36

CHAPTER 3

SIMPLIFIED APPROACH FOR FINITE ELEMENT MODELLING

OF LACED REINFORCED CONCRETE

3.1 INTRODUCTION

Blast loading is different from other loadings not only by the

way it loads the structure but also due to its transient nature. Blast loading is

impulsive, not simply dynamic. Peak pressures are much higher than the

static collapse load of the structure, but their durations generally are

extremely short compared to natural periods of structures and structural

components. The structure has to be designed as a flexible system,

permitting their joints to deform considerably. The stresses in the elements

must be allowed to go beyond the elastic limit, so that the available

deformability in the post-yield region is fully utilised. This approach is

known as the elasto-plastic design approach.

Structural properties of reinforced concrete (RC) elements can be

improved by modifying the concrete matrix by adding fibres and/or by

suitably detailing the reinforcements. Laced Reinforced Concrete (LRC)

consists of continuous bent shear lacings along with longitudinal

reinforcements on both faces of a structural element. Typical detailing of

LRC structural component is shown in Figure 3.1. LRC enhances the

ductility and provides better concrete confinement. Moreover, LRC

38

for obtaining equivalent stresses and strains for various percentages of

longitudinal tensile steel reinforcement is also presented.

3.2 MATERIAL BEHAVIOUR

Resistance-support rotation curve (TM5-1300 Manual) shown in

Figure 3.2 demonstrates the flexural action of a reinforced concrete

element. Initially, the resistance increases with the deflection until yielding

of the reinforcement takes place. In normal concrete elements, at a

deflection corresponding to 1 support rotation, the compression concrete

crushes. For elements with marginal shear reinforcement i.e. without

confinement of concrete, this crushing of the concrete results in failure of

the element. For element with adequate shear reinforcement, either single

leg stirrup or lacing, after yielding of tensile steel, there is a strain

hardening region. Since the reinforcement has become plastic, any further

deflection, can therefore, occur without any further resistance, i.e., the

resistance is constant with increasing deflection.

Figure 3.2 Typical Resistance-Support Rotation curve

for flexural RC elements

1º 2º 4ºSupport Rotation

AB C

39

At 2 degrees support rotation, the element with single legged

stirrups loses its structural integrity and fails. On the other hand, lacing

through its truss action will restrain the reinforcement through its entire

strain hardening region until tension failure of the reinforcement occurs.

While the tri-army manual (TM5-1300 manual) suggests a plastic support

rotation capacity of 12o, the tests conducted by Parameswaran et al (1986)

showed that it varied between 6o to 8o. The results of the above

investigations suggested that a plastic hinge rotation of 4o at end supports

and 8o at all other plastic hinge locations in continuous construction. The

continuous lacings are normally inclined at 45º and 60º to horizontal. The

significance of shear resistance in enhancing the ductility of a flexure

element can be observed. A sudden shear failure is obvious in the event of

inadequate capacity. However, studies have shown that under cyclic loading

the failure rotation is reduced and hence a 4o support rotation has been

suggested for design purposes. LRC construction has equal tension and

compression reinforcement to effectively counter the blast effect.

Continuous inclined transverse reinforcement, in addition to enhancing the

shear behaviour also arrests fragmentation of concrete and reduces flying

debris caused by an external explosion.

3.3 MOMENT-CURVATURE RELATIONSHIP

Moment-curvature relationship for the beam element is

generated using well established procedures using the idealised stress-strain

characteristics of concrete and steel as given in IS 456:2000 and shown in

Figures 3.3(a) and (b) respectively. The modulus of elasticity of concrete Ec

ckf5000for computation of modular ratio is taken as , where fck is the

characteristic cube compressive strength of concrete.

