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Coastal Engineering: Theory and Practice
By solving this équation we get, dl — 3.15 m
D = dl + a - 3.15 + 2.18 = 5.33 « 5.4 m
Increase it by 20%-40%
Z)actuai — 5-4 x 1.4 (assuming as 40%)
■^actual = 7.56 m
-Dactuai = 5.4 x 1.3 (assuming as 30%)
■^actual = 7.02 m
Finding T :
Rp = 6.425d?
= 6.425 x 3.152
Rp = 63.75 kN
We know
Ra - Rp - T = 0, T = 143.67 - 63.75 = 79.91 kN « 80 kN
Finding the zéro shear point
Equating in such a way that shear force is zéro (considering x will be
below Tie Rod)
2.763 + 5.526a: + 1/2 x x2 x 9.75 = 80
4.875a:2 + 5.526a: + (-77.237) = 0
B y solving we get
x = 3.45 m (below tie rod)
Maximum Bending moment
Afmax = 80 x 3.45 - 2.763 x (1/3 + 3.45) - (5.526 x 3.45 x 3.45/2)
- (1/2 x 9.75 x 3.452 x (2/3 x 3.45))
Mmax = 100 kNm
Calculating required section modulus
Z' = Mmax/allowable stress
= 100 x 106/0.67 x 250 (In case of Fe250 grade steel)
= 597 cm3
Coastal Engineering: Theory and Practice
By solving this équation we get, dl — 3.15 m
D = dl + a - 3.15 + 2.18 = 5.33 « 5.4 m
Increase it by 20%-40%
Z)actuai — 5-4 x 1.4 (assuming as 40%)
■^actual = 7.56 m
-Dactuai = 5.4 x 1.3 (assuming as 30%)
■^actual = 7.02 m
Finding T :
Rp = 6.425d?
= 6.425 x 3.152
Rp = 63.75 kN
We know
Ra - Rp - T = 0, T = 143.67 - 63.75 = 79.91 kN « 80 kN
Finding the zéro shear point
Equating in such a way that shear force is zéro (considering x will be
below Tie Rod)
2.763 + 5.526a: + 1/2 x x2 x 9.75 = 80
4.875a:2 + 5.526a: + (-77.237) = 0
B y solving we get
x = 3.45 m (below tie rod)
Maximum Bending moment
Afmax = 80 x 3.45 - 2.763 x (1/3 + 3.45) - (5.526 x 3.45 x 3.45/2)
- (1/2 x 9.75 x 3.452 x (2/3 x 3.45))
Mmax = 100 kNm
Calculating required section modulus
Z' = Mmax/allowable stress
= 100 x 106/0.67 x 250 (In case of Fe250 grade steel)
= 597 cm3