Design of steel structures structures I

Syllabus:
 Introduction
 Tension members – ASD
 Compression members – ASD
 Tension members – LRFD
 Compression members – LRFD
 Design of beams
 Examples on design of beams
 Shear bolted connections
 Eccentric bolted connections
 Welded connections
 Column bases
Codes:

 ASCE 7-16
 AISCE – ASD
 AISCE – LRFD

Design method:
 Allowable design method, ASD
 Load and resistance factor design, LRFD

References
 Jack C. and Stephen F. (2012), Structural steel design, 5th Edition, New Jersey,
U.S.A.
 William T. (2007), Steel design, 4th Edition, Nelson, U.S.A.
 Charles G. and John E. (2009), Steel structures design and behavior, 5th Edition,
New Jersey, U.S.A.

0
Lecture number (1)
Introduction
Steel structures:
Steel frame buildings consist of a skeletal framework which carries all the loads to
which the building is subjected. The sections through three common types of buildings
are shown throuh Fig 1. To Fig. 3. And these are.
1. Single-storey lattice roof building,
2. Single-storey portal frame building, and
3. Medium-rise braced multi-storey building.
These three types cover many of the uses of steel frame buildings such as factories,
warehouses, offices, flats, schools, etc.

Fig. 1: Single-storey lattice roof building with crane

Fig. 2: Single-storey rigid pinned portal

Fig. 3: Multi-storey building

1
Advantages of steel as structural material:
1. High strength per unit volume, that means the weight of structures will be small.
2. Easy to construct, and need short construction time.
3. Ability to connect together by several connections, including bolts and welds.
4. Possible reuse after a structure is disassembled.
5. Steel is ultimate recyclable material.
Advantages of steel as structural material:
In general steel has many disadvantages like, corrosion, fireproofing costs, susceptibility
to buckling, and fatigue. (See the references for more details).
Structural elements:
As mentioned above, steel buildings are composed of distinct elements.
1. Beams and girders, members carrying lateral loads in bending and shear.
2. Ties, members carrying axial loads in tension.
3. Struts, columns or stanchions, members carrying axial loads in compression.
These members are often subjected to bending as well as compression.
4. Trusses and lattice girders-framed members carrying lateral loads. These are
composed of struts and ties.
5. Purlins-beam members carrying roof sheeting.
6. Sheeting rails-beam members supporting wall cladding.
7. Bracing-diagonal struts and ties that, with columns and roof trusses, form vertical
and horizontal trusses to resist wind loads and hence provided the stability of the
building.
Joints connect members together such as the joints in trusses, joints between floor
beams and columns or other floor beams. Bases transmit the loads from the columns to
the foundations.

2
Structural design:
For a given framing arrangement, the problem in structural design consists of.
1. Estimation of loading,
2. Analysis of main frames, trusses or lattice girders, floor systems and bracing to
determine axial loads, shears and moments at critical points in all members,
3. Design of the elements and connections using design data from step 2, and
4. Production of arrangement and details drawings from the designer’s sketches.
Structural steel:
Steel is an alloy consisting of more than 98 percent iron. It also contains small quantities
of carbon, silicon, manganese, sulfur, phosphorus, and other elements. Carbon is the
material has the greatest effect on the properties of steel. Carbon content of steel is 0.5
percent and is normally from 0.2 to 0.3 percent. Structural steels are generally grouped
into several major ASTM classifications:
1. The carbon steels A36, A53, A500, A501 and A529.
2. The high-strength low-alloy steels A572, A612, A913 and A992.
3. The corrosion-resistant high-strength low-alloy steels A242, A588 and A847.
Fig. 4 shows typical stress-strin curves for several different yield stresses steel. And the
minimum yield strength, FY, and minimum tensile strength, Fu, are shown in Table (1).

Fig. 4: Typical stress-strain curves

3
Steel sections:
Steel sections include many types, there are follow:
Rolled sections:
The main rolled sections are shown on Fig. 5.
1. Wide flange or W-shapes: These are sections produced primarily to resist axial
load and bending moment.
2. American standard or S-shapes: These are very efficient sections for resisting
bending moment about major axis.
3. M-shapes: These sections are used for resisting bending moment.
4. HP-shapes: These are sections produced primarily to resist axial load with a high
radius of gyration about the minor axis to prevent buckling in that plane.
5. Equal and unequal angles: These are used for bracing members, truss members
and for purlins, side and sheeting rails
6. Channels or C-shapes: These are used for beams, bracing members, truss
members and in compound members.
7. Structural tees: The sections shown are produced by cutting a universal beam or
column into two parts. Tees are used for truss members, ties, light beams, and
connection.
8. Circular, square and rectangular hollow sections: These sections make very
efficient compression members, and are used in a wide range of applications as
members in roof trusses, lattice girders, in building frames, for purlins, sheeting
rails, etc.
Compound sections:
Compound sections are formed by the following means and shows in Fig. 6.
1. Strengthening a rolled section such as a W-shapes beam by welding on cover
plates.

4
 2. Combining two separate rolled sections, as in the case of the crane girder. The
two members carry loads from separate directions.
3. Connecting two members together to form a strong combined member.
Built-up sections:
Built-up sections are made by welding plates together to form I, H or box members
which are termed plate girders, built-up columns, box girders or columns, respectively
as in Fig. 7. These members are used where heavy loads have to be carried and in the
case of plate and box girders where long spans may be required.

Fig. 5: Metric rolled sections

Fig. 6: Compound sections

5
 Fig. 6: Built-up sections
Loads on building:
Working loads are the actual loads the structure is designed to carry. These are normally
thought of as the maximum loads which will not be exceeded during the life of the
structure. The main loads on buildings may be classified as.
Dead loads:
These are due to 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. However, at the start of
a design, sizes are not known accurately and dead loads must often be estimated from
experience.
Live loads:
These take account of the loads caused by people, furniture, equipment, stock, and etc.
on the floors of buildings and roofs. The values of the floor loads used depend on the
use of the building. Live, snow or rain loads are given in Table 4.1 of ASCE 7 code.
Wind loads:
These loads depend on the location and building size. Wind loads are also in ASCE 7
code.
Design methods:
The AISC specification provides two acceptale methods for designing structural steel
members and their connection. These are allowable stress design (ASD) and load and
resistance factor design. Two methods based on limit states design principles.

6
There are two catogeries of limit states: strength and serviceability. Strength limit state
define the load-carrying capacity, including yielding, fracture, buckling and fatique.
And serviceability define performance, including, deflection, cracking and deterioration.
Nominal strength:
The nominal strength of a member is the theorytical strength, its calculated without
factors of safety.

Loads combinations:
In ASD, the design loads, Ra, due to service loads without factos, must not exceed of the
nominal strength, Rn, divided by safety factor, Ω, greater than 1.0

In LFRD, the design loads, Ra, due to ultimate loads (each service load multiplied by
load facto greater than 1.0), must not exceed of the nominal strength, Rn, multipied by
reduction factor,ϕ, less than 1.0

7
Lecture number (2)
Tension members – ASD
Tension membears are structural elements that are subjected to axial tensile force. They
are used in sevral types of structures and include truss member, braing for buildings and
bridges. Any cross section may be used.
Allowable stress design (ASD) method:
The tension member can be fail by one of two limit states: yielding or fracture. The
allowable tensile stress Ft determined as follow:
When member fail by yielding on gross section,

When member fail by fracture on net section,

The capacity of tension member, when the failure by yeilding is,

The capacity of tension member, when the failure by fracture is,

Where, Ag, is the gross section area, Ae, is the effective area, U, is reduction cofficient,
and An, is the net area.
Net area:
When tension member connected by bolts or revits, hole must be provided at the
connection. As a result the member cross sectional area at the connection is reduced and
the capacity of member may be also reduced depending on the size and locaion of the
holes.

8
In Fig. 1(a) the failure line is along the section A-B, and Fig. 1(b) showing two possible
lines of failure, one throught one hole and other throught diagonal path A-C.
The net area of section in Fig. 1(a) can be computed by,

And the net area of section in Fig. 1(b) can be computed by,

∑ ∑

( )
( )
Where, D, is the diameter of hole, t, is the minimum thickness of connected parts, S, is
the spacing adjacent holes parallel to direction load, and, g, is the gage distance
transverse to the load direction.

