic force-resisting systems comprised solely of moment frames in Seismic Design Categories D, E, and F, the allowable story drift shall comply with the requirements of ASCE 7-05 Section 12.12.1.1. eThere shall be no drift limit for single story structures with interior walls, partitions, ceilings, and exterior wall systems that have been designed to accommodate the story drifts. The structure separation requirement of ASCE 7-05 Section 12.12.3 is not waived. d Structure in which the basic structural system consists of masonry shear walls designed as vertical elements cantilevered from their base or foundation support which are so constructed that moment transfer between shear walls (coupling) is negligible. e Reference 13,2, 13.4 b For
843
13.3 Equivalent Lateral Force Method
When e is greater than emax , the structure is potentially unstable and has to be redesigned. The allowable story drifts are given in Table 13.7. 13.3.7 Overturning
Ground motion can result in overturning of a structure. At any story, the increment of overturning moment in the story under consideration would have to be distributed to the various vertical force-resisting elements, in the same proportion as the distribution of the horizontal shear forces to these elements. The overturning moment at level x, Mx (kip-ft or kN-m), is determined from the following expression: n
Mx =
T
LFi(hi - h x)
(13.19)
i=x
where Fi T
= Portion of hi and hx =height (ft or m) from the base to the level i or x. Overturning moment reduction factor 1.0 for the top 10 stories = 0.8 for the 20th story from the top and below = values between 1.0 and 0.8 determined by a straight line interpolation for stories between the 20th and 10th stories below the top. The seismic base shear, V, is induced at level i. = =
13.3.8 Simplified Analysis Procedure for Seismic Design of Buildings
This procedure can be used for structures in seismic use group I, subject to the following limitations, otherwise either the method in Section 13.2 or this section has to be used. 1. Buildings of light-framed construction not exceeding three stories in height, excluding basement. 2. Buildings of any construction other than light framed, not exceeding two stories in height, excluding basement. The seismic base shear, V, can be computed from the following expression,
V
=
1.2SDS
-R- W
(13.20)
where SDS = Design elastic response acceleration at short periods as determined from Section 13.2 R = Response modification factor from Table 13.4 W = The effective seismic weight of the structure, including the total dead load and other loads listed below. In areas used for storage, a minimum of 25 percent of the reduced floor live load (floor live load in public garages and open parking structures need not be included.) 1. Where an allowance for partition load is included in the floor load design, the actual partition weight of 10 psf of floor area, whichever is greater. 2. Total weight of permanent operating equipment. 3. 20 percent of flat roof snow load where flat snow load exceeds 30 psf (1.44 kN/m 2 ).
The vertical distribution of forces at each level would be computed from the following expression: (13.21)
844
Chapter 13
Seismic Design of Prestressed Concrete Structu(es
where, Wx = The portion of the effective seismic weight of the total structure, W, at story level x.
For structures satisfying this section, the design story drift, Ll, is taken as 1 percent of the story height unless a more exact analysis is made. Table 13.8 gives the requirements for each story resisting more than 35% of the base shear. Table 13.9 outlines the category occupancy cases and conditions where analytical procedures are permitted in the design. 13.3.9 Other Aspects in Seismic Design
The discussion presented in the previous sections is intended only to highlight the most important basic considerations for establishment of the seismic basic shear force values and their distribution over the height of a structure, at all story levels. The scope of this book does not permit more coverage of other essential topics such as modeling, model forces, deflections and drifts, diaphragms, coupling beams, interconnecting shear walls, connections, irregularity of structures, out-of-plane loading, torsion, and foundations. Through a careful review of the details presented, the numerical examples and solving the assignments, the reader becomes well equipped to handle the design requirement aspects of the topics listed. The International Building Code-IBC 2009 (Ref. 13.2) detailed provisions give all the additional provisions and guidance needed for safe complete designs of concrete structures that can successfully resist severe earthquakes. The ensu-
Table 13.8
Requirements for Each Story Resisting More than 35% of the Base Shear-
Lateral Force-Resisting Element
Requirement
Braced Frames
Removal of an individual brace, or connection thereto, would not result in more than a 33 % reduction in story strength, nor does the resulting system have an extreme torsional irregularity (horizontal structural irregularity)
Moment Frames
Loss of moment resistance at the beam-to-column connections at both ends of a single beam would not result in more than a 33% reduction in story strength, nor does the resulting system have an extreme torsional irregularity (horizontal structural irregularity)
Shear Walls or Wall Pier with a height-to-length ratio of greater than 1.0
Removal of a shear wall or wall pier with a height-tolength ratio greater than 1.0 within any story, or collector connections thereto, would not result in more than 33% reduction in story strength, nor does the resulting system have an extreme torsional irregularity (horizontal structural irregularity)
Cantilever Columns
Loss of moment resistance at the base connections of any single cantilever column would not result in more than 33% reduction in story strength, nor does the.resuiting system have an extreme torsional irregularity (horizontal structural irregularity type)
Other
No requirements
* Reference 13.4
845
13.4 Seismic Shear Forces in Beams and Columns of a Frame: Strong Column-Weak Beam Concept Table 13.9
Permitted Analytical Procedures * #
... .!!lll)
ia
111'-
...J~ Seismic Design Category
B,C
D, E,F
cia G)C iae:( >G) .u ::I ... Structural Characteristics
0"0 WLL
II)
G) 'iii
II)~
CIII OC
Q.e:(
g:E
D:::I Cii-=
"u oG)
:i5~
G) II) C"," Ow
!Eo en .......
G)
D:e:(j!! u--::I
.-EoG) ~" II) _
en::E:Q.
Occupancy Category I or II buildings of light-framed construction not exceeding 3 stories in height
p#
P
P
Other Occupancy Category I or II buildings not exceeding 2 stories in height
P
P
P
All other structures
P
P
P
Occupancy Category I or II buildings of light-framed construction not exceeding 3 stories in height
P
P
P
Other Occupancy Category I or II buildings not exceeding 2 stories in height
P
P
P
Regular structures with T<3.STs and all structures of light-frame construction
P
P
P
Irregular structures with T<3.5Ts and having only horizontal irregularities Type 2, 3, 4, or S of Table 32.2-9 or vertical irregularities Type 4, Sa, or Sb of Table 32.2-10
P
P
P
Np#
P
P
All other structures
U
0 G).- ...
. - II)
* Reference 13.4 #
P= Permitted, NP = Not permitted
ing sections will present ACI 318-08 code provisions for proportioning and detailing of reinforced concrete elements that can withstand Seismic loading through conformity with the IEC 2009 requirements.
13.4 SEISMIC SHEAR FORCES IN BEAMS AND COLUMNS OF A FRAME: STRONG COLUMN-WEAK BEAM CONCEPT 13.4.1 Probable Shears and Moments
Shear failure in reinforced concrete members is regarded as brittle failure. Therefore, in designing earthquake-resistant structures, it is important to provide excess shear capacity over and above that corresponding to flexural failure. The ACI 318-05 requirements are based on the strong column-weak beam concept subsequently discussed. Hence, plastification of the critical regions at the ends of the beams will have to be considered as a possible loading condition. The shear force is then computed based on the moment resistances in the developed plastic hinges, labeled as probable moment resistance, M p" developed when the longitudinal flexural steel enters into the hardening stage. Consequently, in the computation of the probable moment resistance, 1.25 fy is used as the stress in the longitudinal reinforcement. This is because the development of inelastic rotation at the faces of the
846
Chapter 13
Seismic Design of Prestressed Concrete Structures
VR
.
M~(t~~b.r VL
(a)
Figure 13.5 Seismic moments and shears at beam ends: (a) sidesway to the left; (b) sidesway to the right
joints is associated with strains in the flexural reinforcement well in excess of the yield strain. As a result, the joint shear force generated by the flexural reinforcement is computed for an increased stress hofy where ho = 1.25, namely, an increase in stress of 25 percent. In order to absorb the energy that can cause plastic hinging, the earthquake resistant frame has to be ductile in part through confinement of the longitudinal reinforcement of the columns and the beam-column joints and in part through the provision of the excess shear capacity previously discussed. Fig. 13.5 shows the deformed geometry of and the moment and shear forces for a beam subjected to gravity loading and reversible side-sway. If the intensity of gravity load is Wu then, ACI 318-08 stipulates: Wu
=
1.2D
+ 1.6L + 1.4E
13.4.1.1 Factored Loads
The IBC (Sec. 1605.2) stipulates the following load combinations; they are comparable to the ACI factored loads in Section 4.11.2: lAD 1.2D + 1.2D + 1.2D + 1.2D + 0.9D ±
1.6L + 0.5(L, or S or R) 1.6L(L, or S) + (f1L or 0.8W) 1.3W + f1L + 0.5(Lr or S or R) 1.0E + (f1 L or f2S) (1.0E or 1.3W)
(13.22)
where, f1 = 1.0 for floors in places of public assembly, for live loads in excess of 100 Ib/ft2 (4,79 kN/m2), and for parking garage live load
=
0.5 for other live loads
f2 = 0.7 For roof configurations (such as saw tooth) that do not shed snow off
the structure 0.2 for other roof configurations = Live load except roof load = Roof live load including any live load reduction = Rain load = Snow load = Wind load =
L L,
R S W
The seismic shear forces are:
13.4 Seismic Shear Forces In Beams and Columns of a Frame: Strong Column-Weak Beam Concept
847
PholO 13.4 Skybridge, Vancouver, Canada. a 202O-fllong cable-stayed bridge and the world's longest transit bridge. (Courtesy Portland Ceme nt Association.)
VL =
M;,L + M;,R 1.20 + 1.6L I + 2
( 13.23)
VR =
M;'L + M;'R I
( 13.24)
1.20
+ 2
1.6L
where I = span, Land R subscripts = left and right ends and M p, = probable moment strength at the e nd of the beam based on stee l re info rcemen t tensile strength o f 1.251"/ and strength reduction factor = 1.0. These instantaneous momen ts, Mw ' should be com· puted on the basis of equilibrium o f momcnts al the joint where the beam mome nts arc equal to the probable moments of resistance. The shea r fo rces in the columns are compu ted in a simila r manner so that the horizon tal shear force, Vr at top and bottom of the column is v ~
,
_M -",,~,_+,--M _,,,,.
"
(13.25)
except that end moments for columns M p,1 and M ptl necd not be greater than the momen ts genera ted by the M p, of beams framin g into the beam -column joint. column height and the subscripts I and 2 indicate the top and bollom col umn end moments respectively as seen in Figure 13.6. The sense o f moments at the joints is shown in Figure 13.7.
,=
13.4.2 Strong Column Weak Beam Concept As previously stated. U.S. seismic codes require that ea nhquake induced energy be dissipated by plastic hinging of the beams rather than the columns. This hypothesis is due to the fact that compression members such as columns have lower ductility than fl exuredominant beams. If the col umns are not stronger than the beams framing into a joint,
848
Chapter 13
Seismic Design of Prestressed Concrete Structures
h
h
h
Mc2
MR2) >7< ~prL2 pr
I
Me1 Me1 S I\
Mc2 SMpt2 (a)
(e)
(b)
Figure 13.6 Seismic moments and shears at column ends: (a) jOint moments (b) sway to right; (c) sway to left.
inelastic action can develop in the column, and if large enough, can cause the column to collapse. Furthermore, the consequence of a column failure is far more severe than a local beam failure. Therefore, the ACI 318-08 Code as well as the mc stipulates "strong columns and weak beams". This is ensured by the following inequality (13.26)
where '2Mcol = sum of nominal flexural strengths of columns framing into joint, calculated for factored axial forces consistent with the direction of lateral forces considered, resulting in lowest flexural strength. '2Mbm = sum of moments, at the face of the joint, corresponding to the nominal flexural strengths of the beams framing into that joint. For a joint subjected to reversible base shear forces, as shown in Fig. l3.7, Eq. l3.26 becomes ( M: + M;;)col
6
2:
"5 ( M: + M;;)b
;Mi.
;M;;
n
n
;M;;
;A("
;M;;
(l3.27)
)
+
)
C ;M,i
U
U ;M;;
;M,i
(a)
(b)
(;M,i + ;M;;)col ~
m(;M,i
+ ;M;;) bm
Figure 13.7 Seismic moment summation at beam-column joint: (a) sidesway to left; (b) sidesway to right.
849
13.5 ACI Confining Reinforcements for Structural Concrete Members
where
0.90 for beams 0.65 for tied and 0.75 for spiral columns. = 0.90 to 0.65 for beam-columns.
=
=
13.5 ACI CONFINING REINFORCEMENTS FOR STRUCTURAL CONCRETE MEMBERS
Members in frames designed for seismic regions can be classified into two categories for proportioning transverse reinforcement as follows: (a) Members with factored axial compressive force Pu not exceeding (Ag
f~/lO)
are
treated as beams. (b) Members with factored axial compressive force Pu greater than (Ag treated as columns.
f~/10)
are
13.5.1 Longitudinal Reinforcement in Compression Members
1. In seismic design, when the factored axial load Pu is negligible or significantly less than Ag f~ 110, the member is considered a flexural member (beam). If Pu > Ag f~ 110, the member is considered beam-column, because it is subjected to both axial and flexural loads as columns and shear walls are. 2. The shortest cross-sectional dimension ~ 12 in. (300 mm). 3. The limitation on the longitudinal reinforcement ratio in the beam-column element is 0.01 :s; Pg = A/Ag ::5 0.06. For practical considerations, an upper limitation of 6 percent is too excessive, because it results in impractical congestion of longitudinal reinforcement. A practical maximum total percentage Pg of 3.5 percent to 4.0 percent should be a reasonable limit. 4. A minimum percentage of longitudinal reinforcement in flexural members (beams) for sections requiring tensile reinforcement. P 2:
3Vfc fy
200 fy
- - 2: -
::5
0.025
(13.28)
But under no condition should the value of P exceed 0.025. The stresses f~ and fy in these expressions are in psi units. All reinforcement has to be continued through the joint. At least two bars have to be continuously provided both at top and bottom. 5. Beams should have at least two of the longitudinal bars continued along both the top and the bottom faces. These bars should be developed at the face of the support. 6. Columns having clear height-to-maximum-plan-dimension ratio offive or less should be designed in shear such that not less than the smaller of (a) and (b) : (a) The sum of the shear associated with development of nominal moment strengths of the member at each restrained end of the clear span and the shear calculated for factored gravity loads; (b) The maximum shear obtained from design load combinations that include modulus E, with E assumed to be twice the modulus prescribed by the governing code for earthquake resistant design. 7. Where design forces have been magnified to account for overstrength of the vertical elements of seismic-force-resisting system, the limit of (Agf~/10) should be changed to (Ag f;/4) and the transverse reinforcement should extend into the discontinued member for at least a distance Id of the largest longitudinal column bar required in the design.
850
Chapter 13 Seismic Design of Prestressed Concrete Structures
Photo 1.3.5 Column localized damage in :1 high-rise frame building. Los Angeles 1994 Earthquake. (Courtesy Po rtland Cemen t Association.)
8. Col umns supporting reactions from discontinuous stiff members, such as walls, should be provided with transverse reinforcement at a spacing,sQ' over the full height beneath the level at which the discontinuity occurs if portion of the faclored axial compressive force in these members related to earthquak e effects exceeded (A , f;/10). Where design forces have been magnified to account for overstrcngth of the vertical elemen ts of the service-force-resisting system, the limit of (A , 1:110) should be changed to (A,J;/4) according to the ACI 318-08 Code. This transverse reinforcement should extend above and below the column as stipulated in this chapter. 9. Main reinforcement should be chosen on the basis of the strong colum n-weak beam concept o f the ACI Code, namely. IM"r 2. 615 I Mn" 10. The nominal moment strength requirements are: (a) M ~ at joint face ~ In M : at that face. (b) Neither the negative nor the positive moment strength at allY section along the span can be less than olle qllarter the maximum moment strength provided at the face of either joint. Hence, at joint face: (13.29,)
at any section: M: 2.
±
(M; )ma.
(13.29b)
/11;
4 (M; )mu
(13.29<)
I
2.
13.5 ACI Confining Reinforcements for Structural Concrete Members
851
11. For coupling beams with aspect ratio [nih < 2, and with factored shear force Vu exceeding 4 Aep has to be reinforced with two intersecting groups of diagonally
Vii
placed bars, symmetrical about the midspan, where Aep = area of concrete resisting shear. 12. Prestressing steel should be unbonded in potential plastic hinge regions. The calculated strain in the prestressing steel under the design displacement procedure should be less than one percent. 13. Prestressing steel should not contribute to more than one quarter of the positive and negative flexural strength at the critical section in a plastic hinge region and should be anchored at or beyond the external face of the joint.
13.5.2 Transverse Confining Reinforcement
Transverse reinforcement in the form of closely spaced hoops (ties) or spirals has to be adequately provided. The aim is to produce adequate rotational capacity within the elastic hinges that may develop as a result of the seismic forces. 1. For column spirals, the minimum volumetric ratio of the spiral hoops needed for
the concrete core confinement cannot be less than the larger of: 0.12!~
(13.30a)
P > -S -
!yt
or Ps
2:
0.45 ( -Ag Aeh
-
)f'!yt
1 ~
(13.30b)
whichever is greater, where Ps = ratio of volume of spiral reinforcement to the core volume measured out to out. Ag = gross area of the column section. Aeh = core area of section measured to the outside of the transverse reinforcement (sq. in.). !yt = specified yield of transverse reinforcement, psi. 2. For column rectangular hoops, the total cross-sectional area within spacing s, can-
not be less than the larger of: (13.31a) or (13.31b) where total cross-sectional area of transverse reinforcement (including cross ties) within spacing s and perpendicular to dimension he' be = cross-sectional dimension of member core measured c.-c. of confining reinforcement, in., as in Figure 13.13. hx = maximum horizontal spacing of hoops or ties on all faces of the column, in. Aeh = cross-sectional area of structural member, measured out-to-out of transverse reinforcement ASh =
852
Chapter 13 Se;sm;c Des;gn of Prestressed Concrete Structures S
= spacing of transverse reinforcement measured along the longitudinal axis
smax
of the member, in. $; one-quarter of the smallest cross-sectional dimension of the member or 6 times the diameter of longitudinal reinforcement, Also
So
S
o
=4+(
14 -
3
hx)
= longitudinal spacing of the transverse reinforcement within length 10 • Its
value should not exceed 6 in. and need not be taken less than 4 in. Additionally, if the thickness of the concrete outside the confining transverse reinforcement exceeds 4 in., additional transverse reinforcement has to be provided at a spacing not to exceed 12 in. The concrete cover on the additional reinforcement should not exceed 4 in. 3. The confining transverse reinforcement in columns should be placed on both sides of a potential hinge over a distance 10 • The largest of the following three conditions governs 10: (a) depth of member at joint face (b) one-sixth of the clear span (c) 18 in. Increase the distance 10 by 50% or more in locations of high axial loading and flexural demand such as at the base of a building. When transverse reinforcement is not provided throughout the column length, the remainder of the column length has to contain spiral of hoop reinforcement with spacing not exceeding the smaller of six times the diameter of the longitudinal bars of 6 in.
4. For beam confinement, the confining transverse reinforcement at beam ends should be placed over a length equal to twice the member depth h from the face of the joint on either side or of any other location where plastic hinges can develop. The maximum hoop spacing should be the smallest of the following four conditions: (a) One-fourth effective depth d. (b) 8 x diameter of longitudinal bars. (c) 24 x diameter of the hoop. (d) 12 in. (300 mm). IBC requires that the spacing of the confining loops in the plasticity zone of the beam not exceed 4 in. Figure 13.13(a) from Ref. 13.14 summarizes typical detailing requirements for a confined column in a monolithic ductile connection and Figure 13.13(b) from Ref. 13.24 for a hybrid precast prestressed assembly. 5. Reduction in confinement at joints: a 50 percent reduction in confinement and an increase in the minimum tie spacing to 6 in. is allowed by the ACI Code, if a monolithic joint is confined on all four faces by adjoining beams with each beam wide enough to cover three quarters of the adjoining face.
6. The yield strength of reinforcement in seismic zones should not exceed 60,000 psi. 13.5.3 Horizontal Shear at the Joint of Beam-Column Connections
Test of joints and deep beams have shown that shear strength is not as sensitive to joint shear reinforcement as for that along the span. On this basis, the ACI Code has assumed the joint strength as a function of only the compressive strength of the concrete and requires a minimum amount of transverse reinforcement in the joint. The effective area Aj within the joint should in no case be greater than the column cross-sectional area.
1, ~I
1
l
853
13.5 ACI Confining Reinforcements for Structural Concrete Members
The minimal shear strength of the joint should not be taken greater than the forces Vn specified below for normal weight concrete, 1. Confined on all faces by beams framing into the joint, Vn
:::;
20A~Aj
(13.32a)
2. Confined on three faces or on two opposite faces, Vn
:::;
15 A~Aj
(13.32b)
3. All other cases, (13.32c) A framing beam in a monolithic joint is considered to provide confinement to the joint only if at least three-quarters of the joint is covered by the beam. The value of allowable Vn should be reduced by 25 percent if lightweight concrete is used. Also, test data indicates that the value in Eq. 13.32c is unconservative when applied to corner joints. Aj = effective cross-sectional area within a joint as in Figure 13.8, in a plane parallel to the plane of reinforcement generating shear at the joint. The reversible seismic forces at the joint are shown in Figure 13.9. The ACI Code assumes that the horizontal shear in the joint is determined on the basis that the stress in the flexural tensile steel = 1.25 fy- Figure 13.9 shows the forces acting on a beam-column connection at the joint.
Effective joint width S b + h Sb+2x
Effective area
Ai Joint depth = in plane of reinforcement generating shear
Reinforcement generating shear
o Direction of forces generating shear
I~
b
Note: Effective area of joint for forces in each direction of framing is to be considered separately. Joint illustrated does not meet conditions to be considered confined because the framing members do not cover at least 3/4 of each of the joints.
Vi
Figure 13.8
Seismic effective area of joint (Ref. 13.1).
1 854
Chapter 13
Seismic Design of Prestressed Concrete Structures
Asfy _ _..._ 0.85 f'cba _ ......._
As (1.25fy )
..
~ ..- - 0.85 f'cba
VU~'fI~~
~ .. - _
A.fy
Figure 13.9 Reversible forces at beam-column jOint connection. (VU = horizontal shear at joint).
For slab-column connections of two-way slabs without bea_m~lab shear reinforcement should be based on Vn ::; 6 b ad provided that v:, 2: 3 V fe' b ad, and should extend at least 4 times the slab thickness from the face of the support.
vt:
13.5.4 Development of Reinforcement
For bars of sizes No.3 through 11 terminating at an exterior joint with standard 90 0 hooks in normal concrete, the development length Cdh beyond the column face, as required by the ACI 318 Code, should not be less than the largest of following: Cdh
f
2: y d b/(65
V"i'c )
(13.33a) (13.33b)
where db = bar diameter (13.33c) The development length provided beyond the column face must be no less than Cd = 2.5 Cdh when the depth of concrete cast in a monolithic joint in one lift beneath the bar ::; 12 in., or Cd = 3.5 C dh when the depth of concrete cast in one lift beneath the bar exceeds 12 in. For lightweight concrete, Cdh for a bar with a standard 90 0 hook should not be less than the largest of 10 db' 7!fJ. in. and 1.25 times the length required by Equation 13.33(a). All straight bars terminated at a joint are required to pass through the confined core of the column or shear wall boundary member. Any portion of the straight embedment length not within the confined core should be increased by a factor of 1.6. 13.5.5 Allowable Shear Stresses in Structural Walls, Diaphragms, and Coupling Beams 1. Structural Walls and Diaphragms
High shear walls, that is, structural walls, with height-to-depth ratio in excess of 2.0 essentially act as vertical cantilever beams. As a result, their strength is principally determined by flexure rather than by shear.
855
13.5 ACI Confining Reinforcements for Structural Concrete Members
Flexural considerations: (a) Displacement-based Approach: For walls or piers continuous in cross section from the base of the structure to the top of the wall and designed to have a single critical section for flexure and axial loads, the compressive zones have to be reinforced with boundary elements with a geometry defined as follows: lw
(13.34a)
but that 'bJh w is taken not less than 0.007. The reinforcement has to extend vertically along the wall a distance not less than the larger of lw or MJ4Vu from the critical section. c = distance from the extreme compression fibers to the neutral axis computed from the factored axial force and nominal moment strength. hw = height of entire wall. 'b u = design displacement (b) Stress-based Approach: This alternative design procedure requires that boundary elements in structural walls have to be provided whenever the extreme fiber compressive stresses exceed 0.20 f~. The boundary elements have to extend along the vertical boundaries of the entire wall and around the edges of openings. They can be discontinued where the computed compressive stress is less than 0.15 f~. The stresses are computed for factored forces using a linearly elastic model and cross-section properties. It should be noted that when boundary elements are required, the wall is essentially detailed in a similar manner in both approaches.
Shear considerations: If the shear wall is subjected to factored in-plane seismic shear forces V uh > Acv A then it should be reinforced with a reinforcement percentage Pv ~ 0.0025. Spacing of the reinforcement each way should not exceed 18 in. center to center. If V uh < Acv A the reinforcement percentage can be reduced to 0.0012 for No.5 bars or less in diameter and 0.0015 for larger deformed bar sizes. Reinforcement provided for shear strength has to be continuous and distributed across the shear plane. At least two curtains of reinforcement are needed in such a wall if the in-plane factored shear forces exceed a value of 2 Acr A where PI = AsJAcv Acv = net area of concrete cross section = thickness x length of section in direction of shear considered. Asv = projection on Acv of area of distributed shear reinforcement crossing the plane Acv.
v'"fc,
v'"fc,
v'"fc.
The nominal shear strength Vn of structural walls and diaphragms of high-rise buildings with aspect ratio greater than 2 should not exceed the shear force computed from: (13.34b) where Pn = ratio of distributed shear reinforcement of a plane perpendicular to the plane
of Acv. For low-rise walls with aspect ratio hwllw less than 2, the ACI Code requires that the coefficient in Eq. 13.34b be increased linearly up to a value of 3 when the hj1w ratio
856
Chapler 13
Seismic Design 01 Prestressed Coocrete Structures
I'hoto 13.6 Northridge, CaHrornia, 1994 earthquake structural failure. (Courtesy Dr. Murat Saatcioglu.)
reaches 1.5 in order to account for the higher shear capacity of low·risc walls. In other words, (13.340)
where Q
c
=2 when " ..11"" ~ 2 and 0 c = 3 when IIjl"" = 1.5; V" =
ing to the development of the nominal flexural strength of the member. The nominal (Jexural strength is detennined considering the most critical factored axial loads incl udi ng earthquake effects. The maximum allowable nominal unit shear strength in structu ral walls is 8Acv A where A ~ is the lotal cross-sectional area (in. 2) previ-
vr;.
ously defined and /; is in psi. However, the nominal shear strength of anyone of the individua l wa ll piers can be permitted to have a maximum value of 10 Arp ~ \.Ii;., where A rp is the cross-sectional area of the individua l pier. 2. Coupling Beams: The provisions fo r allowable shear stresses in coupling beams are as follows. Coupling beams are struct ural elements connecting structural walls to provide additional stiUness and e nergy dissipation. In many cases, geometrical limits result in coupling beams whose depth to clear span ratio is high (Ref. 13.1. 13.2). Hence. they can be con-
857
13.5 ACI Confining Reinforcements for Structural Concrete Members
trolled by shear and subjected to strength and stiffness deterioration in earthquakes. To reduce the extent of the deterioration, the span to depth ratio lnld is limited to a value of 4.0 except in cases of special moment frames in which the width to depth ratio cannot be less than 0.30. Coupling beams should only be used in locations where damage to them would not impair the vertical load carrying capacity of the structure or the integrity of the non-structural components and their connection to the structure (Ref. 13.1). If the factored shear force Vu exceeds 4 X. A cp' two intersecting groups of diagonally-placed bars symmetrical about the midspan have to be used. This requirement can be waived if it can be demonstrated that their stiffness loss does not impair the vertical load carrying capacity of the structure. The nominal shear strength, V n , is determined from the following expression.
