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Dr Fang WANG Senior Research Officer Officer an
oa s,
[email protected]
Outline Member capacities
Strut
Tie
Node
Example
Strut capacity against action force
Strut capacities - Cu=βs0.9f’c A c
βs is strut efficiency factor = 1/(1+0.66(cot)2 with the limit of 0.3≤ βs≤1.0 (considering the transverse tension fields crossing t e strut)
smallest strut angle (with tie element)
Ca acit check criterion
C ≤C*
C* strut force Capacity reduction factor Φc=0.6
(AS 3600)
A c is taken as the minimum cross-sectional cross-sectional area of the strut member
Steel tie Yield
strength capacity - Tu=f sy A st
Capacity check criterion sTu≤T*
T* tie force
Capacity reduction factor Φs=0.8 (AS 3600)
A s is cross-sectional area of steel reinforcement
Node – stress
CCC, CCT, CTT, TTT
Node – stress (cont’d)
Three principal stress
≤
0. f’
Efficiency factor βn =1.0 for CCC, 0.8 for CCT, 0.6 for CTT
Capacity reduction factor Φc=0.6
(AS 3600)
Example – deep beams
Three types of model
Type I – force directly from load point to support, no
Type II – shear transferred by major and minor struts, hanger reinforcement required – carried to the support by a series of minor with hanger reinforcement to
ype –a z Type II – 1≤a/z ≤√ 3 Type III – a/z ≥√ 3
z lever-arm
a z
Type I
Type II
Type III
Example – deep beams
Example
Design of a deep transfer girder
Desi n the 11 meter s an 3.66 meters dee and 600 mm thick cantilevered transfer girder .
1) Strength limit state： *=8000 kN w*=100 kN/m (including self-weight)
2) Serviceability limit state： P=5000 kN w=62.5 kN/m (including self-weight)
Materials
1) fc’=50 Mpa
2) fsy=500 MPa
Example Strut-and tie model is selected as a type II model
based on the a/z=1.13 •
w* is replaced by e uivalent nodal loads. • Node N is separated into two components, N1 2, accommodate the forces from the heavily loaded nc ne struts nto t e wa support.
Example Analysis For the left shear, with a/z= 1.13, the
span is classified as Type II and the angle to the horizontal of the major com ression strut an le is 41.5°. The component of the point load at
support is calculated by subtracting the load at node A (350 kN) from the support react on (4720 N) 8000 kN at node B: 4370 kN goes towards the left su ort reaction，with the balance (3630 kN） carried to the right side support.
Example
Example na ys s The selected design is threefold indeterminate, the
re un anc es can e remove an rep ace w t sets o forces chosen by the designer for a prescribed force distribution: 1) Replace the support at node H with the specified reaction force of 4720 kN upwards. 2) Remove member AI and replace it with a set of statically equivalent equal and opposite forces at nodes A and I of 766 kN.
Example
Analysis
3) proportion the relative components of the load transferred from the 8000 kN point load to each of struts BJ and BK 3630 kN is transferred throu h the ri ht shear s an towards the supports labeled N. (struts BJ and BK shall carry equal vertical components of this force, that is, a
The tie DJ is removed from the analytical model and
replaced with a force pair of 1815 kN at nodes D and J.
Example Left shear span To satisfy the web force of F* AI=776 kN, the area
of steel should satisfy 1） T u= Ast f sy s u
*
s=
.
As≥ F *AI/( s f sy)=776×103/(0.8×500)
=
2
Example Left shear span
Example Left shear span
Example Left shear span
Example Left shear span
Example Left shear span
e n orcemen ra o or
e e s ee s:
p=1940/(2000×600)=0.00162 The dimension of 2.0 m is selected so that the
maximum angle of the minor compression struts does not exceed 75°.
Example Left shear span
e propor on o e ver ca oa carr e rec y to the support from strut BH requires consideration . 1） F *BH=3594/(sin41.5°)=5424 kN 2） F BH，s=(5/8) F *BH=(5/8)5424=3390 kN or serv ce oa s
Example Left shear span
e s ru w rom equa on . , c s ca cu a e 7.2 and 7.3 where Ac=d ct and the strut efficiency is = . 1） C u= β s 0.9 f ’ c Ac 2）
cC u≥ C*
3） =1/(1+0.66(cot θ )) within the limits 0.3≤ β s≤1.0
Example
Example Left shear span
or serv cea y, w an a= gives a bursting force of:
, qua on .