Fig. 1: Failure lines

Reduction coefficient:
̅

Bolted or riveted connections:

- When a tension load is transmitted directly to each of the cross-sectional elements
by fasteners (All elements of the cross section are connected), U=1.0

9
 - Single angle and double angles with two or three fasteners in the direction of
loading, U=0.60
- Single angle and double angles with four or more fasteners in the direction of
loading, U=0.80
- W, M, S, HP and Tee, connected through the flange with three or more fasteners
in the direction of loading, with a width at least 0.67d, U=0.90
- W, M, S, HP and Tee, connected through the flange with three or more fasteners
in the direction of loading, with a width less than 0.67d, U=0.85
- W, M, S, HP and Tee, connected through the web with four or more fasteners in
the direction of loading, U=0.70
Welded connections:
An = Ag
- When a tension load is transmitted directly to each of the cross-sectional elements
by welds, U=1.0
- When the tension load is transmitted only by transverse welds, U = 1.0
- When the tension load is transmitted to a plate by longitudinal welds along both
edges at the end of the plate:
For, 1.5w > L ≥ w, U = 0.75
For, 2w > L ≥ 1.5w, U = 0.87
For, L ≥ 2w, U = 1.00
L, is the length of weld, mm
w, is th plate width (distance between welds), mm

10
Design of tension member:
The design of a tension member involves finding a member with adequate gross and net
area. If the member has a bolted connection, the selection of a suitable gross section
requires an accounting for the area lost because of holes.
The required area of section may be fined based on limit state of yielding by,

Where, Pa, is the design load.

The slenderness ratio limitation of main member will be satisfied if,

The slenderness ratio limitation of secondary member will be satisfied if,

The local slenderness ratio of horizontal member will be satisfied if,

The local slenderness ratio of inclined member will be satisfied if,

The local slenderness ratio of vertical member will be satisfied if,

Where, PD, is the dead load, PL, is the live load, L, is the length of member, r, is the
minimum radius of gyration, and, d, is the minimum depth of section.

11
Example (1):
The figure below shows the Plate used as tension member. Determine the effective area.
If the A36 steel is used, find the tension capacity of Plate. Assume the bolt M24. Use
AISC-ASD method.

Solution:
Net area:
Hole diameter,

Path AD (two holes),

Path ABC (three holes),

∑ ∑

The effective area is,

The all parts of plate are connected, so the reduction coefficient is 1.0

12
From Table (1), A36 steel, FY is 250 N/mm2 and Fu is 400 N/mm2
The capacity of tension member by yielding failure is,

The capacity of tension member by fracture failure is,

The tension capacity of tension member is 270 kN.

Example (2):
Determine the tenstion capacity of a tension member with section W250x67 is shown in
figure below. A572 – Gr50 steel is used for section and Plates. Use AISC-ASD method.

Solution:
Hole diameter,

From Meetic Tables of propeties, the gross sectinal area is 8560 mm2. And flange
thickness is 15.7 mm.
Net area of W250x67:

The effective area is,

Width to depth ratio,

13
From Metric Table, T125x33.5,
̅
The length of connected is 200 mm,
̅

Net area of 2PL10x300:

Failure occurs in plates before W-shape.

From Table (1), A572-Gr50 steel, FY is 345 N/mm2 and Fu is 450 N/mm2
The capacity of tension member by yielding failure is,

The capacity of tension member by fracture failure is,

The tension capacity of tension member is 1148 kN.

Example (3):
The figure below shows the C380x50 used as tension member. Determine the effective
area. If the A36 steel is used, find the tension capacity of Channel. Assume the bolt
M24. Use AISC-ASD method.
Solution:
Net area:
Hole diameter,

14
From Meetic Tables of propeties, the gross sectinal area is 6430 mm2, web thickness is
10.2, and flange thickness is 16.5 mm.
Path ABDG (two holes),

Path ABCDEF (three holes),

∑ ∑

( )

The effective area is,

The all parts of C-shape are connected, so the reduction coefficient is 1.0

From Table (1), A36 steel, FY is 250 N/mm2 and Fu is 400 N/mm2
The capacity of tension member by yielding failure is,

The capacity of tension member by fracture failure is,

The tension capacity of tension member is 965 kN.

15
Example (4):
Design a tension member of length 4.6 m to resist a service dead load of 300 kN and
service live load of 600 kN. Use A36 steel. The connection consists of two raws of bolts
and each raw has 4M20 bolts. Use AISC-ASD method.
Solution:
The design load,

The allowable stress for yielding falure,

From Table (1), A36 steel, FY is 250 N/mm2 and Fu is 400 N/mm2
⁄
Required gross section area,

Assume two un-equal angles connected back to back from long leg, the required area for
one angle,

The available sectins of un-equal angle are:

L152x102x12.7, L127x89x15.9, L203x152x11.1, L203x102x12.7, and L178x102x12.7
Try 2L152x102x12.7:

Determine the net area,

Hole diameter,

Effective area,
Single and double angles with four or more bolts in the direction of load, U is 0.8

16
Tension capacity by fracture failure,

Tension capactiy not satisfactory.

Try 2 L127x89x15.9:

Check for local slenderness,

Assume the member is horizontal,

Not satisfactory,
Try 2 L203x152x11.1:

Check for member slenderness, assume the member is main member,

The radius of gyration from Meteric Tables of properties is,

The shape satisfies all requirements. So use 2L203x152x11.1

17
Problem (1):
The figure below shows the L152x152x12.7 used as tension member. Determine the
effective area. If the A36 steel is used, find the tension capacity of Angle. Assume the
bolt M16. Use AISC-ASD method.

Problem (2):
Repeat solution of problem (1) if L152x152x12.7 connected by weld as in the figure
below. Use AISC-ASD method.

Problem (3):
The figure below shows the C150x17.9 used as tension member. Determine the
effective area. If the A36 steel is used, find the tension capacity of Channel. Assume the
bolt M16. Use AISC-ASD method.

18
Problem (4):
Design and select W-shape to resist a service dead load of 500 kN and a service live
load of 500 kN. The conncetion will be through the flanges with two lines M16 bolts in
each flange as shown in the figure below. Each line contains more than two bolts. The
length of member is 9.0 m. Use AISC-ASD method.

Problem (5):
Design and select an American Astandared Channel shape to resist design load 1500
kN. The length is 4.0 m and there will be two lines of M22 bolts in the web as in figure
below. Use A36 steel. Use AISC-ASD method.

Problem (6):
Use AISC-ASD metheod, design the member AB. If the design load shown in the figyre
below. Neglect the self-weight of member BC.

19
Lecture number (3)
Compression members – ASD
Compression members are structural elements that are subjected only to axial
compressive loads. The load are applied the along a longitudinal axis through the
centeroidal of the meber cross section.
Euler’s theorem:
The stress at which a column buckles decreases as the column become longer. For the
column buckle elastically, its will have to be long and slender. Its critical buckling load,
Pcr, can be computed with the Euler theorem that follows:

This formula usualy is written in defferent form that involves the column,s slenderness
ration. Since,

Substituting this value into the Euler’s formula and deviding both side by cross-sectional
area, the Euler buckling stress is obtained:

( )

Where, Fe, is the Euler stress, L, is the member length, k, is the effective length factor, r,
is the minimum radius of gyration, and, E, is the modulus of elasticity.
The effective length:
The effective length of a column is defined as the distance between points of zero
moment in the column. in the AISC specifications, the effective length of a column is
reffered to as kL, where k, is the effectve length factor. The vlues of effective length
fcator shown in Fig. 1.

20
 Fig. 1: Approximate values of effective length factor, k
Classification of compression section for local buckling:
Compression sections are classified as either a non-slender element or slender elemnt.
AISC defined a section is non-slender when the width to thickness ration does not
exceed of λr. and the section is slender when the width to thickness ratio exceed of λ r.
the limiting values for λr is given in Table (1).
Table (1): Width to thickness ratio, λr for unstiffened axial compression element.
Width to thickness
Description of element λr
ratio
Flanges of rolled I-shapes sections, plates, channels, tees,
outstanding legs of pairs of angles connected with √
continuous contact.
Flanges of built up I-shapes sections, plates or angle legs
√
projecting from built up I shapes sections.

Legs of single angle, Legs of double angles with

√
separators, and all other unstiffened elements.

Web of doubly symmetric I-shapes sections and channels. √

21
Limited slenderness ratio:
AISC recommended a members whose design is based on compressive axial load, the
slendemess ratio preferably should not exceed 200.
Allowable stress:
AISC recommended on the gross section of axially loaded compression member, when
the slendemess ratio less than Cc, the allowable stress is,

( )

( ) ( )

On the gross section of axially loaded compression member with the slendemess ratio
exceed Cc, the allowable stress is,

( )

Capacity of axially loaded compression is,

Design of compressin memner:

The desing of compression member by formula involves a trial and error process. The
allowable stress is not known until member size is selected. Member size may be slected
by required gross section area assuming that the allowable stress is 45% of the yield
stress. After that calculate the slenderness ration and the allowble stress. The try a larger
or smaller section. The required area of section is,

22
Example (1):
Detemine the capacity of a column with section of W310x107. If the column fixed at
lower end and pinned at upper end. The length of column is 4.6 m and A36 steel is used.
Use AISC-ASD method.
Solution:
From Meteric Tables of properties, gross section area is 13600 mm2, the radius of
gyration about major and minor axes are, 134 mm and 77 mm, depth of web is 241 mm,
flange width is 306 mm, web thicknes is 10.9 mm, and flange thickness is 17 mm.
Section classification, from Table (1),

√ √

√ √

Section is non-slender.
Limited slenderness is,

√ √

The column is fixed pinned, from Fig. 1, then th effective length factor is 0.8
Slenderness ration is,

The allowable stress is,

( )

( ) ( )

23
 ⁄

The capacity of compressin member is,

Example (2):
A compression member in a truss with section of L152x102x22.2, if the member has 4.0
m length and A36 steel is used. Detemine the axial compressive load. Use AISC-ASD
method.
Solution:
From Meteric Tables of properties, gross cetion area is 5150 mm2, the radius of gyration
about major and minor axes are, 47.3 mm and 28.2 mm respectively.
Section classification, from Table (1),

√ √

Section is non-slender.
Limited slenderness is,

√ √

The members in trusses are pinned at both ends, from Fig. 1 then th effective length
factor is 1.0
Slenderness ration is,

The allowable stress is,

⁄
( )
24
The axial compressive load of member is,

Example (3):
Design a cmpression member of effective length 3.5 m to resist a service dead load of
200 kN and service live load of 150 kN. A36 steel is used. Use AISC-ASD method.
Solution:
The design load,

Required gross section area,

Try 2L102x89x12.7 back to back from long leg as in figure below.