Vi'c
Vn
=
Vfc Acp
2Avd fy sin a ~ 10 A
(13.35)
where Acp is the cross-sectional area of the beam, and AVd is the total area of reinforcement in each group of diagonal bars in a diagonally reinforced beam. ex = angle between the diagonally placed bars and the longitudinal axis of the coupling beams. For coupling beams when reinforced with intersecting groups of diagonally placed bars, each group has to consist of a minimum of four bars provided in two or more layers. The diagonal bars have to be embedded into the wall not less than 1.25 times the development length ld for steel yield strength fy in tension. Each group of the diagonal bars have to be enclosed in transverse reinforcement having out-to-out dimensions not smaller than b)2 parallel to b w and b w IS in all other sides. The transverse reinforcement should have spacing measured parallel to the diagonal bars not exceeding six times the diameter of the diagonal bar but not exceeding 6 inches along the entire length of the longitudinal bar. The nominal shear Vn is determined from the above expression:
Vn
=
2 AVdfy sin a ~ 10AVfc A cw
where a is the angle between the diagonal bars and the longitudinal axis of the coupling beam. In the case of structural diaphragms, the nominal shear in the diaphragm is limited to Vn ~ 8 Acv A A typical illustration of a diagonally reinforced coupling beam is shown in Fig. 13.10. The diagonally-placed bars have to be developed in tension within the wall and also considered to contribute to the nominal flexural strength of the coupling beam.
Vfc.
AVd = total area of bars in the group
of bars forming one diagonal
Section 1-1
Figure 13.10
Elevation
Coupling Beam with Diagonally Oriented Reinforcement
858
Chapter 13
Seismic Design of Prestressed Concrete Structures
13.6 SEISMIC DESIGN CONCEPTS IN HIGH-RISE BUILDINGS AND OTHER STRUCTURES 13.6.1 General Concepts
The design of concrete structures in seismic regions has to take into consideration the impact of the large reversible seismic horizontal forces that act on a structure during an earthquake. For the main elements of a structure to service such high-intensity forces, the structure must have adequate ductility in the joints of the principal components or in the response of solid vertical elements such as structural shear walls to ground motion. Excessive strength is not necessarily desirable or essential in earthquake-resistant design. Inelastic response can overcome service damage if adequate ductility is available through proper design and confinement. Shear strength has to exceed the flexural strength of the components and joints in order that shear deformations do not occur as a result of significant loss of stiffness and strength (Ref 13.17). The failure due to severe ground motion is accentuated in stories with sudden stiffness changes. The dynamic response of the total structure is determined by the flexible stories. Since loss of stiffness results in large inelastic deformations, such deformations, if of sufficient magnitude, would lead to the collapse of the total structure. Therefore, the design has to proportion the detailing of the members to such a degree that the components can tolerate the expected large inelastic deformations without rupture. Such detailing will be discussed in subsequent sections. In the design of high-rise buildings, a number of analytical tools are usually used to identify the required strength and probable deformations demand (Refs. 13.14, 13.16, 13.17). The required strength is the factored load or required ultimate strength of the component along the lateral load path ("ductile-link") that is expected to absorb the anticipated post-yield deformation. The strength of this "ductile-link" is usually developed by combining load effects (D, L, E, etc.), although moment redistribution may be used to attain a more rational development of the system (Ref. 13.17). The design earthquake load (E) must exceed that required by the lEe or other controlling codes. Typically, the strength of this "ductile-link" is developed from site-specific ground motion studies. The probable deformation demand is the level of deformability likely to be imposed on a structure and most importantly on the component that is expected to deform in the post-yield range during a catastrophic earthquake. Deformation objectives will be many times greater than those associated with the objective strength level. The designer should endeavor to have those post-yield deformations occur where they are likely to create the least potential for collapse of the structure and minimize component damage. It is for this reason that the weak beam/strong column philosophy is adopted in the design of special moment frames, be they constructed of concrete or steel. The brief design examples that are in subsequent sections will presume that the level of strength and deformation required of the "ductile-link" has been determined. The development of the ductile-link is essential to the success of the adopted seismic bracing system, but it is not sufficient. The other members along the lateral load path must be protected so that they do not fail as the ductile-link deforms. This member protection hypothesis is generically referred to as capacity-based design (Ref. 13.16, 13.17). 13.6.2 Ductility of Elements and Plastic Hinging
Ductility is an essential property in structures which have to respond to inelasticity in severe earthquakes. It is measured in terms of strain, displacement, and rotation. High ductility enables a member or a joint to sustain plastic strains without a significant reduction of stress. Hence, large rotations are essential as a measure of curvature if discontinuity, unsustainable displacements, or rupture are to be avoided. Three measures of ductility are identified:
13.6 Seismic Design Concepts in High-Rise Buildings and Other Structures
859
(a) Strain ductility defined by E
i-le = ;-
(13.36a)
y
where
E
Ey
= maximum sustainable strain = yield strain ductility
(b) Curvature ductility defined by:
i-l
(13.36b)
where
i-lLl.=~y
where
~
(13.36c)
= maximum sustainable displacement = ~y + ~p
~y = ~p =
yield displacement plastic displacement
The values of all these ductility factors have to be considerably greater than 1.0 for inelastic behavior to be sustainable. Ductility can effectively be achieved through adequate confinement as stipulated in the ACI 318-08 code (Ref. 13.1) and the International Building Code, IBC 2009 (Ref. 13.2) Due to large rotations, the structure at imposed locations reaches the limit ultimate state through the development of plastic hinges. The plastic hinges generated by seismic action would generally develop close to the side of the column since weak beam-strong column design is generally used, as stipulated in ACI 318 (Ref. 13.1). For the plastic hinge to develop in the beams rather than the columns of a multistory frame, special confinements have to be provided over a beam's length ahead of the columns face, equal to twice the beam depth. Figure 13.11 (a) and (b) schematically demonstrate the imposed locations of the plastic hinges in monolithic construction. Hence, the columns would be large enough to resist the design seismic forces while the beams possess the required ductility to respond to the seismic strains imposed by the earthquake. In the case of using precast ductile moment resisting frames, a hybrid connection can be used and proportioned by a capacity-based design. An example is the Dywidag Ductile Assembly described in Section 13.7.2 or a dual system as in Section 13.7.5, providing a large level of energy dissipation. 13.6.3 Ductility Demand Due to Drift Effect
As the multistory floors drift in response to the horizontal seismic force, the drift increases in the lower levels due to the p-~ effect. The plastic rotation demand increases. If ignoring the p-~ effect results in inelastic drift significantly larger than 1.5 percent of the story height, the drift, with the p-~ influence, would be considerably magnified. In such a case, the plastic rotation demands in both beams and first-story columns would exceed the levels achieved with normal detailing in seismic design (Ref. 13.17). It must be emphasized that design joint deformations associated with shear and bond mechanisms should not result in excessive drift. This is because large shear forces can develop in the beam-column joints under seismic action regardless whether plastic hinges develop close to the column face or ahead in the beam span. In order to prevent
Chapler 13
Seismic Design of Prestressed Concrete Structures
Plastic hinges
----Plastic hinges
t t
(01 ~
>h
X
;;:
~ ~ ~
I·O.5h.
C!:
I
1.5h
>2h
X
Critical section
(bl Figure 13.1 1 Imposed Plastic Hinge Locations: (a) Transfonned hinge location in monolithic construction (b) critical hinge section.
shear fai lure at the joint, both vertical and horizontal shear reinforcemen t is necessary, with the ho rizonta l reinforcement significantly morc than is normally provided by lies or hoops. Also, full anchorage developmen t lengths or bond mechan isms have 10 be ensured in the reinforcement embedded wi thin the beam-col umn join t.
13.7 STRUCTURAL SYSTEMS IN SEISMIC ZONES
In general. three systems are applicable in medium- and high-seism icity zones 1. Struct ural ductile fram es 2. Shear wa ll systems 3. Dual systems, which arc a combi nation o f the two 13.7.1 Structural Ductile Frames
Present bui lding codes when used in high-seismicity zones have generally been limited to situ-cast specia l moment resisting ductile frame and shear walls. From the discussion in Sections 13.6.2 and 13.6.3 it is clear that the beam-column connection is the major part of
861
13.7 Structural Systems in Seismic Zones
the fram e that has to sust'ain large seismically imposed deformations. Both reinforced and mo noli thic prestressed concrete fram es have been designed and built for some time (Ref. 13. 13, 13.14, 13.17). Precast concrete, on the othe r hand, has traditionally been viewed as an assembly of components that attempts to e mulate a situ·cast structure. This approach disregards the advantages presented by the discrete e lements that make up a total precast structure. If, by design , a post·yield defonnalion can be imposed to occur where precast e lements are joined, damage to the structure can be significantly reduced. This is because a weakened plane already exists at the point where a post-yield rotation has (0 be accommodated. A monolithically cast element. on the other hand. must crack, usually along several planes, in o rde r to accommodate the require d rOlat ion. Given this advantage, precast concrete structures can be created capable of surviving earthquakes with lower levels of damage than those created from other materials o r by o the r processes. The use of precast prestressed concrete e le ments in ductile fram e constructio n is comi ng of age. Extensive research is ava ila ble to justify use o f precast c lemen ts safely in ductile beam-column frames in high seismicity zones (Ref. 13.18-13.27). Figure 13.12 from Ref. 13.18 shows a hybrid connection. Th e con nectio n would have well·bonded
Mild steel in
Partial grout location
A
Elevation Fiber reinforced grout
A
Steel angle '\..
B
~~ Lr;JLr;J A
As shown, PTsteel is partially bonded
Sectiorl A-A
Note : Column bars omitted for clarity Detail A Figure 13,12 Precast Hybrid Moment Connection (Ref. 13.18)
Section B-8
862
Chapter 13
Seismic Design of Prestressed Concrete Structures
mild (ductile) steel reinforcing bars at top and bottom of the beam and high-strength prestressing tendons at mid-depth of the beam. The mild steel is intended to dissipate the seismic energy by yielding. The prestressing steel provides the shear resistance from the friction developed by the prestressing force. The system is defined as hybrid because of using two types of reinforcement. The hybrid system is the evolution of an assemblage of precast concrete components by post-tensioning that was first proposed in New Zealand in the early 1970s. The hybrid system was developed largely through an interactive test program (Ref. 13.17, 13.22, 13.23). The objective of the tests performed was to improve upon the energy dissipation characteristics of assemblies connected exclusively by post-tensioning (Ref. 13.24). The basic objectives of the hybrid system are to mainly accomplish the following results: • Balance the restoring force provided by the concentric post-tensioning with the strength developed by the mild steel so that a restorative or self-centering force exists after the earthquake. This should reduce the potential for permanent deformation. • Maintain a strain state in the post-tensioning reinforcement at the deformation limit state that is within the elastic range (fps < 0.9 fpu). • Localize the post-yield deformation so as to cause the post-yield rotation to occur along the interface between the beam and the column. This reduces the potential for nonstructural damage to the beam. Figure 13.13 (a) gives typical detailing of monolithic situ-cast reinforced concrete ductile connection. Figure 13.13 (b) demonstrates typical details of the reinforcement in a hybrid precast frame assembly. The performance of the hybrid moment-resisting beam-column connection has been thoroughly verified through tests in several centers of research as listed in the selected references. The crack widths in all the specimens in both beams and columns were very small, in the 0.04 in. range (Ref. 13.26). Research test results have demonstrated that hybrid precast systems have the following performance capabilities: (a) Can be designed to have the same flexural strength as conventionally reinforced
systems. (b) Have large drift capacity. (c) Dissipate more energy than conventional systems up to 1.5 percent drift. (d) Have concrete in the hybrid system that suffers negligible damage even if the drift is in the range of 6%. Figure 13.14 (a) demonstrates the narrow cracking pattern and negligible damage at 3.5 percent drift while Figure 13.14 (b) shows the contrasting behavior of the monolithically cast assembly. Additionally, studies on large-scale prototype tests have been conducted by Pessiki et al. on precast beam-column non-bonded post-tensioned connections in ductile frames under high-seismic loading (Ref. 13.31). They demonstrate that such assemblages can perform satisfactorily for frames on hard soil conditions. Their tests also indicate that displacement of the frames on medium or soft soil conditions in high-seismicity regions are difficult to reasonably estimate using elastic analysis under the equivalent lateral base force code approach.
p..
p..
~
~ fa ~
Depth {
h
1/6 (clear span)
18 in.
:~lt{DePthh ! 10 ~ 1/6 (clear span)
---L. s
~
::r:r I-r--I
i------l
I I I I
I I I I
I
I
i I ~
I
la}
~
•
~yt
1 .-£. f'
0.45 (ACh -)
I ...
h
Ib)
~I
I: ,,{~~=r~~.mM'
----
10
h
0.12 :,;,
Consecutive crossties engaging the same longitudinal bar have their 90-degree hooks on opposite sides of column
1Sin.
!----i
~I
fa
Ps~ {
• For intermediate moment frames, s $; {8 x smallest long. bar diameter 24 x transverse bar diameter 1/2 x smallest member dimension 12 in.
b
s 0 =4+ 143 h, <61n.>41n.
Ash~ O.09sbc
Ash~
I:E: JJJUi Ash2
t;;;f'c Ag
Xi
AShI
,
1
tc
{ A·>o·''''''IA. - I'.
I :
X,
~
I
c
X, bel
~
I
c
X,
:
I
(e)
Figure 13.13 (a) Typical Detailing of Seismically Reinforced Column (Ref. 13.14) (i) spirally confined, (ii) confined with rectangular hoops, (iii) cross-sectional detailing of ties. Consecutive cross ties should have 90 0 hooks on opposite sides.
co 01 CtJ
e2
864
Chapter 13
Seismic Design of Prestressed Concrete Structures
DebOnded region Concentric I
post-tensioning
71
Mild steel grouted in tubes
(
I
J
)
r
J
I
n 16
Mild steel 3-#6
3-#8
"
;)
t+-t+-+t-+t -
2" clr.
1~120"
I
~"1]L ~C] 8-112"<1> strands BEAM SECTION
Figure 13.13 (b)
COLUMN SECTION Hybrid Frame Assembly (Ref. 13.24)
13.7.2 Dywidag Ductile Beam-Column Connection: DDC Assembly
The DDC assembly was developed by Dr. R. E. Englekirk (Ref. 13.19) and produced by Dywidag Systems International (DSI). It allows the precast concrete beams to be bolted to the column, simplifying construction while at the same time improving seismic behavior. The system seems to satisfy the ductility requirements for both the shear forces at the column joint and the deformation and rotational ductility requirements in high seismicity zones. The design is also simple and easy. The assembly consists of ductile rods embedded in the concrete column. The precast beam contains high-strength (Fymin = 120 ksi) Dywidag Bars® connected to a transfer block. The beam is connected to the column by high-strength steel bolts (H in. -A490). The flexural strength of the beam is limited by the capacity of the ductile rod (Fy = 141 kips). The other components along the load path are designed to the probable strength of the ductile rod (1.25 Fy), a capacity-based approach. Shear is transferred by steel-to-steel friction at the interface (transfer block to ductile rod) and bearing of the head of the ductile rod on the confined concrete of the column. Figure 13.15 illustrates a typical single DDC assembly unit with the high-strength bolts connecting the precast prestressed beams to the assembly embedded in the columns at the joint. Example 13.4 gives the design computational steps for a typical ductile connection in a high-rise frame building. Figure 13.16 gives an example of this application to
"5
13.7 Structural Systems in Seismic Zones
(0)
(b)
FIgure 13.14 Beam-Column Assembly at 3.5 percent Drift (Courtesy Dr. A. E. Englekirk) (a) Precast Connection Assembly (b) Monolithically-cast Assembly
Chapter 13 Seismic Design of Prestressed Concrete Structures
866
Ductile Rod
High Strength
Oywidag Bars®
Transfer
Figure 13.15
DOC Assembly, Tensile Strength at Yield = 282 Kips (Courtesy
Dywidag-Systems International and Dr. A. E. Englekirk)
ct column
R.4 ' xS· xl ' -2 1J.2' for each 2 - - - - - - , rod group
(2)-1318' dywdag thread bars . / .--"----, hex nuts
Precast column
Precast
be,m
Rod ----./
Ductile pocket
Temporary corbel
1'-3" 1 Ill' dia. A490 ,,,,,", ---' pretensioned to 148K each
P/~' dia. dywdag ductile rods
5' dia. shim ELEVATION
Figure 13.16
PLAN VIEW
Beam-Ie-Column Connection showing Dywidag Ductile Connector Details (Ref. 13.19)
It s
13.7 Structural Systems in Seismic Zones
867
Photo 13.7 Willern Center Parking Struct ure. Los Angeles. California (Courtesy Dr. Robert E. EngJekirk).
a park ing garage in high seismicity zoncs showing beams-to-column DDC dctails in a parking garage (Ref. 13. 19). Photo 13.7 shows the ductile garage fr ame structure at completion . 13.7.3 Structural Walls in High-Seismicity Zones (Shear Walls)
Shear walls form effi cient and re liable lateral force resisting systems. They are designed to account for the total lat eral base shear force generated by an earthquake. This condition assumes that the wall has an adequate fo undation, which can transmit deformational actions from the structure to the ground without rocking to any measurable cxtent. They also provide tortional stability to the multi-story system. Figure 13.17 shows a typica l torsional stability arrangement of walls both in the E-W and N-S di rect ions. wit h Figure 13. 17 (b) the torsional stability provided by an interior core.
(aJ
(bJ
Figure 13.17 Torsionally Stable Shear Wall Systems (a) Boundary walls arrangement with concentric resistance center; (b) Core wall system with eccentric resistance cenler.
Chapter 13
868
Seismic Design of Prestressed Concrete Structures
Structural (shear) walls have been successfully used for more than 35 years. They are essentially vertical cantilevers designed to receive lateral forces from diaphragms or coupling beams and then transmit the forces to the ground. The forces in these walls may often be predominantly shear forces for low-rise buildings. Slender walls will also undergo significant bending, mainly flexural stresses. One of the main objectives of the structural analysis is to determine in what proportion the applied wind or seismic forces are distributed among the various shear walls. For the case where no ductile moment frames are present, one can assume that each floor diaphragm displaces in its plane as a rigid body. In such an analysis, the magnitude of lateral displacement becomes the dominant factor in determining the proportion of loads resisted by each wall. If the wall is treated as a deep vertical beam cantilevering from the foundation, shear deformations become a major component of the displacement and have to be taken into account. Based on the analysis by Aswad et al. in Ref. 13.28, it has been shown that the "beam element" method including the shear deformations is quite accurate for evaluating the shear and overturning moments in plan layouts with shear walls. Figure 13.18 shows the modes of failure of structural walls subjected to seismic lateral loading.
Compression
(b)
(a)
Low-rise wall
A
Large
I~
crack""
~~~~ (c)
(d)
(e)
High-rise wall
Figure 13.18 Typical failure modes of structural walls: (a) Shear cracking pattern; (b) compression strut between cracks; (c) fracture of the reinforcement; (d) flexure-shear failure pattern; (e) failure by crushing of concrete.
869
13.7 Structural Systems in Seismic Zones
Figure 13.19 schematically illustrates the drift due to both bending and shear, and Figure 13.20 shows a precast shear wall connection to the foundation using a Dywidag threaded bar connector. For small uplift forces, Figure 13.21 gives a typical welded angle connector to the foundation.
13.7.4 Unbonded Precast Post-Tensioned Walls
Unbonded precast post-tensioned walls are constructed by vertically joining precast wall panels along horizontal connections using post-tensioning reinforcement not bonded to the concrete. Precast concrete walls with substantial initial lateral stiffness can be designed to soften and satisfy estimated nonlinear displacement demands under codespecified design level motion, without yielding in the post-tensioning reinforcement or significant damage in the wall panel (Ref. 13.32-13.33). Figure 13.22 shows a prototype wall from Ref. 13.33 with unbounded tendons prestressing the six structural wall panels. The tests showed that the nonlinear elastic behavior resulted in small inelastic energy dissipation per hysterisis cycle. Because of the small inelastic energy dissipation, larger lateral displacements of the unbounded post-tensioned precast concrete walls are larger than the displacements of convensional reinforced concrete systems. More research is obviously needed in this area particularly for application to designs in highseismicity regions.
Bending deflection only
Sum of shear plus bending deflections X r--+-
t:t=
b
1 Section X-X
Figure 13.19 Schematic of shear wall drift due to bending and shear
3 1/2" 1.0 . spiral t1-~+-- duel sheathing (26ga.)
I 1/4" 0 Ihreadbar
:1---1-- - manufactured grade 150 ksi by Dywidag Grout poll
1_ _ _ _
Folded seam 33 gao metal sheathing (41 mm)
Fle)lible plastic grout tube (I3mm)
5'- 0" minimum
~~~~~ __ _ _ _ Anchor plate ... (Il/2"x S" x S"FB) I~----- Anchor nut
Figure 13.20 Precast shear wall connection to continuous foundation using Dywidag threaded bar connector
.-1:
_
'" recess
.0
.!i " };
.~
@.:.c
• 01 0
f;~a ~.8~
"
"
, ;., 9 ' 0
.0 , 0.. .0
,0
°d:O o· cP~
.: ~= ~ ----~. ~:!t±:L.------
Precast wall (8" min.)
Plan Figure 13.21
" • "" "" • • "" """ •• "" "" • " "" • " •• c:' • •L ________________
"
_
Section
Welded angle connection of precast shear wall to continuous foundation
871
13.7 Structural Systems in Seismic Zones
972 in
I-!'- - 240 in. ---,~ Elevation view #3 spirals
6 in . diameter
!
116 in.
4 In. pitch psp = 1.83% I
_
Tl~~• 0.625 'po
12 in
-L ."
-
!.
!pI::
ap '" 1.485 in 2
in~
\
- -
120in. - - - - - ---I .! Cross-section view
Figure 13.22
2.5
Post· tensioned unbounded precast shear wall (Rei. 13.33)
872
Chapter 13
Seismic Design of Prestressed Concrete Structures
13.8 DUAL SYSTEMS
Ductile frames interacting with shear walls can provide a large level of energy dissipation in a major earthquake. They would also significantly reduce the story drift and the development of pronounced hinges. Since the precast frame primarily deforms in shear due to lateral loading and the wall deforms primarily in flexure with some shear deformations, the combination of both types in a dual system can result in a more efficient structure. Part of the lateral forces in such a system is allocated to the ductile frame. The balance is assigned to the shear wall. In such dual systems, the walls can be either freestanding or connected to the frames by the floor diaphragms or by coupling beams which are continuous beams in their planes connected to the abutting frames. In all systems where nonbonded prestressing is used in high-seismicity regions, it is important that the actual stress in the prestressing reinforcement can achieve and sustain the design ultimate stress level and beyond the yield strength level of 1.25 fpy"
13.9 DESIGN PROCEDURE FOR EARTHQUAKE-RESISTANT STRUCTURES
1. Determine the earthquake seismicity region, namely whether it is in a low, moderate, or high seismicity region and the site classification (A, B, C, D, E, and F) from Table 13.1 2. Determine from the maximum considered earthquake ground motion maps, the maximum spectral response Ss for 0.2 sec and Sl for 1 sec, site-class B, Figure 13.3a and b respectively using the large scale FEMA maps of USGS (Ref. 13.15) 3. Compute for the particular seismic use group (Table 13.3), the design spectral response SDS and SDI from Equations 13.2 and 13.3: 2
SDS =
3" SMS
SDl =
3" SMI
2
where,
SMS = FaSs
where,
SMJ = Fv Sl
There are three seismic use groups I, II, and III with groups II and III structures that require full seismic design consideration. 4. Compute the seismic base shear V = CsW
But Cs cannot exceed CS = SDl/(R/I)Tor less than Cs = 0.044 CDsI
R = Response modification factor from Table 13.4 I = Occupancy importance factor from Table 13.5 T = Fundamental period of vibration of a structure, Sec. 13.3.1, Ta = Cth! where, C T = building period coefficient ranging between 0.035 - 0.020 as given in the text. In cases where moment resisting frames do not exceed twelve stories in height, an approximate period Ta = 0.1 N can be used where N = number of stories. For structures in seismic design categories E or F and for other structures having a spectral response SI;::: 0.6 g, the value of Cs ;::: (0.5S])/(RlI)
873
13.9 Design Procedure for Earthquake-Resistant Structures
5. Vertically distribute the base shear force, V, to forces Fx to the floors above the base level:
x
6. Horizontally distribute the shear Vx
=
2: Fi i=1
where Fi = the portion of the seismic base shear, V, introduced at level i. 7. Tabulate these forces at all story levels. 8. Evaluate the torsional moments, story drift, the P-/:i effect and the overturning moment to ensure they are within permissible limits. 9. Execute a structural frame analysis to determine all shears and moments in the frame beams, columns, shear walls diaphragms and/or coupling beams if these are used to connect shear walls. 10. Proportion members of the ductile moment-resistant frame, that is, all beams, columns, and beam-columns. If the frame is not a ductile moment-resisting frame, the designer has the uneconomical and inefficient alternative of choosing a brittle system using a low Rw factor. 11. Using the strong column-weak beam concept, plastic hinges are assumed to form in the beams.
Seismic beam shear forces
e= beam span, Mpr = probable moment of resistance, and L, R = left and right. Seismic column shear force
where h = column height.
at joint to ensure hinges form in the beams; hence
The nominal moment strengths Mn have to be evaluated and the member proportioned prior to evaluating the seismic beam shear forces.
874
Chapter 13
Seismic Design of Prestressed Concrete Structures
Beam: flexural design, P u insignificant Column: combined bending and axial load P u Beam-column: P u > A/~/10 Shortest cross-sectional dimension;::: 12 in.
