Lecture Notes Structural Steel Design
-
Upload
peter-jean-jacques -
Category
Documents
-
view
96 -
download
5
description
Transcript of Lecture Notes Structural Steel Design
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
1
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.
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
2
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.
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
3
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
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
4
Fig. 1.2 Compound sections
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
5
Fig. 1.3 Rolled and formed sections
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
6
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.
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
7
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)
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
8
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.
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
9
(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.
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
10
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
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
11
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.)
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
12
(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)
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
13
Fig. 2.1 Typical connections
Fig. 2.2 Hexagon head bolt, nut and washer
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
14
Fig. 2.3 Bolts in single and double shear
Fig. 2.4 Failure modes of a bolted joint
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
15
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)
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
16
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
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
17
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
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
18
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
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
19
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
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
20
Fig. 2.9 Bolts in direct shear and tension: approximate analysis
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
21
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
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
22
≤ 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.
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
23
Fig. 2.10 Double-angle splice
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
24
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
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
25
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
Bachelor of Science in Civil Engineering
Structural Steel Design, Structural Dynamics
26
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⇒