Chapter 4.2
BS5950:Part 3: Section 3.1 Simply Supported Composite Beams
Summary:
• • • • • • • •
The procedures for determining the section classification are similar to those for bare steel sections, although some modifications may be made. The moment moment resistance of class class 1 and 2 sections is calculated using plastic analysis, the details depending on the neutral axis position. The moment moment resistance of class 3 sections is calculated using elastic analysis, with due account for creep and special consideration of buildings used mainly for storage. The vertical shear strength is based on that of the bare steel section. The details of the longitudinal shear connection (number and type of connector, and slab reinforcement) are determined on the basis of the longitudinal force transmitted between the steel section and concrete slab. Where insufficient connectors are provided, the beam may be designed on the basis of partial interaction, the moment resistance calculated on the basis of the longitudinal force transmitted between the steel section and concrete slab. Deflection limits are as stated in EC3 for bare steel sections. Concrete cracking can be controlled by ensuring a minimum amount of slab reinforcement and limiting bar size and spacing.
Notes: This material comprises one 60 minute lecture.
Objectives:
•
To outline the design checks which are necessary for both ultimate and serviceability limit states.
•
To describe the procedures for determining the section classification, and the modifications which may be made. To explain the procedures for calculating the plastic moment resistance of class 1 and and 2 sections in relation to neutral axis position. To explain the procedures procedures for calculating the elastic moment resistance of class 3 sections in relation to the method of construction, with special consideration of buildings used mainly for storage. To describe the simplified procedures for checking the vertical shear strength of composite composite beams. To explain how the longitudinal shear connection is designed.
• • • • •
To introduce the concept of partial interaction, and describe how it affects the calculated moment of resistance.
•
To outline the requirements for controlling deflections and concrete cracking at the serviceability limit state.
1
CONTENTS 1 2. 3 4 5 6 7 8
Introduction Advantages and Disadvantages Disadvantages of Composite Construction Construction Methods Effective Width Shear Connection: Full and Partial Interaction Moment Capacity Moment Capacity at High Shear Design of Shear Connections 8.1 Shear Stud Connectors Connectors in Solid Solid Slab 8.2 Shear Stud Connectors Connectors in Composite Slab 8.3 Reduction Factor for Deck Shape 8.4 Full Shear Connection 8.5 Partial Shear Connection 8.6 Minimum degree of shear connection 9 Moment Capacity M c with Partial Shear Connection Based on Plastic Theory 10 Check Capacity At Other Locations 11 Transverse Reinforcement 12 Composite Beams - Serviceability 12.1 Modular Ratio e
α
12.2 Second Moment Moment of Area Ig 12.3 Elastic Section Modulus 12.4 Serviceability Stresses 12.5 Deflections 12.6 Deflection for Partial Composite Composite Section 13 Examples
2
Composite Beams 1 Introduction
Composite construction in buildings is covered by BS5950:Part 3: Section 3.1 which deals with the design of steel beams, usually I section, to act "compositely" with concrete or composite slab by use of shear connectors. If slip at the interface between concrete and steel is free to occur, each component will act independently, as shown in Figure 1. If slip at the interface is eliminated, or reduced, the slab and steel member will act together as a composite composite unit. The resulting increase in strength and stiffness will depend on the extent to which slip is prevented.
Figure 1
Non – compos ite and compos ite beams beams
Various forms of composite section are shown in Figure 2.
Figure Variious forms of composite beams
3
Figure 3 (a) composite beam with composite slab (b) Composite beam supporting precast slab One way of achieving a bond between beam and slab is to weld stud shear connectors on to the upper flange of the steel beam (Figure 3a). These form an anchorage for the concrete slab, preventing any movement in position between the underside of the slab and the beam flange. The concrete slab, which is necessary anyway to support area loads, acquires an additional function; it forms the compression chord of the composite cross-section. The tensile bending stresses are borne by the steel beam. A higher degree of stiffness thus ensures minimal deflection. 2. Advantages and Disadvantages of Composite Construction
Advantages Saving in Steel Weight up to 30 to 50% Greater stiffness leading to shallower steel beams for the same span Increase floor stiffness Increased span length for a given member Rapid construction
Disadvantages Require connectors
Complexity in Design
3 Construction Methods Propped Construction Propping is generally required to support the steel beam until the concrete has hardened. • • Propping at the quarter-span points and mid-span is generally adequate. The props are usually left in place until the concrete slab has developed three-quarter of • its design strength. Unpropped Construction • In order to reduce the time of construction, "unpropped construction" is usually preferred. Initially the steel beam alone resists its own weight, and that of the formwork, wet concrete and placement loads. Other loads are added later and so are carried by the composite member when the concrete is hardened.
4
Beam Span To Overall Depth It is normally found that strength and serviceability design limit states are just satisfied when the ratio of beam span to overall depth (including the concrete or composite slab) is between 18 to 22. This represents the optimum design of simple composite beams. 4 Effective Width
A composite beam behaves structurally as one of the series of parallel T beams with thin wide flanges as shown in Figure 4. The effective width Be of the slab is defined which acts in conjunction with the steel beam. The concrete flange is in compression and the steel section is largely in tension when the beam is subject to sagging moment, as in simply-supported construction. The effective width B e is calculated as followed: • Secondary beams (The slab is perpendicular to the beam span):: Be = 25% the beam length (L/4) but not greater than the beam spacing (b o)
•
•
4.6
Primary beams (The slab & the beam span in the same direction): Be = 25% the beam length (L/4) but not greater than 80% the beam spacing (0.8b o) Edge beams: Half of the above values (L/8), as appropriate, plus any projection of the slab beyond the centre-line of the beam.
where
bo = Average spacing of adjacent beams. L = Beam effective span length
Be e
Be = smaller of L/4 or b o smaller of L/8 or b o /2
bo FIGURE 4
5
bo
Secondary beams Primary beam bo
L bo shown in the figure is the beam spacing for the secondary beam. For primary beam the beam spacing is L (i.e, b o = L) The effective widths are used in both strength and serviceability calculations. For the primary beam (i.e., slab spanning parallel to the steel beam) the factor 0.8 accounts for the co-existing effects of bending in the slab and beam where both span in the same direction and utilise a proportion of the compressive strength of the concrete. The effective span length is taken as the original span length for beams simple supported at both ends. In continuous beams L is taken as 0.8 x span length for end spans, or 0.7 x span length for internal spans.
