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

Explanation:

Statically Determinate and Indeterminate Beams

• In structural engineering, a beam is classified based on how the reactions at the supports can be determined. A statically determinate beam is one where the reactions can be found using only the equations of static equilibrium. Conversely, a statically indeterminate beam requires additional compatibility equations because the number of unknowns exceeds the number of equilibrium equations.
• The stability and determinacy of a beam depend on the types and arrangements of supports and the way loads are applied. For a beam to be statically determinate, the sum of the vertical forces, the sum of the horizontal forces, and the sum of the moments must equal zero, and these conditions must be sufficient to find all unknown reactions without recourse to deformation compatibility.

Types of Beams:

• Simply Supported Beam: A beam supported at both ends with no moment resistance at the supports. This is a classic example of a statically determinate beam.
• Cantilever Beam: A beam fixed at one end and free at the other. This is also statically determinate because the reactions can be found using the equilibrium equations.
• Overhanging Beam: A beam that extends beyond one or both of its supports. It is statically determinate if it does not have additional supports that would introduce indeterminacy.
• Propped Cantilever Beam: A beam fixed at one end and simply supported at the other. This is statically indeterminate.
• Continuous Beam: A beam with more than two supports. This is generally statically indeterminate.
• Fixed Beam: A beam with both ends fixed. This is also statically indeterminate.

Overhanging beam:

• An overhanging beam is a type of beam that has one or both ends extending beyond its support(s). This type of beam is statically determinate if it does not have additional supports that would introduce indeterminacy. The reactions at the supports can be determined using the three equations of static equilibrium (∑Fx = 0, ∑Fy = 0, ∑M = 0).

Explanation:

Statically Indeterminate Beam

Definition: A statically indeterminate beam is a structural element subjected to constraints that cannot be solely determined by static equilibrium equations. This means that the beam has more supports or restraints than required to maintain its equilibrium, leading to additional reactions that are not computable through basic statics alone. Instead, methods such as compatibility conditions, deflection calculations, or advanced structural analysis techniques are needed to solve for these reactions.

Fixed Beam: A fixed beam, also known as a clamped beam, is a type of beam that is rigidly fixed at both ends. This fixing condition imposes both rotational and translational constraints at the support points, resulting in a beam that cannot rotate or translate at its ends. This type of beam is inherently statically indeterminate because the number of unknown reactions exceeds the number of static equilibrium equations available. Specifically, a fixed beam has four unknown reactions: two vertical reactions, one horizontal reaction, and two moments (one at each end), while only three equilibrium equations (∑Fx = 0, ∑Fy = 0, and ∑M = 0) are available. Therefore, additional equations derived from the beam's deflection or compatibility conditions are required to solve for the unknown reactions.

Advantages:

• Increased stiffness and reduced deflection due to the fixed support conditions, making it suitable for applications requiring minimal deformation.
• Enhanced load-carrying capacity because the fixed ends provide additional moment resistance.

Disadvantages:

• Complexity in analysis and design due to the need for additional compatibility conditions or deflection calculations to determine the reactions.
• Potential for higher internal stresses at the fixed ends, which may require careful consideration in the design to avoid material failure.

Applications: Fixed beams are commonly used in construction and mechanical structures where high rigidity and minimal deflection are desired, such as in bridges, buildings, and machine frames.

Important information about other options:

Simply Supported Beam: A simply supported beam is supported at both ends, typically with a pin support at one end and a roller support at the other. This arrangement allows the beam to rotate but not translate at the supports, making it statically determinate. The reactions at the supports can be determined using only the static equilibrium equations, without the need for additional compatibility conditions.

Overhanging Beam: An overhanging beam extends beyond its supports on one or both ends. While it may appear more complex than a simply supported beam, it is still statically determinate if it has only two supports (one pin and one roller), as the reactions can be found using the basic equilibrium equations.

Cantilever Beam: A cantilever beam is fixed at one end and free at the other. It is statically determinate because the reactions (vertical reaction, horizontal reaction, and moment) at the fixed end can be determined using the equilibrium equations. Despite its simplicity in terms of analysis, cantilever beams are commonly used in applications requiring a free end, such as in balconies, overhangs, and certain types of bridges.

