3 Hinged MCQ Quiz - Objective Question with Answer for 3 Hinged - 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

Calculation:

Taking moment about A,

RB × 16 – 30 × 8 × 4 = 0 

∴ RB = 60 kN.

Now, RA + RB = 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

Concept:

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

ILD for beam moment in 3 hinged arch:

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

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 (Mc = 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 = 30 m and Rise, h = 10 m

Vertical concentrated load, P₁ = 5kN kN at x = 6 m from left end

\(\begin{array}{l} {\sum {{F_x} = 0 \Rightarrow {H_A} = H} _B} = H\\ \sum {{F_y} = 0} \Rightarrow {V_A} + {V_B} = 5\\ \sum {{M_B} = 0} \\ {V_A} \times 30 - 5 \times 24 = 0\\ {V_A} = 4 kN\\ {V_B} = 5 - 4= 1kN \end{array}\)

VA = 4 kN

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

\(\begin{array}{l} \sum {{M_c} = 0} \\ {V_A} \times 15 - H \times 10 - 5 \times 9 = 0\\ H \times 10 = 4 \times 15 - 5 \times 9\\ H = 1.5kN \end{array}\)

H = 1.5 kN

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

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:

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.

Concept:

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

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

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

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

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,

Reaction at Support,

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

Taking Moment About C,

Summation,M_C = 0 

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

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

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

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

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

Explanation:

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

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:

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.

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.

Let the reactions be as shown below.

∵ ΣF_Y = 0

⇒ R_A + R_B = 2P

∵ ΣF_X = 0

⇒ H_A = H_B

∵ ΣM_A = 0

⇒ 2P × 5 - R_B × 8 = 0

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

Again,

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

⇒ R_A × 4 - H_A × 1.5 = 0

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

⇒H_A = 2P and H_B = 2P