STRUCTURAL
STEEL
DESIGN
STRUCTURAL STEEL DESIGN
Steel is an alloy with iron and carbon being the primary elements. Generally,
iron-carbon alloys with up to 2.1% carbon by weight are considered steel and
iron-carbon alloys with greater amounts of carbon are cast iron. The main kinds
of structural steel are generally categorized according to the under mentioned
categories of chemical composition: Carbon-manganese steels: The major
chemical ingredients are iron, carbon, and manganese. These are normally
called mild structural steels or carbon steels .Steel is less malleable and harder
than mild steel. Mild steel can be further strengthened through the addition of
carbon. The basic difference is that s/s has very little carbon and is alloyed with
chromium, nickel, molybdenum and other elements to improve its mechanical
and chemical properties Mild steel contains carbon as the alloy, whereas
stainless steel includes chromium. The changes brought about by chromium to
the internal structure of the steel result in properties which gives stainless steel
its name: very high corrosion resistance and a surface which does not stain or
tarnish.
Weathering steel or weather resistant steel are colloquial terms used to describe
structural steels with improved atmospheric corrosion resistance. These steels
are high strength low alloy steels that under normal atmospheric conditions give
an enhanced resistance to rusting compared with that of ordinary carbon
manganese steels No protective coating is needed. It is extensively used for
bridges and has been used externally on some buildings
As with most steels, A36 has a density of 7,800 kg/m3 (0.28 lb/cu in). Young's
modulus for A36 steel is 200 GPa (29,000,000 psi).[2] A36 steel has a Poisson's
ratio of 0.26, and a shear modulus of 75 GPa (10,900,000 psi).
A36 steel in plates, bars, and shapes with a thickness of less than 8 in (203 mm)
has a minimum yield strength of 36,000 psi (250 MPa) and ultimate tensile
strength of 58,000–80,000 psi (400–550 MPa). Plates thicker than 8 in have a
32,000 psi (220 MPa) yield strength and the same ultimate tensile strength of
58,000–80,000 psi (400–550 MPa)
Advantages of Steel as a structural materials
1. high strength
2. uniformity
3. elasticity
4. permanence
5. ductility
6. additions to existing structures
7. ability to be fastened together by several simple devices including welding
8. adaptation to prefabrication
9. speed of erection
10. ability to be rolled into wide variety of sizes and shapes
11. possible reuse after a structure is disassembled
12. scrap value even though not reusable in its existing form.
Disadvantage of steel as a structural materials
1. maintenance costs
2. fireproofing cost
3. susceptibility to buckling
4. fatigue
The properties that need to be considered by designers when specifying steel
construction products are:
•
Strength
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STRUCTURAL
•
•
•
•
STEEL
DESIGN
Toughness
Ductility
Weldability
Durability.
Strength
Yield strength is the most common property that the designer will need as it is the
basis used for most of the rules given in design codes
Toughness
It is in the nature of all materials to contain some imperfections. In steel these
imperfections take the form of very small cracks. If the steel is insufficiently
tough, the 'crack' can propagate rapidly, without plastic deformation and result in
a 'brittle fracture'. The risk of brittle fracture increases with thickness, tensile
stress, stress raisers and at colder temperatures. The toughness of steel and its
ability to resist brittle fracture are dependent on a number of factors that should
be considered at the specification stage.
Ductility is a measure of the degree to which a material can strain or elongate
between the onset of yield and eventual fracture under tensile loading
Weldability
All structural steels are essentially weldable. However, welding involves locally
melting the steel, which subsequently cools. The cooling can be quite fast
because the surrounding material, e.g. the beam, offers a large 'heat sink' and
the weld (and the heat introduced) is usually relatively small. This can lead to
hardening of the 'heat affected zone' (HAZ) and to reduced toughness. The
greater the thickness of material, the greater the reduction of toughness
DURABITY : A further important property is that of corrosion prevention.
Although special corrosion resistant steels are available these are not normally
used in building construction. The exception to this is weathering steel .
The most common means of providing corrosion protection to construction steel
is by painting or galvanizing. The type and degree of coating protection required
depends on the degree of exposure, location, design life, etc. In many cases,
under internal dry situations no corrosion protection coatings are required other
than appropriate fire protection.
Other mechanical properties of structural steel that are important to the
designer include:
•
•
•
•
Modulus of elasticity, E = 210,000 N/mm²
Shear modulus, G = E/[2(1 + ν)] N/mm², often taken as 81,000 N/mm²
Poisson's ratio, ν = 0.3
Coefficient of thermal expansion, α = 12 x 10-6/°C (in the ambient
temperature range).
