Beams MCQ Quiz - Objective Question with Answer for Beams - Download Free PDF

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.

• f_y is the characteristic yield strength of steel.

• A_s is the area of tensile reinforcement.

At ultimate limit state:

• The stress in the tensile reinforcement is typically assumed to be 0.87 fy (the stress corresponding to the yield strength of steel).

• The tensile force in the reinforcement, Tu, is given by: Tu = 0.87 fy As​​

Additional Information

Singly Reinforced Beam:

• 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.

• The beam is subjected to bending, and the tensile reinforcement helps resist the bending stress.

Explanation:

• Moment of Resistance (MR) is the internal resisting moment developed in a structural member (like a beam) due to bending stresses.

• 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:

• Material strength (e.g., yield stress of steel, compressive strength of concrete)

• Section properties (especially the section modulus, Z)

Explanation:

Strain and Stress diagram of the beam

From the picture, the ratio of the depth of the parabolic and rectangular portion block = 4x_u / 7 Divided by 3x_u / 7 = 4/3

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.

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:

\(\frac{{{A_{st\ min}}}}{{b\; ×\; d}} = \frac{{0.85}}{{{f_y}}}\)

\({{{A_{st\ min}}}} = \frac{{0.85\;×\; {{b \;×\; d}}}}{{{f_y}}}\)

\({{{A_{st\ min}}}} = \frac{{0.85\;×\; {{250\; ×\; 400}}}}{{{500}}}\)

\({{{A_{st\ min}}}} = 170 \ mm^2\)

Explanation:

• 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 '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.

Explanation:

Strain and Stress diagram of the beam

From the picture, the ratio of the depth of the parabolic and rectangular portion block = 4x_u / 7 Divided by 3x_u / 7 = 4/3

Explanation:

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.

Concept:

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:

Calculation:

∴ For Fe 250, x_u ≈ 0.53 × 400 = 212 mm.

Concept:

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

b_f = 1000 mm, b_w = 300 mm, d_f = 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

Concept:

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%

Explanation:

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.

Explanation:

Let σ_c and σ_T 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.

σ_C = 15 MPa (given) and σ_t < 25 MPa

Now,

Using above equation (1)

15 = σ_T/2 ⇒ σ_T = 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. σ_T = 25 MPa, σ_c ≤ 15 MPa

Now, from equation (1)

σ_C = 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.

Concept: 

Tensile force = 0.87 × f_y × A_st

Compressive force = 0.36 × f_ck × x_u × 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 x_u is the depth of the neutral axis

f_ck is the characteristic strength of Concrete.

f_y is the yield stress of steel

A_st 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. f_ck = 20 N/mm²

Grade of steel is Fe 415 i.e. f_y = 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

Concept

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.