12. Longitudinal reinforcement Beam-column or columns 0.01 ::; Pg
A
= AS::;
0.06
g
For practical considerations, pg:S; 0.035. Beam (positive reinforcement): 200 fy
2: - -
200 fy
2: - -
Pmin 2: -
3Vi'c fy
Beam (flange in tension): Pmin 2: -
6Vi'c fy
The factor value, 6, in the numerator instead of 3 is because a flange width twice the web width or more is used. where fy is in psi units. P should never exceed 0.025. For proportioning reinforcement in beams, the nominal moment strength requirements are (a) M~ at face of joint;::: !M~ at the face. (b) M~ or M~ at any section;::: ~Mamax at the face.
13. Transverse confining reinforcement (a) Spirals 0.12f~ Ps2:--
fyh
or
Ps 2:
0.45 ( -Ag
-
Aeh
)1'
1 ~ fyt
whichever is greater.
Ag
=
gross area
Aeh = core area to outside of spirals
fyt
= specified yield strength
(b) Rectangular hoops in columns: Total cross-sectional area within spacing s: ASh 2: 0.09 she
2:
ff~
yt
0.3 she ( -Ag Aeh
-
)1'
1 ~ fyt
whichever is greater. ASh = total cross-sectional area of transverse reinforcement (including cross
ties) within spacing s and perpendicular to dimension he he = cross-sectional dimension of column core, in. s = spacing of transverse hoops smax = one-quarter of the smallest cross-sectional dimension or 4 in., whichever is smaller, but not to exceed 6 in.
13.9 Design Procedure lor Earthquake-Resistant Structures
875
Placemem of confin jng re;nforcem elll: Place confining re inforceme nt o n e ithe r side of po te ntia l hinge o ve r a distance the largest of (i) De pth of me mbe r at joint face (ii) One-sixth clear span (iii) 18 in. The spacing of the ties in the ba lance of column heigh t follows no nna l column tie require me nts. (c) Confining reinforcemem ill bemn ends: Should be placed o n a le ngth = 211 on both sides o f the jo int if it is int e rnal ; otherwise, max imum hoop spacing, small est of
(i) One-quarte r effective de pth d (ii) 8 x dia me ter o f lo ngitudina l bar (iii) 24 x dia me ter of hoop (iv) 12 in.
me requires that spacing in ductile (rames a t the plasticity regio n no t exceed 4 in. The ties in the bala nce of the beam spa n follow the sta nda rd shear web re info rcement requirements. If the joint is confined o n all fo ur sides, 50 percent reduction in confine me nt and increase in minimum tie spacing to 6 in. in the columns are allo wed. No smooth bar re inforcement is allowed in seismic structures.
Photo 13.8 NCN B Tower, Charlotte. North Carolina. 9OOO-psi concrete. (Courtesy Portland Cement Association.)
Chapter 13
876
Seismic Design of Prestressed Concrete Structures
14. Beam-column connections (joints): Normal concrete nominal shear strength Vn at a joint: (a) Confined on all faces: Vn ::; 20vfc Aj (b) Confined on three faces or two opposite faces: Vn ~ 15vfc Aj (c) All other cases: Vn ~ 12vfcAj where Aj is effective area at joint (Fig. 13.S). The value of allowable Vn should be reduced by 25% for lightweight concrete. Note from Fig. 13.9 that the horizontal shear in the joint is determined by assuming a stress = 1.25fy in the tensile reinforcement. 15. Development length of reinforcing bars: For bar sizes Nos. 3 to 11 without hooks, the largest of €d = 2.5€dh when concrete below bars::; 12 in. €d = 3.5€dh when concrete below bars
~
12 in.
where for normal-weight concrete €dh ~
f y d b/(65Vfc)
~
Sd b
~
6 in.
When standard 90° hooks are used, Cd = Cdh . Any portion of straight embedment length not within the confined core should be increased by a factor of 1.6.
16. Shear walls: height/depth> 2.0 (i) Minimum Pv = 0.0025 if Yuh > Acv At least two curtains of reinforcement needed if in-plane factored shear force Yuh > 2Acv where Acv = net area of concrete cross section = thickness x length of section in direction of the considered shear. (ii) If extreme fiber compressive stresses exceed O.2f~, shear walls have to be provided with boundary elements along their vertical boundaries and around the edges of openings. (iii) Available Vn = Acv (2 + Pt fy) for hw I€w ~ 2.0. For hw I€w < 2, the factor of 2 inside the parenthesis varies linearly from 3.0 for hwl€w = 1.5 to 2.0 for h)€w = 2.0; Vu =
vfc.
vfc,
vfc
vfc
vfc
Figure 13.23 gives a logic flowchart for the preceeding sixteen steps.
13.10 SI SEISMIC DESIGN EXPRESSIONS compressive strengthf~ ~ 20 MPa Ec =
w/ os 0.043 vfc MPa
Es = 200,000 MPa Equation 13.22
VL
=
Equation 13.23
VR
-
M;rL + M;rR (1.2D + 1.4L) € + 2
_ M;'L + M;;'R _ (1.2D + 1.4L) €
2
877
13.10 SI Seismic Design Expressions
(
)
Start
Determine earthquake seismic region, select IBC seismic coefficients Ss, S" SDS' SOl' R, I, s' Determine period Tby IBC Eqs. 13.13, 13.14 and the n, Ws' Wvalues.
e
n
Compute V = esWand V = Ft + ~ Fi Ft = 0 when T = 0.7 S Ft = 0.07 TV:5 0.25 V. i=l
e
Tabulate base lateral force and each story force Fx = vx V using the summation
evx = -n~~ - - out. Find each story shear and moment where ~ WitT! V = Seismic base shear i=l
Vx =
±
F;
;=1
Execute a structural frame analysis to determine all shears and moments in the frame beams, columns, and shear walls.
Proportion for flexure and revise where necessary the size and main reinforcement of the moment-resistant frame members: beams, and beam-columns (beam - column when
Pu> Ag Af~/10).
Use strong column-weak beam concept, plastic hinges in beams and not columns. L Meol
~
6/5 Mbm at joint. Beams: VL =
VR =
Mp,L + M;rR
e
+
M;rL + MprR
-
e
Columns: Ve
=
1.20
+
1.6L
2 1.20
+
1.6L
2
Mprl + Mpr2 h
Design longitudinal reinforcement. (a) Beam-columns or columns: 0.01
~
Pg ~
For practical considerations Pg ~ 0.035:
As A ~
0.06
9
200 3 Vt:, 6 Vt:, Pmin ~ -,- ~ ~f- (for +M) ~ - , - (for negative region T-beam) y
(b) Beams:
y
y
M~
at joint face ~ 1/2 M; at that face M~ or M;; at any section ~ 1/4 Ma.max at face
1 Figure 13.23 Flowchart for seismic design of ductile monolithic (strong column-weak beam concept) structures.
I
878
Chapter 13
Seismic Design of Prestressed Concrete Structures
Transverse confining reinforcement. (a) Spirals for columns: Ps 2
0.12f~
( Ag
) f~
ch
yt
- t - or 2 0.45 A - 1 T yt
Whichever is greater. (b) hoops for columns: As 2 0.09 5 he
ff'
yt
Ag
2 0.3 5 he ( A
-
1
) f~
T
ch
5 ~
yt
1/4 of smallest cross-sectional dimension or 6 times diameter of longitudinal 14 - x reinforcement or Sx ~ 4 + ( - - 3 - and need not exceed 6 in. or taken less than 4 in.
h)
Use standard tie spacing for the balance of the length. (c) Beams: Place hoops over a length = 2h from face of columns. Maximum spacing: smaller of 5 = 1/4d, 8 db main bar, 24db hoop, or 12 in. If joint confined on all four sides, 50% reduction in confining steel and increase in minimum spacing of ties to 6 in. in columns is allowed. Use the standard size and spacing of stirrups for the balance of the span as needed for shear. IBC requires that maximum spacing not exceed 4 in.
Beam-column connection Goint) Available nominal shear strength 2 applied Vu Confined on all faces:
Vn ~ 20 ~ Vf;, Aj
Confined on three faces or two opposite faces: All other cases:
Vn ~ 12 ~ Vf;,
Vn ~ 15 ~ Vf;,
Aj
Aj
Check development length, normal-weight concrete,
Cdh 2 fy db l(65 Vf;,) 2 8db 2 6 in. Cd = 2.5 C dh for 12 in. or less concrete below straight bar Cd = 3.5 Cdh for> 12 in. in one pour If bars have 90° hooks, C d = C dh • For lightweight concrete, adjust as in the ACI Code.
Design shear wall.
Vuh > 2Acv ~ Vf;,; use two reinforcement curtains in wall. If wall fc> 0.2 ~ Vf;" provide boundary elements. Available Vn = Acv (as
~ Vf;, + Pn fy)
=2.0
For hJC w 2 2.0,
as
For hJ€w = 1.5,
as = 3.0
Interpolate intermediate values of hJ€w' Maximum allowance: Vn = 8Acv ~ Vf;, for total wall
Vn = 1OAcp ~
Vf;, for individual pier
DeSign diaphragms and coupling beams when used as indicated in the text and as detailed in the IBC Code
(
End)
Figure 13.23
Continued
13.11 Seismic Base Shear and Lateral Forces and Moments by the IBC Approach
879
PholO 13.9 Masonry collapse in Los Angeles earthquake, 1994. (Councsy Portland Cement Association.)
Equation 13.24 Equation \3.27
q,:::: 0.9 for beams and 0.7 or 0.75 for columns. Equation 13.28
. . For positIve moment: p 2:
Vi. r1.4 -----r4 J,. ~
J ,.
whcrc/t, f, arc in MPa Equation 13.29(a) A t joi nt face; M ~~' At any section: Equation 13.29(b)
M: 2: ~ (M;)mu
Equation 13.29(c)
M;
M;
I
2:
"4 (M;)m.~
13.11 SEISMIC BASE SHEAR AND LATERAL FORCES AND MOMENTS BY THE IBC APPROACH EXllmple 13.1: A moment-resisling. five-story building with shear walls is idealized as in Figure 13.2. Each nocr has a weigh I W, and a hcight/,;; 9'-6" (2.9 m). Compute the seismic base shear. V. and the ovenurning moment. M, at each siory level in terms of single nocr weight WI ' assuming
880
Chapter 13 Seismic Design of Prestressed Concrete
Structure~
the idealized mass of each floor is Ws' Consider the structure to be a building in seismic occupancy category II, site-class B, design category B and seismic use group II. Given: Response modification factor R = 3.0 Occupancy importance factor 1= 1.25 Use the equivalent lateral force method in the solution.
Solution: (a) Spectral response period and base shear Total building height = 5 x 9.5 = 47.5 ft From the FEMA ground motion maps (Figure 13.3) spectral response accelerations Sl
= 0.42 sec and Ss = 0.85 sec, with a site-B class and 5 percent damping.
Adjusted spectral response accelerations for site class effects: from Tables 13.2(a) and (b), for Sl = 0.42 sec, Fv = 1.0 and for Ss = 0.85 sec, Fa = 1.0 From Equations 13.2(a) and (b), SMS = FaSs = 1.0
X
0.85 = 0.85
SMI = FVS 1 = 1.0
X
0.42 = 0.42
For 5 percent damped design spectral response acceleration using Eqs. 13.3(a) and (b),
The seismic base shear V from Eq. 13.8 is V = Cs W = Cs(5Ws) for the five stories where Ws is the idealized weight of each story. From Table 13.4, the response modification coefficient for ordinary reinforced concrete moment frame is given as R = 3. The occupancy importance factor for building category II from Table 13.5 is: 1=1.25. From Eq. 13.9,
SDS 0.567 Cs = (R/I) = 3/1.25 = 0.236,
But Cs cannot exceed the value:
Cs =
SDI (!j-)T from Eq. 13.10.
For SDl = 0.278 and from Table 13.6, C u :; 1.32. For moment resistant concrete frame systems, a building period coefficient C T = 0.022 will be used in this example. (See Sec. 13.3.1) From Eq. 13.13, the approximate fundamental period, Ta = CT h 3/ 4 = 0.022 (47.5)
3/4
= 0.396 sec
Maximum allowable Ta = CuTa = 1.32 x 0.396 = 0.523 sec For computing the base shear V, T= 1.2 x 0.523 = 0.63. SDI
Cs =
(ti)
I T
0.278
= (_3)0 1.25
= 0.184 sec .63
13.1 1 Seismic Base Shear and Lateral Forces and Moments by the IBC Approach
881
From Eq. 13.1 1. Cs cannot be less than Cs = 0.044 5 0S '" 0.044 x 0.567 = 0.025 Hence, C. E 0. 184 sec controls. :. base shear V ", CsW = CS<5 Ws):::I 0.1 84 x 5 Ws= 0.92 Ws
(b ) Vertical
D /~'lribution
of Forces and O ~'ertum;ng Moments:
From Eqs. 13.15 (a) and (b). the lateral fo rce induced at any story levcl is: F,
k ..,
:=
Cv• V
where, C•., ..
0.630.50 - 0.50 X I.0 + 1.0 '" I.26(by itncar · ·mterpolatl0n .) IV
Since" is constant for all the floors. CyX becomes 7
where i = 5 at the top floor.
~ ,-, •
~ "" I W.
+ 2W$ + 3 W, + 4W, + 5 U1,
0::::
15 U1,
i_I
Lateral force F, = C~.. V = 0.92 CjW, Overturning moment from Eq. 13. 19 is M.. =
•
T
~ FI./,; - II.. ) for the top tcn stories.
,-,
overturning moment reduction factor T = 1.0.
•
Hcnce. M, '" ~ FI (h/ - 11.)
,-,
Computing and tabulating the story fo rces Fi and the overturning moment M, for all stories,
Photo 13.10 O verpass collapse in 197 1 Los Angeles earthquake. (Courtesy Port· land Cement Association.)
1: 882
Chapter 13 Seismic Design of Prestressed Concrete Structures
Floor
Cj
Lateral force Fj =O.92Ws C j
(1)
(2)
(3)
5Ws
5
Story Shear
Story Moment
(4)
(5)
Cs = 15W = 0.333
0. 3064Ws
0
0
4
4 C4 = 15 = 0.267
0.2456Ws
0.3064Ws
0.3064Wsh
3
3 C3 = 15 = 0.200
0.1840Ws
0.5520Ws
0.8584Wsh
2
C2 =
0.133
0.1224Ws
0.7360Ws
1.5944Wsh
1
1 Cj = 15 = 0.067
0.0616W,
0.8584W,
2.4528Ws h
Wall base
Co=O
0
0.9200Ws
3.3728W,h
s
2
15 =
hence seismic base shear V = O.9200Ws • The moments at each story level are tabulated in column (5).
13.12 SEISMIC SHEAR WALL DESIGN AND DETAILING Example 13.2 Design by the ACI 318 Code the reinforcement for a shear wall in a multibay, ductile frame, twelve-story structure (adapted from Ref. 13.9) having a total height hw = 148 ft (45 m) and having equal spans of 22 ft (6.7 m). Except for the ground story, which is 16 ft (4.88 m) high, all other stories have 12 ft (3.67 m) heights. The total gravity factored load on the shear wall is Wu = 4,800,000 Ib (21.4 MN). The factored moment at the base of the wall due to seismic loads from the lateral load analysis of the transverse frames is Mu = 554 X 106 in.-Ib (62.6 MN-m). The maximum axial force on the boundary element is P u = 4,500,000 Ib (20 MN). The horizontal shear force at the base is 885,000 Ib (3940 kN). Given: wall length (horizontally) = 26' - 2" = 26.17 ft = 314 in. (7980 mm) thickness t = 20 in. = 1.67 ft (508 mm) boundary element width = 32 in. (813 mm) depth
=
50 in. (1270 mm)
As = 39 No. 11 bars (39 bars of 35-mm diameter)
in each boundary element f~ =
4000 psi (27.6 MPa), normal weight
fy = 60,000 psi (414 MPa) Use = 0.60 as the strength reduction factor for shear in this example. Solution:
1. Wall geometry and forces: en = 22 ft (6.7 m), ew (horizontal dimension) = 26.17 ft, bWeb = 20 in. = 1.67 ft, and bbound = 32 in. = 2.67 ft. factored Wu
=
4,800,000 Ib (21.4 MN)
Mu = 554
X
106 in.lb (62.6 MN-m)
Pu = 4,500,000 Ib (20 MN)
13.12 Seismic Shear Wall Design and Detailing
883
Pholo n.lI Northridge, California, 1994 earthquake structural failure. (Courtesy Dr. Murat Saatcioglu.)
2. Boundary dt'mt'nr check: COl' ''' 26.17 fI, b = 1.67 fI, p~ '" 4.500,lXX) lb. and M~ '" 550)( Iff> in..lb. Assume that the wall will not be provided with confinement over its entire section. bill 1.67(26.17)1 gross I = = = 2495 ft· g 12 12 A g = 1.67 f~
x 26.17 = 43,7ft 2
P Me = - A ± - ,- ,
c=
26.17 - 2-
x
12 '" 157 in. (3990 mm)
Concrete compressive stress in the wall is
f~
=
4,800,000 554 x l
=::
- 763 - 1682 = -2445 psi (C) (16.8 MPa)
Maximum allowable f~ '" 0.2 f; '" 0.2 )( 4000 = 800 psi (5.52 MPa) in compression if a boundary element is not required. Hence boundary elements are needed subject to the confinement and loading requirements of Section 13.5. 3. Long itudinal and tranSI't'f)'e n lnjoTCt'mt'nt: Check if two curtains of reinforcement are needed, thai is, if in.plane factored shear> 2A .... (Section 13.5.5).
VI;
884
Chapter 13
Seismic Design of Prestressed Concrete Structures
Vu
=
885,000 lb
Acv
=
area bound by web thickness and length of section in direction of shear force
= 20 X 314
=
6280 in. 2
2Acv ~ = 2 X 6280 Y 4000
=
799,400 lb (353 kN)
< Vu = 885,0001b
Hence two curtains of reinforcement are required. min Pv
=
Asv A
= Pn = 0.0025
cv
and
max s
=
18 in.
Acv per ft of wall = 20 X 12 = 240 in. 2 required As in each direction = 0.0025 X 240 = 0.60 in.2/ft Trying No.5 bars (15.8-mm diameter), As = 2(0.31) = 0.62 in. 2 in two curtains. s
=
one bar area required As/12 in.
=
12.4 in. (315 mm)
0.62 0.60/12
<
18 in. limit
O.K.
Uses = 12 in.
Check for shear reinforcement capacity A check is needed in order to determine that the No.5 bars in two curtains at 12 in. c-c both ways are adequate for the wall section to sustain the applied shear force at the base. The shear wall aspect ratio is hw 148 D = -26 = 5.66 "w .17
>
2
Hence from Eq. 13.34 b
=
where = 0.60 in this example; otherwise, refer to the ACI 318-02 Code for other conditions. Acv
=
20(26.17 X 12)
p,
=
2(0.31) 20 X 12 = 0.0026
=
6280 in. 2
available Vn = 0.60 X 6280 (2Y4000 + 0.0026 X 60,000)
= 1,065,000Ib > Vu
=
885,000 lb (4.7 MN
>
required 3.9 MN)
Hence the wall section is adequate. Therefore, use two curtains of No.5 bars spaced at 12 in. c-c in both horizontal and vertical directions.
if acting as a short column under factored vertical forces due to gravity and lateral loads: Pu acting on wall = 4,500,000 lb. From before, b = 32 in., h = 50 in., As = 39 No. 11 bars = 39 x 1.56 = 60.84 in.2 (35,100 mm2) in each boundary element. 60.84 _ 8 _ As _ Ps' - A - 32 X 50 = 1600 - 0.03
4. Boundary element check
g
Pmin = 0.01
<
Ps'
<
Pmax = 0.06 O.K.
The axial load capacity of the boundary element acting as a short column is
= 0.80 X 0.65 [0.85 X 4000 (1600 - 60.84) + 60.84 X 60,000J =
4,619,443 lb
>
Pu = 4,500,000Ib
O.K.
885
13.12 Seismic Shear Wall Design and Detailing
5. Boundary element transverse confining reinforcement: bw = 20 in., bb = 32 in., h or ew =314 in., andAg = 1600in.2• From Eqs.13.31(a) and (b) 0.12f~
p >-fyt
S -
and ASh 2:
0.3Sb e
(:g -1)~~
Jyt
eh
Assume No.5 hoops and crossties spaced at 4 in. c-c. (a) Short direction
bel
=
(1.5 + :6) 46.37 in. 2 (1.5 + :6) = 28.37 in.
50 - 2
b e2 = 32 ACh
= 46.33
=
X
28.37
= 1314 in. 2
_ 0.09f~sbcl _ 0.09 I' Jyt
ASh -
_
ASh -
0.3
X
X
-
4
(core area)
4000 X 4 X 46.37 _ . 2 60000 - 1.08 Ill. ,
(1600 ) 4000 _ . 2 46.37 1314 - 1 60,000 - 0.80 Ill.
X
= 1.08 in. 2 governs. Use three No.5 cross ties, for a total of five legs being provided including the hoop, every 4 in. along the boundary length (wall length ew )' ASh provided = 5 x 0.31 = 1.55 in. 2 , O.K., on the conservative side ASh
(b) Longitudinal direction
b e2
= 28.37 in.,
Aeh
= 1314 in. 2
_ 0.12f~be2s _ 0.12
or
I'
ASh -
X
-
Jyl
_ ASh -
0.3
X
4000 X 4 X 28.37 _ . 2 60000 - 0.91 Ill. ,
4
(1600 ) 4000 _ . 2 28.37 1314 - 1 60,000 - 0.49 Ill.
X
= 0.91 in. 2 (587 mm2 ) controls. With one No.5 crosstie, a total of three legs is provided every 4 in. c-c. ASh provided = 3 x 0.31 = 0.93 in. 2 (600 mm 2 ).
ASh
6. Check for maximum hoop spacing: 1 . S ::; -
4
S ::;
hx =
So ::;
X
32
=
8 Ill.
1 6 times dia. of longitudinal bar = 6 X 18 = 8.25 in. 32 - [2 (I! + ~) + 5 elf) 4 4
=
.
5.4 Ill.
x + ( 14 -3 h ) = 4 + (14 -3 5.4) = 6.9 in. < 6.0 in. within length 10
Maximum spacing of cross-ties or hoops = 4 in. (100 mm)
7. Development of reinforcement: Development length of No.5 horizontal bars assuming no hooks are used within the boundary element: From Eqs. 13.33(a), b, and c, fydb
e
60,000 X 0.625 _
_
dh 2: - - -
65
-
Vi'c
2:
8db
2:
6 in.
=8
65 y'4000 X
0.625
= 5 in.
. 9 Ill.
886
Chapter 13
Seismic Design of Prestressed Concrete Structures
Photo 13. 12 Interfirst Pla7.3, Dallas. Texas, IO,QOO..psi concrete. (Courtesy Porlland Cement Association.) frill = 9 in. (229 mm) governs
el = 3.5e"", = 3.5 x 9 iii 32 in. (8 15 mm) lfbars are straight as in this example. ensure thai development length is provided. If 90° hooks are used. t d ", edit:: 9 in. Note that no lap splices should be allowed for the No.5 horizontal bars.
8. Vuify adequacy of sh~ar wall s~ct;on at iu' base u/ldu combined axial load and btmding in its plane: From before, Actual p. = 4.800.000 lb (Iolal gravity factored load) Actual M. = 554 X
[if in.-Ib,
,= -M. = p.
115 in .. b"'b = 20 in" bbound = 32 in.
p = 4,800,000 = 7"184 6151b
"
0.65
M =
554
x to'
.-',
.
852 X l
"0.65
=
=
1776 in. (45 m)
column action $ = 0.65. beam action ¢I = 0.90 no. o f longitudinal bars in wall plane "" 116 composed of two rows (39 No. II bars) for both boundary elements and two curtains of No.5 bars at 12 in. center-to-center over t_ "" 314 in. total A., in the lateral cross section = 2 x 60.84 + 2 (18 X 0.31 ) :: 132.8 in.2 II, = 2(32 x 50) + 20(314 - 2 x SO) = 7480 in? (4,830JXx) mm1 )
132.8
p = 7480 = 0.0178
>
0.01 and
< 0.06
O.K.
No.5 closed hoops @ 4 in. ~ (16-mm diameter @ 101 mm ~)
No.5 crossties @ 4in.c~
No.5 @ 12- ~ both wavs (15.9-mm diameter @305mm
39 No. 11 bars (39 bars, 35-mm diameter)
39 No. 11 bars
c-c)
• • • • •
• • • • • •
20 in. (508 mm)
50 in. (1270 mm)
50 in.
(1270 mm) 26'-2(7.98 m)
Figure 13.24
co CO
.....
32 in. (812m)
Plan and detailing of shear wall in Example 13.2.
888
Chapter 13
Seismic Design of Prestressed Concrete Structures
7,384,615 Pn
= 4000 x 7480 = 0.247
Enter Figure 9.24(a) in Ref. 13.14 with Pnl(f: Ag) = 0.247 and p = 0.0178. This gives a value of Mnll: Ag h = 0.19. Hence, available Mn = 0.19 x 4000 x 7480 x 314 Mn = 852 X 106 in.-Ib, O.K.
= 1785 x 106 in.-Ib » Required
13.13 EXAMPLE 13.3 STRUCTURAL PRECAST WALL BASE CONNECTION DESIGN A precast structural (shear) wall for a five-story building in a moderate seismicity zone is b = 8 in. (203 mm) thick (Ref. 13.29). The length of the interior wall is 24 ft (7.32 m) and the height of each story is 13'-0" (3.96 m). Structural analysis showed that the wall is subjected to a factored seismic base shear force Vu = LF(Ve) = 157.4 kips (582 kN) and a factored overturning Case II base moment Mu = LF(Me) = 7114 ft-kip (9645 kN-m). The total weight of each floor including attached masses is 2,400 kips (10,675 kN). Design the connection at the base of the wall assuming that it is so reinforced that the neutral axis obtained by trial and adjustment and strain compatibility analysis is c = 17.62 in. (447 mm). Use either welded connection as in Figure 13.21 having a rated capacity of 25 kips per connection (111 kN) or Dywidag rods grade 150 ksi (1034 MPa) with thread bar couplers as in Fig. 13.20. Case I factored overturning moment Mu = 6941 ft-kip (9410 kN-m). Given: Sliding shear friction coefficient fL = 0.60 Maximum allowable horizontal concrete shear interaction stress (sliding friction):
lev
=
I~ =
1200 psi (8.2 MPa) 5,000 psi (34.5 MPa) for shear wall and for grout (dry pack)
Solution: The system forces acting on the structural wall are shown in Figure 13.25. The ACI load factors governing the design are:
• Load Case I LF: = 1.2D + 1.0E + (fl Lor 12 S) • Load Case II LF: = 0.9D ± (1.0E or 1.6W) Usually, Load Case II controls the majority of design cases for gravity walls. Use case II for seismic effects. As given in the problem statement, computer analysis using strain compatibility and trial and adjustment for the reinforcement used in the wall (Ref. 13.29) gave the following factored overturning moment: Seismic overturning moment = 7114 ft-kip (Case II) and a neutral axis depth c = 17.62 in. Actual Seismic Shear: Vu = 157.4 kips Vu 157.4 . Vn = ~ = 0.75 = 209.8 kIPS (930 kN)
Ce = 0.85 l~bl3lc = 1;00 (0.85 X 5000 X 8 X 0.80 X 17.62) = 476.5 kips. From Figure 13.25, the sliding friction contribution, VI = fL Ce = 0.60 X 476.5 = 285.9 kips.