The structural designer arranges and proportions structures and their parts so
that they will satisfactorily support the loads to which they may feasibly be
subjected.
Objective of structural designer
The structural designer must learn to arrange and proportion the parts of his
structures so that they can be practically erected and will have sufficient strength
and reasonable economy. He must be concerned about the safety of the
structure , the cost of materials used and the practicability of fabrication and
erection.
Factor of Safety
The factor of safety of structural member is defined as the ratio of strength of the
member to its maximum anticipated stress.
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Uncertainties Affecting the Safety of Factors
1. Material strength may initially vary appreciably from their assumed values
and they will vary more with time due to creep, corrosion, and fatigue.
2. The method of analysis are often subject to appreciable errors
3. The so called beggaries of nature cause conditions difficult to predict.
4. the stresses produced during fabrication and erection are often severe.
5. The technological changes which affect the magnitude of live loads
6. Although the dead loads of a structure can usually be estimated quite
closely, the estimate of the live load is more inaccurate.
7. Other uncertainties are presence of residual stresses and stress
concentration, variation in dimension of members cross section.
SPECIFICATIONS AND BUILDING CODES
For most structure, the designer is controlled by specifications. Even if he
is not so controlled, he will probably refer to them as guide. Engineering
specifications are developed by various engineering organizations and present
the best opinion of these organizations as to what represent good engineering
practices. Examples of specification governing bodies are:
1. AISC - American Institute of Steel construction
2. ASTM - American Society of Testing Materials
3. AASHTO – American Association of State Highway and
Transportation Officials
4. AWS -American welding Society
5. AREA – American Railway Engineering Association
6. ACI - American Concrete Institute
7. ASEP – Association of Structural Engineers of the Philippines.
which adapted the NSCP (National Structural Code of the
Philippines)
Government Agencies concerned with the safety of public have
established building Codes by which they control construction of various
structures under their jurisdiction.
These codes which are actually ordinances specify:
1. Design Loads
2. Allowable stresses
3. construction type
4. materials quality
Several organizations publish recommended practice for national use. Their
specifications are legally enforceable unless they are embodied in the local
building code, or made part of a particular contract.
Building Codes – reflect historically successful building practices and
communicate to practicing professionals those level of safety considered
acceptable by society.
Specification - represent agreements by individual or groups on standard
practice. It is typically used to develop codes. Quite often, it becomes part of a
building code. It provides essential insight into accepted design and construction
practice in steel and is the single most important resource document for
practicing design professional .It attempt to control production, fabrication and
erection.
TYPE OF LOADINGS
1.Dead load – permanent loads that includes the weight of structure , enclosing
walls, mechanical equipment, interior finishes , etc.
2. Live Load – identified by building codes as lives or transient loads
appropriate for the proposed occupancy.
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STRUCTURAL
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3. Wind Load – regional velocity maps are usually provided in codes along
with those procedures used typically to determine the effect of
building environs on wind intensity and building shape on
distribution on wind pressure. Wind tornadoes and other wind
anomalies are not treated and more precise estimate of wind
effect is often used when severe winds are expected, or the
impact of wind significant from a
design perspective.
4. Seismic Load – a procedure for developing a static elastic loading criteria
intended t model minimum earthquake effects is presented
along with regional intensity map
.
DESIGN METHODS
1. WSD or Allowable Stress Design - Also known as the Elastic Design
Method wherein the working loads are estimated or loads that the
structure may feasibly have to support and proportion the members on the
basis of certain allowable stresses. It is a practice which entails the
designer ensuring that the stresses imposed on the structures owing to
the service load don't exceed the elastic limit.
2. Plastic Design – the theory behind is
the realization of the fact that
steel has ductility to give its reserve strength. In this method, the working
loads are estimated and multiplied by certain load or overcapacity or
safety factors and members are designed on basis of collapse strength.
It is sometimes called limit design or collapse/ design.
3. LRFD or Load Resistance Factor Design – is a design procedure which
combines calculation of ultimate or limit states of strength and
serviceability with a probability based approached to safety.
STEEL SECTIONS
Standard hot-rolled sections
Hot-rolled steel is commonly used to form steel beams and columns on
construction projects. They are created by passing heated steel between large
rollers, which deform it into the required shape, such as; H, I, W, S, C, angles,
tubes, and so on.
