It's a presentation about modeling and analysis using (Strut & Tie model) according to the Egyptian code of design (ECP 203-2007)

1. Strut-and-tie modeling
By : Khaled Sobhy Emam S.M.ASCE
Teaching Assistant – Faculty Of Engineering – Ain Shams University
Cairo – Egypt
Structural Design Engineer

2. Why Strut-and-Tie ?!

3. D-region :
◼ It’s the zone in the member at which the plane section doesn’t still plane
after loading , so the theory of elasticity isn’t valid in these zones .
Positions of D-regions :
1- Change in the geometry of a structural element .
2- Under a concentrated load .
3- Above the support .
D-region covers distance = t , from each side of the D-region .

5. ◼ In D-regions , a discontinuity of the stress distribution occurs .
◼ In D-regions , the load carrying mechanism may be idealized as a truss
made of concrete compression struts and steel ties .
◼ The choice of a strut-and-tie model is a major issue which may be
different from engineer to another for the same structure .
◼ Struts and ties are positions by considering the likely paths of the loads
to the supports and the orthogonal reinforcement pattern .
◼ The forces in the struts and ties are determined by equilibrium .

6. ◼ Strut-and-tie model consists of :
1- Major compression diagonals ( Struts ) .
2- Tension ties .
3- Truss nodes .

8. ◼ The modes of failure are :
1- Yielding in the tension tie .
2- Crushing in one strut when the stress exceeds the effective
compression strength of concrete .
3- The bearing capacity of any node of the truss isn’t sufficient .
If we don’t choose the suitable strut-and-tie model for the problem , the
model may require excessive deformation to reach the fully plastic state

9. ◼ Idealization of the strut divided into three types :
1- Prismatic strut The strut could be idealized as prismatic
compression members as straight lines .
Design of the struts

10. 2- Tapered strut If the effective compression strength at the
two ends of the strut differs due to bearing lengths ; so the strut is
idealized as a uniformly tapered compression member .

11. 3- Bottle-Shaped Sturt If the strut is located in a part of a
member where the width of the compressed concrete at mid-length of the
strut can spread laterally .
- In order to simplify design ; bottle-shaped strut are idealized either as
prismatic or as uniformly tapered .

12. Strength of un-reinforced struts ( Fc )
◼ It’s the smaller value of the compressive strength at the two ends as :
Fc = fcd x Ac
Ac = Cross sectional area of the strut at the strut end under consideration
Fcd = The smaller of :
(1) The effective compressive strength of the concrete in the strut ;
where ;
Fcd = 2/3 x 𝛽𝑠 x
𝑓𝑐𝑢
𝛾𝑐
; 𝛾𝑐 = 1.6

13. ◼ 𝛽𝑠 = It’s a factor which takes into account the stress conditions and the
angle of cracking surrounding the strut .

15. (2) The effective compressive strength of concrete in the nodal zone .
◼ If 𝛽𝑠 = 0.7 ; the axis of the strut shall be crossed by reinforcement
to resist the transfer tensile force resulting from compression
force spreading in the strut , otherwise … use 𝛽𝑠 = 0.6

16. ◼ According to ACI 318 , if fcu ≤ 40 Mpa ; the reinforcement is considered
to be satisfied if the axis of the strut being crossed by layers of
reinforcement satisfies the following equation :
σ
𝐴𝑠𝑖
𝑏𝑖 𝑠𝑖
sin 𝛼 ≥ 0.003
Where Asi = The total area of reinforcement @ spacing “ Si “ in a layer of
reinforcement with bars @ an angle “ 𝛼 " to the axis of the strut .

