Chapter 9 Design of Composite Constructions

Design of Steel Constructions

Third Year Civil

Chapter 9 Design of Composite Constructions


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Third Year Civil

1. Introduction: Composite beams are used in roadway bridges and industrial buildings. They are rarely used in railway bridges due to the dynamic effect of rolling stock. Figure (1) shows various types of composite steel-concrete sections

2. Composite action: Composite action is developed when two load carrying structural members such as concrete floor system and supporting steel beams are integrally connected and deflect as a single unit. The developing of composite action depends on the provisions made to insure a single linear strain from top of concrete slab to the bottom of the steel beam.

Difference between composite and non-compost behavior A) Consider the non-composite beam shown in Figure; -The friction between the slab and the beam is neglected. -The beam and the slab each carry separately a part of the load. -When the slab deforms under vertical loads, its lower surface is in tension and elongates, while the upper surface of the beam is in compression and shortens. Thus, a discontinuity will occur at the plane of contact. Chapter 9 Design of Composite Constructions

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Third Year Civil

-A horizontal slippage resulting from the bottom of the slab in tension and the top the beam in compression. -Only vertical internal forces act between the slab and the beam. -There are two neutral axes, one at the center of gravity of the concrete slab and one at the center of gravity of the steel beam.

of

B) Consider the composite beam shown in Figure; -Horizontal forces (shears) are developed to act on the lower surface of the slab to compress and shorten it, while simultaneously they act on the upper surface of the steel beam to elongate it. -No relative slippage occurs between the slab and the beam. -One axis occurs which lies below that of the slab and above that of the beam.

3. Composite section: A) Steel beam: Rolled section and welded plate girder can be used. B) R.C. slab: The slab may rest on the steel beam or on concrete hunch increasing the capacity of the composite section. The minimum thickness of R.C. slab shall be: For roofs; t >8.0 cm. For floors; t >10.0 cm. for slab carrying moving loads; t >12.0 cm. Chapter 9 Design of Composite Constructions

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Third Year Civil

The concrete thickness must be 10.0 cm at least above the steel beam. Thinner slab thickness has to be provided with hunches as in Fig. 3

Figure (3) Composite section with hunched slab C) Shear connectors: The bond between the concrete slab and the steel beam is not dependable. Therefore, mechanical shear connectors must be provided. They are welded to the top of the steel beam and embedded on the concrete slab to transmit the longitudinal shear and to prevent any slip and to prevent slab uplift. They may take the shapes of the figure below.

4. Advantages of composite beams: 1- Reduction in the weight of steel (20 to 30 %). 2- Shallower steel beams. 3- Increased floor stiffness. 4- Increased span length for a given member. 5- Increased overload capacity.

5. Disadvantages of composite beams: 1- Effect of continuity: As the advantages of composite action is reduced in the area of negative bending moment; with only slab reinforcement to provide continuity. 2- Long-term deflection: This matter becomes problem when live loads are of long duration. This problem can be minimizing by using a reduced effective slab width.

6. Effective width: Chapter 9 Design of Composite Constructions

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Third Year Civil

bE fmax

fx

Figure (6) Non-uniform distribution of compressive stress and effective width

Consider the composite section under stress (Fig. 6), in which the slab is infinity wide. The intensity of the extreme fiber compressive stress, σ x maximum over the steel beam, and decreases non-linearly as the distance from the supporting beam increases. The effective width of a flange for a − − composite member may be taken bE = b f + 2 b where 2b times the maximum stress σ x is equal to the area under the curve for σ x . 1. For interior girders with slab extending on both sides of girders; bE is the least of the following: L/4 bo = spacing between girders from center to center 12 ts + bf 2. For exterior girders with slab extending only on one side of girders; bE is the least of the following: L/12 + bf 0.5 (bo + bf) 6 ts + bf