5 Shear Connection: Full and Partial Interaction
Full Strength shear connection: Sufficient shear studs are provided to develop the plastic strength of the beam. However, this may result in a large number of shear connections, particularly if the dimensions of the cross-section or the material properties are governed by factors other than the strength of the composite beam. Partial Strength Shear Connection: A economical design may be achieved in which the number of shear connections is such that the degree of interaction is just sufficient to provide the required flexural strength.
6
STUDS
6 Moment Capacity Compressive and Tensile Capacities of Concrete and Steel
The tensile capacity of the steel section is obtained from Clause 4.4.2 as:
Rs where A = ρy =
= Aρy
Cross-sectional area of the steel section Design strength of the steel (BS5950:Part 1, Table 5) and it is dependent on the thickness and grade of steel.
For deck laid perpendicular to the beam, the compressive capacity of the concrete slab over its effective width is obtained from Clause 4.4.2 as:
Rc
0.45f c
= 0. 45fcu Be ( Ds − Dp )
Ds - Dp
R c
where Be = the effective width of the slab
f cu = Cube strength of concrete Ds = Overall slab depth D p = Depth of the deck profile.
R s D
7
Be
Be
Steel deck Concrete Slab
Ds
Steel deck Ds
Dp
Parellel Decking
Concrete Slab
Dp
Perpendicular Decking
This equation also conservatively applies for cases where the deck is laid parallel to the beam. B2.2 Plastic Moment Capacity of a Full Composite Beam Only positive moment, full shear connection and compact steel beam, are considered in the following derivation.
The plastic moment capacity of a symmetrical composite section, M c, is derived based on the relative magnitudes of R s and R c which determines the position of the plastic neutral axis (PNA). Design equations, which are given in BS5950:Part 3 Appendix B, are presented in the following order using the case numbers as given in the code. Case 2b: R s
≤ R c (PNA in concrete flange) B
D
e
x D -D p s
s
Compression = R
D p
D Tension = R
s
Neutral axis depth, x Equilibrium of forces Tension = Compression
⇒ Aρy = 0. 45fcu Be x Aρ y x=
0. 45fcu B e ∴ x = R s ( D − D ) because R c = 0. 45fcuBe ( Ds − D p ) s p R c Taking moment about the top of the slab, and substituting for x D x M c = R s ( Ds + ) − R s = R s⎛ ⎜D s + D ⎟ − R s R s (Ds − Dp ) 2 2 ⎝ 2 ⎠ 2R c 8
s
Mc
D R ⎛ = Rs⎜⎝ Ds + 2 − 2R s (Ds − Dp) ⎠⎟ c
where D = Depth of the steel beam, as shown in the figure
Case 2a: R s
> R c ≥ R w (PNA lies in steel flange)
R w is the axial capacity of the web of the steel beam and is equal to R s − 2 R f . R f is the axial capacity of one steel flange of thickness T;
R f = BTρ y
Be Rc
Ds - Dp Ds
Dp
x
ρy
PNA
D
T Neutral axis depth, x From equilibrium:
2ρy Bx + R c
= Rs
⇒ 2ρy Bx = R s − R c x=
− R c R s − R c = 2ρ y B 2 R f / T
Rs
ρ
where R f = resistance of steel flange = BT y
B2.1
Moment about top flange of beam Mc
D = R s 2 + Rc⎜ Dp + ⎝
Ds − D p ⎟ − ( Rs − R c ) x 2 ⎠ 2
or Mc = R s
⎛ Ds + Dp ⎞ (Rs − R c )2 T D ⎟− R c⎜ + 2 R f 4 ⎝ 2 ⎠
9
(Ds-D p)/2+D p
Case 1a:
Rc
< R w (PNA lies in web of steel beam) Be Rc
Ds - Dp Ds
Dp PNA
D
ρy
x
ρy
T
Neutral axis from the centroidal axis of the beam, x 2ρy tx = R c x=
R c 2 ρy t
=
R c 2 R v / d
=
d R c 2 R v
where R = dtρ = resistance of the clear web depth v y Moment about the centroid of the beam Mc
(Ds − D p ) ⎤ ⎡D x = M s + R c⎢ 2 + D p + − R c 2 ⎥ 2 ⎣ ⎦
Mc
⎡ D + 2D p + D s − D p ) ⎤ d R c R c = M s + R c ⎢ − ⎥ 2 2 R v ⎣ ⎦
1 2
⎡ D + Ds + Dp ) ⎤ d R 2c M c = M s + R c⎢ ⎥⎦ − 4 R v 2 ⎣ where Ms = the moment capacity of the steel section R v = the axial capacity of the web of depth d. This differs only slightly from R w (due to the root radius) and may be taken as R w = Dtρy as a good approximation.
The above equations assume that the web is compact (i.e., the depth of web in compression does not exceed 38t ε, where t is the web thickness and ε is 275/ ρy ), and the beam is fully composite. An equivalent expression is also given for a partially composite beam, and it will be discussed in Section 4.8.
10
5.3.4
7 Moment Capacity at High Shear
For beams subject to point load, high shear and moment may co-exist at the point load positions. Where the applied shear force F v exceeds 0.5Pv, the moment capacity of the composite section should be reduced to account for the effects of shear as given in Clause 5.3.4 as: ⎡ 2Fv ⎤2 M cv = M c − ( M c − M f )⎢ − 1⎥ ⎣ Pv ⎦ where Mc = Moment capacity of the composite section Mf = Plastic moment capacity of the section having deducted the web area; Pv = shear stress of 0.6 y times the shear area of the beam when d / t 63
ρ
≤
ε
(Pv is the lesser of the shear capacity and web buckling capacity determined from BS5950:Part 1 Clause 4.2.3. Pv = 0.6ρyDt, as shown in the figure.
