CE 373– Concrete ILecture 6: 4.1 – 4.54.1) Introduction4.2) Diagonal Tension in Homogenous Elastic Beams4.3) RC Beams without Shear Reinforcement4.4) RC Beams with Web Reinforcement4.5) ACI Code Provisions for Shear Design

CE373 – Concrete I Dr. Ammar T. Al-Sayegh

4.1) Introduction Shear failure of RC beams, also called diagonal tension failure, is more

dangerous than flexural failure due to the greater level of uncertainty in predicting it and it’s usually of a catastrophic nature.

If shear reinforcement is not properly designed, shear failure is likely to occur suddenly with no advance warning of distress.

CE373 – Concrete I Dr. Ammar T. Al-Sayegh1

Shear stress in most beams is usually far below the shear strength of concrete. Shear stress, however, when combined with longitudinal flexural stress give diagonal tension stress which can exceed concrete strength.

The shear and bending stresses are given by

→ 𝑣𝑣 = 𝑉𝑉𝑉𝑉𝐼𝐼𝐼𝐼

(3.4)

→ 𝑓𝑓 = 𝑀𝑀𝑀𝑀𝐼𝐼

(3.2)

The average shear stress 𝑣𝑣𝑎𝑎𝑎𝑎 = 𝑉𝑉/𝑎𝑎𝑎𝑎and the maximum 𝑣𝑣𝑚𝑚𝑎𝑎𝑚𝑚 = 1.5𝑣𝑣𝑎𝑎𝑎𝑎 at the neutral axis.

When a small element in the beam is analyzed for shear and bending stresses, principal stresses can be calculated as

→ 𝑡𝑡 = 𝑓𝑓2

± 𝑓𝑓2

4+ 𝑣𝑣2 (3.1)

The principal plane is 𝑡𝑡𝑎𝑎𝑡𝑡−1( ⁄2𝑣𝑣 𝑓𝑓)/2

CE373 – Concrete I Dr. Ammar T. Al-Sayegh2

4.2) Diagonal Tension in Homogenous Elastic Beams

Bending causes vertical cracks near bottom of the beam in zones of large bending.

Shear stresses cause diagonal (inclined) cracks near mid-height.

In rectangular beams, the first cracks are vertical due to bending, then they change direction and become inclined as they reach the level of the N.A. They are named flexure-shear cracks.

In thin-webbed sections (such as I), the “thin” web which is subjected to shear cracks in shear before cracking in the bottom due to bending. These shear cracks are named web-shear cracks.

In such beams, shear cracks propagate and become wider. The interlock between the aggregates helps a lot in resistance.. Eventually, there is sliding across one of the cracks and the beam fails.

CE373 – Concrete I Dr. Ammar T. Al-Sayegh3

4.3) RC Beams without Shear Reinforcement

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Beam just before failure, the shear is resisted across the inclined cracks via three components:

→ The aggregate interlock across the inclined part of the crack.

→ The shear in the compression zone

→ The dowel action in the bottom steel (also shown on the right-hand figure)

All these are lumped in a single factor called 𝑉𝑉𝑐𝑐, the concrete contribution to the shear resistance.

The concrete contribution in web-shear cracks is different than in flexure-shear cracks

→ 𝑣𝑣𝑐𝑐𝑐𝑐 = 2.1 300𝑑𝑑

1/4𝑓𝑓𝑐𝑐′𝜌𝜌

𝑉𝑉𝑑𝑑𝑀𝑀

1/3(4.3c)

The stirrups help in the resistance to the applied shear. The steel contribution depends on the size and spacing s of the stirrups.

Shear reinforcement can also be provided by longitudinal steel by bending up the bars where it is no longer needed to resist flexural tension. In continuous beams, these bent up bars my also provide all the necessary reinforcement for negative moments.

Web reinforcement has no noticeable effect prior to the formation of diagonal cracks.