Explanation:

Initially, taking moment about hinged point C,

ΣΜ_C = 0

A_y × 12 - 100 × 9 - 50 × 3 = 0

⇒ A_y = 87.5 KN

FBD for left half section:

Now taking moment about point B.

ΣM_B = 0

A_x × 6 - 87.5 × 6 + 100 × 3 = 0

⇒ Ax = 37.5 kN

ΣΕ_x = 0

A_x - B_x = 0

⇒ B_x = A_x = 37.5 kN

Explanation:

Statically indeterminate structures are those structures that cannot be analyzed using statics or equations of equilibrium. In such cases, the number unknowns exceeds the number of equilibrium equations available.

Checking each option:

For option (1),

Number of unknown = 3

Number of equilibrium equation = 3

For option (2),

Number of unknown = 3

Number of equilibrium equation = 2

For option (3),

Number of unknown = 2

Number of equilibrium equation = 2

For option (4),

Number of unknown = 2

Number of equilibrium equation = 2

So, (2) is statically indeterminate structures.

Explanation: 

Load: A load is an external effort acting on the structure.

Uniform loading: The load that is either equal or varies uniformly along the element length is given as a uniform load.

Where, UDL = uniformly distributed load, UVL = uniformly variable load

Static loading: Static load is a load that doesn't change over time and there will be no vibrations and no dynamic effects on the member.

Dynamic loading: If the load increases rapidly and changes over time it is known as dynamic loading 

⇒ Impact loading is an example of dynamic loading

Fatigue load: Fatigue loading is primarily the type of loading that causes cyclic variations in the applied stress or strain on a component. Thus any variable loading is basically fatigue loading. 

Under fatigue load, the material fails due to sudden propagation of cracks and fractures

Explanation:

When rod swings to right, Linear acceleration & angular acceleration (∝) comes into picture

Let R = reaction at hinge

Linear acceleration

\(a = \propto r = \frac{L}{2} \times \propto \Rightarrow \propto = \frac{{2a}}{L}\)

and \({\rm{\Sigma }}{M_G} = {I_G} \times \propto\)

\(R\left( {\frac{L}{2}} \right) + P\left( {\frac{L}{6}} \right) = \frac{{M{L^2}}}{{12}}\left( {\frac{{2a}}{L}} \right)\\ \Rightarrow a = \frac{{3R}}{M} + \frac{P}{M}\)           ______________(1)

\(P-R = Ma = M\left( {\frac{{3R}}{M} + \frac{P}{M}} \right)\)

⇒ P – R = 3R + P ⇒ R = 0

Concept:

1) When a rotating body is involved then there comes a force which is known as Centrifugal force (F_c)

It is calculated by using formula

F_c = mr(ω)²

2) Now in the Question, it is mentioned that end Q is just lifted off the ground. So as it is just lifted off the ground there will be no reaction force from that point.

∴ FBD of mobile

M = mass of mobile

m = eccentric mass

r = eccentricity

Hence, Taking a moment about point P = O

∴ Mg × 0.06 - F_c × (0.09) = 0

∴ 90 × 10^-3 × 9.81 × 0.06 = mr(ω)² × (0.09)

90 × 10^-3 × 9.81 × 0.06 = 2 × 10^-3 × 2.19 × 10^-3 (ω)² × 0.09

∴ ω² = 134380.8964

∴ ω = 366.58

\(\therefore \frac{{2\pi N}}{{60}} = 366.58\)

∴ N = 3500 rpm

Explanation:

Let RA and RB be the reactions at points A and B respectively

∑ FY = 0

R_A + R_B = W, RB = W - RA

Balancing the moment about point B.

∑ MA = 0

W × a = R_B × l

W × a = (W - R_A) × l

W × a = (W × l) - (R_A × l)

\(R_A = {W(l-a)\over l}\)

Explanation:

The beam is a structural member that is subjected to transverse loading to its axis.

Continuous beam:

• It is a statically indeterminate multi-span beam on hinged support.
• The end span may be cantilever, may be freely supported or fixed supported.
• It is a beam having more than 2 supports.

