# POINT LOAD STRENGTH INDEX TEST OF ROCK (IS-8764)

|## Aim

To determine point load strength index of rock as per IS-8764

## Equipments

- Point load strength test machine
- A 100 mm scale attached with the loading frame
- Pressure gauge (Capacity 25 kN or 50 kN)

Depending upon the size and shape of test specimen, the point load strength index can be conducted by four different methods. These are

- Diametral Test
- Axial Test
- Block Test
- Irregular lump Test

## 1. Diametral Test

### Procedure

- The diametral test is conducted on rock core sample. Minimum of 10 test specimens are required to find out the average value of point load strength index.
- This test can be conducted on the core specimens which are completely dry or after soaking it for 7 days.
- Measure the total length (
**l**) and diameter (**d**) of the core specimen. Specimen of**l/d=1.5**, are considered to be suitable for this test. - Place the specimen horizontally between two platens in such a way that the distance between the contact point and the nearest free end (
**L**) is at least 0.75times the diameter of the core (**d**). - Measure the distance between two platen contact points (
**D**) with the help of the scale attached with the loading frame. (Note-In case of diametral test, the diameter of the core (**d**) and the distance between two platens (**D**) will be same) - Apply load to the core specimen such that failure occur within 10-60 sec. record the failure load ‘
**P**‘.

### Calculation

Point load strength index (**I _{s}**) =

**(P*1000)/D**Mpa

^{2}Where **P** is breaking load in kN

**D** is the distance between platens in mm

Corresponding point load strength index for the standard core size of 50 mm (**I _{s50}**) diameter is given by the following equation

**I _{s50} = (P*1000)/(D^{1.5}**

**√50)**MPa

Uniaxial compressive strength of rock may be predicted from the following equation

**q _{c} = 22*I_{s50}** Mpa

### Report

The corrected mean value of the point load strength index **I _{s50}** is reported in Mpa.

## 2. Axial Test

### Procedure

- The axial test is conducted on rock core sample of small length. Minimum of 10 test specimens are required to find out the average value of point load strength index.
- This test can be conducted on the core specimens which are completely dry or after soaking it for 7 days.
- Measure the total length (
**l**) and diameter (**d**) of the core specimen. Specimen of**l/d**in between**0.3 to 1.0**, are considered to be suitable for this test. - Place the specimen vertically in between two platens.
- Measure the distance between two platen contact points (
**D**) with the help of the scale attached with the loading frame. Measure the specimen width (**W**) which is equal to the diameter of the core specimen (i.e.**d**). - Apply load to the core specimen such that failure occur within 10-60 sec. record the failure load ‘
**P**’.

### Calculation

Uncorrected Point load strength index (**I _{s}**) =

**(P*1000)/D**Mpa

_{e}^{2}Where **D _{e}**= Equivalent core diameter

**D _{e}^{2} = (4A)/**

**π**

And **A** is calculated using following equation

**A=W*D**

Where **W** is the specimen width in mm

**D** is the distance between platens in mm

**P** is the breaking load in kN

Corrected point load strength index for the standard core size of 50 mm (**I _{s50}**) diameter is given by the following equation

**I _{s50} = (P*1000)/(D_{e}^{1.5}**

**√50)**Mpa

Uniaxial compressive strength of rock may be predicted from the following equation

**q _{c} = 22*I_{s50}** Mpa

### Report

The corrected mean value of the point load strength index **I _{s50}** is reported in Mpa.

## 3. Block or Irregular Lump Test

### Procedure

- This method of test is conducted on rock block or irregular samples. Minimum of 10 test specimens are required to find out the average value of point load strength index.
- This test can be conducted on the core specimens which are completely dry or after soaking it for 7 days.
- Place the specimen horizontally between two platens.
- Measure the distance between two platen contact points (
**D**) with the help of the scale attached with the loading frame. Measure the smallest specimen width (**W**) perpendicular to the load direction. If the sides are not parallel , then ‘**W**’ is obtained from**W**,_{1}**W**and_{2}**W**as show in the figure and calculated as_{3}**W= (W**. Measure the distance between platen contact point and nearest free end (i.e._{1}+W_{2}+W_{3})/3**L**). The distance**L**should be at least**0.5D**. - Apply load to the core specimen such that failure occur within 10-60 sec. record the failure load ‘
**P**’.

### Calculation

Uncorrected Point load strength index (**I _{l}**) =

**(P*1000)/D**Mpa

_{e}^{2}Where **D _{e}**= Equivalent core diameter

**D _{e}^{2} = (4A)/**

**π**

And **A** is calculated using following equation

**A=W*D**

Where **W** is the specimen width in mm

**D** is the distance between platens in mm

**P** is the breaking load in kN

Corrected point load strength index for the standard core size of 50 mm (**I _{s50}**) diameter is given by the following equation

**I _{l50} = (P*1000)/[(D*W)^{0.75}**

**√50)]**Mpa

Uniaxial compressive strength of rock may be predicted from the following equation

**q _{c} = 15*I_{s50}** Mpa

### Report

The corrected mean value of the point load strength index **I _{s50}** is reported in Mpa.

## Note

See the figure below to find out which test is a valid test and which are invalid.

if we get the diameter 43mm, but in the formula the standard is 50mm so how to correct this please can you share any procedure…