lecture 29 strut tie modeling example fw 18oct
TRANSCRIPT
Lecture 29 Strut and tie modelingLecture 29 Strut-and-tie modeling
Dr Fang WANG
Senior Research Officer
M i R d WAMain Roads, WA
Outline
Member capacities
Strut
Tie
Node
Example
Strut capacity against action force
Strut capacities - Cu=βs0.9f’cAc
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
h )the strut)
smallest strut angle (with tie element)
Capacity check criterion cCu≤C*Capac ty c ec c te o cCu C
C* strut force
Capacity reduction factor Φc=0.6 (AS 3600)
Ac is taken as the minimum cross-sectional area of the strut
member
Steel tie
Yield strength capacity - Tu=fsyAst
Capacity check criterion sTu≤T*
T* tie force
d f 8 (AS 6 ) Capacity reduction factor Φs=0.8 (AS 3600)
As is cross-sectional area of steel reinforcement
Node – stress
CCC, CCT, CTT, TTT
Node – stress (cont’d)
Three principal stress
Hydraulic or non-hydraulic: 3≤ cβn0.9f’cAc Hydraulic or non hydraulic:
Efficiency factor βn =1.0 for CCC, 0.8 for CCT, 0.6 for CTT
Capacity reduction factor Φc=0.6 (AS 3600)
3≤ cβn0.9f cAc
Example – deep beams
Three types of model Type I – force directly from load point to support, no
consideration of effect of vertical web reinforcementconsideration of effect of vertical web reinforcement Type II – shear transferred by major and minor struts,
hanger reinforcement required
Type III shear is Type III – shear is carried to the support by a series of minor compression struts
a
z
compression struts with hanger reinforcement to
Type I a/z ≤ 1
Type I
Type I – a/z ≤ 1 Type II – 1≤a/z ≤√3 Type III – a/z ≥√3 a shear span
Type II a shear span z lever-arm
Type III
Example – deep beams
Example Design of a deep transfer girder
Design the 11 meter span, 3.66 meters deep and es g e e e spa , 3.66 e e s deep a d600 mm thick cantilevered transfer girder.
LoadingLoading
1) Strength limit state:P*=8000 kNw*=100 kN/m (including self-weight)
2) Serviceability limit state:P=5000 kNw=62.5 kN/m (including self-weight)
Materials1) 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 equivalent nodal loads.q
• Node N is separated into two components, N1
and N to betterand N2, to better accommodate the forces from the heavily loaded i li d i h llinclined struts into the wall support.
ExampleAnalysis 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 compression strut angle is 41.5°.p g
The component of the point load at node B that is taken to the left sidenode B that is taken to the left side support is calculated by subtracting the load at node A (350 kN) from the
i ( k )support reaction (4720 kN)8000 kN at node B: 4370 kN goes towards the left support reaction,with the ppbalance (3630 kN) carried to the right side support.
Example
Example
A l iAnalysis The selected design is threefold indeterminate, the
d d i b d d l d ith t fredundancies can be removed and replaced with sets of 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 through the right shear span g g ptowards the supports labeled N. (struts BJ and BK shall carry equal vertical components of this force, that is, a vertical component of 1815 kN each)vertical component of 1815 kN each)
The tie DJ is removed from the analytical model and l d i h f i f k d dreplaced 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) Tu=Astfsy
2) T ≥T* ( =0.8,AS 3600)2) sTu≥T ( s 0.8,AS 3600)
As≥F*AI/( sfsy)=776×103/(0.8×500)=1940 mm21940 mm
Example
Left shear span
Example
Left shear span
Example
Left shear span
Example
Left shear span
Example
Left shear span
R i f t ti f th ti t l i Reinforcement ratio for the tie steel is:
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
Th ti f th ti l l d i d di tl The proportion of the vertical load carried directly to the support from strut BH requires consideration of bursting forces The force carried by BH is:of bursting forces. The force carried by BH is:
1) F*BH=3594/(sin41.5°)=5424 kNfor ultimate loadsfor ultimate loads
2) FBH,s=(5/8)F*BH=(5/8)5424=3390 kNf i l dfor service loads
Example
Left shear span
Th t t idth (d ) i l l t d f ti 7 1 The strut width (dc) is calculated from equation 7.1, 7.2 and 7.3 where Ac=dct and the strut efficiency is βs=0 54:βs 0.54:
1) Cu=βs 0.9 f ’ c Ac
2) cCu≥C*
3) βs=1/(1+0.66(cot θ)) βs ( ( ))within the limits 0.3≤βs≤1.0
Example
Example
Left shear span
F i bilit ith t 1/2 E ti 7 14 For serviceability, with tan a=1/2, Equation 7.14 gives a bursting force of:
Tb=C tan a
where C is the compression force in the strutand the deviation is taken as:
1) tan a=1/2 for serviceability) y2) tan a =1/5 for strength
lb=3390×1/2=1695 kNlb 3390 / 695 N
Example
Left shear span
Li iti th i t i th b ti t l Limiting the maximum stress in the bursting steel to fs=200 Mpa for a moderate level of crack control the area of reinforcement crossing normalcontrol, the area of reinforcement crossing normalto the cracking plane is:
A =1695×103/200=8480 mm2As=1695×103/200=8480 mm2
Example
Left shear span
F t th ith t 1/5 E ti 7 14 i For strength, with tan a=1/5, Equation 7.14 gives abursting force of:
T*b=5424×1/5=1085 kN
and an area of steel of:
As≥T*b/( sfsy)=1085×103/(0.8×500)=2710 mm2
d h f h b i f i l dand therefore, the bursting forces at service loadscontrol the design.
