Arches MCQ Quiz in বাংলা - Objective Question with Answer for Arches - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept:

Arches:

An arch is a curved beam in which horizontal movement at the support is wholly or partially prevented. Hence, there will be horizontal thrust induced at the supported. The shape of an arch does not change with loading and therefore some bending may occur.

Types of Arches:

• Three hinged arches
• Two hinged arches
• Fixed arches

Explanation:

The equation of a parabolic three-hinged arch

\({\bf{y}} = \frac{{4{\bf{h}}}}{{{{\bf{l}}^2}}}{\bf{x}}\left( {{\bf{l}} - {\bf{x}}} \right)\)

Where,

h = Rise of the arch

l = Span of the arch

y = rise of the arch at distance x from endpoint.

Explanation:

Arcade:

• An arcade is a set of contiguous arches that are all supported with columns or piers.
• An arcade on the outside of a building normally creates a walkway, while an arcade on the inside can be found on the lowest part of a wall.

​

Impost:

• The projecting course, provided on the upper part of a pier or abutment to stress the springing line is called impost

​Keystone:

• Keystone is the stone at the apex of the arches.
• Keystone plays a role in distributing all weight down the side support blocks in the columns.
• The task of the keystone here is to provide balance and strength distribution in terms of its position at the apex of the structure.

Voussiors:

• A voussoir is a stone block that’s wedge-shaped, Voussoirs are used to make specific buildings like arches and tombs.
• The rounded form of an arch is tricky to build, and the wedge shape of voussoirs helps them fit.

​Haunch:

• The haunch, where the thrust of the arch is usually greatest, is at a point located about one-third the distance between the springer and keystone.

​Spandrel:

• It is the triangular walling, enclosed by the extrados of the arch, a horizontal line from the crown of the arch, and a perpendicular line from the springing of the outer curve.

Intradros/Exrados:

• The interior curved part of the arch is called intrados.
• The inner surface of an arch is called a soffit. Soffit and intrados are used synonymously
• While the exterior curved part is called the extrados.

Concept:

A three hinged arch is a determinant member and hence there won’t be any additional stress due to changes in temperature, however there will be an increase in the crown height with the increase in temperature and subsequently the horizontal reaction at the support also decreases with the increase in crown height. The equations are mentioned below-

\(\Delta h = \frac{{{{\rm{L}}^2} + 4{{\rm{h}}^2}}}{{4{\rm{h}}}} \times {\rm{\alpha \Delta T}}\),

So, with the increase in temperature the crown height increases (crown is the top most point of the arch)

Also \(\frac{{{\rm{\Delta H}}}}{{\rm{H}}} = - \frac{{{\rm{\Delta h}}}}{{\rm{h}}}\), Hence with the increase in height of the arch the horizontal reaction at the supports decreases.

A linear arch is a compression member in which the shear force and bending moment is zero throughout the span and it is subjected to normal thrust only.

This arch has a parabolic shape and when subjected to a uniform horizontally distributed vertical load, then from the analysis of cables it follows that 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.

Concept:

From symmetry, \({V_A} = {V_B} = \frac{{Wl}}{2}\)

\(\sum {M_C} = 0,\)

\({H_A} \times h + W \times \frac{\ell }{2} \times \frac{\ell }{4} = {V_A} \times \frac{\ell }{2} = \frac{{W{\ell ^2}}}{4}\)

⇒ \({H_A} = \left( {\frac{{W{\ell ^2}}}{4} - \frac{{W{\ell ^2}}}{8}} \right) \times \frac{1}{h} = \frac{{W{\ell ^2}}}{{8h}}\)

Important Points

You Should not miss this:

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,

Concept:

There are three types of arches:

1. Three hinged arch

• It is a statically determinate structure.
• There are 4 unknown reactions, 3 equations of equilibrium, and one extra equation of equilibrium (M_c = 0) at the hinge.

2. Two hinged arch

• It is a statically indeterminate structure to the first degree.
• There are 4 unknown reactions and only 3 equations of equilibrium.

3. Fixed arch

• It is a statically indeterminate structure to the third degree.
• There are 6 unknown reactions and only 3 equations of equilibrium.

Calculation:

Given,

Three hinged parabolic arch

Span, L = 16 m and Rise, h = 4 m

Vertical concentrated load, P = 80 kN at x = 4 m from left end

\(\begin{array}{l} {\sum {{F_x} = 0 \Rightarrow {H_A} = H} _B} = H\\ \sum {{F_y} = 0} \Rightarrow {V_A} + {V_B} = 80\\ \sum {{M_B} = 0} \\ {V_A} \times 16 - 80 \times 12 = 0\\ {V_A} = 60kN\\ {V_B} = 80 - 60 = 20kN \end{array}\)

V_A = 60 kN

\(\sum {{M_c} = 0} \)

\(\begin{array}{l} \sum {{M_c} = 0} \\ {V_A} \times 8 - H \times 4 - 80 \times 4 = 0\\ H \times 4 = 60 \times 8 - 80 \times 4\\ H = 40kN \end{array}\)

H = 40 kN

Concept:

The influence line diagram for thrust in parabolic arch:

The horizontal thrust due to a concentrated load W at a distance x from end A is given by:

\({\rm{H}} = \frac{5}{8}\frac{{\rm{W}}}{{{{\rm{L}}^3}{\rm{h}}}}{\rm{x}}\left( {{\rm{L}} - {\rm{x}}} \right)\left( {{{\rm{L}}^2} + {\rm{xL}} - {{\rm{x}}^2}} \right)\)

Which is a 4° parabolic equation.

Now, to draw influence line diagram for horizontal thrust, W = 1 kN

\({\rm{H}} = \frac{5}{{8{\rm{h}}{{\rm{L}}^3}}}{\rm{x}}\left( {{\rm{L}} - {\rm{x}}} \right)\left( {{{\rm{L}}^2} + {\rm{xL}} - {{\rm{x}}^2}} \right)\)

\(\Rightarrow {\rm{H}} = \frac{5}{{8{\rm{h}}}}{\rm{x}}\left( {1 - \frac{{2{{\rm{x}}^2}}}{{{{\rm{L}}^2}}} + \frac{{{{\rm{x}}^3}}}{{{{\rm{L}}^3}}}} \right)\)

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,

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

Calculation:

ΣF_y = 0,

⇒ V_A + V_B = 30 kN..........(i)

ΣF_x = 0

⇒ H_A = H_B = say H

ΣM_C = 0 (From left)

⇒ -H_A × 4 + V_A × 6 = 0

⇒ H_A = 1.5V_A..................(ii)

ΣM_B = 0

⇒ V_A × 12 + H_A × 2 - 30 × 6 = 0

⇒ 6V_A + H_A = 90.............(iii)

Put the value of H_A from (ii) in equation (iii)

⇒ 7.5V_A = 90

⇒ V_A = 12 kN

From (ii), H_A = 18 kN

∴ Resultant reaction at A, \({R_A} = \sqrt {{V^2}_A + {H^2}_A} = \sqrt {{{12}^2} + {{18}^2}} \)

R_A = 21.63 kN

You Should not miss this:

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,

Explanation:

Three hinge arch is a statically determinate structure.

As we know, 

No. of support reactions, R = 4

Equations of static equilibrium, r_e = 3

Additional equation due to released reaction at hinge at crown (∑ M = 0), r_r= 1

Degree of static determinacy, D_s = R - (re + rr) = 4 - (3 + 1) = 0.