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Structural Steelwork EurocodesDevelopment of

A Trans-national ApproachCourse: Eurocode 3Module 7 : Worked ExamplesLecture 20 : Simple braced frame

Contents: 1. Simple Braced Frame

1.1 Characteristic Loads1.2 Design Loads Fd = γF Fk1.3 Partial Safety Factors for Strength

2. Floor Beam - Fully Restrained 2.1 Classification of Cross-section

2.1.1Flange buckling2.1.2 Web buckling

2.2 Shear on Web2.3 Deflection Check2.4 Additional Checks if Section is on Seating Cleats2.5 Crushing Resistance2.6 Crippling Resistance2.7 Buckling Resistance2.8 Summary

3. Roof Beam – Restrained at Load Points3.1 Initial Section Selection3.2 Classification of Cross Section

3.2.1 Flange buckling3.2.2 Web buckling

3.3 Design Buckling Resistance Moment3.4 Shear on Web3.5 Deflection Check3.6 Crushing, Crippling and Buckling3.7 Summary

4. Internal Column4.1 Loadings4.2 Section properties

Structural Steelwork Eurocodes – Development of a Trans-National ApproachWorked examples Design of a 3-storey unbraced frame

15/02/07 2

4.3 Classification of Cross-Section4.3.1 Flange (subject to compression)4.3.2 Web (subject to compression)

4.4 Resistance of Cross-Section4.5 Buckling Resistance of Member4.6 Determination of Reduction Factor χy4.7 Determination of Reduction Factor χz

5. External Column5.1 Loadings5.2 Section properties5.3 Classifcation of Cross-Section

5.3.1 Flange (subject to compression)5.3.2 Web (subject to compression)

5.4 Resistance of Cross-Section5.5 Buckling Resistance of Member5.6 Determination of Reduction factor χy5.7 Determination of Reduction factor χz

6. Design of Cross-Bracing6.1 Section Properties6.2 Classification of Cross-Section6.3 Design of Compression Member

6.3.1 Resistance of Cross-section 6.3.2 Design Buckling Resistance6.3.3 Determination of Reduction Factor χ?

6.4 Design of Tension Member6.4.1Resistance of Cross-Section

7. Concluding Summary


15/02/07 3

1. Simple Braced Frame

The frame consists of two storeys and two bays. The frames are at 5 m spacing. The beam span is 7,2 m. The height from column foot to the beam at floor level is 4,5 m and the height from floor to roof is 4,2 m. It is assumed that the column foot is pinned at the foundation.

7,2 m

4,2 m

4,5 m

7,2 m

Roof Beam

Floor Beam

InternalColumn

ExternalColumn

Figure 1 Typical Cross Section of Frame

It is assumed that resistance to lateral wind loads is provided by a system of localised cross-bracing, and that the main steel frame is designed to support gravity loads only.The connections are designed to transmit vertical shear, and it is also assumed that the connections offer little, if any, resistance to free rotation of the beam ends.With these assumptions, the frame is classified as ‘simple’, and the internal forces and moments are determined using a global analysis which assumes the members to be effectively pin-connected.

6.4.2.1(2)

5.2.2.2

1.1 Characteristic LoadsFloor: Variable load, Qk = 3,5 kN/m2 Permanent load, Gk = 8,11 kN/m2

Roof: Variable load, Qk = 0,75 kN/m2 Permanent load, Gk = 7,17 kN/m2

1.2 Design Loads Fd = γF Fk 2.2.2.4(1)Floor: Gd = γG Gk. At ultimate limit state γG = 1,35 (unfavourable)Gd = 1,35 x 8,11 = 10,95 kN/m2

Qd = γQ Qk. At ultimate limit state γQ = 1,5 (unfavourable)Qd = 1,5 x 3,5 = 5,25 kN/m2

2.2.2.4(2)Table 2.22.2.2.4(2)Table 2.2


15/02/07 4

Roof: Gd = γG Gk. At ultimate limit state γG = 1,35 (unfavourable)Gd = 1,35 x 7,17 = 9,68 kN/m2

Qd = γQ Qk. At ultimate limit state γQ = 1,5 (unfavourable)Qd = 1,5 x 0,75 = 1,125 kN/m2

2.2.2.4(2)Table 2.22.2.2.4(2)Table 2.2

The steel grade selected for beams, columns and joints is Fe360.(fy = 235 N/mm2)

Table 3.1

1.3 Partial Safety Factors for StrengthThe following partial safety factors for strength have been adopted during the design:• Resistance of Class 1,2 or 3 cross-section, γM0 = 1,1• Resistance of member to buckling, γM1 = 1,1• Resistance of bolted connections, γMb = 1,25

2.3.3.2(1)

5.1.1(2)5.1.1(2)6.1.1(2)

The following load case, corresponding to permanent and variable actions (no horizontal loads) is found to be critical.

2. Floor Beam - Fully RestrainedThe beam shown in Figure 2 is simply supported at both ends and is fully restrained along its length.For the loading shown, design the beam in grade Fe360, assuming that it is carrying plaster, or a similar brittle finish.Fd = γG Gk + γQ QkDesign load, Fd = (5 x 1,35 x 8,11) + (5 x 1,5 x 3,5) = 81 kN/m

Table 2.1

7,2 m

81 kN/m

Figure 2 Loading on Fully Restrained Floor Beam

Design moment, MF L

8Sdd

2

=

Where MSd is the design moment in beam span,Fd is the design load = 81 kN/m, and L is the beam span = 7,2m.

M81x7,2

8525 kNmSd

2

= =

Design shear force, V F L2

81x7,22

292 kNSdd= = =

To determine the section size it is assumed that the flange thickness is less than 40 mm so that the design strength is 235 N/mm2, and that the section is class 1 or 2.