Common terminology of steelwork sections include:
•
•
•
•
Flange: The top and bottom horizontal elements of the 'I'. This resists
most of the bending moment experienced.
Web: The vertical element. This resists shear forces.
Root radius: The radius of the weld point at which the flange meets the
web.
Toe radius: The radius of the outside edge of the flange.
The standard sections available for use in structural steelwork are given in:
1. W shape or W-section( wide flange section)
will norm ally
I-Beam,( W flange ) or Universal beam (UB)
An I Beam, also known as an H beam or a universal beam,
has two horizontal elements, the flanges, with a vertical
element as the web. The web is capable of resisting shear
forces, while the horizontal flanges resist most of the beam’s
bending movement. The I shape is very effective at carrying
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STRUCTURAL
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DESIGN
shear and bending loads in the web’s plane. The construction
industry widely uses I beams in a variety of sizes.
Proven to be the most economical beam section and have largely replaced
channels and S sections for beams.
W 200 x 42
Linear weight, w = 42 kg. / m
Depth , d = 205 mm ( approximately )
UB 203 x 133 x 25 – A universal beam of nominal
dimensions 203 mm deep, 133 mm wide, and weight 25 kg/m.
2. American Standard Beam (S-Shaped)
Generally known as an S beam, the American standard beam has a rolled
section with two parallel flanges, all connected by a web. The flanges on Sshaped beams are relatively narrow. it is the first beam section rolled in America.
S 250 x 38 (American Standard beam) Linear weight, w = 38 kg. / m
Depth , d = 250 mm ( approximately )
3. Channel (C-Shaped) or ( Channel section)
Structural C channels, or C beams, also known as parallel flange channels (PFC)
have a C-shaped cross section. Channels have top and bottom flanges, with a
web connecting them. C-shaped beams are cost-effective solutions for short- to
medium-span structures. Channel beams were originally designed for bridges,
but are popular for use in marine piers and other building applications
Channels are sometimes used as beams for light loaded beams such as
purlins. It has a very little resistance to lateral forces and need to be braced.
C 10 x 30
Linear weight, w = 30 lb. / ft.
Depth , d = 10 inches ( approximately )
4. Angle (angle bar)
Angle beams take an L shape, with two legs that come together at a 90-degree
angle. Angle beams come in equal or unequal leg sizes. An unequal leg L beam
may have one leg of 2x2x0.5 and one leg of 6x3x0.5, for example. L beams are
typically used in floor systems because of the reduced structural depth
L 150 x 100 x 6 unequal angle
L 150 x 150 x 6 equal angle
5. Bearing Pile (H-Shaped)
When builders can’t find a structure on a shallow foundation, they use bearing
piles to design a deep foundation system. Bearing piles are H-shaped to
effectively transfer loads through the pile to the tip. Bearing piles work best in
dense soils that offer most resistance at the tip. Individual piles can bear more
than 1,000 tons of weight.
H-Piles are often driven into the ground and used for deep foundations to
support structures in commercial construction, such as buildings and bridges. ...
H-piles are also commonly used for "soldier pile and lagging" construction, in
which steel piles and timber lagging is used to create earth retention
HP 12 x 74 ( bearing pile section) it made wit the regular W rolls but with
thicker webs to provide better resistance to the impact driving
.
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STRUCTURAL
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Linear weight, w = 74 lb. / ft.
Depth , d = 12 inches ( approximately )
6... M 8 x 6.5 ( Miscellaneous section) one of the group of doublysymmetrical
H-shape members which cannot by dimension be classified as W , S,
HP section
Linear weight, w = 6.5 lb/ft.
Depth , d = 8 inches ( approximately )
7. MC 18 x 58 (Miscellaneous channel) –Those cannot be classified as
a C shape by dimension.
Linear weight, w = 58 lb. / ft.
Depth , d = 18 inches ( approximately )
8. A tee beam, or T beam, or WT or Structural Tee is a load-bearing beam
with a T-shaped cross section. The top of this cross section is the flange, with the
vertical web below. Tee beams can withstand large loads but lack the bottom
flange of the I Beam, giving it a disadvantage in some applications
WT 18 x 140 - can be obtained by splitting a W 36 x 280.
9. Solid steel bars these units can be square, rectangular, circular in cross
section can either be a mild steel , stainless steel or aluminum.
Linear weight, w = 140 lb. / ft.
