3,4 &5 .DESIGN OF BEAM – SHEAR, DEFLECTION AND CRACK

When loads applied to beams produce not only bending moment but also internal shear forces. In the reinforced concrete beams, the primary longitudinal bending reinforcement is usually considered first. This leads to the size of the section and the arrangement of the reinforcement to provide the necessary moment resistance. Limits are placed on the amount of bending reinforcement to ensure that if failure were ever to occur, it would gradually, giving warning to the occupants.

Once the primary longitudinal reinforcement has been determined, then the reinforced concrete beams are designed to resist the shear forces resulting from the various combinations of ultimate loads. Most of shear failure is frequently sudden and brittle, hence the design for shear must ensure that the shear strength equals or exceeds the flexural strength at all points in the beam. The manner in which shear failure can occur varies widely depending on the dimensions, geometry, loading and properties of the members.

3.1 Shear Failure Patterns

The figures below show the possible modes failure of shear in beam.

Case I : av / d > 6

Case II : 2 < av / d > 6

1

d

v

av

av

d

v

Case III : av / d < 2

Figure 3.1

3.2 Shear Resistance

Taylor (1974) did research on shear resistance in reinforced concrete beam without shear reinforcement. He stated that in a reinforced concrete beam without shear reinforcement the shear is carried by a combination of three main components. These are :-

(i) concrete in the compression zone(ii) dwelling action of tensile reinforcement(iii) aggregate interlock across flexural cracks

Figure 3.2

Total shear resistance given by the beam,

Vc = Vcz + Va + Vd

The actual relationship between all forces is complex and is difficult to analyse theoretically. However BS 8110, simplified the analysis and design for shear reinforcement have been developed.

2

d

v

av

Va = interlocking between aggregates (35 – 50%)

Concrete in compression

Vcz = shear in compression zone 20 – 40%

tensile steel

Vd = dwelling action (35 – 50%)

v

3.3 The Truss Analogy

The analysis and design of a reinforced concrete beam in shear is based on the concept of an equivalent truss:

Figure 3.3

Applying the method of section for a stirrup spacing = d

Tensile force in stirrup = Shear at the section

0.95 fyv Asv = V

In practice some of the shear is taken by the concrete so that

0.95 fyv Asv = V – vc bvd

where Asv = the cross – sectional area of the legs of the stirrup

d = the effective depth

fyv = the characteristic tensile stress for the links or stirrups

V = the shear force

vc = the ultimate shear stress resistance of the concrete

Let V = v bvd

Where v is the average shear stress at the section

Then

0.95 fyv . Asv = (v - vc) bv d

3

T

d

d

v

o

o o

o

b

Links

With a stirrup spacing reduced from d to sv the force in the stirrups is reduced proportionately, so that

0.95 fyv . Asv =

i.e.

[Link or stirrup sizes are usually smaller diameter and they are often of mild steel to minimize the radii of bends.]

Shear resistance of a Given Section – Concrete + Stirrups

Total shear resistance, V = vbd

As shown

Therefore re-arranging

Total shear resistance

Minimum Shear Reinforcement

For mild steel (fy = 250 N/mm2)

For high yield steel (fy = 460 N/mm2)

4

Eqn. Table 3.7 BS 8110

Ultimate Shear Resistance of the Concrete, v c

Value of vc are given in Table 3.8 of BS 8110 and can be calculated from the formula:

where As = area of longitudinal shear reinforcement that continues a distance at least d beyond the section considered.

= partial safety factor = 1.25

> 3

< 1

If the characteristic strength of the concrete is greater than 25 N/mm2 vc can be multiplied by (fcu/25)1/3 but fcu should not be taken as greater than 40 N/mm.

To resist the shearing forces, bars may be bent up near the support as shown in figure 5.3. The bent-up bars and the concrete in compression are considered to act as an analogous lattice girder and the shear resistance of the bars is determined by taking a section XX through the girder.

From the geometry of part (a) of the figure, the spacing of the bent-up bars is

and the section XX the shear resistance of the single bar is

where Asb is the cross-sectional area of the bent-up bar.

5

AnchorageLength

(a) Single System

x

d

'd

x

sbA sbA

cotcot'dds

Figure 3.4

For a multiple system of bent-up bars, as in part (b) of the figure, the shear resistance is increased proportionately to the spacing sb. Hence

or

equation 4BS 8110

The angles and should both be greater than, or equal to 45o and the code requires that the spacing sb has a maximum value of 1.5d. With = = 45o and

, equation 5.5 becomes

6

(b) Multiple System

sbA

sb

and this arrangement is commonly referred to as a double system.