t
D
Example 1: Design a simply supported composite beam using
BS5950:Part3.1
Design data: Beam
3.0m Grade 43 Universal Beams Span = 12.0m Spacing of steel beams = 3.0m Concrete slab depth = 125mm Concrete Grade = 30
3.0m
12m
Loading Dead Load = 15.0kN/m Imposed Load = 16.0KN/m Bending Moment Design Moment =
Be = 3m
Ds
2 . ×15+16 . ×16)×122 M= WL = (14 =839kN 8 8
11
Effective width of the concrete flange Be is the lesser of b o = 3.0m or L/4 = 3.0m Assuming that the natural axis is in the concrete flange i.e., R c > R s where R s = Aρy
⎛ ⎝
D R s ⎞ − (D − Dp )⎟ ⎠ 2 2R c s Substituting for D p = 0, Ds = 125mm and Mc = Rs⎜ Ds +
Rc = 045 . × 30 × 3000 × (125) × 10−3 = 5063kN we have D R s Mc = R s 0125 . + − (0125 . ) ⎝ ⎠ 2 2 × 5063 Trial and error design Procedure: (1) select a steel section, and find R s
= Aρy from the section table
(2) check R c = 5063kN > R s (3) evaluate M c from Eq. (2) and check that M c is greater than 839kNm. Section
D (m)
R s (kN)
M c (kNm)
UB406 x 178 x 67 UB406 x 178 x 74 *UB457 x 191 x 74
0.4094 0.4128 0.4572
2350 2620 2620
741 783.4 841.8
*Work Example: Select UB 457 x 191 x 74 R s = 2620kN < R c = 5063kN, i.e., neutral axis lies in the slab Mc
04572 . 2615 ⎞ . . ) =841.8kNm>839kNm OK = R s⎛ + 2 − 2 × 5063 (0125 ⎝ 0125 ⎠
Check for High Shear Note that high shear does not co-exist with maximum moment, therefore moment reduction is not required.
The design assumes that serviceability criteria do not control and the beam is fully composite. 8 Design of Shear Connections
5.4.6
8.1 Shear Stud Connectors in Solid Slab The characteristic resistance Qk of a headed shear stud with the dimensions and properties given in Clause 5.4.6 (Table 5) as shown below.
12
The design capacities of shear connectors to resist longitudinal shear are taken as 80% of their characteristic resistance in sagging (positive) moment regions and 60% in the hogging (negative) moment region. Design Capacity = Q = 0.8Q k 5.4.3 For lightweight aggregate concrete, the characteristic resistance should be reduced and taken as 90% the value given in Table 5. For lightweight concrete: Q = 0.9(0.8Q k ) 8.2 Shear Stud Connectors in Composite Slab There is a potential reduction in the resistance of the shear connectors when used in a composite slab consisting of deck profile.
In principle, the smooth flow of force into concrete depends on the projected angle from the base of the connector to the top of the adjacent profile, as shown in the figure below.
13
8.3 Reduction Factor for Deck Shape The following reduction factor, k, are to be applied to the characteristic resistance in Table 5 of the code. (i) Decking perpendicular to beam
5.4.7.2
h •
⎡ b r ⎤⎡ h k = 085 . ⎢ ⎥⎢ ⎣ D p ⎦⎣ D p •
⎤ − 1⎥ ≤ 10 . ⎦
For two shear connectors per trough:
⎡ b ⎤⎡ h k = 0.6⎢ r ⎥⎢ ⎣ D p ⎦⎣ Dp •
D p
For one shear connector per trough:
br
⎤ − 1⎥ ≤ 08 . ⎦
h
For three shear connectors per trough:
⎡ br ⎤⎡ h ⎤ − 1 ≤ 0.6 ⎣ D p ⎥⎦⎢⎣ Dp ⎥⎦
D p
k = 05 .⎢
br 14
where
b r is the average trough width (for trapezoidal deck profile) or minimum trough width (for re-entrance deck profile). h is the as-welded height of shear connector In all cases h ≥ D p + 35 and h ≤ 2 D p . The lower limit is to ensure an adequate projection of the head of the stud above the trough. The upper limit is to avoid the above formulae becoming unconservative. Restrictions on br Restrictions are placed on the dimension b r of that may be used in the above equations when shear connectors are welded non-centrally in the troughs. For shear connectors placed in the "unfavourable" location such that the zone of the concrete in compression in front of the stud is maximised, then b r = 2 e where e = minimum distance from the centre of the shear connector to the mid-height of the adjacent web of the profile, as shown in the following figures. For shear connectors placed in pairs, but in off-set pattern, alternatively on the unfavourable and favourable sides on the trough, b r may be determined as for centrally loaded shear connectors.
15
5.4.7.3
(ii) Decking parallel to beam
For b r / D p
≥ 1. 5; k = 1. 0
< 1. 5 ⎡ b ⎤⎡ h ⎤ − 1 ≤ 10 k = 0.6⎢ r ⎥⎢ . ⎣ D p ⎦⎣ D p ⎥⎦
For b r / D p
Design capacity of shear connector with decking Normal weight concrete R = (0.8Qk ) k Light weight concrete R = 0.9 (0.8Q k ) k k = reduction factor for decking effect.
8.4 Full Shear Connection In order to develop full composite action, the longitudinal force R q to be transferred by the shear connectors should exceed the smaller of R c or R s. This is known as "full shear connection."
Rq
> Smaller of
Rs or R c
where R q = n s ( kQ k ) ns = number of shear connectors between the points of zero and maximum moment k = reduction factor for decking Qk = characteristic resistance from Table 5. If R q is less than the smaller of R c or R s then the beam should be designed as "partial shear connection". The moment capacity of the section needs to be evaluated to account for the effects of partial connection. Example 2: From Example 1, determine the number of shear studs required for full shear connection.
M = 839kNm
Design Data:
UB457x191x74 280kN Span = 12.0m Concrete slab depth = 125mm Concrete Grade = 30 Shear Studs: ∅19mm, 95mm long Loading Dead Load = 15.0kN/m Imposed Load = 16.0KN/m
280kN
16
Design moment = 839kNm Design Shear force = 280kN For UB 457 x 191 x 74
Rs
= Aρy = 2615kN
Smaller of R c and R s is 2615kN. Capacity of shear connector (19mm diameter and 95mm long)
Q k = 100 kN
Table 5
=
Design capacity Q = 0.8 Q k 80 kN No. of connectors per half span = 2615/(80) = 32.7 Use 34 connectors with two connectors per trough in pairs with spacing as shown 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34
8.5 Partial Shear Connection
If the choice of cross-section is governed by considerations other than ultimate flexural resistance, it may be unnecessary to provide as many shear connectors as required by full shear connection. Indeed such “partial connection” may be unavoidable where a slab is constructed with profiled sheeting. If this spans onto the steel section, the number of connectors may be limited by only being able to place them in the troughs of the profile. The reduced interaction between the slab and the steel beam will result in a reduction in the load carrying capacity of the composite member. Where ductile shear connectors are used, the resistance moment of the critical cross-section of the beam M cr may still be calculated by means of plastic theory but now taking account of the reduced value of the compressive force that will be developed in the concrete flange.
17
Mcr /Mc Ms
R q Mr
Min. of R c or R s
Ms/Mc
Mc
Na/N p
8.6 Minimum degree of shear connection
To avoid brittle failure of shear connector due to excessive slip, it is necessary to impose a minimum degree of shear connection, dependent of the span L of the beam. Let the degree of shear connection, N a/N p, be defined as the actual of shear connections provided (Na) divided by the number required for full shear connection (N p). Limits are placed on the degree of shear connections
5.5 .2.
