Computational Modeling of Concrete Structures

Gianluca Cusatis

The Lattice Discrete Particle Model for the Simulation ofReinforced Concrete Elements

Tokyo Institute of Technology

July 27, 2018

1 / 11 July-2018 :: Gianluca Cusatis

Presentation Outline

1 Simulation of Reinforced Concrete

2 / 11 Simulation of Reinforced Concrete July-2018 :: Gianluca Cusatis

Rebar model and its interaction with concrete

Rebar model

σs ={Esεs εs ≤ εyfy εs > εy

with εy = fy/Es

Rebar Concrete Interaction

f(t) = φ(u)v(t)

v(x, t) = vc(x, t)− vr(x, t)

φ(u) = Penalty Stiffness

3 / 11 Simulation of Reinforced Concrete July-2018 :: Gianluca Cusatis

Under-Reinforced Concrete Beam Behavior

0 5 10 15 20 250

5

10

15

20

25

30

35

Mid−span displacement [mm]

Mid−s

pan

mom

ent [

KN

.m]

Exp.fy=600 MPa

fy=540 MPa

−36−30−24−18−12 −6 0 60

4

8

12

16

20

24

28

Stress [MPa]

Dep

th [c

m]

UncrackedCrackedUltimate

-19.7

2.24

0 5 10 15 20 250

5

10

15

20

25

30

35

Mid Span Displacement [mm]

Mid

span

Mom

ent [

KN

.m]

ExperimentalSimulated

c) d)

a)

120 cm

280 cm

10 cm 10 cm

b)

-35.7

1.11

-3.41

3.36

Uncracked Cracked Yielding

Uncracked

Cracked Yielding

Reinforcement and Stirrup detail

Yielding

0.0

0.0 0.0

fy

fy

Figure: a) Geometry; b) moment-displacement curve; c) stress distribution atmid-span; and d) average stress distribution at mid-span.

4 / 11 Simulation of Reinforced Concrete July-2018 :: Gianluca Cusatis

Ductile Flexural Failure

BeamReinforcement Mn Pmin Pmax Psim

Top Bottom Stirrups [KNm] [KN] [KN] [KN]B1 (2) No. 3 (2) No. 3 No. 3 @ 150 mm 33.81 41.81 46.71 42.5B2 (3) No. 6 (2) No. 3 No. 3 @ 150 mm 158.43 194.83 218.85 208.17B5 (5) No. 8 (2) No. 3 No. 4 @ 150 mm 386.41 475.96 533.79 528B6 (5) No. 8 (3) No. 8 No. 4 @ 150 mm 419.57 515.99 579.60 536.16

5 / 11 Simulation of Reinforced Concrete July-2018 :: Gianluca Cusatis

Ductile Flexural Failure, Cont.

6 / 11 Simulation of Reinforced Concrete July-2018 :: Gianluca Cusatis

Ductile Flexural Failure, Cont.

Mid Span Displacement [mm]10 14 18 22 26 30 34

Mid

Spa

n Lo

ad [K

N]

500

525

550

575

600

625

1

2 3

4 5

6 7 8 9 10

1

2

3

4

5

6

7

8

9

10

Axial Strain [x10-3]

2.464 2.468 2.469 2.470

7 / 11 Simulation of Reinforced Concrete July-2018 :: Gianluca Cusatis

Over-Reinforced Concrete Beams

Beam L× b× h [mm] a [mm] As [mm2]Group IL11×15 4600×110×150 1960 400L11×30 4600×110×300 1960 800L11×60 4600×110×600 1900 1600Group IIS5×7 1150×55×75 480 100

M11×15 2300×110×150 960 400L22×30 4600×220×300 1900 1600Group IIIS11×15 1150×110×150 460 400M11×30 2300×110×300 930 800L11×60 4600×110×600 1900 1600

8 / 11 Simulation of Reinforced Concrete July-2018 :: Gianluca Cusatis

Over-Reinforced Concrete Beams, Cont.

c)

a)

b)

c) -1.0 -0.75 -0.50 -0.25 0.0 Principal Stress [MPa] Axial Force [N]

-24 80 160 240 320 395

Figure: a) Before failure but after the elastic stage; b) at the onset of failure(maximum load) showing concrete crushing; c) rebars axial force andcompressive principal stress distribution before failure showing the arch actionmechanism.

9 / 11 Simulation of Reinforced Concrete July-2018 :: Gianluca Cusatis

Over-Reinforced Concrete Beams, Cont.

with d = h, µ = 3Pua/(bd2), µ = µ0(1 + d/D0)

−1/2

10 / 11 Simulation of Reinforced Concrete July-2018 :: Gianluca Cusatis

Over-Reinforced Concrete Beams, Cont.

Group BeamPu [KN] (δPu [%]) Pred. Error [%] µu [MPa]

Exp. LDPM ACI318 LDPM ACI318 Exp. LDPM ACI318

IL11×15 12.4 (19.4) 10.9 (1.8) 10.9 -11.4 -11.9 44.4 39.3 39.1L11×30 50.8 (6.2) 39 (2.9) 43.6 -23.1 -14.2 45.6 35 39.1L11×60 159.1 (11.1) 142.4 (1.5) 179.8 -10.5 9.6 34.6 31 39.1

IIS5×7 5.6 (20.8) 6.1 (3.3) 5.6 9.8 -0.2 39.2 43 39.1

M11×15 22.8 (7.9) 22.6 (1.5) 22.2 -1 -2.4 40.1 39.7 39.1L22×30 84.2 (2.2) 85.8 (1.3) 89.9 1.9 6.7 36.7 37.4 39.1

IIIS11×15 52.8 (44.3) 65.5 (3.7) 46.4 24 -12.2 44.5 55.2 39.1M11×30 101.9 (9.1) 92.4 (1) 91.8 -9.3 -9.9 43.4 39.3 39.1L11×60 159.1 (11.1) 142.4 (1.5) 179.8 -10.5 9.6 34.6 31 39.1

Groupµ0 [MPa] D0 [mm]

Exp. Num. Exp. Num.I 55.6 43.4 324 501II 44.1 44.6 540 553III 80.8 87.7 97 68

µ = µ0(1 + d/D0)−1/2

11 / 11 Simulation of Reinforced Concrete July-2018 :: Gianluca Cusatis