Structural

My Notes
Engineers Design of Steel Structures
CONCEPTS IN STRUCTURAL STEEL DESIGN
Introduction

The structural design of buildings, whether of structural steel or reinforced concrete, requires the
determination of the overall proportions and dimensions of the supporting framework and the
selection of the cross-sections of individual members. The engineer and architect will collaborate
throughout the design process to complete the project in an efficient manner. The architect decides
how the building should look; the engineer must make sure that it doesn’t fall down. The first priority
of the structural engineer is safety; which follows the serviceability and economy. An economical
structure requires an efficient use of materials and construction labor.

A good design requires the evaluation of several framing plans which consists of different
arrangements of members and their connections.

Structural Design may be defined as a mixture of art and science, combining the experienced
engineer’s intuitive feeling for the behavior of structure with a sound knowledge of the principles of
statics, dynamics, mechanics of materials, and structural analysis, to produce a safe, economical
structure that will serve its intended purpose.

Design is a process by which an optimum solution is obtained.

Purpose of Structural Analysis

Structural analysis is the process of determining the internal forces and deformations of structural
elements in a structure due to specified loads so that each member can be designed rationally or
the state of safety of existing structures can be checked. It is the foundation of structural
engineering works that the strength of whole structure depends. Without proper structural analysis,
the structure might experience deformations even in the small degree of loads or uneconomical
due to over sizing of members.

In the design of structures, it is necessary to start with a concept leading to a configuration which
can be analyzed. This is done so members can be sized and the needed reinforcement determined,
in order to: (a) carry the design loads without excessive deformations (serviceability or working
condition); and (b) prevent collapse before a specified overload has been placed on the structure
(safety or ultimate condition).

Stages of Structural Engineering

Most of structural engineering projects are consists of repetitive stages namely modeling,
analysis/design, and interpretation of the result of analysis and design.
Modeling involves the creation of a structural model that exhibits the characteristics of actual
structure. As much as possible, the model shall consists of all structural components of the actual
structure, from frame element (beam and column) to plate and shell (wall, floor and folded plates),
diaphragms and all rigid links shall be provided if necessary and all none-structural elements shall
be regarded as loads.
It is proper for the structure to assume the actual condition of support. The support might be pinned,
roller, fixed or other conditions necessary to simulate the actual behavior of the structure. Due to
improper application of support conditions, structure are experiencing deformations that leads
usually to unprecedented damage if not total collapse.

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Sectional properties (Ax, Ix, Iy and Iz) and material constants (E, density, Poisson’s Ratio, etc) are
also important for analysis of the structure. Thus, assumed properties shall be established prior to
the analysis of the structure. This will involve experience on the part of the analyst to minimize time
and effort.
All loads from simple joint, member and element loads to complex such as seismic, wind or moving
loads shall be carefully incorporated to simulate the actual behavior of the structure. Special
treatment shall be emphasized on live loads for pattern and/or checkered loading.
Consider the analysis of a simple beam. Usually, we are encountering a simple problem as such.
In order to visualize the problem, we are usually having simple sketch to represent the actual
condition. The sketch includes the length of the member, loading (whether uniform or concentrated
force) and support condition. The sectional properties of this beam are first assumed in order to
compute for the weight of the structure and also for the analysis to be performed. Thus, a simple
model has been established!
As soon as the model is completed, analysis comes next. The behavior of the structure under
loading condition shall be investigated. Analysis might be linear static or dynamic depending on
code or structural requirement.
Under this analysis, a force or moment envelope will be created for design purposes. The structural
components shall be design for different loading conditions. A designer cannot pick the maximum
value in each loading condition and altogether use it for design. Such situation will not occur.
After the design is completed, a final analysis shall be made and the actual behavior is observed.
If the behavior does not conform to the existing code requirements such as deflection, some
adjustment shall be made until its behavior becomes acceptable.