T b=C tan a where C is the compression force in the strut and the deviation is taken as: 1) tan a=1/2 for serviceability 2) tan a =1/5 for strength
l =3390×1/2=1695 kN
Example Left shear span
m ng e max mum s ress n e urs ng s ee to f s=200 Mpa for a moderate level of crack , to the cracking plane is: s
Example Left shear span
or s reng , w an a= bursting force of:
, qua on .
g ves a
T * b=5424×1/5=1085 kN and an area of steel of:
As≥T * b/( s f sy)=1085×103/(0.8×500)=2710 mm2 an t ere ore, t e urst ng orces at serv ce oa s control the design.
Example Left shear span
as y, e m n mum urs ng re n orcemen s calculated. At the point in the loading that the , concrete and to be transferred to the bursting reinforcement is (E uation 7.16):
T b，cr =0.7 t l b f ’ct w ere
s
e
c ness o
e s ru an
f ’ct=0.36( f ’c)1/2 T b，cr =0.7×600×4055×0.36×501/2×10-3 =2710 mm2
Example Left shear span
e u ma e con on, w an a= , qua on 7.14 gives a bursting force in strut BH:
T * b=5424×1/2=2712 kN . b,cr , required. The area is calculated as: b
Example Left shear span
ompar ng s an s,min, reinforcement governs.
e m n mum urs ng
The area of bursting reinforcement normal to strut GH is:
Ast=10 840 mm2 The reinforcement ratio is:
pw=As /l bt=10840/(4055×600)=0.00445
Example Left shear span
e orce across e urs ng p ane s ma n a ne orthogonal reinforcement is placed parallel and 1) pwh= pw sinθ=0.00445×sin41.5=0.00295 2) pwv= pw cosθ=0.00445×cos41.5=0.00333 The vertical web reinforcement in the left shear
span, considering bursting reinforcement and hanger reinforcement, is:
pwv=0.00162+0.00333=0.00495
Example Left shear span
op ng wo ayers one ayer a eac ace o mm diameter bars, with a total bar area across the 2, spacing requirement of ： . Use two layers N20 bars at 200 mm spacing for the
ver ca re n orcemen n
e e s ear span.
Example Left shear span
or e or zon a re n orcemen , pwh= . . Using two layers of 16 mm diameter bars ( As=400 2
s≤400/(0.00295×600)=226 mm Use two layers N16 bars at 220 mm spacing for the
vertical reinforcement in the left shear span.
Example Right shear span
orces n e anger mem ers , an similar and it is decided to distribute the
are .
For design, the maximum force F *FM=2015 kN is As≥ F *FM/(
3/(0.8×500) )=2015 10 f × s sy = mm
or: s≤620/(0.00420×600)=246 mm
Example Right shear span
o ma c e e s ear span, s ec e o use two layers of N20 bars at 200 mm centers. Although there is no horizontal steel required for equilibrium, a minimum of 0.2 per cent is adopted , ductility and for shrinkage and temperature re uirements. Two N16 bars at 220 mm centers satisfy this
of pwh=0.00303.
Example Compression members
ew o e compress on s ru s s o a ne y rearranging Equations 7.1, 7.2 and 7.3 such that,
The resultin minimum strut widths are iven in
the Tables for the stinger and web struts. With the struts ro ortioned the model is revised
slightly as sketched in the following Figure.
Example
Example Bottom tensile reinforcement
e max mum orce carr e y e o om s ee s 5131 kN and occurs directly beneath the load . The area of horizontal bottom steel required is: Ast≥ F *IJ/(
3/(0.8×500) )=5131 10 f × s sy =12 830 mm2
This can be supplied with 16N32 bars in four
layers with four bars per layer.
Example Bottom tensile reinforcement
s anger re n orcemen s progress ve y introduced into each of the left and right shear , reduced provided proper anchorage requirements are maintained.
Example Top tensile reinforcement
erg an suppor e op or zon a s ee carries a force of 4992 kN requiring a steel area of:
Ast≥ F *FG/( s f sy)=4992×10 /(0.8×500) =12 480 mm2 This can be supplied with 16N32 bars placed in
four layers with four bars per layer. As for the bottom steel, the negative moment
reinforcement can be progressively reduced as the anger stee s ntro uce .