Example
Left shear span
L tl th i i b ti i f t i Lastly, the minimum bursting reinforcement is calculated. At the point in the loading that the bursting crack forms the force carried by thebursting crack forms, the force carried by the concrete and to be transferred to the bursting reinforcement is (Equation 7.16):( q )
Tb,cr=0.7 t lb f ’ct
h t i th thi k f th t t dwhere t is the thickness of the strut and
f ’ct=0.36(f ’c)1/2
Tb,cr=0.7×600×4055×0.36×501/2×10-3
=2710 mm2
Example
Left shear span
At th lti t diti ith t 1/2 E ti At the ultimate condition, with tan a=1/2, Equation 7.14 gives a bursting force in strut BH:
T*b=5424×1/2=2712 kN
As T* >0 5T minimum web reinforcement is As T*b>0.5Tb,cr, minimum web reinforcement is required. The area is calculated as:
Example
Left shear span
C i A d A th i i b ti Comparing As and As,min, the minimum bursting reinforcement governs.
The area of bursting reinforcement normal to strut GH is:
Ast=10 840 mm2
The reinforcement ratio is:
pw=As/lbt=10840/(4055×600)=0.00445
Example
Left shear span
Th f th b ti l i i t i d if The force across the bursting plane is maintained if orthogonal reinforcement is placed parallel and normal to the axis of the member such that:normal to the axis of the member such that:
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
Ad ti t l ( l t h f ) f 20 Adopting two layers (one layer at each face) of 20 mm diameter bars, with a total bar area across the 600 mm thick section of 620 mm2 gives a bar600 mm thick section of 620 mm , gives a bar spacing requirement of:
s≤620/(0 00495×600)=209 mms≤620/(0.00495×600)=209 mm
Use two layers N20 bars at 200 mm spacing for the ti l i f t i th l ft hvertical reinforcement in the left shear span.
Example
Left shear span
F th h i t l i f t 0 00295 For the horizontal reinforcement, pwh=0.00295. Using two layers of 16 mm diameter bars (As=400 mm2) gives a bar spacing requirement of:mm ) gives a bar spacing requirement of:
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
F i th h b DJ EK d FM Forces in the hanger members DJ, EK and FM are similar and it is decided to distribute the reinforcement for each over a length of two metersreinforcement for each over a length of two meters.
For design, the maximum force F*FM=2015 kN is used giving a reinforcement area of:used giving a reinforcement area of:
As≥F*FM/( s fsy)=2015×103/(0.8×500)5040 2=5040 mm2
or:
s≤620/(0.00420×600)=246 mm
Example
Right shear span
T t h th l ft h it i d id d t To match the left shear span, it is decided to 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 to ensure that the strut efficiencies are obtained forto ensure that the strut efficiencies are obtained, for ductility and for shrinkage and temperature requirements.equ e e ts.
Two N16 bars at 220 mm centers satisfy this requirement giving a horizontal reinforcement ratiorequirement giving a horizontal reinforcement ratio of pwh=0.00303.
Example
Compression members
Th idth f th i t t i bt i d b The width of the compression struts is obtained by rearranging Equations 7.1, 7.2 and 7.3 such that,
The resulting minimum strut widths are given in g gthe Tables for the stinger and web struts.
With the struts proportioned, the model is revised W t t e st uts p opo t o ed, t e ode s ev sedslightly as sketched in the following Figure.
Example
Example
Bottom tensile reinforcement
Th i f i d b th b tt t l i The maximum force carried by the bottom steel is 5131 kN and occurs directly beneath the load columncolumn.
The area of horizontal bottom steel required is:
Ast≥F*IJ/( s fsy)=5131×103/(0.8×500)=12 830 mm2
This can be supplied with 16N32 bars in four layers with four bars per layer.
Example
Bottom tensile reinforcement
A h i f t i i l As hanger reinforcement is progressively introduced into each of the left and right shear spans the main bottom steel can be progressivelyspans, the main bottom steel can be progressively reduced provided proper anchorage requirements are maintained.
Example
Top tensile reinforcement
At th i ht h d t th t h i t l t l At the right hand support the top horizontal steel carries a force of 4992 kN requiring a steel area of:
3 Ast≥F*FG/( s fsy)=4992×103/(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 h l i i d dhanger steel is introduced.
Example