Frame Truss and Beam MCQ Quiz - Objective Question with Answer for Frame Truss and Beam - 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:

Sin θ = \(\frac{F_1}{F_{AB}}\), 
∑F_y = 0 ⇒ F₁ = mg
∴ Sin θ = \(\frac{mg}{F_{AB}}\)
⇒ F_AB = mg × Cosec θ

Explanation:

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 (For a node or joint, there will be three unknown but dependent support reactions). so if
  1. m = 2j - 3 then system is statically determinate
  2. m < 2j - 3 then system is collapsed
  3. m > 2j - 3 then system is statically indeterminate
• where, m = number of members, j = total number of joints (including those attached to the supports)

Explanation:

Degree of Freedom:

• The DOF of a mechanism refers to the number of independent parameters required to completely specify the configuration of the mechanism in space.
• According to Kutzback's equation of DOF

DOF = 3(n - 1) - 2j - h

where, n = Number of links, j = Number of joints, h = Number of higher pairs in mechanism.

The physical interpretation of DOF:

• DOF < 0, Overstructure/Superstructure
• DOF = 0, structure /Frame/Truss
• DOF > 0, Mechanism

• DOF of a mechanism predicts the possible number of output with respect to a given input.
• DOF also predicts the number of inputs required in order to obtain a constrained mechanism or number of links that should be controlled as input in order to have a single output.

Concept:

The force in the member AB will be determined by method of joints, i.e. by analyzing each joints.

For this we will be fixing some rule, they are

• for every joint we will consider the unknown forces are going away from the joint
• and the  forces are considered positive.
• So after determination, if one forces comes positive then, it will mean that it is tensile and if it comes negative then it would mean that it is compressive.

Calculations:

Taking moment about A, we get - 

R_C × 6 = 5 × 6 sin 60

⇒ R_C = 5 sin 60

⇒  RC = \(\frac{{5\sqrt 3 }}{2}\)

Analyzing joint C:

R_C + R_CB sin 60 = 0

⇒ RCB sin 60 = - RC 

⇒ RCB × \(\frac{{\sqrt 3 }}{2}\) = - \(\frac{{5\sqrt 3 }}{2}\)

⇒ RCB = - 5 kN [-ve denotes that the force will be toward the joint, hence it will be compressive in nature]

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

Concept: 

A plane truss or frame consists of several bars laying in one plane and connected by hinges or pins at their ends so as to provide a stable configuration.

A perfect frame is that which is composed of members just sufficient to keep it in equilibrium, when loaded, without any change in its shape.

A perfect frame should satisfy the following expression:

m = 2j - 3 where m = Number of members, and j = Number of joints.

An imperfect frame is one that does not satisfy the above equation (m = 2j - 3).

• If m < 2j - 3 then the frame is called a deficient frame.
• If m > 2j - 3 then the frame is redundant.

Calculation:

Given:

m = 5, j = 4

m = 2j - 3, 5 = 2 × 4 - 3

Hence, The given frame is perfect frame

Explanation:

The frame of reference: If we are observing any moving body or anybody at rest with respect to any moving object or from any object at rest then the object is called a frame of reference.

There are two types of frame of reference:

Inertial frame of reference:

• The frame of reference having zero acceleration is called the Inertial frame of reference.
• This frame of reference will be either at rest or will be moving with a constant velocity.
• Newton’s law is valid in this frame of reference.
• Example- A bus moving with constant velocity.

The non-inertial frame of reference:

• The frame of reference having non-zero acceleration is called a non-inertial frame of reference.
• Newton’s law is not valid in this frame of reference.
• For example: If we are observing an object from a freely falling object then this will be a non-inertial frame of reference because the freely falling body has some acceleration.

Explanation:

Rigid-jointed frames

• In a rigid jointed frame, the joints are considered to rotate as a whole.
• These are framed structures in which the members transmit applied loads by axial, shear, and bending effects.
• Rigid-joints (moment connections) are designed to transfer axial and shear forces in addition to bending moments between the connected members.

Concept:

Lami's theorem states that if three co-planar, concurrent forces acting at a point are in equilibrium, each force is proportional to the sine of the angle between the other two forces.

Consider three forces FA, FB, FC acting on a particle or rigid body making angles α, β and γ with each other.

From Lami's theorem:

\(\frac{A}{{\sin \alpha }} = \frac{B}{{\sin\beta }} = \frac{C}{{\sin\gamma }}\)

Calculation:

Given:

W = 200 N.

\(\frac{W}{\sin 120}=\frac{{{F}_{xy}}}{\sin 120}=\frac{{{F}_{yz}}}{\sin 120}\)

\(\frac{200}{\sin 120}=\frac{{{F}_{xy}}}{\sin 120}=\frac{{{F}_{yz}}}{\sin 120}\)

∴ F_xy = F_yz = 200 N.

If at a joint 3 members meet out of which 2 are collinear then force in 3^rd member will be zero provided there is no load/reaction at that joint.

∴ F_GC = 0, F_FB = 0 and F_HD = 0

i.e. forces in the member GC, FB, HD is equal to zero

Now, considering joint E:

∵ F_EH = R_E

There is no horizontal force.

∴ F_DE = 0

Also considering joint D, we can say that force in member CD = 0

Explanation:

Taking moment about point P

F ×  L - T _RS × L = 0

T RS = F

Force in member RS  = F (compressive)

Now FBD of point R

Taking summation of force in Y direction 

-T_PR sin45° + T_RS = 0

-TPR sin45° + F = 0

T_PR = F √ 2

Forces in the members PR  F √ 2 (tensile)