Beams MCQ Quiz in বাংলা - Objective Question with Answer for Beams - বিনামূল্যে ডাউনলোড করুন [PDF]

Explanation:

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

Explanation:

Continuous beam:

• It is a beam having more than 2 supports.
• The intermediate supports are always subjected to bending moment which is negative (Hogging case)
• If end supports are simply supported, the bending moment at end supports will be zero i.e MA = 0 & MC = 0
• If end supports are fixed, the bending moment at end supports will not be zero.
• Continuous beams are those that rest over three or more supports.

Explanation:

Point load is that load which acts over a small distance. Because of concentration over a small distance, this load can may be considered as acting on a point. Point load is denoted by P and symbol of point load is arrow heading downward (↓).

A uniformly distributed load is that whose magnitude remains uniform throughout the length.

Uniformly Varying Load (Non – Uniformly Distributed Load): It is that load whose magnitude varies along the loading length with a constant rate.

Uniformly varying load is further divided into two types:

• Triangular Load: whose magnitude is zero at one end of span and increases constantly till the 2nd end of the span.

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• Trapezoidal Load: which is acting on the span length in the form of trapezoid. Trapezoid is generally form with the combination of uniformly distributed load (UDL) and triangular load.

Explanation:

To remain in equilibrium all the forces or moments acting on the body should be balanced by the counter-reaction force or moment. 

Here in the given diagram, there is a moment at centroid which must be balanced by the moments produced by the reaction forces at the support to attain equilibrium.

Free body diagram of given configuration:

Configuration (P):

Since the reaction forces are concurrent at the centre of the plate, which cannot produce a balancing moment about the centre. Hence it will not be in equilibrium.

Configuration (Q):

the reaction forces are acting at the corner and generate a counterclockwise moment at the centre. Hence It will be in equilibrium.

Configuration (R)

the reaction forces acting at two corners along the side of the square plate will produce a counterclockwise moment at the centre to balance the moment and thus will remain in equilibrium.

Configuration (S)

In this configuration, one reaction force is concurrent to the centre and thus will not generate any moment at the centre.

The other reaction force will generate a counterclockwise moment at the centre and will balance the moment at the centre. Hence it will be in equilibrium.

Calculation:

For equilibrium of the structure,

Here,

H_A - Horizontal reaction at 'A' = 0 (Given)

R_A and R_B - Vertical reaction at 'A' and 'B' respectively.

Taking moment about A,

ΣM_A = 0 

R_B × \(\frac{L}{2}\) - \(\left[ \frac{1}{2}\times w \times L\times \frac{L}{3}\right]\) = 0

⇒ R_B = \(\frac{wL}{3}\)

Also,

ΣF_V = 0 

⇒ R_A = \(\left[ \frac{wL}{2}\right]\) - R_B

⇒ R_A = \(\left[ \frac{wL}{2}\right]\) - \(\left[ \frac{wL}{3}\right]\) = \( \frac{wL}{6}\)

Explanation:

\(BM\;at\;centre = W{\left( {\frac{L}{2} + a} \right)^2} \times \frac{1}{2} - W\left( {\frac{L}{2} + a} \right) \times \frac{L}{2}\)

\(=\;W \times \left( {a + \frac{L}{2}} \right)\left( {\frac{a}{2} - \frac{L}{4}} \right)\)

BM will change sign if \(\frac{a}{2} - \frac{L}{4} < 0\)

L > 2a

Concept:

At static equilibrium

Net force = 0

At dynamic equilibrium

Net moment about any point = 0

Moment of a force = (F × L) Nm

where F = Force, L = perpendicular distance of point of force from point about which moment is to be calculated

Calculation:

Given:

L₁ = 3 m, L₂ = 7 m, net upward force = 5 N

F₁ + F₂ = 5 N

Also

Let Net moment about point of action of 5 N force = 0

F₁ ×  3 = F₂ × 7

\(∴ F_1 = \frac 73 F_2\)

Now

F1 + F2 = 5 N

\(\frac 73 F_2 +F_2= 5~N\)

\(\frac {10}3 F_2 = 5\)

∴ F₂ = 1.5 N

and F₁ = 5 - F₂ = 5 - 1.5 = 3.5 N

Concept:

For equilibrium, sum of horizontal force = 0

∴ R_BH + R_AH = 0

Sum of vertical force = 0

R_BV - R_AV = P

Stress on  \(AB = \frac{{Force\;on\;AB}}{{Area\;of\;AB}} = \frac{{{R_{BV}} - {R_{AV}}}}{{\frac{\pi }{4}{d^2}}}\)

\({\sigma _{AB}} = \frac{{4P}}{{\pi {d^2}}}\)

Explanation:

1. Railway sleepers: These are the wooden or concrete components on which the rail-tracks are arranged with proper gauge. The load from rails when train passes, is taken by these sleepers. Railway sleepers with rail tracks form Overhanging beam.
2. Portico Slabs: A portico or porch is an area with a roof by the front door of a home. Either columned or roof-only, porticos are places to prepare for going out or coming in from the elements. Roof-only porticos (without columns) are usually installed over side doors.

          Porch, roofed structure, usually open at the sides, projecting from the face of a building and used to protect the entrance. So, these are Cantilever beams.

3. Roof Slab: A slab is a structural element, made of concrete, that is used to create flat horizontal surfaces such as floors, roof decks and ceilings. A slab is generally several inches thick and supported by beams, columns, walls.
4. Bridge: Based on the type of support bridges are classified as:

• Simply supported beam
• Cantilever beam
• Overhanging beam
• Continuous beam
• Fixed beam

Portico slabs are always Cantilevered, but in bridges support condition varies based on terrain, geographical conditions and desired load capacity.

Explanation:

The strength of two beams of the same material can be compared by the section modulus values.

The beam is stronger when section modulus is more, the strength of the beam depends on section modulus.

The strength of the beam also depends on the material, size, and shape of cross-section. 

Section modulus of a beam can be expressed as

 \({\sigma _b} = \frac{M}{Z}\)

Therefore, 

\(Z = \frac{M}{{\sigma _b}}\)

where Z is the section modulus and found out as \(Z = \frac{I}{y}\)

\(Z=\frac{I}{y}=\frac{\frac{\pi d^4}{64}}{\frac{d}{2}}={\frac{\pi d^3}{32}}\)

Strength of the beam ∝ Z ∝ d3