Lecture Notes Structural Steel Design

Bachelor of Science in Civil Engineering

Structural Steel Design, Structural Dynamics

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PART I STRUCTURAL STEEL DESIGN

1. DESIGN OF STRUCTURAL STEEL ELEMENTS

1.1 Design methods

Steel design may be based on three design theories

(1) Elastic design;

(2) Plastic design; and

(3) Limit state design.

Elastic design is the traditional methods using the permissible stresses. Steel is almost

perfectly elastic up to the yield point.

Plastic theory developed to take account of behaviour past the yield point is based on

finding the load that causes the structure to collapse. Then the working load is the collapse

load divided by a load factor.

Limit state design has been developed to take account of all conditions that can make the

structure become unfit for use in accordance with BS 5950: The Structural Use of Steelwork

in Building. Part 1 – Code of Practice for Design in Simple and Continuous Construction:

Hot Rolled Sections.

1.2 Properties of structural steel

Steel is produced in three strength grades – 43, 50 and 55.

The stress-strain curves for the three grades of steel are shown in Figure 1.1. Plastic design

is based on the horizontal part of the stress strain shown in Figure 1.2.

1.3 Limit states for steel design

1.3.1 Ultimate limit states

(1) Strength (including general yielding, rupture, buckling and transformation into a

mechanism);

(2) Stability against overturning and sway;

(3) Fracture due to fatigue;

(4) Brittle fracture.



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When the ultimate limit states are reached, the whole structure or part of it collapses.

1.3.2 Serviceability limit states

(5) Deflection;

(6) Vibration (for example, wind-induced oscillation);

(7) Repairable damage due to fatigue;

(8) Corrosion and durability.

The serviceability limit states, when reached, make the structure or part of it unfit for

normal use but do not indicate that collapse has occurred.

All relevant limit states should be considered, but usually it will be appropriate to design on

the basis of strength and stability at ultimate loading and then check that deflection is not

excessive under serviceability loading.

1.4 Working and factored loads

1.4.1 Working loads

The working loads (also known as the specified, characteristic, unfactored or nominal loads)

are the actual loads the structure is designed to cry. These are normally thought of as the

maximum loads which will not be exceeded during the life of the structure. In statistical

terms, characteristic loads have a 95 per cent probability of not being exceeded. The main

loads on buildings may be classified as:

(1) Dead loads. The weights of floor slabs, roofs, walls, ceilings, partitions, finishes,

services and self-weight of steel. When sizes are known, dead loads can be calculated

from weights of materials or from the manufacturer’s literature.

(2) Imposed loads. The loads caused by people, furniture, equipment, stock, etc. on

the floors of buildings and snow on roofs. The values of the floor loads used depend

on the use of the building. Imposed loads are given in BS6399: Part 1 for various type

of buildings, or in the Hong Kong Building (Construction) Regulations by the H. K.

Government.

(3) Wind loads. These loads depend on the location and building size. Wind loads are

given in the Hong Kong Code of Practice on Wind Effects (1983).

(4) Dynamic loads. These are caused mainly by cranes.



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1.4.2 Factored loads (ultimate loads) for the ultimate limit states

Factored loads are used in design calculations for strength and stability.

( ) ( )fγfactorloadoverallrelevantloadnominalorworkinglaodFactord ×=

The overall load factor takes account of the unfavourable deviation of loads and the

reduced probability that various loads acting simultaneously.

It also allows for the uncertainties in the behaviour of materials and of the structure.

Fig. 1.1 Stress-strain diagrams for structural steels



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Fig. 1.2 Compound sections



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Fig. 1.3 Rolled and formed sections



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Fig. 1.4 Built-up section

Fig. 1.5 Cold-rolled section

The fγ factors are given in Table 2 of BS 5950: Part 1 and some of the factors are given

in the following Table 1.1.

Table 1.1 Overall load factors fγ (Part of Table 2 of BS5950)

The factored loads should be applied in the most unfavourable manner.



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1.5 Serviceability limit state of deflection

Deflection under serviceability loads of a building or part should not impair the strength or

efficiency of the structure or its components or cause damage to the finishing.

The serviceability loads used are the unfactored imposed loads except in:

Dead + imposed + wind. Apply 80 per cent of the imposed and wind load.

The most adverse combination of loads should be assumed in defection calculation.

Table 1.2 Deflection limit (part of Table 5 of BS5950)



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1.6 Design strength of materials

Table 1.3 of BS5950 Design strengths yp for steel

For rolled sections the thickness should be taken as the specified flange thickness from

BS4.

Note that the material strength factor mγ , is taken as 1.0 in the code.

