Shear And Diagonal Tension 2o466h

Shear and Diagonal Tension

Types of Shear Failure Diagonal Tension Failure -usually occur when the shear span is greater than 3d or 4d.

Shear-Compression Failure -occurs when the shear span is from d to 2.5d.

Splitting or True Shear Failure -occurs when the shear span is less than the effective depth d.

Basic code requirements factored shear force 𝑉𝑢 shall be equal or less than the design shear ∅𝑉𝑛 𝑽𝒖 ≤ ∅𝑽𝒏

Eq. 4-1

where ∅ = 0.85 𝑽𝒖 = 𝑽𝒄 + 𝑽𝒔

Eq. 4-2

Shear Strength Provided By Concrete, 𝑉𝑐 For Nonprestressing For subject to shear and flexure only

For subject to axial compression

At sections where factored torsional moment 𝑇𝑢 exceeds

For subject to shear and flexure only,

For subject to significant axial tension

where N, is negative for tension. Quantity N,j A8 shall be expressed in MPa. In

However, Vc shall not be taken greater than

Shear Strength Provided By Reinforcement When factored shear force V, exceeds strength q, Vc, shear reinforcement shall be provided to satisfy Eq. 4-1 and Eq. 4-2. The shear strength provided by the stirrups is given by the following but 2 shall not be taken greater than 3 𝑓′𝑐𝑏𝑤 (a) When shear reinforcement perpendicular to axis of member is used

where Av is the area of shear reinforcement within a distance s. Av = 2 Ab for a U stirrup

(b) When inclined stirrups are used as shear reinforcement

where 𝛼 is the angle between inclined stirrups and longitudinal axis of member. (c) When shear reinforcement consist of a single bar or a single group of parallel bars, all bent up at the same distance from the

TYPES OF SHEAR REINFORCEMENT According to Section 5.11.5.1 of the Code, shear reinforcement may consist of: a) stirrups perpendicular to axis of member, and b) b) welded wire fabric with wires located perpendicular to axis of member. For nonprestressed , shear reinforcement may also consist of: a) stirrups inaking an angle of 45° or more with longitudinal tension reinforcement, b) longitudinal reinforcement with bent portion making an angle of 30° or more with the longitudinal tension reinforcement, c) combinations of stirrups and bentlongitudinal reinforcement, and d) spirals.

Design Yield Strength Of Stirrups According the Section 5.11.5.2. the design yield strength of shear reinforcement shall not exceed 415 MPa. Stirrups and other bars or wires used as shear reinforcement shall extend to a distance d from extreme compression fiber and shall be anchored at both ends to develop the design yield strength of reinforcement.

Spacing Limits Of Shear Reinforcement, s According to Section 5.11.5.4 of the Code, the spacings of shear reinforcement placed perpendicular to axis of shall not exceed d/2 in nonprestressed and (3/4)h in prestressed , nor 600 mm. Inclined stirrups and bent longitudinal reinforcement shall be so spaced that every 45° line, extending toward the reaction from middepth of member d/2 longitudinal tension reinforcement, shall be crossed by at least one line of shear reinforcement. 1

When V, exceed3 𝑓′𝑐𝑏𝑤 𝑑maximum spacing given by the above limits shall be reduced by one-half.

Minimum Shear Reinforcement According to Section 5.11.5.5 of the Code, a minimum area of shear reinforcement shall be provided in all reinforced concrete flexural (prestressed and nonprestressed) where factored shear force 𝑉𝑢 exceeds one-half the shear strength provided by concrete ∅𝑉𝑐 except: (a) Slabs and footings (b) Concrete joist construction defined by Sec. 5.8.11 (c) Beams with total depth not greater than 250 mm, 2-1/2 times thickness of flange, or 1/2 the width of web, whichever is greatest.

Critical Section For Beam Shear According to section 5.11.1.3 of NS, the maximum factored shear force V,. at s may be computed in accordance with the following conditions provided that: (a) the reaction, in direction of the applied shear, introduces compression into the end regions of member, and (b) no concentrated load occurs between the face of the and the location of the critical section.

1. For non-prestessed , sections located less than a distance d from face of may be designed for the same shear Vu as that computed at a distance d.

2. For prestressed member, sections located less than a distance h/2 from face of may be designed for the same shear V, as that computed at a distance h/2.

STEP BY STEP PROCEDURE

1. Calculate the factored shear force Vu

Can either be in the critical section Or within any point in the beam

2. Calculate the shear strength provided by the concrete (Vc).

Use the appropriate formula for Vc. Typical formula is Vc = 0.17λ(√f’c)bwd

3. Check whether stirrups are necessary or not.

4. Calculate the shear strength provided by the stirrups.

5. Determine the required spacing of the stirrups.

6. Draw the shear reinforcement details for the beam

If Vu > ΦVc, stirrups are necessary, proceed to Step 5. If Vu < ΦVc but Vu > Φ(0.5Vc), use minimum area of stirrups If Vu > ΦVc, beam section is capable to carry the factored shear Vn = Vs + Vc If Vs < 0.66√f’c ,proceed to Step 5 Otherwise, adjust the size of the beam S = Avfytd/Vs Check with the required maximum spacing If Vs < 0.66√f’c , Smax = d/2 or 600 mm (get smaller) If Vs > 0.66√f’c , Smax = d/4 or 300 mm (get smaller)

Avmin = 0.062(√f’c)bwS/fyt > 0.35bwS/fyt

Sample Beam Reinforcement Details

SAMPLE PROBLEM

Problem. A rectangular beam has the following properties b = 320 mm Stirrup diameter = 10 mm d = 570 mm fyt = 275 MPa f'c = 24 MPa Determine the following: a. Required spacing of stirrups when the required shear strength is 90 kN. b. Required spacing of stirrups when the required shear strength is 130 kN.