Transverse reinforcement, or web reinforcement, is placed to intercept the propagation of the diagonal crack. It is best to orient the bars to be perpendicular to the crack, but it is more practical to place vertical stirrups.

Stirrups can be of different shapes (Nos. 10 to 16), and they are more concentrated in zones of larger shearing forces (smaller spacing near supports).

CE373 – Concrete I Dr. Ammar T. Al-Sayegh5

4.4) RC Beams with Web Reinforcement

CE373 – Concrete I Dr. Ammar T. Al-Sayegh6

When cracks form, web reinforcement augments concrete resistance to shear in four ways:

1. Shear is resisted by stirrups.

2. The stirrups prevent further cracking allowing concrete to resist shear.

3. The stirrups prevent the widening of cracks so the two crack faces stay in contact and allow for interface force 𝑉𝑉𝑖𝑖 to develop.

4. The stirrups tie the longitudinal reinforcement in the main bulk of the concrete providing a restraint against splitting of the concrete along the longitudinal reinforcement and increasing the share of the shear force resisted.

Failure in shear takes place when stirrups start to yield.

Each stirrup traversing a crack exert a force 𝐴𝐴𝑎𝑎𝑓𝑓𝑎𝑎 on the given portion of the beam where 𝐴𝐴𝑎𝑎 is the area of the bar and 𝑓𝑓𝑎𝑎 is the stress in that bar.

CE373 – Concrete I Dr. Ammar T. Al-Sayegh7

ACI assumes that the cracks propagates at 45 degrees. If the horizontal distance along the crack is 𝑑𝑑, and the stirrups are spaced at distance 𝑠𝑠, then the number of stirrups crossed by the crack is approximately 𝑑𝑑/𝑠𝑠. The total force resisted by the stirrups across the inclined crack is

→ 𝑉𝑉𝑠𝑠 = 𝐴𝐴𝑣𝑣𝑓𝑓𝑦𝑦𝑑𝑑𝑠𝑠

The shear resistance of the concrete is

→ 𝑉𝑉𝑐𝑐 = 0.16 𝑓𝑓𝑐𝑐′ + 17 𝜌𝜌𝑉𝑉𝑑𝑑𝑀𝑀

𝑎𝑎𝑑𝑑 ≤ 0.29 𝑓𝑓𝑐𝑐′𝑎𝑎𝑑𝑑 (4.3a)

The nominal shear strength is→ 𝑉𝑉𝑛𝑛 = 𝑉𝑉𝑐𝑐 + 𝑉𝑉𝑠𝑠 (4.7a)

The shear stress can be calculated by dividing by 𝑎𝑎𝑑𝑑 as

→ 𝑣𝑣𝑛𝑛 = 𝑉𝑉𝑛𝑛𝑏𝑏𝑑𝑑

(4.7b)

Location of the critical section for shear design is shown in the figure above.

ACI code 11.2.1 allows for the calculation of the concrete contribution to the shear strength as:

→ 𝑉𝑉𝑐𝑐 = 0.17𝜆𝜆 𝑓𝑓𝑐𝑐′𝑎𝑎𝑤𝑤𝑑𝑑

→ 𝜆𝜆 = 1.0 for normal weight concrete, and 𝜆𝜆 = 𝑓𝑓𝑐𝑐𝑐𝑐

0.56 𝑓𝑓𝑐𝑐′≤ 1.0 otherwise.

According to ACI, the design for shear must be based on the relation

→ 𝑉𝑉𝑢𝑢 ≤ ∅𝑉𝑉𝑛𝑛 (4.10)

Where→ 𝑉𝑉𝑢𝑢: Factored applied shear load

→ 𝑉𝑉𝑛𝑛 = 𝑉𝑉𝑐𝑐 + 𝑉𝑉𝑠𝑠 (4.7a)

→ ∅ = 0.75

CE373 – Concrete I Dr. Ammar T. Al-Sayegh8

4.5) ACI Code Provisions for Shear Design

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