7. Computation of section modular “n”: The section properties of a composite section can be computed by the transformed section method. The concrete slab is transformed in to an-equivalent steel. The concrete area is reduced by using a slab width = bE / n

n=

Es Modulus of elasticity of steel = Ec Modulus of elasticity of cocrete

Table (1) Values of modular ratio n ( E.S.S) Chapter 9 Design of Composite Constructions

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Third Year Civil

Concrete strength Fcu kg/ cm2 250 300 400 ≥500

Ec

t / cm2 220 240 280 310

n 10 9 8 7

8. Methods of design: In design of composite section, two different methods of erection are to be considered; I) When no temporary intermediate supports are used under steel beam during pouring and setting of concrete slab. - The steel beam carries the dead load. - The composite section carries the additional live load and any load applied after setting of concrete. II) When effective temporary supports are used under the steel beam during pouring and setting of concrete. Method (I) is more economic than method (II) although the steel section in method (II) is smaller than in method (I).

9. Concrete in tension zone: If the neutral axis of the composite section falls inside the concrete slab, the cooperation of concrete in tension is neglected. The tensile stress in the concrete shall not exceed the following: Concrete strength Fcu, kg/ cm2 250 300 400 500 Tensile stress, t / cm2 17 19 23 27 The neutral axis of the composite section is to be calculated from the following equation if the cooperation of concrete in tension is neglected.

Figure (7) Concrete in tension zone

10 Shear Connectors: The horizontal shear that develops between the concrete and the steel beam during loading must be resisted so the composite section acts monolithically. The bond that developed between the slab and the steel beam can not be dependable to provide the required interaction. Neither can the frictional force developed between the slab and the steel beam. Instead mechanical shear connectors welled to the top of the steel beam and concerted with the slab must be provided.


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Third Year Civil

10.1 Types of Shear Connectors: A) Non-Rigid shear connectors: Headed and non-headed Stud connectors, hooked studs, spiral connectors and stirrup connectors. They are transmitting sheer forces by bond stresses. B) Rigid Shear Connectors: Channels, angles, I-section, or Z-section connectors are transmitting the sheering forces by bearing stresses.

10.2 Resistance of shear Connectors: This section applies to the calculation of the allowable horizontal shear force Rsc , in tons, for one connector. The value of Rsc computed from the following equation shall not exceed the allowable horizontal load, Rw provided by the connector connection to the beam flange.

10.2.1 Stud sheer connectors: The allowable shear force Rsc, in tons, for one stud connector shall be computed as follows: Rsc = .0054 Asc (Fcu Ec)^.5 ≤ 0.58 Asc Fy ≤ Rw Where Fy: the yield stresses of stud connector ≥ 3.4 t/cm2 Asc: cross-sectional area of stud connector cm2 Ec: modulus of elasticity of concrete t/cm2 Fcu: concrete compressive strength Kg/cm2. Chapter 9 Design of Composite Constructions

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Third Year Civil

10.2.2 Channel shear connector: The allowable shear force Rsc, in tons, for one channel shear conn. shall be computed as follows: Rsc = .0038 (tf + 0.5 tw) Lc (Fcu Ec)^.5 ≤ Rw 10.2.3Angle shear connector: The allowable shear force Rsc in ton for one angle shear connector shall be computed as follows: Rsc = .0056 Lc tc^0.75 (Fcu)^(2/3) ≤ Rw 10.3 Spacing of Connectors: Longitudinal spacing of connectors “e” shall not be greater than the following: 60.0 cms or 3 ts (slab thickness) or 4*height of connector

10.4 Design loads of Connectors: I) If the construction is done by method (I), (no temporary supports), the dead load is carried by the steel beam and the shear stress due to live load only is carried by the connectors. But to allow for shrinkage and creeping effect; The design loads for connectors are; (0.5 Dead Load + Live Load) II) If the construction is done by method (II), (by using temporary supports), The design loads for connectors are; Total (Dead Load + Live Load)

