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

Explanation:

In reinforced concrete design, the design shear strength of concrete increases with the percentage of tensile reinforcement (pt). Here's the reasoning behind it:

1. Tensile Reinforcement in Shear Resistance:

   • The primary function of tensile reinforcement in a beam is to resist bending, but it also helps in resisting shear forces.

   • When tensile reinforcement increases, it helps to improve the overall shear strength of the beam because it contributes to the beam's ability to resist shear stresses, particularly in preventing diagonal tension cracks.

2. Concrete and Reinforcement Interaction:

   • Concrete alone has limited shear strength, but with increased tensile reinforcement (especially in cases of high shear demand), the interaction between the concrete and reinforcement improves.

   • This interaction helps in better resisting shear forces, thus increasing the overall shear strength of the beam.

3. Increasing Tensile Reinforcement:

   • When the tensile reinforcement percentage increases, it provides more resistance to the formation of shear cracks and helps in carrying more shear force.

   • This results in a direct increase in the shear strength of the concrete beam.

Thus, the shear strength of concrete increases with an increase in the percentage of tensile reinforcement, and this is why the correct answer is 3.

 Additional Information

• The percentage of tensile reinforcement should not be excessive, as other factors such as ductility and the potential for early cracking come into play. But up to a certain limit, increasing the reinforcement enhances the beam’s shear strength.

Explanation:

• According to IS 456:2000, the spacing of vertical stirrups (shear reinforcement) in a reinforced concrete beam should not exceed 0.75 times the effective depth (d) of the beam.
• This ensures that the shear reinforcement is adequate to resist shear forces and prevent shear failure.

Additional Information Key Points from IS 456:2000:

1. Shear Stress: The shear stress in a beam increases near the supports, so the shear reinforcement must be more concentrated at these locations.

2. Stirrup Spacing: The code stipulates that the spacing of vertical stirrups (shear reinforcement) should not exceed 0.75 times the effective depth (d) of the beam to ensure that shear is effectively resisted across the entire length of the beam.

3. Shear Reinforcement Distribution: In general, stirrups should be placed closer together near the supports where shear is higher, and the spacing can increase as you move toward the middle of the beam where shear stress is lower.

Explanation:

• Stirrups are closed-loop steel bars placed perpendicular (or sometimes inclined) to the longitudinal bars in RCC beams.

• Their primary function is to resist shear forces and hold the longitudinal bars in position.

• They are most effective in preventing diagonal shear cracks near the supports.

Additional Information

• Compression bars: Resist compressive forces in columns or compression zones of beams, not shear.
• Lateral ties: Used in columns to confine the core concrete and prevent buckling of vertical bars, not for shear in beams.
• Longitudinal bars: Resist bending (flexural tension and compression), not shear.

Concept:

Spacing of verticle stirrups is given by,

\({{S}_{v}}=\frac{{{A}_{sv}}~\times 0.87~{{f}_{y~}}\times ~d}{V}\)

Maximum spacing = 0.75 d or 300 mm whichever is less

Calculation:

Area of 2 legged vertical stirrups of 8 mm diameter bar is,

\({{A}_{sv}}=2\times \frac{\pi }{4}~{{\left( 8 \right)}^{2}}=100.53~m{{m}^{2}}\)

1. \({{S}_{v}}=\frac{{{A}_{sv}}~\times 0.87~{{f}_{y~}}\times ~d}{V}=\frac{100.53~\times 0.87~\times 415~\times 550}{58\times 1000}=~344.18~mm~c/c\)

2. Maximum spacing = 0.75d = 0.75 × 550 = 412.5 mm c/c or 300 mm

∴ Providing 300 mm spacing of verticle stirrups.

According to IS 456:2000, clause 26.5.1.5,

The maximum spacing of shear reinforcement measured along the axis of the member shall not exceed 0.75 d for vertical stirrups where d is the effective depth of the section under consideration. In no case shall the spacing exceed 300 mm.