Table 3.1


15/02/07 5

The design bending moment, MSd, must be less than or equal to the design moment resistance of the cross section, Mc.Rd:MSd ≤ Mc.Rd

Mc.Rd = Mpl.y.Rd =W fpl y

M0γWhere Wpl is the plastic section modulus (to be determined),fy is the yield strength = 235 N/mm2, andγM0 is the partial material safety factor = 1,1.

5.4.5.1(1)

Table 3.15.1.1(2)

Therefore, rearranging:

W Mf

525x10 x1,1235

2457 cmpl.requiredsd M0

y

33= = =

γ

Try IPE 550Section properties:Depth, h = 550 mm, Width, b = 210 mmWeb thickness, tw = 11,1 mm Flange thickness, tf = 17,2 mmPlastic modulus, Wpl = 2787 cm3

5.4.5.1

This notation conforms with Figure 1.1 in Eurocode 3: Part1.1.2.1 Classification of Cross-section

As a simply supported beam is not required to have any plastic rotation capacity (only one hinge required), it is sufficient to ensure that the section is at least class 2 to develop the plastic moment resistance.

5.35.3.2 and Table 5.3.1

Figure 3 shows a typical cross-section for an IPE.IPE sections have been used in this example to reflect the European nature of the training pack.

tw

t

d

c

f

Figure 3 A Typical Cross-Section

2.1.1 Flange buckling

Class 1 limiting value of c/tf for an outstand of a rolled section is 10ε.ε = 235 / fy and fy = 235 N/mm2, therefore ε =1.

Calculate the ratio ct f

, where c is half

the width of the flange = 105 mm, and tf is the flange thickness = 17,2 mm (if the flange is tapered, tf should be taken as the average thickness).ct

10517,2

6,10f

= =

Table 5.3.1 (Sheet 3)


15/02/07 6

2.1.2 Web buckling

Class 1 limiting value of d/tw for a web subject to bending is 72ε.ε = 235 / fy and fy = 235 N/mm2, therefore ε =1.

Calculate the ratio dtw

, where d is the depth between root radii = 467,6

mm and tw is the web thickness = 11,1 mm.dt

467,611,1

42,1w

= =

ct

10f

< ε and dt

72w

< ε

∴ Section is Class 1 and is capable of developing plastic moment.


Table 5.3.1 (Sheets 1 and 3)

2.2 Shear on WebThe shear resistance of the web must be checked. The design shear force, VSd, must be less than or equal to the design plastic shear resistance, Vpl.Rd:

VSd ≤ Vpl.Rd

Where Vpl.Rd is given by Af / 3

vy

M0γ

5.4.6

For rolled I and H sections loaded parallel to the web,Shear area, Av = 1,04 h tw,fy is the yield strength = 235 N/mm2, andγM0 is the partial material safety factor = 1,1.

5.4.6(4)

Table 3.15.1.1(2)

∴ =V1,04ht f

3xpl.Rdw y

M0γ= =

1,04 x 550 x 11,1 x 2353 x 1,1x10

783 kN3

This is greater than the shear on the section (292 kN).The shear on the beam web is OK.If the beam has partial depth end-plates, a local shear check is required on the web of the beam where it is welded to the end-plate.

V Af / 3

pl.Rd vy

M0

=γ

where Av = twd, andd is the depth of end-plate = (for example) 300 mm.

V 11,1 x 300 x 2353 x 1,1 x 10

411 kNpl.Rd 3= =

This is greater than the shear on the section (292 kN).The local shear on the beam web is OK.


15/02/07 7

Other simple joints may be used instead, e.g. web cleat joints or fin plate joints.

A further check is sometimes required, especially when there are significant point loads, cantilevers or continuity, to ensure that the shear will not have a significant effect on the moment resistance. This check is carried out for the moment and shear at the same point. The moment resistance of the web is reduced if the shear is more than 50% of the shear resistance of the section. With a uniform load, the maximum moment and shear are not coincident and this check is not required for beams without web openings.

5.4.7(3)

2.3 Deflection CheckEurocode 3 requires that the deflections of the beam be checked under the following serviceability loading conditions:• Variable actions, and• Permanent and variable actions.Figure 4 shows the vertical deflections to be considered.

4.2

δ

δδ

δ

L

0

max

1

2

Figure 4 Vertical Deflections

δ0 is the pre-camber (if present),δ1 is the deflection due to permanent actions,δ2 is the deflection caused by variable loading, andδmax is the sagging in the final state relative to the straight line joining the supports.For a plaster or similar brittle finish, the deflection limits are L/250 for δmax and L/350 for δ2. Deflection checks are based on the serviceability loading.

Table 4.1 Figure 4.1

For a uniform load δ =5

384x F L

EIk

3

y

where Fk is the total load = Qk or (Gk + Qk) as appropriate,L is the span = 7,2 m,E is the modulus of elasticity (210 000 N/mm2), andIy is the second moment of area about the major axis = 67120 x 104

mm4.

3.2.5


15/02/07 8

For permanent actions, Fk = 5 x 8,11 x 7,2 = 292 kN.Therefore, deflection due to permanent actions,

δ 1 = =5 x 292x10 x 7200

384 x 210 000 x 67120x1010,1 mm

3 3

4

For variable actions, Fk = 5 x 3,5 x 7,2 = 126 kN.Therefore, deflection due to variable actions,

δ 2 = =5 x 126x10 x 7200

384 x 210 000 x 67120x104,3 mm

3 3

4

The maximum deflection, δ δ δmax 1 2+ = 10,1+ 4,3 = 14,4 mm=

Deflection limit for δ 2L

3507200350

20,6 mm= = =

The actual deflection is less than the allowable deflection: 4,3 mm < 20,6 mm

Deflection limit for δ maxL

2507200250

28,8 mm= = =

The actual deflection is less than the allowable deflection: 14,4 mm < 28,8 mm ∴ OK.