Criteria for optimum design:

 Minimum cost,
 Minimum weight,
 Minimum construction time,
 Minimum labor,
 Minimum cost of manufacture of owner’s product, and
 Maximum efficiency of operation of owner

Design Procedure

The design procedure may be considered to be composed of two parts – functional design and
structural framework design.

Functional Design Structural Framework Design

Ensures that the intended results are achieves The selection of the arrangement and sizes of
such as (a) adequate working areas and structural elements so that the service loads
clearances; (b) proper ventilation and/or air may be safely carried, and displacements are
conditioning; (c) adequate transportation within acceptable limits.
facilities, such as elevators, stairways, and
cranes or material handling equipment; (d)
adequate lighting; and (e) aesthetics.

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Outline of Iterative Design Procedure

1. Planning. Establishment of the functions which the structure must serve. Set criteria
against which to measure the resulting design for being an optimum.
2. Preliminary structural configuration. Arrangement of the elements to serve the functions in
step 1.
3. Establishment of the loads to be carried.
4. Preliminary member selection. Based on the decisions of step 1, 2, and 3, selection of
member sizes to satisfy an objective criterion, such as least weight or cost.
5. Analysis. Structural analysis involving modeling the loads and the structural framework to
obtain internal forces and any desired deflections.
6. Evaluation. Are all strength and serviceability requirements satisfied and is the result
optimal? Compared the result with predetermined criteria.
7. Redesign. Repetition of any part of the sequence 1 through 6 found necessary or desirable
as a result of evaluation.
8. Final decision. The determination of whether or not an optimum design has been achieved.

Loads

The determination of the loads on structure or structural elements will be subjected is an estimate.
Even if the loads are well known at one location of structure, the distribution of load from element
to element throughout the structure usually requires assumptions and approximations.

 Dead loads – a fixed-position gravity load. The weight of the structure is considered dead
load, are as attachments to the structure.

 Live loads – gravity loads acting during the service of the structure, but varying in
magnitude and location. Examples are human occupants, furniture, movable equipment,
vehicles, and stored goods.

 Wind loads – exerts as a pressure or suction on the exterior surfaces of a building. Because
of the relative complexity of determining wind loads, wind is usually considered a separate
category of loading. Because lateral loads are most detrimental to tall structures, wind
loads are usually not as important for low buildings, but uplift on light roof systems can be
critical.

 Earthquake loads – another special category and need to be considered only in those
geographic locations where there is a reasonable probability of occurrence. A structural
analysis of the effects of an earthquake requires an analysis of the structure’s response to
the ground motion produced by the earthquake. Similar methods are sometimes used in
which the effects of the earthquake are simulated by a system of horizontal loads, similar
to those resulting from wind pressure, acting at each floor level of the building.

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STRUCTURAL STEEL PROPERTIES

Steel Shapes

In the design process outline above, one of the objectives is the selection of the appropriate cross
sections for the individual members of the structure being designed. Most often, this selection will
involve choosing a standard cross-sections shape that is widely available rather than requiring
fabrication of a shape with unique dimensions and properties.

Figure 1.1 – Hot Rolled Shapes

It is almost always economical to choose rolled shape sections. In this manufacturing process (hot-
rolling) molten steel is taken from an electric furnace and poured into a continuous casting system
and allowed to pass through a series of rollers that squeeze the material into a desired cross-
sectional shape. Figure 1.1 shows commonly used cross sections.

The W-shape (also called wide-flange shape) consists of two parallel flanges separated by a single
web. It has two axes of symmetry. A typical designation would be “W18x50”, where W indicates
the type of shape, 18 is the nominal depth parallel to the web, and 50 is the weight in pounds per
foot of length.

S-shape like W-shape has two parallel flanges, a single web and two axes of symmetry. The flange
of the S-shape is narrow compared with W-shape. Its inside face of the flanges slope with respect
to the outside faces. An example of its designation is “S18x70”, with the S indicating the type of

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shape, and the two numbers giving the depth in inches and the weight in pounds per foot. This
shape was formerly called an I-beam.