From Meteric Tables of properties, gross cetion area is 2260 mm2, the radius of gyration
about major and minor axes are, 31.5 mm and 26.5 mm, length of leg is 102 mm,
thickness is 12.7 mm, Ix, is 2.24x106 mm4, Iy, is 1.59x106 mm4, and, x, is 25.4 mm.
Section classification, from Table (1),

√ √

Section is non-slender.

25
 ( ) ( )

√ √

√ √

Limited slenderness is,

√ √

Slenderness ration is,

The allowable stress is,

( )

( ) ( )

The axial compressive load of member is,

Use 2L102x89x12.7 for compression member.

26
Repeat the solution of example (3):
Try C310x46
From Meteric Tables of properties, gross cetion area is 5880 mm2, the radius of gyration
about major and minor axes are, 120 mm and 28.3 mm, flange width is 93.2 mm, flange
thickness is 17.8 mm, depth of web is 239 mm, and web thickness is 9.4 mm.
Section classification, from Table (1),

√ √

√ √

Section is non-slender.
Slenderness ration is,

The allowable stress is,

The axial compressive load of member is,

Use C310x89x46 for compression member.

27
Problem (1):
Detemine the capacity of a compression member with section consist of two Channel
2C250x42.4. If the member is pinned at two ends and the length of column is 5.0 m.
A36 steel is used. Use AISC-ASD method.

Problem (2):
A column section consist of HP200x54 and 2PL12x200 as in the figure below. It has
effective length of 5.0 m, and A992 steel used for HP and PL. Use AISC-ASD method,
determine the capacity of column.

Problem (3):
Design and select a suitable rolled section of a column. The column is subjected to a
service dead load of 400 kN and a cervice live load of 250 kN. It has length of 5.0 m. If
the column is fixed at two ends and A992 steel is used. Use AISC-ASD method. Try
HP-shape and W-shape.

28
Lecture number (4)
Tensin members - LRFD
Tension membears are structural elements that are subjected to axial tensile force. They
are used in sevral types of structures and include truss member, braing for buildings and
bridges. Any cross section may be used.
Nominal strength:
The tension member can be fail by one of two limit states: yielding or fracture. The
nominal strength is determined as follow:
When member fail by yielding on gross section,

When member fail by fracture on net section,

In load and resistance factor design method, the factored tensile load is compare to
design strength. The design strength is,

The design load does not exceed the design strength, then,

Wher, ϕt, is the reduction factor for tension member, taken 0.90 for yielding and 0.75 for
fracture, FY, is the yield stress, Fu, is the ultimate stress, Ag, is the gross section area,
and Ae, is the effective area, U, is reduction cofficient, and An, is the net area.
Net area:
When tension member connected by bolts or revits, hole must be provided at the
connection. As a result the member cross sectional area at the connection is reduced and
the capacity of member may be also reduced depending on the size and locaion of the
holes.

29
In Fig. 1(a) the failure line is along the section A-B, and Fig. 1(b) showing two possible
lines of failure, one throught one hole and other throught diagonal path A-C.
The net area of section in Fig. 1(a) can be computed by,

And the net area of section in Fig. 1(b) can be computed by,

∑ ∑

( )
( )
Where, D, is the diameter of hole, t, is the minimum thickness of connected parts, S, is
the spacing adjacent holes parallel to direction load, and, g, is the gage distance
transverse to the load direction.

Fig. 1: Failure lines

Reduction coefficient:
̅

Bolted or riveted connections:

- When a tension load is transmitted directly to each of the cross-sectional elements
by fasteners (All elements of the cross section are connected), U=1.0

30
 - Single angle and double angles with two or three fasteners in the direction of
loading, U=0.60
- Single angle and double angles with four or more fasteners in the direction of
loading, U=0.80
- W, M, S, HP and Tee, connected through the flange with three or more fasteners
in the direction of loading, with a width at least 0.67d, U=0.90
- W, M, S, HP and Tee, connected through the flange with three or more fasteners
in the direction of loading, with a width less than 0.67d, U=0.85
- W, M, S, HP and Tee, connected through the web with four or more fasteners in
the direction of loading, U=0.70
Welded connections:
An = Ag
- When a tension load is transmitted directly to each of the cross-sectional elements
by welds, U=1.0
- When the tension load is transmitted only by transverse welds, U = 1.0
- When the tension load is transmitted to a plate by longitudinal welds along both
edges at the end of the plate:
For, 1.5w > L ≥ w, U = 0.75
For, 2w > L ≥ 1.5w, U = 0.87
For, L ≥ 2w, U = 1.00
L, is the length of weld, mm
w, is th plate width (distance between welds), mm

31
Design of tension member:
The design of a tension member involves finding a member with adequate gross and net
area. If the member has a bolted connection, the selection of a suitable gross section
requires an accounting for the area lost because of holes.
The required area of section may be fined based on limit state of yielding by,

Where, Pu, is the design load.

The slenderness ratio limitation of main member will be satisfied if,

The slenderness ratio limitation of secondary member will be satisfied if,

The local slenderness ratio of horizontal member will be satisfied if,

The local slenderness ratio of inclined member will be satisfied if,

The local slenderness ratio of vertical member will be satisfied if,

Where, PD, is the dead load, PL, is the live load, L, is the length of member, r, is the
minimum radius of gyration, and, d, is the minimum depth of section.

32
Example (1):
The figure below shows the Plate used as tension member. Determine the effective area.
If the A36 steel is used, find the design strength of Plate. Assume the bolt M24. Use
AISC-LRFD method.

Solution:
Net area:
Hole diameter,

Path AD (two holes),

Path ABC (three holes),

∑ ∑

The effective area is,

The all parts of plate are connected, so the reduction coefficient is 1.0

33
From Table (1), A36 steel, FY is 250 N/mm2 and Fu is 400 N/mm2
The nominal strength of tension member by yielding failure is,

The nominal strength of tension member by fracture failure is,

The design strength of tension member by yielding failure is,

The nominal strength of tension member by fracture failure is,

The design strength of tension tension member is 405 kN.

Example (2):
Determine the design strength of a tension member with section W250x67 is shown in
figure below. A572-Gr50 steel is used for section and Plates. Use AISC-LRFD method.

Solution:
Hole diameter,

From Meetic Tables of propeties, the gross sectinal area is 8560 mm2. And flange
thickness is 15.7 mm.
Net area of W250x67:

34
The effective area is,
Width to depth ratio,

From Metric Table, T125x33.5,

̅
The length of connected is 200 mm,
̅

Net area of 2PL10x300:

Failure occurs in plates before W-shape.

From Table (1), A572-Gr50 steel, FY is 345 N/mm2 and Fu is 450 N/mm2
The nominal strength of tension member by yielding failure is,

The nominal strength of tension member by fracture failure is,

The design strength of tension member by yielding failure is,

The nominal strength of tension member by fracture failure is,

The design strength of tension tension member is 1721 kN.

35
Example (3):
The figure below shows the C380x50 used as tension member. Determine the effective
area. If the A36 steel is used, find the design strength of Channel. Assume the bolt M24.
Use AISC-LRFD method.
Solution:
Net area:
Hole diameter,

From Meetic Tables of propeties, the gross sectinal area is 6430 mm2, web thickness is
10.2, and flange thickness is 16.5 mm.
Path ABDG (two holes),

Path ABCDEF (three holes),

∑ ∑

( )

The effective area is,

The all parts of C-shape are connected, so the reduction coefficient is 1.0

36
From Table (1), A36 steel, FY is 250 N/mm2 and Fu is 400 N/mm2
The nominal strength of tension member by yielding failure is,

The nominal strength of tension member by fracture failure is,

The design strength of tension member by yielding failure is,

The nominal strength of tension member by fracture failure is,

The design strength of tension tension member is 1446 kN.

