Test: Beam - Civil Engineering (CE) MCQ

Side face reinforcement:
Side face reinforcement shall be provided as per CI. 26.5.1.3 and 26.5.17(6) of IS 456:2000

• Beam depth > 450 mm (if beam subjected to torsion)
• Beam depth > 750 mm (if beam not subjected to torsion)
• Provide @ 0.1% OR 0.001 times of web area and distribute it equally on both side faces

NOTE:
In question it is not mentioned, wether the beam is subjected to torsion or not, so we have selected the most appropriate option, as correct option for beam subjected to torsion ie 450 mm is not given in the solution. We have to go for 750 ie for beam not subjected to torsion.

Beam is a structural member subjected to transverse loads that is loads perpendicular to its longitudinal axis. The mode of deflection of beam is primarily by bending.

Moment of resistance for under reinforced beam is given by:
MOR = 0.87× f_y × A_st(d - 0.42X_u)
Where, X_u = Depth of neutral axis, A_st = Area of reinforcement
Depth of neutral axis is calculated as:
Compression force = Tension force
0.36 f_ck b X_u = 0.87 f_y A_st
Calculation:
Given,
d = 450 mm, b = 300 mm
M25 and Fe415

We know that,
Depth of Neutral Axis (x_u):

Ulyimtae MOR is given as:
M_u = 0.87 f_y.A_st.(d - 0.42X_u)
M_u = 0.87×415× 1472.62 (450- 0.42 × 196.9)
M_u =  195290598.7 N-mm
M_u = 195.29 kN-m 
Working MOR = 
= 195.29/1.5
Working MOR = 130.19 kN-m

Structural members subjected to bending accompanied by large axial compressive loads at the same time are known as beam-column. A beam-column differs from column only by presence of eccentricity of load application, end moment, transverse load.

Assuming a simply supported reinforced concrete beam carrying uniformly distributed load of intensity w kN/m over the span length 'L'.
The maximum shear force would be produced at the supports and the shear force at the center of the span is zero.
∴ The maximum shear resistance is required at the ends of the supports and the minimum at the center of the span.
Shear reinforcement shall be provided in any of the following forms:
a) Vertical stirrups
b) The bent-up bar along with stirrups
c) Inclined stirrups

Important Points

• Stirrups are provided to take up the shear stress in the rectangular beam.
• Spacing is the distance between each consecutive stirrup.
• And we know from earlier that the shear force acts more near the support
• So to take up that shear stress, more stirrups need to be present at the support rather than the middle of the beam.

Note that, the shear reinforcement increases as the spacing among the stirrups decreases and vice versa.
As the shear strength requirement is more at the supports than the center, the spacing of stirrups decreases towards the end of the beam.

Girt is horizontal member fastened to and spanning between peripheral column of industrial buildings. It is used to support wall cladding such as corrugated metal sheet.

To calculate the uniform distributed load (UDL) that a simply supported beam can carry without exceeding the maximum bending stress, we use the bending stress formula:

σ = (M * y) / I

Where:
- σ is the bending stress,
- M is the moment at the point of interest,
- y is the distance from the neutral axis to the outermost fiber (half of the beam depth for a symmetrical I-section),
- I is the moment of inertia of the beam section.

Step 1: Moment of inertia and maximum bending stress
Given values:
- Moment of inertia, I = 1 x 10^8 mm^4,
- Maximum bending stress, σ = 100 N/mm²,
- Beam depth, d = 300 mm,
- Span of the beam, L = 6 m = 6000 mm.

For a simply supported beam under a uniformly distributed load (UDL), the maximum moment occurs at the center and is given by:

M = (w * L²) / 8

Where:
- w is the UDL in N/mm,
- L is the span of the beam.

Step 2: Applying the bending stress formula
the maximum bending stress occurs at the outermost fiber, which is at a distance of y = d / 2 = 300 / 2 = 150 mm.

Using the bending stress formula:

σ = (M * y) / I

Substitute for M:

σ = ((w * L²) / 8) * y / I

Substitute the known values:

100 = ((w * (6000)²) / 8) * 150 / (1 * 10⁸)

Step 3: Solve for w
Now, let's solve for w:

100 = ((w * 36000000 * 150) / (8 * 10⁸))

100 = (w * 5400000000) / (8 * 10⁸)

100 = 6.75 * w

w = 100 / 6.75 = 14.81 N/mm

Step 4: Convert to kN/m
Since w is in N/mm, we need to convert it to kN/m.

w = 14.81 N/mm = 14.81 * 1000 N/m = 14.81 kN/m

Thus, the UDL that the beam can carry is 14.81 kN/m.

Final Answer:
The correct option is (c) 14.81 kN/m.

Lintels are beam members used to carry wall loads over wall openings for doors, windows, etc.

• A minimum area of tension steel is required in flexural members (like beams) in order to resist the effect of loads and also control the cracking in concrete due to shrinkage and temperature variations.
• Minimum flexural steel reinforcement in beams: CI. 26.5.1.1 of IS 456:2000 specify the minimum area of reinforcing steel as:

= 0.34% for Fe 250
= 0.205% for Fe 415
= 0.17% for Fe 500
For flanged beams, replace 'b' with the width of web 'b_w'
Important Points

• The maximum area of tension steel in beams(Intension beams as well as compression beam) provided as per IS 456:2000 = 4% of gross area
• The minimum area of tension steel in the slab as per CI. 26.5.2 of IS 456:2000 
  • A_stmin = 0.15% of gross area for Fe 250
  • A_stmin = 0.12% of gross area for Fe 415

Confusion Points

• Minimum flexural steel reinforcement in the slab is based on shrinkage and temperature consideration and not on strength consideration because, in slabs, there occurs a better distribution of loads effects unlike in beams, where minimum steel requirement is based on strength consideration.

As per IS 456 : 2000,
a) Minimum % tension reinforcement:

b) Maximum % tension reinforcement:
p_max = 4%of gross area
Calculation:
Minimum percentage of steel (for Fe500)

= 0.2%
∴ Minimum percentage of steel (for Fe415) is 0.2