Cable when loaded takes shape of bending moment diagram that would be formed if same set of load is applied on to simply supported beam of same length and properties. Thus, shape of cable is a funicular polygon.

Two hinged arches is an indeterminate structure. We can calculate vertical reactions by using ?M = 0 and ?V = 0 but horizontal reaction cannot be computed by any of equilibrium equations. Thus, two hinged arches is an indeterminate structure.

Calculation:

Taking equilibrium of structure

V_A + V_B  = 30 KN       -----------(1)

Taking moment about A

$\sum M_A=0$

V_B × 45 - 10 × 15 - 20 × 30 = 0

V_B = 16.667 KN

From equation (1)

V_A = 30 - 16.667 = 13.333 KN

Cable is a flexible member and hence it cannot resist a moment. The bending moment is zero everywhere in cable.

BM_D = 0

$V_B\times 15-H\times 5=0$

$16.667\times 15-H\times 5=0$

H = 50 KN

BM_C = 0

$V_A\times 15-H\times y=0$

$13.333\times 15-50\times y=0$

y = 4 m

It is indeterminate to 3 degrees.

?H = 0
H = \(\frac{?M.y dy}{?y^2 dy}\)
Where, y=\(\frac{4 H x (L-x)}{L^2}\)
Hence, H = \(\frac{25WL}{128H}.\)

Statement A – False

Temperature effect on 3 Hinged arches:

1. No stresses are induced in three-hinged arch due to temperature changes alone.
2. Due to temperature changes, rise of arch increases or decreases, and horizontal thrust of arch carrying load will also get changed.
3. There is a decrease in horizontal thrust due to rise in temperature.

Statement B - False

The degree of flexibility is directly related to degree of static indeterminacy of arch.

For arches, degree of static indeterminacy is calculated as:

D_s = number of external reactions – Number equations available

For 3 Hinged arches:

3 Hinged arch is hinged at 3 points (two supports and a third at crown).

Number external reactions, r = 4 ( 2 for each hinge)

Number equations available = 3( equilibrium equations)  + 1 ( at crown BM is zero due to hinge)

D_s = 4 – 4 = 0

∴ 3 Hinged arch is statically determinate.

For 2 Hinged arches:

2 Hinged arch is hinged at 2 points only.

Number external reactions, r = 4 (2 for each hinge)

Number equations available = 3(equilibrium equations)  

D_s = 4 – 3 = 1

 ∴ 2 Hinged arch is statically indeterminate to degree 1.

Based on above, we can say that a 3 hinge arch is more flexible than a 2 hinge arch.

Concept:

We know for a cable with supports at same level subjected to udl has a Horizontal reaction $H=\frac{wl2}{8}$

And we know that, 

Maximum Tension occurs at support (Let's suppose support A) which is given by T_max= √( H²+ V²)

Minimum Tension occurs at dip which is given by T_min= Horizontal reaction at support.

Where H_A and V_A are horizontal and vertical reactions at support.

The vertical reaction is given by $V=\frac{wl}{2}$

Calculation:

Given:

UDL (w) = 10 kN/m and Length of cable = 100 m

As loading and structures are symmetrical 

 $V_A=V_B=\frac{100\ast 10}{2}=500kN$

$H_A=H_B=\frac{w{l}^2}{8h}=\frac{10\times {100}^2}{8\times 8}=1562.5kN$

∴ The horizontal reaction at support = 1562.5 kN

With reference to suspension cable bridge, following points to be remember:

1. The form and magnitude of ordinates of cable curve remain unchanged even after application of live loads. By this assumption, cable always remains parabolic with same central sag. Obviously, stiffening girder must deflect under live loads, pulling cable downwards along with it.

2. The cable is perfectly flexible. The wire ropes and parallel wire cables are obviously quite flexible and have little flexural stiffness.

3. The stiffening girder is straight, its moment of inertia is constant and it is tied to cable throughout its length. This assumption takes into account deformation of only chord members of truss and neglects deformation of web members.

Due to its geometry, shear force always comes out to be zero in 3 hinged arches.