Depth , d = 18 inches ( approximately )
10. Hollow Steel Section (HSS)
HSS is a metal profile that has a hollow, tubular cross section. HSS units can be
square, rectangular, circular, or elliptical. HSS structures are rounded, with
radiuses that are about twice the thickness of the wall. Engineers commonly use
HSS sections in welded steel frames for which units experience loading in
different directions.
11. Open web joist or bar - a type of beam section which is commonly used to
support floor and roof slab. It is a light shop fabricated parallel chord truss and
particularly economical for long and light loads.
12. Steel sheet piles are long structural sections with a vertical interlocking
system that creates a continuous wall.,
Z-shaped sheet piles the single piles
are shaped roughly like a horizontally
stretched Z.
Flat sheet piles are formed in circles and
arcs to create gravity cells.
U sheet piles retain soil and water just like
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Z piles with one important difference:
U piles have the interlock on the neutral axis.
pan shaped cold form sheet piles are much smaller
than most other sheet piles and are only intended
for short, lightly loaded walls.
Pipe
Structural steel pipes are important for a variety of construction applications,
lending strength and stability. Pipes are hollow, cylindrical tubes that come in a
variety of sizes. Engineers often use steel pipes to meet the needs of water, oil,
and gas industry projects. A pipe is a tubular section or hollow cylinder, usually
but not necessarily of circular
Steel pipes are long, hollow tubes that are used for a variety of purposes. They
are produced by two distinct methods which result in either a welded or seamless
pipe.
STRESS - STRAIN
Typical Stress diagram for Mild Steel
Hooke’s Law – in elastic materials deformation , are directly proportional to the
stress or in other words stress is directly proportional to the displacement.
Proportional Limit – the largest stress for which Hooke’s Law applies or the
highest point on the straight line potion of stress-strain diagram.
Elastic limit –The largest stress to which a material can withstand without being
permanently deformed. This term is synonomous with the
proportional limit, that’s the reason that for structural steel it is also
term as proportional elastic limit.
Yield Point – The stress at which there is a decided increase in the elongation
or stain without a corresponding increase in stress.
Elastic Strain - The strain that occurs before yield point.
Plastic Strain . the strain that occurs after yield point.
Strain-Hardening – a range following after the plastic stain where additional
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stress is necessary to produce additional strain.
Necking - a sharp reduction in the cross section of a member taking place
followed by failure.
Elastic Range. Is the line portion in the stress-strain diagram of steel.
THREE MAJOR GROUP CLASSIFICATION OF STEEL
Steel – is an alloy of iron and non metallic element carbon .its properties can be
greatly changed by varying the quantities of carbon present and by adding other
elements such as silicon , nickle, manganese, and copper.
1. CARBON STEEL - have a principal strengthening agents carefully
controlled by the quantities of carbon and manganese.
FOUR CATEGORIES DEPENDING ON CARBON PERCENTAGE
. 1. Low Carbon Steel < 0.15%
2. Mild carbon steel 0.15 –0.29 % ( structural carbon steel falls into this
category)
3 .Medium Carbon steel - 0.30 – 0.59%
4. High Carbon Steel 0.60 – 1.7%
A 36 - the most common structural steel which has a yield stress of 36 Ksi.
A 500 & A501 both applying for tubing
A 529 - pertaining to some thinner plate and shape.
2. HIGH STRENGTH LOW ALLOY STEEL
This steel obtain their higher strengths and other properties by the addition of
one or more alloying agent such as columbium, vanadium, chromium, silicon,
copper nickle and others. Its yield stress range from low 40 ksi to high 70
ksi and have much greater atmospheric corrosion resistance than carbon
steel.
3. QUENCHED AND TEMPERED ALLOY STEEL
These steels have alloying agent in excess of those used in carbon steels
and they are heat treated by quenching and tampering to obtain strong and
tough steels with yield strength in the 80Ksi to 110 ksi. Range.
Load Combination:
LRFD - Load and Resistance Factor Design
γiQi ≤ ϕRn
Qi = load effect (a force or moment)
γi= load factor
Rn =nominal resistance, or strength
Φ = resistance factor
ϕRn = design strength
Ru ≤ ϕRn ← Required strength = sum of factor load effect ( forces of moment)
Load Combination
Combination 1:
Combination 2:
Combination3:
Combination 4:
Combination 5:
Combination 6:
Combination 7:
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1.4D
1.2D + 1.6L + 0.5(Lr or S or R)
1.2D + 1.6(Lr or S or R) + (0.5L or 0.5W)
1.2D + 1.0W + 0.5L + 0.5(Lr or S or R)
1.2D ± 1.0E + L + 0.2S
0.9D ± 1.0W
0.9D ± 1.0E
STRUCTURAL
STEEL
DESIGN
In combination 3 ,4, and 5 , the load factor on L can be reduced to 0.5 if L is no greater than 100
psf, except for garages or places of public assembly.