40

(a) Concrete (b) Reinforcement Steel

Figure 3.3 Stress-strain behaviour

The moments and corresponding curvatures are calculated for

the stages shown in Figure 3.4 and mentioned below for all the elements:

Figure 3.4 Moment-curvature relationship

Stage 1. When the concrete in tension starts cracking, i.e., strain in

concrete reaches strain corresponding to a stress of ckf7.0

(Mcr and cr

fY

Strain0.002 0.0035

0.67fck

Strain

)

Mcr

cr 0.8fy fy c u

M0.8fy

Mfy

Mu

Curvature

41

Stage 2. When the strain in steel reaches the strain corresponding to

the stress of 0.8 times fy, where fy is the yield stress of the

reinforcement steel (M0.8fy and 0.8fy

Stage 3. When the strain in steel reaches the strain corresponding to

the stress of f

)

y (Mfy and fy

Stage 4. When the strain in concrete in the extreme compressive fibre

reaches a value of 0.0035 (M

)

c and c

Stage 5. Ultimate stage (M

)

u and u)

Until cracking (stage 1), the transformed moment of inertia, ‘I t’,

is used for calculating the moment and curvature. The cracking moment

‘Mcr

2D

IfM tcr

cr

’ for the beam element is given by

(3.1)

where

crf = modulus of rupture = 0.7 ckf ,

ckf = characteristic cube compressive strength of concrete,

tI = transformed moment of inertia,

and D = overall depth

The curvature, cr is given by

tc

crcr IE

M(3.2)

where

cE = modulus of Elasticity of concrete = 5000 ckf

The values for the moments and the corresponding curvatures at

other stages (2, 3 and 4) are computed using linearity of strain profile across

42

the section and the equilibrium of tensile and compressive forces. Concrete

below neutral axis is neglected.

Moment is obtained by the following equation:

dTjM (3.3)

where

M = moment at any stage,

T = total tensile force = C,

C = total compressive force,

andd

j = lever arm

= distance between centres of compression and

tension forces

The curvature is calculated using the relation,

dxscccc (3.4)

where

d = effective depth,

x = depth of neutral axis,

cc = strain in extreme concrete fibre,

and s = strain in steel

Beyond stage 4, a hinge is formed at mid span and the beam

starts to rotate. Therefore, there is no increase in load carrying capacity of

the beam, but deflection is relatively high. Curvature at failure is restricted

to 4º support rotation, based on the experimental results. Failure curvature,

u is taken as least of

43

c

se

c

csseceu d

2d

2d

(3.5)

where

ce = failure compressive strain of concrete including

the effect of confinement,

se = average failure tensile strain in steel,

cs = failure strain of compression steel with heavy web

steel,

and cd = distance between centre of gravity of tension and

compression steel

se is computed from

p

gpslse l2

ll2(3.6)

where

sl = percentage elongation over the gauge length at

failure,

pl = length of plastic hinge taken as equal to effective

depth, ‘d’,

gl = gauge length,

and = average percentage elongation outside the gauge

length at failure

Angle of lacing has an influence on the strain in compression

steel. When the lacing angle is more, length of compression reinforcement

between the transverse rods is less reducing the probability of buckling of

reinforcement and ensuring plastification under compression.

44

3.4 EQUIVALENT STRESS-STRAIN RELATIONSHIP FOR

LRC FLEXURAL ELEMENTS

Multi-linear moment-curvature relationship as shown in

Figure 3.4 can easily be obtained based on sectional properties. Most of the

finite element analysis including non-linear analysis requires definition of a

non-linear stress-strain variation for the material used. Conventionally, steel

and concrete are modeled as separate entities and their material models are

used directly. This leads to complexities in modeling and analysis of these

reinforced concrete structural elements. In this study, a single material

model using an equivalent stress-strain relationship for LRC beams is

proposed and is derived from the moment–curvature relationship as shown

in Figure 3.5.