Upper bound of VI = [(wall thickness b) (N.A. depth c) (allowable horizontal frictional stress lev)]
889
13.13 Example 13.3 Structural Precast Wall Base Connection Design
Applied
Vu
I I I I I
X4..
--4 X c
~i==~~,~V<2~~~~~~-±I f'c
1...._ _~y,._ _ _..J'
Additional connectors i for shear resistance, I .c:(
i
where: V1 = Il C c
z
(a) Equilibrium at a joint
1
~T2T CC = (O.85f'clb~1c compression block
Grouted post-tensioning or mild steel 3 1/4" duct for post-tensioning rods or equivalent mild steel bars
--H-.........._ .....H
x4-
r--:--
--4 X
Effective d
-----i o
b
L
lLb 1 ~ b
L------,/'l
Typical elevation of multistory wall
Section X-X (b) Typical cross section of a wall
Figure 13.25
VI
Net V 2
Equilibrium forces and stresses at base of structural wall
= 1000 (8.0 x 17.62 x 1200) = 169.2 kips ~ controls ( < 285.9 kips). 1
=
Actual Vn - VI
=
209.8 - 169.2
=
40.6 kips.
Hence, required horizontal force contribution for designing the connection at the wall base is 40.6 kips. If a welded connection is used with the given rated shear capacity of 25 kips, two welded connections have to be used per wall, which is the minimum per panel (see Figure 13.21). If Dywidag connector and grouted post-tensioning is used throughout the wall height, use Dywidag type A 722, Grade 150 ksi connectors with minimum yield strength of 120 ksi (Jpu = 150 ksi and effective pull after seating losses = 0.75 in. minimum). Grout the horizontal joint as in Figure 13.20. The vertical flexural reinforcement in the precast 8-in. wall panel elements is a typical 6 x 6 - W5 x W5 welded wire fabric reinforcement for such standard flexural wall design.
890
Chapter 13
Seismic Design of Prestressed Concrete Structures
It should be noted that the design engineer has to consult with the local pre casters for the appropriate connection configuration and its rated capacity.
13.14 DESIGN OF PRECAST PRESTRESSED DUCTILE FRAME CONNECTION IN A HIGH-RISE BUILDING IN HIGH-SEISMICITY ZONE USING DYWIDAG DUCTILE CONNECTION ASSEMBLY (DOC) Example 13.4 Design a typical ductile precast prestressed concrete moment-resisting frame connection in a thirty-nine-story high rise building in a high-intensity seismic zone with design data from Ref. 3.21. Use the Dywidag ductile connector assembly (DDC) described in Section 13.7.2 and Figure 13.15, namely, 282 kips capacity per single assembly. The frame analysis output for this connection gave a factored moment Mu =1,150 ft-kip (1559 kN-m) and a post-yield rotation Sp = 3.0 percent. The floors are post-tensioned slabs with 200 psi stress limit in the slab concrete at service. The 5 1/2 in. thick slab panels were 18 ft x 27 ft on the average, prestressed both ways. Given: Precast beam span (L): 18 feet (5.49 cm) 15 feet (4.57 m) Clear Span (Lcl): Story Height: 9ft 8 in. (2.95 m) Column Size: 36 in. X 36 in. (914 mm X 914 mm) 30 in. X 36 in. (762 mm X 914 mm) : [b X h] Beam Size: Center to Center of Ductile Rods: 2.33 ft (710 mm) : [d-d'] Objective Strength 1150 ft-kips (1559 kN-m) : [Mu] Objective Post-Yield Rotation: 3% (Sp) Load intensity on precast beam: = 5000 psi (34.5 MPa), normal weight Slab WD = 0.70 K/ft fy = fyt = 60,000 psi (414 MPa) fpu = 270,000 psi (1861 MPa) WL = 0.16 K/ft fps ~ 0.90 fru fpe = 162,000 psi (1117 MPa) Es = 29,000,000 psi (200,000 MPa) Eps = 28,000,000 psi (193,000 MPa) Maximum concrete stress,/c' at post-tensioning =1000 psi (6.9 MPa)
f:
Note that '11.0 fy = 1.25 fy due to probable seismic increase in the longitudinal reinforcement strain at the joint well beyond the yield strain, as stipulated in the ACI 318 and IEC 2009 codes (see Sec. 13.4.1). Solution:
1. Determine nominal capacity of a double DDC connection. Ty = 2(282) = 564 kips
Mn = Tid - d') = 564(2.33) = 1314 ft-kips Mu = <\>Mn = 0.9(1314) = 1183 ft-kips
> 1150 ft-kips
As discussed in Sec. 13.6 and 13.7, it is important to note that a capacity-based approach is being used in order to develop the strength required along the seismic load path.
2. Determine the shear imposed on the connector at probable demand of the DDC assembly. Beam WD =
30 X 36 144 (0.150) = 1.13 K/ft
WD = 1.13 k/ft. (beam)
+ 0.70 k/ft. (slab) = 1.83 k/ft.
891
13.14 Design of Precast Prestressed Ductile Frame Connection
Photo 13. 0 Paramount Apartments. San Francisco, CA : the first hybrid precast prestressed concrete moment-resistant 39 floor high-rise frame building in high seismicity zone. 2002_ Ref. 13.2 1 and Examples 13.4 and 13.5. (Photo 1.1 7 lists acknowledgements for bot h photos. courtesy Charles Pankow Builders Ltd.)
The subscriptspr denote the probable seismic force. shear, or mo ment. From Eq. 4.3 I(e). controlling U = 1.2 D + I.OE + I.OL Lei = clear span Vbp, = ( =
MA
+ MB) + {1.2( 1.83) + 1.0(0. 16)] T L"
Lei
1.25(2)(1314) 15
+ 17.67
= 237 ki ps
where M", and M s are the probable flexural strengths that can be developed in the beams at the column face ( 1.25 Mit). Dead and live loads are factored . Required frictio n factor between transfer block and face of ductile rod.
f.
=
Vb 237 l.25T, = (1.25)564 ... 0.34
Note that class-A slip critical connectors develop a friction factor oCO.33 (A ISC Specifications) bUI this includes a safety factor of about 40 percent (Ref. 13.16). Observe also that the available friction increases in direct proportion to the te nsile load developed in the ductile rod.
892
Chapter 13
Seismic Design of Prestressed Concrete Structures
3. Check the induced bearing pressure (p) under a ductile rod at the probable strength of the ductile rod Shear/boit
P
V bpr
=4
237
.
= 4 = 59.3 kIPS
59.3 X 2.95
= 2.95
.
= 6.81 kSI
Note that the shear transfer mechanism is assumed conservatively to flow through the compression node (Figure 13.26). The bearing area under the ductile rod is confined on all sides. On the open face where it meets the beam, an oversized washer is provided to accomplish this objective. Allowable nominal bearing pressure
Pallow =
1.7 t;
= 1.7(5) = 8.5 ksi Pallow
= 0.7(8.5) = 6.0 ksi
4. Design of shear reinforcement for the beam. Vbpr
=
237 kips V bpr
vu
(see step 2) 237
.
= bd = 30(33) = 0.24 kSI
It should be noted that since inelastic behavior will not occur in the beam, the ability of the concrete to carry shear is not diminished (Ref. 13.17). Hence, Vc = 2Vf'c = 2Y5000 = 141 psi = 0.141 ksi. V upr
V
s
=--v
=
~:~~
c
- 0.141 = 0.179 ksi
14"
Shear fan
Shear fan -+1++---..
2.95" sq.
------,- 25" min. angle
Concrete fill in blockout ELEVATION
No.4 @ 4V2in. SECTION
Figure 13.26 Shear transfer mechanism in the discontinuous region of a DDC@ frame beam (Ref. 13.21)
893
13.14 Design of Precast Prestressed Ductile Frame Connection
0.4(60)
Avfyt
s=
-----;;t; = 0.179(30)
.
= 4.47 m. c./c.
Provide #4 closed U-stirrups (ties) at 4! in. c./.c. as shown in Fig. 13.26 to cover the shear fans zone. It is suggested that the first two stirrups should be hoop sets and include an inner hoop set to provide lateral support for the flexural bars (see Figure 13.26). The first hoop set should be placed at the edge of the block out. The shear fan describes the shear transfer mechanism in this discontinuous region. Shear transfer in the Discontinuous Region A shear fan describes the shear transfer mechanism in this discontinuous region. Capacity of one #4 tie set:
v. = Avfyt = 0.4(60) = 24 kips . d 237 Number ReqUIre = 24 == 10 Provide five double-#4 closed U-stirrup sets within the shear fan region.
5. Load transfer mechanism within the joint. The bearing plate on the interior end of the ductile rod develops the tensile strength of the rod in bearing. Joint behavior is significantly improved because this load transfer mechanism does not slip, as a conventional bar will as it debonds when subjected to load reversals. Bearing on the end plate: AoR =
1.25C~4) = 176.5 kips
Bearing stress p = 1~~5 = 6.3 ksi == 6.0 ksi, O.K. where the DDS plate area = 28 in. 2
v
1\
~
Transfer ties (capacity-T)
,J><
C~
>
""
~
II
m
-
n
~
II
"'-.
"" "
;>
""',
n. ~,
-f1
11
U
v Figure 13.27 13.21 )
~T
11+
tt ~
Strut mechanism ±=~ '",-
1\
Load path within the beam-column jOint of a DOC assembly (Ref.
894
Chapter 13
Seismic Design of Prestressed Concrete Structures
Photo 13.14 Failure of piers at Hanshin Highway Bridge, Kobe Earthquake, Japan . 1995 (Courtesy Professor Mcgurn; Tominaga, Kyoto University. Japan)
Lo(u! Irtllls/er with;" a beam-column joint It is accomplished through the activation of a strut mechanism and II truss mechanism ( Ref. 13.17, 13.22). As seen from Figure 13.27: The strut mechanism is activated by bearing o n the column face in the case of the compression side at the peak load. On the tension side the load is delivered 10 the truss mechanism. Since the stirrer load path is. at least initially. through the strut mechanism. it is advisable to provide a sufficient number of lies in the vicinity of the ductile rod to transfer the developed rod force to the node region. T he suggested tie force (7:') is:
TT = 4 (;\.,R) = 4 ( 176.4) "' 706 kips
where ;\" :: 1.25 for probable increase of25 %. A T= 7: = IUin?
Use seven triple-#5 tie sets 7(3 x 0.62) = 13.02 in. 2 > I U in. 2, OK- Illree ubove allli jOllr below the ductile rods. Tie sets should be located within a 65° angle of the duct.ile rod bearing area.
Joint Shear Slfess-AC1318 Code basis
VI. Vb", = LJ 2""
V
'pr
V&p,L h
1.25(1314) . 7.5 = 219klps 219(18) 9.67
~ --~--- =408ki ps
Vp.
= 2;\"T, -
vjIt
= -
Vp. AI
V,,,,
1002
~ --
36(36)
= 2(t.25X564) -
408 = 1002 kips
= 0773ksi
vJIo ..n- = I~v'fl ::: 15(0.75)V5000 = O.80ksi
>
0.773 ksi,O.K.
6. Column Design Criu rion Design of the column follows standard sit u-cast concrete design procedures using the strong column-weak beam requirement in the ACI Code:
895
13.15 Design of Precast Prestressed Ductile Frame Connection
Alternatively, and consistent with the capacity design approach, the moment demand can be developed directly from probable column shears.
Mnc
hc
2:
2
(Vcp,)
where he is the clear height of the column.
Mnc =
9.~7 (408)
= 1972 ft-kips Once the nominal axial load Pn is entered from the computer analysis output, the column is designed in the usual manner as outlined in Chapter 8.
7. Post-yield deformability The portion of the ductile rod, which was designed to absorb post-yield deformations, is 9 in. long. The elongation required of the ductile rod (~e) is:
(d - d') . 2 = 0.03(14) = 0.42 m.
~e
= 9p
E p
= 0.42 9 = 0047 .
The strain associated with the fracture of the ductile rod is in excess of 50 percent or 0.50, hence O.K.
13.15 DESIGN OF PRECAST PRESTRESSED DUCTILE FRAME CONNECTION IN A HIGH-RISE BUILDING IN HIGH-SEISMICITY ZONE USING A HYBRID CONNECTOR SYSTEM Example 13.5: Design a typical moment-resisting connection for the ductile frame building in Example 13.4 using a hybrid system as described in Section 13.7.1. Use both well-bonded mild reinforcing bars at top and bottom of the frame beams with debonding at the column face and concentric post-tensioning steel reinforcement at their mid-depth. Given: Beam size: 24 in. x 36 in. (610 mm x 914 mm) 5000 psi (34.5 MPa), normal weight fy = 60,000 psi (414 MPa) fpu = 270,000 psi (1861 MPa) fps 5, 0.90 fpu fpe = 162,000 psi (1117 MPa) Es = 29,000,000 psi (200,000 MPa) Eps = 28,000,000 psi (193,000 MPa) Maximum concrete stress'!e, at post-tensioning = 1000 psi (6.9 MPa)
f~ =
Solution: The design procedure in this example differs in that the beam width here does not need to be a function of the hardware as is the case for the DDC in the previous example. It is important to note in this example also that a capacity-based approach is used in order to develop the strength required along the seismic load path.
1. Determine the amount of post-tensioning required. The post-tensioning should be capable of resisting about 60 percent of the moment demand. The balance is to be resisted by the mild steel reinforcement.
896
Chapter 13
Mu
Seismic Design of Prestressed Concrete Structures =
1150 ft-kips
Mu
M = n
1150 = - - = 1278 ft-kips 0.9
Mnps = 0.6(1278) = 767 ft-kips
Assume that the effective lever arm is about 16 in. (h12 - a12) = 1.33 ft. Mnps 767 . Tnps = ( h ) = -1 = 576 kips a .33 2 2 Aps = Tnps = 576 = 3.56 in.2 fpe 162 Use 0.6 in. diameter strands, Aps = 0.213 in. 2 I strand.
. ReqUired number of strands, N
3.56
= 0.213 = 16.7
It should be noted that the selection of the amount of post-tensioning is fairly arbitrary. The objective is to provide a restoring force (see design objectives) and satisfy objective strength requirements. The designer may select a 17-strand tendon or opt to use a 19-strand tendon. The latter will be used in this example since a 19-strand tendon anchorage is available and would be used for either choice. Also observe that posttensioning strands are more cost effective than Grade 60 reinforcing when developed as described in Figure 13.13(b). The consequence associated with the use of proportionately higher amounts of post-tensioning is the loss of some energy dissipation.
Aps
=
19(0.213)
=
4.05 in. 2
Tps = Apsfpe = 4.05(162) = 656 kips _
Tps
656
_
_ 64 . . 3 m.
a - 0.85 f~ b - 0.85(5)24 h Mnps = Tps ( "2
(from data, b = 24 in.)
a) = 656(1812- 3.21) = 813 ft-kIps.
-"2
Check compressive stress on the concrete = Pc lAc P e = Tps = ~ = 0.76ksi Ac Ag 24(36)
Note that the post-tensioning stress level in the concrete should not exceed 1000 psi. This is because stress compatibility with the other components of the structure will become more of a problem and system shortening is likely to be excessive. In the building being designed and used in this example (Ref. 13.21), the floor slab is post-tensioned and the level of effective post-tensioning in the floor will be on the order of 200 psi (or less).
2. Determine the amount of mild steel required to attain the objective strength. Mns = Mn - Mnps = 1278 - 813 = 465 ft.kips T
- Mns ns - d - d'
465 - 186 k'IpS (33 - 3)/12 -
(d' = 3 in.)
_ Tns _ 186 _ . 2 As 60 - 3.1 m.
J; -
Hence, provide four No.8 bars (As = 3.16 in.2/Mns = 474 ft.-kips). Available Tns = 3.16 X 60 = 190 kips
3. Limit state at ultimate strength. The original design guidelines for this construction system (Ref. 13.18) require that the flexural overstrength (probable strength) provided by the mild steel be less than
897
13.15 Design of Precast Prestressed Ductile Frame Connection
that provided by the post-tensioning steel. This provision is intended to produce the objective of a self-restoring bracing system. The mild steel reinforcing bars are de bonded at the beam-column interface as in Figure 13.28. This is because the precast beam will tend to separate or "lift-off" the column, introducing large strains in the mild steel in this region. The extent of the debonding determines the strain/stress induced in the mild steel when the deformation limit state is reached. It should be noted that as the strain limit in the mild steel reinforcing bars is approached, their stress level becomes on the order of approximately 105 ksi. The probable strain in the mild steel would be comfortably less than the strain limit state at the deformation limit state. Determine the mild steel stress at a post-yield rotation of 3% (see stress-strain diagram in Figure 2.18 and Ref. 13.19). The unbonded mild steel length includes a region of probable debonding of 2.75 db on either side of the intentionally debonded region. If the intentionally debonded length is 6 in., the total debonded length becomes:
2.75d b 2.75d b Region where debonding is expected
Intentionally debonded region
, - - - Mild steel
d'
'---'1--------1
:=3
Post-tensioning
Figure 13.28 Reinforcing elongation at post-yield rotation limit state hybrid beam subassembly (Ref. 13.21)
898
Chapter 13
Seismic Design of Prestressed Concrete
Lu + 5.5d b
Structu~es
= 6 + 5.5(1) = 11.5 in.
The strain states in the mild and post-tensioned steel are developed from the post-yield deformation state of the connection in Fig. 13.28 and the stress-strain diagram of the mild steel reinforcement in Figure 2.16 or 2.18b of Chapter 2. The process is an iterative trialand-adjustment procedure since the location of the neutral axis and steel strain states are mutually dependent. A reasonable estimate of the neutral axis is possible. Tps will increase by about 25% to about 200 ksi. Tns will increase by about 67 percent (100/60), which is the increase transmitted to the concrete at the compression side due to the probable seismic load. The compressive force that must be resisted by the concrete is: (a) Mild steel reinforcement For equilibrium, Tns = C; C
= 1.25 Tps + 1.67 Tns - C; = 1.25Tps + 0.67Tns = 1.25(656)
+ 0.67(190)
947
a = 0.85(5)(24) c
=
!.!...- = 9.3 ~I
=
820 + 127 = 947 kips
. 9.3 Ill.
=
11.6 in.
=
0.80
Assume c = 11.6 in. for this trial cycle d su = 6p (d - c)
_ dEs -
= 0.03(33
- 11.5)
= 0.645 in.
(elongation of the mild steel)
d su _ 0.645 _ 6 Lu + 5.5d - 11.5 - 0.05 b
= 0.056 + Ey = 0.058 fsu == 100 ksi
Esu
(Refer to Figure 2.16 or Figure 2.18b)
(b) Prestressing steel reinforcement
d pu = 6p (~ -
c)
=
0.03(18 - 11.6)
=
0.195
(elongation required of the posttensioning tendon)
0.195 18(12)(0.5) = 0.0018
dIps
=
dEpuEps
=
0.0018(28,000)
!psu = Ipse + d!pe = 162 + 50
=
=
50 ksi
212 ksi
<
0.9 !pu (Refer to Figure 2.16 or Figure 2.18b)
= As fsu = 3.16 (60 + 40) = 3.16(100) = 316 kips Tpsu = Aps !psu = 4.05(212) = 859 kips C = Tpsu + Tsu - C~ = 859 + 316 - 190 = 985 kips Tsu
a
=
965' 985 0.85(5)(24) = . Ill.
c
=
9.65 0.8
h . h ence 0 .K .) = 12 in. (wh'IC h'IS CIose to tee = 11 .6 Ill.,
Mpr = Tpsu
(~ - ~) + (Tsu -
= 212(4.05)(18 - 4.83)
< ~)
Tns
d-
+ Tns (d - d')
+ 40(3.16)(33 - 4.83) + 60(3.16)(30)
= 11,308 + 3561 + 5688 = 20,560 in.-kips = 1713 ft-kps
899
13.15 Design of Precast Prestressed Ductile Frame Connection
Msu = 3561
+ 5688 = 9249 in.-kips < Mpsu = 11,308 in.-kips
. Mpr 1713 Overstrength factor achIeved = M = 1278 = 1.34, O.K. n
Verification of = 0.90 value: ddt = 12/33 = 0.363 < 0.375 in Figure 4.45, hence the section is tension controlled, and the value of = 0.90 is verified.
4. Verify that the capacity of the joint is sufficient. Comparing with the ACI 318 Code requirements, the ratio of probable strength (Mpr) to nominal strength (Mn) is 1.33 for the hybrid beam as opposed to the 1.25 stipulated by the ACI for conventional cases. This would probably be the conclusion resulting from the analysis of a conventionally reinforced frame beam at a post-yield rotation demand of 3 percent. Tps - pr Ts - pr
= 1.25(162)(4.05) = 820 kips =
1.25(60)(3.16)
Ts = Asfy = 60
X
=
3.16 = 190 kips
Tps - pr + 0. 25Ts
0.85f~b
a= Mpr
=
Tps - pr
867 0.85(5)(24)
(~ - ~) + Ts(d -
= 820 (18 - 4.25) =
237 kips
5.
= 8.
lll.
d') + 0.25Ts (d -
~)
+ 190 (33 - 3) + 47 (33 - 4.25)
18,326 in.-kips
Mpr 18,326 . Vb - pr = L/2 = 9(12) = 170 kips Vb-prL 170(18) . V c - pr = - h - = ~ = 316 kips Vi
+ 2Ts- pr - V c - pr = 820 + 2(237) - 316 = 978 kips
=
Tps - pr
Vi 978 . vi = Ai = 36(36) = 0.75 kSI
Vi-allow = 15~ = 0.75(15)Y5000 = 795 psi = 0.80 ksi
> 0.75 ksi, O.K.
5. Design the shear reinforcement for the beam. Since post-yield behavior is anticipated in the end of the beam, the procedures used for the design of a monolithically cast "special" concrete frame beam are required. From step 4, Vb - pr = 170 kips V b - pr 170 . Vb-pr = ~ = 24(33) = 0.215 kSI (a) Hinge region:
Disregard in the hinge region the shear strength of the plain concrete, namely, V c = O. VS =
0.215 ksi
Trying two No.4 stirrup sets, Av = 0.20 in. 2/0ne leg of a stirrup. fytAv 0.75(60)(4)(0.20) . s=~= 0.215(24) = 6.98 lll.
where s is the maximum spacing of stirrups and Av is the stirrup area.
900
Chapter 13 Seismic Design of Prestressed Concrete Structures
- .J
r-
DebOnded region Concentric LPost-tensioningll
Mild steel grouted in tubes
(
J
J
)
r I
n
24")
Mild steel 4-#8
@
t-2"
clr.
4-'8~~!, I~ ~ #4
1-
1+-1+-+1-+-1 -
:3
I
..t '- '-
-
COLUMN SECTION
BEAM SECTION Figure 13.29 ample 13.5
28"
~C:3
L
19-6/10"0 strands
7 in. c/c
n
Ir
Schematic of the hybrid ductile beam-column connection in Ex-
Provide two No.4 closed V-stirrups (hoops) at 7 in. center-to-center in the hinge region. (b) Outside the hinge region: Assume double No.4 V-stirrups, Av = 0.40 in. 2 per stirrup. Vc =
2v1"; = 141 psi = 0.141 ksi Vb-pr
Vs
= --
Vc
0.215 . = --5 - 0.141 = 0.146 kSl 0.7
fyAv 60(2 X 0.40) . s = vsb = 0.146(24) = 13.69 m. c./c.
Provide double No.4 V-stirrups at 14 in. center-to-center outside the hinge region which do not necessarily have to be hooped. Figure 13.29 schematically shows details of the ductile beam-column connection designed in this example. The first three sets of stirrups at 12 in. c. toc.
SELECTED REFERENCES 13.1 ACI Committee 318, Building Code Requirements for Structural Concrete (AC! 318-08) and Commentary (AC! 318R-08). American Concrete Institute, Farmington Hills, MI, 2008, pp. 465.
Selected References
901
13.2 International Code Council, International Building Codes 2009 (IBC), Joint UBC, BOCA, SBCCI, Whittier, CA, 2009, 667 p. 13.3 International Conference of Building Officials, Uniform Building Code (UBD), Vol. 2, ICBO, Whittier, CA, 1997. 13.4 ASCE Standard 7-05, Minimum Design Loads for Buildings and Other Structures, American Society of Civil Engineers, Reston, VA, 2005, pp. 110-125. 13.5 Norris, H. c., Hansen, R. J., Holley, M. J., Biggs, J. M., Namyet, S., and Minami, K., Structural Design for Dynamic Loads, McGraw-Hill, New York, 1959. 13.6 Englekirk, R. E., and Hart, G. c., Earthquake Design of Concrete Masonry Buildings, Vols. I & II, Prentice Hall, Upper Saddle River, N.J., 1982. 13.7 Schneider, R. R., and Dickey, W. L., Reinforced Masonry Design, Prentice Hall, Upper Saddle River, N.J.,1994. 13.8 Clough R. W., "Dynamic Effects of Earthquakes," Proc. ASCE, Vol. 86, ST4, New York, April 1960, pp. 49-65. 13.9 Ghosh, S. K., "Special Provisions for Seismic Design," PCA Publication, Notes on AC/3l8-89 Code, Chapter 31, Portland Cement Association, Skokie, IL, 1990, pp. 31-1 to 31-81. 13.10 Derecho, A. T., Fintel, M., and Ghosh, S. K., "Earthquake-resistant Structures," Ch. 12 in Handbook of Concrete Engineering, 2nd ed., Van Nostrand Reinhold, New York, 1985, pp. 411-513. 13.11 Borg, S. E., Earthquake Engineering-Damage Assessment and Structural Design, John Wiley & Sons, New York, 1983. 13.12 Wakabayashi, M., Design of Earthquake-resistant Buildings, McGraw-Hill, New York, 1986. 13.13 Naja, W. M., and Bane, C. T., "Seismic Resisting Construction" Chapter 26, in E. G. Nawy, editorin-chief, Concrete Construction Engineering Handbook, 2nd Ed. CRC Press, Boca Raton, FL, 2008, 1560 p. (CH 32, pp. 1-70). 13.14 Nawy, E. G., Reinforced Concrete-A Fundamental Approach, 6th Ed., Prentice Hall, Upper Saddle River, N.J., 2009, pp. 936. 13.15 Federal Emergency Management Agency (FEMA), "NEHRP Recommended Provisions for Seismic Regulations for New buildings and Other Structures," FEMA 302, Part I & II, Building Seismic Safety Council, Washington, D.C., 1998. 13.16 Englekirk, R. E., Steel Structures-Controlling Behavior through Design, John Wiley & Sons, New York, 1994. 13.17 Pauley, T., and Priestely, M. J. N., Seismic Design of Reinforced Concrete and Masonry Walls, John Wiley & Sons, New York, 1992,744 pp. 13.18 Cheok, G. S., Stone, W. C. and Nakaki, S. D., "Simplified Design Procedure for Hybrid Precast Concrete Connection," NISTIR Report No. 5765, National Institute of Standards and Technology, Gaithersburg, MD, February 1996, 82 pp. 13.19 Englekirk, R. E., "An Innovative Design Solution for Precast Prestressed Concrete Buildings in High Seismic Zones," Vol. 41 No.4, PCl Journal, Precast/Prestressed Concrete Institute, Chicago, IL, July/August 1996, pp. 44-53. 13.20 Ghosh, S. K., Nakaki, S. W., and Krishnan, K., "Precast Structures in Regions of High Seismicity: 1997 UBC Design Provisions," Vol. 42, No.6, PCI Journal, Precast/Prestressed Concrete Institute, Chicago, IL, November-December 1997, pp. 76-93. 13.21 Englekirk, R. E., Design of Paramount Apartments: Thirty-Nine Story Ductile Frame Precast Prestressed Concrete Building in San Francisco. Private Communication, 1999. 13.22 Englekirk, R. E., and Llovet, D., "Cyclic Test of Cast-in-Place High Strength Beam-Column Joints," Concrete International, American Concrete Institute, Farmington Hills, MI, 1999. 13.23 Choek, G. S., and Lew, H. S., "Model Precast Beam-to-Column Connections Subject to Cyclic Loading," PCI Journal, Vol. 38, No 4, July-August 1993, Chicago, IL: Precast/Prestressed Concrete Institute, Chicago, IL, July/August, pp. 80-100. 13.24 Choek, G. S., and Stone, W. c., and Kunnath, S. K., "Seismic Response of Precast Prestressed Concrete Frames with Hybrid Connections," ACI Structural Journal, American Concrete Institute, Farmington Hills, MI, September/October 1998, pp. 527-546.
