RD-E: 4702 Splitting Tensile Test (Brazilian Test)

Using a splitting tensile test to calculate input for material LAW24.

A splitting tensile test is also known as a Brazilian test. It is a typical test used for concrete material characterization carried out using procedure adhering to ASTM D3967, “Standard Test Method for Splitting Tensile Strength of Intact Rock Core Specimens”. A load is applied to a concrete cylinder with its axis normal to the loading direction. A tensile stress develops in the center. The force is slowly increased until the specimen fails by an extension fracture along the loading plane. Then the tensile strength is computed from this force of failure.

The splitting tensile test aims to evaluate failure limits. It is difficult to apply uniaxial tension to a concrete specimen. Therefore, the tensile strength of the concrete material is determined by indirect test methods such as a Split Cylinder Test or even a Flexure Test. It should be noted that both methods give a higher value of tensile strength than the uniaxial tensile strength. This will be explained in the current example. This test will be modeled, and test results will be used for numerical calibration of the material law.

Options and Keywords Used

Concrete material law (/MAT/LAW24 (CONC))
Brick elements
Solid property (/PROP/TYPE14 (SOLID))
Boundary condition (/BCS)
Imposed displacement (/IMPDISP)

Input Files

The following input file is used in this example:

<install_directory>/hwsolvers/demos/radioss/example/47_concrete_test/Splitting_tensile_test/

Model Description

Since this test is quasi-static, a concrete cylinder is crushed with slow velocity. It is the standard test, to evaluate the tensile strength of concrete. This test could be performed following IS:5816-1970.

A cylinder of the concrete specimen is placed horizontally between the loading surfaces of a compression testing machine (Figure 2). The compression load is applied diametrically and uniformly along the length of the cylinder until the failure of the cylinder along the vertical diameter. A uniform distribution of the pressure load is created by using strips of plywood which are placed between the specimen and loading plates of the testing machine. This also reduces the magnitude of the high compressive stresses near the points of application of this load, concrete cylinders split into two halves along the vertical plane due to indirect tensile stress generated by Poisson's effect.

Assuming the concrete specimen behaves as an elastic body, a uniform lateral tensile stress acting along the vertical plane causes the failure of the specimen. This can be calculated from the following formula for the splitting tensile strength: (1)

f_{s t} = \frac{2 F_{\max}}{π L D}

Where,

$L$: Cylinder length
$D$: Diameter
$F_{\max}$: Ultimate force that leads to failure causing the specimen to split into two halves

The Radioss concrete material LAW24 is designed to work with only a few mandatory parameters. The other parameters are optional. If the optional parameters are not entered default values are calculated using typical properties of concrete material.

The numerical implementation is based on work by Han & Chen. ¹ It defines an Ottosen failure envelope. ² It can be fully determined by providing 4 failure points which are described with 5 parameters. The mandatory input is compression strength

f_{c}

. The four other ones are optional parameters. They are written as a ratio of compression strength which allows the following default values that are typical for concrete:

Direct Tensile Strength: $f_{t} = 0.05 f_{c}$
Biaxial Compression Strength: $f_{b} = 1.2 f_{c}$
Confined Compression Strength (tri-axial test): $f_{2} = 4.0 f_{c}$
Confined pressure: $s_{0} = 1.25 f_{c}$

Splitting Tensile Test

If only splitting tensile test data is available, then all of the failure parameters

f_{c}, f_{t}, f_{b}, f_{2}, s_{0}

are not available.

Experience provides that $f_{s t}$ is related to loading force $F$ on the cylinder. This value is sometimes used as an estimation of the direct tensile strength $f_{t}$ which is one of the input parameters. It can be observed that $f_{t} < f_{s t}$ .

Elastic theory enables to write: (2)

{\begin{matrix} σ_{1} (d) = \frac{2 F}{π L D} {(\frac{D^{2} - 4 d^{2}}{D^{2} - 4 d^{2}})}^{2} \\ σ_{2} (d) = - \frac{2 F}{π L D} {(\frac{4 D^{2}}{D^{2} - 4 d^{2}} - 1)}^{2} \end{matrix}

Where,

$F$: Loading force
$D$: Cylinder diameter
$L$: Length
$d$: Position on the diameter

This formula is maximized at the center, where

d

=0 to obtain: (3)

{\begin{matrix} σ_{1}^{\max} (t) = \frac{2 F (t)}{π L D} \\ σ_{2}^{\max} (t) = - 3 σ_{1}^{\max} \end{matrix}

The loading path direction and stress state in this test are different than a typical uniaxial tensile test. The material in this test undergoes both compression and tension. Compression is due to the loading force over the cylinder, and tension is due to the Poisson effect.

Another significant finding from this theoretical result is to observe that the direct tensile strength is lower than splitting tensile strength (Figure 4).

Fusco suggested for conventional concrete buildings the relation $f_{t} = 0.85 f_{s t}$ but other authors found the relation $f_{t} = 0. 66 f_{s t}$ . ⁴

Using the theoretical loading path from the elastic hypothesis with Han & Chen failure surface and the values for concrete:

$\frac{f_{t}}{f_{c}} = \frac{1}{10}$ ; $\frac{f_{b}}{f_{c}} = \frac{s_{0}}{f_{c}} = \frac{6}{5}$ ; $\frac{f_{2}}{f_{c}} = 4$

leads to the following estimation: (4)

f_{t} = \frac{1}{\sqrt{2}} f_{s t} \approx 0.71 f_{s t}

This estimation is of course dependent on the failure envelope which is shaped by all 5 parameters. Changing their values will change the $\frac{f_{s t}}{f_{t}}$ ratio.

