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1
Design for Torsion by ACI 318-11
Basile G. Rabbat
Consultant Mt. Prospect, IL
September 25- 26, 2013
Learning Objective
Determine when design for torsion is mandated by Code
Distinguish between: Equilibrium Torsion Compatibility Torsion
Design torsion reinforcement Detail torsion reinforcement
2!
3!
Two-Way Flat Plate w/Spandrel Beams (13.6.3)
3!
Design spandrel for torsion
2
4!
Torsion in Canopy Spandrel Beam
Design for Torsion
11.5 Design for Torsion
Based on Thin-Wall Tube, Space Truss Analogy
Applies to: R/C & P/S Hollow and Solid Sections
Tc = 0 5!
Solid vs. Hollow Section Strength
6!Source: MacGregor, ACI 318-11 Ref. 11.31
3
R11.5 Thin-Wall Tube Analogy
7!
!
8!
11.5 Design for Torsion
Compute Tu (from analysis) If Tu < Tcr /4 (threshold torsion)
Ignore torsion Cracking at = 4 fc If Tu Tcr /4 Design for torsion
Equilibrium torsion? Compatibility torsion?
Design torsion reinforcement Transverse & longitudinal reinf.
!
11.5.1(a) Threshold Torsion
Nonprestressed Concrete
Threshold torsion (Tcr /4)
Tu < !" fc' Acp
2
pcp
!
"##
$
%&&
9!
4
11.5.1(b) Threshold Torsion
Prestressed Concrete
Tu < !" fc' Acp
2
pcp
!
"##
$
%&& 1+
fpc4" fc
'
fpc = compressive stress in concrete, after prestress losses, at centroid of cross section
10!
11.5.1(c) Threshold Torsion
Nonprestressed w/Axial Force
Tu < !" fc' Acp
2
pcp
!
"##
$
%&& 1+
Nu4Ag! fc
'
Nu = factored axial force normal to cross section occurring simultaneously with Vu or Tu; positive for compression and negative for tension
11!
12!
11.5.1 Acp & pcp
Members cast monolithically with a slab Effective flange width per 13.2.4
h (h hf ) 4hf
hf
bw
Acp = Shaded area pcp = Perimeter of shaded area
5
11.5.1 Threshold Torsion
Determine Acp and pcp
Neglect overhanging flange(s) when:
Acp2 pcp
Beam w/o flange(s)
Acp2 pcp
Beam with flange(s) <
13!
11.5.1 Threshold Torsion
Hollow sections
Replace Acp with Ag in equations for threshold torsion
14!
15!
R11.5.2.1 Equilibrium Torsion
Design torque may not be reduced, because moment redistribution is not possible
6
16!
R11.5.2.2 Compatibility Torsion
Design torque for spandrel beam may be reduced because moment redistribution is possible
17!
11.5.2.2 Compatibility Torsion
Twist
Tcr Torque
Tn
Tu,comp = 4 fc Acp2 pcp
= Tu,comp
18!
11.5 Design for Torsion - Recap
Compute Tu (from structural analysis)
If Tu < Tcr /4 Ignore torsion
Tu Tcr /4 Design for torsion Equilibrium torsion? Design for Tu (analysis) Compatibility Torsion?
Design for Tu,comp = Tcr (compatibility) Redistribute T = Tu Tu,comp = Tu - Tcr
7
11.5.2.2 Compatibility Torsion, Tu,comp
19!
Tu ! !4" fc' Acp
2
pcp
"
#$
%
&' 1+
fpc4" fc
'
Tu ! !4" fc' Acp
2
pcp
"
#$
%
&' 1+
Nu4Ag" fc
'
Tu ! !4" fc' Acp
2
pcp
"
#$
%
&'
R/C w/Axial Force
P/S
R/C
Note: Do not replace Acp with Ag
11.5.2.3 Slab/Spandrel Torsion
Determine torsional moment distribution through analysis
Assume torsion from slab to spandrel to be uniformly distributed
20!
11.5.2.4-.5 Critical Section
Distance from support R/C d
P/S h/2
Exception: Concentrated torque within distance d (R/C), or h/2 (P/S)
21!
8
Torsional Stresses Shear Stresses
11.5.3.1 Adequacy of Solid Section
22!
11.5.3.1 Adequacy of Solid Section
Circular interaction
23!
Vubwd
!
"#$
%&
2
+Tuph1.7Aoh
2
!