902
Chapter 13
Seismic Design of Prestressed Concrete Structures
13.25 Priestly, M. J. N. and Marcie, G. A, "Seismic Tests of Precast Beam-to-Column Joint Subassemblages with Unbonded Tendons," PCI Journal, Vol. 41 No.1, Precast/Prestressed Concrete Institute, Chicago, IL, January-February, 1996, pp. 64-81. 13.26 Stone, W. c., Choek, G. S., and Stanton, J. F., "Performance of Hybrid Moment-Resisting Precast Beam-Column Concrete Connections Subjected to Cyclic Loading," Title 92-S22, ACI Structural Journal, American Concrete Institute, Farmington Hills, MI, March-April, 1995, pp. 229-249. 13.27 Stanton, J., Stone, W. c., and Choek, G. S., "A Hybrid Reinforced Precast Frame for seismic Regions," PCI Journal, PrecastlPrestressed Concrete Institute, Chicago, IL, March-April, 1997, pp. 20-32. 13.28 Aswad, G. S., Djazmati and Aswad, A, "Comparison of Shear Wall Deformations and forces Using Two Approaches," PCI Journal, Vol. 44, No.1, Precast/Prestressed Concrete Institute, Chicago, IL, January-February, 1999, pp. 34-46. 13.29 Aswad, A, Private Communication and "Analysis of Reinforced or Post-Tensioned Shear Walls," SHEARWAL Computer Program, Version 2.1, Jacques and Aswad Inc., Denver, CO, 1990. 13.30 Cleland, N. M., "Design for Lateral Resistance with Precast Concrete Shear Walls," Col. 42, No.5, PC! Journal, PrecastlPrestressed Concrete Institute, Chicago, IL, September-October 1997, pp. 44-63. 13.31 Pessiki, S. et al. "Seismic Analysis, Behavior and Design of Unbonded Post-Tensioned Precast Concrete Frames." Report No. EO-97-02, Dept. of Civil and Environmental Engineering, Lehigh University, Bethlehem, PA, November 1997, pp. 1-315. 13.32 Pessiki, S. et al. "Analytical Modeling and Lateral Load Behavior of Unbonded Post-Tensioned Precast concrete walls," Reports No. EO-96-02, Dept. of Civil and Environmental Engineering, Lehigh University, Bethlehem, PA, November 1996, pp.1-192. 13.33 Pessiki, S. et al. "Seismic Design and Response evaluation of Unbonded Post-Tensioned Precast Concrete Walls" Report No. EO-97-01, Dept. of Civil and Environmental Engineering, Lehigh University, Bethlehem, PA, November 1997, pp. 185. 13.34 Englekirk, R. E., "Design-Construction of the Paramount -A 30 Story Precast Prestressed Concrete Apartment Building," PCI Journal, Precast/Prestressed Concrete Institute, Chicago, IL, July-August 2002, pp. 56-69 13.35 Nawy, E. G., "Discussion - The Paramount Building," PCI Journal, Precast/Prestressed Concrete Institute, Chicago, IL, November-December 2002, p. 116 13.36 ACI Committee 340, ACI Design Handbook-Design of Structural Reinforced Concrete Elements with the Strength Design Method, ACI SP-17, American Concrete Institute, Farmington Hills, MI, 1997, pp. 482. 13.37 Englekirk, R. E., "Seismic Design of Reinforced and Precast Concrete Buildings," Pub I. John Wiley and Sons, Inc., 2003, pp. 825.
PROBLEMS FOR SOLUTION 13.1 A 3 x 18 panel ductile, moment-resistant frame category-II site-class B frame building has a ground story 15 ft. high (4.6 m) and ten upper stories of equal height of 11 '-6" (3.5 m). Compute the seismic base shear V and the overturning moment at each story level in terms of the weight Ws of each floor. Use the equivalent lateral force method in the solution. Given:
Sl
= 0.34 sec, Ss = 0.90 sec R = 5,
Ws per floor = 2400 kips (9560 kN) 13.2 A moment-resisting ductile frame building is located in a high-seismic-intensity zone. The earthquake forces are resisted equally as a dual system by the ductile frame and a monolithic reinforced concrete shear wall over the total height of the building. The geometry of the structure is given below. Design the shear wall assuming that the magnitude of the loads, forces and moments applied to the wall are 110 percent of the values used in Ex. 13.2. Given:
floors have slabs of thickness hf = 7 in. (178 mm) clear beam spans in both longitudinal and transverse directions = 20'-0" (6.1 m) shear wall base length lw = 25 ft (39.6 m)
903
Problems for Solution
shear wall height hw = 130 ft (39.6 m) f~ =
5000 psi, normal weight (34.5 MPa)
fyv = fyh
=
60,000 psi (414 MPa)v
Sketch the wall reinforcement. 13.3 A precast shear wall in a moderate seismicity zone for a six story frame building has a wall thickness of 10 in. (254 mm). The height of each story is 11'-6" (3.5 m) and the wall segments extend the height of the building and prestressed vertically. The wall is subjected to a factored seismic base shear Vu = 210 kips (1048 kN) and a factored overturning moment Me = 8000 ft-kip (12,150 kN-m) for Case II loading as controlling in this case (gravity load dominant). The total weight of each floor including any attached masses is 2800 kips (12,454 kN). Design the connection at the base of the wall assuming that the neutral axis obtained by strain-compatibility analysis is c = 21.5 in. (546 mm). Given: sliding coefficient f.L = 0.60 beams size: 24 in. x 28 in. (610 mm x 711 mm) effective beam spans: 22 ft 6 in. (6.9 m) allowable horiz. shear stress fev = 1200 psi (8.3 MPa) f~ =
5000 psi, normal weight (34.5 MPa)
fy
fyt
=
=
60,000 psi (414 MPa)
Use a Dywidag connector system. 13.4 Design a ductile precast prestressed concrete moment-resistant connection of a ductile frame in a high intensity seismic zone subjected to a factored seismic moment Mu = 1350 ft-kip (1831 kN-m) and a post-yield rotation 6p = 2.75 percent. The frame precast beams have spans of 20'-0" and the clear spans are 17'-4" (5.3 m). Each story height is 9'-0" (2.74 m). Use the Dywidag Ductile Connection assembly (DDC) in your solution. Given: column sizes: 38 in. x 38 in. (965 mm X 965 mm) center to center Ductile Rods, (d - d'): 27 in. (686 mm)
n=
5000 psi, normal weight concrete (34.5 MPa)
13.5 Design the moment resisting connection in Problem 13.4 as a hybrid connection using both mild
steel and prestressing post-tensioned reinforcement. Given: f~ =
5000 psi, normal weight concrete (34.5 MPa)
fy = fyt = 60,000 psi (414 MPa) fpu = 270,000 psi (1862 MPa) fps = < 0.90 fpu (determine from compatibility analysis) fpe = 160,000 psi (1103 MPa) Es = 29,000 ksi (200,000 MPa) Eps = 28,000 ksi (193,000 MPa)
fe = 1000 psi (6.9 MPa) maximum concrete stress at post-tensioning
UNIT CONVERSIONS, DESIGN INFORMATION, PROPERTIES OF REINFORCEMENT Table A- 1 Conversion to International System of Units (SI)
To convert 'rom
10
Multiply by
L~ngth
inch (in.)
millimeter (mm) meier (m)
inch (in.) fool (ft )
meier (m) meter (m)
yard (yd)
A".
square meier (sq m)
square fool (sq fI) square inch (sq in .) square inch (sq in.) square yard (sq yd) Volume
square millimeter (sq mm)
cubic inch (cu in.) cubic fool (cu fl) cubic yard (cu yd) gallon (gal) Can. liquidgallon (gal) C'l.n. liquid-
25.4 0.0254 0.3048
0.9144 0.09290 645.2
square meter (sq m) square meier (sq m)
OJXX)6452
cubic meter (cu m)
O.CKXXl1639
cubic meier (cu m)
0.02832 0.7646
cubic meier (cu m) liter cubic meIer (cu m)
0.8361
4.546
0.004546
"Rcficclions"-High slrength polymer concrete sculpture at Rutgers University. Work by R. H. Karol. the civil engineering class of 1982, and the author. 905
906 Table A-1
Appendix A
Unit Conversions, Design Information, Properties of Reinforceme.nt
1
1
I
Continued To convert from
gallon (gal) u.s. liquid* gallon (gal) U.S. liquid* Force kip kip pound (lb) pound (lb) Pressure or Stress kips/square inch (ksi) pound/square foot (psf) pound/square inch (psi) pound/square inch (psi) pound/square foot (psf) Mass pound (avdp) ton (short, 2000 lb) ton (short, 2000 lb) grain tonne (t) Mass (weight) per Length kip/linear foot (kIf) pound/linear foot (plf) poundllinear foot (plf) Mass per volume (density) pound/cubic foot (pef) pound/cubic yard (pcy) Bending Moment or Torque inch-pound (in.-Ib) foot-pound (ft-Ib) foot-kip (ft-k) Temperature degree Fahrenheit (deg F) degree Fahrenheit (deg F) Energy British thermal unit (Btu) kilowatt-hour (kwh) Power horsepower (hp) (550 ft lb/sec) Velocity mile/hour (mph) mile/hour (mph) Other Section modulus (in. 3) Moment of inertia (in.4) Coefficient of heat transfer (Btu/ft2/hfOF) Modulus of elasticity (psi) Thermal conductivity (Btu-in.lft 2/hfOF) Thermal expansion (in.lin.lDF) Areallength (in.2/ft) *One u.s. gallon equals 0.8321 Canadian gallon. **A pascal equals one newton/square meter.
to
liter cubic meter (cu m) kilogram (kgf) newton (N) kilogram (kgf) newton (N) megapascal(MPa)** kilopascal (kPa)** kilopascal (kPa)** megapascal (MPa)** kilogram/square meter (kgf/sq m) kilogram (kg) kilogram (kg) tonne (t) kilogram (kg) kilogram (kg) kilogram/meter (kg/m) kilogram/meter (kg/m) newton/meter (N/m)
Multiply by
3.785 0.003785 453.6 4448.0 0.4536 4.448 6.895 0.04788 6.895 0.006895 4.882 0.4536 907.2 0.9072 0.00006480 1000 0.001488 1.488 14.593
kilogramlcubic meter (kg/cu m) kilogramlcubic meter (kg/cu m)
16.02 0.5933
newton-meter newton-meter newton-meter
0.1130 1.356 1356
degree Celsius (C) degree Kelvin (K) joule U) joule U) watt (W) kilometer/hour meter/second (mls) mm3 mm4 W/m2/DC MPa Wm/m 2fOC
mm/mm/DC mm 2/m
tc = (tF - 32)/1.8 tK = (tF + 459.7)/1.8
1056 3,600,000 745.7 1.609 .04470 16.387 416.231 5.678 0.006895 0.1442 1.800 2116.80
Table A-2
Recommended Minimum Floor Live Loads*
Uniformly Distributed Loads Occupancy or Use
Apartments (see Residential) Armories and drill rooms Assembly halls and other places of assembly: Fixed seats Movable seats Platforms (assembly) Balcony (exterior) On one- and two-family residences only and not exceeding 100 sq ft Bowling alleys, poolrooms, and similar recreational areas Corridors: First floor Other floors, same as occupancy served except as indicated Dance halls and ballrooms Dining rooms and restaurants Dwellings (see Residential) Fire escapes On multi- or single-family residential buildings only Garages (passenger cars only) For trucks and buses use AASHTO lane loads (1) Grandstands (see Stadium and arena bleachers) Gymnasiums, main floors and balconies Hospitals: Operating rooms, laboratories Private rooms Wards Corridors, above first floor CD
o .....
Uniformly Distributed Loads Live Load (psf)
150
60 100 100 100
60 75
100 100 100 100
40 50
100
60 40 40 80
Occupancy or Use
Hotels (see Residential) Libraries: Reading rooms Stack rooms (books & shelving at 65 pef) but not less than. Corridors, above first floor Manufacturing: Light Heavy Marquees and canopies Office buildings: Offices Lobbies File and computer rooms require heavier loads based upon anticipated occupancy Penal institutions: Cell blocks Corridors Residential: Dwellings (one- and two-family) Uninhabitable attics without storage Uninhabitable attics with storage Habitable attics and sleeping areas All other areas Hotels and multifamily houses: Private rooms and corridors serving them Public rooms and corridors serving them
Live Load (psf)
60 150 80 125 250 75 50 100
40 100
10 20 30
40 40 100
i
Table A-2
Continued Concentrated Loads
Uniformly Distributed Loads Occupancy or Use
Schools: Classrooms Corridors above first floor Sidewalks, vehicular driveways, and yards, subject to trucking (2) Stadiums and arena bleachers (3) Stairs and exitways Storage warehouse: Light Heavy Stores: Retail: First floor Upper floors Wholesale, all floors Walkways and elevated platforms (other than exitways)
Live Load (psf)
40 80
250 100 100 125 250
100 75
125 60
Load (Ib)
Location
300 200
Elevator machine room grating (on area of 4 sq in) Finish light floor plate construction (on area of 1 sq in) Garages Office floors Scuttles, skylight ribs, and accessible ceilings Sidewalks Stair treads (on area of 4 sq in at center of tread)
(4)
2000 200 8000
300
(1) American Association of State Highway and Transportation Officials. (2) AASHTO lane loads should also be considered where appropriate. (3) For detailed recommendations, see Assembly Seating, Tents and Air Supported Structures, ANSIINFPA 102-1978 [Z20.3]. (4) Floors in garages or portions of buildings used for storage of motor vehicles shall be designed for the uniformly distributed live loads shown or the following concentrated loads: (1) for passenger cars accommodating not more than nine passengers, 2000 pounds acting on an area of 20 in. 2 (2) mechanical parking structures without slab or deck, passenger cars only, 1500 pounds per wheel; (3) for trucks or buses, maximum axle load on an area of 20 in. 2
*Source: American National Standard ANSI A58.1-1982. Local building codes take precedence.
5"·6%
tt
rTsngin.
Appendix A Table A-3
909
Unit Conversions, Design Information, Properties of Reinforcement Dead Weights of Floors, Ceilings, Roofs, and Walls
Floorings
Weight (psf)
Normal weight concrete topping, per inch of thickness Sand-lightweight (120 pef) concrete topping, per inch Lightweight (90-100 pef) concrete topping, per inch ~ in. hardwood floor on sleepers clipped to concrete without fill 1! in. terrazzo floor finish directly on slab 1! in. terrazzo floor finish on 1 in. mortar bed 1 in. terrazzo finish on 2 in. concrete bed i in. ceramic or quarry tile on ! in. mortar bed i in. ceramic or quarry tile on 1 in. mortar bed i in. linoleum or asphalt tile directly on concrete i in. linoleum or asphalt tile on 1 in. mortar bed i in. mastic floor Hardwood flooring, ! in. thick Subflooring (soft wood), i in. thick Asphaltic concrete, H in. thick
12 10 8
5 19 30 38 16 22 1 12 9 4
2! 18
Ceilings
! in. gypsum board
2 2!
~
in. gypsum board i in. plaster directly on concrete i in. plaster on metal lath furring Suspended ceilings Acoustical tile Acoustical tile on wood furring strips
5 8 2
1 3
Roofs
Ballasted inverted membrane Five-ply felt and gravel (or slag) Three-ply felt and gravel (or slag) Five-ply felt composition roof, no gravel Three-ply felt composition roof, no gravel Asphalt strip shingles Rigid insulation, per inch Gypsum, per inch of thickness Insulating concrete, per inch Walls
4 in. brick wall 8 in. brick wall 12 in. brick wall 4 in. hollow normal weight concrete block 6 in. hollow normal weight concrete block 8 in. hollow normal weight concrete block 12 in. hollow normal weight concrete block
16 6! 5! 4 3 3 ! 4 3 Unplastered
One side plastered
Both sides plastered
40 80 120 28 36 51 59
45 85 125 33 41 56 64
50 90 130 38 46 61 69
910 Table A-3
Appendix A
Unit Conversions, Design Information, Properties of Reinforcement
Continued
Walls
4 in. hollow lightweight block or tile 6 in. hollow lightweight block or tile 8 in. hollow lightweight block or tile 12 in. hollow lightweight block or tile 4 in. brick 4 in. hollow normal weight block baking 4 in. brick 8 in. hollow normal weight block backing 4 in. brick 12 in. hollow normal weight block backing 4 in. brick 4 in. hollow lightweight block or tile backing 4 in. brick 8 in. hollow lightweight block or tile backing 4 in. brick 12 in. hollow lightweight block or tile backing 4 in. brick, steel or wood studs, i in. gypsum board Windows, glass, frame and sash 4 in. stone Steel or wood studs, lath, i in. plaster Steel or wood studs, ~ in. gypsum board each side Steel or wood studs, 2 layers ~ in. gypsum board each side
Unplastered
One side plastered
Both sides plastered
19 22 33 44 68 91 119 59 73 84 43 8 55 18 6 9
24 27 38 49 73 96 124 64 78 89
29 32 43 54 78 101 129 69 83 94
Table A-4
Area of Bars in a 1-Foot-wide Slab Strip
Cross Section Area of Bar, As (or A;,), in. 2 Bar Size Spacing, in.
#3
#4
#5
#6
#7
#8
#9
#10
#11
4
0.20
7
0.19 0.18 0.17 0.16 0.15 0.14 0.13 0.13 0.12 0.11 0.11 0.10 0.09 0.09 0.08 0.08 0.07
0.93 0.83 0.74 0.68 0.62 0.57 0.53 0.50 0.47 0.44 0.41 0.39 0.37 0.35 0.34 0.32 0.31 0.29 0.27 0.25 0.23 0.22 0.21
1.80 1.60 1.44
6!
0.60 0.53 0.48 0.44 0.40 0.37 0.34 0.32 0.30 0.28 0.27 0.25 0.24 0.23 0.22 0.21 0.20 0.18 0.17 0.16 0.15 0.14 0.13
1.32 1.17
6
0.33 0.29 0.26 0.24 0.22
2.37 2.11 1.90 1.72 1.58 1.46 1.35 1.26
3.00 2.67 2.40 2.18 2.00 1.85 1.71 1.60 1.50 1.41
3.81 3.39 3.05 2.77 2.54 2.34 2.18 2.03 1.91 1.79 1.69 1.60 1.52 1.45
4.68 4.16 3.74 3.40 3.12 2.88 2.67 2.50 2.34 2.20 2.08 1.97 1.87 1.78 1.70 1.63 1.56 1.44 1.34 1.25
4! 5
5!
n 8
8! 9
9! 10
1O! 11
1I! 12 13 14 15 16 17 18
CD ......
......
1.06 0.96 0.88 0.81 0.75 0.70 0.66 0.62 0.59 0.56 0.53 0.50 0.48 0.46 0.44 0.41 0.38 0.35 0.33 0.31 0.29
1.31 1.20 1.11 1.03 0.96 0.90 0.85 0.80 0.76 0.72 0.69 0.65 0.63 0.60 0.55 0.51 0.48 0.45 0.42 0.40
1.19 1.12 1.05 1.00 0.95 0.90 0.86 0.82 0.79 0.73 0.68 0.63 0.59 0.56 0.53
1.33 1.26 1.20
1.14 1.09 1.04 1.00 0.92 0.86 0.80 0.75 0.71 0.67
1.39 1.33 1.27
1.17 1.09 1.02 0.95 0.90 0.85
1.17 1.10 1.04
#14
#18
Spacing, in. 4
4!
6.00
5.40 4.91 4.50 4.15 3.86 3.60 3.38 3.18 3.00 2.84 2.70 2.57 2.45 2.35
2.25 2.08 1.93 1.80 1.69 1.59 1.50
9.60 8.73 8.00 7.38 6.86 6.40 6.00 5.65 5.33 5.05 4.80 4.57 4.36 4.17 4.00 3.69 3.43 3.20 3.00 2.82 2.67
5
5! 6
6! 7
n 8
8! 9
9! 10 10! 11
1I! 12 13 14 15 16 17 18
912
Appendix A Table A-5
Unit Conversions, Design Information, Properties of Reinforcement
Properties and Design Strengths of Prestressing Strand and Wire
Seven-Wire Strand. fpu = 270 ksi Nominal Diameter. in.
3/8
7/16
1/2
9/16
0.600
Area. sq in.
0.085
0.115
0.153
0.192
0.215
Weight. plf
0.29
0.40
0.53
0.65
0.74
0.7 fpu Aps. kips
16.1
21.7
28.9
36.3
40.7
0.75 f.u A••• kips
17.2
23.3
31.0
38.9
43.5 46.5 58.1
0.8 fpu Aps. kips
18.4
24.8
33.0
41.4
fpu Aps. kips
23.0
31.0
41.3
51.8
Seven-Wire Strand. fpu = 250 ksi 1/4
5/16
3/8
7/16
1/2
0.600
Area. sq in.
0.036
0.058
0.080
0.108
0.144
0.215
Weight. plf
Nominal Diameter. in.
0.12
0.20
0.27
0.37
0.49
0.74
0.7 fpu Aps. kips
6.3
10.2
14.0
18.9
25.2
37.6
0.8 fpu Aps. kips
7.2
11.6
16.0
21.6
28.8
43.0
fpu Aps. kips
9.0
14.5
20.0
27.0
36.0
53.8
Three- and Four-Wire Strand. fpu = 250 ksi Nominal Diameter. in.
1/4
5/16
3
3/8
7/16
3
3
4
Area. sq in.
0.036
0.058
0.075
0.106
Weight. plf
0.36
No. of wires
0.13
0.20
0.26
0.7 fpu Aps. kips
6.3
10.2
13.2
18.6
0.8 fpu Aps. kips
7.2
11.6
15.0
21.2
fpu Aps. kips
9.0
14.5
18.8
26.5
Prestressing Wire Diameter. in.
0.105
0.120
0.135
0.148
0.162
0.250
'0.276
Area. sq in.
0.0087
0.0114 0.0143
0.0173
0.0206 0.0246 0.0289 0.0302 0.0491
0.0598
Weight. plf
0.030
0.039
0.049
0.059
0.070
0.177
0.083
0.192
0.098
0.196
0.10
0.17
0.20
Ult. strength. fpu. ksi
279
273
268
263
259
255
250
250
240
235
0.7 fOI ADs. kips
1.70
2.18
2.68
3.18
3.73
4.39
5.05
5.28
8.25
9.84
0.8 fou ADs. kips
1.94
2.49
3.06
3.64
4.26
5.02
5.78
6.04
9.42
11.24
fpu Aps. kips
2.43
3.11
3.83
4.55
5.33
6.27
7.22
7.55
11.78
14.05
Appendix A
913
Unit Conversions, Design Information, Properties of Reinforcement Table A-6
Properties and Design Strengths of Prestressing Bars
Smooth Prestressing Bars. 'ou = 145 ksi*
7/S
1
1 1/S
11/4
1 3/S
Area. sq in.
0.442
0.601
0.785
0.994
1.227
1.485
Weight. plf
1.50
2.04
2.67
3.38
4.17
5.05
0.7 fpu Aps. kips
44.9
61.0
79.7
100.9
124.5
150.7
0.8 fpu Aps. kips
51.3
69.7
91.0
115.3
142.3
172.2
fpu Aps. kips
64.1
87.1
113.8
144.1
177.9
215.3
Nominal Diameter. in.
3/4
Smooth Prestressing Bars. fou = 160 ksi*
3/4
7/S
1
1 lIS
1 1/4
1 3/S
Area. sq in.
0.442
0.601
0.785
0.994
1.227
1.485
Weight. plf
1.50
2.04
2.67
3.38
4.17
5.05
0.7 fpu Aps. kips
49.5
67.3
87.9
111.3
137.4
166.3
O.S fpu Aps. kips
56.6
77.0
100.5
127.2
157.0
190.1
fpu Aps , kips
70.7
96.2
125.6
159.0
196.3
237.6
1 1/4
1 1/4
13/S
Nominal Diameter, in.
Deformed Prestressing Bars Nominal Diameter, in.
5/S
1
1
Area. sq. in.
0.28
0.85
0.85
1.25
1.25
1.58
Weight. plf
0.98
3.01
3.01
4.39
4.39
5.56
157
150
160'
150
160'
150
Ult. strength. fpu. ksi
0.7 fpuAps. kips
30.5
89.3
95.2
131.3
140.0
165.9
08 fpuAps. kips
34.8
102.0
108.8
150.0
160.0
189.6
fpu Aps,kips
43.5
127.5
136.0
187.5
200.0
237.0
Stress·strain characteristics (all prestressing bars): For design purposes, following assumptions are satisfactory: 29,000 ksi Es fy ; 0.95 fpu ;
'Verify availability before specifying
914
Appendix A
Table A-7
Unit Conversions, Design Information, Properties of Reinforcement
Moments in Beams with Fixed Ends Moment at A
Loading
A
1/2
I..
t
(1)~
P
1/2
.~
I
I..