ASD: Allowable Stress Design ( f
F)
.
= Required strength
= Nominal strength
= safety factor
= 5/3 or 1.67 involving yielding or compression buckling
= 2.00 involving rupture
= allowable strength
Load Combination:
Combination 1:
Combination 2:
Combination3:
Combination 4:
Combination 5:
Combination 6a:
Combination 6b:
Combination 7 and 8:
D
D+L
D + (Lr or S or R)
D + 0.75L + 0.75(Lr or S or R)
D ± (0.6W or 0.7E)
D ± 0.75L + 0.75(0.6W) + 0.75(L r or S or R)
D ± 0.75L + 0.75(0.7E) + 0.75 S
0.6D ± (0.6W or 0.7E)
Relationship between ϕ and
ϕ=
Example:
A compression member is subjected to the following loads:
Dead Load = 109 kips
Floor live load = 46 kips
Roof live load = 19 kips
Snow
= 20 kips
a.
Determine the controlling load combination for LRFD and the corresponding load
factor.
b. If the resistance factor Φ is0.90. what is the required nominal strength?
c. Determine the controlling load combination for ASD and the corresponding service
load factor.
d. If the safety factor is 1.67. what is the required nominal strength based on the
required service load strength?
Combination 1:
1.4D = 1.4 ( 109) = 152.6 kips
Combination 2:
1.2D + 1.6L + 0.5(Lr or S or R)= 1.2(109) + 1.6(46) + 0.5(20) = 214.4
kips
Combination3:
1.2D+1.6(Lr or S or R)+ (0.5L or 0.5W)=1.2(46) + 1.6(20) + 0.5(46)=185.8
kips
Combination 4:
1.2D + 1.0W + 0.5L + 0.5(Lr or S or R)
Combination 5:
1.2D ± 1.0E + L + 0.2S
Combination 6:
0.9D ± 1.0W
Combination 7:
0.9D ± 1.0E
a) The rest of the combination will yield a smaller load ,therefore combination 2 controls
=214.4 kips
b) Ru ≤ ϕRn , 214.4 ≤.0.90Rn
Rn = 238 kips. (required nominal strength)
c.) ASD Load Combination:
Combination 1:
D =109 kips
Combination 2:
D + L = 109 + 46 = 155 kips
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STRUCTURAL
Combination3:
Combination 4:
Combination 5:
Combination 6a:
Combination 6b:
Combination 7 and 8:
STEEL
DESIGN
D + (Lr or S or R) = 109 + 20 = 129 kips
D + 0.75L + 0.75(L r or S or R) =109 +.75(46) +0.75(20)= 158.5 kips
D ± (0.6W or 0.7E)
D ± 0.75L + 0.75(0.6W) + 0.75(Lr or S or R)
D ± 0.75L + 0.75(0.7E) + 0.75 S
0.6D ± (0.6W or 0.7E)
The rest of the combination will yield a smaller load ,therefore combination 4 controls = 158.5
kips
d.) from ASD relationship:
.
.
Rn = 265 kips
TENSION MEMBERS
Tension members are found in bridge and roof trusses , towers , bracing system ,
and in situations where they are used as tie rods. The circular rod is the simples
form of tension member but connecting it to many structure appear to be a little
bit difficult.
Other type of tension members
1. angles
5. Box section
2. double angle
6. wide flange
3. structural tee
7. s - shape
4. Flat bar
8. built- up section
Members consist of more than one section , such as built-up and box section
should be tied together with tie plate ( also called tie bars ) located at various
interval or perforate cover plates serve to hold various pieces in their correct
position.
T
Areqd =
Ft
Allowable Tensile Stresses and loads
a tension member without holes or threads has a tensile strength ;
Fy
T=
Ag = Ft Ag
S.F
When tension member s have holes punch in them for rivets and bolts, then;
T = Ft An
Although the maximum strength of a member is often controlled by yielding
strength at its net section sometimes the gross section away from the holes may yield
before the ultimate stress, Fu , is reached. This means that the actual failure may occur
when the yield stress is reached on the gross section.
