MB 2 TEST 2: SPLITTING TENSION TEST (BS EN 12390-6: 2009) Table 2. Result of Splitting Tension Test Cylinder Specimen Tensile Average Tensile Diameter,d Height, L Failure Load, F Age(Day) strength, f (mm) (mm) (kN) Number (MPa) Strength (MPa) si 28 100.00 201.00 135.10 4.28 S2 28 102.50 199.00 107.40 3.35 3.73 S3 28 105.00 200.00 117.50 3.56 Formula: 2 x F F = ct πxLxd Where Fa= Tensile splitting strength (MPa or N/mm?) F = Maximum load (N) L = Length of the line of contact of the specimen (mm) d = Designated cross-sectional dimension (mm) Specimen 1: F 2 x 135.10 x10 TI x 201.00 x 100.00 %3D = 4.28 MPa 2 x 107.40 x10³ TE x 199.00 x 102.50 Specimen 2: Fa %3D = 3.35 MPa Specimen 3: Ft 2 x 117.50 x10 TE x 200.00 x 105.00 = 3.56MPA 4.28 +3.35+3.56 Average tensile strength = 3.73 MPa %3D 3 Based on the result obtained above, explain in detail the effect of curing on the development of compressive strength.

MB 2 TEST 2: SPLITTING TENSION TEST (BS EN 12390-6: 2009) Table 2. Result of Splitting Tension Test Cylinder Specimen Tensile Average Tensile Diameter,d Height, L Failure Load, F Age(Day) strength, f (mm) (mm) (kN) Number (MPa) Strength (MPa) si 28 100.00 201.00 135.10 4.28 S2 28 102.50 199.00 107.40 3.35 3.73 S3 28 105.00 200.00 117.50 3.56 Formula: 2 x F F = ct πxLxd Where Fa= Tensile splitting strength (MPa or N/mm?) F = Maximum load (N) L = Length of the line of contact of the specimen (mm) d = Designated cross-sectional dimension (mm) Specimen 1: F 2 x 135.10 x10 TI x 201.00 x 100.00 %3D = 4.28 MPa 2 x 107.40 x10³ TE x 199.00 x 102.50 Specimen 2: Fa %3D = 3.35 MPa Specimen 3: Ft 2 x 117.50 x10 TE x 200.00 x 105.00 = 3.56MPA 4.28 +3.35+3.56 Average tensile strength = 3.73 MPa %3D 3 Based on the result obtained above, explain in detail the effect of curing on the development of compressive strength.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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