14
Moment at center
Moment at B
B
- PI 8
+ PI 8
- PI 8
~I
P
~
(2)~
p
1/3
'3'~' 1
1/3
f
- P/a 2 (1- a)
-Pla(1-a)2
1/3 -l
~
- 2 PI 9
+ PI 9
- 2 PI 9
~
- 5 PI 16
+ 3 PI 16
- 5 PI 16
(5)~ 1111111111I111111 ~
-WI 12
+WI 24
-WI 12
- WI (1 + 2a- 2a 2 ) 12
+ WI (1 + 2a + 4 a 2 ) -----24
-WI (1 + 2a-2a 2 ) 12
t t tl / P
~.II'
P
1/4
P
1/4
4
(4)
~I
W
W
(6)~~111111111::~ a/
al
r----=-1 W/2
W/2
(7)~ IIII a/
'8,{IIIII"i I"
112
W
r----=---1
~ ~
-WI (3a-2a 2 ) 12
+ W/a 2 6
- W/a (6- 8a +3a 2 ) 12
-W/(3a-2a 2 ) 12
- W/a 2 (4- 3a) 12
--l--!/~-----J
'9'~
- 5 WI 48
"0'~~
-WI 10
W = Total load on beam
+ 3 WI 48
- 5 WI 48
-WI 15
Table A-a
Camber (Deflection) and Rotation Coefficients for Prestress Force and Loads'
Prestress Pattern
(1)
!---/-----I -C.G - -I~
1=t -
(2)
~
p+ MI
~ MI
Mt2 16 EI
+-
+--
Mt2 16 EI
+-
Mt2 8 EI
+2 EI
-28
NP 48 EI
+--
Nt2 16 EI
-16 EI
b(l-b)Nt2 2 EI
Nt - b (l-b) 2 EI
+--
4
r/-,)M
M = Pe
-C.G.--
~/l
M = Pe
r--- I -----I
(4)
e':L_ ~~-' C.<;>.
p
I.
r
112
12
M(
1
3 EI
-6ET
':cik f ;}' I-!!LI
l)M
MI
6 EI
MI
+-
MI
-3EI
MI
r--/--j
4 Pe'
N=-I-
t I 112
112
~
+--
N/2
N
.,
c
~
CD .... en
1
M (.
~'---l
(3)
1
+t
1--/---1
l--c.G·--~e
5
Equivalent loading
iIi M = Pe
End Rotation
Camber
Equivalent Moment or load
t--'--j N=~ bt
WN
NtJ
+
b (3-4b 2 ) NP 24 EI
+
2
....co0'1 Table A-a
Continued
ellP,LC.~.--=/ ~~~
(6'
1---1---1
I.
112
.1.
112
j---/-----+j
P w = 8 Pe'
p
1---1 -----i
I
I e - ' I.. 1/2 .1. '
=8Pe '
p
w
112 ~
=~
P
w 10.5· b) 1 w w 1O.5.b) 1
b
(10)
I~1----+1 I =i e'I __
•.
C.G.
4Pe' -w =(O.5·b)P
I
w
j.ELI
r--/~=--:t ~l'
(11)
I
C.G._.
~
r--
f4
1- - - ,
I
I
P
,
w 10.5 . b) b
w=(O.~P~)/2 w
r-~~ ~ I bl i ~
w
~
I---Ij
rTT'"w, ~U.L fllllw",
i"
bl
t--
1/2
w/
J
24 EI
f4
+
_
3
+384EI
~
7 w/ J 384EI
7 wP
9 w/ J
+ 384 EI
- 384 EI
+(1-b)(1-2b)w[3 24 EI
(l-b) (l-: b)w[3 24
768EI
r5 b 2J wf4 La - 2' (3-2b J48 EI
~ ~ ~-b(2-W ~ 48EI __ 8 4
[5 b 2 48EI --4'(3-2b) 16
[
w/ 3
l
f7 --+b(2-b 8
2~ 48EI wP
I
j......!.!.=-.j
~I' I~
~-
9 w/
w +7:8 EI
+tlllllllw 1.1 1/2 1/2 . • •
8 Pe' .. w w= p . 1/2 . • ••
P
P
_
+ 24 EI
w
!---/----! I ILllllli~. 112
P
c.Gla·/
/+==:~/=--:t -I~' C.G.-- - - . eJ-r-IJ?lJ 1M-!
(9'
wI"
+ 384 FI
.1
(n P je--- 1 - - - 1 ........ C.G._ ell ~ J 112 1/2 (8)
5w/4
+111 f , , 111114
w
r~_~(3_2b2~ ~6 J 4
w/
4
48EI
r2._b(2_b2~~ 4;~31 t~+b(2-b)2J 4;~JI
L8
__~____________~____________-L_____________ Iw'=t-~(O~.5_·lb)~~==I=/2::~~bl~ .~
* The tabulated values apply to the effects of prestressing. By adjusting the directional notation, they may also be used for the effects of loads. For patterns 4-11, superimpose on 1, 2, or 3 for other C.G. locations.
Appendix A Table A-9
Unit Conversions, Design Information, Properties of Reinforcement
917
Presumptive Bearing Capacity (tons/ft 2 )
Type of Soil
Massive crystalline bedrock, such as granite, diorite, gneiss, and trap rock Foliated rocks, such as schist or slate Sedimentary rocks, such as hard shales, sandstones, limestones, and siltstones Gravel and gravel-sand mixtures (GW and GP soils) Densely compacted Medium compacted Loose, not compacted Sands and gravely sands, well graded (SW soil) Densely compacted Medium compacted Loose, not compacted Sands and gravely sands, poorly graded (SP soil)
Bearing Capacity
100
40
15 5 4
3 3~
3
21
Densely compacted Medium compacted Loose, not compacted Silty gravels and gravel-sand-silt mixtures (GM soil)
3
Densely compacted Medium compacted Loose, not compacted Silty sand and silt-sand mixtures (SM soil) Clayey gravels, gravel-sand--clay mixtures, clayey sands, sand-clay mixtures (GC and SC soils) Inorganic silts, and fine sands; silty or clayey fine sands and clayey silts, with slight plasticity; inorganic clays of low to medium plasticity; gravely clays; sandy clays; silty clays; lean clays (ML and CL soils) Inorganic clays of high plasticity, fat clays; micaceous or diatomaceous fine sand or silty soils, elastic silts (CH and MH soils)
2~
2
H 2 2
1 1
Table A-10
co ..... Q)
Standard Wire Reinforcement
W&DSize
Smooth
W31 W30 W28 W26 W24 W22 W20 W18 W16 W14 W12 W11 WlO.5 WlO W9.5 W9 W8.5 W8 W7.5 W7 W6.5 W6 W5.5 W5 W4.5 W4 W3.5 W3 W2.9 W2.5 W2 W1.4
Area (in. 21ft of width for various spacings)
u.S. Customary
Deformed
D31 D30 D28 D26 D24 D22 D20 D18 D16
Dl4 D12 D11
DlO D9 D8 D7 D6 D5 D4
Nominal diameter (in.)
Nominal area (in.2)
Nominal weight (Iblft)
2
3
4
6
8
10
12
0.628 0.618 0.597 0.575 0.553 0.529 0.504 0.478 0.451 0.422 0.390 0.374 0.366 0.356 0.348 0.338 0.329 0.319 0.309 0.298 0.288 0.276 0.264 0.252 0.240 0.225 0.211 0.195 0.192 0.178 0.159 0.135
0.310 0.300 0.280 0.260 0.240 0.220 0.200 0.180 0.160 0.140 0.120 0.110 0.105 0.100 0.095 0.090 0.085 0.080 0.075 0.070 0.065 0.060 0.055 0.050 0.045 0.040 0.035 0.030 0.029 0.D25 0.020 0.014
1.054 1.020 0.952 0.934 0.816 0.748 0.680 0.612 0.544 0.476 0.408 0.374 0.357 0.340 0.323 0.306 0.289 0.272 0.255 0.238 0.221 0.204 0.187 0.170 0.153 0.136 0.119 0.102 0.098 0.085 0.068 0.049
1.86 1.80 1.68 1.56 1.44 1.32 1.20 1.08 0.96 0.84 0.72 0.66 0.63 0.60 0.57 0.54 0.51 0.48 0.45 0.42 0.39 0.36 0.33 0.30 0.27 0.24 0.21 0.18 0.174 0.15 0.12 0.084
1.24 1.20
0.93 0.90 0.84 0.78 0.72 0.66 0.60 0.54 0.48 0.42 0.36 0.33 0.315 0.30 0.285 0.27 0.255 0.24 0.225 0.21 0.195 0.18 0.165 0.15 0.135 0.12 0.105 0.09 0.087 0.075 0.06 0.042
0.62 0.60 0.56 0.52 0.48 0.44 0.40 0.36 0.32 0.28 0.24 0.22 0.21 0.20 0.19 0.18 0.17 0.16 0.15 0.14 0.13 0.12 0.11 0.10 0.09 0.08 0.07 0.06 0.058 0.05 0.04 0.028
0.465 0.45 0.42 0.39 0.36 0.33 0.30 0.27 0.24 0.21 0.18 0.165 0.157 0.15 0.142 0.135 0.127 0.12 0.112 0.105 0.097 0.09 0.082 0.075 0.067 0.06 0.052 0.045 0.043 0.037 0.03 0.021
0.372 0.366 0.336 0.312 0.288 0.264 0.24 0.216 0.192 0.168 0.144 0.132 0.126 0.12 0.114 0.108 0.102 0.096 0.09 0.084 0.078 0.072 0.066 0.06 0.054 0.048 0.042 0.036 0.035 0.03 0.024 0.107
0.31 0.30 0.28 0.26 0.24 0.22 0.20 0.18 0.16 0.14 0.12 0.11 0.105 0.10 0.095 0.09 0.085 0.08 0.075 0.07 0.065 0.06 0.055 0.05 0.045 0.04 0.035 0.03 0.029 0.025 0.02 0.014
Center-to-center spacing (in.)
1.12
1.04 0.96 0.88 0.80 0.72 0.64 0.56 0.48 0.44 0.42 0.40 0.38 0.36 0.34 0.32 0.30 0.28 0.26 0.24 0.22 0.20 0.18 0.16 0.14 0.12 0.116 0.10 0.08 0.056
\~
ft
7.Qa11itY"&i!w'<·M t"witt'.'·"'C f
Appendix A
Unit Conversions, Design Information, Properties of Reinforcement
f.
=
f. =
f tQ - 7.5
«
919
for normal weight concrete
f tQ - 6.4« for sand·lightweight concrete
0.50r------,,-~--,-~---,--~--_r--~--T_--~_,----_,------~
0.451-+---H----\li-----'H---t--t--+--t-I---t----f
.....
-
~
.....
'" \I o-+_ _ _
0.3b I---'l:--~--~i--+_-I---'\_---f\_--\
~
~
0.301----fit
0.251----I---rli-~-;.t-__\'--;:\_-+_-t_+-~r-+----I
... ..........
0<
0.20~---+-----~--~--~~~--~--~--~--~--~------~
0.151----I-----1i-----t-~.-~___"'I.--_T_-->t:-;_~r-++_---I
0.101----+----1------1i----I-+~~"'r-f~~tr_~;_+_--f
0.051----+------1i----I----t-J,...---t--~~~¥c+-I
O~
0.30
____
~~
0.40
____
~
______
0.50
Figure A-1
~
______. .
0.60
~
0.70
____
~
0.80
Effective moment of inertia
____
~
______a
0.90
1.00
Appendix A
920
Unit Conversions, Design Information, Properties of Reinforcement
Section
Geometric section modulus, Zs, in.3
B~J
bh 2 4
Shape factor
1.5
Lbj
X-J(
y
t
141
axis.:
w bt (h - t) + - (h - 2t)2 4 y-yaxis: b2t
kbJ
1.55 (approx)
(h - 2t)w 2 4
2+
III
1.12 (approx)
I t(ave.)
h
W
bt (h - t) +
I
t
w(h - 2t)2 4
1.12 (approx)
b
f J
0
3
h 6
h
al
h:
[1 _ (1 _
th 2 for t
~
1.70
~t) 3]
16
311"
[ ( 112.), J 1- 1-
1 _ (1
1.27 for t
h
~
_ ~ty
h
t
Dr
W...
.
1
h
j
bh 2 4
[1-(1- 2:)(1_
~t)2]
1.12 (approx) for thin walls
Lbj
0 t
.:
:
I
bh 2 12
1
J
Lbj
Figure A-2
Geometric section moduli and shape factors
2
Appendix A
921
Unit Conversions, Design Information, Properties of Reinforcement 270
I
E. = 28,000 ksi
f
250
/
~
270 ksi strand
~
250 ksi strand
V
k~
230
~
:a::l!
..J. m
210
~
190
170
II
150
o
.005
.015
.010
strain
.020
.025
(in./in.)
EpI
These curves can be approximated by the following equations: Epa E;;; 0.008:
fp, = 28,000 ep1 (ksi)
Epa> 0.008:
0.058 _ 0006
<
250 ksi strand: fps
248 -
270 ksi strand: fps
268 - Eps ~'~~;065
Eps
.
0.98 fpu (ksi)
<
0.98 fpu (ksi)
Figure A-3 Typical stress-strain curve, 7-wire stress-relieved and low-relaxation prestressing strand
.030
922
Appendix A
1 = K • ••
Unit Conversions, Design Information, Properties of Reinforcement , 3(1 - 13.)2(13.)(a. - I) K,. = I + (a. - 1)13. + - - - - - - I + 13.(a. - I)
(-!...12 b h') w
3.0
",
L
V
.. V
~ V'
V
~
V
"""'"
V
V t3 h :: / ./ ~ OII!!!I ~ ~ il I""'" ~ 0.41/ ~ V ~ ~ ~ ,...... hf
,..,..
2.5
~ ~ ~ ..,. V ./ ~ ~ ~ -" V 1t>.I~ /' ./ ~ ~ /. ~V / ~ V ~~ V
2.0
)~ V
!:
:.::
~V V ~~ V
~ ~ rr1J
1.5
~
..
V
I
JV II/
-,
~
I-
h
I 1I
~
/
~
V
5
a
20
15
10 b=b/bw
Example: For the T-beam shown, find the moment of inertia I.: a.
h
= b/b = w
13. = hJ/h
143/15
= 9.53
I..
= 8/36 = 0.22
Interpolating between the curves for 13. = 0.2 and 0.3, read K,. = 2.28 bwh' 15(36)' 1 = K = 2.28 - - = 133,000 in.' • •• 12 12
Figure A-4
143in~.1
~ Wii~1
Gross moment of inertia of T sections
I··· f
36in.
Appendix A
Unit Conversions, Design Information, Properties of Reinforcement
923
240 200 160
(')
0
x
·wc. I CIl CIl
~
U5
fps
140 120 100
Mild Steel
fs fy
.... .... ....
-
40 0 0.02
(in.lin.) Strain
Figure A-5 Stress-strain diagram for prestressing steel strands in comparison with mild steel bar reinforcement
924
Appendix A
P
2:
Unit Conversions, Design Information, Properties of Reinforcement
)1'fyt
0.45 (A ~ - 1 ---'-
Ae
= 40,000 psi f~ = 6000 psi
and
2:
f~
0.12-
fyt
fyt
A/Ae
f~
= 4000 psi
= 60,000 psi f~ = 6000 psi
fyt f~
= 8000 psi
f~
= 4000 psi
f~
= 8000 psi
Ps
1.1
1.4 1.5
0.012 0.012 0.014 0.D18 0.023
0.018 0.018 0.020 0.027 0.034
0.024 0.024 0.027 0.036 0.045
0.008 0.008 0.009 0.012 0.015
0.012 0.012 0.014 0.D18 0.023
0.016 0.016 0.018 0.024 0.030
1.6 1.7 1.8 1.9 2.0
0.027 0.032 0.036 0.041 0.045
0.041 0.047 0.054 0.081 0.068
0.054 0.063 0.072 0.027 0.090
0.018 0.021 0.024 0.027 0.030
0.027 0.032 0.036 0.041 0.045
0.036 0.042 0.048 0.054 0.060
2.1 2.2 2.3 2.4 2.5
0.050 0.054 0.058 0.063 0.067
0.074 0.081 0.088 0.094 0.101
0.099 0.108 0.117 0.126 0.135
0.033 0.036 0.039 0.042 0.045
0.050 0.054 0.058 0.063 0.067
0.066 0.072 0.078 0.084 0.090
1.2
1.3
Figure A.6
Volumetric ratio of spiral reinforcement Ps for concrete confinement (Ref. 13.36).
Pc
A sh 2: 2: _
0.3 (Ag - 1)
she
ACh
= 40,000 psi f~ = 6000 psi
and
2:
I'
0.09---'-
fyt
fyt
A/Aeh
f~
= 4000 psi
= 60,000 psi f; = 6000 psi
fyt f;
= 8000 psi
f;
= 4000 psi
f;
= 8000 psi
Ps
1.1
1.4 1.5
0.009 0.009 0.009 0.012 0.015
0.014 0.014 0.014 0.018 0.023
0.D18 0.018 0.018 0.024 0.030
0.006 0.006 0.006 0.008 0.010
0.009 0.009 0.009 0.012 0.D15
0.012 0.012 0.012 0.016 0.020
1.6 1.7 1.8 1.9 2.0
0.018 0.021 0.024 0.027 0.030
0.027 0.032 0.036 0.041 0.045
0.036 0.042 0.048 0.054 0.060
0.012 0.014 0.016 0.018 0.020
0.018 0.021 0.024 0.027 0.030
0.024 0.028 0.032 0.036 0.040
2.1 2.2 2.3 2.4 2.5
0.033 0.036 0.039 0.042 0.045
0.050 0.054 0.058 0.063 0.067
0.066 0.072 0.078 0.084 0.090
0.022 0.024 0.026 0.Q28 0.030
0.033 0.036 0.039 0.042 0.045
0.044 0.048 0.052 0.056 0.060
1.2
1.3
Figure A.7
Area percentage of rectilinear hoop reinforcement Pc for concrete confinement (Ref. 13.36).
Appendix A
Mpr
= Kpr
F= bd 2/12000
Fft-kips
p =A/bd
fy = 60,000 psi
fy = 40,000 psi p
925
Unit Conversions, Design Information, Properties of Reinforcement
f~ =
4000 psi
f~ =
6000 psi
f~ =
8000 psi Kpr
f~ =
4000 psi
f~ =
6000 psi
f~ =
8000 psi
(psi)
0.003 0.004 0.005 0.006 0.007
147 194 241 287 332
148 196 244 291 338
148 197 245 293 341
218 287 354 420 484
220 291 361 430 498
221 293 365 435 505
0.008 0.009 0.010 0.011 0.012
376 420 463 506 547
384 430 475 520 565
388 435 482 528 574
547 608 667 725 781
565 630 695 758 821
574 641 709 775 840
0.013 0.014 0.015 0.016 0.017
588 628 667 706 744
609 652 695 737 779
619 664 709 753 797
835 888 939 988 1036
882 942 1001 1059 1116
905 969 1032 1094 1155
0.018 0.019 0.020 0.021 0.022
781 817 853 888 922
821 862 902 942 981
840 884 926 969 1011
1082 1126 1169 1210
1171 1226 1279 1332 1383
1216 1276 1335 1393 1450
0.023 0.024 0.025 0.026 0.027
956 988 1020 1051 1082
1020 1059 1097 1134 1171
1053 1094 1135 1176 1216
1433 1482 1530 1577 1623
1506 1562 1616 1670 1723
0.028 0.029 0.030 0.031 0.032
1112 1141 1169 1197 1224
1208 1244 1279 1314 1349
1256 1295 1335 1373 1412
1668
1776 1827 1878 1928 1976
0.033 0.034 0.035
1250 1275 1300
1383 1417 1450
1450 1487 1525
Figure A.a
Probable seismic moment strengths for beams (Ref. 13.36).
926
Appendix A Unit Conversions, Design Information, Properties of Reinforcement
CD
I
w~
x x'
•
~rumpet
..
Duct
/
.
C
l
Plate Anchorage SO
Flat Anchorage FA Tendon Size
Flat Anchor
Transltlon Pocket Former
Duct
Combination Plate Anchorage SO
3-0.6" or 4-0,5" 4-0.6" or 5-0,5'
Tendon Size
3-0.6" or 4-0,5" 4-0.6" or 5-0,5"
4-0.6"
5-0.6" or 7-0,5" 6-0.6" or 8-0,5" 7-0.6" or 9-0,5"
0
10 \ 255
13 \ 330
Combln. 0 Plate
4-15/16 \ 125
5-5116 \ 135
5 \ 127
5-7/8 \ 150
6-1/2 \ 165
6-11/16\ 170
E
4 \ 100
4 \ 100
E
5-112\ 140
6-5/16\ 160
9 \ 229
7-1/16 \ 180
8-1/16 \ 205
8-1/2 \ 215
f-
2-1/4 \ 57
2-1/4 \ 57
H
1-5/8 \ 41
1-5/8 \ 41
2 \51
1-9/16 \ 40
1-3/4 \44
1-3/4 \ 44
3-3/4 \ 95
3-3/4 \ 95
Transltion
K
12-1/4 \ 310
--
L
4-1/2 \ 115
8-5/8 \ 220
A
10-314 \ 275
13-3/4 \ 350
A
B
4-1/2 \ 115
4-7/8 \ 124
B
C
5-1/2\140
5-7/8 \ 148
C
101
1125
1 \ 25
102
3175
3 \ 75
Pocket Former
Duct
0
2-9/16 \ 65
2-15/16\75
2·13/16\72
3-318 \ 85
L
11-3/8 \ 290
10-7/16 \ 265
11 \ 280
14 \355
6-112\ 165
6-1/2\165
7-1/16 \ 179
7-1/16 \ 180
7-7/8\200
7-7/8 \ 200
7-5/161185
7-5/161185
9-7/8 \ 251
8-1/41210
9-7/16 \ 240
9·7/16 \ 240
3-15/16 \ 100
3-15/16 \ 100
5-118 \ 130
3-15/16 \ 100
4-5116\110
4-5/161110
10
1-9/16140
1-13/16146
2151
2-1116\52
2-7/16\62
2-7/16 \ 62
L2
8\200
8 \200
4 \ 100
8 \200
8\200
8\200
15-15/16 \ 405 15-15/16 \ 405
All dimensions are nominal and are expressed in inch \ mm. Technical data subject to change.
Figure A-9 Dywidag flat anchorage for prestressing strands (Courtesy Dywidag Systems International).
Appendix A Unit Conversions, Design Information, Properties of Reinforcement
L
927
--=l-_.-
L,
-~.
~,----
l,
«cou lSl lSl lSl
,
In 1\1 1\1
PE-Trumpel Ducl Coupler DuCI
-----'
Wedges
Anchoraga Size Min. Block-out Dis. A
Wedge Plate
Trumpet
7-0.6"or 9-0.5"
9..Q.6" or 12-0.5"
12..Q.6"or 15-0.5"
15-0.6' or 20-0.5"
19-0,6' or 27-0.5'
27-0,6" or 37-0.5"
37..Q.6'or 59-0.5"
7 \ 179
8 \203
9 \229
10 \254
11 \ 279
12 \ 305
13-1/2 \ 343
16 \ 407
12-3/8 \ 314
13-7/16 \ 341
15-314 \ 400
20 \ 508
22-5/8 \ 575
25-3/16 \ 640
27-5/8 \ 702
35 \890
B
5-15/16 \ 150
6-11/16 \ 170
7-112 \ 190
8-5/8 \ 220
9-7/8 \ 250
11 \280
12-3/8 \ 315
14-1/8 \ 360
0
3-9116 \ 90
3-7/8\98
4-7/16 \ 113
5-1/16 \ 128
5-13116 \ 148
6-3/8 \ 162
7-1/2 \ 190
8-1/2 \ 220
H
3-9/16 \ 90
3-15116 \ 100
4-15/16 \ 125
7-1/161180
7-7/81200
8-5/8 \ 220
9-7116 \ 240
12-1/2 \ 320
C
5-1/81130
5-1/8 \ 130
5-1/2 \ 140
6-5/16 \ 160
7-1/16 \ 180
7-7/8 \ 200
9-7/16 \ 240
10-2/3 \ 270
E
2 \ 50
1-9116 \ 40
1-11/16 \ 43
1-11/16143
2\ 50
2-3/16 \ 55
2-15/16175
3-1/2190
L1
8-7/8 \ 225
9-1/2 \ 241
10-13116 \ 275
12-7/8 \ 327
14-3/4 \ 375
16-1/2 \ 419
18-1/8 \ 460
22-1/2 \ 600
Transition Length Anchor Dla.
5-0.6" or 7-0.5"
#4\ 15M
#4115M
#4115M
#5\ 15M
#5115M
#5\ 15M
#6 \ 20M
# 7 \22M
Grade 400MPa
60KSI\ 400MPa
60 KSI \ 400MPa
60 KSII 400MPa
60KSI\ 400 MPa
60 KSI \ 400MPa
60 KSI \ 400MPa
60 KSI\ 400 MPa
60KSI\ 400 MPa
Pitch
1-7/8150
1-7/8 \ 50
1-7/8 \ 50
2-1/4155
1-7/8150
1-7/8 \ 50
2-1/4155
2-3/8160
Rebar Spiral' Size
10\ 255
10-1/21265
10-5/8 \ 270
14 \355
14-3/4 \ 365
15 \380
16-5/8 \ 420
18 \460
7-314 \ 190
9\ 230
9-1/21240
11-1/41285
12-1/2 \ 315
14-1/21365
171430
22 \ 560
ID
2 \ 50
2-318 \ 60
3 \ 75
3-3/8 \ 85
3-3/4 \ 95
4 \ 100
4-1/2 \ 115
5-1/8 \ 130
Duct Coupler L2
8 \200
8 \200
81200
81200
81200
8 \200
81200
12 \300
0.1211.5
0.17\2.1
0.28 \ 3.46
0.35 \ 4.39
0,44 \ 5.48
0.47 \ 5,80
0.58 \ 7.25
0.72 \ 8.90
J
00 Duct
Grout Requirements gal/It \ I/m
Spiral required in "local anchor" zone, III dimensions are nominal and are expressed in inch \ mm. Technical data subject to change.
Figure A-10 Dywidag multiple anchorage for prestressing strands (Courtesy Dywidag Systems International)
SELECTED TYPICAL STANDARD PRECAST DOUBLE TEES, INVERTED TEES, HOLLOW CORE SECTIONS, AND AASHTO BRIDGE SECTIONS
929
930
AppendllC B
StrIInd PMtem DH'gMlIon
A
y, y, 1, 1,
. 20.985 In.' V ,720 In.' 17.15 In. In. In. ' .13 in. In.' in.' in.' 4.119 In.'