Table 4.1

Table 4.1

The calculated deflections are less than the limits, so no pre-camber is required. It should be noted that if the structure is open to the public, there is a limit of 28 mm for the total deflection of δ1 + δ2 (neglecting any pre-camber) under the frequent combination, to control vibration. This is based on a single degree of freedom, lumped mass approach. For the frequent combination the variable action is multiplied by ψ, which has a value of 0,6 for offices.

4.3.2(2)Lecture 3, section 6.2

2.3.4(2)

2.4 Additional Checks if Section is on Seating CleatsThere are cases where the beams may be supported on seating cleats, or on other materials such as concrete pads. A similar situation arises when a beam supports a concentrated load applied through the flanges. In these cases the following checks are required:• Crushing of the web• Crippling of the web• Buckling of the webGenerally, these checks need only be carried out for short beams or beams with concentrated loads. For the sake of completeness, these checks are included in this worked example.

5.7.35.7.45.7.5

The following detailed checks are for an 85 mm stiff bearing. (The actual length of stiff bearing will depend on the detail of the connection - see Figure 5.7.2)


15/02/07 9

2.5 Crushing Resistance 5.7.3The crushing resistance is given by

R(s s )t f

y.Rds y w yw

M1

=+

γwhere ss is the length of stiff bearing = 85 mm,tw is the web thickness = 11,1 mm,fyw is the yield strength of the web = 235 N/mm2,γM1 is the partial material safety factor for buckling resistance = 1,1, andsy is the length over which the effect takes place, based on the section geometry and the longitudinal stresses in the flange.

Table 3.15.1.1(2)5.7.3(1)

sy = 2tf (bf/tw)0,5 (fyf/fyw)0,5 [1 - (σf.Ed /fyf)2 ]0,5

At the support, the stress in the beam flange, σf.Ed, is zero, fyf = fyw but the value of sy is halved at the end of the member.

5.7.3(3)

s2 x 17,2 x (210 / 11,1)

275 mmy

0,5

= =

∴ Crushing resistance, Ry.Rd =+

=(85 75) x 11,1 x 235

1,1x10379,4 kN3

This is greater than the reaction (292 kN).The crushing resistance is OK

2.6 Crippling ResistanceThe crippling resistance is given by

R0,5t (Ef ) [(t / t ) 3(t / t )(s / d)]

a.Rdw

2yw

0,5f w

0,5w f s

M1

=+

γ

where tw is the thickness of the web = 11,1 mm,E is the modulus of elasticity = 210 000 N/mm2,fyw is the yield strength of the web = 235 N/mm2,tf is the flange thickness = 17,2 mm,ss is the length of stiff bearing = 85 mm, but ss is limited to a maximum of 0,2d (467,6 mm x 0,2 = 93,5 mm),d is the depth between root radii = 467,6 mm, andγM1 is the partial material safety factor buckling resistance = 1,1.

R0,5 x 11,1 (210000 x 235) [(17,2 / 11,1) 3(11,1 / 17,2)(85 / 467,6)]

1,1x10a,Rd

2 0,5 0,5

3=+

= 626 kN

This is greater than the reaction (292 kN).The crippling resistance is OK.

5.7.4(1)

3.2.5Table 3.1

5.7.4(1)

5.1.1(2)


15/02/07 10

2.7 Buckling ResistanceThe buckling resistance is determined by taking a length of web as a strut. The length of web is taken from Eurocode 3, which in this case, gives a length:

b 0,5(h s ) a s2eff

2s2 0,5 s= + + + but ≤ [h2 + ss

2]0,5

where h is the height of the section = 550 mm,ss is the length of stiff bearing = 85 mm, anda is the distance to the end of the beam = 0 mm.

b 0,5(550 85 ) 852

320,5 mmeff2 2 0,5= + + =

5.7.5

Figure 5.7.3

Provided that the construction is such that the top flange is held by a slab and the bottom by seating cleats, against rotation and displacement, the effective height of the web for buckling should be taken as 0,7 x distance between fillets.l = 467,6 mm x 0,7 = 327 mm

Radius of gyration for web, i t12

11,112

w= = = 3 2,

Slenderness of the web, λ = = =li

3273,2

102

Non-dimensional slenderness of the web, λλλ

β=1

A0,5

Where λ1 = 93,9 ε = 93,9 x 1 = 93,9, and βA = 1

∴ = =Non - dimensional slenderness of the web, λ10293 9

1 09,

,

Using buckling curve c, the value of the reduction factor, χ may be determined from Table 5.5.2.Reduction factor, χ = 0,49

Buckling resistance of a compression member, NAf

b.RdA y

M1

=χβ

γA is the cross-sectional area = beff tw,fy is the yield strength = 235 N/mm2, andγM1 is the partial material safety factor for buckling resistance = 1,1.

N 0,49 x 1 x 320,5 x 11,1 x 2351,1x10

372,4 kNb.Rd 3= =

This is greater than the reaction (292 kN).The buckling resistance is OK.

5.7.5(4)

5.5.1.4(3)

5.5.1.2

5.7.5(3)

5.7.5(3) and Table 5.5.2

5.5.1.1(1)

Table 3.15.1.1(2)

2.8 SummaryThe trial section IPE 550 is satisfactory if the section is on a stiff bearing 85 mm long. If it is supported by web cleats or welded end plates, the web checks, except for shear, are not required and the section is again satisfactory.The beam is satisfactory.


15/02/07 11

3. Roof Beam - Restrained at Load PointsThe roof beam shown in Figure 5 is laterally restrained at the ends and at the points of application of load. The load is applied through purlins at 1,8 m spacings.

Internal point load = 1,8 [(5 x 1,35 x 7,17) + (5 x 1,5 x 0,75)] = 97,2 kNExternal point load = 0,9 [(5 x 1,35 x 7,17) + (5 x 1,5 x 0,75)] = 48,6 kN

It is assumed that the external point loads will be applied at the end of the beams, and will contribute to the maximum shear force applied to the end of the beam, and the moment induced in the column due to the eccentricity of connection.