10.5 Design of Connector’s Pitch “e”: The sheering force are max at the supports and tend to be zero at mid span, more shear connectors would be required near the support than at mid span. The longitudinal shear force per one cm length of a beam = Q.S / Icomp. = Q [Ac.Yc] / Icomp. Where Ac = area of concrete = be.ts / n Yc = distance between C.A of concrete section And C.A. of composite section The total horizontal shear to be transmitted by one connector is D D = Q [Ac.Yc] / Icomp * e e = D [ Icomp./ Q Ac Yc] e is inversely proportional to Q. The connectors are arranged at small intervals near the support and at bigger intervals near the middle.

Example I: Chapter 9 Design of Composite Constructions

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Third Year Civil

Design a composite section for interior floor beam of an office building, Fcu = 250 Kg/cm2 n = 10 Steel used ST- 37 --> fs = 0.64 fy = 1.54 t/cm2 Beam span = 7 m. Beam spacing =2.0 m. Slab thickness =10 cm Live load = 250 kg/m2 Flooring =100 kg/m^2 considering the following: a) The construction is made without shoring. b) The construction is made using temporary supports. Loads: Dead loads:

Wconcrete slab = 2.5 * 0.1 *2.0 O.w of steel beam to be assumed 50 Kg/m Wd.l = 0.55 t/m\ Md.l = 0.55*(7) (7)/8

=0.50 t/m´. =0.05 t/m´ =3.37 t.m

Live loads + Flooring:

Wl.l

Ml.l.

=4.29 t.m

Md+l+fl=4.29 + 3.37

=7.66 t.m

=0.35*2.0 =0.7t/m

Dead loads + Live loads + Flooring:

The composite action reduces the weight of steel (20 to 30 %), then; Zrequired = (Md+L+fl / fs)(1- 0.3) = 348 cm3 Choose IPE-270 Effective width: be

be (transformed)

Ix=5790 cm4 Z= 429 cm3

A=45.9 cm2

bf=13.5 cm

= bo = bf + 12 ts = 13.5+12*10 = L/4

= 200 = 133.5 = 175.0

cm cm cm

= bE / n

= 13.35

cm

controls

Composite section properties: Y = {13.35*10*(5+27) + 45.9*(27/2)} / [13.35*10+45.9] = 27 cm I (composite) = 5790 + 45.9(13.5)^2 + 13.35*(10)^3/12 + 13.35(10)(5)^2 = 18605 cm4 yt = 10 cm yb = 27 cm Zt = 18605/10 = 1860.5 cm3 Zb = 18605/27 = 689 cm3

Yt

Y Yb

4. Check of stresses: Chapter 9 Design of Composite Constructions

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Third Year Civil

a) Case I: Construction is done without temporary shores: The steel section sustains dead loads, while the composite section sustains live + floor loads Mdl = 3.37 t.m f (top of steel) =f (bottom of steel = 3.37*100 / 429 = 0.785 t/cm2 The additional stresses after pouring and setting of concrete are; fc (top of concrete) = Mll / Zt (composite) =429/(1860.5*10) = 0.023 t/cm2 = 23 kg/cm2 < 250/4 Kg/cm2 (safe) Where the stresses in the concrete slab is I/n times the stresses on the transformed section ft (bottom of steel) = Mll / Zb (composite)

=

429/689

=

Total maximum tensile stresses on steel

=fst + f (bottom of steel) = 0.785 + 0.622 =

0.622

t/cm2

1.4 < 1.54

t/cm2 Safe

b) Case II: with temporary shores: All loads are resisted by the composite section fc (top of concrete) ft (bottom of steel)