Modulus of elasticity 2mmkN205E =

Design methods for buildings

The design of building must be carried out in accordance with one of the methods given

below:

(1) Simple design. The connections of members of the structure are assumed to be pin

jointed for analysis. Bracing or sheer walls are necessary to provide resistance to

horizontal loading.

(2) Rigid design. The connections are assumed to be capable of developing full

continuity (full moment connection). The analysis may be made using either elastic or

plastic methods.



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(3) Semi-rigid design. Practical joints are capable of transmitting some moment.

(4) Experimental verification.

In practice, structures are designed to either the simple or rigid methods of design.



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2. CONNECTIONS

Types of connections

Connections are needed to join two parts together. Some typical connections are shown on

Figure 2.1.

Connections are made by:

Bolting – ordinary bolts in clearance holes and friction-grip bolts;

Welding-fillet and butt welds.

2.1 Bolted Connection

2.1.1 Ordinary bolts

Bolts, nuts and washers

The ‘black’ hexagon head bolt shown in Figure 2.2 with nut and washer is commonly used

structural fastener. The bolts are in two strength grades as specified in BS 4190:

Grade 4.6: Mild steel : Yield stress = 235 N/mm2

Grade 8.8: High-strength steel : Yield stress = 627 N/mm2

The main diameters used are:

16, 20, (22), 24, (27) and 30 mm

The sizes shown in brackets are not preferred.

2.1.2 Direct shear joints

Bolts may be arranged to act in single or double shear, as shown in Figure 2.3.

Requirements for bolts in normal conditions are set out in Section 6.2 of BS 5950, Part 1.:

1. Minimum spacing = 2.5 (bolt diameter);

2. Maximum spacing in the direction of stress = 14t

where t is the thickness of the thinner plate connected.

3. Minimum edge and end distance for a rolled, machine-flame ct or planed edge =

1.25D

where D is the hole diameter.

4. Maximum edge distance is = 11tε



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where ( )0.5

yp275ε =

Nominal diameters of holes for ordinary bolts

= bolt diameter plus 2 mm for up to 22 mm diameter.

= bolt diameter plus 3 mm for larger diameter bolts.

A shear joint can fall in the following four ways as shown in Figure 2.4:

1. By shear on the bolt shank;

2. By bearing on the member or bolt;

3. By tension in the member;

4. By shear at the end of the member.

2.1.3 Design of bolted connection

(1) Effective area resisting shear sA .

Where threads occur in the shear plane:

Ts AA =

where areastresstensileAT =

When threads do not occur n the shear plane:

AA s =

where diameter.nominaltheonbasedareashankA =

(2) Shear capacity sP of a bolt:

sss ApP =

where ps = shear strength given in Table 32 in the ode. (see Table 2.1).

(Note that long joints are not discussed. Refer to the code.)



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(3) Bearing capacity should be taken as lesser of –

Capacity of the bolt Pbb = d t pbb

where d = nominal diameter of the bolt

t = thickness of the connected plate

pbb = bearing strength of the bolt given in Table 32 in the code.

(See Table 2.1.)

Capacity of the connected plate:

Pbs = d t pbs

bsp t e 0.5≤

where pbs = bearing strength of the connected parts given in Table

33 in the code. (see Table 2.1.)

e = end distance

Then check on 0.5 e t pbs ensures that the plate does not fail by end shear as shown in Fig.

4(c).

Table 2.1 Ordinary bolts in clearance holes (Table 32, 33 BS 5950)



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Fig. 2.1 Typical connections

Fig. 2.2 Hexagon head bolt, nut and washer



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Fig. 2.3 Bolts in single and double shear

Fig. 2.4 Failure modes of a bolted joint



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2.1.4 Direct tension joints

Tension capacity of bolts

Joints with bolts in direct tension are shown in Figure 2.5.:

Tension capacity of an ordinary bolt

Pt = ptAt

where pt = tension strength from Table 32 in the code

= 195 N/mm2 for Grade 4.6 bolts

= 450 N/mm2 for Grade 8.8 bolts

At = tensile stress area. (at thread)

Fig. 2.5 Bolts in Tension

Tension values for Grade 4.6 bolts are given in Table 2.2.

Table 2.2 Load capacity (Grade 4.6 bolts and Grade of 43 steel)



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2.1.5 Eccentric connections

There are two principal types of eccentrically loaded connections as shown in Figure 2.6:

(1) Bolt group in direct shear and torsion; and

(2) Bolt group in direct shear and tension.

2.1.6 Bolts in direct shear and torsion

In Figure 2.6(a) the moment is applied in the plane of the connection and the bolt group

rotates about its center of gravity. A linear variation of loading due to moment is assumed.

The direct shear is divided equally between the bolts and the side plates re assumed to be

rigid.

In Figure 2.7(a), the load P is applied at an eccentricity e.

The bolts A, B, etc. are at distances r1, r2, etc. from the centroid of the group.

