53

B2 Composite beam – arrangement of shear connectors in solid slab

1. Purpose of example

The purpose of the example is to show the calculation and the arrangement of the headed stud connectors in the case of the simple supported composite beam. The class of cross-section is such that we can calculate the resistance moment of the cross-section of the beam by means of rigid plastic theory. The thickness of the concrete slab is hc = 14 cm. The essence of the example is the calculation and the arrangement of the headed stud connectors in the solid slab. Therefore, the design of the beam is conducted more simply without considering creep and shrinkage of the concrete. Spacing and arrangement of the studs is determined according to the diagram of shear forces or by the shear flow, which is calculated according to the theory of elasticity which assumes a linear relationship between the shear force and the longitudinal shear force. The compressive normal force in the concrete flange Nc is transmitted to the composite beam by means of the headed studs.

2. Static system, cross-section and actions

Figure B2.1 Static system and cross-section Actions Permanent action

54 B Composite beams

- concrete slab and steel beam = 11,1kg kN/m Variable action (including partitions) - variable and partitions = 15,0kq kN/m

3. Properties of materials

Concrete strength class: C 25/30 = 25ckf N/mm2

25= = = 16,71,5

ckcd

c

ff N/mm2

Structural steel: S355 = 355ykf N/mm2

355= = = 3551,0

ykyd

M

ff N/mm2

Shear connectors: ductile headed studs = 450uf N/mm2 = 19d mm = 100sch mm

100= = 5,3 > 4,019

schd

= 1,0

=RdP 73,7 kN

4. Ultimate limit state

4.1 Design values of combined actions and design values of effects of actions

The design load of governed combination of actions is:

The calculation of effects of actions is:

Example B2 55

4.2 Effective width of concrete flange

The distance between the centres of the rows of shear connectors is b0 = 0 because there is only one row of shear connectors.

The effective cross-section of the concrete flange is shown in Figure B2.2.

Figure B2.2 Effective cross-section of the concrete flange

4.3 Plastic resistance moment of composite cross-section

Since the main purpose of the example is to show the calculation and the arrangement of the headed stud connectors, a simplified verification for the ultimate limit state contains: a) Check of the resistance of the composite cross-section to bending and b) Check of the resistance of the composite cross-section to vertical shear. The calculation of the plastic resistance moment Mpl,Rd can be performed in two ways: (a) by calculating the position of the plastic neutral axis, xpl, (b) directly without calculating xpl.

56 B Composite beams

a) By calculating of the position of the plastic neutral axis xpl If , ,>c f pl aN N , the plastic neutral axis is located in the thickness hc of the concrete of the slab:

, ,>c f pl aN N

Figure B2.3 Effective cross-section of the concrete flange with dimensions The plastic neutral axis lies a distance xpl below the top of the concrete flange:

When the plastic neutral axis lies within the concrete slab, the plastic resistance moment Mpl,Rd may be determined from:

b) Directly without calculating xpl

the plastic neutral axis is located in the slab depth. The effective cross-section of the concrete flange, with dimensions, is shown in Figure B2.3.

Example B2 57

The design value of the plastic resistance of the structural steel section to normal force is:

The resistance of the effective area of the concrete flange acting compositely with the steel section is:

Since Npl,a < Nc,f the plastic neutral axis lies within the concrete slab. The lesser of the two values, Npl,a and Nc,f, is governed so that Na = Nc,f = Npl,a. The bending moment Mc, which takes the concrete flange, is:

The plastic resistance moment of the composite section Mpl,Rd is:

Figure B2.4 Stress blocks for calculating the resistance of the composite cross-section

Criterion:

, so the resistance moment of the composite cross-section is verified.

58 B Composite beams

4.4 Vertical shear resistance

The shear buckling resistance of the web should be verified, for the unstiffened web when:

where:

= 1,2 , the factor defined in EN 1993-1-5

Remark The resistance of the composite beam to vertical shear is normally taken as the shear resistance of the steel section according to clause 6.2.6, EN 1993-1-1, which gives:

, , ,0

( / 3)= = V y

pl Rd pl a RdM

A fV V

For rolled I- and H-sections, if the load is applied parallel to the web, the shear area is calculated as:

= – 2· · + ·( + 2· )V a a f f wA A b t t t r , but not less than · ·w wh t

Since , the condition is satisfied. The shear buckling resistance of the web need not be verified.

Example B2 59

The governed shear area AV is shown in Figure B2.5, and it is:

Figure B2.5 Governed shear area

= 1,2

Criterion:

d h tw

Thus, AV = cm2 The design plastic shear resistance of the steel section is:

Since , the condition is satisfied.

60 B Composite beams

4.5 Check of resistance of headed stud connectors

The design value of the compressive normal force in the concrete flange Nc is transferred to the composite beam by means of the headed studs, as shown in Figure B2.6.