= [Mdl + Mll] / Zt(composite) = [7.66*100] / (1860.5*10) =0.041

t/cm2 = 41 kg/cm2 < 250/4 Safe

= [Mdl + Mll] / Zb(composite) = [766] / 689 =1.11

t/cm2< 1.54 t/cm2

Case I

Case II

0.023

0.041

0.23 10

0.785

27.5

Safe

=

+ 27

27.5

0.785

0.622

1.4

1.11

Design of Shear Connectors: Chapter 9 Design of Composite Constructions

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Third Year Civil

Use stud type shear connectors The allowable load on one stud connector is calculated as follows; Rsc = .0054 Asc (fcu Ec)^.5 ≤ 0.58 Asc Fy ≤ Rw Asc = π /4 *(1.2)^2 = 2.01 cm^2 = 2.5 ton Rsc = 0.0054 * 2.01* (250 * 220)^.5 0.58 Asc Fy = 0.58 * 2.01 * 2.4 = 2.8 ton Rw = π /4 *( φ)^2 * 0.2 Fu = 1.44 ton

govern

a) If the construction is done by method (I), (no temporary supports), The design loads for connectors Q Shear flow

= (0.5 Dead Load + Live Load + flooring load) =0.5*0.55 + 0.7 =0.975 t/m = W*L/2 = 0.975*7.0 / 2 = 3.4 ton = Q*S / I(composite) = 3.4 * [13.35*10*5.0] / [18605] = 0.122 t / cm

Use two studs per raw: = R / τ = 1.44 * 2 / 0.122 = 23.7 cm Spacing e E max = 3*10 = 30 cm or 4*height of connector = 4*4* φ = 25.6 cm Use two stud connectors φ 16 @ 23 cm b) If the construction is done by method (II), (by using temporary supports), The design loads for connectors are =Total (Dead Load + Live Load + flooring loads) = 0.55 + 0.7 = 1.25 t/m = W*L/2 = 1.25*7.0 / 2 = 4.375 ton Q Shear flow

= Q*S / I(composite) = 4.375 * [13.35*10*5.0] / [1860.5] = 0.157t / cm

Use two studs per raw: Spacing e = R / τ = 1.44 * 2 / 0.157 = 18.3 cm E max = 3*10 = 30 cm or 4*height of connector = 4*4* φ = 25.6 cm Use two stud connectors φ 16 @ 18 cm Check of deflection:

δ l.l

= [5/384]*[(Wl.l * (L) ^4) / (E * Icomp.)] = [5/384]*[(.005* (700) ^4) / (2100 * 18605)] = 0.4 cm = L /1750 < L / 300


safe

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Third Year Civil

Composite Columns The objective of this section is to introduce composite columns, to describe their behavior and to explain the design method for axially loaded composite columns.

TYPES OF CROSS-SECTION FOR COMPOSITE COLUMNS AND THEIR ADVANTAGES Figure 1 shows typical cross-sections of composite columns together with the dimensional notation used in Egyptian code. The sections can be classified into two groups: • •

Totally and partly encased sections. Concrete filled sections in which the concrete is hidden

All cross-sections are symmetrical about both axes and in addition can be reinforced. There are many advantages associated with the use of composite columns: small cross-sections, for example, can be designed to withstand high loads; similarly, sections with different resistance, but identical external dimensions, can be produced by varying steel thickness, concrete strength and additional reinforcement. Thus the outer dimension of a column can be held constant over a number of floors in a building, simplifying architectural detailing. Economic efficiency also results from the use of concrete - a low cost material - and from the time saved by using the highly developed connection techniques of steelwork construction. With concrete filled profiles (Figure 1 d-f), the steel section serves as formwork during casting. Concrete filled sections, therefore, provide the opportunity to erect the steel frame of a building and afterwards fill the cross-sections by pumping in the concrete. By so doing the time of erection can be reduced. The protective steel casing also allows the concrete to achieve greater strength; in the case of concrete filled circular hollow profiles, for example, the effect of confinement by the steel leads to an increase in overall resistance. The influence of creep and shrinkage of the concrete can usually be neglected for these sections. However, this influence must be considered for concrete encased profiles (Figure 1a-c). The complete encasement of the steel section by concrete (Figure 1a), generally fulfils the technical requirements for high classes of fire protection without any additional measures. For partly encased sections (Figure 1b and c), as well as for concrete filled sections, these requirements can be achieved using additional reinforcement. Partly encased sections have the advantage that they can be produced quite simply by casting the concrete whilst the steel section lies horizontally; 24 hours later the column can be turned around and further concrete added, the formwork to the wet concrete being provided by the steel profile. For sections similar to that shown in Figure 1b the tendency for the concrete to drop out while turning the column must be avoided by suitable means, such as stud connectors. Another important advantage of these partly encased sections is that there is still a considerable area of steel available for connections, even after concreting.