Example (4):
Design a tension member of length 4.6 m to resist a service dead load of 300 kN and
service live load of 600 kN. Use A36 steel. The connection consists of two raws of bolts
and each raw has 4M20 bolts. Use AISC-LRFD method.
Solution:
The design load,

From Table (1), A36 steel, FY is 250 N/mm2 and Fu is 400 N/mm2
Required gross section area,

Assume two un-equal angles connected back to back from long leg, the required area for
one angle,

The available sectins of un-equal angle are:

L152x102x12.7, L127x89x15.9, L203x152x11.1, L203x102x12.7, and L178x102x12.7

37
Try 2L152x102x12.7:

Determine the net area,

Hole diameter,

Effective area,
Single and double angles with four or more bolts in the direction of load, U is 0.8

The desing strength by fracture failure,

( )
Check for local slenderness,
Assume the member is horizontal,

Not satisfactory,
Try 2 L203x152x11.1:

( )

The radius of gyration from Meteric Tables of properties is,

The shape satisfies all requirements. So use 2L203x152x11.1

38
Problem (1):
The figure below shows the L152x152x12.7 used as tension member. Determine the
effective area. If the A36 steel is used, find the design strength of Angle. Assume the
bolt M16. Use AISC-LRFD method.

Problem (2):
Repeat solution of problem (1) if L152x152x12.7 connected by weld as in the figure
below. Use AISC-LRFD method.

Problem (3):
The figure below shows the C150x17.9 used as tension member. Determine the
effective area. If the A36 steel is used, find the design strength of Channel. Assume the
bolt M16. Use AISC-ASD method.

39
Problem (4):
Design and select W-shape to resist a service dead load of 500 kN and a service live
load of 500 kN. The conncetion will be through the flanges with two lines M16 bolts in
each flange as shown in the figure below. Each line contains more than two bolts. The
length of member is 9.0 m. Use AISC-ASD method.

Problem (5):
Design and select an American Astandared Channel shape to resist design load 1500
kN. The length is 4.0 m and there will be two lines of M22 bolts in the web as in figure
below. Use A36 steel. Use AISC-ASD method.

Problem (6):
Use AISC-ASD metheod, design the member AB. If the design load shown in the figyre
below. Neglect the self-weight of member BC.

40
Lecture number (5)
Compression members – LRFD
Compression members are structural elements that are subjected only to axial
compressive loads. The load are applied the along a longitudinal axis through the
centeroidal of the meber cross section.
Euler’s theorem:
The stress at which a column buckles decreases as the column become longer. For the
column buckle elastically, its will have to be long and slender. Its critical buckling load,
Pcr, can be computed with the Euler theorem that follows:

This formula usualy is written in defferent form that involves the column ,s slenderness
ration. Since,

Substituting this value into the Euler’s formula and deviding both side by cross-sectional
area, the Euler buckling stress is obtained:

( )

Where, Fe, is the Euler stress, L, is the member length, k, is the effective length factor, r,
is the minimum radius of gyration, and, E, is the modulus of elasticity.
The effective length:
The effective length of a column is defined as the distance between points of zero
moment in the column. in the AISC specifications, the effective length of a column is
reffered to as kL, where k, is the effectve length factor. The vlues of effective length
fcator shown in Fig. 1.

41
 Fig. 1: Approximate values of effective length factor, k
Classification of compression section for local buckling:
Compression sections are classified as either a non-slender element or slender elemnt.
AISC defined a section is non-slender when the width to thickness ration does not
exceed of λr. and the section is slender when the width to thickness ratio exceed of λ r.
the limiting values for λr is given in Table (1).
Table (1): Width to thickness ratio, λr for unstiffened axial compression element.
Width to thickness
Description of element λr
ratio
Flanges of rolled I-shapes sections, plates, channels, tees,
outstanding legs of pairs of angles connected with √
continuous contact.
Flanges of built up I-shapes sections, plates or angle legs
√
projecting from built up I shapes sections.

Legs of single angle, Legs of double angles with

√
separators, and all other unstiffened elements.

Web of doubly symmetric I-shapes sections and channels. √

42
Limited slenderness ratio:
AISC recommended a members whose design is based on compressive axial load, the
slendemess ratio preferably should not exceed 200.
Nominal strength of compression member:
AISC recommended, the nominal strength of compression member is,

For LRFD, the design strength is,

The factored compressive load is compare to design strength as,

Where, Fcr, is the critical stress, Ag, is the gross section area, and, ϕc, is the reduction
factor for compressin member, its taken 0.85.
AISC recommended that the values of critical stress depends on the slenderness ration
as follow:

( )

( )

43
Design of compressin memner:
The desing of compression member by formula involves a trial and error process. The
allowable stress is not known until member size is selected. Member size may be slected
by required gross section area assuming that the critical stress is 60% of the yield stress.
After that calculate the slenderness ration and the allowble stress. The try a larger or
smaller section. The required area of section is,

Example (1):
Detemine the design strength of a column with section of W310x107. If the column
fixed at lower end and pinned at upper end. The length of column is 4.6 m and A36 steel
is used. Use AISC-LRFD method.
Solution:
From Meteric Tables of properties, gross section area is 13600 mm2, the radius of
gyration about major and minor axes are, 134 mm and 77 mm, depth of web is 241 mm,
flange width is 306 mm, web thicknes is 10.9 mm, and flange thickness is 17 mm.
Section classification, from Table (1),

√ √

√ √

Section is non-slender.
The column is fixed pinned, from Fig. 1, then th effective length factor is 0.8
Slenderness ration is,

44
 √ √

The Euler’s stress is,

⁄
( )

The critical stress is,

( ) ( ) ⁄

The nominal srength of compression member is,

The design strength of compression member is,

Example (2):
A compression member in a truss with section of L152x102x22.2, if the member has 4.0
m length and A36 steel is used. Detemine the nominal strength of compression member.
Use AISC-LRFD method.
Solution:
From Meteric Tables of properties, gross cetion area is 5150 mm2, the radius of gyration
about major and minor axes are, 47.3 mm and 28.2 mm respectively.
Section classification, from Table (1),

√ √

Section is non-slender.

45
The members in trusses are pinned at both ends, from Fig. 1 then th effective length
factor is 1.0

Slenderness ration is,

√ √

The Euler’s stress is,

⁄
( )

The critical stress is,

( ) ( ) ⁄

The nominal srength of compression member is,

The design strength of compression member is,

Example (3):
Design a cmpression member of effective length 3.5 m to resist a service dead load of
200 kN and service live load of 150 kN. A36 steel is used. Use AISC-LRDF method.
Solution:
The design load,

Required gross section area,

46
Try 2L102x89x12.7 back to back from long leg as in figure below.

From Meteric Tables of properties, gross cetion area is 2260 mm2, the radius of gyration
about major and minor axes are, 31.5 mm and 26.5 mm, length of leg is 102 mm,
thickness is 12.7 mm, Ix, is 2.24x106 mm4, Iy, is 1.59x106 mm4, and, x, is 25.4 mm.
Section classification, from Table (1),

√ √

Section is non-slender.

( ) ( )

√ √

√ √

Slenderness ration is,

47
 √ √

The Euler’s stress is,

⁄
( )

The critical stress is,

( ) ( ) ⁄

The nominal srength of compression member is,

The design strength of compression member is,

Use 2L102x89x12.7 for compression member.

Repeat the solution of example (3):
Try C310x46
From Meteric Tables of properties, gross cetion area is 5880 mm2, the radius of gyration
about major and minor axes are, 120 mm and 28.3 mm, flange width is 93.2 mm, flange
thickness is 17.8 mm, depth of web is 239 mm, and web thickness is 9.4 mm.
Section classification, from Table (1),

√ √

48
 √ √

Section is non-slender.

Slenderness ration is,

√ √

The Euler’s stress is,

⁄
( )

The critical stress is,

( ) ( ) ⁄

The nominal srength of compression member is,

The design strength of compression member is,

Use C310x89x46 for compression member.

49
Problem (1):
Detemine the capacity of a compression member with section consist of two Channel
2C250x42.4. If the member is pinned at two ends and the length of column is 5.0 m.
A36 steel is used. Use AISC-ASD method.

Problem (2):
A column section consist of HP200x54 and 2PL12x200 as in the figure below. It has
effective length of 5.0 m, and A992 steel used for HP and PL. Use AISC-ASD method,
determine the capacity of column.

Problem (3):
Design and select a suitable rolled section of a column. The column is subjected to a
service dead load of 400 kN and a cervice live load of 250 kN. It has length of 5.0 m. If
the column is fixed at two ends and A992 steel is used. Use AISC-ASD method. Try
HP-shape and W-shape.

