---
title: "Splitting tensile strength of cylinder (400 kN)"
subject: "Concrete Technology and Masonry Structure"
chapter: 5
key: "semester_5/concrete_technology_and_masonry_structure:splitting-tensile-cylinder-400kn"
source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/splitting-tensile-cylinder-400kn
site: BCT Engineering
---

# Splitting tensile strength of cylinder (400 kN)

Concrete Technology and Masonry Structure · Chapter 5: Testing of concrete and quality control

- Asked in: 2081 Bhadra, 2 marks
- Video: [Split Tensile Strength](https://www.youtube.com/watch?v=VKqE_WTtBUc) — Concrete Technology LAB
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/splitting-tensile-cylinder-400kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-5

## Question

Calculate the splitting tensile strength of a concrete cylinder 200 mm $\times$ 150 mm diameter under the standard splitting test, if the load shown by the testing machine is 400 kN.

## Short answer

Splitting tensile strength f_sp = 2P/(pi L d). With P = 400 kN = 400000 N, L = 200 mm, d = 150 mm, f_sp = 2*400000/(pi*200*150) = 8.49 N/mm^2.

## Answer

**Formula (IS 5816):** $f_{sp} = \dfrac{2P}{\pi L d}$

Given $P = 400$ kN $= 400\,000$ N, length $L = 200$ mm, diameter $d = 150$ mm.

$$
f_{sp} = \frac{2\times400000}{\pi\times200\times150} = \frac{800000}{94247.8} = 8.49\ \text{N/mm}^2
$$

**Answer: splitting tensile strength = 8.49 N/mm$^2$ (MPa).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which formula is used for the splitting tensile strength?
   A. P/(L d)
   B. 2P/(pi d^2)
   C. Pl/(bd^2)
   D. 2P/(pi L d)
   Answer: D. fsp = 2P/(pi L d). (Key point: Formula)

2. What is the splitting tensile strength for P = 400 kN, L = 200 mm, d = 150 mm?
   A. 8.49 N/mm2
   B. 4.24 N/mm2
   C. 16.98 N/mm2
   D. 2.12 N/mm2
   Answer: A. 2 x 400000/(pi x 200 x 150) = 8.49 N/mm2. (Key point: Result)

3. What load value must be substituted so that stress comes in N/mm2?
   A. 400 N
   B. 400000 N
   C. 40000 N
   D. 400 kg
   Answer: B. Convert kN to N and use dimensions in mm. (Key point: Units)

## Close variants set in other exams

- [Splitting tensile strength of cylinder and cube (500 kN)](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Asplitting-tensile-cylinder-cube-500kn) — Concrete Technology and Masonry Structure, chapter 5, asked 2×

## Related in Concrete Technology and Masonry Structure

- [Splitting tensile test and its advantage](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Asplitting-test-advantage) — Concrete Technology and Masonry Structure, chapter 5
- [Tensile strength tests of concrete](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Atensile-strength-tests) — Concrete Technology and Masonry Structure, chapter 5, asked 2×
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- [Corner stresses of 400 x 800 column (200 kN)](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Acolumn-stress-400x800-200kn) — Concrete Technology and Masonry Structure, chapter 8, asked 7×

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