Example 3.1

Maximum shear stress,

from Table 3.8 or

vc is derived from the expression

7

o

oo

o

2 T 20

1252 T 12

6m

Ult. design load, w = 10KN/m

250V = 30 KN

V = 30 KN

minimum links for the whole length of beam

Try R8, Asv = 101

Use

Example 3.2

8

R8 - 175

a

a

o

o o

o2 T 12

2 T 20

R8 - 175

5m

160KN

160KNoo

d = 450

250

oo

Ult. design load w= 64 KN/m

4 T 20

Ultimate shear stress at support

The size of the beam is ok.

The ultimate shear stress is less than the maximum shear stress.

9

shear reinforcement is required

Try R8, Asv = 101mm2

Use link R8 – 100

Nominal reinforcement

Try R8, Asv = 101 mm2

Sv = = 24 mm < 0.75d (337.5)

Use links R8-225

Shear resistance R8-225

v = 0.43 + 0.62 = 1.05 N/mm2

V = vbd

= 1.05 x 250 x 450 = 118.0 KN

10

or

3.4 Summary of design procedures for shear reinforcement according to BS8110

(1) Calculate ultimate shear stress

(2) Check or , which is the lesser.

(3) Determine from Table 3.8 BS 8110 or from formula.

where

(4) If v < , for member of minor structural shear

11

160.0KN

118.0KN

a

118.0KN

160.0KN160

118

5.2

a

m84.1a R8 - 100

3.68m 0.66m

R8 - 225R8 - 100

0.66

reinforcement can be omitted but for beam ofstructural importance minimum links should be provided.

(5) If , minimum links should be provided

(6) If or links should be provided

(7) Choose a link size and spacing

(8) Plot shear force diagram with shear resistance of concrete and minimum links. Specify arrangement.

The spacing of the links,

Deflection

Having analysed the section at ultimate limit state and calculated the necessary reinforcement for the moment and shear, it is however necessary to consider to check that the serviceability limit state are satisfied. The three principal criteria regarding serviceability are that, when a member is subjected to the forces and moments that arise from working load, deflection, vibration and cracking must not be excessive.

However BS 8110 stated that for reinforced concrete members, the general requirements to be regarded as acceptable limit are (a) that the final deflection (including all time-dependent effects such as creep and shrinkage as well as those of temperature) of each horizontal member below the supports must not exceed span/250, and (b) that the deflection occuring after the construction of a partition or the application of a finish should not exceed the lesser of span/350 or 20 mm for non-brittle partition, and span/500 or 20 mm for brittle materials.

That can be expressed as follow:

12

a1

a2

a1 = deflection before erection of partitions, etc.a2 = deflection after erection of partitions

a2 > or 20 mm whichever is the lesser

a1 + a2 >

Example 3.3

Deflection

Refer Example 3.1

13

6m

W=10 kN/m

o o

KNm.XWL

M 0458

610

8

22

mm.Asreq 4490mm.Asprov 0629

basic = 20 (Table 3.9, simply supported

rectangular section)

The design service stress in the tension reinforcement in a member may be estimated from,

Modification factor for tension reinforcement

14

2 T 12

250

o o

Modification factor for compression reinforcement

M. F = 1.19

a = 0.053

limit

actual

actual > limit

Deflection is not satisfactory. Hence, it is advisable to increase the size of the beam.

15

1.14

0.5

1.20

0.75

0.22

0.06

0.72

Cracking

As with deflection, any crack in reinforced concrete will not be excessive. Excessive cracking and wide deep cracks affect durability and this leads to corrosion of reinforcement although strength may not be affected. BS 8110 states that for reinforcement concrete cracking should be kept within reasonable bounds.

Cracking is controlled by specifying the maximum distance between bars in tension. The spacing limits are specified in clause.3.12.11.2. The clause indicates that in normal condition the internal or external exposure the bar spacing given will limit crack widths to 0.3mm. The rules are asBars diameter less than 0.45, the largest bar in the section should be ignored except when considering bars in the side faces the beams.

1. The clear horizontal distance s1 between bars or groups near the tension face beam should not be greater than the values given in Table 3.28, the code which are given by the expression.

16

S2

o o

oo

o

oo

oS2

Sb

anti – crackreinforcement

where S1 > value in Table 3.28S2 > ½ S1

Sb > 250mm

Example 3.4

(a)

Refer 3.12.11.2.3 BS 8110

Clear horizontal distance between bars in tension,

Allowable clear distance between bars

17

oo

S1

oo o

S1 S1

o o Sb

oo

oo

yS1

y

450

225

2 T 16

2 T

R8 - 150

30

= 155 ( Table 3.28, fy = 460 and no redistribution 0%)

Clear distance between the face of the beam and nearest longitudinal bar in tension

Allowable distance

= 155 x 0.5 = 77.5

satisfied

(b)

Clear distance between horizontal tension bars

18

ooo oo

1500mm

300mm

5 T 25

R10 - 200

40 2 T 20

S2 25

S1 25

S1 25

O.K

h > 250mm

Bars should be provided in side faces of beam to control cracking.

Distance between bars

Minimum size of bar

= 11.4mm

Provide T12 @ 200 mm

19