Rigid Walls

Rigid walls are used to model the plywood used in the test to apply the load. Nodes on the cylinder the width of the plywood are included as secondary nodes of the rigid wall. A high friction value is set to prevent the concrete cylinder from sliding.

Loading Pressure

A compressive load is created by applying displacements in opposite directions to both rigid walls using the imposed displacement option /IMPDISP. The imposed displacement uses a linear function large enough to split the cylinders in half.

Solid Properties

q_a =1e-20 and q_b =1e-20
I_solid = 24
I_HKT = 2
All other property values use the default options

Material Data

Units: mm, ms, g, MPa

The concrete material data is:

Initial density = 0.0024 $[\frac{g}{m m^{3}}]$
Concrete elasticity Young’s modulus $E_{c} = 61000 [MPa]$
Poisson’s ratio $ν = 0. 17$
Concrete uniaxial compression strength from test $f_{c} = 58 [MPa]$
Concrete uniaxial tension strength is 0.05 $f_{c}$ so the ratio is defined as $\frac{f_{t}}{f_{c}} = 0.05$
All other parameters can be left as default in LAW24 because the default values are representative of generic concrete materials.

Some measurement from splitting tensile test data:

$\begin{array}{l} D = 150 m m \\ L = 300 m m \\ F_{\max} = 280100 N \\ f_{s t} = \frac{2 F_{\max}}{π L D} = 3.96 M P a \end{array}$

Starting with the nominal values:

Tensile strength $f_{t} = \frac{1}{\sqrt{2}} f_{s t} = \frac{1}{\sqrt{2}} \times 3 .96$ with $f_{c} = 58 \Rightarrow \frac{f_{t}}{f_{c}} = 0.0 5$
Biaxial Compression Strength: $f_{b} = 1.20 f_{c}$ by default.
Confined Compression Strength (tri-axial test): $f_{2} = 4 .00 f_{c}$ by default
Confined pressure: $s_{0} = 1.25 f_{c}$ by default This leads to the following material card input file.

Results

The maximum force measured from the rigid wall is 275612 N which is close to the maximum force from the test of 280100 N. The elastic theory equations at d=0 predict that the principal stress P1 at the center of the cylinder should be 3.90 MPa. The principal stress P2 at the center of the cylinder should be -11.7 MPa.(5)

{\begin{matrix} σ_{1}^{\max} (t) = \frac{2 F (t)}{π L D} \approx \frac{2 * 275612}{3.14 * 300 * 150} \approx 3.90 \\ σ_{2}^{\max} (t) = - 3 σ_{1}^{\max} \approx - 3 * 3.90 \approx - 11.70 \end{matrix}

The stress of the center elements, Element ID 39000 and 39520, show that the results match the analytical results of 3.96 MPa.

From the displacement plot, the Poisson's effect is shown at the top and bottom elements. The center of the cylinder is under tensile loading until the cylinder is cracked.

The numerical result is consistent with the theoretical solution. However, there is a slight deviation when approaching the failure limit due to the crack opening and damage in adjacent finite elements (nonlinearities). The next diagram shows the principal stress of some elements from the simulation.

Conclusion

The splitting tensile strength $f_{s t}$ measured from the Brazilian test is bigger than the tensile strength $f_{t}$ from a direct tensile test $f_{t} < f_{s t}$ . Using LAW24 with mostly default values, the tensile strength $f_{t}$ is about 0.71 times the splitting tensile strength $f_{s t}$ . Only the maximum force $F_{\max}$ from the splitting tensile test and $f_{c}$ from a cube compression test are needed as input to LAW24.

From a splitting tensile test, the parameter

\frac{f_{t}}{f_{c}}

in LAW24 could be determined. (6)

F_{\max}^{\exp e r i m e n t} - > f_{s t}^{a n l y t i c a l} - > f_{t}^{a n l y t i c a l} - > \frac{f_{t}}{f_{c}}

Next, compare the experiment and simulation force and strength.(7)

\begin{array}{l} F_{\max}^{\exp e r i m e n t} ~ F_{\max}^{s i m u l a t i o n} \\ f_{s t}^{a n l y t i c a l} ~ f_{s t}^{s i m u l a t i o n} \end{array}

¹ D.J. Han, W.F. Chen “Plasticity model for concrete in Mechanics of Materials”, North Holland

² Ottosen N.S. “Nonlinear Finite Element Analysis of Concrete Structures” Ris. National Laboratory DK 4000 Roskilde Denmark, May 1980

³ FUSCO, P.B. Concrete structures - Fundamentals of structural design. McGraw-Hill, 1976, São Paulo.

⁴ A. Ghaffar, M. A. Chaudhry and M. Kamran Ali - A new Approach for measurement of tensile strength of concrete - Journal of Research, Bahauddin Zakariya University, Vol.16, No.1, June 2005, pp. 01-0