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2081 Bhadra · 2 marks

Splitting tensile strength of cylinder (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.

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.

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).

✓ Answer checked by Claude Sonnet 5.5

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