For the loading shown, design the beam in grade Fe360 steel.

7,2 m

48,6 kN 48,6 kN97,2 kN 97,2 kN 97,2 kN

1,8 m 1,8 m 1,8 m 1,8 m

A B C D E

Figure 5 Beam Restrained at Load PointsReactionsVSd (at supports) = [(2 x 48,6) + (3 x 97,2)] / 2 = 194,4 kNDesign MomentFigure 6 shows the distribution of bending moments.

1,8 m 1,8 m 1,8 m 1,8 m0 0

262,4 kNm 262,4 kNm

349,9 kNmFigure 6 Bending Moment Diagram

Moment at mid-span (maximum)MSd = [(194,4 - 48,6) x 3,6] - (97,2 x 1,8) = 349,9 kNm


15/02/07 12

3.1 Initial Section SelectionAssume that a rolled I beam will be used and that the flanges will be less than 40 mm thick. For grade Fe360 steel, fy = 235 N/mm2. Because the beam is unrestrained between load points, the design resistance, Mc.Rd, of the section will be reduced by lateral torsional buckling. The final section, allowing for the buckling resistance moment being less than the full resistance moment of the section, would have to be determined from experience.

Table 3.1

Try IPE O 450Section properties:Depth, h = 456 mm, Width, b = 192 mmWeb thickness, tw = 11 mm Flange thickness, tf = 17,6 mmPlastic modulus, Wpl = 2046 cm3

5.4.5.1

This notation conforms with Figure 1.1 in Eurocode 3: Part1.1.

3.2 Classification of Cross-SectionAs a simply supported beam is not required to have any plastic rotation capacity (only one hinge required), it is sufficient to ensure that the section is at least class 2 to develop the plastic moment resistance.

5.35.3.2 and Table 5.3.1

3.2.1 Flange buckling

Class 1 limiting value of c/tf for an outstand of a rolled section is 10ε.ε = 235 / fy and fy = 235 N/mm2, therefore ε =1.

Calculate the ratio ct f

, where c is half the width of the flange = 96 mm,

and tf is the flange thickness = 17,6 mm (if the flange is tapered, tfshould be taken as the average thickness).ct

9617,6

5,5f

= =


3.2.2 Web buckling

Class 1 limiting value of d/tw for a web subject to bending is 72ε.ε = 235 / fy and fy = 235 N/mm2, therefore ε =1.

Calculate the ratio dtw

, where d is the depth between root radii = 378,8

mm and tw is the web thickness = 11,0 mm.dt

378,811,0

34,4w

= =

ct

10f

< ε and dt

72w

< ε

∴ Section is Class 1.




15/02/07 13

3.3 Design Buckling Resistance MomentThe design buckling resistance moment of a laterally unrestrained beam is given by:

M =W f

b.RdLT W pl.y y

M1

χ β

γ

in which χLT is the reduction factor for lateral torsional buckling, from Table 5.5.2, for the appropriate value of λ LT , using curve a for rolled sections.

5.5.2

Table 5.5.25.5.2(4)

In this case full lateral restraint is provided at the supports and at the load points B, C and D. In general, all segments need to be checked, but in this case they are all of equal length. The segments B - C and C - D are subject to the most severe condition, but with symmetrical loading only one segment needs to be checked.

Segment B - CThe value of λLT can be determined using Annex F. For segment B - C it is assumed that the purlins at B and C provide the following conditions:• restraint against lateral movement,• restraint against rotation about the longitudinal axis (i.e. torsional/twisting restraint), and• freedom to rotate in plan.i.e. k = kw = 1,0

Annex F

F.1.2(2)

For this example, the general formula for λLT has been used, as the section is doubly symmetric and end-moment loading is present.The following formula for λLT may be used:

( )λ LT

LT

1LT

L / i

CL / a

=

+

0 52 0 25

125 66

,

,

,

where L is the length between B and C = 1800 mm,Iz is the second moment of area about the z - z axis = 2085 x 104 mm4,Iw is the warping constant = 998 x 109 mm6,Wpl.y is the plastic modulus about the y - y axis = 2046 x 103 mm3, andIt is the torsion constant = 89,3 x 104 mm4.

Equation F.15

i I IW

2085x10 x 998x10(2046x10 )

47,2 mmLTz w

pl.y2

0,254 9

3 2

0,25

=

=

=

F.2.2(3)

a II

998x10109x10

957 mmLTw

t

0,5 9

4

0,5

=

=

=

F.2.2(1)


15/02/07 14

Note: These properties will probably be tabulated in handbooks.See also appendix A at the end of this example.C1 is the correction factor for the effects of any change of moment along the length, L.ψ = 262,4/349,9 = 0,75, k = 1, therefore C1 = 1,141 Table

F.1.1Substituting into the above equation:

( )λ LT =

+

=1800 47 2

1141 11800 957

25 66

34 60 5

2 0 25/ ,

,/,

,,

,

Non-dimensional slenderness, λλλ

βLTLT

1w

0,5=

Where λ1 = 93,9 ε = 93,9 x 1 = 93,9,βw = 1 for class 1 sections.

Therefore, λ LT0,5= =

34 693 9

1 0 0 37,,

( , ) ,

5.5.2(4)

5.5.2(5)

For rolled I sections, buckling curve a should be used. From Table 5.5.2, the reduction factor, χLT = 0,96. (This represents a 4% strength reduction due to moment)

Table 5.5.3Table 5.5.2

Wpl.y is the plastic modulus about the y - y axis = 2046 x 103 mm3,fy is the yield strength of the steel = 235 N/mm2, andγM1 is the partial material safety factor for buckling resistance = 1,1.

