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.

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Solve the Modulus of Rupture Given: Span length of the beam(L) = 18 inches Width of the beam(b) = 6 inches Depth of the beam(d) = 6 inches Maximum load applied(P) = 9950 lb

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If the specimen is originally square and has dimensions of a = 4 in., b = 3 in., and a thickness of t = 0.4 in., determine its new dimension a' after the load is applied. Express your answer to six significant figures and include appropriate units. a' = 4.00737 μA Submit Previous Answers Request Answer Part B X Incorrect; Try Again; 5 attempts remaining 0 in Determine its new dimension b' after the load is applied. Express your answer to six significant figures and include appropriate units. μᾶ b= 3.00333 ? in Submit Previous Answers Request Answer ? X Incorrect; Try Again; 5 attempts remaining

How do I find the specimen’s final length and diameter