Explanation

Minimum shear reinforcement in beams is provided to ensure the following:

• To prevent sudden failure of the beam due to loss of bond between steel and concrete due to bursting of concrete cover.
• To prevent brittle shear failure due to diagonal principal tension.
• To prevent cracks due to shrinkage of concrete and thermal stresses.
• To hold the main reinforcement and increases the confinement.

Out of the above given, the most critical is diagonal principal tension.

∴ Minimum shear reinforcement in beams is provided in the form of stirrups to resist the principal tension is a correct statement.

Note:

To resist extra shear due to live load or any other load and shear cracks, we need to analyse the shear capacity of the cross-section and accordingly, needs to design for shear reinforcement. 

Explanation:

• The above figure shows the shear failure of simply supported and cantilever beams without shear reinforcement. The failure plane is normally inclined at an angle of 30^o to the horizontal.
• However, in some situations, the angle of failure is more steep either due to the location of the failure section closed to a support or for some other reasons. Under these situations, the shear force required to produce failure is increased.

Concept

Equivalent Shear force (Ve) is given by

\({V_e} = V + \dfrac{{1.6T}}{B}\)

Where, T = Torque

B = Width of the section, V = shear force

Calculation:

Given,

B = 250 mm = 0.25 m,

V = 140 kN, T = 20 kN-m

Equivalent shear force is given by,

\({V_e} = V + \dfrac{{1.6\:\times \:T}}{B}\)

\({V_e} = 140 + \dfrac{{1.6 \times 20}}{{0.25}}\) = 268 kN

Additional Information

Equivalent Moment (Me) is given by

\({M_e} = M + \dfrac{T}{{1.7}}\left( {1 + \dfrac{D}{B}} \right)\)

Where, M = Bending moment, D = Overall dept, T is Torque

Concept:

Diagonal tension:

Diagonal Tension is the reason for the cracks in shear failure. The diagonal tension crack originates at an angle of 45° and breaks the beam into two pieces. It is present in both the brittle and ductile material.

Shear reinforcement is required to prevent the propagation of diagonal cracks.

The diagonal tension cracks develop in the beam when the strength of the beam in diagonal tension is less than the strength in flexural tension. On this type of failure, the cracks start to develop due to the flexure at mid-span. For the prevention of diagonal tension failure, the shear link is provided.

According to IS 456: 200, the shear reinforcement is must be provided

1. To resist the diagonal tension or principal tension. In the beam diagonal failure occurs when the shear span is higher than three times the value of effective depth.
2. To avoid the formation of tension cracks due to any reason given below

• Due to diagonal tension
• Due to the sudden increase in stress
• Due to shrinkage and creep
• To improve the ductility of member
• To improve the rate of main reinforcement

Concept:

Recent laboratory experiments confirmed that reinforced concrete in beams has shear strength even without any shear reinforcement. This shear strength (τc) depends on the grade of concrete and the percentage of tension steel in beams and this is known as the permissible shear stress in concrete.

On the other hand, the shear strength of reinforced concrete with the reinforcement is restricted to some maximum value τcmax depending on the grade of concrete.

The amount of shear reinforcement to be provided is determined to carry a shear force Vus equal to

Vus = Vu – τc ×  b × d

Where b is the breadth of rectangular beams or bw in the case of flanged beams.

d = effective depth of beam

Vus = Design shear force

Vu = Factored shear force and τc = permissible shear stress in concrete.

Calculation:

Given,

b = 300 mm           

d = 500 mm

Vu = 270 kN = 270,000 N

τc = 0.4 MPa

Design shear force is 

Vus = 270,000 – (0.4 × 300 × 500) 

V_us = 210,000 N = 210 kN 

Explanation:

The shear strength of reinforced concrete with the reinforcement is restricted to some maximum value τcmax depending on the grade of concrete.

Table 20 of IS 456

Stipulates the maximum shear stress of reinforced concrete in beams τcmax as given below in Table. Under no circumstances, the nominal shear stress in beams τv shall exceed τcmax given in the table for different grades of concrete.

Diagonal tension:

Diagonal Tension is the reason for the cracks in shear failure. The diagonal tension crack originates at an angle of 45° and breaks the beam into two pieces. It is present in both the brittle and ductile material.