Table 3.15.1.1(2)

The design buckling resistance moment for segment B - C is:

M =W f x 1 x 2046x10 x 235

1,1 x 10kNmb.Rd

LT W pl.y y

M1

3

6

χ β

γ= =

0 96419 6

,,

5.5.2

Mb.Rd = 419,6 kNm > MSd = 349,9 kNm, therefore the section is satisfactory.


15/02/07 15

3.4 Shear on WebThe maximum shear occurs at the supports, VSd = 194,4 kN.The design shear resistance for a rolled I section is:

( )V

1,04ht fpl.Rd

w y

M0

=/ 3

γ

where h is the height of the section = 456 mm,tw is the web thickness = 11 mm,fy is the yield strength of the steel = 235 N/mm2, andγM0 is the partial material safety factor for the resistance of the cross-section = 1,1.

V 1,04 x 456 x 11 x 235x 1,1 x 10

643 kNpl.Rd 3= =3

VSd = 194,4 kN < Vpl.Rd = 643 kN, therefore the section is satisfactory.

5.4.6(1)

5.4.6(4)

Table 3.15.1.1(2)

Inspection shows that VSd < (Vpl.Rd / 2), so there is no reduction in moment resistance due to the shear in the web.

5.4.7(2)

3.5 Deflection CheckEurocode 3 requires that the deflections of the beam be checked under the following serviceability loading conditions:

• Variable actions, and• Permanent and variable actions.

4.2

For a general roof, the deflection limits are L/200 for δmax and L/250 for δ2. Deflection checks are based on the serviceability loading.

Table 4.1 Figure 4.1

Consider the deflection from the permanent loading.For a point load, distance a from the end of the beam:

Central deflection, δ = −

F aEI

L16

a12

k

y

2 2

where Fk is the value of one point load = (7,17 x 5 x 1,8) = 64,5 kN,E is the modulus of elasticity = 210 000 N/mm2,Iy is the second moment of area about the major axis = 40920 x 104

mm4,L is the span of the beam = 7,2 m, anda is the distance from the support to the adjacent load = 1,8 m.


=

64,5x10 x 1800210 000 x 40920x10

7200 180012

mm3

4

2 2

164 0,

3.2.5(1)


15/02/07 16

For a central point load:

Central deflection, δ =F L

48EIk

3

y

where Fk is the value of one point load = (7,17 x 5 x 1,8) = 64,5 kN,L is the span of the beam = 7,2 m,E is the modulus of elasticity = 210 000 N/mm2, andIy is the second moment of area about the major axis = 40920 x 104

mm4.

Central deflection, δ = =64,5x10 x 7200

48 x 210 000 x 40920x10mm

3 3

4 5 8,

3.2.5(1)

Total deflection due to permanent loading, δ1 = 5,8 + (2 x 4,0) = 13,8 mmConsider the deflection from the variable loading.

For a point load, distance a from the end of the beam:


F aEI

L16

a12

k

y

2 2

where Fk is the value of one point load = (0,75 x 5 x 1,8) = 6,75 kN,E is the modulus of elasticity = 210 000 N/mm2,Iy is the second moment of area about the major axis = 40920 x 104

mm4,L is the span of the beam = 7,2 m, anda is the distance from the support to the adjacent load = 1,8 m.


=

6,75x10 x 1800210 000 x 40920x10

7200 180012

mm3

4

2 2

160 4,

3.2.5(1)

For a central point load:

Central deflection, δ =F L

48EIk

3

y

where Fk is the value of one point load = (0,75 x 5 x 1,8) = 6,75 kN,L is the span of the beam = 7,2 m,E is the modulus of elasticity = 210 000 N/mm2, andIy is the second moment of area about the major axis = 40920 x 104

mm4.

Central deflection, δ = =6,75x10 x 7200

48 x 210 000 x 40920x10mm

3 3

4 0 6,

3.2.5(1)


15/02/07 17

Total deflection due to variable loading, δ2 = 0,6 + (2 x 0,4) = 1,4 mmTherefore, the total central deflection, δmax = δ1 + δ2 = 13,8 + 1,4 = 15,2 mm.

The limit for δ2 is L/250 = 7200/250 = 28,8 mm. The limit for δmax = L/200 = 7200/200 = 36 mm. 13,8 mm < 28,8 mm and 15,5 mm < 36 mm, therefore the deflections are within limits and no pre-camber of the beam is required.

Table 4.1 and Figure 4.1

3.6 Crushing, Crippling and BucklingIf the beam is supported on seating cleats, the checks for web crushing, crippling and buckling must be made. To satisfy the assumptions made in the design, both flanges must be held in place laterally, relative to each other. If seating cleats are used then the top flange must be held laterally. There is no requirement to prevent the flanges from rotating in plan, as k has been taken as 1,0.

5.7.1

3.7 SummaryAll Eurocode recommendations are satisfied, therefore this beam is satisfactory.The beam is satisfactory.

4. Internal ColumnThe internal column shown in Figure 7 is subject to loads from the roof and one floor. Design the column for the given loading, in grade Fe360 steel, as a member in simple framing.


15/02/07 18

4.1 LoadingsAt roof level, the applied axial load = =2 x (54 x 7,2)

2389 kN

At first floor level, the applied axial load = =2 x (81 x 7,2)2

583 kN

∴ Maximum load, from the first floor to the base, = 389 + 583 kN = 972 kN

4,2 m

4,5 m

Floor

Roof

InternalColumn

Figure 7 Internal Column

Consider the column from ground floor to first floor.The size of the column must be determined from experience and then checked for compliance with the Eurocode rules.