Figure 3.5 Derivation of equivalent stress-strain curve

b – breadth D – depthAs – area of tension reinforcementAsc – area of compression reinforcementfy – yield stress of steelfck – concrete Cube strengthM – moment

– curvature

Sectional Properties(D, Ast, Asc, fy, fck)

Moment – curvature relationship (M- curve)

Equivalent stress-strain curve of the - curve)

RC Beam Theory

– stress– strain

k – factordi – depth where strain is equal to ii – stage

Dd

kMM

dDd

1dD

D1

12dD

bM

MMM

1i1i

'1i

1i

1i1i

1ii

21i

2'i

'1i

'ii

2D

ii

From the above relations, i is evaluated

45

Procedure for deriving equivalent stress-strain relationship for

characteristic strength of concrete, fck = 25 MPa and yield strength of

reinforcement steel, fy = 415 MPa from the moment-curvature relationship

obtained for various stages mentioned in the previous section is described

below.

The stress-strain characteristics are hypothesised to be multi-

linear. The condition to be satisfied by the equivalent stress-strain

relationship is,

D2

deitc (3.7)

where

c = compressive strain in concrete,

t = strain at level of tensile steel,

and ei = equivalent strain

Equivalent stress at first stage corresponding to concrete

cracking (Figure 3.6(a)), 1 is obtained by

2cr

1bDM6 (3.8)

where

b = width of the section,

and crM = cracking moment

Equivalent stress at second stage mentioned in previous section,

2 is obtained as follows:

Moment in second stage is M2

'crM

. This moment is summation of

moment due to portion 1 ( ) and moment due to portion 2 ( '2M ) as

shown in Figure 3.6(b).

46

'2

'cr2 MMM (3.9)

where'crM = moment corresponding to portion 1 of cross-

section where strain reaches 1 ,

'2M = moment corresponding to portion 2

Figure 3.6 Stress and strain distribution across depth

'crM is evaluated to be

21

cr'cr D

dMM (3.10)

where

1d is depth of cross-section where strain reaches 1 as shown in

Figure 3.6(b).

Taking into consideration the nonlinear variation of stress, '2M is

evaluated to be

(a) 1st stage (c) Ultimate Stage

(d) Stage ‘i’(b) 2nd Stage

d1D

2

cr

1

cr

Beam cross-section

Strain variation Stress variation

Portion 2

Portion 2

Portion 1

2

cr

i

1i

i i

1i

i

1i

Strain variation Stress variation

D

1 cr 1 cr

cr

Beam cross-section

Strain variation Stress variation

u

u u

u

Strain variation Stress variation

D

cr

di-1

47

1

11

12

21

2'2 dD

d1dD

D112

dDbM (3.11)

Equations (3.10) and (3.11) are substituted in Equation (3.9) and

1 is obtained.

Equivalent stress at ith stage is obtained by extending the above

procedure. Moment corresponding to ith stage can be expressed as'i

'1ii MMM (3.12)

where'

1iM = moment corresponding to portion 1 of

cross-section where strain reaches i 1 ,

and 'iM = moment corresponding to portion 2

Moment '1iM is evaluated as

21i

1i'

1i DdkMM (3.13)

where ‘k’ is a factor to account for nonlinear variation of stress across the

cross-section. Value of ‘k’ varies between 1 for second stage and 0.67 for

ultimate stage corresponding to fully plastic (Figure 3.6(c)). Therefore, ‘k’

is evaluated as

1k for second stage and,

1i

21i

MbDk for all other stages (3.14)

Moment 'iM is evaluated to be

1i

11i

1ii

21i

2'i dD

d1dD

D112

dDbM (3.15)

where

48

1id = depth where strain is equal to 1i as shown in Figure 3.6(d),

and 1i = stress at stage ‘i-1’

Substituting Equations (3.13) and (3.15) in Equation (3.12),

stress at stage ‘i’, i is obtained.

Equivalent strain is obtained by

2Diei (3.16)

where

ei = equivalent strain at stage ‘i’,

and i = curvature at stage ‘i’ given by Equation (3.4)

The equivalent stress corresponding to various percentages of

longitudinal tensile steel is computed, and a regression analysis is carried

out to determine the equivalent stresses and strains for various percentages

of steel at five stages mentioned in the earlier section and given in

Table 3.1.