Figure B2.6 Transfer of compressive normal force in concrete flange Nc by means of studs

Calculating the plastic resistance moment of the composite section Mpl,Rd is conducted so that the minimum force of Nc and Npl,a is multiplied by the lever arm z. In this example, the lesser force is Npl,a so that:

The design value of the compressive normal force in the concrete flange Nc can be reduced in the ratio of the degree of utilization (MEd/Mpl,Rd):

where red Nc is the reduced compresive normal force in the concrete flange. Since the design resistance of each stud is = 73,7RdP kN, the required number of studs can be determined for half span L/2 as:

In the elastic design, the shear connectors are spaced in accordance with the shear flow with a triangular distribution. If the span of the beam is divided into the several ranges where the studs are distributed at equal intervals, this leads to the idea that is shown in Figure B2.7. The total number of studs, n, is shared between

Example B2 61

lengths of the ranges in proportion to the areas of the design shear force diagram of the considered ranges. This option results in uniform resistance in each length. The longitudinal shear resistance is made to match the peak value of shear flow over that length and the following assumptions are valid: - the peak shear flow within each length does not exceed the design longitudinal

shear resistance per unit length by more than 10%, - the total design longitudinal shear over the length does not exceed the total

design resistance for this length. This simplified model is shown in Figure B2.7.

Figure B2.7 Stepped form of shear flow The ranges in proportion to the areas of the shear force diagram are calculated as follows:

62 B Composite beams

When determining the spacing of the studs, it is necessary to comply with the following conditions: - the minimum spacing of studs in the direction of the shear force eL 5d, but not

greater than six times the thickness of the concrete slab or 800 mm, - the minimum spacing of studs in the direction transverse to the shear force eq

Clause 6.6.1.3(5), EN 1994-1-1, refers to the calculation of the longitudinal shear flow according to elastic theory. This assumes the use of the expression:

where: I is the ideal second moment of area of the cross-section, S is the ideal first moment of area of the concrete slab or the steel section about

the elastic neutral axis. Thus, the distribution of the shear flow is obtained as shown in Figure B2.8.

Figure B2.8 Distribution of shear flow

L/2 = 5500

vL,Ed (kN/m)

2,5d. The arrangement of studs and the spacing of the studs are determined according to the diagram of the longitudinal shear flow. The half span of the beam is divided into three regions with the same areas of the shear flow. The lengths of these regions are 18%, 24% and 58% of the half span, L/2. The procedure is carried out as follows. The reduced compressive normal force in the concrete flange is:

Example B2 63

The diagram of the elastic shear flow is triangular, and the design value of the shear flow at the first support is:

Since we are dealing with the half span of the beam divided into three regions with the same shear flows, the model for calculating the lengths of the regions is formed as shown in Figure B2.9.

Figure B2.9 Model for calculating the lengths of the regions In the first step, we calculate the length of the first region, a, and 2

,L Edv according to Figure B2.10.

Figure B2.10 Model for calculating the length of the first region The first equation is obtained from the calculation of the area of the triangle of the length x:

3,L Edv2

,L Edv1

,L Edv

A1 = A2 = A3 = Nc/3 = 1976/3 = 659 kN

L/2 = 5500

a b c

A1 A2 A3

2,L Edv1

,L Edv

L/2 = 5500

a x

A1 A2+A3

64 B Composite beams

The second equation is obtained from the ratio of the sides of triangles according to Figure B2.10:

Substituting the expression 1 2

, ,· = 5,5·L Ed L Edv x v into 2, · = 2636L Edv x , and rearranging

gives:

The area of the first region from Figure B2.10 can be given in the following form:

Analogously, the value of 3

,L Edv is calculated, and then the other percentages of the lengths of the regions 2 and 3: 24% and 58% (Figure B2.11). In each region, 10 studs are installed which are spaced uniformly over the length of the regions. In regions 1 and 2 there are five pairs of studs, while in region 3 there are 10 studs in a row. The arrangement of the studs is shown in Figure B2.11. Another option is that the half span of the beam is divided into two regions. If each region took 50% of the longitudinal shear force, the lengths of the regions should

Substituting the previously calculated values in the above expression and the rearranging gives: a = mm In relation to the half span of the beam, the length of the first region a is:

Example B2 65

be divided in a ratio of 30% : 70% of the half span, L/2. The calculation procedure is the same as in the previous case.

Figure B2.11 Arrangement of stud connectors with regard to the shear flow diagram

4.6 Check of the longitudinal shear resistance of the concrete flange

The transverse reinforcement in the slab is designed for the ultimate limit state so that premature longitudinal shear failure or longitudinal splitting is prevented. Figure B2.12 shows the critical sections for the failure of the concrete flange due to longitudinal shear. For these sections it is necessary to implement the verification.

Figure B2.12 Critical sections for failure of the concrete flange due to longitudinal shear

5. Commentary

The example illustrates the arrangement of the studs according to clause 6.6.1.3(5), EN 1994-1-1, which relates to “longitudinal shear calculated by elastic theory”. Clause 6.6.1.3(1)P and to a lesser extent 6.6.1.1(2)P, refer to the

Longitudinal shear resistance of concrete flange

90

10

100

32

Dimensions of stud

19

802

2 2

2

11

100 140

2

2

140100

66 B Composite beams

“spacing” of the studs and the “appropriate distribution” of the longitudinal shear. The interpretation of “appropriate” depends on the applied method of analysis and the ductility of the studs. The spacing of the studs in accordance with clause 6.6.1.3(5), i.e. the studs being spaced “elastically”, can be applied generally. The more appropriate use of uniform spacing requires the studs to satisfy clause 6.6.1.3 (3), which implies, but does not require, the use of plastic resistance moment. The studs must be “ductile”, as defined in clauses 6.6.1.1(4)P and 6.6.1.1(5)P. This is normally achieved by satisfying clause 6.6.1.2.