For spans up to 10m, Na/N p ≥ 0.4. For spans between 10 to 16m, Na/N p > (L-6)/10, ≥ 0.4 where L is the beam span in metres. For spans greater than 16m, full shear connection should be used. The moment capacity need o be calculated based on either the plastic method or by the simplified method. Moment Capacity
The reduced interaction between the slab and the steel beam will result in a reduction in the load carrying capacity of the composite member. Where ductile shear connectors are used, the 18
resistance moment of the critical cross-section of the beam M Rd may still be calculated by means of plastic theory givenin the subsequent sections. but now taking account of the reduced value of the compressive force that will be developed in the concrete flange. The relation between M rc and degree of shear connection is qualitatively given by the convex curve ABC in the above Figure where M s and M c are the design plastic resistances to sagging bending of the structural steel section alone, and of the composite section with full shear connection, respectively. Alternatively, a conservative value of M rc may be determined by the straight line AC in the above Figure:
M rc
= M s + ( M c − M s )
N a N p
9
Moment Capacity M c with Partial Shear Connection Based on Plastic Theory A ppendix B Partial shear connection applies when R q is less than R c and R s .
For a compact section with equal steel flanges, the moment capacity of a composite section, Mc, is determined by the following:
Case 3a: R q
< R w (PNA in web)
B.2.3
Be
y Ds
D p
Rq Rq
ρy
=
PNA
x
D
ρy
ρy 2ρy R q Ms
xt 2ρy = R q and R v = tdρy R q d R q x= = t 2ρ y 2 R v y Also, R = R q c Ds − D p R ⇒ y = q ( Ds − D p ) R c Take moment about the centroid of steel section Mc = Ms +R q [D/2+Ds-y/2]-R q x/2 Substituting for x & y and rearranging terms:
19
2 R q ⎛ Ds − Dp R q D d ⎜ ⎟ − R 4 M c = M s + R q⎢ + Ds − ⎥ 2 R 2 ⎠⎦ v ⎣ c ⎝
R w =Dt ρy is the axial capacity of the web between the extreme fibre of the flanges. R v is the axial capacity of the web of depth d. R v may be taken as equal to R w. Case 4: R q
≥ R w (PNA in steel flange)
Be Ds - D p D p
y R q
x
Ds
ρy
D
R q 2ρy
PNA
R s -R q
R s
ρy
x=
y=
− R q R s − R q in which R f = 2BTρy = 2ρ y B 2 R f / T
Rs
R q R c
( Ds
− Dp )
Moment about top of steel flange D y ⎞ x ⎛ M c = R s + R q ⎜ Ds − ⎟ − (R s − R q ) 2 2 ⎠ 2 ⎝ 2 R q (Ds − D p ) ⎞ (R s − R q ) T ⎛ ⎜ ⎟ M c = R s + R q ⎜ Ds − ⎟− 2 R c 2 R f 4 ⎝ ⎠
D
The last term in this equation is small, and can be neglected. Note that the expression in case 3a assume that the web is compact (see Table 2 of BS5950:Part 3 for web classification) 76ε If d > , the web is non-compact.
t
1 − R q / R v
A modified expression for M c is given in BS5950: Part 3 Clause B2.3 Case 3 (b) when the web is non-compact.
10
Check Capacity At Other Locations
20
For beams subject to uniform loads, no check is required on moment capacity at other locations except at the maximum moment location. For beams subject to concentrated loads, the number of shear connectors calculated shodl be distributed to ensure that all intermediate locations have adequate moment capacity. The following formula may be used (5.4.5.5): Ni = N p (Mi-Ms) / (Mc-Ms) Where Ni = number of shear connectors between the intermediate load point and the adjacent support. N p = number of shear connectors provided. Mi = moment at the intermediate point i. Ms = moment capacity of the steel section. Mc = moment at the composite section
N2 – N1
N1
N3 =0
N2 N1
Ms
N2 – N1 M1
M2
21
N2 – N1
N1
Example: Moment capacity of partial composite beam
Moment capacity of steel section M s = 278 kNm Moment capacity of full composite section M c = 612 kNm No. of connectors from zero to max. moment, N p = 38
15
23
35-15 = 23 M1 = 410
15
M2 = 565
N1
N2
Ni = Np (Mi-Ms) / (Mc-Ms) = 38(410-278)/(612-278) = 15 N2 = 38
11 Transverse Reinforcement
Composite design requires a shear transfer from the shear connectors to the effective width of the slab. It is therefore needed to prevent shear failure of concrete surrounding the connectors. The following figure shows the critical planes for shear failure, i.e., planes 1-1, 2-3-3-2. Transverse Reinforcement 1
1 3
3 2
2
1
1
Figure: Potential shear planes through slab. The shear resistance per unit length of shear plane in a solid slab of normal concrete is given by ν r = 0.7A sv f y + 0.03ηA cv f cu + V p subject to an upper limit of
22
νr = 0.8ηA cv f cu + V p f cu = characteristic cube strength of concrete in N/mm ≤ 40 N/mm ) f y = yield strength of the reinforcement Asv = steel reinforcement area per unit length of the beam, crossing the shear plane Acv = cross-sectional area per unit length of beam of the concrete shear surface under consideration. = 1.0 for normal weight concrete η = 0.8 for lightweight concrete. V p = the contribution of the profile steel decking if applicable. = thickness of steel deck x design strength of steel deck 2(
2
Shear to be Resisted The total longitudinal shear per unit length V to be resisted should be determined from V = NQ/S N = no. of shear connectors in a group S = longitudinal spacing C/C of shear connectors Q = 0.8Qk Qk = characteristic resistance of shear connector from Table 5 of BS5950Part 3.1
The designer must ensure that V ≤ Vr .