Table 3.1 Expressions for equivalent stress and strain

Stage Equivalent stress, N/mm Equivalent strain2

1 0.6533pt 0.00014+3.52 13.196 p t 0.0003 p-0.4781 t+0.00133 13.476 p t 0.0003 p+0.6949 t+0.00324 13.476 p t 0.0091p+1.6949 t

-0.3131

5 13.476 p t 0.035+2.0p t maxtmin ppp– percentage of steel,

Using the equations given in Table 3.1, equivalent stress-strain

curves for various percentages of longitudinal tensile steel are plotted as

49

shown in Figure 3.7. The dotted line corresponds to an equivalent stress-

strain for beam A2 having cross section of 150 mm x 300 mm and p t=1.34.

These curves can be used for unconfined and normally confined concrete by

restricting the maximum value of strain corresponding to confined concrete.

Figure 3.7 Equivalent stress-strain curve for various percentages

of tension steel

3.5 NUMERICAL STUDIES

Details of LRC beams, which were tested under static loading

are given in Table 3.2. Dimensions of the beams are 150 mm x 300 mm.

Simply supported span of the beams A2, C1, D1 and E1 are 2.8 m, while

beams LRC/8/60 and LRC/10/60 have a simply supported span of 2.7 m.

Typical arrangement of test set-up is shown in Figure 3.8.

LRC beams are modelled using beam elements (BEAM23) in

ANSYS. Equivalent stress-strain relationship of LRC beams as explained in

the previous section are obtained for each beam. Inelastic isotropic multi-

0

10

20

30

40

50

60

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035

Strain

pt=0.5

unconfined concrete

confined concrete

LRC

pt=1.0

pt=1.5

pt=2.0

pt=2.5

pt=3.0

pt=3.5

pt=4.0

pt=1.34

50

linear material model is used in analysis. Two point loading is applied on

the model. Nonlinear analysis with 10000 load steps is carried out.

Deflection at the center of the beam is obtained and load-deflection curve is

plotted.

Table 3.2 Details of LRC beams tested under static load

BeamTop

Longitudinal reinforcement

Bottom Longitudinal reinforcement

Lacing reinforcement

A2 3 16 mm 3 16 mm 45 lacing 8mm bar (2pairs)

C1 3 20 mm 3 20 mm 60 lacing 8mm (1 Pair) plus 10 mm (single)

D1 2 25 mm +2 16 mm

2 25 mm +2 16 mm

60 lacing 12 mm bar (2 pairs)

E1 3 12 mm 3 20 mm 60 lacing 8mm (1 Pair) plus 10 mm (single)

LRC/8/60 2 25 mm 2 25 mm 60 lacing 8 mm

LRC/10/60 2 25 mm 2 25 mm 60 lacing 10 mm

Six beams namely, beams A2, C1, D1, E1, LRC/8/60 and

LRC/10/60, are analysed. In beam A2, 45 lacing scheme is used, while in

other beams, 60 is used. In beam E1, areas of compression and tension

reinforcement are different. Thus, the proposed equivalent stress-strain

model is applied for different conditions. Comparison of load-deflection

curve of experiments with that of finite element analysis for beams A2, C1,

D1, E1, LRC/8/60 and LRC/10/60 are shown in Figure 3.9.