50
Lecture number (6)
Design of beams
Beams are structural members support transverse loads and are therefor subjected to
bending memnet, shear force and deflection. The cross sectional of beams include the
W-shape, M-shape and S-shape, also C-shape is used for beam light weight.
Bending stress and the plastic moment:
From elementary mechanics of materials, the bending stress at any point can be found
from the bending theory,

Where, M, is the bending moment, y, is the distance from the neutral axis to the point of
interest, and I, is moment of inertia. The maximum stress occurs at the extreme fiber,
there are two maximum stress: compressive and tensile stress. For semmetry section,
these two stresses will be equal in magnitude. Then,

⁄
Where, C, is the distance from neutral axis to the extreme fiber, and, S, is the elastic
section modulus, is taken fom the Tables of properties of AISC. For structural steel, the
maximum stress does not exceed the yield stress, then,

Where, My, is the yield moment.

In Fig. 1, a simply supported beam with concentrated load at midspan is shown a stages
of loading. Once yielding begins, the distribution of stress on the cross section will no
longer be linearand yielding will progress from the extreme fiber toward the neutral
axis. In Fig. 1(c), yielding has just begun. In Fig. 1(d), the yielding has progressed in th
web, and in Fig. 1(e) the entire cross section hass yielded. When stage, e, has been
reached, any ferther increase in the load will cause cllapse. Aplastic hing is said to have
form at the center of beam.
51
 Fig. 1: Stress distribution of beam section
The plastic moment capacity, which is moment required to form the plastic hinge can be
computed from the stress distribution in Fig. 2. Assuming that, Ac, is the area of
compression, and, At, is the area of tension.

Fig. 2: Stress distribution plastic stage

From equilibrium forces,

The plastic neutral axis devides the cross section into two equal areas. From which, the
plastic and elastic neutral axes are the same.
The plastic moment is determined by taking moment of couple forces about the plastic
neutral axis as follow,

Where, A, is the cross sectional area, a, is the distance between centeroids of the two
half area, and, Z, is the plastic modulus of section, its taken from Tables of properties.

52
Classificasion of section:
AISC classifies cross sectional section as compact, non-compact, or slender depending
on the values of width thickness ratio, λ. If the width thickness, λ, not exceed the uppet
limit of compact category, λp, the section is compact. And width thickness ratio, λ, is
greater than upper limit of compact category and not exceed upper limit of non-compact
category, λr, the section is non-compact. And the section is said slender, when the width
thickness ratio, λ, is greater than upper limit of non-compact category, λr. The values of
upper limits compact and non-compct categories are shown in Table (1).
Table (1): Width thickness ratios of sections
Shapes Element λ λp λr

Rolled I shapes Flange √ √

√

Rolled C shapes Flange √ √

√

Rolled I, C shapes Web √ √

Flange √ √
√
Built up, I shapes
Web √ √

Built up, plate √ √

Outstand, plate, angle N/A √

53
Nominal moment strength:
The nominal moment is the lowest value obtained according to the limit states of
yielding: flange local buckling (FLB), web local buckling (WLB), and lateral-torsional
buckling (LTB).
The nominal moment, Mn, shall be determined as follows for each limit state:

For the limit states of flange and web local buckling:

For the limit states of flange and web local buckling:

( )( )

( )
Where, Mr, is the limiting buckling moment, Fr, is the residual stress, its taken 70
N/mm2 for rolled section and 115 N/mm2 for welded section.
Nominal moment for limit state lateral torsional buckling:

* ( )( )+

√ √ ( )
( )
54
When, Lb, is the lateral un-braced length, Lp, limiting un-braced length for Mp, Lr,
limiting un-braced length for Mr, Cb, is the bending coeffecient dependent on moment
gradient, and, Fcr, is the critical stress.

√
√
( ) ( )

Where, Mmax, is the absolute value of maximum moment of un-braced length, MA, is the
absolute value of moment at quarter point of the un-braced length, MB, is the absolute
value of moment at centerline of the un-braced length, MC, is the absolute value of
moment at three-quarter point of the un-braced length, and, X1 and X2 are the torsional
properties.
The design moment, Md, must be not less than the required ultimate moment, Mu.

Where, ϕb, is the reduction factor for moment, and taken 0.90
Nominal shear strength:
From elementary mechanics of materials, the sear stress is,

Where, fv, is the shear stress at the point of interest, V, is the vertical shear, A, is the
area of cross section between the point of interest and the top or bottom of section, y', is
the distance from the center of area to the neutral axis, I, is the moment of inertia, and,
b, is the width of section.
Fig. 1, shows distribution of shear stress for I-shapes. Clearly, the web will completely
yield before the flange begin to yield. Because of this, yielding of the web represents
one of the shear limit states. Taking the shear yield stress 60 percent of the tensile yield
stress, we can writte the shear stress formula for the stress in the web at failure as:

55
Where, Aw, is the area of web. The nominal strength corresponding to this limit state is
therefore,

Fig. 1: Distribution of shear stress

AISC specification modified the nominal shear formual to covering both beams with
stiffened webs and with un-stiffened webs as follow,

The desing shear strength, Vd, must be not less than the required shear strength, Vu.

Where,ϕv, is reduction factor for shear, taken 0.90, and cv, is the critical shear stress to
shear yield stress. The value of, cv, depends on whether the limit state is web yielding,
web inelastic buckling, or web elastic buckling. Its determine as follow:
The limit state is yeilding,

The limit state is inelastic web buckling occurs,

√ √

56
 √

( )

The limit state is elastic web buckling occurs,

( )

Where, kv, is the buckling coefficient, and determined as follow:

For beam with stiffened web, when

( )

For beam with stiffened web, when

( )

For beam with un-stiffened web, when

Where, a, is the distance bettween stiffeners.

For special case of hot-rolled I-shapes:

57
Deflection:
Vertical deflection due to applied loads must be calculated using any of the available
methods.
The formulas of special cases of beams are shown in Table (3).
The Vertical deflection of steel beams is usually limited to certain maximum values
shown in Table (2).
Table (2): Deflection limits.
Type of member L.L D.L+L.L W.L
Roof beams
L/360 L/240 L/360
- Supporting plaster ceiling
L/240 L/180 L/240
- Supporting non-plaster ceiling
L/180 L/120 L/180
- Not supporting a ceiling
Floor beams L/360 L/240 N/A

Table (3): The maximum vertical deflection

58
Lecture number (7)
Examples on design of beams
Example (1):
The beam shown in the figure below is a W410x46 of A992 steel. Its support a
reinforced concrete floor slab that provides contiuous lateral support to the compression
flange. The service dead load is 6.6 kN/m. this load is additional dead load on beam, it
does not include the self-weight of beam. The service live load is 8.0 kN/m. Check the
section for moment, shear and deflection.

Solution:
Design load,
( ) ⁄
Maximum moment at midspan,

Maximum shear at support,

Allowable deflection,

From Table (1), A992 steel, the yield stress, is 345 N/mm2

59
From Metric Tables of properties:

⁄ ⁄

Section classification:

√ √

√ √

So section is compact.
The nominal moment of section is,

The design momeent of section is,

The moment is satisfactory.

Nominal shear strength:
Depth to thickness web ratio is,

√ √

Design shear strength is,

Section is satisfactory for shear.

60
Deflection:
Deflection due to servise live load is,

Deflection due to servise dead and live loads is,

The section is not satisfactory for deflection.

Example (2):
Unrestrained beam has a section of W360x101. The A572-Gr50 steel used. If the beam
subjected to uniformly distributed load and length of span is 6.0 m. Use AISC-LRFD
method to determine the nominal moment.
Solution:
From Table (1), A572-Gr50 steel, the yield stress, is 345 N/mm2
From Metric Tables of properties:

⁄ ⁄
Section classification:

√ √

√ √

So section is compact.
The nominal moment of section is,

61
Limiting un-braced length for Mp is,

√ √

Limiting un-braced length for Mr is,

√ √ ( )
( )

√ √ ( )
( )
Un-braced length is 6.0 m.

A beam is subected to uniformly distributed load and compression flange free to rotate,
the bending coeffecient is 1.14

The nominal moment is,

* ( )( )+

( ) ( )

[ ( )( )]

The nominal moment of section is 610 kN.m

Example (3):
Select a standard hot-rolled shape of A36 steel for a simply supported beam with span
of 9.0 m. The beam has continuous support a uniform service live load of 70 kN/m. Use
AISC-LRFD method
Solution:
The design load is,
⁄
From Table (1), A36 steel, the yield stress, is 250 N/mm2
62
Maximum moment at midspan,

Assume a section is compact,

The required plastic modulus is,

Try W610x174
⁄

Maximum shear at support,

Allowable deflection,

From Metric Tables of properties:

Section classification:

√ √

√ √

63
So section is compact.
The nominal moment of section is,

The design momeent of section is,

The moment is satisfactory.

Nominal shear strength:
Depth to thickness web ratio is,

√ √

Design shear strength is,

Section is satisfactory for shear.

Deflection:
Deflection due to servise live load is,

Deflection due to servise dead and live loads is,

The section is not satisfactory for deflection.

64
Problem (1):
A simply supported beam with span 14.0 m is laterally supported at its ends and is
subjected to a service dead load of 5.8 kN/m including the wieght of beam and a service
live load of 14.6 kN/m. if A36 steel used. Check the W360x134 for moment, shear, and
deflection. Use AISC-LRFD method
Problem (2):
The bea shown in the figure below must support must support two concentrated loads of
90 kN each at quarter points. Lateral support procided at the ends of the beam. Use A36
steel, select suitable section of beam. Use AISC-LRFD method.