4.2 Section PropertiesTry an HE 240 A Grade Fe360 (terminology in accordance with EC3)h = 230 mm b = 240 mmtw = 7,5 mm tf = 12 mmd/tw = 21,9 c/tf = 10A = 7680 mm2 Iy = 77,63 x 106 mm4

Iw = 328 x 109 mm6 Iz = 27,7 x 106 mm4

It = 41,6 x 104 mm4 Wpl.y = 745 x 103 mm3

Wel.y = 675 x 103 mm3 iy = 101 mmiz = 60 mm

i I IW

27,7x10 x 328x10(745x10 )

63,6 mmLtz w

pl.y2

0,256 9

3 2

0,25

=

=

=

F.2.2(3)

a II

328x1041,6x10

888 mmLtw

t

0,5 9

4

0,5

=

=

=

F.2.2(1)

All the above properties can be obtained from section property tables.


15/02/07 19

4.3 Classification of Cross SectionThis section is designed to withstand axial force only. No moment is applied as the connecting beams are equally loaded.

5.3

4.3.1 Flange (subject to compression)Class 1 limiting value of c/tf for an outstand of a rolled section is 10ε.ε = 235 / fy where fy = 235 N/mm2, therefore ε = 1.From section properties, c/tf = 10


4.3.2 Web (subject to compression)Class 1 limiting value of d/tw for a web subject to compression only is 33ε.From section properties, d/tw = 21,9c/tf ≤ 10ε and d/tw ≤ 33ε∴Class 1 section.

Table 5.3.1(Sheet 1)Table 3.5.1 (Sheets 1 and 3)Class 1 section

4.4 Resistance of Cross-Section 5.4.4It is highly unlikely that the resistance of the cross-section will be the critical case - it is generally the buckling resistance that governs the suitability of a cross-section. For the sake of completeness, the check is included in this worked example. The resistance of the cross-section will only be critical if a short, stocky column is used.For members in axial compression, the design value of the compressive force, NSd, at each cross-section shall satisfy NSd ≤ Nc.Rd

5.4.4(1)

For a class 1 cross-section, the design compression resistance of the cross-section, Nc.Rd, may be determined as:

NAf

c.Rdy

M0

=γ

where A is cross-sectional area = 7680 mm2,fy is the yield strength = 235 N/mm2, andγM0 is the partial material safety factor = 1,1.

N 7680 x 2351,1x10

1641 kNc.Rd 3= =

NSd = 972 kN, therefore Nsd ≤ Nc.Rd. The section can resist the applied axial load.

5.4.4(2)

Table 3.15.1.1(2)


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4.5 Buckling Resistance of MemberA class 1 member should be checked for failure due to flexural and lateral torsional buckling. Here, since My = Mz = 0, only failure due to flexural buckling needs to be checked.The design buckling resistance of a compression member shall be taken as:

NAf

b.RdA y

M1

=χβ

γ

where χ is the reduction factor for the relevant buckling mode,βA = 1 for class 1 cross-section,A is the cross-sectional area = 7680 mm2,fy is the yield strength of the steel = 235 N/mm2 , andγM1 is the partial material safety factor for buckling resistance = 1,1.

The magnitude of the reduction factor, χ depends on the reduced slenderness of the columns. χ is the lesser of χy and χz, where χy and χzare the reduction factors from clause 5.5.1 for the y-y and z-z axes respectively. Values of χ for the appropriate value of non-dimensional slenderness, λ may be obtained from Table 5.5.2.

5.5.1.1(1)

5.5.1.1(1)

Table 3.15.1.1.(2)


β=

1A

0,5

Where the slenderness, λ =li

l is the column buckling length, andi is the radius of gyration about the relevant axis.The braced frame is designed as a simple “pinned” structure. Therefore, the buckling length ratio l/L is equal to 1 - the buckling length is equal to the system length.

λ π ε1y

0,5Ef

93,9=

=

5.5.1.2(1)

5.5.1.4(3)

5.5.1.5(2)

5.5.1.2(1)

4.6 Determination of Reduction Factor, χySlenderness, λy = l/iy = 4500/101 = 44,6λ1 = 93,9ε = 93,9 x 1 = 93,9

Non-dimensional slenderness, λλ

λβy

y

1A

0,5 0,544,693,9

x 1 0,47=

=

=

5.5.1.4(3)

From Table 5.5.2, using buckling curve b, the reduction factor, χy = 0,89 Table 5.5.3 Table 5.5.2


15/02/07 21

4.7 Determination of Reduction Factor, χzSlenderness, λz = l/iz = 4500/60 = 75λ1 = 93,9ε = 93,9 x 1 = 93,9


βyz

1A

0,5 0,57593,9

x 1 0,8=

=

=

From Table 5.5.2, using buckling curve c,the reduction factor, χz = 0,6622

5.5.1.4(3)

Table 5.5.3 Table 5.5.2

Therefore χ = χz = 0,6622.Design buckling resistance of member:

NAf 0,6622 x 1 x 7680 x 235

1,1x101086 kNb.Rd

A y

M13= = =

χβ

γThe design buckling resistance of the member is greater than the applied load (972 kN), therefore the column is satisfactory.The column is OK.

5.5.1.1(1)

5. External ColumnThe external column shown in Figure 8 is subject to loads from the roof and one floor. Design the column for the loading given below, in grade Fe360 steel, as a member in simple framing.

5.1 LoadingsAt roof level, the applied axial load = =

(54 x 7,2)2

194 kN

At first floor level, the applied axial load = =(81 x 7,2)

2292 kN

∴ Maximum load, from first floor to base, = 194 + 292 kN = 486 kN

The beams in the frame are designed to span from column centre to column centre, therefore all axial load is applied at the mid-point of the column. No moment due to eccentricity of applied load is therefore applied to the column. See Annex H


15/02/07 22

4,2 m

4,5 m

First Floor

Roof

Figure 8 External ColumnConsider the column from ground floor to first floor.The size of the column must be determined from experience and then checked for compliance with the Eurocode rules.

