University of Western AustraliaSchool of Civil and Resource Engineering 2004

6. Prestressed Concrete :

Ultimate shear strength

of beams

• Approach to shear strength design

• Inclined web crushing

• Inclined web fracture -

• flexure-shear resistance

• web shear resistance

• Concrete and steel ‘contributions’ to shear

strength

APPROACH TO SHEAR STRENGTH DESIGN

• Shear fracture can occur without warning, and . .

• Shear resistance is variable, even for identical beams, so . .

• We design such that shear resistance ALWAYS

exceeds the bending strength of the member, and

• A lower strength reduction factor is used: f = 0.7

• In prestressed beams, shear strength at any section is

affected by the bending moment at that section. So we

need to estimate this bending moment as part of the

design.

• Typically, ultimate shear strength includes contributions

from concrete and steel (stirrups) so that Vu = Vuc + Vus

Firstly consider a section’s response to shear force . .

A B

CD

B’

C’

A B

CD

Basics: Diagonal Tension and Compression:

Shearforce V

AC extends to AC’.

So there is a tensile strain in direction AC’.

If stress is high enough, tensile fracture could occur.

Also BD contracts to B’D.

So there is a compressive

strain in direction B’D.

If stress is high enough,

crushing could occur.

Prestressing can inhibit both of these potential failures.

Consider first ‘hold-up’ enhancement . . .

Shear strength design

P

x

MV

Pv

Ph

qx = e’x

P

Ph = P

Pv = Pqx

Forces on concrete - length

of beam from support to x.

Components of prestress

force P, at x from support.

The vertical component Pv = Pqx is a vertical force which

‘holds-up’ the right hand concrete, hence enhances the shear

strength in this case.

[This is not always the case, e.g. in continuous beams.]

‘Hold-up’ enhancement of shear resistance :

Now consider inclined web crushing . . .

Shear strength design

INCLINED WEB CRUSHING

Just as for reinforced concrete, diagonal web crushing provides an upper

bound to the crushing capacity of a prestressed section. Except for the

hold-up force, the prestress does not increase the web crushing capacity.

The inclined web capacity is conservatively estimated by

V u.max = 0.2 f ’c bv do + Pv

bw

Groutedduct

dd

Note the terms:

bv = bw - 0.5 Sdd and bw = web width.

For grouted duct - for ungrouted = 1.0

d0 = distance from compression face to

outermost tendon or rebar, but not less

than 0.8D.

Pv = hold-up force.

If fV u.max < V*, then change

of design may be needed.

web-shear web-shear

flexure -shear

flexure -shear

flexure

flexure

Types of Cracking -

actual and potential:

We have already dealt with flexure (bending) cracking. Now we must

address the two types of diagonal cracking called ‘flexure-shear’ and

‘web-shear’cracking, and the ultimate shear resistance.

We can expect these patterns of cracking to be inhibited by

prestress, but by how much?

Let’s see . . .

flexure -shear

INCLINED WEB FRACTURE

Inclined Web Fracture:

Flexure-Shear Resistance: V ucf

This must be considered wherever a flexural crack, perhaps caused by a

previous loading, or by shrinkage, or any other reason, may occur. Hence we

must consider every section along the length of the beam. In reality, we

concentrate on selected sections and several intermediate sections.

The flexure-shear resistance is designated Vucf , and is one of the two values

of Vuc to be estimated.

Vucf is comprised of two additive resistances:

1. the shear force at the section when a crack at the section just starts to open (termed initiating force V0 ), and

2. the force required to allow a diagonal crack to fully propagate from

the flexural crack (termed propagating force Vp ).

Vucf is also augmented by the hold-up force P vSo …

. . . Vucf - Pv = V0 + Vp

i.e. Vucf = Pv + Vo + Vp

Pv = P e’x as before. Vo is explained thus :

UBMD and Mo

diagram

USFD

x

M*

Mo (Mdec)

V*

At section x from support :

M* is factored design moment

Mo is decompression moment

V* is factored shear force corresponding to M*

So V0 is a simple proportion

of V* :

V0 = ( M0 / M*) V*

. . . which leaves Vp to be explained . . .

Inclined Web Fracture : Flexure-shear

. . . Vp is estimated by the ‘beta’ formula from

reinforced concrete, with some modifications :

Vp = b1 b2 b3 bv do {( Ast + Apt ) f ’c / ( bv do )} 1/ 3

b1 = depth factor.

b2 = axial load factor, but not including prestress ! !

b3 = ‘support’ factor.

bv allows for reduction due to ducts, if any.

Ast + Apt includes both tendons and rebars which are in

the tensile zone at ultimate load.

So now we have it all:

Flexural shear Resistance Vucf = Pv + Vo + Vp

of which all terms have been explained.

Inclined Web Failure : Flexure-shear

Near the supports of simply supported beams, web-shear cracking may occur

before flexure-shear cracking. This possibility is very serious indeed, since

sudden fracture, without warning, could ensue. So, quite separately from

consideration of flexure-shear, we must examine web-shear, but only at sections

where tensile cracking is not expected to exist at ultimate load.