So AISC specification provides that;
Ft = .60 Fy
or .50 Fu
T=0.60Fy Ag
or
0.50FUAe
For A36 steel , Fu = 58 ksi
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Net Areas
The term “ net cross-sectional area “ or simply “net area” refers to the
gross cross-sectional area of a member minus any holes, notche, or other
indentation. Structural steels to be fabricated which is connected with rivets or
bolts the hole is usually punched 1/8” (3 mm) larger than diameter of the rivet or
bolt. The connection of tension members should be arranged so that no
eccentricity is present.
.Example:
In the figure shown, determine the net
area of the 3/8 in. plate. The plate is
connected at its end with two lines of ¾
inches bolt.
An =(
3 1 3
3
)(8) –(2)( + ) ( ) = 2.34 in2
4 8 8
8
Effect of staggered holes
Should there be more than row of bolt or rivet holes in a member it is
usually desirable to stagger them in order to provide as large a net area as
possible at any one section to resist the load.
To determine the critical net area considering the effect of the zigzag
s2
Section using the expression
where s is the longitudinal spacing (or
4g
pitch) of any hole and g is the traverse spacing
Example 1:
Determine the critical net area of the ½ -in thick plate
Shown, using the AISC specification . The holes are
for ¾ in bolt.
Solution
The net width for each case.
7
ABCD = 11 – (2)   = 9.25 in
8
 7   32 
ABCEF = 11 – (3)   +
= 9.125 in (controls)
 8   ( 4)(3) 
 7   32 
ABEF = 11 – (2)   + 
= 9.625 in
 8   ( 4)(6) 
An = (9.125)( 1/2)=4.56 in2
Example 2.
Determine the pitch which will give a net
area DEFG equal to the one along
ABC. The holes is for 3/4 in bolt.
Solution:
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7
ABC = 6 – (1)   = 5.125 in
8
 s2 
 7   s2 
DEFG = 6 –(2)   +
=4.25+  
8
 8   ( 4)(2) 
 
But ;
ABC = DEFG
5.125 =
s2
8
;
s =2.65 in
s2
rule is merely an approximation or simplification of the
4g
complex stress
variation which occurs in members with staggered
arrangement of bolts and rivets. Steel specifications can only provide minimum
standards and designers will have to logically apply such information to
complicated situation which the specification could not cover in their attempts at
brevity and simplicity.
The
Example 3:
Determine the net area of the W 12 x 16
Shown in the figure,
assuming the bolt holes are for
1-in bolts . Use AISC specification.
Solution:
From AISC handbook;
A = 4.71 in2
tw=0.220 in
d= 12 in
 1
ABDE = 4.71 – (2)  1  (0.220) = 4.21 in2
 8
 22 
 1
 (0.220) = 4.11 in2
ABCDE = 4.71 – (3)  1  (0.220)+(2) 


 8
 ( 4)(3) 
Example 4:
Determine the net area along the route
ABCDEF for the C 15 x 33.9 shown.
Holes are for 3/4 bolts.
Solution:
From the AISC handbook, the
Properties of channel C 15 x 33.9 are:
d= 15” , tf= 0.65” bf = 3.6 “, tw= 0.40 , A= 9.96 in2
7
7
App. Net area ABCDEF = 9.96- (2)   (0.65 )- (2)   (0.40)+
8
8
 32   0.65 + 0.40

+(2) 
 ( 4)( 4.6)  
2


 32 

 (0.40)
 ( 4)(9) 



 = 8.736 in2


EFFECTIVE NET STRESS
When a member other than a flat plate or bar is loaded in axial tension
until failure occurs across its net section its actual tensile failure stress will
probably be less than the coupon tensile strength of steel. This will normally be
the case unless all the various element which make up the section are connected
so stress is transferred uniformly across the section. The reason for the reduced
Engr. Joselito B. Padayhag
Licensed Civil Engineer
Licensed Master Plumber
STRUCTURAL
STEEL
DESIGN
strength of the member is the concentration of shear stress, called shear slag.
in the vicinity of the connection. In such a situation the flow of tensile stress
between the full member cross section and the smaller connected cross section
is not 100 % effective. As a result the AISC (1.14.2.2) specifies that the effective
net area , Ae., of such a member is to be determined by multiplying its net area,
An, by a reduction coefficient, Ct , as follows.
