3 Hinged MCQ Quiz - Objective Question with Answer for 3 Hinged - Download Free PDF

Last updated on Jun 13, 2025

Latest 3 Hinged MCQ Objective Questions

3 Hinged Question 1:

The equation of a parabolic arch of span 'L' and central height h is given by

  1. \(y = 8h \times (L - x) / L^2\)
  2. \(y = 3h \times (L - x) / L^2\)
  3. \(y = h \times (L - x) / L^2\)
  4. \(y = 4h \times (L - x) / L^2\)

Answer (Detailed Solution Below)

Option 4 : \(y = 4h \times (L - x) / L^2\)

3 Hinged Question 1 Detailed Solution

Explanation:

  • An arch is a curved structural element that spans an opening and supports loads from above by transferring them as compressive forces along its curve to the supports (abutments) at the ends.

  • Arches work mainly in compression, unlike beams which primarily resist bending.

  • They are widely used in bridges, doorways, tunnels, and buildings due to their ability to carry heavy loads efficiently.

  • The equation of a parabolic arch of span 'L' and central height h is given by \(y = 4h \times (L - x) / L^2\)

Additional InformationTypes of Arches:

  • Three hinged arches
  • Two hinged arches
  • Fixed arches

F1 Abhayraj Anil 05.02.21 D2

3 Hinged Question 2:

At which of the following points of a three -hinged arch is the bending moment equal to zero?

  1.  In the middle third of the span
  2. At the supports 
  3. at all three hinges
  4. At the quarter span 

Answer (Detailed Solution Below)

Option 3 : at all three hinges

3 Hinged Question 2 Detailed Solution

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.

3 Hinged Question 3:

Consider the following statements with respect to the temperature effect on a load-carrying, three-hinged arch.

a) There is a decrease in horizontal thrust due to rise in temperature.

b) There is an increase in horizontal thrust due to rise in temperature.

c) No stresses are produced in the arch due to change in temperature.

Which of the above statement(s) is/are correct? 

  1. Only b 
  2. Only a
  3. Only a and b
  4. Only a and c

Answer (Detailed Solution Below)

Option 1 : Only b 

3 Hinged Question 3 Detailed Solution

 

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

3 Hinged Question 4:

A parabolic 3 hinged arch (AB) carries a u.d.l. of 30 kN/m on the left half of the span. It has a span of 16 m and a central rise of 3 m. What is the resultant thrust at 'A' (left hand end)?

  1. 271 kN
  2. 141 kN
  3. 241 kN
  4. 171 kN

Answer (Detailed Solution Below)

Option 3 : 241 kN

3 Hinged Question 4 Detailed Solution

F1 Engineering Arbaz 16-10-23 D16

Calculation:

Taking moment about A,

RB × 16 – 30 × 8 × 4 = 0 

∴ RB = 60 kN.

Now, RA + R= 30 × 8 = 240

RA = 240 - 60 = 180 kN

Taking moment about C, Mc = 0

180 × 8 – 30 × 8 × 4 - HA × 3 = 0

HA = 160 kN

Now, Resultant force at A will be F

\( F = \sqrt{R^{2}+H^{2}}\)

\( F = \sqrt{180^{2}+160^{2}}\)

F = 241 kN

3 Hinged Question 5:

The maximum ordinate under the crown hinge of influence line diagram for the horizontal thrust of three hinged arch is equal to

Where 'L' is span of the arch and 'h' is the height of the crown hinge.

  1. L/4h
  2. L/8h
  3. L/2h
  4. L/6h

Answer (Detailed Solution Below)

Option 1 : L/4h

3 Hinged Question 5 Detailed Solution

Concept:

For a 3 hinged arch due to horizontal thrust 'H':

F1 A.M Madhu 09.07.20 D13

ILD for beam moment in 3 hinged arch:

F1 A.M Madhu 09.07.20 D14

Bending moment in an arch is due to the combined action of beam moment & moment due to horizontal thrust. ILD for BM:

F1 A.M Madhu 09.07.20 D15

Top 3 Hinged MCQ Objective Questions

A three-hinged parabolic arch of span 20 m and rise 4 m carries a concentrated load of 150 kN at 4 m from left support 'A'. Calculate the vertical reaction and the horizontal thrust, respectively, at support 'A'.

