Beams MCQ Quiz - Objective Question with Answer for Beams - Download Free PDF
Last updated on May 9, 2025
Latest Beams MCQ Objective Questions
Beams Question 1:
Which of the following correctly represents the ultimate tensile force (Tu ) in the tensile reinforcement (As ) of a singly reinforced beam at the ultimate limit state of flexure, considering the characteristic yield strength of steel (fy )?
Answer (Detailed Solution Below)
Beams Question 1 Detailed Solution
Explanation:
At the ultimate limit state of flexure for a singly reinforced beam, the ultimate tensile force (Tu) in the tensile reinforcement (As) is derived from the stress-strain relationship of steel in tension at its yield strength.
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fy is the characteristic yield strength of steel.
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As is the area of tensile reinforcement.
At ultimate limit state:
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The stress in the tensile reinforcement is typically assumed to be 0.87 fy (the stress corresponding to the yield strength of steel).
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The tensile force in the reinforcement, Tu, is given by: Tu = 0.87 fy As
Additional Information
Singly Reinforced Beam:
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A singly reinforced beam is a beam where the tensile reinforcement is placed in the bottom (tensile region) of the beam, while the compression reinforcement is either absent or negligible.
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The beam is subjected to bending, and the tensile reinforcement helps resist the bending stress.
Beams Question 2:
The Moment of Resistance (MR) in a beam is defined as:
Answer (Detailed Solution Below)
Beams Question 2 Detailed Solution
Explanation:
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Moment of Resistance (MR) is the internal resisting moment developed in a structural member (like a beam) due to bending stresses.
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It is equal in magnitude and opposite in direction to the external bending moment at a section in equilibrium.
Additional Information
MR is the maximum internal bending moment a section of a beam or structural member can resist without failing.
It depends on:
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Material strength (e.g., yield stress of steel, compressive strength of concrete)
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Section properties (especially the section modulus, Z)
Beams Question 3:
The ratio of the depth of the parabolic and rectangular portion block at the limiting state of collapse of a singly reinforced section is:
Answer (Detailed Solution Below)
Beams Question 3 Detailed Solution
Explanation:
Strain and Stress diagram of the beam
From the picture, the ratio of the depth of the parabolic and rectangular portion block = 4xu / 7 Divided by 3xu / 7 = 4/3
Beams Question 4:
The spacing between two vertical stirrups in a rectangular RCC beam is:
Answer (Detailed Solution Below)
Beams Question 4 Detailed Solution
Explanation:
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.
Beams Question 5:
Determine the minimum area of tension reinforcement permitted by IS 456: 2000 for the given (in figure) rectangular singly reinforced RCC beam. Reinforcement steel grade is Fe500.
Answer (Detailed Solution Below)
Beams Question 5 Detailed Solution
Concept:
Minimum tension reinforcement:
\(\frac{{{A_{st\ min}}}}{{b\; ×\; d}} = \frac{{0.85}}{{{f_y}}}\)
Where,
Ast min = Minimum tension reinforcement, b = width of the beam, d = depth of the beam, fy = yield strength of the steel, D = overall depth
Maximum tension reinforcement:
As per IS 456 maximum area of tension reinforcement in the beam shall not exceed 4% of the total area i.e. 0.04bD.
Calculation:
Given, b = 250 mm, d = 400 mm, D = 440, fy = 500 N/mm2
Minimum tension reinforcement:
\(\frac{{{A_{st\ min}}}}{{b\; ×\; d}} = \frac{{0.85}}{{{f_y}}}\)
\({{{A_{st\ min}}}} = \frac{{0.85\;×\; {{b \;×\; d}}}}{{{f_y}}}\)
\({{{A_{st\ min}}}} = \frac{{0.85\;×\; {{250\; ×\; 400}}}}{{{500}}}\)
\({{{A_{st\ min}}}} = 170 \ mm^2\)
Top Beams MCQ Objective Questions
The minimum tension reinforcement in beam should not be less than______.
Answer (Detailed Solution Below)
Beams Question 6 Detailed Solution
Download Solution PDFExplanation:
- 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:
\(\frac{{{A_{st\min }}}}{{bd}} \ge \frac{{0.85}}{{{f_y}}}\)
or, \({p_{t\min }} \ge \frac{{85}}{{{f_y}}}\)
= 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 'bw'
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
- Astmin = 0.15% of gross area for Fe 250
- Astmin = 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.
