Lecture 3.3 Lecture 3.3 Shear Strength Shear Strength of Slabs - - PDF document

Reinforced Concrete II Hashemite University

• Dr. Hazim Dwairi

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The Hashem ite University Departm ent of Civil Engineering

Lecture 3.3 Lecture 3.3 – – Shear Strength Shear Strength

• f Slabs
• f Slabs

Dr Hazim Dwairi Dr Hazim Dwairi

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• Dr. Hazim Dwairi
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Shea r Strength of Sla bs Shea r Strength of Sla bs

• In two

In two-

• way floor systems, the slab must

way floor systems, the slab must h d t thi k t i t b th h d t thi k t i t b th have adequate thickness to resist both have adequate thickness to resist both bending moments and shear forces at bending moments and shear forces at critical section. There are three cases to critical section. There are three cases to look at for shear. look at for shear.

– One One-

• way Slabs supported on beams

way Slabs supported on beams

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– Two Two-

• Way Slabs without beams

Way Slabs without beams – Shear Reinforcement in two Shear Reinforcement in two-

• way slabs without

way slabs without beams. beams.

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One One-

• w a y Shea r: Sla bs w ith

w a y Shea r: Sla bs w ith Bea m s Bea m s

Two Two-

• way slabs supported on beams (one

way slabs supported on beams (one-

• way

way shear): shear): shear): shear): The critical section for shear is found at The critical section for shear is found at d distance distance from the column, where: from the column, where:

⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = bd f V

c c

6

'

φ φ

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The supporting beams are stiff and are capable of The supporting beams are stiff and are capable of transmitting floor loads to the columns. transmitting floor loads to the columns.

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⎟ ⎠ ⎜ ⎝ 6

One One-

• w a y Shea r: Sla bs w ith

w a y Shea r: Sla bs w ith Bea m s Bea m s

The ultimate shear force is calculated using the triangular and The ultimate shear force is calculated using the triangular and trapezoidal areas. If no shear reinforcement is provided, the trapezoidal areas. If no shear reinforcement is provided, the t ape o da a eas

• s ea

e

• ce

e t s p o ded, t e t ape o da a eas

• s ea

e

• ce

e t s p o ded, t e shear force at a distance shear force at a distance d from the beam must be: from the beam must be:

c u@d

V V φ ≤

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Where: Where:

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⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = d l w V 2

2 u u@d

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Shea r in Sla bs w ithout Bea m s Shea r in Sla bs w ithout Bea m s

• There are two types of shear that need to be

There are two types of shear that need to be addressed addressed addressed addressed

– One One-

• way shear or beam shear at distance

way shear or beam shear at distance d from the from the column column – Two Two-

• way or punch out shear which occurs along a

way or punch out shear which occurs along a truncated cone. truncated cone.

Two-way shear

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One-way shear

One One-

• w a y shea r (Bea m Shea r)

w a y shea r (Bea m Shea r)

One One-

• way shear considers critical section a

way shear considers critical section a di t di t d f th l d th l b i f th l d th l b i distance distance d from the column and the slab is from the column and the slab is considered as a wide beam spanning considered as a wide beam spanning between supports. between supports.

⎟ ⎞ ⎜ ⎛ ≤ bd f V V

c '

φ φ

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⎟ ⎟ ⎠ ⎜ ⎜ ⎝ = ≤ bd V V 6

c u@d

φ φ

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Tw o Tw o-

• w a y Shea r: Critica l Section

w a y Shea r: Critica l Section

Two Two-

• way shear fails along a truncated cone or

way shear fails along a truncated cone or pyramid around the column The critical section is pyramid around the column The critical section is pyramid around the column. The critical section is pyramid around the column. The critical section is located located d/2 d/2 from the column face, column capital, from the column face, column capital,

• r drop panel.
• r drop panel.

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Tw o Tw o-

• w a y Shea r: Critica l Section

w a y Shea r: Critica l Section

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Tw o Tw o-

• w a y Shea r: Concrete Shea r

w a y Shea r: Concrete Shea r Strength Strength

• For Slabs and footings,

For Slabs and footings, Vc is the smallest of a, b is the smallest of a, b and c: and c: and c: and c:

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Where: bo = perimeter of critical section β = ratio of long side of column to short side αs = 40 for interior columns, 30 for edge columns and 20 for corner columns.

Tw o Tw o-

• w a y Shea r:

w a y Shea r: β Ev a lua tion Ev a lua tion

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Augm enting Shea r Strength Augm enting Shea r Strength

• For slabs which don’t meet the condition for

For slabs which don’t meet the condition for shear one can either: shear one can either: shear, one can either: shear, one can either:

– Thicken the slab over the entire panel. Thicken the slab over the entire panel. – Use a drop panel to thicken the slab adjacent to the Use a drop panel to thicken the slab adjacent to the column. column. – Increase Increase bo by increasing the column size, or by by increasing the column size, or by adding a fillet or shear capital around the column. adding a fillet or shear capital around the column.

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– Add shear reinforcement. Add shear reinforcement. Reinforcement can be done by shear heads, anchor bars, Reinforcement can be done by shear heads, anchor bars, conventional stirrup cages and studded steel strips. conventional stirrup cages and studded steel strips.

