- 2081 Bhadra · 4 marks
Platen restraint effect and failure modes
Short answer
Platen friction restrains lateral expansion, creating triaxial compression that increases measured strength. Cubes (h/d=1) are fully restrained, failing in an hour-glass shape. Cylinders (h/d=2) have a free central zone, failing by vertical splitting or shear. Lubricating platens reduces restraint.
Describe the platen restraint effect on concrete with typical failure modes of test specimen.
Platen restraint is the lateral confinement given to the ends of a concrete specimen by the steel platens of the testing machine. Friction between platen and concrete stops the specimen ends expanding sideways (Poisson's effect), putting the ends in triaxial compression, which raises the measured strength.
Cube (h/d = 1) Cylinder (h/d = 2)
======= platen ======= ======= platen =======
|\ /| |\ /|
| \ / | | \ / |
| \ / | | \ / |
| \ / | | | | |
| \ / | | | | |
| / \ | | | | |
| / \ | | | free | |
| / \ | | | zone | |
| / \ | | | | |
|/ \| | | | |
======= platen ======= | | | |
| / \ |
| / \ |
|/ \|
======= platen =======
Hour-glass (double cone) Cones at the ends, columnar
failure; whole cube is splitting or shear in the
restrained free central zone
(sketch not to scale)
- Cube: both ends are restrained, so the sides spall off and the remaining cone or pyramid shapes (hour-glass) are left. The restraint extends through the full cube, so a higher strength is read.
- Cylinder with h/d = 2: the central zone is free of restraint, so failure is by vertical splitting or a diagonal shear plane, and the strength is closer to uniaxial.
- Remedies: lubricate the platens (grease, rubber or PTFE sheets) to cut friction (specimens then split into columns and the strength is lower), use brush platens, and test the cube on its cast sides, not on the top surface.
- Effect: strength increases as h/d decreases; this is why cube strength is about 1.25 times cylinder strength.
✓ Answer checked by Claude Sonnet 5.5
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