RCC : Shear, torsion & Bond and Development length

A. LIMIT STATE OF SHEAR

1. INTRODUCTION

RCC structures are generally subjected three type of shear stresses namely Flexural shear, punching shear and torsional shear. Diagonal tension arises in the section due to the shear forces which results in formation of cracks in concrete as concrete is weak in tension. Hence, limit state of shear must be checked in design of section. Shear reinforcement is provided in the form of vertical stirrup or bent up bars.

Following are the major mechanism in which shear failure takes place:

(i) Diagonal Tension: At the support of simply supported beam, bending stresses are zero and the shear stresses are maximum. Thus, tensile stresses generated along a plane inclined at 45˚ to the horizontal. As concrete is weak in tension, cracks start to develop at an angle of 45˚.

(ii) Flexural Shear: This type of cracks occurs due to large bending moment and small shear force. Cracks are normally inclined at 90˚ to the horizontal.

(iii) Diagonal Compression: This type of failure also takes place along with diagonal tension in heavily reinforced section. It is characterised by crushing of concrete.

NOMINAL SHEAR STRESS

The average shear stress can be calculated using the following formula:

Where,

V_u = ultimate shear stress at the section

b = width of the section

d = effective depth of the section

For beams with varying depth

Where,

β = inclination of flexural tensile force to the horizontal.

M_u = factored bending moment at the section.

Bond, Anchorage and Development Length

Bond stress 

Where V = Shear force at any section

d = Effective depth of the section

∑ p = Sum of all perimeter of reinforcement

n = Number of reinforcement

 diameter of reinforcement

Permissible bond stress

As per IS 456 : 2000

• These value of bond stress is for plain bar in tension.
• For deformed bar the above value should be increased by 60%.
• For bar in compression the above value should be increased by 25%.

Development Length (L_d)

Shear stress

Shear stress

(a) For Homogeneous beam

where,

q = shear stress at any section

V = shear force at any section

 Moment of area of section above the point of consideration

I = Moment of inertia of section

(b) For Reinforced concrete beam

(i) Shear stress above NA

(ii) Shear stress below NA 

Nominal shear stress, 

Design shear strength of concrete  without shear reinforcement as per IS 456: 2000  depends on

(i) Grade of concrete

(ii) Percentage of steel,

Where,

A_st = Area of steel

b = Width of the Beam

d = Effective depth of the beam

Maximum shear stress  with shear reinforcement is

Minimum shear reinforcement (As per IS 456 : 2000)

This is valid for both WSM and LSM

where,

A_SV = Area of shear reinforcement

S_V = Spacing for shear reinforcement

Spacing of shear reinforcement

Maximum spacing is minimum of (i), (ii) and (iii)

(i) 

(ii) 300 mm

(iii) 0.75 → For vertical stirrups

d → For inclined stirrups

where, d = effective depth of the section

(a) Critical section X-X at d from the face of the support

(b) Critical section X-X at the face of the support

The above provisions are applicable for beams generally carrying uniformly distributed load or where the principal load is located beyond 2d from the face of the support.

Vertical stirrups:

Shear force V_S will be

Resisted by shear Reinforcement provided in 'd' length of the beam,

where, A_SV = Cross-sectional area of stirrups

S_V = Centre to centre spacing of stirrups

Inclined stirrups: or a series of bars bent-up at different cross-section:

Bent up Bars:

• Single or a group of bent up bars are provided at distance  from support.

Generally bar should not be bent up beyond a distance l/4 from the support. Where l = length of span.

 B. LIMIT STATE OF TORSION

1. INTRODUCTION

In RCC structures generally torsion is accompanied by flexure and shear. It occurs if the line of action of force does not passes through the shear centre. Torsion is divided into two categories:

(i) Primary Torsion: Developed due to eccentric loading.

(ii) Secondary Torsion: Torsion is induced in the structure due to rotation of a member which depends on torsional stiffness of the member.

1. DESIGN OF TORSION

Torsion induces shear stress in the member which causes warping of non-circular section. Diagonal tension develops in the section which causes torsional cracking in the section. Torsional strength is improved by providing longitudinal and transverse reinforcement. The longitudinal reinforcement helps in reducing cracks through dowel action while transverse reinforcement helps in resisting shear stresses.

As per IS 456 the effect of torsional stresses is divided into

(i) Equivalent Shear

(ii) Equivalent moment

2.1. Equivalent Shear

V_e = Equivalent shear force

V_u = Shear force

T_u = Torsional moment

B = Width of the section

The equivalent nominal shear stress is calculated by dividing the equivalent shear force divided by area of the section.

2.2.  Longitudinal reinforcement

The longitudinal tension reinforcement should be designed to carry equivalent bending moment of

Where,

M_u = Flexural moment

M_t =  

T_u = Torsional moment

D = Overall depth of the section

b = width of the section

• If numerical value of M_t exceeds the numerical value of M_u, compression longitudinal reinforcement is provided to resist the bending moment equal to (M_t – M_u).
• The reinforcement should be provided close to the corners.
• If the depth of the section exceeds 450 mm, the area of reinforcement equal to 0.1% of web area should be distributed equally on both faces.

2.3.  Transverse Reinforcement

As per Is 456, transverse reinforcement is provided in the form of two legged closed hoops. The area of transverse reinforcement is obtained by the following formula:

Where,

T_u = Torsional moment

V_u = Shear force

s_v = Spacing of shear reinforcement

b₁ = centre to centre distance between corner bar in the direction of width

d₁ = centre to centre distance between corner bar in the direction of depth

b = width of the member

f_y = Characteristics strength of stirrup reinforcement

τ_ve = equivalent nominal shear stress

τ_c = shear strength of concrete

Note: The distribution of transverse reinforcement should be such that the spacing should be a minimum value of x₁,  or 300 mm where x₁ and y₁ are short and long dimension of stirrup.

x₁ = b₁ + Diameter of longitudinal bar + Diameter of stirrup

y₁ = d₁ + Diameter of longitudinal bar + Diameter of stirrup

You can avail of BYJU’S Exam Prep Online classroom program for all AE & JE Exams:

BYJU’S Exam Prep Online Classroom Program for AE & JE Exams (12+ Structured LIVE Courses)

You can avail of BYJU’S Exam Prep Test series specially designed for all AE & JE Exams:

BYJU’S Exam Prep Test Series AE & JE Get Unlimited Access to all (160+ Mock Tests)

Thanks

Team BYJU’S Exam Prep

Download  BYJU’S Exam Prep APP, for the best Exam Preparation, Free Mock tests, Live Classes.