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Third Year Civil

REQUIREMENTS In order to qualify as a composite column, the following requirements shall be fulfilled: • The characteristic 28- day cube strength of concrete, fcu, shall not be less than 250 Kg/cm2, nor greater than 500 kg/cm2. • Total cross sectional area of steel section shall not be less than four percent of the gross column area. • As ≥ 4% Ac • Concrete encasement shall be reinforced with longitudinal bars and lateral stirrups to restrain concrete and to prevent cover spalling. The cross section area of the longitudinal bars and lateral stirrups shall not be less than .02cm2 per cm of bar spacing. The spacing of lateral ties shall not exceed 2/3 b, or 30 cm, whichever is smaller. Chapter 9 Design of Composite Constructions

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Third Year Civil

LOCAL BUCKLING FAILURE In order to qualify the composite column design, it must be ensured that premature failure of the thin parts of the cross-section, due to instability, cannot occur. • For steel rectangular or circular tubing filled with concrete instability can be prevented by using a limiting ratio of wall dimension to wall thickness. These limiting ratios must be; • t ≥ b √ (Fy / 3Es) • t ≥ d √ (Fy / 8 Es)

for rectangular tubing. for circular tub.

• For completely encased steel parts, verification of local buckling resistance is not necessary. For larger steel parts, e.g. flanges in Figure 1a, sufficient concrete cover must be provided in order to avoid splitting of the concrete. The minimum concrete cover in this case must not be less than 40mm or 1/6 of the dimension of the steel part. For cross-sections according to Figure 1a it follows: • 40mm ≤ cz ≥ b/6.

RESISTANCE OF CROSS-SECTIONS TO AXIAL LOADS The allowable compressive axial stress, Fc, for axially loaded composite column shall be computed on the steel sectional area utilizing modified radius of gyration, yield stress and young’s modulus, rm, Frm and Em respectively.

Fc = (0.58 - α Fym λ²) Fym

for inelastic buckling, λ ≤ 100

Fc = 3.57 Em / λ²

for elastic buckling, λ >100

Where:

Fym

=

Fy + c1 Fyr (Ar / As) + c2 fcu (Ac / As)

Em

=

Es + c3 Ec (Ac /As)

α

=

(0.58*10^4 Fym – 3.57 Em) / (10^4 Fym) ²

λ

=

KL rm

KL / rm

= =

buckling length, bigger of in-plane and out-of-plane buckling lengths. radius of gyration of the steel shape, pipe or tubing. rm ≥ 0.3 bc for steel shapes incased in concrete

As, Ac and Ar are the areas of the structural steel, the concrete and the reinforcement, respectively. Fym, Fy and Fyr are the modified yield stress, the steel structure yield stress, and the yield stress of the reinforcing bars, respectively.

c1 , c2, c3 = 0.70, 0.48, 0.20 for concrete encased sections. c1 , c2, c3 = 1.00, 0.68, 0.40 for concrete filled pipes or tubing. Example for an axially loaded composite column: Chapter 9 Design of Composite Constructions

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Third Year Civil

Design a composite column with length 8.0m, N= -100 tons, fcu = 300 Kg/cm². Use square hollow section. Compare the results with the non-composite hollow square column.