5.2 Section PropertiesTry an HE 200 A Grade Fe360h = 190 mm b = 200 mmtw = 6,5 mm tf = 10 mmd/tw = 20,6 c/tf = 10A = 5380 mm2 Iy = 36,92 x 106 mm4

Iw = 108 x 109 mm6 Iz = 13,36 x 106 mm4

It = 21,0 x 104 mm4 Wpl.y = 429 x 103 mm3

Wel.y = 389 x 103 mm3 iy = 82,8 mmiz = 49,8 mm

i I IW

13,36x10 x 108x10(429x10 )

52,9 mmLtz w

pl.y2

0,256 9

3 2

0,25

=

=

=

F.2.2(3)

a II

108x1021,0x10

717 mmLtw

t

0,5 9

4

0,5

=

=

=

F.2.2(1)

All the above properties can be obtained from section property tables.


15/02/07 23

5.3 Classification of Cross SectionThis section is designed to withstand axial force only.

5.3

5.3.1 Flange (subject to compression)Class 1 limiting value of c/tf for an outstand of a rolled section is 10ε.ε = 235 / fy where fy = 235 N/mm2, ∴ ε = 1.10ε = 10 x 1 = 10From section properties, c/tf = 10


5.3.2 Web (subject to compression)Class 1 limiting value of d/tw for a web subject to compression only is 33ε.ε = 235 / fy where fy = 235 N/mm2, ∴ ε = 1.33ε = 33 x 1 = 33From section properties, d/tw = 20,6

Table 5.3.1(Sheet 1)

c/tf ≤ 10ε and d/tw ≤ 33ε∴Class 1 section.


5.4 Resistance of Cross-Section 5.4.4It is highly unlikely that the resistance of the cross-section will be the critical case - it is generally the buckling resistance that governs the suitability of a cross-section. For the sake of completeness, the check is included in this worked example. The resistance of the cross-section will only be critical if a short stocky column is used.For members in axial compression, the design value of the compressive force, NSd, at each cross-section shall satisfy NSd ≤ Nc.Rd

5.4.4(1)

For a class 1 cross-section, the design compression resistance of the cross-section, Nc.Rd, may be determined as:

NAf

c.Rdy

M0

=γ

where A is cross-sectional area = 5380 mm2,fy is the yield strength = 235 N/mm2, andγM0 is the partial material safety factor = 1,1.

N 5380 x 2351,1x10

1149 kNc.Rd 3= =

NSd = 486 kN, therefore Nsd ≤ Nc.Rd. The section can resist the applied axial load.

5.4.4(2)

Table 3.15.1.1(2)


15/02/07 24

5.5 Buckling Resistance of MemberA class 1 member should be checked for failure due to flexural and lateral torsional buckling. Here, since My = Mz = 0, only failure due to flexural buckling needs to be checked.The design buckling resistance of a compression member shall be taken as:

NAf

b.RdA y

M1

=χβ

γwhere χ is the reduction factor for the relevant buckling mode,βA = 1 for class 1 cross-section,A is the cross-sectional area = 5380 mm2,fy is the yield strength of the steel = 235 N/mm2 , andγM1 is the partial material safety factor for buckling resistance = 1,1.The magnitude of the reduction factor, χ depends on the reduced slenderness of the columns. χ is the lesser of χy and χz, where χy and χzare the reduction factors from clause 5.5.1 for the y-y and z-z axes respectively. Values of χ for the appropriate value of non-dimensional slenderness, λ may be obtained from Table 5.5.2.

5.5.1.1(1)

5.5.1.1(1)

Table 3.15.1.1.(2)


β=

1A

0,5


l is the column buckling length, andi is the radius of gyration about the relevant axis.The braced frame is designed as a simple “pinned” structure. Therefore, the buckling length ratio l/L is equal to 1 - the buckling length is equal to the system length.

λ π ε1y

0,5Ef

93,9=

=

5.5.1.2(1)

5.5.1.4(3)

5.5.1.5(2) Annex E Figure E.2.1

5.5.1.2(1)

5.6 Determination of Reduction Factor, χySlenderness, λy = l/iy = 4500/82,8 = 54,3λ1 = 93,9ε = 93,9 x 1 = 93,9

Non-dimensional slenderness, λλ

λβy

y

1A

0,5 0,554,393,9

x 1 0,58=

=

=

5.5.1.4(3)

From Table 5.5.2, using buckling curve b, the reduction factor, χy = 0,84 Table 5.5.3 Table 5.5.2


15/02/07 25

5.7 Determination of Reduction Factor, χzSlenderness, λz = l/iz = 4500/49,8 = 90,4λ1 = 93,9ε = 93,9 x 1 = 93,9


βyz

1A

0,5 0,590,493,9

x 1 0,96=

=

=

From Table 5.5.2, using buckling curve c,the reduction factor, χz = 0,55

5.5.1.4(3)


Therefore χ = χz = 0,55.Design buckling resistance of member:

NAf 0,55 x 1 x 5380 x 235

1,1x10632,2 kNb.Rd

A y

M13= = =

χβ

γThe design buckling resistance of the member is greater than the applied load (486 kN), therefore the column is satisfactory.The column is OK.

5.5.1.1(1)

6. Design of Cross-BracingAll horizontal loading will be resisted by bracing. For the purpose of illustration assume this will be present on every other frame (i.e. at 10 m spacing). It is more likely that bracing will be located at each end of the building or perhaps in a stair/lift well. The forces may therefore be greater than here but the principles would remain the same. For the loading shown, design the bracing members in grade Fe360 steel.

Characteristic wind load, Qk = 1,6 kN/m2.Design wind load, Qd = γQ Qk

At ultimate limit state, γQ = 1,5 (unfavourable), Qd = 1,5 x 1,6 = 2,4 kN/m2.Therefore, the total load (per m height of frame) = 10 x 2,4 = 24 kN/m.

2.2.2.4(2)Table 2.2


15/02/07 26

7,2 m

4,2 m

4,5 m

24kN/m

Figure 9 Wind Load on FrameIt is assumed that the uniformly distributed load acts as two point loads on the frame.

