Beams - Free MCQ Practice Test with solutions, GATE Civil Engineering Mechanics

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:

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.

When a simply supported beam is subjected to a point load, L at the center of the length of the beam then there will be the generation of two equal vertical reactions, L/2 at the ends.
By drawing a bending moment diagram of this beam, it is clearly indicated that the maximum bending moment in the beam occurs at the center of beam length and its value is [(L/2) × (l/2)] or L × l/4 kN-m

There are different types of beam:

Overhanging beam:

• A beam having its end portion extended beyond the support is known as an overhanging beam.
• A beam may be overhanging on one side or on both sides.

Cantilever beam:

• A beam fixed at one end and free at the other end is known as a cantilever beam.

Simply Supported Beam:

• A beam supported at its both ends is known as a simply supported beam.

Fixed beam:

• A beam whose both ends are fixed is known as a fixed beam.

Continuous beam:

• A beam supported on more than two supports is known as a continuous beam.

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

• Cantilever beam
• Simply supported beam
• Overhanging beam

Statically indeterminate beams

• Propped cantilever beam:
• Fixed beam
• Continuous beam

• 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
• 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

∴ N = 3500 rpm

When rod swings to right, Linear acceleration & angular acceleration (∝) comes into picture
Let R = reaction at hinge
Linear acceleration

 ...(1)

Equilibrium Conditions:

Calculation:
Given:

Now, we know that
Balancing vertical forces by 

R_P + R_Q = 0 ... (1)
Now,
Taking moment about point 'P'

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)

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 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 which causes cyclic variations in the applied stress or strain on a component. Thus any variable loading is basically a fatigue loading. 
Under fatigue load, the material fails due to sudden propagation of cracks and fractures

From differential equations of elastic curve

Where y = deflection of beam.
As the I = Area moment increases, the deflection will reduce.