  1. VA = 40 kN and HA = 80 kN
  2. VA = 75 kN and HA = 120 kN
  3. VA = 80 kN and H= 50 kN
  4. VA = 120 kN and HA = 75 kN

Answer (Detailed Solution Below)

Option 4 : VA = 120 kN and HA = 75 kN

3 Hinged Question 6 Detailed Solution

Download Solution PDF

Calculation:

F1 Abhishek M 02-11-21 Savita D2

ΣV = 0, VA + VB = 150 kN

ΣH = 0, HA = H= H

At support B, ΣMB = 0

V× 20 - 150 × 16 = 0

VA = 120 kN

V= 150 - 120 = 30 kN

At central hinge, ΣMC = 0

VA × 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.

A three-hinged parabolic arch has a span of 30 m and the central rise is 5 m. It is subjected to a point load of 40 kN at a distance of 20 m from the right hinge. Calculate the vertical reaction component at its left support.

  1. 35.35 kN
  2. 40 kN
  3. 13.13 kN
  4. 26.67 kN

Answer (Detailed Solution Below)

Option 4 : 26.67 kN

3 Hinged Question 7 Detailed Solution

Download Solution PDF

Explanation:

F2 Ankita.S 26-02-21 Savita D1

Given:

L = 30 m

h = 5 m

∑FH = 0 

HA = HB = H

∑FV = 0 

RA + RB = 40 kN     ---(1)

∑MB = 0 

RA × 30 - 40 × 20 = 0

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

RA = 26.67 kN

A three-hinged parabolic arch subjected to a load w at L/4 distance from left support. The rise of the crown is 4 m, span is 20 m and w is 80 kN, the horizontal thrust at the support is

  1. 40 kN
  2. 25 kN
  3. 50 kN
  4. 80 kN

Answer (Detailed Solution Below)

Option 3 : 50 kN

3 Hinged Question 8 Detailed Solution

Download Solution PDF

For the given condition:

F1 A.M Madhu 26.05.20 D56

From equilibrium equation:

∑V = 0,

Taking moment about point A,

VB × 20 – 80 × 5 = 0

Hence, VA = 60 kN and VB = 20 kN

Taking MC = 0 from left portion(AC),

VA × 10 - 80 × 5 – H × 4 = 0

∴ H = 50 kN

A parabolic three hinged arch ABC is supporting Uniformly Distributed Load of 500 N/m over its entire span of 100 m. The centre point ‘B’ is vertically 25 m high from supports A and C. The reactions shall be ______.

  1. 50 kN horizontal and vertical reactions at each support
  2. 25 kN horizontal and 50 kN vertical reactions at each support
  3. 50 kN horizontal and 25 kN vertical reactions at each support
  4. 25 kN horizontal and vertical reactions at each support

Answer (Detailed Solution Below)

Option 4 : 25 kN horizontal and vertical reactions at each support

3 Hinged Question 9 Detailed Solution

Download Solution PDF

Explanation:

F17 Abhishek M 17-4-2021 Swati D16

From the figure,

VA + VC = 0.5 X 100 = 50 KN

Bending moment at point A     ∑ MA = 0

VC X 100 – (0.5 X 100 X 100/2) = 0

VC = 25 KN

So,

VA = 25 KN

In parabolic three hinges arch Bending moment is zero.

So, Stand at B and look left,

Bending moment at B = 0

VA x 50 – HA x 25 – (0.5x 50 x 50/2) = 0

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

HA = 25 KN

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

A three-hinged parabolic arch of span ‘l’ and rise ‘h’ is subjected to u.d.l. of intensity ‘ω’, then the horizontal thrust at the supports is 

F1 A.M Madhu 10.07.20 D8

  1. \(\frac{{\omega {l^2}}}{{8h}}\)
  2. \(\frac{{\omega l}}{h}\)
  3. \(\frac{{\omega l}}{{8{h^2}}}\)
  4. \(\frac{{\omega hl}}{8}\)

Answer (Detailed Solution Below)

Option 1 : \(\frac{{\omega {l^2}}}{{8h}}\)

3 Hinged Question 10 Detailed Solution

Download Solution PDF

Concept:

F1 A.M Madhu 10.07.20 D9

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

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

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

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

For a three-hinged parabolic arch, what will be the ratio L/R to satisfy H= W?

quesOptionImage2328

  1. 0.50
  2. 1.50
  3. 2.00
  4. 4.00

Answer (Detailed Solution Below)

Option 4 : 4.00

3 Hinged Question 11 Detailed Solution

Download Solution PDF

Concept:

  • A three-hinged arch, which is usually made from steel or timber, is Statically Determinate.
  • Unlike Statically indeterminate arches, they are not affected by differential settlement or temperature changes.
  • The three-hinged arch structures have three natural hinges as the name implies. the two support are hinged, and another internal hinge is usually located at the crown.

