Concept:

As loading is UDL and shape of arch is parabolic so re will be no moment in arch rib.

But as temperature increases in a two hinged arch (degree of indeterminacy one), horizontal thrust will increase.

Moment due to horizontal thrust is – Py.

Top most part is called crown. Or options are name of or part.

Explanation:

The horizontal thrust in cable of a parabolic profile is given as:

$H=\frac{w{L}^2}{8h}$

Where,

W is load intensity in KN/m

L is horizontal span of cable

h is central rise of given cable.

Based on above formula, it can be said that when one unit load moves from A to B, L and h will remain constant and refore, ILD  for horizontal thrust is constant through its horizontal span with load  W = 1 KN.

For calculating ordinate of ILD for H in cable put W = 1 KN or  wL = 1 kN

Or

Ordinate = L/8h

Concept:

Initially horizontal thrust is towards right at A.

When support B sinks.

∑ M_B = 0

V_A × 1 – H_A × δ  = 0

$\Rightarrow H_A=\frac{V_A\cdot l}{\delta }$

So, H_A is acting towards left

So overall horizontal thrust will decrease.

Mistake Points When support sinks/settles horizontal thrust acts in opposite direction, hence horizontal thrust gets decreased. 

Alternate Method:

The equation for horizontal thrust in a 2-hinged parabolic arch is

$H=\frac{\int \frac{M_{x}ydx}{EI}+\alpha tl}{\int \frac{y^{2}dx}{E{I}_C}+\frac{l}{AE}+k}$

Where A = Horizontal thrust

E = Modulus of elasticity of arch material

I = moment of inertia of Arch cross-section

A = Area of cross-section of arch

α = Coefficient of rmal expansion

t = Difference of temperature

l = length of arch, k = yielding of supports

Here re is no clear mention of temperature Change, (t = 0),

so horizontal thrust will decrease as support sinks, i.e k has some value.

And H inversely proportional to k

Maximum tension in cable is square root of sum of squares of maximum vertical and maximum horizontal loads.
Maximum Vertical Load ? V = \(\frac{WL}{2} \)
Maximum Horizontal Load ? H = \(\frac{WL^2}{8H} \)
Therefore, Maximum Tension is \(\frac{WL}{2}(\sqrt{1+\frac{L^2}{16H^2}}) \)

Explanation:

Cable:

• A cable is a structural element that can resist only tension. it cannot resist shear, bending, and compressive forces.
• A cable structure can be defined as a structure in which a cable or cable system is used as a structural element to withstand initial load. Cable means any structural element whose diameter is very small in relation to its length.
• If we consider cross-section of a cable, stress is same throughout cross-section. This type of stress is called net stress. One of reasons for high efficiency of se structures is net tensile stress. One of disadvantages of cable structures is problem of vibration due to wind and earthquakes. This problem occurs due to lightness of structures against lateral stresses such as wind.
• Cables are often used in engineering structure support and to transmit loads from one member to anor. When used to support suspension bridges, cables from main load-carrying element in structure.

Stone brick is very good in compression but weak in tension.

Explanation:

(i) Cable is a structural element that can resist only tension. It cannot resist shear, bending, and compressive forces.

(ii) The shape of a cable suspended between two supports is defined by  help of applied load. And  shape of cable is equivalent to shape of bending moment diagram of simply supported beam subjected to same loading as cable.

Given, Beam supports two concentrated loads

Following are figure for SFD & BMD

As Bending Moment Diagram is Trapezoidal 

Option 3 is correct.

Additional Information

(iii) If a concentrated load is acting on a cable, shape will be triangle because bending moment diagram of simply supported beam subjected to concentrated load is triangular. 

(iv) Similarly, if cable subjected to Horizontal UDL, n BMD of simply supported beam has parabolic shape and hence, shape of cable is also parabolic.

(v) If UDL on cable is not horizontal or due to its self weight, cable will take Catenary shape.

(vi) The girder in a suspension bridge transmits to its supports require a little load.

In particular, if 2-hinge arch has a parabolic shape and it is subjected to a uniform horizontally distributed vertical load, n only compressive forces will be resisted by arch. Under se conditions, arch shape is called a funicular arch because no bending or shear forces occur within arch.

Bending moment, Mx = 0

Radial shear, Sx = 0

Normal thrust, Nx ≠ 0

Important Point:

?H = 0
H = \(\frac{?M.y dy}{?y^2 dy}\)
Y = \(\sqrt{R^2- x^2} ? \sqrt{R^2-(\frac{L^2}{2})} \)
Hence, H = \(\frac{W}{\pi}\)