51

Figure 3.8 Arrangement of test setup for LRC beams

Figure 3.9 Load-deflection response of LRC beams

It can be seen that results are found to be in good agreement and

thus validating the proposed approach of modeling LRC structural element

with equivalent stress-strain curve. Peak load and deflections obtained from

the present study and experimental study are tabulated in Table 3.3. The

deformation predicted are in the range of L/30 to L/10. At such high

deformation, failure strains particularly in concrete, the effects of cover

Span

CONNECTED TO HYDRAULIC LOADING PLANT

JACK

100

LOAD CELL

Loading pointLoading point

LRC beam

SupportSupport

0

50

100

150

200

250

300

350

400

0 50 100 150 200 250

Deflection at centre, mm

Beam A2 - Exp.Beam A2 - Num.Beam C1 - Exp.Beam C1 - Num.Beam D1 - Exp.Beam D1 - Num.Beam E1 - Exp.Beam E1 - Num.Beam LRC/8/60 -Exp.Beam LRC/8/60 -Num.Beam LRC/10/60 -Exp.Beam LRC/10/60 -Num.

52

have a pronounced influence. For the purposes of design, after obtaining the

failure deformation, a suitable factor can be used to avoid collapse.

Table 3.3 Peak loads and deflections

BeamPeak load (kN) Yield deflection

(mm)Ultimate deflection

(mm)Num. Exp. Num. Exp. Num. Exp.

A2 159.49 160.00 30.90 30.40 248.55 170.00

C1 163.73 165.00 25.45 29.20 204.88 210.00

D1 250.00 252.08 31.40 26.55 171.59 170.00

E1 185.00 190.00 36.05 28.00 195.00 106.00

LRC/8/60 347.41 363.00 18.55 18.00 128.74 62.00*

LRC/10/60 340.21 350.00 19.76 17.00 123.23 57.00*

*Experiment stopped due to limitation in test set-up

Ductility factors are calculated and are tabulated as shown in

Table 3.4. As seen from the results, the proposed simplified material model

is able to predict the peak load and ductility factors for LRC beams. It may

be noted that for beams LRC/8/60 and LRC/10/60, the ductility factors

from the experimental study are 3.444 and 3.353 respectively, while that

due to present study is 6.94 and 6.23. This is due to the fact that, the

experiments were stopped due to limitations in the test set-up (hydraulic

jack has reached its maximum capacity) and ultimate deflection was not

reached. For beams C1, D1 and E1, the maximum rotation is restricted to 4°

in present study, while during experiment it would have gone slightly

beyond this value. For beam E1, there is difference in areas of tension and

compression reinforcements. This may be the reason for a lower ductility

factor predicted by the present study.

53

Table 3.4 Ductility factors of LRC beams

BeamDuctility factor

Numerical ExperimentalA2 8.04 5.59

C1 8.05 7.71

D1 5.47 6.40

E1 5.41 3.79

LRC/8/60 6.94 3.44*

LRC/10/60 6.23 3.35**Experiment stopped due to limitation in test set-up

Variation of strain at mid-span in concrete surface and tension

steel with moment are plotted as shown in Figure 3.10. From this, strain

profile for various stages of loading, namely, 60 kN, 100 kN and 165 kN for

beam C1 are plotted and compared with measured values of strain during

experiments. As it can be seen from Figures 3.11 and 3.12, the proposed

model is able to predict the values of strain in concrete surface and tension

reinforcement to an acceptable level of accuracy. Crack patterns plotted

from the tested beams are shown in Figure 3.13 for beams A2, D1,

LRC/8/60 and LRC/10/60.

Figure 3.10 Variation of strain in Beam C1

0

20

40

60

80

0 0.002 0.004 0.006 0.008 0.01

Strain

Strain in Concrete surfaceStrain in tension reinforcement

54

Figure 3.11 Average strain profile for concrete surface strain of

beam C1

Figure 3.12 Average Strain profile for tension reinforcement strain of

beam C1

0

1000

2000

3000

4000

5000

0 400 800 1200 1600 2000 2400 2800

Span, mmC S i

60 kN - Exp.60 kN - Num.100 kN - Exp.100 kN - Num.165 kN - Exp.165 kN - Num.

0

1000

2000

3000

4000

5000

0 400 800 1200 1600 2000 2400 2800Span, mm

60 kN - Exp.60 kN - Num.100 kN - Exp.100 kN - Num.165 kN - Exp.165 kN - Num.