The coordinates of each bolt are (x1,y1), (x2,y2), etc.

Let FT = the force due to the moment on bolt A, the farthest bolt.

Then the force on a bolt at r2 = FT r2 / r1

and so on for all the other bolts.

The moment of resistance of the bolt group:

.............../rrrF/rrrFrFM 133T122T1TR ++=

.....)..........rrr)(/rF( 3

3

2

2

2

11T ++=

∑= 2

1T r)/rF(

∑∑ += )x)(/rF( 22

1T y

= applied moment = P. e

The load FT due to moment on the maximum loaded bolt A is

( )∑ ∑+= 22

1T yx/r e PF



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The load FS due to direct shear is

FS = P/ (No. of bolts)

The resultant load FR on bolt A can be found graphically, as shown in Figure 2.7(b).

The algebraic formula can be derived by referring to Figure 2.7(c).

Resolve the load FT vertically and horizontally to give

Vertical load on bolt A = FS + FT cos φ

Horizontal load on bolt A = FT sin φ

Resultant load on bolt A

[ ]0.52

TS

2

TR )cosF(F)sin(FF ϕϕ ++=

[ ]0.5

TS

2

T

2

S cosF2FFF ϕ++=

The size of bolt required can then be determined from the maximum load on the bolt.

Fig. 2.6 Eccentrically loaded connections



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Fig. 2.7 Bolt group in direct shear and torsion

2.1.7 Bolts in direct shear and tension

In Figure 2.6(b) the bolts are in combined shear and tension

The following must be satisfied:

SS PF ≤

TT PF ≤

1.4/PF/PF TTSS ≤+

where PS = shear capacity of the bolt

= pS AS

FS =Factored applied shear

PT = tension capacity

= pt AT

FT = factored applied tension



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The bearing capacity checks must also be satisfactory.

Approximate method:

A bracket subjected to a factored load P at an eccentricity e is shown in Figure 2.9(a).

The center of rotation is assumed to be at the bottom bolt in the group. The loads vary

linearly as shown on the figure, with the maximum load FT in the top bolt.

The moment of resistance of the bolt group is:

MR = 2 [FT y1 + FT y22

/ y1 +…]

= (2 FT / y1) [y12 + y2

2 +…]

= (2 FT / y1) Σ y2

= P e

The maximum bolt tension is:

FT = P e y1 / (2 Σ y2)

The vertical shear per bolt:

FS = P / (No. of bolts)

A bolt size is assumed and checked for combined shear and tension as described above.

Fig. 2.8 Interaction diagram: bolts in shear and tension



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Fig. 2.9 Bolts in direct shear and tension: approximate analysis



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Example 2.1

The joint shown in Figure 2.10 is subjected to a tensile working dead load of 85 kN and a

tensile working imposed load of 95 kN. All data regarding the member and joint are shown

in the figure. The steel is Grade 43 and the bolt Grade 4.6. The gross area of the angles is

12.7 cm2 per angle. Check the joint.

Factored load = (1.4 x 85) + (1.6 x 95) = 271 kN (Table 1 & 2 of BS 5950)

Strength of bolts.

Bolts A and B are in double shear, 8mm and 10mm bearing thickness

Bolts C and D are in single shear, 8mm bearing thickness

Single shear capacity on threads (Table 1 & 2 for 20-mm diameter bolts)

PS = pS AS

= 160 x 245 = 39200 = 39.2 kN

Double shear capacity on threads

= 2 PS = 2 x 39.2 = 78.4 kN

Bearing capacity of bolts on 10 mm splice plate

Pbb = d t pbb

= 20 x 10 x 460 = 92000 N = 92 kN

Bearing capacity of plate on 10 mm splice plate with 30 mm end distance

Baring strength pbs = 460 N/mm2 from Table 2.1:

Pbs = d t pbs = 20 x 10 x 460 = 92000 N = 92 kN

≤ 0.5 e t pbs = 0.5 x 30 x 10 x 460 = 69000 N = 69 kN

i.e. Pbs = 69 kN

Bearing capacity of bolts on 8 mm cover plate or angle

Pbb = d t pbb

= 20 x 8 x 460 = 73600 N = 73.6 kN

Bearing capacity of plate on 8 mm cover plate or angle with 30 mm end distance

Baring strength pbs = 460 N/mm2 from Table 2.1:

Pbs = d t pbs = 20 x 8 x 460 = 73600 N = 73.6 kN



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≤ 0.5 e t pbs = 0.5 x 30 x 8 x 460 = 55200 N = 55.2 kN

i.e. Pbs = 69 kN

Bolt capacity of bolts A and B (2 bolts)

Critical values are: double shear for bolt A (= 78.4), end sher in 10mm plate for bolt B (=

69)

Bolt capacity of bolts C and D (4 bolts)

Critical values are: single shear for bolts C and D (= 39.2)

Total bolt capacity – two bolts are in double shear and four in single shear

= (78.4) + 69 + (4 x 39.2) = 304.2 kN

Strength of the angles. (Refer to tension member design notes)

Gross area = 12.7 cm2 per angle.