Ae= CtAn
Effective Net Area
Type of members
Minimum
Number
of
Fasteners
Per line
Special
requirement
Effective
Net Area
Ae
(a) full length tension
members having all cross-sectional elements connected to transmit
the tensile force
1
None
An
(b) Short tension member fittings, such as splice plates, gusset plates, or
beam-to column fittings
1
None
(c ) W , M , or rolled
shapes
3
flangewidth
2

sec tiondepth 3
Connection is to
flange or flanges
An
But not to
exceeding
0.85Ag
0;90 An
(d) structural tees cut from sections meeting requirements of (c) above
(e) W , M,
or S shapes
not meeting
the
conditions of
(c ) , and
other
shapes,
including
built-up
sections
,
having
unconnected
segments
not in the
plane of the
loading
(f) All shapes in (c) ,(d) or (e)
3
Connection is to
flange
or flanges
0;90 An
3
none
0;85 An
2
none
0;75 An
Example :
Determine the allowable tensile load a W 10 x 45 with two lines of ¾ in bolts in
each flange can support using A36 steel and the AISC specification. There assumed to
be at least three bolts in each line and the bolts are not staggered with respect to each
other.
Solution:
From AISC manual, for W 10 x 45 ( Ag= 13.3 in2 , d= 10.10 in, bf= 8.02 in., tf=
0.620 in )
(a) T = 0.60FyAg = (0.60 ) (36) (13.3) = 287.3 kips
b) An = 13.3 –(4)(7/8) (0.620) = 11.13 in2
Ct = 0.90 since bf > 2/3 d
Ae= Ct An = (0.90 ) 11.13) =10.02 in2
T = 0.50 FuAe = (0.50) ( 58 ksi)(10.02 in2) = 290.6 kips
Allowable T = 287.3 kips
DESIGN OF TENSION MEMBERS
The designer has to consider in the design of tension member the following
properties:
1.compactness
2. dimension that fits into structure with reasonable relation to the
Engr. Joselito B. Padayhag
Licensed Civil Engineer
Licensed Master Plumber
STRUCTURAL
STEEL
DESIGN
dimension of the member
3. connection as to many parts of the section as possible to minimize
slag.
Slenderness ratio of a member – is the ratio of its unsupported length to its least radius
of gyration.
AISC recommendation for slenderness ratio;
For main members = 240
For bracing and other secondary members = 300
AASHTO recommendation for slenderness ratio
For main members = 220
For bracing and other secondary members = 240
Steps in the designing of tension members:
1. Satisfy minimum gross area that must be at least equal to the following.
T
min A g =
0.60Fy
2. Satisfy the min value of Ae that must be at least
T
min Ae =
0.50Fu
And since Ae = Ct An the minimum value of An is
min A e
T
min An =
=
0.50Fu C t
Ct
min Ag =
T
+ estimated hole areas
0.50Fu C t
l
r
determine the trial, slenderness ratio, =
actual un sup ported length
recommende d slendernes s ratio
3. From that min Ag , and slenderness ratio , find a section you desired from
the manual that has an area not less than computed Ag , and determine
the other important properties of the selected section.
Example :
Select a W 12 section to resist a 390 k tensile load using A36 steel and the AISC
specification. The member is to be 30 ft. long and is to be connected through its
flanges only with at least three 7/8 in. bolts in each line. Assume that there can
be as many as four bolt holes at any cross- section ( two in each flange.).
Solution:
(a) min A g =
T
390
=
= 18.06 in2 (11,652 mm2 )
(
0
.