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Shea r Reinforcem ent Shea r Reinforcem ent

• Shear

Shear-

• heads

heads: consist of steel I : consist of steel I-

• beams or channel

beams or channel welded into four cross arms to be placed in slab above a welded into four cross arms to be placed in slab above a column Does not apply to external columns due to column Does not apply to external columns due to

• column. Does not apply to external columns due to
• column. Does not apply to external columns due to

lateral loads and torsion. lateral loads and torsion.

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Shea r Reinforcem ent Shea r Reinforcem ent

• Anchor Bars

Anchor Bars: consists of steel reinforcement rods or : consists of steel reinforcement rods or bent bar reinforcement . bent bar reinforcement . bent bar reinforcement . bent bar reinforcement .

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Shea r Reinforcem ent Shea r Reinforcem ent

• Conventional Stirrup Cages

Conventional Stirrup Cages:

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h ≥ 150 mm h ≥ 16ds

h =

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Shea r Reinforcem ent Shea r Reinforcem ent

• Headed Shear Studs

Headed Shear Studs:

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Nom ina l Shea r Strength Nom ina l Shea r Strength

• Shear reinforcement consisting of bars or wires

Shear reinforcement consisting of bars or wires and single and single-

• or multiple
• r multiple-
• leg stirrups shall be

leg stirrups shall be itt d i l b d f ti h h i itt d i l b d f ti h h i permitted in slabs and footings where h is permitted in slabs and footings where h is greater than or equal to 150mm & 16d greater than or equal to 150mm & 16ds

d b f V V V

c

• '

s c n

5 . ≤ + = d b f V V V

c

• '

s c n

58 . ≤ + =

Conventional Stirrup Cage Shear Head Reinforcement

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S d f A V d b f V

c y v s

• '

c

17 . = ≤

fc

• s

c n Reinforcement

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Shea r Exa m p le Problem Shea r Exa m p le Problem

Determine the shear Determine the shear reinforcement required for an reinforcement required for an reinforcement required for an reinforcement required for an interior flat panel considering interior flat panel considering the following: the following: Vu= 865kN, = 865kN, slab thickness = 220 mm, slab thickness = 220 mm, d = 190 mm, d = 190 mm, f’ f’c

c = 21 MPa,

= 21 MPa,

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fy= 420 MPa, and column is = 420 MPa, and column is 500 x 500 mm. 500 x 500 mm.

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Shea r Exa m p le Problem Shea r Exa m p le Problem

• Compute the shear terms find b

Compute the shear terms find b0 for for Vc

( )

mm b 760 , 2 190 500 4

• =

+ =

Smaller

• f

= 0.51 = 0.39 =

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=

kN V d b f V

c

• c

c

8 . 594 ) 2760 )( 190 )( 21 )( 33 . ( 75 . 33 .

'

= = = φ φ φ

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Shea r Exa m p le Problem Shea r Exa m p le Problem

• V

Vu =865 =865 kN kN > 594.8 > 594.8 kN kN, so , so Sh i f t i d!!! Sh i f t i d!!!

• Shear reinforcement is need!!!

Shear reinforcement is need!!!

• Compute maximum allowable shear

Compute maximum allowable shear φVn

kN V V V 8 . 393 , 1 ) 2760 )( 190 ( 21 58 .

s c n

= = + = φ k k φ

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Shear reinforcement can be used Shear reinforcement can be used

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kN V kN V

u

865 8 . 1393

n

= > = φ

Shea r Exa m p le Problem Shea r Exa m p le Problem

• Use shear heads or studs

Use shear heads or studs C t l th C t l th ‘ ‘ ’ d b t d b t d

• Compute length

Compute length ‘ ‘a’ a’ covered covered by studs by studs

( )

a width column b 2 4

• +

= d b f V

• c

21 17 75 000 865 17 .

' u = φ

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mm a a 024 , 1 190 ) 2 500 ( 4 21 17 . 75 . 000 , 865 = × + × × × =

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Shea r Exa m p le Problem Shea r Exa m p le Problem

Total length = a + d = 1024+190 =1214 mm Total length = a + d = 1024+190 =1214 mm S 1250 S 1250 Say = 1250 mm Say = 1250 mm

• Determine shear reinforcement

Determine shear reinforcement

kN V V V

c u s

3 360 1 793 75 / 865 − = φ

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kN side Per V kN

s

08 . 90 4 / 3 . 360 3 . 360 1 . 793 75 . / 865 = = = − =

Shea r Exa m p le Problem Shea r Exa m p le Problem

• Determine shear reinforcement

Determine shear reinforcement U φ10 l d ti A 10 l d ti A 2*78 5 150 2*78 5 150

2

• Use

Use φ10 closed stirrups A 10 closed stirrups Av =2*78.5=150mm =2*78.5=150mm2

mm S V d f A S

s y v

9 132 190 420 150 = × × = =

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direction each stirrups closed mm USE mm d S mm S / 90 @ 10 95 2 / 9 . 132 1000 08 . 90

max

φ = = = × =