Shear reinforcement is required to prevent propagation of diagonal cracks.

The diagonal tension cracks develop in the beam, when the strength of the beam in diagonal tension is less than the strength in flexural tension. On this type of failure, the cracks start to develop due to the flexure at mid-span. For the prevention of diagonal tension failure, the shear link is provided.

According to IS 456: 200, the shear reinforcement is must to be provided

1. To resist the diagonal tension or principal tension. In the beam diagonal failure occurs when the shear span is higher than three times the value of effective depth.
2. To avoid the formation of tension cracks due to any reason given below

• Due to diagonal tension
• Due to the sudden increase in stress
• Due to shrinkage and creep
• To improve the ductility of member
• To improve the rate of main reinforcement

Flexural crack:- In flexural tension failure, under the reinforced condition, the steel yielding first, after that concrete crack and the cracks goes up to a neutral axis from the compressive side of the beam. These types of cracks are avoided by providing the steel and concrete inappropriate manner that neither the section is under reinforced nor over reinforced.

Dowel crack:- It occurs in the road pavement when the concrete pavement is a

A dowel is provided a mechanical connection between concrete slabs.

The bar increases load transfer efficiency and reduces its joint stress, deflection. Hence the dowel crack is avoided by providing an adequate length of dowel bar.

Splitting crack:- This type of crack occurs due to insufficient steel reinforcement and low concrete quality. This type of crack is generally found in concrete columns with non-uniform width.

Concept: 

Bends and Hooks Forming End Anchorages ( As per IS 2502:1963 )

Here

k = 2 for Mild Steel

k = 3 for Medium Tensile Steel

k = 4 for Cold-worked Steel

As per IS 2502:1963, P-6, Table-II

H = Hook allowance taken as 9d, 11d, 13d, and 17d for k values 2, 3, 4 and 6 respectively and rounded off to the nearest 5 mm, but not less than 75 mm.

B = Bend allowance is taken as 5d, 5.5d, 6d, and 7d for k values 2, 3, 4 and 6 respectively and rounded off to the nearest 5 mm, but not less than 75 mm.

∴ Extra length for one hook = 9d

Concept:

Nominal shear stress (τvu)

= \({V_{ueq}\over bd}\)

Where V_ueq = Equivalent shear force, b = Width of the beam, and d = effective depth of the beam.

Vueq = Vu + \({1.6T_u\over b}\), (For LSM)

Calculation:

Given

Shear force (V_u) = 20 kN

Torque (T_u) = 9 kN-m

Width of the beam (b) = 300 mm

Total depth (D) = 425 mm

Effective cover (d') = 25 mm

Effective depth (d) = D - d' = 400 mm

Equivalent shear force (V_ueq)

= V_u + \({1.6T_u\over b}\)

Equivalent shear force (Vueq)

= 20 + \({1.6\times 9\over 0.3}\) = 68 kN

Equivalent shear force (Vueq) = 68 kN

Nominal shear stress (τ_vu)

= \({V_{ueq}\over bd}={68\times 10^3\over 300\times 400}\)

Equivalent Nominal shear stress (τvu) = 0.566 N/mm²

Equivalent Nominal shear stress (τvu) = 0.566 MPa

Concept:

The minimum shear reinforcement in the form of stirrups shall be provide such that:

\(\frac{{{{\rm{A}}_{{\rm{sv}}}}}}{{{\rm{b}}{{\rm{S}}_{\rm{v}}}}} \ge \frac{{0.4}}{{0.87{\rm{\;fy}}}}\)

Where,

A_sv = Total c/s area of stirrup legs effective in shear, S_v = Spacing along the length of member, b = width of beam, and f_y = characterastics strength of stirrups reinforcement

Calculation:

Given: b = 205 mm, S_v = 435 mm, and f_y = 410 N/mm²

\(\therefore {{\rm{A}}_{{\rm{sv}}}} = \frac{{0.4 \times 205 \times 435}}{{0.87 \times 410}} = 100{\rm{\;m}}{{\rm{m}}^2}\)