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Tensile strength tests of concrete

Short answer

Tensile strength is 7–12% of compressive strength and controls cracking, shear, and bond. The splitting tension test on a cylinder gives fsp = 2P/(pi L d) and is about 0.5–0.6 sqrt(fck). The flexure test on a beam (modulus of rupture) gives fcr = Pl/(b d^2) for third-point loading; IS 456 gives fcr = 0.7 sqrt(fck) MPa. Direct tension is lower and rarely used.

Describe in detail the tensile strength tests of concrete (also describe the importance of tensile strength).

Importance of tensile strength

Tensile strength of concrete is only about 7 to 12% of its compressive strength, but it controls cracking, shear, bond and the design of pavements, water tanks and prestressed members. Concrete is not designed to carry tension, but the tensile strength decides when the first crack appears. It is needed for crack control, shear and bond design, pavement thickness, water-retaining structures (IS 3370) and prestress release.

Direct tension test

A dog-bone or cylindrical specimen with embedded end grips is pulled axially. It is rarely used, because it is hard to avoid eccentricity and secondary stresses from the grips, and the result scatters. Direct tensile strength is about 0.35fck0.35\sqrt{f_{ck}} to 0.45fck0.45\sqrt{f_{ck}} (MPa).

Splitting tension test (Brazilian test, IS 5816)

A standard cylinder (150 mm dia × 300 mm long) is laid horizontally between the platens of the testing machine, with thin plywood or fibre strips (about 3 mm thick and 15 to 25 mm wide) along the contact lines. The compressive line load is increased at a rate of 1.2 to 2.4 N/mm2^2/min (splitting tensile stress) until the cylinder splits along the vertical diameter. The load applied gives a nearly uniform tensile stress across the loaded plane:

fsp=2PπLdf_{sp} = \frac{2P}{\pi L d}

where PP = failure load, LL = length and dd = diameter of cylinder. Splitting strength is usually a little higher than the direct tensile strength and is about 0.5fck0.5\sqrt{f_{ck}} to 0.6fck0.6\sqrt{f_{ck}} (MPa).

Advantages: simple, uses the same cylinders and compression machine as the compression test, gives a more uniform and reproducible result than direct tension, with no grip problem.

Flexure test (modulus of rupture, IS 516)

A plain concrete beam 150 × 150 × 700 mm is supported on a 600 mm span and loaded at the third points (two-point loading) at 0.7 N/mm2^2/min extreme-fibre stress rate. If the crack occurs in the middle third:

fcr=P lb d2f_{cr} = \frac{P\,l}{b\,d^{2}}

If the crack is outside the middle third but within 5% of the span from the load point, fcr=3Pa/(bd2)f_{cr} = 3Pa/(bd^2) (aa = distance of crack from the nearer support); otherwise the result is rejected. For single (centre) point loading, fcr=3Pl/(2bd2)f_{cr} = 3Pl/(2bd^2). IS 456 (Cl. 6.2.2) gives the flexural strength fcr=0.7fckf_{cr} = 0.7\sqrt{f_{ck}} N/mm2^2. The modulus of rupture is greater than direct tension (by about 1.5 to 2 times) since the stress distribution is assumed linear while the actual stress block is plastic.

The three results are related: direct tension < splitting < flexural.

✓ Answer checked by Claude Sonnet 5.5

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