The ratio of the depth of the parabolic and rectangular portion block at the limiting state of collapse of a singly reinforced section is:
Answer (Detailed Solution Below)
Beams Question 7 Detailed Solution
Download Solution PDFExplanation:
Strain and Stress diagram of the beam
From the picture, the ratio of the depth of the parabolic and rectangular portion block = 4xu / 7 Divided by 3xu / 7 = 4/3
The breadth of rib in a T-beam should at least be equal to the __________ depth of rib.
Answer (Detailed Solution Below)
Beams Question 8 Detailed Solution
Download Solution PDFExplanation:
RCC T-beam:
(i) The beam consists of a flange and a rib in the form of a T, generally made of RC concrete or metal is known as T-beam.
(ii) The top part of the slab which acts along the beam to resist the compressive stress is called flange.
(iii) The part which lies below the slab and resists the shear stress is called the rib.
Dimension of T-beam:
- The effective width of the flange is adopted as the minimum of c/c distance of the nearby ribs or beams.
- The overall thickness of the slab crossing over the beam is taken as flange thickness.
- The breadth of the rib is taken down earth ground. It should be adequate to hold the steel zone in it, effectively. It might be taken as between 1/3 to 2/3 of the rib depth of the beam.
- The depth of T-beam is taken between 1/10 to 1/20 of the span.
The limiting depth of neutral axis for a beam having effective depth of 400 mm with Fe 250 grade steel is:
Answer (Detailed Solution Below)
Beams Question 9 Detailed Solution
Download Solution PDFConcept:
The ratio of limiting depth of neutral axis to the effective depth of the beam is given by
\(\frac{{{x_u}}}{d} = \frac{{{\rm{Max\;strain\;concrete}}}}{{\frac{{{\rm{Grade\;of\;steel}}}}{{{\rm{Modulus\;of\;Elasticity\;of\;steel}} \times {\rm{F}}.{\rm{O}}.{\rm{S}}}} + 0.002 + {\rm{Max\;strain\;concrete}}}}\)
Note:
Without getting into the tedious calculation,
We know the standard values of the ratio of limiting depth of neutral axis to the effective depth (k) of the beam for different steel sections as following:
Grade of Steel |
Fe 500 |
Fe 415 |
Fe 250 |
‘k’ value |
0.46 |
0.48 |
0.53 |
Calculation:
∴ For Fe 250, xu ≈ 0.53 × 400 = 212 mm.
An isolated T-beam is used as a walkway. The beam is simply supported with an effective span of 6 m. The effective width of the flange for the cross-section shown in the figure is:
Answer (Detailed Solution Below)
Beams Question 10 Detailed Solution
Download Solution PDFConcept:
Effective Width of Flange |
||
Monolithic Beams |
T - Beams |
\({b_f} = {b_w} + \frac{{{l_o}}}{6} + 6{d_f}\) |
L - Beams |
\({b_f} = {b_w} + 0.5\left( {\frac{{{l_o}}}{6} + 6{d_f}} \right)\) |
|
Isolated Beams |
T - Beams |
\({b_f} = {b_w} + \frac{{{l_o}}}{{\frac{{{l_o}}}{b} + 4}}\) |
L - Beams |
\({b_f} = {b_w} + \frac{{0.5 \times {l_o}}}{{\frac{{{l_o}}}{b} + 4}}\) |
Where
bf = effective width of flange
lo = distance between points of zero moments in the beam
bw = breadth of the web
df = thickness of the flange
b = actual width of the flange
For the given isolated T beam
Calculation:
Given,
The beam is Isolated T - Beam
From the figure
bf = 1000 mm, bw = 300 mm, df = 150 mm, b = 1000 mm, lo = 6 m = 6000 mm
\({b_f} = {b_w} + \frac{{{l_o}}}{{\frac{{{l_o}}}{b} + 4}}\) ⇒ \({b_f} = 300 + \frac{{6000}}{{\frac{{6000}}{{1000}} + 4}} = 900\;{\rm{mm}}\)
∴ Effective Flange width is 900 mm
What is the minimum area of tension reinforcement in beams when Fe 415 is used?
