Design for Shear for Prestressed

Concrete Beam

Introduction

The behaviour of prestressed beams at failure in shear is

distinctly different from their behaviour in flexure.

The beam will tend to fail abruptly without sufficient warning

and the diagonal cracks that develop are considerably wider

than the flexural cracks.

Shear forces result in shear stress. Such a stress can result in

principal tensile stresses at the critical section which can

exceed the tensile strength of the concrete

When the tensile strength of the concrete is exceeded cracks

will formed.

Cracking Patterns & Failure Modes Cracking in prestressed concrete beams at ultimate load

depends on the local magnitudes of moment and shear.

In regions where the moment is large and shear is small,

vertical flexural cracks appear after the normal tensile stress in

the extreme concrete fibres exceeds the tensile strength of

concrete. This type of cracks shown as type A in figure.

Where both the moment and shear force are relatively large,

flexural cracks which are vertical at the extreme fibres become

inclined as they extend deeper into the beam owing to the

presence of shear stresses in the beam web. These inclined

cracks, which are often quite flat in a prestressed beam are

called flexure-shear cracks and are designated crack type B.

Cracking Patterns & Failure Modes If adequate shear reinforcement is not provided, a flexure-shear

crack may lead to a so-called shear-compression failure, in which the area of concrete in compression above the advancing inclined crack is so reduced as to be no longer adequate to carry compression force resulting from flexure.

Another type of inclined crack sometimes occurs in the web of a prestressed beam in the regions where moment is small and shear is large, such as the cracks designated type C adjacent to discontinuous support and near the point of contraflexure in the figure.

In such locatio, high principal tensile stress may cause inclined cracking in the mid-depth region of the beam before flexural cracking occurs in the extreme fibres. These cracks are known as web-shear cracks and occur most often in beams with relatively thin webs.

Cracking pattern

Effect of Prestressing in Shear

The longitudinal compression introduced by prestress delays

the formation of each of the crack types shown previously. The

effect of prestress on the formation and direction of inclined

cracks can be seen by examining the stresses acting on a small

selement located at the centroidal axis of the uncracked beam as

shown in figure of next page.

Effect of Prestressing in Shear

Using a simple Mohr’s circle construction, the principal

stresses and their directions are well establish.

When the principal tensile stress 1 reaches the tensile

strength of concrete, cracking occurs and the crakcs form in

the direction perpendicular to the direction of 1.

When the prestress is zero, 1 is equal to shear stress and acts

at 45o to the beam axis. If diagonal cracking occurs, it will be

perpendicular to the principal tensile stress.

Effect of Prestressing in Shear

When the prestress is not zero, the normal compressive stress

( = P/A) reduces the principal tension 1. the angle between

the principal stress direction and the beam axis increases and

consequently if cracking occurs, the inclined crack is flatter.

Prestress tehrefore improves the effectiveness of any transverse

reinforcement (strirrups) that may be used to increase the

shear strength of a beam.

With prestress causing the inclined crack to be flatter, a larger

number of the vertical stirrup legs are crossed by the crack and

consequently a larger tensile force can be carried across the

crack.

Shear

Analysis

Uncracked

Vco

Cracked

Vcr

Uncracked Sections

Small element A at the centroidal of a simply supported

prestressed concrete member subjected to compressive stress,

fcp and a shear stress, fs.

and from Mohr circle,

The allowable principal tensile stress is given in BS8110 as

For a rectangular section

Uncracked Sections

Where

Vco = design ultimate shear resistance of a section uncracked in

flexure

fcp = design compressive stress at centroidal axis due to prestress = Pe/A

ft = maximum design principle tensile stress,

bv = breadth of the member or for T, I and L beams used width of the

web

If grouted duct is present in the web

bv = bw – 0.67dd (dd = diameter of duct)

Cracked Section

Clause 4.3.8.5 BS8110 gives the following empirical equation

for the ultimate shear resistance of a section cracked in flexure:

Mo = moment which produces zero stress at extreme tension

fibre

fpt = level of prestress in concrete at the tensile face

the value of Vcr should be taken as less than 0.1bvdfcu

Cracked Section

where

m = 1.25

Should not be taken as greater than 3

Should not be taken as less than 0.67 for

members without shear reinforcement

Should not be taken as less than 1 for members with shear

reinforcement providing a design shear resistance of ≥ 0.4

N/mm2

or using Table 3.8

For characteristic concrete strengths greater than 25N/mm2, the values

in Table 3.8 may be multiplied by (fcu/25)1/3 . The value of fcu should not

be taken as greater than 40.

Design Ultimate Shear Resistance, Vc

According to Clause 4.3.8.3 BS8110

Vc = Vco at uncracked section (M < Mo)

Vc = is the smaller of Vco and Vcr at cracked section (M ≥ Mo)

For deflected tendon, the vertical component of the prstressing

will help to resist the shear force.