60
)(36)
0.60Fy
(b) Ct = 0.90 from the table of effective net area
Assume flanges are about 5/8 in (15.9 mm) thick after some study of
the W 12 section in the AISC manual
T
min Ag =
+ estimated hole areas
0.50Fu C t
=
c. min r=
Engr. Joselito B. Padayhag
Licensed Civil Engineer
Licensed Master Plumber
390
5
+ (4)(1.0)  = 17.44 in2 (11,252 mm2)
(0.50)(58)(0.90)
8
l
30x12l
=
=1.50 in (38.1 mm)
240
240
STRUCTURAL
STEEL
DESIGN
Axis - X
Axis - Y
Designation
W
kg/m
A
mm2
d
mm
tw
mm
bf
mm
tf
mm
rt
mm
Ix
×103 mm4
Sx
×103 mm3
rx
mm
Iy
×103 mm4
Sy
×103 mm3
W 12 × 79
117.74
14,968
314.50
11.90
306.80
18.70
84.07
275,545
1,752
135.68
89,906
586
77.50
W 12 × 72
107.08
13,613
311.20
10.90
305.80
17.00
83.57
248,490
1,597
135.11
81,165
531
77.22
W 12 × 65
96.93
12,323
307.80
9.90
304.80
15.40
83.31
221,851
1,442
134.18
72,424
475
76.66
W 12 × 58
86.27
10,968
309.60
9.10
254.30
16.30
69.09
197,710
1,277
134.26
44,537
350
63.72
W 12 × 53
79.16
10,064
306.30
8.80
253.90
14.60
68.83
176,898
1,155
132.58
39,875
314
62.95
W 12 × 50
74.60
9,484
309.60
9.40
205.20
16.30
55.12
163,995
1,059
131.50
23,434
228
49.71
W 12 × 45
66.99
8,516
306.30
8.50
204.30
14.60
54.61
145,681
951
130.79
20,812
204
49.44
W 12 × 40
59.88
7,613
303.30
7.50
203.30
13.10
54.36
129,032
851
130.19
18,356
181
49.10
Try W12 x 65( A=19.1 in2 , tf=.605 in , ry = 3.02 in )
Try W12 x 65 ( A=12,323 mm2 , tf=.15.4 mm , ry = 83.81 mm )
Investigation:
T= (0.60)(36)(19.1) = 412.6 k > 390 k
OK
Ae = CtAn =0.90[19.1-4(1.0)(0.605)] = 15.01 in2
T= (0.50)(58)(15.01) = 435.3 k > 390 k
l
r
=
30x12l
= 119 < 240
3.02
OK
OK
Example2:
Design a 9 ft single angle tension member to support a total tensile load of 65.8
k. The member is to be connected to one leg only with 7/8 in bolts ( at least three
in a line) Assume that only one bolt is located at any one cross section . Use the
AISC specification and A 36 Steel.
(a) min A g =
T
65.8
=
= 3.05 in2
(
0
.
60
)(
36
)
0.60Fy
b.) Ct = 0.85 from the table of effective net area
T
min An =
+ estimated hole areas
0.50Fu C t
= 2.67 in2
min An =
c.) min r=
=
=0.45
From the manual ,the lightest angle that has an area of
A=3.05 is L5 x 3 ½ x 3/8 which has a least r =0.762 > 0.45.
Example of built up tension member.
Two C 12 x 30s 12 inches back to back as shown, have been
selected to support a 280 kips tensile load. The member is 30 ft
long ,consist of A 36 steel, and has one 7/8 in bolt through each
channel flange. Using the AISC specification and assuming that
Ct = 0.85 determine if the member is satisfactory.
Solution:
Engr. Joselito B. Padayhag
Licensed Civil Engineer
Licensed Master Plumber
STRUCTURAL
STEEL
DESIGN
Using C 12 x 30s (A= 8.82 in2, tf =0.501 in.)
a) T= 0.60 Fy Ag = (0.60) (36)(2 x 8.82) = 381.0 k > 280 OK
b) Ct = 0.85
An =(2) (8.82) –4(1.0)(.501) =15.64 in2
T = 0.50Fu Ae + (0.50) (58) (.85)(15.64) = 385.5 k>280 OK
Slenderness ratio:
Ix = (2) (162) = 324 in4
Iy = (2)(5.14)+(2)(8.82)(5.33)2 =511 in4
324
= 4.29 in
17.64
l (30x12)
=
=83.9 < 240
r
4.29
rx =
OK
RIVETED / BOLTED CONNECTION
SHEAR FAILURE
a.) Shear failure of the bolt
Ss =
Ab = cross-sectional area of the rivets or bolt
Ss = Shearing stress of the rivets or bolt
T = applied load parallel to the area of the bolt.
b.) Tear of plate section through rivet/ bolt holes.
T = Ft An
Efficiency =
1. ) A tension member is made up of 12 mm thick steel plate 250 mm wide by a
lap joint made with three rows of 25 mm bolts snug fit in drilled holes and arrange
in the pattern shown. The permissible stresses in the bolts and plates are 138
MPa for tension in the net section, 110 MPa for shear for shear and 220 MPa
for bearing.
a) What is the permissible tension on the plate
b. What is the efficiency of the joint in the plate.
Engr. Joselito B. Padayhag
Licensed Civil Engineer
Licensed Master Plumber
STRUCTURAL
Engr. Joselito B. Padayhag
Licensed Civil Engineer
Licensed Master Plumber
STEEL
DESIGN