23
Calculation of Shear Plane
Example: Light-weight concrete Grade 30 to be used K = 0.8 = reduction facor due to metal decking
Ds = 130mm
D p = 50mm
Design shear force V = NQ/S N = 2 studs per rib Q = 0.8 (0.9 x 0.8Q k ) = 58 kN (0.8 is the rduction factor for decking perpendicular to the beam; 0.9 is rduction fator for light-weight concrte; Qk = 100 kN for 19mm stud) 24
Metal Decking
Spacing of stud = 375 mm For intermediate beam, there are two shear planes For each shear plane 1-1 as shown in the figure V = 309/2 = 155 N/mm Sehar Resistance ν r = 0.7A sv f y + 0.03ηA cvf cu + V p 2
Acv = (80 + 50/2) x 1mm = 105 mm / mm η = 0.8 f cu = 30 N/mm2 For sheeting continuous across the beam: 2 V p = t p x py = 1 x 280 N/mm = 280 N/mm 2
2
Assume A142 mesh: Asv = 142 mm /m or 0.142 mm /mm, 2 Fy = 460 N/mm 0.7Asvf y = 0.7 x 0.142 x 460 = 46 N/mm 0.03ηAcvf cu = 0.03 x 0.8 x 105 x 30 = 76 N/mm
ν r = 0.7 A sv f y + 0.03ηA cv f cu + V p = 46 + 76 + 280 = 402 N/mm
νr = 0.8ηA cv
f cu
+ Vp = 0.8 x 0.8 x 105 (30) 0.5 + 280 = 648 N/mm
Therefore Vr = 402 N/mm > V = 155 N/mm OK However, for edge beam, there is only one shear plane. In this case V = 155 x 2 = 310 N/mm, hence same reinforcement can be used for the edge beam.
12 Composite Beams - Serviceability 12.1 Modular Ratio
αe
The modular ratio
4.1
α e is defined as the ratio of the elastic moduli of steel to concrete . α e = α s + ρ ι (α ι − α s )
where
α ι is the modular ratio for long term loading; α s is the modular ratio for short term loading; ρι is the proportion of the total loading which is long term. α ι and α s can be obtained from Clause 4.1, Table 1 of BS5950 Part 3.1.
• •
Imposed loads on floors should be assumed to be two-thirds short term and one-third long term in building of normal usage. Storage loads and loads which are permanent in nature should be taken as long term
25
Appropriate values of
α e for office-type buildings are 10 and 15 for normal and lightweight
concrete respectively.
αe = Es / Ec 12.2 Second Moment of Area I g
The second moment of area of the section (often termed inertia) is used in • Establishing the moments and forces in the structures in elastic global analysis of continuous beams. Calculation of deflections. • The second moment of area is normally derived assuming the concrete is uncracked, although allowance can be made for the effects of concrete cracking in as given in the code Clause 5.2.3. The composite section is “transformed” into an equivalent steel section by dividing the crosssectional area of concrete by the appropriate modular ratio αe as shown in the figure. Take moments about the upper surface of the concrete gives a neutral axis depth, defining the part of the composite section in compression of
yg
=
Aα e ( D + 2 Ds ) + Be ( D s 2[ Aα e
− D p )2
+ Be ( Ds − Dp )] Be/αe Stress <
Ds D p
yg
Equivalent Steel Area D Stress less than ρy
Elastic Stress Distribution
Transformed Section
The second moment of area of the uncracked composite section I g as in Appendix B.3 is given as:
26
Ig
= Ix +
− D p ) 3 AB e ( D s − D p )(D + D s + D p ) 2 + 12α e 4[ Aα e + B e ( D s − D p )]
Be ( D s
where Ix is the second moment of area of the steel section of cross-section area, A. e is the ratio of the elastic moduli of steel to concrete (modular ratio). This may be obtained
α
from Clause 4.1, Table 1. This expression ignores the contribution of the concrete within the ribs of the decking which is conservative for the case where the ribs run parallel to the beam.
12.3 Elastic Section Modulus The elastic section modulus is used to determine the stresses in the section at the serviceability limit state. If the concrete is uncracked, the elastic section modulus is (i) At the lower fibre of the bottom Steel flange Ig Zs = D + Ds − y g
(ii) At the upper surface of the concrete slab I gα e Zc = yg where yg is the depth of the elastic neutral axis below the upper surface of the concrete slab such that:
yg
=
Aα e ( D + 2 Ds ) + Be (Ds 2[ Aα e
− Dp )2
+ Be ( Ds − Dp )]
Checking of cracked and uncracked section Appendix B.4 of the code gives alternative expressions for the cases where the concrete is cracked. In most cases, the concrete will be uncracked at the serviceability limit state. 2 If A < (Ds − Dp ) Be
(D + 2D p )αe 2 If A ≥ ( Ds − Dp ) Be ( D + 2D p )αe
the section is cracked it is uncracked.
B4.1
27
12.4 Serviceability Stresses The stresses in the steel and concrete are determined using the elastic section moduli calculated from the above step for the case where the beam is subject to unfactored imposed loads. For unpropped beams add the steel stresses to those calculated for the steel section subject to the self-weight of the floor. Compare the total steel stresses with the stress limit of y (Clause 2.4.3).
ρ
For propped beams, determine the stresses in the concrete due to self weight of the floor applied to the composite section. Compare the total concrete stress due to self weight and imposed loads with the stress limit of 0.5 f cu (Clause 2.4.3). Compare the steel stresses with
ρy .
At the extreme steel fibre, the bending stress is
f bs =
M Zs
< ρy
At the extreme concrete fibre, the bending stress is
f bc
M
= Z < 05 . f cu c
12.5 Deflections Deflection limits for beams are specified in BS5950:Part 1. Deflections are calculated for the case of unfactored imposed loads applied to the composite section. It is also necessary to consider the effect of deflection of the steel beam due to self weight of the floor in the unpropped construction.
For unpropped construction: Total deflection = Deflection of composite section due to unfactored imposed load + Deflection of steel beam due to self-weight of the floor. Deflection due to self weight may be offset by precambering of the steel beam. For propped construction. Total deflection = Deflection of composite section due to unfactored imposed load For the case where the composite beam is subject to unfactored imposed loading, the beam deflection is calculated using the second moment of area given in the previous section Compare this deflection with a limit of Span/360 typical of buildings of normal usage. It may be necessary to reduce this deflection limit for beams supporting cladding or heavy partition. 12.6 Deflection for Partial Composite Section For partial shear connection, the deflection of the beams should be determined from Clause 6.1.4 which is given below:
For propped construction:
δ = δ c + 0. 5(1 − N a / N p )(δ s − δ c )
For unpropped construction:
δ = δ c + 0. 3(1 − N a / N p )(δ s − δ c ) 28
where
δ s is the deflection of the steel beam acting alone δ c is the deflection of a composite beam with full shear connection for loading. N a = degree of shear connection N p
Clause 5.4.4
N p = number of shear connectors required for full composite N a = actual number of shear connectors provided
For full composite beams N a = 1 N p
δ = δc
29
the same
13 Examples 13.1 Example A From Example 1, determine the serviceability limit state of the simply-supported beam during (a) construction stage (b) composite stage.