55

(a) Crack pattern on Beam – D1

(b) Crack pattern on Beam – A2

(c) Crack pattern on Beam LRC/8/60

(d ) Crack pattern on Beam LRC/10/60

Figure 3.13 Crack patterns observed in beams during testing

3.6 APPLICATION OF THE PROPOSED MODEL TO LRC

SLAB

Proposed model is extended for application to LRC slab. Slab is

of dimension 3 m x 2m and of thickness 100 mm. Main reinforcement

consists of 12 mm bars spaced at 80 mm c/c in shorter direction and 8

mm bars spaced at 80 mm c/c in longer direction. 8 mm lacings are

provided at a spacing of 80 mm c/c. Lacings run along shorter direction for

a width of 1.5 m and along the longer direction for a width of 0.75 m at

either ends. Reinforcement details in the LRC slab is shown in Figure 3.14.

56

Slab is simply supported on all four sides and subjected to uniform pressure

loading.

Figure 3.14 Reinforcement details in LRC slab

Equivalent stress-strain relationship is calculated per meter width

of the slab. Plate elements are used to model the LRC slab. Nonlinear

analysis is carried out. Load-deflection behaviour at the centre of the slab is

plotted along with experimental results as shown in Figure 3.15.

Experimental results are in good agreement with the finite element results.

Comparison of deflection profiles along short span for various load steps

are shown in Figure 3.16, while that for long span are shown in Figure 3.17.

Strain in slab is plotted in Figure 3.18 for various stages of loading.

Formation of yield line can be seen from the contour plot. From the

comparison of numerical and experimental results, this model can be

adopted for analysis of structures / structural components made of LRC.

57

(a) at midspan

(b) at quarter point* Experiment was stopped at this stage due to limitation in test set-up

Figure 3.15 Load-deflection behaviour for LRC slab

0

200

400

600

800

1000

0 50 100 150 200Deflection, mm

Present study

Experimenta l

0

200

400

600

800

1000

0 10 20 30 40 50 60Deflection, mm

Present Study

Experimenta l

*

*

Extrapolated from experimental data

58

*Beyond 800 kN only central dial gauge was available

Figure 3.16 Deflection profiles along short span

*Beyond 800 kN only central dial gauge was available

Figure 3.17 Deflection profiles along long span

-80

-60

-40

-20

00 500 1000 1500 2000

Span, mm

584 kN - Exp. 608 kN - Num.752 kN - Exp. 728 kN - Num.790 kN - Exp. 788 kN - Num.829 kN - Exp. 799 kN - Num.900 kN - Exp. 817 kN - Num.

-80

-60

-40

-20

00 500 1000 1500 2000 2500 3000

Span, mm

584 kN - Exp. 608 kN - Num.752 kN - Exp. 728 kN - Num.790 kN - Exp. 788 kN - Num.829 kN - Exp. 799 kN - Num.900 kN - Exp. 817 kN - Num.

59

Load = 100 kN Load = 250 kN

Load = 532 kN Load = 824 kN

Figure 3.18 Strain in Y-direction at various load steps for LRC slab

3.7 SUMMARY

This chapter presents a simplified approach to solve problems of

equally reinforced RC / LRC structural elements subjected to flexure.

Equations for obtaining the equivalent stress and strain characteristics for

equally reinforced LRC beams under flexure have been derived retaining

the moment-curvature characteristics. The proposed approach and

simplified material model using equivalent stress-strain behaviour is able to

predict the peak load and ductility factors satisfactorily for LRC beams. The

above stress-strain characteristics can be easily adopted in any finite

element software using multi-linear inelastic isotropic material model. The

proposed approach is extended for solving a LRC slab subjected to uniform

60

distributed loading. The model can be used for ordinary reinforced concrete

by modifying the ultimate stress and strain values. This model can be

extended for steel-concrete composite flexural elements.