The total shear resistance then becomes :

Vc + Pesin where is the angle of inclination of the

prestressing tendon

For parabolic profile, e(x) = (-4/L2)x2 + (4/L)x

(x) = (-8d/L2)x + (4/L) in radian

where = abs(es – ems)

Parabolic Profile

Shear Design Steps

1. Draw the shear force and bending moment diagram

2. Check maximum allowable shear stress (Clause 4.3.8.2)

v = V/bvd 0.8fcu or 5N/mm2

3. Plot Mo on the BMD. Mo will varies along the span if the

tendon profile is not straight. Determine the cracked

(M ≥ Mo) and uncracked (M < Mo)

4. Calculate Vco and Vcr

5. Determine the ultimate shear resistance of the prestressed

beam as follows :

cracked region : Vc = Vcr or Vco + Pesin

uncracked region : Vc = Vco + Pesin

Shear Design Steps

6. If V 0.5Vc, shear reinforcement is not required (Clause 4.3.8.6)

7. If 0.5Vc < V Vc + 0.4bvd, nominal shear reinforcement is required. Refer Cl. 4.3.8.7, use

8. If V > Vc + 0.4bvd, shear reinforcement is required (Cl. 4.3.8.8). Use

dt is the depth from the extreme compression fibre either to the longitudinal bars of to the centroid of the tendons, whichever is greater.

9. Spacing of shear reinforcement, Sv the lesser of 0.75dt or 4 x web thickness, but when V > 1.8Vc, the max spacing should be reduced to 0.5dt.

10. The lateral spacing of the individual legs of the links provided at a cross-section should not exceed dt.

Example 1

A prestressed concrete T-beam shown in figure is simply

supported over a span of 28m, has been designed to carry in

addition to its own weight, a characteristic dead load of 4kN/m

and a characteristic imposed load of 10kN/m. The beam is pre-

tensioned with 14 nos 15.7mm diameter 7-wire super strands

(Aps = 150mm2) but due to debonding only 7 of the strands are

active at a section 2m from support. The effective prestressing

force, Pe for these 7 strands is 1044kN. Use the following data:

fcu = 50 N/mm2, A = 5.08 x 105 mm2, I = 134x109 mm4,

y2 = 912mm, fpu = 1770 N/mm2, fyv = 250 N/mm2,

Design the section for shear.

Solution

1. Loads

Selfweight, Wsw = 24 kN/m3 x 0.508 m2 = 12.19 kN/m

Ultimate load, W = 1.4(12.19 + 4) +1.6(10) = 38.67 kN/m

2. Shear and moment at 2m from support

V (x) = W(0.5L-x)

V(2) = 38.67 (0.5x28 – 2) = 464 kN

M(x) = 0.5W(Lx – x2)

M(2) = 0.5x38.67 (28x2 – 22) = 1005 kNm

Solution

Calculate Mo

e = 814mm, d = 1500-98 = 1402mm

< M = 1005 kNm

Section is cracked in flexure

Solution

Calculation of Vco

Solution

Calculation of Vcr

… Ok

Use 400/d = 1

fcu = 50 N/mm2 > 40 N/mm2 ; use fcu = 40 N/mm2

520 kN

And Vcr > 0.1bvdfcu = 174 kN … ok

Solution Shear resistance provided by the concrete, Vc

Vc = smaller between Vco (420 kN) and Vcr (520 kN)

Vc = 420 kN

Design of shear reinforcement

V = 464 kN , 0.5Vc = 210kN , Vc +0.4bvd = 518 kN

Where 0.5Vc < V < Vc +0.4bvd

Only nominal links required

Using 10mm diameter stirrup/links, Asv = 157mm2

< 0.75dt = 0.75(1500-98) = 1051.5mm

Use R10-450mm c/c

Example 2

The beam shown below supports an ultimate load, including

selfweight of 85kN/m over a span of 15m and has a final prestress

force of 2000kN. Determine the shear reinforcement required.

Use the following data:

fcu = 40N/mm2, A = 2.9 x 105 mm2,

I = 3.54 x 1010 mm4 , Aps = 2010 mm2, fpe/fpu = 0.6

Solution

Draw BMD and SFD

M(x) = 0.5w(Lx-x2) & V(x) = w(0.5L – x)

Where w = 85 kN/m

Check maximum allowable shear stress (Cl. 4.3.8.2)

V = 637.5 kN

< 0.8fcu or 5 N/mm2 … OK

Solution

Plot Mo on BMD

Mo vary along the length of beam since the tendon profile is

parabolic which produce the different eccentricity along the

length of beam.

Eccentricities of tendon along parabolic tendon profile given by:

Where = 425mm

Solution

Distribution of M and Mo along beam span

Variation of V along beam span

Solution

Nominal Shear Reinforcement

Use R8,

Shear reinforcement

< 0.75dt =0.75(908) = 681mm

Use R8 – 350 mm

V-Vc = 107.12 kN

< 0.75dt =0.75(772) = 579mm

Use R8 – 150 mm