Design Data: UB457x191x74 Span = 12.0m Beam spacing = 3.6m Concrete slab depth = 125mm Concrete Grade = 30 Desity of concrete = 24 kN/m 3 Shear Studs: ∅19mm, 95mm long
3.6m beams
3.6m
Loading (1) During construction (Unpropped): 12m Dead Load 3 Floor (Concrete slab) 0.125m x 24 kN/m x 3.6m =10.8kN/m -3 Steel Beam 74kg/m x 9.81 x 10 = .73kN/m Total Dead Load =11.53kN/m 2 Imposed construction loads = 0.50kN/m x 3.6m = 1.8kN/m (2) Composite Stage (After construction) Total Dead Load (including floor, beam, ceiling etc.) = 15.0kN/m ......................................in which 13kN/m is long term loading. = 16.0KN/m Total Imposed Load (including partitions etc.) .....................in which (1/3x16kN/m = 5.33kN/m) is long term loading. Serviceability Deflection Deflection of beam at the construction stage:
δ=
5WL4 384 EI
δ =
5 × 11.53 × 120004
= 45.5mm > L/360 = 33.33mm 4 384 × 205000 × 33400 × 10 Precambering for dead load is required. Deflection of beam at the composite stage:
α s = 6 α ι = 18
α e = α s + ρι (α ι − α s ) ρι =
Long term loading
Total Loading long term loading = dead load + 1/3 Imposed load 15 + 5.33 ρ ι = = 0.656 15 + 16 ⇒ α e = 13.9
30
= Ix +
Ig
− D p ) 3 AB e ( D s − D p )(D + D s + D p ) 2 + 12α e 4[ Aα e + B e ( D s − D p )]
Be ( D s
A = 95cm2 Ix = 33400cm4 Be = 3600mm D s = 125mm D p = 0 D = 457.2mm
= 33400 + 4473 + 49522 = 87395 cm 4
Ig
δ=
5WL4 384EIg
W = Imposed load + additional dead Load after construction = 16kN/m + (15-6.13) = 24.9 5 × 24.9 × 120004 δ= 384 × 205000 × 87395 × 104
= 37.4 mm
Total deflection for fully composite beam: Construction stage (assumed pre-cambering for 45mm).... Composite stage .............................................................. .. Total Deflection
0 mm 37.4mm = 37.4 mm =
L
321
This is slightly higher than the allowable limit of L/360. Check for serviceability stresses To check whether the section is cracked or uncracked. A = 9500mm2 Be = 3600mm D s = 125mm D p = 0 D = 457.2mm
( Ds − D p ) 2 Be ( D + 2 D p )α e Since A >
=
1252 × 3600 (457.2 + 0)13.9
= 7376 mm2
− D p )2 Be elastic neutral axis is in the steel member and the section is ( D + 2 D p )α e
( Ds
uncracked. Bending stress in steel section at construction stage: 2 . × 122 M = WL = 613 . kNm = 1103 8
8
M 110.3 × 106 = = 7556 . N / mm2 Zx 1460 × 103 Stresses in steel and concrete at composite stage Depth of neutral axis below top of the concrete flange: Bending stress f bf =
yg
=
y g
Aα e ( D + 2 D s ) + B e ( D s 2[ Aα e
=
− D p )2
+ Be ( Ds − Dp )]
95 × 13.1(45.72 + 2 × 12.5) + 360(12.5) 2 2[95 × 13.9 + 360(12.5)]
=13.83cm
(i) At the lower fibre of the Steel flange Zs
=
Ig D + Ds
=
− yg
87395 45.72 + 12.5 − 13.83
(ii) At the upper surface of the concrete slab I g α e = 87395 × 13.1 = 96938cm3 Zc = 13.83 y g
31
= 2173cm3
W = Imposed load + additional dead Load after construction W = 24.9kNm 2 2 Bending moment M = WL = 24.9 × 12 = 4482 . kNm
8
8
Bending stress in concrete f bc
=
M Z c
=
448.2 × 106 96938 × 10
3
= 4.64 N / mm2 < 0.5 fcu = 15 N / mm 2
Bending stress in steel f bs
=
M Z s
=
448.2 × 106 2173 × 10
3
= 206 N / mm2
Total stress in steel: Construction Stage...............................................75.56 N/mm 2 Composite Stage...................................................206 N/mm 2 Which is greater than
ρy = 275 N / mm 2
NG!
Use Grade 50 steel or larger steel section!!
32
282 N/mm2
13.2 Example B The floor layout consisting of simply supported composite beams of 12m span length is shown below. Design the beam for strength and serviceability.
Floor Plan
3m beams
Unpropped Construction 3m
Floor Span
12m Materials: Structural steel Grade 50 Concrete:
ρy = 355 N / mm 2 E = 205000N/mm2 Lightweight Grade 30 fcu = 30 N / mm 2
Density 1800 kg / m 3
300mm 150mm
Ds Dp
95mm 1mm thick steel deck 19mm diameter stud
Deck Slab depth
D s = 130 mm
Profile height
D p = 50 mm
Trough spacing = 300mm Average trough width = 150mm Design strength of profile steel sheets: ρyp = 280 N / mm 2 Thickness t p = 1 mm Shear connectors 19mm diameter studs 95mm as welded length
33
Floor Loading: (i) Concrete slab and steel decking Assuming slab thickness of 80 + 25 =105mm Weight of concrete = 1800 x 9.81 x 105 x 10 -6 = 1.85 kN/m2 Weight of steel deck = 0.15 kN/m2 Total weight = 2.00 kN/m2 (ii) Construction Stage Floor = 2.00 kN/m2 Steel beam = 0.22 kN/m2 Total dead load = 2.22 kN/m2 Imposed construction load 0.50 kN/m 2 (iii) Composite Stage Floor = 2.00 kN/m2 Steel beam = 0.22 kN/m2 Ceiling = 0.50 kN/m2 Total dead load = 2.72 kN/m2 Total Imposed load (including partitions) 6.0kN/m 2 Effective Width L/4 = 12000/4 = 3000 mm , b o = 3000mm Be ⇒ = 3000mm Select initial size of steel beam At composite stage Design load W = (1.6 x 6 + 1.4 x 2.72 ) x 3 = 40.2 kNm Design shear force Fv = 40. 2 × 12 / 2 = 241 kN
Design moment M =
WL2 8
=
40. 2 × 12 2 8
= 724 kNm
Assuming that the plastic neutral axis lies in the concrete slab i.e., R c > R s Full shear connection: M = R ⎛ D + D − R s ( D − D ) c s p ⎝ s 2 2R c s Resistance of concrete flange R c = 0. 45fcuBe (Ds − Dp )
D p Rc
B2.1
= 50, Ds = 130mm
= 0. 45 × 30 × 3000 × (130 − 50 ) ×10 −3 = 3240 kN
⎛ ⎝
D R s (0130 . − − 0.05) ⎠ 2 2 × 3240 D R s Mc = R s 0130 . (0.08) + − ⎝ 2 2 × 3240 Mc = R s 0130 . +
(1)
Normally the depth of the beam may be taken approximately as Span length / 20
≈ 12000 / 20 = 600 mm
By trial and error, check R c
> R s , and Mc from Eq. (1) is greater than 724kNm.