17. ◼ In order to increase the strength of the strut ; we can reinforce it with
compression reinforcement that satisfies the following conditions :
1- The compression reinforcement should be placed within the strut and
parallel to its axis .
2- The reinforcement should be properly anchored .
3- The reinforcement should be enclosed in ties or spirals satisfying the
conditions applied to column .
The strength of reinforced struts ( Fc ) :
Fc = 2/3 x 𝛽𝑠 x
𝑓𝑐𝑢
𝛾𝑐
x Ac + As x
𝑓y
𝛾𝑠
𝛾𝑠 = 1.3 & 𝛾c = 1.6

19. ◼ Strength of the tie ( Tud ) :
Tud = As x
𝑓𝑦
𝛾𝑠
Where ; Tud = Design tension force
As = Cross sectional area of steel 𝛾𝑠 = 1.15
◼ The width of the tie is determined to satisfy safety conditions for
compressive stresses at nodal points for struts and ties meeting at that
node …. The width should be ≤ 70% of the width of the largest strut
connected to the tie at the node .
Design of ties

20. ◼ Developed length of RFT .
Lbd = 2C + node width

21. ◼ The compressive strength of concrete of the nodal zone depends on
many factors :
1- The tensile strain from intersecting ties .
2- Confinement provided by compressive reactions .
3- confinement provided by transverse reinforcement .
Design of nodal zone

22. ◼ Types of nodal zone :
1) CCC
2) CCT
3) CTT
4) TTT

23. ◼ Strength of the nodal zone ( Fcu ) :
Fcu = 2/3
𝑓𝑐𝑢
𝛾𝑐
x 𝛽𝑛 x Acn
Acn = The area of the face of the nodal zone taken perpendicular to the
direction of strut .
𝛾𝑐 = 1.6
𝛽𝑛 = Depends on the type of nodal zone , where :
- CCC ….. 𝛽𝑛 = 1
- CCT …. 𝛽𝑛 = 0.8
- CTT & TTT …. 𝛽𝑛 = 0.6

24. ◼ In order to determine the height of the node ( U ) :
(1) In case of using one row of bars w/o sufficient development length :
U = zero

25. (2) In case of using one row of bars with sufficient development length :
U = ø + 2C

26. (3) In case of using multi-rows of bars with sufficient development length :
U = ø + 2C + ( n – 1 ) S
n = number of bars
S = spacing

27. As long as resulting model satisfies equilibrium and the struts
, ties and nodal zones satisfy the provisions ; the structure
should develop the ultimate strength required

28. ◼ For the following transfer girder , fcu = 30 Mpa , fy = 400 Mpa
Design this girder using strut-and-tie method
Example

29. Check bearing stress :
Fbu =
1500 𝑥 103
450 𝑥 500
= 6.667 Mpa
At column ; fce = 2/3 x 30/1.6 x 1 = 12.5 Mpa
At support ; fce = 2/3 x 30/1.6 x 0.8 = 10 Mpa
∴ fce > fbu …… SAFE

30. ◼ Establish strut-and-tie model
a
a

31. As … Fbc = Fad = Capacity
∴ 2/3 x fcu/𝛾𝑐 x 1 x b x ws = 2/3 x fcu/ 𝛾𝑐 x 0.8 x b x wt
∴ wt = 1.25 ws
Sec ( a – a )
Fbc x ( y ) = 1500 x 103 x 1900
2/3 x 30/1.6 x 1 x 500 x ws ( 2200 – ws/2 – 1.25ws/2 ) = 1500 x 103 x
1900

32. Ws = 235.7 mm ………. Wt = 295 mm
∴ Ws = 250 mm & Wt = 300 mm
Y = 1925 mm
Fbc = Fad = 1480.52 KN
σ 𝑥 = 0 …… Fab = 2107.6 KN
For tie …..
Fad = As x fy/𝛾𝑠 ……… As = 4256.5 mm2 ….. USE 9 ø 25

33. For Fab ….
From equilibrium … Fab = 2107.6
For AB :
Graphically ; wst = 495.9 mm
wsb = 531 mm
∴ Wst ≠ Wsb ….. Assume bottle-shaped strut
Fuab = 2/3 x 0.7 x 30/1.6 x 500 x 495.9 = 2180.36 KN > 2107.6 KN
SAFE

34. Provide min. web reinforcement , where
Min. vertical web RFT. … Asv = 0.0025 bs
Min. horizontal shear RFT … Ash = 0.0015 bs

35. ◼ Design of reinforced concrete structures … Prof. Mashhour Ghoneim
◼ ECP 203 - 2007
References

36. ◼ For contact : engkhaled_sobhy94@yahoo.com