Age | Strength percentage |
1 days | 16 % |
3 days | 40 % |
7 days | 65 % |
14 days | 90 % |
28 days | 99 % |
Longitudinal type 'e' in diagonal AC = (AC '- AC) / AC = (AC ’- AE) / AC = EC ’/ AC ----------- (1) When CE is specified from C to AC '. Since CC’s are very small, consider Angle ACB = angle ACB = 450 So EC '= CC'cos450 = CC' / √2 Long-term type ‘e’ = CC ’/ AC√2 = CC ’/ √2.BC.√2 = tan Φ / 2 = Φ / 2 = e S / 2 ------------------ (2) Where, Φ = CC ’/ BC represents the shear type (e S) For shear pressure τ and modulus of rigidity G, shear type (e S) = τ / G ------------ (3) Setting shear type (e S) = 2. Longitudinal type Longitudinal stiffness of diagonal AC = τ / 2 ----------- (4) Strain in a disconnected AC is also provided by = Strain due to the tensile stress in the AC - strain due to the compressive stress in the BD = τ / E - (–τ / m E) = τ / E (1 + 1 / m) ------------- (5) From equals (4) and (5), we obtain τ / 2G = τ / E (1 + 1 / m) or E = 2G (1 + 1 / m) or E = 2G (1 + µ) ------------- (6) Relationships between E, G and K: With reference to the (1) and (6) relationships obtained above, E = 2G (1 + µ) = 3K (1- 2 µ) Subtracting 1 / m from these two E expressions, we have E = 9KG / (G + 3K) Finally; E = 2G (1 + µ) = 3K (1 - 2 µ) or E = 9KG / (G + 3K) |