Arches MCQ Quiz - Objective Question with Answer for Arches - Download Free PDF

Explanation:

Three-Hinged Arch Bending Moment

In a three-hinged arch, the bending moment at certain points is zero due to the structural configuration and load distribution. Specifically, the bending moment is zero at the hinges because these points are free to rotate and cannot resist any moment. Thus, they act as moment release points.

Analyzing the Given Options

1. "In the middle third of the span." (Not necessarily)

   • While the bending moment can be low in the middle third of the span, it is not guaranteed to be zero unless under specific loading conditions.

2. "At the supports." (Correct for many structures, but not necessarily for three-hinged arches)

   • The bending moment at supports is zero for simply supported beams, but three-hinged arches specifically have zero moments at the hinges.

3. "At all three hinges." (Correct)

   • In a three-hinged arch, the hinges are designed to allow rotation and thus cannot resist bending moments. Hence, the bending moment is zero at these points.

4. "At the quarter span." (Not necessarily)

   • The quarter span is a general point of interest for bending moment calculations, but it does not inherently have a zero bending moment in a three-hinged arch.

Temperature changes significantly affect the horizontal thrust in a three-hinged arch because of its structural geometry and thermal expansion behavior.

1. Effect of Rise in Temperature: A decrease in horizontal thrust would only occur if there were a contraction in the arch (e.g., due to cooling). However, during heating, the thrust increases.

So, statement a is incorrect

2. Statement B is correct, as The horizontal thrust is directly proportional to the thermal expansion, and a rise in temperature results in an increase in horizontal thrust.

3. Statement C is incorrect as The restraints at the supports of the arch create stresses due to thermal expansion or contraction, meaning that temperature changes do produce stresses in the arch

Explanation:

Three hinge arch is a statically determinate structure.

As we know, 

No. of support reactions, R = 4

Equations of static equilibrium, re = 3

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

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

Explanation:

• Arches and lintels are found wherever it is necessary to support a wall etc.
• Over an opening which is usually not more than about two meter wide.
• They provide support in a completely different way.
• Lintel is preferred to arches to carry the load because arches require strong abutments to withstand arch thrust.
• In arches and lintels, the soffit refers to the underside or the lower surface of the architectural element.
• For arches, the soffit is the curved or flat underside of the arch, while for lintels, it is the bottom surface of the horizontal structural element. 

Calculation:

ΣV = 0, V_A + V_B = 150 kN

ΣH = 0, H_A = H_B = H

At support B, ΣM_B = 0

V_A × 20 - 150 × 16 = 0

V_A = 120 kN

V_B = 150 - 120 = 30 kN

At central hinge, ΣM_C = 0

V_A × 10 - H × 4 - 150 × 6 = 0

H × 4 = 120 × 10 - 150 × 6

H = 75 kN

So, the vertical reaction at A and the horizontal thrust are 120 kN and 75kN respectively.

Explanation:

Given:

L = 30 m

h = 5 m

∑F_H = 0 

H_A = H_B = H

∑F_V = 0 

R_A + R_B = 40 kN     ---(1)

∑M_B = 0 

R_A × 30 - 40 × 20 = 0

\(R_A = \dfrac{80}{3} = 26.67 \ kN\)

R_A = 26.67 kN

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,

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

For the given condition:

From equilibrium equation:

∑V = 0,

Taking moment about point A,

V_B × 20 – 80 × 5 = 0

Hence, V_A = 60 kN and V_B = 20 kN

Taking M_C = 0 from left portion(AC),

V_A × 10 - 80 × 5 – H × 4 = 0

∴ H = 50 kN

Explanation:

Masonry arch:

• The structural efficiency of the masonry arch is attributed to the curvature of the arch.
• Which transfers vertical loads laterally along the arch to the abutments at each end.
• The transfer of vertical forces gives rise to both horizontal and vertical reactions at the abutments.
• The curvature of the arch and the restraint of the arch by the abutments cause a combination of flexural stress and axial compression.
• The arch depth, rise, and configuration can be manipulated to keep stresses primarily compressive.
• Brick masonry is very strong in compression, so brick masonry arches can support the considerable load and transfers this load by horizontal thrust.
• Due to its heavyweight and short span, the load is not transferred by bending or radial shear

​ 

3 Hinge or 2 hinge arch, transfer load by bending moment, radial shear, and horizontal thrust, and hence considered superior to masonry arch.

Explanation:

Three hinge arch :

• In the case of a three-hinged arch, we have three hinges, two at the support and one at the crown and there are four reaction components in the three-hinged arch.
• One more equation is required in addition to three equations of static equilibrium for evaluating the four reaction components.
• Taking moment about the hinge of all the forces acting on either side of the hinge can set up the required equation, making 3  hinge arch determinate.

Two hinge arch :

• In the two hinge arch, we have 2 hinges at two support of the arch and have four reaction or unknown components, as in a 3 hinge arch.
• But here we don't have an additional equation to solve the structure and hence our 2 hinge arch is indeterminate by degree 1.

​Hinge less/ Fixed arch :

• In the fixed arch, both the support are fixed. So, in a fixed arch six reaction forces ( One moment extra on each support ).
• But, we have only 3 equilibrium equations. So, the degree of indeterminacy of fixed arch becomes 6 - 3 = 3. 

​​

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:

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

Explanation:

From the figure,

V_A + V_C = 0.5 X 100 = 50 KN

Bending moment at point A     ∑ M_A = 0

V_C X 100 – (0.5 X 100 X 100/2) = 0

VC = 25 KN

So,

V_A = 25 KN

In parabolic three hinges arch Bending moment is zero.

So, Stand at B and look left,

Bending moment at B = 0

V_A x 50 – H_A x 25 – (0.5x 50 x 50/2) = 0

25 X 50 - H_A x 25 – (0.5 x 50 x 50/2) = 0

H_A = 25 KN

So, in parabolic 3 hinged arch 25 kN horizontal and vertical reactions at each support.