- Assume fc = 0.58 fy = 1.4 t/cm²→Area req. = D.F. / 1.4 = 100 / 1.4 = 70 cm² → rm = 800/ 180 = 4.44 cm - For λ = KL / rm ≤ 180 Use square hollow section

250*250*8

- r = 9.8 > (0.3*25 = 7.5 cm)

As = 77.1 cm²

r = 9.87 cm

→ rm = 9.87 cm

In order to qualify as a composite column, As must be ≥ 4% Ac - Ac = (25 – 0.8*2)² = 547 cm² - As / Ac = 77.1/547 = 0.14 > 4% ok In order to qualify the composite column design, it must be ensured that premature failure of the thin parts of the cross-section, due to instability, cannot occur. for square tub • t ≥ d √ (Fy / 3 Es) = 250 √ (2.4 / 3 *2100) = 4.8 mm < 8mm ok - tmin = t ≥ d √ (Fy / 3 Es) The allowable compressive stress, Fc, for axially loaded composite column shall be computed on the steel sectional area utilizing modified radius of gyration, yield stress and young’s modulus, rm, Frm and Em respectively.

-

c1 , c2, c3 = 1.00, 0.68, 0.40 for concrete filled pipes or tubing. Fym = Fy + c1 Fyr (Ar / As) + c2 fcu (Ac / As) Fym = 2.4 + 0 + 0.68*0.3* (547 / 77.1) = 3.847 t/cm² Em = Es + c3 Ec (Ac /As) 2100 + 0.4*240* (547 /77.1) = 2780 t/cm² Em = α = (0.58*10^4 Fym – 3.847 Em) / (10^4 Fym) ² = 8.83*10^-6

- λ

=

KL / rm

= 800 / 9.87 = 81.6 < 100

- Fc = [0.58 – 8.83*10^-6 * 3.847*(81.6)² ] * 3.847= 1.33 t/cm² - fc = 100 / 77.1 = 1.297 t/cm²

= Fc safe

Comparison of the max strength of composite and non- composite column: - λ

=

KL / rm

= 800 / 9.8 = 81.6 < 100

- Fc = 1.4 – 0.000065 ( 81.6)² = 0.967 t/cm² - Max compression force for non composite hollow section = 0.967*77.1=74.55 - Max compression force for composite column = 1.33*77.1 = 102 - Percentage of additional loads sustained by the composite column = (102 – 74.55) / 74.55 = 37.5%


tons tons

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Third Year Civil

Composite Beam Columns Composite members subjected to bending in addition to axial compression shall be proportioned to satisfy the following interaction equation: fca/ Fc + (fbcx / 0.72Fy) A1 + (fbcy / 0.72Fy) A2 ≤ 1.0 A1 = Cmx / (1 – fca/ Femx) ≥ 1.0 A2 = Cmy / (1 – fca/ Femy) ≥ 1.0 Femx = 3.57 Em / λx² Femy = 3.57 Em / λy²

BEARING IN COMPOSITE COLUMNS In load bearing regions it has to be ensured that the individual components of the cross-section (concrete and steel) are loaded according to their resistances, so that no significant slip between them occurs. Header plates can be used in single-storey columns; these represent the ideal form of load introduction. Steel collars are welded onto the sides of concrete filled profiles, onto which the flange of the crossing beam may be fixed after concreting. For continuous composite columns special detailing for load transfer is necessary. The connections shown in Figure 4 have proved economical and efficient for this purpose. Figure 4a shows details of headed studs in the webs of I-profiles. This arrangement promotes additional load transfer, which increases the resistance of the connection. The introduction of gusset plates, punched through the steel section into concrete filled hollow profiles (Figure 4b), activates three dimensional stresses in the concrete and increases the resistance of the connection.


