Lecture 29 Strut and tie modelingLecture 29 Strut-and-tie modeling

Dr Fang WANG

Senior Research Officer

M i R d WAMain Roads, WA

fang.wang@mainroads.wa.gov.au

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