This question was previously asked in

BPSC AE: Paper 5 (Civil Engineering) 2018 Official Paper

Option 2 : is decreased

CT 1: Current Affairs (Government Policies and Schemes)

54560

10 Questions
10 Marks
10 Mins

**Concept:**

Initially the 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 the support sinks/settles the horizontal thrust acts in the opposite direction, hence the horizontal thrust gets decreased.

**Alternate Method:**

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

\(H\; = \;\frac{{\smallint \frac{{{M_x}ydx}}{{EI}}\; + \;\alpha tl}}{{\smallint \frac{{{y^2}dx}}{{E{I_C}}}\; + \;\frac{l}{{AE}}\; + \;k}}\)

Where A = Horizontal thrust

E = Modulus of elasticity of the arch material

I = moment of inertia of Arch cross-section

A = Area of cross-section of the arch

α = Coefficient of thermal expansion

t = Difference of temperature

l = length of arch, k = yielding of supports

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

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

**And H inversely proportional to k**