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Third Year Civil

Example for Design of Composite Beam-Column:

Design a composite rectangular column with buckling length Lx = 8.0m, Ly = 4.0m, N= -25.0 tons, Mx = 15 t.m, fcu = 300 Kg/cm². -

Assume fbcx = 0.5 fy = 1.2 t/cm² →Zx required = Mx / 1.2 = 1500/1.2 = 1250 cm³ For λ = KL / rm ≤ 180 → rx = 800/ 180 = 4.44 cm → ry = 800/ 180 = 2.22 cm

Use rectangular hollow section 400*200*10 - As = 112 cm² Zx = 1132 cm³ rx = 14.2 > 0.3 bc > 0.3*40 = 12 rmx

= 14.2 &

rx = 14.2 cm

ry = 8.34 cm

ry = 8.34 > 0.3 bc > 0.3*20 = 6

→ rmy = 14.2

In order to qualify as a composite column, As must be ≥ 4% Ac - Ac = 38*18 = 648 cm² - As / Ac = 112/648 = 0.16 > 4% ok In order to qualify the composite column design, it must be ensured that premature failure of the thin parts of the cross-section, due to instability, cannot occur. • t ≥ b √ (Fy / 3 Es) for rectangular tub - tmin = t ≥ b √ (Fy / 3 Es) = 200 √ (2.4 / 3 *2100) = 3.9 mm < 10mm ok The allowable compressive axial stress, Fc, for axially loaded composite column shall be computed on the steel sectional area utilizing modified radius of gyration, yield stress and young’s modulus, rm, Frm and Em respectively. -

-

c1 , c2, c3 = 1.00, 0.68, 0.40 for concrete filled pipes or tubing. Fym = Fy + c1 Fyr (Ar / As) + c2 fcu (Ac / As) = 2.4 + 0 + 0.68*0.3* (684 / 112) = 3.65 t/cm² Em = Es + c3 Ec (Ac /As) = 2100 + 0.4*240* (684/112) = 2686 t/cm² α = (0.58*10^4 Fym – 3.57 Em) / (10^4 Fym) ² = 8.69*10^-6

-

λx

=

KL / rmx

= 800 / 14.2 = 56.3

-

λy

=

KL / rmy

= 400 / 8.34 = 48

-

Fc = [0.58 – 8.69*10^-6 * 3.567*(56.3)² ] * 3.65

= 2.016 t/cm²

-

fc = Design force / Area of steel column

= 25 / 112

= 0.2232 t/cm²

-

fbcx

= 1500/1132 = 1.325 t/cm²

-

fc/Fc = 0.2232 / 2.016

-

fca/ Fc + (fbcx / 0.72Fy) A1 + (fbcy / 0.72Fy) A2

-

0.2232/ 2.016 + (1.325 / 0.72*2.4) *1.0

-

= Mx / (Zx of steel column)


= 0.11 < 0.15 →A1 = 1.0 ≤ 1.0

= 0.877

< 1.0 safe

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Third Year Civil

Exercise for Design of Composite Steel Concrete Structures 1) Design a simply supported beam with span 12m and arranged every 3m. Use steel ST 37, a 12cm thick slab with fcu 250 kg/cm2. Assume a live load of 300 kg/m2 and a flooring of 100 kg/m2. Consider the following cases; abc-

Use a composite I-section with a lower cover plate and use channel shear connectors. (Temporary shores are used). Use a composite I-section without a lower cover plate and use stud shear connectors (no temporary shoring are used). Use a non-composite I-section. Compare the results by problem I-2.

2) Find the maximum compression capacity of a concrete- filled hollow steel tubing on the basis of composite design. The column has a buckling length of 6m. The cross-section is square tube 200*200*8 , steel St 37, and fcu of concrete is 300 Kg/cm2. 3) Check the adequacy of a concrete-filled rectangular steel tube 250*250*10, steel st 37, to be used as a beam column. The column has a buckling length lx=6m, ly = 3m. The column is subjected to an axial compression force = 50 ton. Fcu of concrete is 250 Kg/cm2.


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Chapter 9 Design of Composite Constructions

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