We do this by examining the principal tensile stress in the web:

RECTANGULARSECTION

bw

y

Bendingstress

sy

Shearstress

ty

GENERAL

PRISMATICSECTION

bw

y sy ty

sy =

P/A + Pey / I - My / I

and

ty = V Q / (I bw )

A note about Q

. . .

Inclined Web Failure : Web-shear resistance: Vucw

. . . Q is the second moment of area, taken about the

centroidal axis, of the area lying beyond the level being

considered, for example :

Centroidal axis

To estimate t at centroidal axis:

A y’

Q = A y’

Centroidal axis

To estimate t at junction of

web and tensile flange :

A y’

Q = A y’

The appropriate Q is then used in t y = V Q / (I bV )

Web-shear resistance: Vucw

t

q = angle of principal plane to the horizontal - remember that there is no shear

stress on a principal plane - ands1 = principal stress (+ = compression) on this plane.

Forces on triangle

a b

c

q

t.1

ttan q

s tan q

s1

cos q

s1

s1 tan q

MV

P P

t

t

t

ss

q

unitelement

a b

c

Stresses on unit element

SH=0: s tan q- t- s1 tan q = 0

SV=0: - ttan q - s1 = 0

Eliminate tan q : t2 = s12 - ss1

Web shear crack occurs when s1 reaches 0.33(f’c)0.5

So Vt = I bv t/ Q where t2 = s12 - ss1

in which s1 = - 0.33 (f ’c)0.5

Web-shear resistance

Mohr’s Circle representation :

t

t

ss

0

0

ss1

Directstress

s/2

t

Shear stress

+ = compression,- = tension

0

Unit element

subjected to stress s

horizontally, and 0

vertically.

s1 is maximum

principal tensile stress

at tensile fracture.

So t is shear stress

corresponding to

maximum principal

tensile stress.

This solves to t2 = s12 - s s1, as before. (s1 is negative!)

Note how the maximum shear strength is enhanced by the

magnitude of s : higher s => higher t.

Web-shear resistance

But at what depth y does the highest principal tensile stress

occur ? Consider these diagrams :

Bendingstress s

Shearstress t

y

y

It is not apparent at which

depth y the combination

of shear stress and

compressive stress are

critical. So . . .

For rectangular web, we

check at centroidal axis,

and . . .

For general section, we

check at both centroidal

axis, and at flange to

web intersection, and

adopt the most critical

value.

Inclined Web Failure : Web-shear

Calculate V*

at section

Calculate Vu.max

V* > fVu.max ?yes

Change section

no

Pv (hold-upforce) +

Vo (crack initiation force) +

Vp (crack propagation force)

= Vucf

M* > Mcr ?yes

V uc = Vucf

no

Pv (hold-up force) +

min Vt (centroidal)

Vt (flange-web junction)

= Vucw

{ }

Vuc = min Vucf

Vucw

CONCRETE

CONTRIBUTION

TO SHEAR

STRENGTH

flexure-

shear

web-

shear

This logic chart

shows how to

calculate Vuc

{ }

Can we use stirrups to

increase shear capacity as

for reinforced concrete ?

Yes, and in the same manner !

Remember that stirrups are always

required if the beam, whether

reinforced or prestressed, is deeper

than 750 mm.

Recall the truss model which

describes how stirrups act as

vertical members, in conjunction

with horizontal chords . . .

d0

s

How shear steel works :

Stirrups act as vertical tension members in the fictitious truss.

Stirrups also hold the crack closed, up to their yield capacity.

So we design thus: fVu >= V*

which means f(Vuc + Vus) >= V*

from which fVus >= V* - fVuc

STEEL CONTRIBUTION

TO SHEAR STRENGTH

Beam: V* <= f V uc

AND

D <= max (250mm, bv/2)yes

No shear

steel required

no

V* <= 0.5 f Vuc

Slab : V* <= f V uc

yes

no

yesD > 750 mm

no

no

0.5 f V uc < V* <= f V u.min

yesProvide minimum

steel, using

s max = min(0.75D, 500mm)

yes

no

Provide calculated

steel, using

s max = min(0.5D, 300 mm)

Steel ‘contribution’ to

shear strengthThis logic chart shows how to

select shear steel and spacing :

Shear Steel Contribution to Shear strength :

Vus = (Asv f sy.f d0/s) cot qv

For design purposes, we commonly use the following form to

select the required stirrups:

Asv / s > = Vus / (f sy.f d0 cot qv)

Remember there are limitations on the spacing s of stirrups.

Also qv lies between 30o and 45o, as given by

qv = 30o + (V* - fV u.min) / (fV u.max - f V u.min) 15o

A sv.min = 0.35 bv s / f sy.f where f sy.f in MPa.

. . . and of course V u = V uc + V us

SUMMARY

• Design to ensure that shear failure does not occur

under the loading that produces ultimate bending failure.

• Check for web crushing i.e. ensure V* <= f Vu.max

• Check for diagonal tension fracture i.e.

V* <= f{Vuc + Vus}

• Vuc is Vucf if section is cracked by ultimate moment or

{Vucf , Vucw} min otherwise.

• Always use shear steel if D > 750 mm. This includes

longitudinal steel top and bottom.

• f = 0.7 for all shear calculations.