2 Hinged MCQ Quiz - Objective Question with Answer for 2 Hinged - Download Free PDF

Explanation:

In a two-hinged parabolic arch, temperature rise causes the arch to expand horizontally. Since the supports are hinged and cannot move, this thermal expansion leads to:

• Increased horizontal thrust to maintain the equilibrium.

• This is due to the development of internal compressive forces along the arch axis.

Additional Information

• Two-hinged arches are statically indeterminate.

• When temperature increases, the crown tries to rise due to thermal expansion.

• But the hinges resist this deformation → hence internal force builds up → horizontal thrust increases.

Concept:

Reaction locus for a two-hinged arch:

The reaction locus for a two-hinged arch is the locus of the point of Intersection of the two resultant reactions at the supports as a point load moves on the span of the arch.

a) For a two-hinged Semicircular Arch: The reaction locus is a straight line parallel to the line joining abutments and at a height of \(\pi R\over 2\) above the base.

b) For a two-hinged Parabolic Arch: The reaction locus for a parabolic arch is a curve whose equation is \(y={1.6hl^2\over l^2+lx-x^2}\).

Hence option (1) is correct.

Explanation:

Concept-

Two-hinged arch is the statically indeterminate structure to degree one. Usually, the horizontal reaction is treated as the redundant and is evaluated by the method of least work.

In the early part of the nineteenth century, three-hinged arches were commonly used for the long span structures as the analysis of such arches could be done with confidence. However, with the development in structural analysis, for long span structures starting from late nineteenth century engineers adopted two-hinged and hinge less arches.

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

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

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 arch

α = Coefficient of thermal expansion

t = Difference of temperature

l = length of arch

So when the temperature increases the temperature difference increases, so the horizontal thrust will also increase.

Explanation:

Two Hinged Arch:

Two hinge arch is an indeterminate structure to first degree. Hence to analyse it one additional compatibility condition is required. Vertical reaction can be determined by taking moment about either end. The horizontal thrust at each support may be determined from the condition that the horizontal displacement of the either end with respect to the other is zero.

∴ δU/δH = 0

Horizontal thrust for two hinge arch is given by,

\(H = \frac{{\int {My\frac{{dy}}{{dx}}} }}{{\int {{y^2}\frac{{ds}}{{EI}}} }}\)

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

Bending moment, Mx = 0

Radial shear, Sx = 0

Normal thrust, Nx ≠ 0

Important Point:

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

Concept:

For two hinged semi-circular arches:

Horizontal thrust is given by:

\({{\rm{H}}_{\rm{A}}} = {{\rm{H}}_{\rm{B}}} = \frac{{\smallint \frac{{{{\rm{M}}_{\rm{x}}}{\rm{ydx}}}}{{{\rm{E}}{{\rm{I}}_{\rm{c}}}}}}}{{\smallint \frac{{{{\rm{y}}^2}{\rm{dx}}}}{{{\rm{E}}{{\rm{I}}_{\rm{c}}}}}}}\)

Where,

Mx = Moment at A – A’, y = vertical distance,

E = Modulus of elasticity, and

 I = Moment of inertia of c/s of the arch.

When  W load acts at the crown, then

Horizontal thrust (H) = w/π

∴ H is independent of the radius of the arch.

So, the ratio will be 1: 1: 1.

Note

Important Points:

 For a Two Hinged Semicircular arch subjected to concentrated load (W) at any other point which makes an angle α with the horizontal, then the horizontal thrust,

\(H = \left( {\frac{W}{\pi }} \right){\sin ^2}\left( \alpha \right)\;\)

Concept:

The locus of the reaction of a two hinged semi-circular arch is a straight line whereas the locus of the reaction of a two-hinged parabolic arch is a parabolic curve.

For two Hinged Semi-circular Arch:

Reaction locus is straight line parallel to the line joining abutments and height at πR/2.

For two Hinged Parabolic Arch:

\(y = PE = \frac{{1.6h{L^2}}}{{{L^2} + Lx - {x^2}}}\)

Concept:

For two hinged semi-circular arches:

Horizontal thrust is given by:

\({{\rm{H}}_{\rm{A}}} = {{\rm{H}}_{\rm{B}}} = \frac{{\smallint \frac{{{{\rm{M}}_{\rm{x}}}{\rm{ydx}}}}{{{\rm{E}}{{\rm{I}}_{\rm{c}}}}}}}{{\smallint \frac{{{{\rm{y}}^2}{\rm{dx}}}}{{{\rm{E}}{{\rm{I}}_{\rm{c}}}}}}}\)

Where,

Mx = Moment at A – A’, y = vertical distance,

E = Modulus of elasticity, and

 I = Moment of inertia of c/s of the arch.

When  W load acts at the crown, then

Horizontal thrust (H) = w/π

Important Points

For a Two Hinged Semicircular arch subjected to concentrated load (W) at any other point which makes an angle α with the horizontal, then the horizontal thrust,

\(H = \left( {\frac{W}{\pi }} \right){\sin ^2}\left( \alpha \right)\;\)

Concept:

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

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

Moment due to horizontal thrust is – Py.

Concept:

The locus of the reaction of a two hinged semi-circular arch is a straight line whereas the locus of the reaction of a two-hinged parabolic arch is a parabolic curve.

For two Hinged Semi-circular Arch:

Reaction locus is straight line parallel to the line joining abutments and height at πR/2.

For two Hinged Parabolic Arch:

\(y = PE = \frac{{1.6h{L^2}}}{{{L^2} + Lx - {x^2}}}\)

Concept:

For two hinged semi-circular arches:

Horizontal thrust is given by:

\({{\rm{H}}_{\rm{A}}} = {{\rm{H}}_{\rm{B}}} = \frac{{\smallint \frac{{{{\rm{M}}_{\rm{x}}}{\rm{ydx}}}}{{{\rm{E}}{{\rm{I}}_{\rm{c}}}}}}}{{\smallint \frac{{{{\rm{y}}^2}{\rm{dx}}}}{{{\rm{E}}{{\rm{I}}_{\rm{c}}}}}}}\)

Where,

M_x = Moment at A – A’, y = vertical distance,

E = Modulus of elasticity, and

 I = Moment of inertia of c/s of the arch.

When  W load acts at the crown, then

Horizontal thrust (H) = w/π

∴ H is independent of the radius of the arch.

So, the ratio will be 1: 1: 1.

Note

Important Points:

 For a Two Hinged Semicircular arch subjected to concentrated load (W) at any other point which makes an angle α with the horizontal, then the horizontal thrust,

\(H = \left( {\frac{W}{\pi }} \right){\sin ^2}\left( \alpha \right)\;\)

Concept-

Two-hinged arch is the statically indeterminate structure to degree one. Usually, the horizontal reaction is treated as the redundant and is evaluated by the method of least work.

In the early part of the nineteenth century, three-hinged arches were commonly used for the long span structures as the analysis of such arches could be done with confidence. However, with the development in structural analysis, for long span structures starting from late nineteenth century engineers adopted two-hinged and hinge less arches.

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

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

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 arch

α = Coefficient of thermal expansion

t = Difference of temperature

l = length of arch

So when the temperature increases the temperature difference increases, so the horizontal thrust will also increase.

Concept:

If two-hinged parabolic is subjected to uniformly distributed load of intensity q per unit length.