OOUBLETEE
,,5_ .... 0. ...~ No.
---T_
Selected Typical Standard Precast Double Tees, Inverted Tees, Hollow Core Sections
um_
oIatrand (10)
8'-0" x 24" Nonnal Welghl Concret.
108-01
t L No. ~"""""'_ DIameter oIlInwld In
8 · ~·
,. r-------~~~c_------
lett.
"
5 314· ... ,..
,.
&M loads 5hoWn h:tIde dead IoItd ol to psi 10' IJfIfopp«J ~ hJ 15 psi lOr 1fJPp«J~. RMMintIM is 1M load.
'" ,... ..,,,
"27
I.' "
... " ,. "
'.063
WI
Long-timB camb8n: /ncAJde ~ dead load bCII 0:» not ohcWQ IIw load.
In.'
I- 3 314-
4 '.()"
'"
n
""
52
VIS ...
in.
1.• '
""
t'" 5,000 psi
173 - sate superimposed aervice load, psi 0.5 - Estimated earnbel' at erection, In. 0.7 - Eslimatecliong-time camber, In.
\. = 270,000 psi
80T24 Table 01 saf. superimposed service load (psf) and cambers St"""
.....
~
....
...
"'", 108-01 ,~,
,...."
" " 11 .15 11.15
•. 15 1.15
" ... e .15
7.15 14.15
. . . . . . . ,. ,. " " " " " ,.,., ,,.. ,,.. ,., ,,.,. ,., ,.", ,.", ,.,.,", ,,"..• ,,"...• ,.,.,, .. •• 'r:..I.' ';'].. . :!.., ,,....• ..,..r.:.• ,,...I.,.,• ,.. ". .. . . ." ,... ,.
32
,.
,.
173 \47 126 108
Sp.n, fl
42
50
54
50
70
72
92
OJ
'~ 0.7 '~ 0-7 '~ 0.8
No Topping
OJ 0.'
OJ
116 J~ 0.8 '~ 0.8 0.1 0.11 1.0
1.0
1.0
"
56
~ ~
57
1.0
'~ '~ ',~
79
'rl
70
1j ' .1
62
32
I.' I.' I.' I.' I.'
U
U
OJ
U
:t.O
I.e
" 'f; u'~~~i~~r, ~~~ ~~~;, ';, '-' '-'
U
2.3
U
2.2
2.1
I.
1.3 1.0 0.5
.." .. ,." ,...
B~f:nrs:~~~~ 7 2.5 2.3 2.1 U loS U 0.1
"5."...
r7
55
• .20 13.65
2.3
I.'
80T24 + 2 2" Normal Weight Topping
Table of sate superimposed service load (pst) and cambers Strand
'.
Plttern
'e 14.15 14.15 11.15
...., ...., ,...., ,~,
11 .15
11 .15 14.65
Spen, fl. 262830323436384042444848505254 183 149 122 100 82 66 53 42 0.4 0.4 0.4 O.S 0.$ O.S 05 O.S 0.4 0.4 0.' 0.' 0.' 0.3 0.3 0.2 175 147 123 103 86 72 o .~ 0.1 0.1 0.7 0.7 0.7 o .~ 0.1 0.1 O.S 0.$ 0.4
8.15 14.40
601264
O.S 0.0
60 49 39 0.7 0.7 0.7 0.3 0.2 0.0
lMl~I~113
0.7 0.1 0.1 0.1
5858
~
~
0..9 0..9 0.1 0.1
81
~
~
~
~
,.0 1.0 1.0 1.0 1.0 1.0 0.7 0.7 0.1 O.S 0.3 0.1
190 11» 143 124 101 s;s au 69 59 51 1.1 U 1.3 U 1.5 1..5 1.8 1.1 1.1 1.1 1.1 1.1 12 1.2 U 1.1 0.9 oa 0 ,1 0.4
43 1.1
'~ 'f.:'~ lX 1.5 1.5 1.4 1.3
if O.S
...
7.15 14.15
ro U
2.~ 1.0
r.
0,7
0.1
~
0.1
74 65
2.5 1.0
,3.10
Str8llg/tl bas«J 011 strain cotTJNItbIlity; botIom ~ Iimi!ed to 12,1f;:
ShadBd v&be:s require relN54I stnJngths higher than 3500 p.sL
Figure B-1
8"--Q" x 24 Double Tee (Courtesy PCI, Ref. 4.9) W
2.5 0.7
57 49 2.5 25 0.3 0.1
Appendix B
Selected Typical Standard Precast Double Tees, Inverted Tees, Hollow Core Sections
Str.od P.nem OHlgllltlon
tL
""""w.... '-""'-' I
No. 01 depression points
DiMleter at strlnd In 16lhs
y, y,
I
Beca.Ise IhBM tnfs . . prtIIopp9d and .,. typ/clJIIy II5«J In pMdng ,trvctvr6S, ,ate loads shown do nollncJrJde any S~(· /mpos«I dsad obads. Loads shown at8 IIw load. LOIIfI· ,inNI cambers do ""' Include
1"
\J
......
~ WI
I
K"
.'
A
10'·0" x 26"
108-01
... . ... .
Section Properties
PRETOPPED DOUBLE TEE
No. 01___ SlrancI(10) ,r-== , 5. 0.....-
931
.'
.. 30.716 In.' 30.716 In.' 20.29 20.29 n 5.71 5.71 n 1,514 .' 1,514 n ' 5>" .' 5.379 .' pi! 550 pi!
..•• . .. .. ••
"8
VIS •
n ' .05
55
••
' .05
••
~ . 5,000 psl
198 - Safe superimposed service load. psI 0.4 - Estimated camber .terectlon.ln. 0.5 - Estimated IonQ-llme camber. In.
f,. •
270,000 ps;
10DT26 Tabfe of sate superimposed service load (pst) and cambers
...... .... Strand
8&S
-,-, 19'"
14.2. 14.2.
12.211 12.211
88-., "... 12.211
I08-Dl
,...., 148-Dl
-
.
30
32
,.
161 133 109 90 0.' 0.' 0.' 0.' 0.' 0.' 0.' 0.' 0.' 0.' IS9 142 119 05 0.' 0.' 0.1 0.' ~ 1 170 0.7 0.1 1.0 1.1
,.
. . . ,.. .
" 74
0.' 0.' 100 0.' 0.' 146 O.i \.2
38
42
0.' 0.'
49 39 30 0.' 0.' 0.' 0.' 0.' 0.'
50
"
54
58
"
39 0.'
~
. ..
SO
82
. . ..
"
83 O. " O. ,.
57 30 0.' 0.' 05 0.' 0.' 0.' 0.1 0.' 125 107 91 18 0.1 1.0 ..0 ..0 "0 ' .0 0.' 1.3 1.3 U U
0.'
'~ '~ loS 1.11
10.211 ".211
No Topping
Spen, ft.
... ,". ,.", "
U
' .0
rl ~ ~ Ir~ ,.,• "u 1.1 1.8 I II 1.8 1.7 ,. 13C 118 103 79 SOil" 1.' 1.1 1.1 1.1 1.1 1.7 1.11 U ,.
.. . .." .... ...,. ,..... .. " ,".• " " " ,.•
':i 'r.i '?~ 1.7 1.7 1.8
." 17.04
r.:
U
1.43
U
20 2.8
16.79
711
U
2.8
32
u
20 87 50 2.7
1.1
'.0
2,5
2A
10LDT26
Tabte of safe superimposed service load (pst) and cambers Strand
e.
-,
14.211
14.211 12.211 12.211
88-01
,.01 1~1
148-01
...•••
38 33
' .1 2>
"...
12.211
28
40
42
44
46
175 146 123 104 88 74 53
53 0.1
«
36
0.6 0.1
30 32 34 36 38 0.7 0.7 0.11 0.11 0.1 1.0 1.0 1.\ 1.2 1.2 183 158 133 113 97 0.1 0.1 1.0 1.0 1.\ 1. \ 1..2 1.3 U 1.5
0.1 1.3 1.3 83 71 1.1 12 1.1 1.1 1~ 1~ 1~! rl20 105 1.3 1.4 1.15 1.1 1.1 U U 2.1 2.2 2.3
0.9 1.2 61 12 1.6
92
1.8 2.4
No Topping
Spenoft. 50 52
48
54
56
5& 60 6264
H
68
70
41 3.1 3.11 52 4.0 5.\
3.0 S.I 47 4.0 5.0
36
31 2.8 3.4 41 3.6 4.7
72
30 O.i 0.11 1.2 1.1 52 44 37 31 1.2 1.2 U 1.0 1.11 1.11 1.5 1.4 80 70 61 53 46 39 34 !.II LSI 1.1 1.1 1.l1 1.8 1.7 2.4 2.5 2.5 2.5 2.5 2.4 2.3
10.211 " .211
....
lQ.1 13 83 )..0
." 17.04
3.1
4.0 4.1
7,03
74 68 58 32 3.3 3.3 34 42 4.3 4.3 4.3 73 40 5.3
52 3.S 4.2 85 40 5.3
Strength baSlldon s/rllltl compatN/y: bonom IfHlSion 6mIred to 12,p;; ShadId vaIue.s require ffIINH SlT8fIgIhs hi(}her than 3500 psi.
Figure B-2
10'-0" x
26~
Double Tee (Courtesy PCI, Ref. 4.9)
46 32 4.0 58 4.0 52
37 3.6 4.3
932
Appendix B Selected Typical Standard Precast Double Tees, Inverted Tees, Hollow Core Sections
PRETOPPED DOUBLE TEE
Strand Pattern Designation r-::==NO. of strand (10) • . - S = straight 0 = depressed
,I ,
Section Properties Normal Weight
12'-0" X 34"
108-01
L- No. of depression points Diameter of strand in 16ths
Because these units are pretopped and are typically used in parking structures, safe loads shown do not include any superimposed dead loads. Loads shown are live load. Long-time cambers do not include live load.
I:
12'-0" 3'-0"·1-
-il~:"'lh
6'-0"
l]\ ,.
CHAMFER
f~
176 - Safe superimposed service load, psf 0.8 -Estimated camber at erection, in. 1.1-Estimated long-time camber, in.
y, Sb S.
li +
wt
T
-l L
Key
A I Yb
VIS
978 = 86,072 25.77 8.23 3,340 = 10,458 1,019 85 2.39
Ughtweight
in2 978 in2 In' 86,072 in' in. 25.77 in. in. 8.23 In. in" 3,340 in' in" 10,458 in" plf 781 plf psf 65 psf in. 2.39 in.
4';'''
= 5,000 psi 270,000 psi
fpu =
12DT34 Table of safe superimposed service load (pst) and cambers (in.) Strand Pattern
Span,ft
e., in. ecjln.
128-01
14.10 22.52
148-01
12.91 22.27
168-01
12.77 22.02
188-01
11.38 21.77
208-01
10.27 21.52
228-01
9.36 21.27
No Topping
42
46
48
58 59
60 51
62 43
64
79
56 69
36
66 30
0.9 1.2
0.8 1.2
0.8 1.1
0.7 1.0
0.7 0.9
0.6 0.7
0.5 0.6
50 176 155 135 119 104
52
54
91
0.9 1.2
0.9 1.2
0.8 1.1
44
0.8 1.1
0.9 1.2
0.9 1.2
65 146 129 114 101 1.0 1.0 1.1 1.1 1.1 1.1 18:r 1.4
1.4
1.4
1.5
1.5
196 174 155 1.2 1.7
1.3 1.7
1.5
89
78
68
1.1 1.5
1.1 1.5
1.1 1.5
23 110
1.3 13:r 1.4 1.4 1.8 1.9 1.9
1.4 1.9
97
1.4 1.9
86
1.4 1.9
60
1.0 1.4
2.1
2.1
2.2
2.2
2.2
76
67 1.4 1.8
2.2
131 118 106 1.8 2.5
1.9 2.5
44 0.9 1.1
1.4 1.9
178 160 143 128 115 10;191 1.5 1.5 1.6 1.6 1.7 1.7 1.7 2.0
52 1.0 1.3
1.9 2.5
59
1.3 1.7
82
1.7 2.2
95
32
26 0.4 0.5
45 1.1 1.5
64
57
1.5 2.0
1.5 1.9
98
2.1 2.8
52
1.2 1.6
2.1 2.8
76
1.9 2.4
88
2.1 2.8
72
0.6 0.8
73
1.9 2.5
109
38
0.7 1.0
70
1.6 2.1
85
1.9 2.5
68
68 1.8 2.3
79
2.1 2.7
39
1.0 1.3
50
1.4 1.8
61
1.7 2.2
71
2.1 2.6
74
33
0.8 1.0
43
1.2 1.6
54
1.6 2.1
76
78
80
82
84
86
28
0.6 0.7
38
1.1 1.4
32
27
0.9 1.1
0.6 0.7
41
1.5 4:1 1.3 1.9 1.7
36
1.1 1.4
64
57
50
44
2.0 2.5
1.9 2.4
1.7 2.2
1.6 2.0
31 0.9 1.1
39
1.4 1.7
26
0.6 0.7
34
29
1.1 1.4
0.8 1.0
12LDT34 Table of safe superimposed service load (pst) and cambers (in.) Strand Pattern
e., In. ec,ln.
128-01
14.10 22.52
148-01
12.91 22.27
168-01
12.77 22.02
188-01
11.38 21.77
208-01
10.27 21.52
228-01
9.36 212.7
No Topping
Span,ft 42 44 46 48 50 52 193 171 152 135 120 107 1.3 1.8
1.4 1.9
1.5 2.0
1.5 2.0
1.6 2.1
1.6 2.1
54 56 95. 85
58 76
60 67
62 59
64 52
66 46
1.7 2.2
1.7 2.2
1.7 2.2
1.7 2.2
1.6 2.1
1.6 2.1
1.7 2.2
182 162 1:r3O 117 105 1.7 1.8 1.9 2.0 2.0 2.1 2.3
2.4
2.5
2.6
2.7
2.8
94 2.1 2.8
65
2.1 2.8
191 172 155 139 126 114 2.1 2.8
2.2 3.0
2.3 3.1
2.4 3.2
2.5 3.3
76 2.1 2.8
93
2.6 10:1 2.7 2.7 3.4 3.5 3.6
144 131 119 108 2.7 3.7
2.8 3.8
2.9 3.9
3.0 4.0
68
2.1 2.8
84
2.7 3.5
98
3.1 4.1
61 2.1 2.7
76
2.7 3.5
89
3.2 4.1
102 3.5 4.6
68 40 1.5 2.0
54
2.1 2.7
68
2.7 3.5
81
3.2 4.2
93
3.6 4.7
70 35
72
74
30
26
1.4 1.8 48 2.0 2.6
1.2 1.6
1.0 1.3
61
2.7 3.4
73
3.2 4.1
85
3.6 4.7
42
78
1.8 2.3
33
28
1.6 2.1
1.4 1.8
55
49
44
2.6 3.3
2.5 3.2
1.9 2.4
66
3.2 4.0
77
3.6 4.7
37
76
2.4 3.1 54 3.1 6;1 3.0 3.9 3.7
70
3.6 4.7
64
3.6 4.6
!lU
{'oj
4.1 5.3
4.1 5.2
39
2.3 2.9
48 2.9 3.6
58
3.5 4.4 til 4.1 5.2
Strength based on strain compatibility; bottom tension limited to 121['; see pages 2-2-2-6 for explanation. Shaded values require release strengths higher than 3500 psi.
Figure 8-3
Pretopped 12'-0" x 34" Double Tee (Courtesy PCI, Ref. 4.9)
80
82
34
30
2.1 2.7
1.9 2.4
43
2.8 3.4
52 3.4 4.2
~~ 5.0
39
2.6 32
47
3.3 3.9
84
86
26
1.6 2.1
34
2.4 3.0
42 3.1 3.7
30
2.1 2.7
38
2.9 3.5
82
50
45
3.9 4.8
3.8 4.6
3.6 4.2
Appendix B Selected Typical Standard Precast Double Tees, Inverted Tees, Hollow Core Sections
933
INVERTED TEE BEAMS 6"
I"
1'·6"
~I
6" Normal Weight Concrete
-1---,------.
Section Properties
2'·6"
f'c = 5,000 psi fpu = 270,000 psi % in. diameter low-relaxation strand
1 (In.')
y.
Z.
Z,
wt
(in.)
(In.')
(in.')
(plf)
456 576 648 720 792 912
15,240 26,352 41,824
8.74 10.50 12.22
1,744 2,510 3,423
1,354 1,952 2,650
475 600 675
62.400 88,678 121.923
14.00 15.82 17.47
4.457 5.605 6.979
28116 984 32/16 1.056 36116 1,128
162.161 210.199 266,627
19.27 8.415 21.09 9,967 22.94 11,623
h,/h.
(in.)
(In.)
30lT20 30lT24 30lT28
20 24 28
1218 12112 16112
301T32 30lT36 30lT40
20/12 24112 24/16
301T44 30lT48 30lT52
32 36 40 44 48 52
30lT56 30lT60
56 60
40/16 1,200 332,032 24.80 13,388 44/16 1,272 406.997 26.68 15.255
Designation
.c
A (in.')
h
750 825 950 6.557 1,025 7.811 1.100 9.175 1,175 10,642 1,250 12.215 1,325 3.467 4,394 5.412
1. Check local area for availability of other sizes. Key
8,428 0.4 0.2 -
2. Safe loads shown include 50% dead load and 50"'{' live load. 800 psi top tension has been allowed, therefore additional top reinforcement is required.
Safe superimposed service load, plf Estimated camber at erection, in. Estimated long-time camber, in.
3. Safe loads can be significanffy increased by use of structural composite topping.
Table of safe superimposed service load (plf) and cambers Span, ft. De81gnation
No.
e 18
Strand
30lT20
14
6.65
30lT24
17
7.67
30lT28
20
9.06
30lT32
23
10.50
30lT36
24
12.32
30lT40
30
12.92
30lT44
30
14.73
301T48
33
16.17
30lT52
36
17.62
30lT56
39
19.06
30lT60
42
20.49
20
22
24
26
28
30
8.428 6,736 5.485 4,533 3.792 3,204 2,730 0.4 0.5 0.6 0.7 0.9 1.0 1.1 0.3 0.2 0.2 0.2 0.3 0.3 0.3 9,736 7,942 6.578 5,516 4,673 3.994 0.9 0.4 0.5 0.6 0.7 0.8 0.2 0.2 0.2 0.2 0.3 0.3 9,087 7,643 6,497 5,573 0.6 0.7 0.8 0.6 0.2 0.2 0.3 0.3 8.647 7.436 0.7 0.7 0.3 0.2
34
36
38
40
42
2.3422.020 1.2 1.3 0.3 0.3 3,4372,976 1.1 1.0 0.3 0.3 4.8164,189 1.0 0.9 0.3 0.3 6,445 5.623 0.8 0.9 0.3 0.3
1,751 1.4 0.3 2.592 1.2 0.3 3,664 1.1 0.3 4.935 1.0 0.3
1,523 1.4 0.3 2.269 1.2 0.3 3.219 1.2 0.3 4.352 1.1 0.4
1,332 1.5 0.3 1,993 1.3 0.3 2,839 1.2 0.3 3,855 1.2 0.4
1.167 1.6 0.3 1,755 1.4 0.2 2.513 1.3 0.3 3,426 1.2 0.4
32
44
46
48
50
1.024 1.6 0.2 1.550 1.370 1.212 1,073 1.5 1.4 1.5 1.5 0.2 0.2 0.1 0.0 2,334 1,990 1,776 1,588 1.4 1.4 1.5 1.5 0.3 0.3 0.2 0.3 3,055 2,732 2,448 2.201 1.4 1.3 1.5 1.5 0.4 0.4 0.4 0.3
9.492 8.243 7,207 6.340 5.605 4.978 4,439 3,971 3.563 3,205 0.7 0.7 0.8 0.9 1.0 1.0 1.1 1.2 1.3 1.3 0.3 0.2 0.2 0.3 0.3 0.3 0.3 0.3 0.3 0.3 9,077 7,994 7,077 6,295 5.621 5,037 4.528 4.081 0.8 0.8 0.9 1.0 1.1 1.2 1.2 1.3 0.3 0.3 0.4 0.3 0.4 0.4 0.4 0.4 9,659 8,564 7,629 6,825 6.1275,5194.985 0.7 O.B 0.9 1.0 1.2 1.0 1.1 0.3 0.3 0.3 0.3 0.3 0.3 0.3 9,222 8.262 7,431 6.705 6,068 0.9 1.0 1.0 1.1 0.8 0.3 0.3 0.3 0.3 0.3 9,836 8.858 8.004 7.255 1.0 0.9 0.9 1.1 0.3 0.3 0.3 0.3
2.892 1.4 0.3 3,687 1.4 0.4 4,514 1.2 0.3 5,506 1.2 0.3 6.594 1.1 0.4
9.407 8,538 7,770 1.0 1.0 1.1 0.3 0.4 0.4 9,917 9,036 1.0 1.0 0.4 0.3
Figure 8-4
Inverted Tee Beam Sections (Courtesy PCI. Ref. 4.9)
Appendix B
934
Selected Typical Standard Precast Double Tees, Inverted Tees, Hollow Core Sections.
HOLLOW-CORE SLABS Section Properties - normal weight concrete
Oy·Core
Trade name: DY-Core® Licensing Organization: Dy-Core Systems, Inc., Vancouver, British Columbia
10000001 t·;·;.·~ . . ·.'. . /~
[6060001 0'
'~'.··'4·.· :-':0: .~'.' ',:~': ',~.. :.,',~>:.:.~
With 2" topping
width
x
IO. .O. .O. Oj -f
Untopped
Section
b •. · ....
A In.'
y.
in.
I in.'
wt
depth
y.
psf
in.
4'-0" x 6" 4'-0" x 6" 4'-0" x 10' 4'-0' x 12' 4'-0' x 15"
151 190 216 262 269
3.11 3.95 5.10 6.34 7.34
663 1,566 2,692 4,675 8,701
40 51 56 71 76
4.54 5.54 6.60 6.01 9.36
I wt In.' pst 1,552 65 3,130 76 5,097 63 7,623 96 13,776 103
-:::';'
'.0. 0 . 0 . 0.' Note: All sections not available from all producers. Check availability with local manufacturers.
Section Properties - normal weight concrete
Oynaspan
Trade name: Dynaspan® Equipment Manufacturers: Dynamold Corporation, Salina, Kansas
fO.O.O.O.o.o.oi Untopped
Section
t6:b~(5(5.6.6.cA
With 2" topping
width
x
io.o.o.O.O.O.O.O.O.O.O.O.O.O.~
t'.O.O.O.O.O.O.O.O.O.O.O.o.o.o.i · ..,. ' J ",.·.6., , . . ;.d· b ·, '. , . . . . . . . . . . . . . . . . . .
·1··.
C
y.
depth
A In.'
in.
4'-0"x4" 4'-0" x 6" 4'-0' x 8" 4'-0' x 10" 8'-0"x6" 8'-0' x 6" 8'-0' x 10" 8'-0· x 12"
133 165 233 260 336 470 532 615
2.00 3.02 3.93 4.91 3.05 3.96 4.96 5.95
I
in.' 235 706 1,731 3,145 1,445 3,525 6,422 10,505
wt
y.
psf
in.
35 43 61 68 44 61 69 80
3.08 4.25 5.16 6.26 4.26 5.17 6.28 7.32
Note: All sections not available from all producers. Check availability with local manufacturers.
Figure 8-5
Hollow Core Slab Sections (Courtesy PCI, Ref. 4.9)
I in.'
wt pst
689 60 1,543 68 3,205 86 5,314 93 3.106 69 6,444 86 10,712 94 16,507 105
Appendix B
Selected Typical Standard Precast Double Tees, Inverted Tees, Hollow Core Sections
935
AASHTO I-Beams
n
02
BI
I'
-I
03 04 01
01
I.
B2
I.
Type I-IV
B2
_I
Type V-VI
Dimensions (inches) Type
Dl
D2
D3
D4
D5
D6
Bl
B2
B3
B4
B5
B6
I
28.0
4.0
0.0
3.0
5.0
5.0
12.0
16.0
6.0
3.0
0.0
5.0
II
36.0
6.0
0.0
3.0
6.0
6.0
12.0
18.0
6.0
3.0
0.0
6.0
III
45.0
7.0
0.0
4.5
7.5
7.0
16.0
22.0
7.0
4.5
0.0
7.5
IV
54.0
8.0
0.0
G.O
9.0
8.0
20.0
26.0
8.0
6.0
0.0
9.0
V
63.0
5.0
3.0
4.0
10.0
8.0
42.0
28.0
8.0
4.0
13.0
10.0
VI
72.0
5.0
3.0
4.0
10.0
8.0
42.0
28.0
8.0
4.0
13.0
10.0
Properties Type
Area in.2
in.
Inertia in.4
Weight kip/ft
Ybotlom
Maximum Span,* ft
I
276
12.59
22,750
0.287
48
II
369
15.83
50,980
0.384
70
III
560
20.27
125,390
0.583
100
IV
789
24.73
260,730
0.822
120
V
1,013
31.96
521,180
1.055
145
VI
1,085
36.38
733,320
1.130
167
'Based on simple span, HS-25 loading and f~ =7,000 psi. Figure 8-6 (a)
AASHTO/PCI Standard Bridge Sections (Courtesy PCI, Ref. 12.11)
,
I,
936
Appendix B
Selected Typical Standard Precast Double Tees, Inverted Tees, Hollow Core Sections
AASHTO I-Beams
I
I
36"
28"
lL--.J
L'---------l Type I
Type II
2" (TYP.)
r
I
54"
45"
LL----J Type III
2"(TYP.)
l:::HY Type IV
63"
.......... ........... ........... ..
........... .. ........... Type V
Figure 8--6 (b)
Type VI
Possible Strand Arrangement for Sections in Figure C-6 (a)
I
Appendix B
Selected Typical Standard Precast Double Tees, Inverted Tees, Hollow Core Sections
AASHIO-PCI Bulb-Tees 3.5"
3'-6"
I'
2" ~ k-.-.~ 2"
H
I
::::.--.
1'-4"
~
~
~"t-
Hw
10"
4,5
6"
"r
11
6"
I,
.~ 2'- 2"
.I
Properties H in.
Hw in.
Area
BT-54
54
36
659
BT-63
63
45
BT-72
72
54
Type
in.
Weight kip/ft
Maximum Span,* ft
268,077
27.63
0.686
114
713
392,638
32.12
0.743
130
767
545,894
36.60
0.799
146
•
Ul.
2
Inertia in.4
Ybouom
*Based on simple span, HS-25 loading and ( = 7,000 psi.
Figure 8-7 (a)
AASHTO/PCI Bulb Tees (Courtesy PCI, Ref. 12.11)
937
938
Appendix B
Selected Typical Standard Precast Double Tees, Inverted Tees, Hollow Core Sections
AASHTO-PCI Bulb-Tees 2"(TYP.)