The angles are connected through both legs.

Net area = 2 (1270 – 2 x 22 x 8) = 1836 mm2

Design strength py = 275 N/mm2 (Table 1.3)

Capacity Pt = Ae py

= 275 x 1636 / 103 = 504.9 kN 271 kN O.K.⇒

Splice plate and cover plate (Refer to tension member design notes)

Effective area Ae = Ke (net are)

= 1.2 [(95 – 22) 10 + (140 – 44) 8] = 1798 mm2

≤ gross area = 2070 mm2

Capacity Pt = Ae py

= 275 x 11798 = 494300 N = 494.3 kN 271 kN O.K.⇒

The splice is adequate to resist the applied load.



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Fig. 2.10 Double-angle splice



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Example 2.2

Check that the joint shown in Figure 2.11 is adequate. All data required are given in the

figure.

All bolts: grade 4.6, all steel: grade 43.

Factored Load = (1.4 x 60) + (1.6 x 80) = 212 kN

Moment N = (212 x 525) / 103 = 111.3 kNm

Bolt group Σx2 = 12 x 250

2 = 750 x 10

3

Σy2 = 4 (35

2 + 105

2 + 175

2) = 171.5 x 10

3

Σx2 +Σy

2 = 921.5 x 10

3

Cos φ = 250 / 305.16 = 0.819

Bolt A is the bolt with the maximum load:

Load due to moment

FT = P e r1 / (Σx2 +Σy

2)

= (111.3 x 103 x 305.16) / (921.5 x 10

3) = 36.85 kN

Load due to shear

FS = P / (No. of bolts)

= 212 / 12 = 17.67 kN

Resultant load on bolt

FR = [FS2

+ FT2

+ 2 FS FT cos φ] 0.5

= [17.672

+ 36.852

+ (2 x 17.67 x 36.85 x 0.819)] 0.5

= 52.31 kN

Single-shear value of 24 mm diameter ordinary bolt on the threads

PS = pS AS

= 160 x 353 = 56400 N = 56.4 kN 52.31 kN O.K.⇒

Universal column flange thickness = 17.3 mm

Side-plate thickness = 15 mm

Minimum end distance = 45 mm



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Bearing capacity of the bolt

Pbb = d t pbb

= 24 x 15 x 460 / 103

= 165.5 kN 52.31 kN O.K.⇒

Bearing capacity of the plate

Pbs = d t pbs

= 24 x 15 x 460 / 103 = 166 kN

≤ 0.5 e t pbs

= 0.5 x 45 x 15 x 460 / 103 = 155 kN

⇒ use Pbs = 155 kN

52.31 kN O.K.⇒

The strength of the joint is controlled by the single shear value of the bolt.

Overall joint capacity = single-shear capacity of the bolt

= PS = 56.4 kN 52.31 kN

The joint is satisfactory.

Fig. 2.11 Example: bolt group in direct shear and torsion



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Example 2.3

A bolded bracket connection subjected to a ultimate load of 143 kN is shown in Figure

2.12.

Check the adequacy of the connection.

All bolts: grade 4.6, all steel: grade 43.

Strength of bolts: (Table 2.2 for 20-mm diameter bolts)

Single shear capacity on threads

PS = pS AS = 160 x 245 = 39200 = 39.2 kN

Tension capacity

PT = pt AT = 195 x 245 = 47.8 kN

Conditions: FS ≤ PS and FT ≤ PT

FS / PS + FT / PT ≤ 1.4

Use approximate method of analysis for conservative results.

Σy2 = (90

2 + 180

2 + 270

2) = 113.5 x 10

3 mm

2

The maximum bolt tension is at bolts y1 from the lowest bolts:

FT = P e y1 / (2Σy2)

= 143 x 200 x 270 / (2 x 113.5 x 103)

= 34 kN ≤ PT O.K.⇒

The vertical shear per bolt:

FS = P / No. of bolts

= 143 / 8 = 17.9 kN ≤ PS O.K.⇒

FS / PS + FT / PT ≤ 1.4

FS / PS + FT / PT = 17.9 / 39.2 + 34 / 47.8

= 0.457 + 0.17 = 1.17

≤ 1.4 O.K.⇒

ry.satisfacto is connection The⇒

Lecture Notes Structural Steel Design

Documents

design methods steel

plastic design

elastic design

working loads

limit state design

design theories

wind loads

characteristic loads