Answer (Detailed Solution Below)
Beams Question 11 Detailed Solution
Download Solution PDFConcept:
As per IS 456 : 2000,
a) Minimum % tension reinforcement:
\({{\rm{p}}_{{\rm{min}}}} = \frac{{0.85}}{{{{\rm{f}}_{\rm{y}}}}} \times 100\% \)
b) Maximum % tension reinforcement:
pmax = 4%of gross area
Calculation:
Minimum percentage of steel (for Fe500)
\({{\rm{p}}_{{\rm{min}}}} = \frac{{0.85}}{{415}} \times 100\%\)
= 0.2%
∴ Minimum percentage of steel (for Fe415) is 0.2In case of deep beam or in thin webbed R.C.C members, the first crack formed is-
Answer (Detailed Solution Below)
Beams Question 12 Detailed Solution
Download Solution PDFExplanation:
It is diagonal tension which is responsible for crack formation near the supports. The shear force is quite high and the bending stresses are very small. Hence principal tensile stress is nearly equal to shear stress.
A simply supported beam of overall depth 150 mm has neutral axis at 50 mm from extreme top fibre. Permissible flexural stresses in extreme compression and extreme tension fibre are restricted to 15 MPa and 25 MPa respectively. If the load on the beam is increased gradually, then:
Answer (Detailed Solution Below)
Beams Question 13 Detailed Solution
Download Solution PDFExplanation:
Let σc and σT are the Compressive and tensile stresses developed in concrete at top and bottom fibre respectively due to given load ‘P’
Since, stress distribution is Linear
\(\frac{{{σ _c}}}{{50}} = \frac{{{σ _T}}}{{100}}\)
⇒ \({σ _c} = \frac{{{σ _T}}}{2}\) ---(1)
Case 1:
Let an increasing the load, P (gradually), stresses in extreme, compression fibre reaches to its maximum permissible but in extreme tension fibres do not.
i.e.
σC = 15 MPa (given) and σt < 25 MPa
Now,
Using above equation (1)
15 = σT/2 ⇒ σT = 30 MPa > 25 MPa
Hence, our assumption is wrong.
Case 2:
Now, assume that stress in external bottom fibre tensile stress reduce and maximum permissible value but in extreme top fibre do not
i.e. σT = 25 MPa, σc ≤ 15 MPa
Now, from equation (1)
σC = 25/2 = 12.5 MPa < 15 MPa, Hence.
⇒ compressive stress in extreme top fibre is 12.5 MPa which is less then its permissible value, when, extreme bottom fibre reaches to its maximum value. Therefore, it can be concluded that failure of section will be due to stresses in extreme fibre reaching to maximum permissible tensile stresses, on increasing the load gradually.
A rectangular beam section of 300 mm width and 500 mm effective depth has tensile reinforcement of 1346 mm2. The beam M is constructed using M20 grade concrete and Fe 415 steel. Calculate the neutral axis.
Answer (Detailed Solution Below)
Beams Question 14 Detailed Solution
Download Solution PDFConcept:
Tensile force = 0.87 × fy × Ast
Compressive force = 0.36 × fck × xu × b
Actual neutral axis of the beam is given by
Compressive force = Tensile force
0.36 × fck × xu × b = 0.87 × fy × Ast
\(\;{x_u} = \frac{{0.87 \times {f_y} \times {A_{st}}}}{{0.36 \times {f_{ck}} \times b}}\)
Where xu is the depth of the neutral axis
fck is the characteristic strength of Concrete.
fy is the yield stress of steel
Ast is an Area of tensile reinforcement
b,d is the width and depth of the beam
Calculations:
Given,
Width of the beam = 300 mm
The effective depth of the beam = 500 mm
Area of steel = 1346 mm2
Grade of concrete is M20 i.e. fck = 20 N/mm2
Grade of steel is Fe 415 i.e. fy = 415 N/mm2
\(\;{x_u} = \frac{{0.87 \times {f_y} \times {A_{st}}}}{{0.36 \times {f_{ck}} \times b}}\)
\({x_u} = \frac{{0.87 \times 415 \times 1346}}{{0.36 \times 20 \times 300}}\)
= 225 mm
Diagonal tension in beam _____
Answer (Detailed Solution Below)
Beams Question 15 Detailed Solution
Download Solution PDFConcept
The concept of "diagonal tension" is mainly associated with shear stresses within a beam. Shear stress across the depth of the beam usually follows a parabolic distribution - close to zero at the top and bottom surfaces and maximum at (or near, but not typically exactly at) the neutral axis.