Test: Design for Shear and Torsion - Civil Engineering (CE) MCQ

Curved cables are advantageous in reducing the effective shear and together with the horizontal compressive prestress, reduce the magnitude of the principal tension and the effect of shear is to induce tensile stresses on diagonal planes and prestressing is beneficial since it reduces the magnitude of the principal tensile stress in concrete.

• As per IS 456 CI. D-1, Restrained slab is the slab whose corners are prevented from getting lifted up and are provided with suitable reinforcement to resist torsion and are called a restrained slab.
• All four edges of the slab are assumed to be rigidly tied with the beams or walls underneath and the edges may be either continuous (Fixed) or discontinuous.
• A corner restrained in the slab reduces the bending moment and deflection in the middle of the slab just like a fixed beam which reduces the mid-span moment and deflection as compared to simply supported beams.

Important Points
As per IS 456 (B 1.8, B 1.9, B 1.10), Torsion reinforcement is provided as:
B 1.8 Torsion reinforcement is required at the corner where both edges are discontinuous
Reinforcement is provided in 4 layers..
size of mesh = 1/5
Area of steel in each layer = (3/4) x A_st (+ve)
B 1.9: Torsion reinforcement is required where at least one edge is discontinuous
Provided in 4 layers
size of mesh = 1/5
Area of steel in each layer = (3/8) x A_st (+ve)
B 1.10: Where both edges are continuous no need to provide torsional reinforcement.

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

The various codes recommend empirical relations to estimate the ultimate shear resistance of the section by considering the flexure shear and web shear cracking modes and the design shear resistance should exceeded the ultimate shear due to the transverse loads and if not, suitable transverse reinforcements are designed to resist the balance shear force.

As per clause 41.3.1, IS 456:2000, The expression for equivalent shear is given by:
 Where V_e = Equivalent Shear, V_u = Nominal Shear, T_u = Torsional Moment, b = Width of beam
Important Points
The equivalent bending moment, M_el given by 
, Where M_el = Equivalent bending moment, M_u = Bending moment at the cross section, T_u = Torsional moment, D = Overall depth of beam, b = Width of beam.

The shear strength of reinforced concrete with the reinforcement is restricted to some maximum value τ_cmax depending on the grade of concrete.
Table 20 of IS 456
Stipulates the maximum shear stress of reinforced concrete in beams τ_cmax as given below in Table. Under no circumstances, the nominal shear stress in beams τ_v shall exceed τ_cmax given in the table for different grades of concrete.

Important Points
When the shear stress in concrete exceeds these values it leads to brittle failure due to diagonal compression.

In members with thin webs such as I and T sections nominal shear reinforcements have to be provided to prevent cracking due to variations in temperature the provisions for design of shear reinforcements prescribed in British, American and Indian standard have been dealt with in the case off structural concrete members subjected to torsion, shear stresses develop depending upon the type of cross section and magnitude of torque and the shear stresses in association with the flexural stresses may give rise to principal tensile stress the value of which when it exceeds tensile strength of the concrete results in the development of cracks on the surface of the member.

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% of web area and distribute it equally on both side faces

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

• 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.
• 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.

Pre tensioned or post tensioned members with bonded tendons develop bond stresses between steel and concrete when the sections are subjected to transverse shear forces due to the rate of change of moment along length of the beam and in the case of type 1 and 2 members, which are uncracked at service loads, the flexural bond stresses developed are computed by considering the complete section.