"#$
%&
2
' !Vcbwd
+8 fc'!
"#$
%&
Torsional Stresses Shear Stresses
11.5.3.1 Adequacy of Hollow Section
24!
9
11.5.3.1 Adequacy of Hollow Section
Linear interaction
25!
Vubwd
!
"#$
%&+
Tuph1.7Aoh
2
!
"#$
%&' !
Vcbwd
+8 fc'!
"#$
%&
11.5.3.3 Thin Wall Section
If t < (Aoh /ph ) Replace
26!
Tuph1.7Aoh
2
!
"##
$
%&& with
Tu1.7Aoht
!
"##
$
%&&
Space Truss Analogy Torsional Cracks Spiral Around Section
27!Courtesy : Dr. Michael Collins, University of Toronto
10
28!
Concrete Shell Spalls
Courtesy : Dr. Michael Collins, University of Toronto 28!
Stress Trajectories Tension in Hoop
Compression in Concrete
Compression in Concrete
Tension in Hoop
Outside of Concrete
Source: Collins & Mitchell, Prestressed Concrete Structures
Concrete Shell Spalls
29!
11.5.3 At Maximum Torque
Concrete shell spalls
Aoh = Area within stirrup centerline
Assume Ao = 0.85Aoh
30!
11
Fig. R11.5.3.6(b) Definition of Aoh
31!
32!
T
Longitudinal Bar
xo
yo
Stirrups Cracks
= 30o to 60o
V1 V2
V3
V4
R11.5.3.6(a) Space Truss Analogy
Concrete Compression Diagonals
For R/C = 45o For P/S = = 37.5o
33!
11.5.3.6 Transverse Reinforcement
Atfyt
yocot
s
yo = center-to-center length of closed stirrup
V2
Atfyt
V2 =qyo =T2Ao
yo =Atfytsyo cot!
12
34!
11.5.3.6 Transverse Reinforcement
Atfyt
yocot
s
yo = center-to-center length of closed stirrup
V2
Atfyt
Tn =2AoAtfyts
cot!
where Ao = 0.85Aoh = 0.85xoyo
11.5.3.6 Torsional Moment Strength
Tn Tu
35!
= 0.75 (9.3.2.3)
Tn = 2Ao At fyt
s cot (11-21)
11.5.3.6 Transverse Reinforcement
Determine transverse reinforcement required for torsion
36!
=
Ao = 0.85Aoh
At s
Tu (or Tu,comp) 2Aofyt cot
13
11.5.5.2 Minimum Transverse Reinf.
37!
Av At s
+ Minimum = 0.375 f c bws
fyt
25bw fyt 2s
Av = 2fyt d
2s Vu Vc
38!
11.5.3.7 Longitudinal Reinforcement
N2 /2
N2 /2
N2
D2 V2
yo cos
yo
N2 =V2 cot! =A!fy
V2 =Atfytsyo cot!
Afy/2
39!
11.5.3.7. Longitudinal Reinforcement
N2 /2
N2 /2
N2
D2 V2
yo cos
yo
A! =Atsphfytfycot2!
ph = perimeter of centerline of exterior stirrup/hoop
Afy/2
14
40!
11.5.3.7 Longitudinal Torsional Reinf.
A =
s
ph
fyt
fy
cot2 (11-22) At
41!
11.5.5.3 Minimum Longitudinal Reinf.
A,min =
5 fc Acp
fy ph
s
At
fyt
fy
At s
25bw fyt
11.5.3.6-.7 Torsion Reinforcement
42!
h
bw
At @ s
A"
15
Determine longitudinal reinforcement required for flexure
Combine longitudinal reinforcement required for torsion with that required for flexure A added to Aflexure on tension side A may be reduced in flexural compression
zone of member by Mu/(0.9dfy) 43!
11.5.3.9 Details of Longitudinal Reinforcement
11.5.6.2 Details of Longitudinal Torsional Reinforcement
44!
Minimum one longitudinal bar in every corner
12 (typ.)
11.5.6.2 Details of Longitudinal Torsional Reinforcement
45!
16
db s/24
db 3/8
46!
11.5.6.2 Details of Longitudinal Torsional Reinforcement
Extend torsion reinforcement beyond theoretical cut-off point a distance of (bt + d)
47!
11.5.6.3 Details of Torsion Reinforcement
11.5.6 Spacing of Transverse Reinf.
ph /8
12 in.
d/2
48!
17
Questions?
49!