Section D (m) R s (kN) (Grade 50 Steel) UB356 x 171 x 67 0.364 3032 *UB457 x 191 x 67 0.4537 3032 2 Note design strength for Grade 50 steel is 355N/mm 34
M c (kNm) 834.4 968
*Select UB 457 x 191 x 67 Grade 50 Since (R c = 3240kN) > (R s = 3032kN), neutral axis lies in the slab 0.4537 3032 ⎛ ⎞ M c = 3032⎜ 0.13 + (0.08) ⎟ = 968kNm > 724kNm OK − 2 2 × 3240 ⎝ ⎠ Check for Shear
Fv =
WL
=
(1. 4 × 2. 72 × 3 + 1. 6 × 6 × 3) × 12
= 241kN 2 2 0.5Pv = 05 . (0.6Dtρyw ) = 05 . × 821 = 411kN > 241kN Note that high Shear does not coincide with the maximum moment. Shear Connection R c = 3240 kN Rs = Aρy = 3032 kN
Smaller of R c and R s is 3032kN. Capacity of shear connector (19mm diameter and 95mm long) in lightweight concrete 5.4.6
Q k = 0. 9 × 100kN = 90kN Design capacity Q = 0.8 Q k = 72 kN
Reduction factor for deck profile ⎞ b ⎛ h k = 0.60 r ⎜ − 1⎟ ≤ 08 . for two studs per rib D p ⎝ D p ⎠
5.4.7.2
b r = Average
trough width = 150 mm h = overall height of the stud = 95 mm
k = 0.60
150 ⎛ 95 − 1 = 162 . > 08 . 50 ⎝50 ⎠
∴k = 0.8 Resistance of a shear connector = 0.8 x 72 = 57.6kN For full composite, no. of connectors per half span required = 3032/(57.6) = 52 Since the trough spacing is 300 mm, no. of connectors that can be accommodated in half span, assuming two connectors per trough = 2 x 6000/300 = 40 as shown below.
C L
2 studs @300mm half span length = 6000mm
35
C L
Degree of shear connection = N a
N p = No. of actual connectors provided = 40 No. of connectors in full composite 52 N a ( L − 6 ) ≥ = 0. 6 ≥ 0. 4 N p 10
= 0. 77 OK
Therefore, the beam needs to be redesigned for partial shear connection. Redesign for Beam Moment Capacity with Partial Shear Connection
Resistance of overall web depth R w
R w = 3032 −
= R s − R f
2 × 189. 9 × 12. 7 × 355
1000 40 × 57. 6 = 2304kN
As R q = Since R q Mc
B2.1
= 1320 kN
> R w , plastic neutral axis is in flange R ( Ds − D p ) ( R − R )2 ⎟− s q T 2 R f 4 ⎠ × × −3 =
D = R c 2 + R q⎜ Ds − R q ⎝ c
B2.3 B2.3
R f = 189. 9 × 12. 7 355 10 856 kN 4536 . 2304 80 (3032 − 2304)2 12.7 ⎛ Mc = 3032 × + 2304 130 − × − ⎝ 2 3240 2 ⎠ 856 4 = 920kNm > M = 724kNm
OK
Serviceability Deflection Deflection of beam at the construction stage due to dead load: 5WL4
δ= δ=
384 EI 5 × (2.22 × 3) × 12000 4 384 × 205000 × 29410 × 10
4
= 29.8 mm < L/360 = 33.33mm
Deflection of beam at the composite stage:
α e = α s + ρ ι (α ι − α s ) α s = 10 α ι = 25 for lightweight concrete
Long term loading:
Table 1
2.72 kN/m 2 2.00 kN/m 2 4.72 kN/m2
Dead load 1/3 Imposed Load
8.72 kN/m2
Total Loading 4. 72 ρι = = 0. 541 8. 72
⇒ α e = 10 + 0. 541(25 − 10 ) = 18.1
36
Ig
= Ix +
Be ( Ds − D p ) 3 12α e
+
ABe ( Ds − D p )( D + Ds + D p ) 2 4[Aαe
+ Be ( Ds − Dp )]
A = 85.4cm2 Ix = 29410cm4 Be = 3000mm D s = 130mm D p = 50 D = 453.6mm 3 3000 × 80 8540 × 3000 × 80(453.6 + 130 + 50) 2 4 + I g = 29410 × 10 + 12 × 18.12 4(8540 × 18.12 + 3000 × 80)
δ=
= 80900 cm4
5WL4 384EI g
Total imposed load W = (6+0.5) x 3m = 19.3kN/m 2 Dead Load due to ceiling is small (0.5kN/m ) and it is neglected in the service check. 5 × (19.3) × 12000
= 31.7mm 4 384 × 205000 × 80900 × 10 Deflection when steel beam is acting alone 80900 × 31.7 = 87.2mm δ s = 29410 31.7 + 0.3(1 - 0.77) (87.2 – 31.7) = 35.5 mm δ c
=
4
δ=
Total deflection for partial composite beam: Construction stage (no pre-cambering, and unpropped).... Composite stage .............................................................. .. Total Deflection
29.8mm 35.5mm = 65.3mm =
L
184 This is probably unacceptable. However, we will proceed with the calculation. Check for serviceability stresses To check whether the section is cracked or uncracked. A = 8540mm2 2 ( Ds − D p ) 2 Be 80 × 3000 = = 1914mm2 ( D + 2 D p )α e (453.6 + 2 × 50)18.12
− D p )2 Be elastic ( D + 2 D p )α e
( Ds
Since A >
neutral axis is in the steel member and the concrete is
uncracked. Bending stress in steel section at construction stage: M=
WL2 8
=
2. 22 × 3 × 12 2 8
= 120 kNm
For 457 x 191 x 67 UB Grade 50 M 120 × 106 Bending stress f bf =
Zx
=
1300 × 103
B.4
= 92 N / mm 2
Stresses in steel and concrete at composite stage since A > ( Ds − Dp )2 Be ( D + 2 D p )α e Depth of neutral axis below top of the concrete flange: 37
yg
y g
=
Aα e ( D + 2 D s ) + Be ( D s − D p ) 2 2[ Aα e
=
+ Be ( Ds − D p )]
8540 × 19.3(453.6 + 2 × 130) + 3000(80) 2 2[8540 × 18.12 + 3000(80)]
=169 mm
(i) At the lower fibre of the Steel flange 80900 × 10 Ig = Zs = D + D s − y g 453.6 + 130 − 169