Problem (3):
The figure below shows a part of floor system from Hospital building. Design beam B4
as simply supported, full laterlly restrained, using AISC-LRFD method. Take Finishing
load on slab is 1.25 kN/m2.

65
Lecture number (8)
Shear bolted connections
Connections consist of affected elements of connected members (e.g. leg angle, beam
web, column flange), connecting elements (e.g. gusset plate, angle, Tee), and connectors
(welds, bolts, rivets). These components shall be selected with design strength exceeds
the required strength determined by structural analysis for factored loads acting on the
structure.
Types of bolts:
There are several types of bolts as in Fig. 1, that can be used for connection steel
members. They are classified by ASTM into ordinary bolts and hight strength bolts.
Ordinary bolts are olso called A307 bolts made from carbon steel similar to A36 steel.
They are available in diameters from 12.7 to 38 in 3.2 mm increments and length up to
200 mm. They are used for in light structures subjected to static loads. They are used for
bearing-type connections.
High strength bolts are made from alloy steel and have tensile strength two or more
times those of ordinary bolts. There are two basic types, A325 bolts and A490 bolts.
Sometimes high strength are needed with diameters greater than 38 mm and length
larger than 200 mm. They are used for bearing-type or friction-type connections.
The diameters of bolts, areas of bolts, and diameters of holes are shown in Table 2, and
the ultimate strength, tensile strength, and shear strength of bolts are shown in Table 3.

Fig. 1: Structural bolts

66
Minimum and maximum distance between bolts:
AISC recommended that the mimimum and maximum distance between bolts are shown
in Table (4).
Classifications of connections:
Connection may be classified into the following types:
According to types of bolts, connecections classified into: bearing-type connections and
friction-type connections.
According to loads transfere, the connections slassified into: shear connections, when
loads transfere through the center of bolts group, eccentric connections, when the
applied loads are eccentric, and moment connections, when the connection subjected to
shear and moment.
Failures of bolted joints:
Fig. 2, shows sevreal ways of failure can be occurs in bolted connections. These are
follow:
1. Single shear failure on bolts,
2. Tension failure on the plates,
3. Bearin failure on bolts and/or plates,
4. Shearing failure of a par of connected member, and
5. Double shear failure on bolts.

Fig. 2: Types of failure on bolted connections

67
Nominal strengths of bolts in bearing-type connections
The nominal shear strength of bolt in single shear is,

The nominal shear strength of bolt in double shear is,

The bearin strength of bolt and connected part is,

Bearing strength is independent of the type of bolt, because the stress under
consideration is on the connected part. The nominal bearing strength is,

The nominal tensile strength of bolt is,

Where, Fnv, and, Fnt, are the nominal shear stress and tensile stress of bolt, taken from
Table (3), Fu, is the ultimate tensile strength of connected part, taken from Table (1), A b,
is the cross sectional area of bolt, taken from Table (2), d, is the diameter of blt, t, is the
thickness of connected part, and Lc, is the clear distance, in the direction parallel to the
applied load, its shown in Fig. 3.

Fig. 2: Clear distance between bolts

Design strength of bolt:
The design strength of bolt can be determined by reduced the nominal strength as,

Where, ϕ, is the reduction factor, taken 0.75 for shear and bearing strength.

68
Nominal shear strength of gusset plate:
The nominal shear strength of gusset plate by yielding failure is,

The nominal shear strength of gusset plate by fracture failure is,

Where, FY, is the yield stress of gusset plate, Fu, is the ultimate tensile stress of gusset
plate, Ag, is the gross sectional area of gusset plate, and A nv, is the net shear area of
gusset plate.
Example (1):
A 4.0 m long of tension member with section L102x76x6.4 and its connection are
shown if the figure below subjected to a service dead load of 40 kN and a service live
load of 90 kN. The thickness of gusset plate is 6.4 mm. Use A307 bolts and A36 for
angle and gusset plate. If the connection is bearing-type conncetion, check the
connection using AISC-LRFD method.

Solution:
Design load is,

From Table (1), the yield stress and ultimate tensile stress are 250 N/mm2 and 400
N/mm2 repectively, From Tabe (2), the area of bolt is 198 mm2, diameter of bolt is 15.9
mm, and hole diameter is 18 mm. and From Table (3), the nominal shear stress is 165
N/mm2.
69
Load carry by bolt is,

Design shear strength of bolt is,

Design bearing strength of edge bolt is,

Design bearing strength of other bolts is,

For connection,

Shear and bearing is satisfactory.

From, Table (4), the minimum and maximum distance for edge bolt are 22 mm and 77
mm. and the minimum and maximum spacing center to center of bolt is 42 mm and 90
mm respectively. Then,

Spacings of bolts are satisfactory.

Check for tension member capacity:
From Metric Tables of properties, gross sectional area is 1090 mm2,
Design strength by yielding failure is,

70
Net area is,

Effective area is,

Single and double angle with for or more bolts, the factor of effective area is 0.8

Design strength by fracture failure is,

The connection is satisfactory for all types of failures.

Example (2):
The C200x27.8 shown in figure below has beaan selected to resist a service dead load of
80 kN and service live load of 240 kN. Its to be attached to a 9.5 mm gusset plate with
M22-A325 bolts. A36 steel is used. Assume the connection is bearing-type connection,
design the connection. Use AISC-LRFD method.

Solution:
Design load is,

From Table (1), the yield stress and ultimate tensile stress are 250 N/mm2 and 400
N/mm2 repectively, From Tabe (2), the area of bolt is 388 mm2, diameter of bolt is 22.2
mm, and hole diameter is 24 mm. and From Table (3), the nominal shear stress is 330
N/mm2.
Design shear strength of bolt is,

71
Design bearing strength of bolt is,

The design strength per bolt based on shear.

Required number of bolts is,

Use 6 bolts in two rows, arranged in 3 bolts in each row.

From, Table (4), the minimum and maximum distance for edge bolt are 28 mm and 108
mm. and the minimum and maximum spacing center to center of bolt is 60 mm and 126
mm respectively. Then,

Design bearing strength of edge bolt is,

Design bearing strength of other bolts is,

72
For connection,

Check for tension member capacity:

From Metric Tables of properties, gross sectional area is 3550 mm2,
Design strength by yielding failure is,

Net area is,

Effective area is,

Factor of effective area is,
̅

Design strength by fracture failure is,

The connection is satisfactory for all types of failures.

73
Problem (1):
A double angle-shape, 2L152x152x15.9 is connected to a 16 mm gusset plateas shown
in figure below. Determine the maximum total service load that can be applied if the
ration of dead load to live load is 8.5. the bolts are M22-A325 bearing type-bolt.
A572Gr50 steel is used for angle, and A36 steel is used for gusset plate. Use AISC-
LRFD method.

Problem (2):
The 2L102x102x6.4 shown in figure below has beaan selected to resist a service dead
load of 30 kN and service live load of 600 kN. Its to be attached to a 9.5 mm gusset
plate with M20-A325 bolts along line a-b. A36 steel is used. Assume the connection is
bearing-type connection, design the connection. Use AISC-LRFD method.

74
Lecture number (9)
Eccentric bolted connections
Friction-type connection:
Allmost all bolted connection ith srtandard size holes are design as bearing-type
connection. On some cases, particularly in a bridges, use fricition-type connection also
called slip critical-connection. AISC specification states that the nominal slip resistance
of a friction-type connection shall be determined with the expression:

Where, µ, is the mean slip coefficient, Du, is the ratio of mean actual pretension to the
specified minimum pretension, hsc, is the hole factor, Tb, is the minimum bolt tension
from Table (5) and Ns, is the number of slip planes. For standard hole size, the reduction
factor for critical slip resistance taken 1.0.
Eccentric connections:
Eccentrically loaded bolts group are subjected to direct shear and torsional shear.
Eccentrically is quite obvious in shown in Fig. 1, where a beam is connected to a
column with a plate. It’s obvious that this connection must resist some moment, because
the center of gravity of the load from the beam does not coincide with the reaction from
the column.

Fig. 1: Eccentric connection

75
Analysis of eccentric connection by elastic method:
In Fig. 2, the shear areas and load are shown separate from the column and plate. The
eccentric load, P, can be replaced with the same load acting at the centroid and the
couple, M, as shown in Fig. 3.