11 -1 t
2 "(TYP.)
54"
2" (TYP.)
BT-54
63"
BT-63 Figure 8-7 (b)
BT-72 Possible Strand Arrangement for Sections in Figure C-7 (a)
939
Appendix B Selected Typical Standard Precast Double Tees, Inverted Tees, Hollow Core Sections
Deck Bulb-Tees w
~1
I" 6" 3"
2"
VAR.
H H
w
3"
6"
I.
2'-1 n
./
Dimensions and Properties H in.
H. in.
W in.
Area
in. 2
Inertia in. 4
Ybouom
Weight
Maximum
in.
kip/ft
Span*
ft
35
53
65
15
33
45
48
677
101,540
21.12
0.75
100
72
823
116,071
23.04
0.91
78
96
967
126,353
24.37
1.07
65
48
785
294,350
31.71
0.87
145
72
931
335,679
34.56
1.03
121
96
1,075
365,827
36.63
1.19
105
48
857
490,755
38.55
0.95
168
72
1,003
559,367
41.95
1.11
148
96
1,147
610,435
44.46
1.27
130
"Based on simple span, HS-25 loading and
Figure 8-8 (a)
(= 7,000 psi.
AASHTO/PCI Shallow Bridge Deck Bulb Tees (Courtesy PCI, Ref. 12.11)
CD
Deck Bulb-Tees
"'c"
,-
48"-96"
I
2" (TYP.)
,
r
T 35"
L
,
48"-96"
53"
~ •• : : •• ~
I
2" (TYP.)
DeckBT-35 DeckBT-53
I-
48"-96"
'/
35"
.......... DeckBT-65 Figure 8-8 (b)
Possible Strand Arrangement for Sections in Figure C-8 (a)
"~• • __~
, ;;'ill;' ;,;"
.
;I
nWWtWrWHW'''tt!'[
1
F'
Appendix B Selected Typical Standard Precast Double Tees, Inverted Tees, Hollow Core Sections
AASHTO Box Beams
~J Fi5"
w
5"~
5.5" H
=0 5.5"
t
~~g
IS' min•• 1
r----I I I ____ _ L
--I
Typical Longitudinal Section
Typical Keyway Details DimensiolU (inches) W Type
BI-36 BI-48 BII-36 BII-48 BIII-36 BIII-48 BIV-36 BIV-48
H
36 48 36 48 36 48 36 48
Properties
Type
.
27 27 33 33 39 39 42 42
.
Area JR.
y........ in.
Inertia in.•
BI-36 560.5 13.35 50.334 BI-48 65941 692.5 13.37 16.29 BI1-36 620.5 85.153 BII-48 752.5 16.33 110,499 19.25 BIII·36 680.5 131.145 BIII-48 812.5 19.29 168.367 BIV-36 710.5 20.73 158.644 BIV-48 20.78 842.5 203.088 "Based on simple span, HS-25 loading and fe = 7,000 psi.
Figure 8-9 (a)
314"
Weight kip/ft 0.584 0.721 0.646 0.784 0.709 0.846 0.740 0.878
Max. Span' ft 92 92 107 108 120 125 124 127
AASHTO/PCI Bridge Box Girders (Courtesy PCI, Ref. 12.11)
941
942
Appendix B Selected Typical Standard Precast Double Tees, Inverted Tees, Hollow Core Sections
AASHTO Box Beams
J
~i
I_ i~"
48"
r ........... . 22 Spa<. @ 2"
rr(
}
1 :::::::::::::::::::::::
J---:::36"=---1 _ ~1~pac'~I~"
liD'll
I1 ............... . .................
TypeBI·48
Type BI·36
................. T· .
2" (TYP.)
33"
1
r
.......................
:/
39"
1
Type BIII·48
r
·· · :
.. . : .. .
·· ··
.................. ................. Type BII·36
................. T
'0' 1 33"
......................... .......................
:0:
·· ... ··· . ·: ..: ·· .. ··· ... ................. . ............... .
............... .. T· . Type BIII·36
·
42"
'0' 1 42"
1
:....................... ,,'------.../ ....................... 1Ype BIV·48
Figure 8-9 (b)
.
·· · ··
. .. .
:
:
·· .. ·· ... ·· . ................. . ............... . TypeBIV-36
Possible Strand Arrangement for Sections in Figure C-9 (a)
1
INDEX
AASHTO: 71, 74,130,140,161,165,168, 169,220,742,744,750,758,764,780, 820 design, concrete bridges, 742 LFD, 744, 748 LRFD specifications, 753, 754, 756 sections, 116-119 standard specifications, 741 truck loads, 745 Abeles, P.W., 6, 28, 72, 220, 321, 416, 713, 741 Acceleration, 829 maps, 830, 831 ACI, 71, 79,130,161,185,187,200,413, 415,443,557,559,704,741,744,759, 763,820 ACI Committee 318: 71, 104, 148, 152, 219,235,298,302,336,519,552,573, 629,705,715,849,851,900 ACI Committee 435: 72, 104, 220 ACI-ASCE, 74, 80, 82,104,228 Aggregates shrinkage of, 48 Allowable deflections, 453, 454 Allowable stresses concrete, 59, 60 steel, 59, 60 Analysis, flexural, 107 Anchorage zone design, 141, 145 ANSI, 182, 220 ANSI!ASCE, 705 Arthur, P.D., 416 ASCE, 740, 741, 820
ASTM, 70, 71 ASTM Specification, 50 Aswad, A., 868, 902 Aswad, G.S., 868, 902 Badie, S.S., 821 Baishya, M.C, 821 Baker, A.L.L., 181, 220, 416, 408 Balaguru, P.N., 71, 220 Balancing effects, 15 Bardhan-Roy, B.K., 28, 70, 72, 220, 416, 715,741 Bars prestressing, 53, 54 reinforcing, 50, 51 Base restraint, 672 Basic concepts of prestressing, 7 Bazant, 497 Beams constant eccentricity, 113 continuous, 341, 358, 405 variable eccentricity, 115 weak beam-strong column concept, 847 Bearing stresses, 634, 638 Biaxial bending, 531 Billington, D.P., 713, 724, 740 Blair, H., 708, 741 Blair, K., 630 Borg, S.E., 901 Branson, D.E., 49, 71,104,427,451,496 Breen, J.E., 144,220 Bridges AASHTO sections, 935-942 943
944
Index
Bridges (cant.) cable-stayed, 3, 7,11,506 segmental, 3, 7 Brondum-Nielsen, T., 741 Burdet, 0., 144,220 Burns, N.H., 28, 220, 630 Bursting stresses, 140 Cantilever factor, 677 CEB,74 CEB-FIP,130 Center-of-pressure line (C-line), 13 Chakrabarti, P., 571, 630 Chan, T.e., 658 Chen, B., 39, 41, 72 Chen, Ke., 219, 235, 336 Chiang, J.y', 219, 479, 497 Choek, G.S., 901, 902 Circular prestressing, 70 Circular tanks, 660 Cleland, N.M., 902 C-line, 12,21,347,350,362 Clough, D.P., 659 Clough, RW., 901 Cohn, M.Z., 104,408,416,479,497,630 Collins, M.P., 148, 149,220,287,289,290, 292,337,762,763,820 Columns behavior, 501 load contour method, 530 modified, 540 nonslender, 508 P-.:l effect, 523 reciprocal load method for biaxial bending, 540 slender, 515, 525 strength of, 507 Combined slide and pin, 679 Compatibility, 198, 298 Composite beams elastic flexural stresses in, 158, 162,242 section properties of, 115-119 shear transfer in, 242 Composite members, 635 Composite sections, 118, 152, 165, 242 Compression-controlled state, 198, 759 Concordant tendon, 354 Concrete, 31 compressive strength, 32, 36, 39 creep properties, 43 modulus of rupture, 422 shear strength, 35
stress-strain diagram, 36 stress-strain properties of, 467 tensile strength, 33 Confining reinforcement, 849 transverse, 849, 851 Connections dapped-end beam, 640 details of, 651 tolerances for, 634,635 Continuous beams, 340, 341 Corbels, 263, 647 Corbel design by strut-and-tie method, 327,329 Corley, W.G., 408, 416 Cornell, e.A., 182, 220 Coupling beams, 856 Crack control, 479, 480 Cracking beam behavior before cracking, 420, 423 beams, 427,443,479 inclined, 228 longitudinal,140 tension members, 548 Crack width calculation of, 708 Crack widths calculation of, 480, 481 tank walls, 708 tolerable, 485 Creasy, L.R, 661, 677, 694, 740 (Ref. 11.2) Creep coefficient, 46 concrete, 43 in long-term deflections, 437, 444, 454 losses, 80 Crom, J.S. Sr., 660 Cross, H., 630 Curing, 47 Curvature beams, 423,429,440,452 creep, 450 instantaneous, 419, 426 Danesi, R, 408, 416 Decompression, 110, 111 Deep beams by strut-and-tie method, 320, 322,325 Deflections allowable, 453, 454 of flat plates, 610 incremental, 429
945
Index
time-step method for, 448 of two-way slabs and plates, 617 Deformation of concrete, 33 Derecho, AT., 901 Design aids, 905 Design information, 903 Diagonal shear cracking, 228 Diaphragms, 841, 854, 857 Dikshit, O.P., 630 Dilger, W.H., 104, 496 Dill, RE., 5 Djazmati, c., 868, 902 Dobell, c., 28 Doehring, C.W., 5 Double-T beams, 116, 117 Drift, 841, 842 Dual systems, 872 Ductile frames, 858, 860, 890, 895 Ducts, 66 Dywidag ductile connection assembly (DDC), 65, 864, 890 Earthquake ground motion, 826, 829 Eccentricity limits of, 122 variation of, 115 Economics of prestressed concrete, 4 Effective prestress, 73, 93, 97, 100 Elastic deformation, 75 Elastic modulus of concrete, 36 of steel, 56 Elastic shortening losses, 75 Ellingwood, B., 182, 220 Empty tank technique, 671 End-block reinforcement, 173 Englekirk, RE., 864, 866, 867, 897, 901, 902 (Ref. 13.21, 13.22) Equivalent frame method for deflection, 563, 610, 611 for flexural analysis, 543, 547 Equivalent lateral force method in seismic design, 837 Equivalent load method, 16,22,347,348, 351,352 Ezeldin, A, 220 Factored loads, 846 Failure shear compression, 231
Federal Emergency Management Agency (FEMA),901 Federal Highway Administration, 220 FIP, 74 First-order second-moment (FOSM) method,182 Fixity at wall foot, 695, 736 Fixity of dome support, 712 Flange width, effective, 161 Flat dome, 712 Flat-plate slabs, 592 Flexural strength of beams, 178 Flexural stresses, 59, 106 allowable in concrete, 59 allowable in steel, 59 Flexure-shear cracking, 225 Form factor, 677 Frames, indeterminate, 379 Freely sliding, 659, 662, 664, 692 Freudenthal, AM., 71 Freysinnet, E., 6, 28 Frictional losses, 85 Furlong, R.W., 408, 416, 659 Gadebeku, B.K., 220 Gadebeku, C., 144, 145 Galambos, T.V., 182,220 Gas load, 662 Gergely, P., 220 Gerwick, B.C. Jr., 28, 220, 361, 416, 550 Gesund, H., 630 Ghali, A, 104,496,740 Ghosh, S.K., 901 Goodkind, H., 219 Gouwens, AJ., 630 Grosco, J., 406, 416 Grouting tendons, 64 Gustafson, D.P., 630 Guyon, Y., 6, 28, 220 Hemabom, R, 630 Hemispherical dome, 709, 712 Hewett, W.H., 5, 660 High-rise buildings, 855, 858, 895 Hinges, plastic, 401, 405, 408 Historical development of prestressing,S Hoffman, E.S., 630 Hoop stress, 673 Horizontal base shear, 837 Horizontal interface shear, 768 Hsu, T.T.C., 285, 288, 295, 302, 337, 762, 821
Ind~x
946
Hsu, C.T.T., 525, 526, 536 Huang, P.T., 220, 479,480,497 Hung, T.Y., 630
LRFD AASHTO design, concrete bridges, 742 Lund,J.,5
Indeterminate beams, 341, 363,401 frames, 379, 401 Influence coefficients, 403, 405, 669, 670 Initial prestress, 10, 11 Institute of Structural Engineers, 28 International Building Code (IBC2009), 413,416,751,826,828,901 Itaya, R., 630
Magnel, G., 6, 28 Maher, M.H., 39, 41, 72 Mansfield, E.H., 622, 630 Marshall, W.T., 220 Martin, L.D., 496 Mast, R.F., 773, 818, 819, 821 (Ref. 12.16) Mattock, AH., 143, 218, 235, 332, 408, 416, 659 McGee, W.D., 332 McGregor, J.G., 182,220 Mehta, P.K, 71 Membrane coefficients, 677, 680-693 Mikhailov, V., 6 Mindess, S., 71 Mirza, S.A, 659 Mitchell, D., 148, 149,220,287,289,290, 291,337,762,763,820 Modified load contour method for biaxial bending, 542 Moment distribution, 351, 352 carryover factor, 597, 598 Moment magnification factor for columns, 520 Moment of inertia, effective, 427 Moment redistribution, 361 Morgan, N., 630 Moriadith, F.L., 652 Multipliers for deflection, 446
Jack, hydraulic, 63 Jackson, P.H., 5 Johansen, KW., 616 Kern of the cross-section, 130, 134, 135 Kudlapur, S.T., 821 Kunnath, S.K, 862, 901 (Ref. 13.24) Ledges, 647 Lee, J.y., 821 Leonhardt, F., 6 Lew, H.S., 901 Lift slabs, 861 Limit analysis, 401 Lin, T.Y.,6,28,104,220,416,550,630 Linear elastic anchorage zone analysis, 144 Linear transformation, 354 Liquid load, 661 Llovet, D., 901 Load -and -resistance-factor-design (LRFD) method, 182 Load balancing, 22 for beams, 347, 358 for slabs, 567, 577 Load-balancing method for beams, 15 Load combinations, 846 Load contour method of biaxial bending, 540 Load factors, ACI, 185 Losses anchorage seating, 88 creep, 80 elastic shortening, 76 friction, 85 relaxation, 78, 80 shrinkage, 83 Loss of prestress, 73
Naaman, AE., 28, 220, 759, 773, 774, 778, 821 Naja, W.M., 834, 861, 901 Nakaki, S.D., 861, 901 (Ref. 13.18) Nawy, E.G., 28, 39, 41, 62, 66, 71, 72,104, 144,145,202,219,220,266,336,408, 479,480,496,550,573,574,708,741, 751,763,821,901 Neville, AM., 71 Nilson, AH., 220, 416, 451, 496, 630 Nonprestressed steel, 50 in beams, 50 in columns, 50 in slabs, 51, 52 Norris, H.C, 901 Occupancy, seismic, 833, 834 One-way slabs, 570 Overturning, 843
947
Index
Park, R, 416 Partial prestressing, 179 Paulay, J., 416 Pauley, T., 901 PCI,74,83,104,117,120,130,220,245, 259,303,336,337,416,446,496,550, 635,640,651,659,661,704,741,780, 786,803,821 connection details, 651-658 deflection multipliers, 454, 468 Period of vibration, 827 Pessiki, S., 862, 869, 871, 902 (Refs. 13.31-13.33) Pinned wall foot, 671, 695, 697 Plastic hinges, 305, 308, 401 Plasticity equilibrium torsion, 293 Plates carryover factors, 596, 597 fixed-end moments, 570 torsional stiffness, 564 Poisson's ratio for concrete, 668, 671, 675 Popovices, S., 70 Portland Cement Association, 71, 678, 740 Post-tensioning, 62 Post-Tensioning Institute, 71,104,144, 219,550,629,659,704,741,821 Potyondy, G.J., 62, 72, 202, 219, 408, 416, 474,480,496 Precast prestressed frames in high seismicity zones, 852, 864 Precast tanks, 704, 705 Preload,663,665,715 Pressure, center of, 13 Pressure grouting, 64 Preston, RL., 630, 708, 741 Pre tensioning, 61 Priestley, M.J.N., 902 Primary moments, 344 Principal stresses in beams, 232 Probable shears and moments, 845 Profile of tendons, 129, 132, 133 Rabbat, B.G., 287, 337 Radial shear, 662, 672 Ramakrishnan, V., 416 Reciprocal load method for biaxial bending, 540 Redistribution of elastic moments, 358 Reinforcement index, 198,200 Relative stiffness, 558, 564 Relaxation losses, 78, 80
of steel, 78 Restrained wall foot, 662 Reynolds, C.E., 569, 570, 571, 630 Rice, P.F., 630 Ring force, 674, 676 Roberts, c., 144, 220 Roll, F., 71 Roof,708 Ross, AD., 71 Rotation, 402 Safety, 742 structural, 181 Salek, F., 408, 416 Sanders, D., 144, 220 Sauer, J.A, 71 Sawyer, H.A, 408 Schneider, RR, 897 Scordelis, AC., 630 Secondary moments, 344 Section modulus, 108 Seismic design, 824 Seismic forces, 751 Semisliding, 733 Serviceability, 418, 420, 453, 479 Shaikh, AF., 447, 451, 496 Shear, 223 Shear design by ACI code, 228, 235, 240, 246 Shear friction theory, 264, 266 Shear-moment transfer, 583 Shear reinforcement in beams, 234, 266 in slabs, 582, 585 Shear wall, 855, 879 design, 882 Shells, 660,664,709 Shrinkage of concrete, 48 losses, 83 SI conversion table, 905-906 Siess, c.P., 234, 336 Sign convention, 107,680 Slabs deflection of two-way, 586 flat-plate, 592 shear-moment transfer, 583 types of, 554, 555 Slender columns ACI design method for, 525 Smulski, E., 416 Soogswang, K., 235, 336
, Index
948
Sozen, M.A., 220, 234, 336 Spalling zone, 141 Span-depth ratios, flat plates, 570 Spectral response, 829 Standard sections, 115 Stanton, J.F., 861, 902 Steedman, J.e., 569, 570, 571, 630 Steel low-relaxation, 54 relaxation of, 56, 78 strands, 54 stress-relieved, 54 wires, 54 Steiner, G.R., 5 Stirrups, 241, 248, 250 Stone, W.e., 220, 901, 902 Story height drift, 841 Strain, 36, 43 Strain compatibility analysis, 209 strain limits design approach, 191,759, 760 strain limits, 191, 192, 759 Strength concrete, 31, 35,59 flexural, beams, 188 nominal, 181, 186 of prestressing steel, 53, 54 reduction factor, 507 reduction factors, ACI, 185 reinforcing bars, 50, 51, 52 tanks, 705 Stress block, rectangular, 190 Stress distribution, 183 Stress-strain properties of concrete,36,37,421,467 prestressing steel, 51, 56, 441 reinforcing bars, 51 Strong column-weak beam concept, 845, 847,873 Structural (shear) walls, 854 Strut-and-tie anchorage zone design, 144 Strut stress path in post-tensioned anchorage zone, 144 Strut-and-tie hypothesis for corbels and deep beams, 317,318,322,324 Subsoil support, 699 Support displacement method, 344-348 Tadros, M.K., 104, 425, 496, 741, 821 Tanks circular prestressing, 69, 704, 715
influence coefficients, 680-693 moments, 663 radial stress, 663 roof dome, 708 toe, 699, 700 vertical prestressing, 736 wall, 679, 716, 717 Taylor, F.W., 416 Tendon profile, concordant, 354 Tensile strength concrete, 31 prestressing steel, 54 reinforcing bars, 50, 51 Tension-controlled state, 192, 759 Tension members, 500 behavior of, 546 design of, 548 linear, 548 Thickness of walls, 706, 707 Thompson, S.E., 416 Thornton, K., 416 Time-step method, incremental, 448 Timoshenko, S., 668, 671, 713, 740 Torsion in beams, 286, 298, 308 compression field theory, 289 cracking, 290, 292 design for, 298, 304 equilibrium in element theory, 294 reinforcement requirements, 303 skew-bending theory, 285 space-truss analogy theory, 287 Transformation, linear, 355 Ukadike, M.M., 71, 266, 336 Unbonded post-tensioned precast concrete precast shear walls, 867, 888 wall-base bolting, 870 wall base connection, 888 wall base grouting, 889 Unified design approach (See Strain limits design approach) Uniform Building Code (UBC), 413, 416, 829,901 Vertical moments, 679, 696 Vertical prestressing, 679, 715 Vessey, J.y', 630, 708, 741 (Ref. 11.14) Wakabayashi, M., 622, 696 Walters, D.B., 630
i
949
Index
Water-cement ratio, 48 Web reinforcement, minimum, 238 Web-shear cracking, 231 Wheel load distribution, 748 Wheen, R.J., 552 Wilhelm, W.J., 552 Wind loads, 751 Wire mesh, welded, 51 Wires for prestressing, 54, 55, 56
Woinowsky-Krieger, S., 668, 671, 713, 740 Wollman, G., 144, 220 Yong, Y.K., 144, 145,220 Young, J.F., 71 Zhu, R.H., 821 Zia, P., 285, 336 Zwoyer, E.M., 234, 336
Vj = factored shear force at section due to exter-
Vn wu
x Xl
Y YI
Yl a
a
nally applied loads occurring simultaneously with Mmax' = nominal shear strength. = factored load per unit length of beam or per unit area of slab. = shorter overall dimension of rectangular part of cross section. = shorter center-to-center dimension of closed rectangular stirrup. = longer overall dimension of rectangular part of cross section. = distance from centroidal axis of gross section, neglecting reinforcement, to extreme fiber in tension. = longer center-to-center dimension of closed rectangular stirrup. = total angular change of prestressing tendon profile in radians from tendon jacking end to any point x. = ratio of flexural stiffness of beam section to flexural stiffness of a width of slab bounded laterally by centerlines of adjacent panels (if any) on each side of the beam. Echlh
EeJ, average value of a for all beams on edges of a panel. ~a = ratio of dead load per unit area to live load per unit area (in each case without load factors). ~d = ratio of maximum factored dead load moment to maximum factored total load moment, always positive. ~ = a ratio of clear spans in long to short direction of two-way slabs. 'Vf = fraction of unbalanced moment transferred by flexure at slab-column connections. 'VI' = factor for type of prestressing tendon. = 0.55 for fp/fpu not less than 0.80 = OAO for fp/fpu not less than 0.85 = 0.28 for fp/fpu not less than 0.90
am =
'V v
= fraction of unbalanced moment transferred by eccentricity of shear at slab-column connections. = 1 - 'Vf
0ns
= moment magnification factor for frames
braced against sidesway, to reflect effects of member curvature between ends of compression member. Os = moment magnification factor for frames not braced against sidesway, to reflect lateral drift resulting from lateral and gravity loads. curvature friction coefficient.
f.L
= time-dependent factor for sustained load. Plrho) = ratio of nonprestressed tension reinforcement. = AJbd p' = ratio of nonprestressed compression rein~IXi)
forcement
= A:lbd Ph = reinforcement ratio producing balanced strain
conditions. Pp
PI
e
w
= ratio of prestressed reinforcement. = Ap/bdp = AjAcv; where A,·v is the projection on Acv of area of distributed shear reinforcement crossing the plane of Acv. = angle of compression diagonals in truss analogy for torsion.
= strength reduction factor. = PI/t:.
w' = p'fylf; .. wI'
=
p/p /!:. ,
wpw,ww,w w
= reinforcement indices for flanged sections computed as for w,w p ' and w' except that b shall be the web width, and reinforcement area shall be that required to develop compressive strength of web only.
To convert from Length inch (in.) inch (in.) foot (ft) yard (yd) Area square foot (sq. ft) square inch (sq. in.) square inch (sq. in.) square yard (sq yd) acre (A) Volume cubic inch (cu in.) cubic foot (cu ft) cubic yard (cu yd) gallon (gal) Can. liquid* gallon (gal) Can. liquid* gallon (gal) U.S. liquid* gallon (gal) U.S. liquid* Force kip kip pound (lb) pound (lb) Pressure or Stress kips/square inch (ksi) pound/square foot (psf) pound/square inch (psi) pound/square inch (psi) pounds/sq uare foot (psf) Mass pound (avdp) ton (short, 2000 Ib) ton (short, 2000 Ib) gram tonne (t) Mass (weight per Length) kip/linear foot (kif) pound/linear foot (plf) pound/linear foot (plf) Mass per volume (density) pound/cubic foot (pcf) pound/cubic iard (pcy) gallon per ydoz per yd 3 Bending Moment or Torque inch-pound (in.-Ib) foot-pound (ft-Ib) foot-kip (ft-k) Temperature degree Fahrenheit (deg F) degree Fahrenheit (deg F) Energy British thermal unit (Btu) kilowatt-hour (kwh) Power horsepower (hp) (550 ft Ib/sec) Velocity mile/hour (mph) mile/hour (mph) Other Section modulus (in. 3 ) Moment of inertia (in.4) Coefficient of heat transfer (Btu/ft2/h/oF) Modulus of elasticity (psi) Thermal conductivity (BTU-in./ft 2/h/oF) Thermal expansion in.iin.fOF Area/length (in.2/ft) *One U.S. gallon equals 0.8321 Canadian gallon ** A pascal equals one newton/square meter
to
multiply by
millimeter (mm) meter (m) meter (m) meter (m) square meter (sq m) square millimeter (sq mm) square meter (sq m) square meter (sq m) hectare (ha) = 10,000 sq m cubic cubic cubic liter cubic liter cubic
meter (cu m) meter (cu m) meter (cu m) meter (cu m) meter (cu m)
kilogram (kgf) newton (N) kilogram (kgf) newton (N) megapascal (MPa)** kilopascal (kPa)** kilo pascal (kPa)** megapascal (MPa)** kilogram/square meter (kgf/sq m) kilogram kilogram tonne (t) kilogram kilogram
(kg) (kg) (kg) (kg)
25.4 0.0254 0.3048 0.9144 0.09290 645.2 0.0006452 0.8361 0.4047 0.00001639 0.02832 0.7646 4.546 0.004546 3.785 0.003785 453.6 4448.0 0.4536 4.448 6.895 0.04788 6.895 0.006895 4.882 0.4536 907.2 0.9072 0.00006480 1000
kilogram/meter (kg/m) kilogramlmeter kg/m newton/meter (N/m)
1488 1.488 14.593
kilogram/cubic meter (kg/cu m) kilogram/cubic meter (kg/cu m) Kg/m 3 Kg/m 3
16.02 0.5933 4.985 0.037
newton-meter newton-meter newton-meter
0.1130 1.356 1356
degree Celsius (C) degree Kelvin (K) joule U) joule U) watt (W) kilometer/hour meter/second (m/s) mm 3 mm 4 W/m 2/oC MPa Wm/m 2/oC mm/mm;oC mm 2/m
tc = (tF - 32)/1.8 tK = (tF+459.7)i1.8
1056 3,600,000 745.7 1.609 0.4470 16,387 416,231 5.678 0.006895 0.1442 1.800 2116.80