= 1950 cm 3
(ii) At the upper surface of the concrete slab I g α e = 80900 × 19. 3 Zc = = 92400 cm 3 169 × 10 −1 yg Additional service load after composite = 6 + 0.5 = 6.5 kN/m
2
2 2 Bending moment M = WL = (6.5 × 3) × 12 = 351kNm
8
8
Bending stress in concrete f bc
=
M Zc
=
324 × 10 6 92400 × 10
3
= 3. 51 N / mm 2 < 0.5 fcu = 15 N / mm 2
Bending stress in steel f bs
=
M Zs
=
324 × 10 6 1950 × 10
3
= 166. 2 N / mm 2
Total stress in steel: Construction Stage................................................92.0 N/mm 2 Composite Stage...................................................166.2 N/mm 2 258.2 N/mm2 As 258.2 N/mm2 is less than ρy = 355 N / mm 2 OK
13.3 Example C The floor plan in Fig. 1 shows the layout of the secondary composite beams of 8m span length, simply supported at their ends. The beams are spaced at 3.0m centres. The steel decking acts compositely with the concrete slab and the deck ribs run perpendicular to the secondary beams. The materials used for the composite beams are Grade 43 steel and Grade 25 normal weight concrete. Shear studs of 19mm diameter and 95mm welded length are to be used to provide the composite action. Based on the following loading data: • Slab weight of concrete and decking = 2.1 kN/m 2 • Weight of Steel beam (approximated) = 0.22 kN/m 2 • Imposed Construction Load = 0.50 kN/m 2 • Ceiling, services, and raised floor after construction = 0.7 kN/m 2
(additional dead load) = 5.0 kN/m 2 • Total imposed load after construction (a) Design a composite beam section and determine the number of shear connectors required to carry the factored loading. (b) Determine whether propped construction method is required if the maximum service deflection is limited to L/360 (consider dead and imposed load)..
38
Secondary beams
120
Deck span L = 8m
70
160
50 A
300
A
Secti on A-A Di mensi ons are i n mm
3m
3m
3m
3m
3m
Floor Plan Figure 1 Solution: Floor Loading: (i) Construction Stage Dead Load = 2.1 + 0.22 = 2.32 kN/m 2 Imposed Load = 0.50 kN/m 2 (ii) Composite Stage Dead Load = 2.1 + 0.22 + 0.7 = 3.02 kN/m 2 Imposed Load = 5.0 kN/m 2 Effective Width L/4 = 8000/4 = 2000 mm , b o = 3000 mm Be = 2000 mm Select initial size of steel beam At composite stage Design load W = (1.6 x 5 + 1.4 x 3.02 ) x 3 = 36.7 kNm Design shear force Fv = 36. 7 × 8 / 2 = 146. 8 kN
36. 7 × 82
= = 293. 6 kNm 8 8 Resistance of concrete flange R c = 0. 45fcu Be ( Ds − D p ) D p = 50, Ds = 120 mm Design moment M
=
WL2
B2.1
R c = 0. 45 × 25 × 2000 × (120 − 50) × 10−3 = 1575kN Assuming that the plastic neutral axis lies in the concrete slab R c > R s Full shear connection: M c
D R s = Rs⎛ − D ( Ds − D p ) s+ ⎝ 2 2R c
D R s ⎛ − (0120 . − 0.05)⎠ ⎝ 2 2 × 1575 ⎛ . + D − 2.222 × 10−5 R M c = R s 0120 s ⎝ ⎠ 2 Mc = R s 0120 . +
(1)
Select a steel section and check the neutral axis depth lies with the concrete slab, i.e., R c > R s , and M c from Eq. (1) is greater than 293.6kNm. 39
Section (Grade 43)
D (m)
UB254 x 146 x 37 UB305 x 165 x 40 UB356 x 127x 39
0.2595 0.3038 0.3529
R s (kN) < R c 1304 1419 1359
M c (kNm) 287.9 341.8 361.8
Select UB356 x 127x 39 R c = 1575 kN R s = Aρy = 1359 kN Smaller of R c and R s is 1359 kN. Capacity of shear connector (19mm diameter and 95mm long) in normal weight concrete is Q k = 95kN Design capacity Q = 0.8 Q k = 76kN Reduction factor for deck profile k = 0.6
b r ⎛ h ⎜ D p ⎝ D p
⎞ − 1⎟ ≤ 0.8 for two stud per rib ⎠
b r = Average trough width = 170 mm D p = 50 mm h = overall height of the stud = 95 mm 170 ⎛ 95 ⎞ k = 0.60 − 1 = 1836 . > 08 . 50 ⎝ 50 ⎠ ∴k = 0.8 Resistance of a shear connector = 0.8 x 76 =60.8 kN For full composite, No. of connectors per half span required = 1359/(60.8) = 22.4 = 23 Since the trough spacing is 300mm, max. no. of connectors can be accommodated (assuming two connectors per trough) = 2x 4000/300 = 27. i.e., full composite is possible!! Total number of studs per span for full composite action = 2 x 23 = 46 studs OK Serviceability Deflection Deflection of beam at the construction stage: 5WL4 W = 2.32 x 3 kN/m I = 10100 cm 4 δ= 384 EI 5 × ( 2. 32 × 3) × 8000 4 δ= = 17. 93 mm 384 × 205000 × 10100 × 10 4 Deflection of beam at the composite stage: α e = α s + ρι (α ι − α s ) α s = 6 α ι = 18 for normal weight concrete
Long term loading:
Dead load 1/3 Imposed Load
3.02 kN/m 2 5/3 = 1.67 kN/m 2 4.69 kN/m2 5 + 3.02 = 8.02 kN/m 2
Total Loading 4. 69 ρ ι = = 0. 585 8. 02 ⇒ α e = 6 + 0. 585 (18 − 6 ) = 13. 0 40