Fig. 2: Eccentric load, P

Fig. 3: Eccentric load and replacement load

In Fig. 4(a), the load is centric, and each bolt can be assumed to resist an equal shear of
the load, the shear force on each bolt is,

In Fig. 4(b), the couple act on the group of bolts, and the shear stress in each bolt can be
found from the torsion formula as follow,

76
Where, N, is the number of bolt, d, is the distance from the centroid of the bolts group to
the point where the stress is being computed, and J, is the polar moment of inertia of the
bolts about the centroid.
If the parallel axes theorem is used and the polar moment of inertia approximately is,

∑ ∑

Provided all bolts have the same area, the shear stress on ach bolt is,

∑
And the shear force in each bolt caused by the couple is,

∑ ∑ ∑

Fig. 4: Analysis of eccentric load on bolts

The horizontal and vertical components for each bolt from direct shear are,

Where, Px, and, Py, are the horizontal and vertical components of eccentric load on
connection as shown in F. 5.
77
 Fig. 5: Components of force on bolt caused by couple
The horizontal and vertical components for each bolt caused by the couple can be found
as follow. In terms of x-y coordinates of the centers of bolts, where the origin the
coordinate system is at the centroid of group bolts. The components are,

∑ ∑ ∑ ∑

∑ ∑ ∑ ∑
And the resultant shear force on each bolt is,

√( ) ( ) √(∑ ) (∑ )

Example (1):
For friction-type connection is shown in the figure below. The service axial loads are
120 kN of dead and 180 kN for live. Determine the number of M24-A325 slip critical
bolts in standard-size holes needed for limit state. If the slip coefficient is 0.3 and A36
steel is used.

78
Solution:
Design load is,

From Table (1), the ultimate tensile stress is 400 N/mm2, from Tabe (2), the area of bolt
is 507 mm2, diameter of bolt is 25.4 mm, and hole diameter is 27 mm, from Table (3)
the nominal shear stress is 330 N/mm2. and from Table (5) and A325 bolts, the
minimum bolt tension is 205 kN.
Design strength of critical slip of bolt is,

Required number of bolts is,

Use 6 bolts in two rows, arranged in 3 bolts in each row.

From, Table (4), the minimum and maximum distance for edge bolt are 30 mm and 192
mm. and the minimum and maximum spacing center to center of bolt is 68 mm and 224
mm respectively. Then,

Design bearing strength of edge bolt is,

79
Design bearing strength of other bolt is,

Bearing strength for connection is,

Design bearing strength of 6 bolt is,

( ) ( )
Example (2):
Determine the load on critical bolt for the bearing-type connection shown in the figure
below. Use the elastic analysis method.

Solution:
Taking moment about lower bolts line to determine the centroid of the group bolts,

80
 ̅
The horizontal and vertical components of load are,

√ √

√ √
Taking moment of forces about centroid of bolts group to determine the couple moment,
( ) ( )
Components of direct shear force in each bolt are,

Components caused by couple in each bolt are,

∑ ∑ ( ) ( )

∑ ∑

∑ ∑
For the critical bolt, the components caused by couple are,

The resultant shear force on critical bolt is,

√(∑ ) (∑ ) √

81
Example (3):
Eccentric connection, connected beam to column as in the figure below, its subjected to
ultimate load. Use M20-A325 bolts and the A36 steel for plate, beam and column, select
the suitable number of bolts if the connection is bearing-type connection. Use AISC-
LRFD method.

Solution:
From Table (1), the ultimate tensile stress is 400 N/mm2, from Tabe (2), area of bolt is
285 mm2, diameter of bolt is 19.1 mm, and hole diameter is 22 mm, and from Table (3),
the nominal shear stress is 330 N/mm2. And ssume thickness of gusset plate is 9.5 mm.
Design shear strength of bolt is,

Design bearing strength of bolt is,

82
Assume edge distance and spacing center to center are 75 mm for each bolt.
From, Table (4), the minimum and maximum distance for edge bolt are 26 mm and 114
mm. and the minimum and maximum spacing center to center of bolt is 50 mm and 133
mm respectively. Then,

Centroid of bolt groups is,

̅
̅
The horizontal and vertical components of load are,

Taking moment of forces about centroid of bolts group to determine the couple moment,
( )
Components of direct shear force in each bolt are,

Components caused by couple in each bolt are,

∑ ∑ ( ) ( )

For the critical bolt, the components caused by couple are,

∑ ∑

83
 ∑ ∑

The resultant shear force on critical bolt is,

√(∑ ) (∑ ) √

Not satisfactory, so increase the nimber of bolts.

Centroid of bolt groups is,

̅
̅
The horizontal and vertical components of load are,

Taking moment of forces about centroid of bolts group to determine the couple moment,
( )
Components of direct shear force in each bolt are,

Components caused by couple in each bolt are,

∑ ∑ ( ) ( )

84
For the critical bolt, the components caused by couple are,

∑ ∑

∑ ∑

The resultant shear force on critical bolt is,

√(∑ ) (∑ ) √

Design bearing strength of bolts is,

Design bearing strength of connection is,

Use 6M20-A325 Bolts.

85
Problem (1):
Determine the load on critical bolt for the bearing-type connection shown in the figure
below. Use the elastic analysis method. And also check the shear and bearing of bolt.

Problem (2):
Use AISC-LRFD method; determine the required number of M22-A325 bolts in
standard size holes for the bearing-type connection shown in the figure below. A36 steel
is used for connected parts

86
Lecture number (10)
Welded connections
Structural welding is a process wherby tha parts to be coonected are heated and fused,
with suplementary molten metal added to the joint.
Classification of welds:
Three sepatate classifications of welds are available. These classifications are based on
types of welds, the positins of the welds, and the types of joints used.
Type of weld:
The main two types of welds are the fillet welds and the grove welds. In addition there
are plug and slot welds, which are not a common in structural work. These four types of
welds are shown in Fig. 1.

Fig .1: Four types of structural welds

87
Position fo weld:
Welds are refer to as flat, horizontal, vertical, or overhead. These types of welds are
shown in Fig .2.

Fig. 2: Weld position

Type of joint:
Welds cn be further classified acoording to the type of joint used. These types are butt,
lab, tee, edge, corner, etc. as in Fig. 3.

Fig. 3: Type of weld joints

Adgantages and disadvantages:
Take assignment
Fillt welds:
Fillet welds are stronger in tension and compression than they are in shear, so the
controlling fillet weld stresses given by various specifications are shearing stresses. The
design and analysis of fillet welds is based on the assumption that the cross section of
the welds is a 45o right triangle as shown in Fig. 4

88
 Fig. 4:
For a given length of weld, L, and size of weld, w, subjected to a load, P, the critical
shearing stress is,

If the weld ultimate shearing stress, FW, is used in this equation, the nominal load
capacity of the weld can be written as,

The strength of a fillet welds depends on the weld metal used. That its function of the
type of electrode. AISC recommended that, the ultimate shearing stress, FW, in a fillet
weld is 0.6 times the tensile strength of the weld metal, denote, F EXX. The nominal stress
is therfore,

For LRFD, the design strength of fillet weld is,

The nominal shear strength of the base metal will be based on either the limit state for
yielding or limit state for fractue. The shear yield and ultimate stresses are taken as 0.6
times the tensile yield and ultimate stresses. AISC gives the shear yield strength as

And the shear fracture strength is,

89
The design shear yield is,
( ) ( )
And the design shear fracture is,
( ) ( )
The design strengths per unit length are,

( ) ( )
( ) ( )
AISC recommended, that:
The minimu size of a fillet weld is 6 mm,
Maximum size of a fillet weld is,

Minimum length of a fillet weld not less that four times size of weld.
Maximum length of a fillet weld is not exceed βL, β is,

( )

Where, t, is the thickness of connected part, and, L, is the actual length of the weld.
Eccentric welded connection:
Ecentric welded connection are analyzed in much the same way as bolted connection,
except that unit lengths of weld replaced idividual bolts in the computations. The
analysis can be done by elastic method as follow:
The load on the eccentric bearing-type connection shown in Fig. 5 may be considered to
act in the plane of the weld. The calculated load can then be multiplied by 0.707w times
the weld size to obtain the actual load.
An eccentric load in the plane of the weld subjects the weld to both direct shear and
torsional shear. The direct shear stress is,

90
 Fig. 5: Analysis of eccentric welded connection
Where, L, is the length of weld. If a rectangular components are used,

Where, Px, and Py, are the componants of the applied load. The shearing stress csused by
the couple is found with the torsion formula,

Where, d, is the distance from the centroid to the shear area to the point where the stress
being computed, J, is the polar moment of inertia, and, I x, and, Iy, are the rectangular
moment of inertia of the shear area.
Fig. 6 shows the stress at the upper right-hand corner of the given weld. In term of
rectangular components,

Fig. 6: Sear stress caused by couple

91
Once all rectangular components have been found, they can be added vectorially to
obtain the resultant shearing stress at the point of interest.

√(∑ ) (∑ ) √( ) ( )

Example (1):
A PL12.7x200 of A36 steel is used as a tension member and is to be connected to 9.5
mm thick gusset plate, as shon in the figure below. Design a weld to devolop the full
tensile capacity ot the member. Use AISC-LRDF method and E70xx electrodes.

Solution:
Design strength of tension member based on yielding is,

Design strength of tension member based on fracture is,

Use the minimum size of fillet weld.

