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[Supes]

sa 1 ts

Given data:

Uniform load Wo= 12KN/m

Length of support (L1) =3

Length of support (L2) =2

From the equation VA+VB =5x12= 60KN

Lenght = Diameter

L1=d1 , L2=d2

L1/L2 = ✓d1/d2 =✓ 3/2 =1.224

Therefore L1+L2=5

1.22L1+L2=5

L2=5/1.224=2.24m

Similarly then L1=2.75 and L2=2.24

From the equilibrium reactions , taking moments from A is

VB x 5 - H x 3 - 12 x 5 x 5/2 =0

VB = H x 3+150/5 = 0.6H+30.......eq (1)

From equilibrium reactions , taking moments from C,

VB x L1 - H x 3 - 12 x L1 x L1/2 =0

VB x 2.75 -H x3 -(64x64)=0

2.75VB = 3H+ 45.37

VB = 1.09H + 16.49......eq (2)

Substituting eq(1) and eq (2) is

VB = 1.09H + 16.49 - 0.6H +30

H = 27.57 KN

The equation (2) is substituting the value of H

VB = 1.09 H + 16.49

= 46.54 KN

From the equation of VA and VB is 60KN , then VA = 13.45 KN

Similarly , the tension of the cable at support B ( max) is

T(max) = ✓(H2 + VB2)

= ✓(27.5 72 + 46.54 2 )

=54.09 KN

Similarly , the tension of the cable at support A ( min) is

T(min) = H

=27.57 KN .

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