Top load = (24 x 2,1) = 50,4 kN.Middle load = (24 x 2,1) + (24 x 2,25) = 104,4 kN.

Assume that all horizontal load is resisted by the bracing only.

Therefore, the load in the top brace = 50,4 / cos 30,3º = 58,4 kN, and the load in the bottom brace = (104,4 +50,4)/ cos 32º = 182,5 kN.


15/02/07 27

104,4kN

7,2 m

4,2 m

4,5 m

50,4kN

58,4 kN

182,5 kN

Figure 10 Equivalent Point Wind Loads and Loads Within Bracing

Design the bottom brace, as this carries the greater load.

Try a CHS 175 x 5,06.1 Section Properties

Depth of section, d = 175 mm, Thickness, t = 5 mmArea of section, A = 2670 mm2, Ratio for local buckling, d/t = 35,Radius of gyration, i = 60,1 mm.

6.2 Classification of Cross-SectionAs the bracing is axially loaded, check the section classification is at least class 1, 2 or 3. Figure 11 shows a typical CHS cross-section.

5.4.4(1)a

t d

Figure 11 Typical CHS Cross-Section


15/02/07 28

Class 1 limiting value of d/t for a tubular section is 50ε2.ε = 235 / fy where fy = 235 N/mm2, ∴ ε = 1.50ε2 = 50 x 1 = 50From section properties, d/t =35, therefore the section is Class 1.

Table 5.3.1(Sheet 4)

6.3 Design of Compression MemberThe bracing members need to be checked as axially loaded.

6.3.1 Resistance of Cross-SectionIt is highly unlikely that the resistance of the cross-section will be the critical case - it is generally the buckling resistance that governs the suitability of a cross-section. For the sake of completeness, the check is included in this worked example. The resistance of the cross-section will only be critical if a short, stocky column is used.

The applied axial load, NSd, must be less than the design compressive resistance of the cross-section, Nc.Rd.

Applied axial load, NSd = 182,5 kN.

Design compressive resistance of cross-section, Nc.Rd = NAf

pl.Rdy

M0

=γ

Where A is the cross-sectional area = 2670 mm2,fy is the yield strength of the steel = 235 N/ mm2, andγM0 is the partial material safety factor = 1,1.

N 2670 x 235x 10

kNpl.Rd 3= =11

570 4,

,

The design compressive resistance of the cross-section, Nc.Rd = 570,4 kN, is greater than the applied axial load, NSd = 182,5 kN. Therefore the section is satisfactory.

5.4.4(1)

Table 3.15.1.1(2)

6.3.2 Design Buckling Resistance A class 1 member subject to axial compression should be checked for failure due to buckling.The design buckling resistance of a compression member shall be taken as:

NAf

b.RdA y

M1

=χβ

γwhere χ is the reduction factor for the relevant buckling mode,βA = 1 for class 1 cross-section,A is the cross-sectional area = 2670 mm2,fy is the yield strength of the steel = 235 N/mm2 , andγM1 is the partial material safety factor for buckling resistance = 1,1.Values of χ for the appropriate value of non-dimensional slenderness, λmay be obtained from Table 5.5.2.

5.5.1.1(1)

5.5.1.1(1)

Table 3.15.1.1.(2)


15/02/07 29


β=

1A

0,5


l is the member buckling length, andi is the radius of gyration.

The bracing is designed as a simple “pinned” member. Therefore, the buckling length ratio l/L is equal to 1 - the buckling length is equal to the system length.Length of member = ( )4 5 7 2 85002 2, ,+ = mm

λ π ε1y

0,5Ef

93,9=

=

5.5.1.2(1)

5.5.1.4(3)

5.5.1.5(2) Annex E Figure E.2.1

5.5.1.2(1)

6.3.3 Determination Of Reduction Factor, χSlenderness, λ = l/i = 8500/60,1 = 141λ1 = 93,9ε = 93,9 x 1 = 93,9


β=

=

=1

A0,5 0,5141

93,9x 1 1,50

5.5.1.4(3)

From Table 5.5.2, using buckling curve b, the reduction factor, χ = 0,3422.


Design buckling resistance of member:

NAf 0,3422 x 1 x 2670 x 235

1,1x10195,2 kNb.Rd

A y

M13= = =

χβ

γThe design buckling resistance of the member is greater than the applied load (182,5 kN), therefore the bracing is satisfactory.

5.5.1.1(1)


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6.4 Design of Tension MemberWhen the wind load is applied in the opposite direction, the bracing will be loaded in tension. The section therefore needs to be checked, for the same magnitude of loading, to ensure it is also satisfactory in tension.

104,4kN

7,2 m

4,2 m

4,5 m

50,4kN

58,4 kN

182,5 kN

Figure 12 Equivalent Point Wind Loads and Loads Within Bracing

6.4.1 Resistance of Cross-SectionThe applied axial load, NSd, must be less than the design tension resistance of the cross-section, Nt.Rd.

Applied axial load, NSd = 182,5 kN.

Design tension resistance of the cross-section, Nt.Rd = NAf

pl.Rdy

M0

=γ

where A is the cross-sectional area = 2670 mm2,fy is the yield strength of the steel = 235 N/ mm2, andγM0 is the partial material safety factor = 1,1.

N 2670 x 235x 10

kNpl.Rd 3= =11

570 4,

,

The design tension resistance of the cross-section, Nt.Rd = 570,4 kN, is greater than the applied axial load, NSd = 182,5 kN. Therefore the section is satisfactory.

5.4.3(1)

Table 3.15.1.1(2)

The bracing fulfils all the Eurocode requirements for members in tension and in compression, and is therefore satisfactory.

The frame is satisfactory for all EC3 checks


15/02/07 31

7. Concluding Summary

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