Design strength per unit length is,
⁄
Base metal shear yield strength is,
⁄
Base metal shear fracture strength is,
⁄
The weld strength of 0.935 kN/mm is govern.

92
Required length of weld is,

Transeverse length is 200 mm, and the length of longitudinal weld is,

Use 6mm, fillet E70xx electrodes.

Example (2):
Determine the size of weld required for the bearing-type connection in the figure below.
The service dead load is 45 kN, and service live load is 135 kN. A36 steel is used for the
plate, and A992 steel is used for the column. Use AISC-LRFD method.

Solution:
Design load is,

Taking the moment of area about y-axis to determine the centroid coordinate,

93
̅

The couple moment is,

( )
Moment of inertia about x and y axes are,

( )

( ( ) )

( )
Direct shear stress per unit length is,

Shear caused by couple is,

The resultant shear sress is,

√( ) ( )

√( ) ( ) ⁄
Design strength per unit length is,
⁄
Required size f weld is,

Use 6 mm fillet weld, E70xx electrodes.

94
Problem (1):
Determine the maximum service load that can be applied if the live load to dead load
ratio is 3.0. investigate all limit states. All structural steel is A36 and the weld is a 6 mm
fillet weld with E70xx elctrodes. Note that the tension member is a double angle shape,
and both of the angles are welded as shown in figure below. Use AISC-LRFD method.

Problem (2):
Design a welded connection for an MC230x35.6 of a A572-Gr50 steel connected to a
9.5 mm gusset plate. The gusset plate is A36 steel and E60xx electrodes. Show your
results in sckech, complete with dimension. Use AISC-LRFD method.

Problem (3):
Determine the size of weld required for the bearing-type connection in the figure below.
The ultimate load is 40 kN. A36 steel is used for column and plate, and E60xx
electrodes. Use AISC-LRFD method.

95
Lecture number (11)
Column bases
When steel column is supported by a footing, it is necessary for the column load to be
spread over a suffecient area to keep the footing from being overstressed. Loads from
steel columns are transferred through a steel base plate to fairly large area of the footing
below. The base plates of steel columns can be welded directly to the columns, or they
can be bolted by means of some type of bolted or welded angles. These connection
method are illustrated in Fig. 1.

Fig .1: Methods of connected column bases

Fig .2: Axially loaded column base

96
To analyze the base plate shown in Fig. 2, note than the column is assume to apply a
total load to the base plate, Pu. then the load is assumed to be transmitted uniformy
through the plate to footing. AISC recommended that, the maximum moment in base
plate occur at distances 0.8bf and 0.95d a part.
Plate area:
The design strength of the concrete in bearing beneath the base plate must at least equal
the load load to be carried. When the base plate covers the entire area of the concrete,
the nominal bearing strength of the concrete is,

Where, f'c, is the compressive steel of concrete and A1, is the area of the base plate.
AISC permits the nominal strength to be increased as follow,

Where, A2, I the maximum area of the portion of the supporting concrete, and the
quantity of square root of (A2/A1) is limitedt to maximum value of 2. The ultimate load
does not exceed the design strength of base plate as,

Wher, ϕc, is the reduction factor of bearing, taken 0.65. arter controlling value of, A1,
can be determined the plate dimensions, B, and, N, as follow:
√

97
Plate thickness:
To determine the required plate thickness, t, moments are taken in the two directions as
through the plate were cantelivered out by the dimensions, m, and, n. refere to Fig. 2,
the moments in the two directions are:

In general, when, L, is the maximum value of, m, and, n, the ultimate moment is,

The nominal moment of unit length of base plate is,

( )

The design stength of base plate is,

The ultimate moment does not exceed the design moment, then,

The rquired thickness of base plate is,

√ √ √

Where, ϕb, is the reduction factor for bending, taken 0.90.

98
Moment-resisting column base plate:
Column bases are design to resist bending moment as well as axial load as in Fig. 3. An
axial load causes compression between a base plate and supporting footing, a moment
increases the compression on one side and decreases it on the other side.

Fig. 3: A column base subjected to axia load and moment

The pressure underneath the plate is distributed as in Fig. 4. It can be calculated by the
following equation.

The eccentricity due to moment does not exceed one sixth th length of base plate, N.

Wher, p, is the pressure underneath the plate, Pu, is the ultimate axial load, Mu, is the
ultimate moment, B, and, N, are the dimensions of base plate, and, e, is the eccentricity
due to moment.
The nominal moment of base plate is,

The ultimate moment caused by axial load and moment in two direction does not exceed
the design moment.

99
The required thickness of base plate is,

Fig. 4: Pressure distribution unerneath plate

Example (1):
Design a base plate of A36 steel for a W310x97 column of A992, that support the loads
890 kN dead load and 1330 kN live load. The concrete has compressive strength of 21
N/mm2, and the footing has has the dinensions 3.0 m by 3.0 m. Use LRFD method.
Solution:
Design load is,

From Metric Tables of properties, the depth of section is 308 mm, and flange width is
305 mm.
Required area of base plate is,

100
The dimensions of base plate are,

√ √

Check the bearing strength of concrete,

The lever arm of moment is,

101
The required thickness of base plate is,

√ √

Use PL40x400x400 A36

Example (2):
Design a moment resisting base plate to support a W360x179 column with an ultimate
axial load of 2760 kN and an ultimate bending moment of 305 kN.m. use A572-Gr50
steel and a concrete footing with 21 N/mm2. Use AISC-LRFD method.
Solution:
Design load is,

From Metric Tables of properties, the depth of section is 368 mm, and flange width is
373 mm.
Try PL500x700
The eccentricity is,

The resultant located in the midle third of the plate.

Pressure underneath the base plate is,

⁄ ⁄
⁄
The maximum pressure was less than bearing stress of concrete, and the minimum
pressure still compression. There are satisfactory.
102
An equation of pressure is,

( ))

The moment in longitudinal direction:

Taking moment to right at center of right flange,
( )
( )

The moment in transverse direction:

Average pressure in transverse direction is,

103
Required thickness of base p;ate is,

√ √

Use PL65x500x600
Problem (1):
Design a base plate of A36 steel for a W310x226 column of A572-Gr50, that support
the loads 850 kN dead load and 1900 kN live load. The concrete has compressive
strength of 21 N/mm2, and assume, A2 equal to A1. Use AISC-LRFD method.
Problem (2):
Design a moment resisting base plate to support a W310x179 column with an ultimate
axial load of 2400 kN and an ultimate bending moment of 150 kN.m. use A572-Gr50
steel and a concrete footing with 21 N/mm2. Use AISC-LRFD method.

104
Table (1): Properties of structural steel
Steel type ASTM FY (N/mm2) Fu (N/mm2)
Carbon 250 A36 400
240 A53 Gr B 415
228 Gr A 310
Carbon steel

290 A500 Gr B 400

345 Gr C 430
250 Gr A 400
A501
345 Gr B 485
345 G50 450
A529
380 G55 485
290 Gr42 415
345 Gr50 450
High strength low alloy steel

380 A572 Gr55 485

415 Gr60 520
450 Gr65 550
345 Gr2 485
A618
345 Gr3 450
345 Gr50 415
415 Gr60 520
A913
450 Gr65 550
480 Gr70 620
345 A992 450
290 435
low alloy steel
high strength
Corrosion
resistance

A242 315 460

345 480
A588 345 485
A847 345 485
E, is 200x10 N/mm and. G, is 200x103 N/mm2
3 2

Table (5): Minimum bolt pretension, (kN)

Nominal size A325 A490 Nominal size A325 A490
M16 91 114 M27 267 334
M20 142 179 M30 326 408
M22 176 221 M36 475 595
M24 205 257 ////////// ////////// //////////

105
Table (2): Bolts diameters
Nominal Bolt size Area Hole diameter
size (M) mm Ab (mm2) mm
M16 15.9 198 18
M20 19.1 285 22
M22 22.2 388 24
M24 25.4 507 27
M27 28.6 641 30
M30 31.8 792 33
M36 34.9 958 39

Table (3): Strengths of bolts for bearing-type connections

Description of bolt Fu (N/mm2) Fnt (N/mm2) Fnv (N/mm2)
A502 Grade 1 rivet 415 310 170
A502 Grade 2 and 3 rivet 550 415 228
A307 bolt 415 310 165
A325M bolt (threads included) 828 620 330
A325M bolt (threads excluded) 828 620 415
A490M bolt (threads included) 1035 780 415
A490M bolt (threads excluded) 1035 780 520
A449M bolt (threads included) 620 465 248
A449M bolt (threads excluded) 620 465 310

Table (4): Minimum and maximum distance between bolts

Minimum spacing center to center should not
Nominal size Edge distance
be less than 2.66d
M mm
Maximum edge distance is 12 times the
M16 22 thickness of the connected part, but not more
M20 26 than 150 mm
M22 28 Maximum spacing center to center for member
M24 30 subjected to corrosion is 14 times the
M27 34
thickness of thinner plate, not to exceed 180
M30 38
M36 46 mm

106