In three-hinged Arch,

  quesImage7586

Reaction at Support,

\({R_A} = \;{R_{B\;}} = \;\frac{W}{2}\)

Taking Moment About C,

Summation,MC = 0 

\({R_A} \times \frac{L}{2} - {H_A} \times R = 0\)

\({H_A} = H = \;\frac{{WL}}{{4R}}\)

If H = W then,
\(\;\frac{L}{{4\;R}} = \;\frac{H}{W}\)

\(\frac{L}{R} = 4 \times \frac{H}{W} = 4 \times 1 = 4\)

\(\frac{L}{R} = 4\)

A symmetrical parabolic arch of span 19 m and rise 4 m is hinged at the springings. It supports a uniformly distributed load of 1.5 tonnes per meter run of the span. The horizontal thrust in tonnes at each of the springing is -

  1. 20.63
  2. 16.92
  3. 8.5
  4. Zero

Answer (Detailed Solution Below)

Option 2 : 16.92

3 Hinged Question 12 Detailed Solution

Download Solution PDF

Explanation:

F1 Abhishek M 12.4.21 Pallavi D1

Horizontal thrust at springings A and B = \(\frac{{w{L^2}}}{{8h}}\)

Where, L = 19 m

h = 4 m

\({H_A} = \frac{{w{L^2}}}{{8h}}\)

\({H_A} = \frac{{1.5 \times {{19}^2}}}{{8 \times 4}}\)

HA = 16.92 tonnes

The equation of a parabolic arch of span 'I' and rise 'h' is given by:-

  1. \({\bf{y}} = \frac{{{\bf{h}}}}{{{{\bf{l}}^2}}}{\bf{x}}\left( {{\bf{l}} - {\bf{x}}} \right)\)
  2. \({\bf{y}} = \frac{{3{\bf{h}}}}{{{{\bf{l}}^2}}}{\bf{x}}\left( {{\bf{l}} - {\bf{x}}} \right)\)
  3. \({\bf{y}} = \frac{{2{\bf{h}}}}{{{{\bf{l}}^2}}}{\bf{x}}\left( {{\bf{l}} - {\bf{x}}} \right)\)
  4. \({\bf{y}} = \frac{{4{\bf{h}}}}{{{{\bf{l}}^2}}}{\bf{x}}\left( {{\bf{l}} - {\bf{x}}} \right)\)

Answer (Detailed Solution Below)

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

3 Hinged Question 13 Detailed Solution

Download Solution 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

F1 Abhayraj Anil 05.02.21 D2

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.

A three hinged arch is

  1. Statically determinate
  2. Statically determinate or indeterminate depending upon loading
  3. Statically indeterminate because of central hinge
  4. Determinate if the springs are at the same level

Answer (Detailed Solution Below)

Option 1 : Statically determinate

3 Hinged Question 14 Detailed Solution

Download Solution PDF

Explanation:

15.03.2018.035

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.

A three-hinged symmetrical arch is loaded as shown in the figure below. Which one of the following is the magnitude of the correct horizontal thrust?

F1 Corrected   Akhil.P 12-08-21 Savita D1

  1. 2.66 P
  2. 2 P
  3. 1.5 P
  4. 0.75 P

Answer (Detailed Solution Below)

Option 2 : 2 P

3 Hinged Question 15 Detailed Solution

Download Solution PDF

Concept:

A three-hinged arch is a statically determinate structure and its reactions are calculated using equilibrium equations only. It is mounted on two hinged support and contains one hinge anywhere in the arch rib, generally at the crown which makes the structure determinate.

F1 N.M Deepak 08.04.2020 D6

Let the reactions be as shown below.

F1 N.M Deepak 08.04.2020 D7

∵ ΣFY = 0

⇒ RA + RB = 2P

∵ ΣFX = 0

⇒ HA = HB

∵ ΣMA = 0

⇒ 2P × 5 - RB × 8 = 0

\( \Rightarrow {{\rm{R}}_{\rm{B}}} = \frac{{10{\rm{P}}}}{8}{\rm{and\;}}{{\rm{R}}_{\rm{A}}} = \frac{{3{\rm{P}}}}{4}\)

Again,

∵ ΣMC = 0 (Because it is an internal hinge)

⇒ RA × 4 - HA × 1.5 = 0

\( \Rightarrow {{\rm{H}}_{\rm{A}}} = \frac{{3{\rm{P}}}}{{1.5}}\)

HA = 2P and HB = 2P

Get Free Access Now
Hot Links: teen patti vip teen patti royal - 3 patti teen patti lucky teen patti rules