Fixed beam:

• It is a beam with ends restrained from rotation.
• The figure shows a fixed beam subjected to uniformly distributed load throughout.
• It is a statically indeterminate beam on Fixed support.

Overhanging beam:

• It is defined as a beam that has its one or both ends stretching out past its support.
• It can have any number of supports.
• It is a statically determinate multi-span beam on hinged support.

Simply supported beam:

• A simply supported beam is a type of beam that has pinned support at one end and a roller support at the other end.
• It is a statically determinate multi-span beam on hinged support.

Propped cantilever beam:

• The propped cantilever beam is a beam with one end fixed and the other end simply supported.
• It comes under statically indeterminate beams.

Cantilever beam: 

• A beam fixed at one end and free at the other end is known as a cantilever beam.
• It comes under statically determinate beams.

From the above,

Statically determinate beams

1. Cantilever beam
2. Simply supported beam
3. Overhanging beam

Statically indeterminate beams

1. Propped cantilever beam:

2. Fixed beam

3. Continuous beam

Calculation:-

Given:-

P = 20 kN, θ = 60° 

\(\cos \theta = \frac{{base}}{{{\rm{Hypotenuse}}}}\)

\(\cos \;60^\circ = \frac{{OX}}{{OP}}\)

\(\frac{1}{2} = \frac{{OX}}{{20}}\)

OX = 10 kN

Concept:

Equilibrium Conditions:

\(\sum {{\rm{F}}_{\rm{v}}} = 0{\rm{\;and\;}}\sum {{\rm{M}}_{\rm{P}}} = 0\)

Calculation:

Given:

Now, we know that

Balancing vertical forces by 

\(\sum {{\rm{F}}_{\rm{v}}} = 0\)

R_P + R_Q = 0       ............ (1)

Now,

Taking moment about point 'P'

\(\sum {{\rm{M}}_{\rm{P}}} = 0\)

1 = R_Q × 1 ⇒ R_Q = 1 kN (upward)

Then, from equation (1), we get

R_P = 0 - 1 

∴ R_p = -1 kN = 1 kN (downward)

EXPLANATION:-

A beam is said to be in static equilibrium when the beam is initially at rest and remains at rest when subjected to a system of forces and couples.

Generally, there are 3 equilibrium equations for a planer structure.

The conditions of zero resultant force and zero resultant couple can be expressed as:

\(\sum Fx=0\;\)

\(\sum Fy=0\;\)

\(\sum Mz=0\;\)

where F_x represents forces along the horizontal or x-axis, F_y represents forces along the vertical or y-axis, and M_z represents the sum of moments taken around any point on the beam.

Degree of indeterminacy = Total number of unknown forces - Total number of equilibrium equations

Eg. Statically indeterminate structure

Here, the total number of unknown forces = 7

total number of equilibrium equations = 3

So, the degree of indeterminacy = 7 - 3 = 4, which is greater than zero so it is a statically indeterminate structure.

Eg. Statically determinate structure

Here, the total number of unknown forces = 3

Total number of equilibrium equations = 3

So, the degree of indeterminacy = 3 - 3 = 0, so it is a statically determinate structure.

Concept:

• Truss: A framework composed of members joined at their ends to form a rigid structure.
• Planar truss: In these types of trusses, members lie in a single plane.
• The basic element of a planar truss is a triangle i.e. three members are joined at three points.
• For a planar truss, as there are three unknown support reactions. so if
• m = 2j - 3 then system is statically determinate
• A truss hinged at one end, supported on rollers at the other as shown in figure.
• Therefore, the net resultant reaction at the hinged end A is,
• \(R=\sqrt{V_A^2+V_B^2}\) i.e. the reaction at the hinged end will be Resultant of horizontal and vertical

Method for the resolution for the resultant force:

• Resolve all the forces horizontally and find the algebraic sum of the horizontal components.
• Resolve all the forces vertically and find the algebraic sum of the vertical components.

The resultant of the above both can be given below as,

\(R = \sqrt {{{\left( {\sum H} \right)}^2} + {{\left( {\sum V} \right)}^2}} \)

The resultant force will make an angle with the horizontal can be given as,

\(tan\theta = \frac{{\sum V}}{{\sum H}}\)