The angle shapes are available in either equal-leg or unequal-leg. A typical designation would be
“L6x6x¾” being the L indicating the type of shape, the next two numbers are the length of the two
legs measured from the corner, or heel, to the toe at the other end of the leg, and the thickness
which is the same for both legs. For unequal-leg angle, the longer leg dimension is always given
first.

C-shape has two flanges and a web, with one axis of symmetry. It carries a designation such as
“C9x20”. Similar with W- and S- shapes, the first number is the total depth in inches parallels to
the web and the second number the weight in pounds per linear foot. Like the S- shape, the
inside face of the flanges are sloping.

The Structural Tee is produced by splitting an I-shaped members at middepth (sometimes called
Split-tee). The prefix of the designation is either WT, ST, or MT, depending on which shape is the
“parent”. For example, WT18x105 has a nominal depth of 18 inches and a weight of 105 pounds
per foot, and is cut from a W36x210.

The M-shape (miscellaneous shapes) has two parallel flanges and a web, but it does not fit
exactly into either the W or S categories. The HP shape, used for bearing piles, has parallel
flange surfaces, approximately the same width and depth and equal flange and web thicknesses.

Figure 1.2 Other Shapes

Other frequently used cross-sectional shapes are shown in Figure 1.2. Bars can have circular,
square, or rectangular cross sections. If the width of a rectangular shape is 8 inches (200 mm) or
less, it is classified as a bar. If the width is more than 8 inches, the shape is classified as plate.
The usual designation for both is the abbreviation PL followed by the thickness in inches, the
width in inches, and the length in feet and inches; ex. PL 3/8 x 5 x 3’-2 ½ “. Plates and bars are
available in increments of 1/16 inch.

Hollow shapes are produced either by bending plate material into desired shape and welding the
seam or by hot-working to produce a seamless shape. The shapes are categorized as steel pipe,
round HSS, and square HSS (Hollow Structural Sections).

Round HSS are designated by outer diameter and wall thickness, expressed to three decimal
places; ex. HSS 8.625 x 0.250. Square and rectangular HSS are designated by nominal outside
dimensions and wall thickness, expressed in rational numbers; ex. HSS 7 x 5 x 3/8.

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Steel pipe is available as standard, extra-strong, or double-extra-strong, with designation such as
Pipe 5 Std., Pipe 5 x-strong, or Pipe 5 xx-strong, where 5 is the nominal outer diameter in inches.
The different strengths correspond to different wall thickness for the same outer diameter. For
pipes whose pipes whose thicknesses do not match those in the standard, extra-strong, or
double-extra-strong categories, the designation is the outer diameter and wall thickness in inches,
expressed to three decimal places; ex., Pipe 5.563 x 0.500.

Figure 1.3 Built-up Shapes

Sometimes a standard shape is augmented by additional cross-sectional elements, as when a

cover plate is welded to one or both flanges of a W-shape (see Figure 1.3). Building up sections
is an effective way of strengthening an existing structure that is rehabilitated or modified for some
use other than the one for which it was designed. Sometimes a built-up shape must be used
because none of the standard rolled shapes are large enough. These can be I-shaped sections,
with two flanges and a web, or box sections, with two flanges and two webs. The components
can be welded together and can be designed to have exactly the properties needed. Built-up
shapes can also be created by attaching two or more standard rolled shapes to each other, such
as a pair of angles placed back-to-back and connected at intervals along their length. This is
called double-angle shape.

Strength of Various Steel

Property A36 A572 Gr. 50 A992

Yield Point (Fy), min. 36 ksi 50 ksi 50 ksi

(250 MPa) (340 MPa) (340 MPa)

Tensile Strength (Fu), min. 58 to 80 ksi 65 ksi 65 ksi

(400 – 550 MPa) (450 MPa) (450 MPa)

Yield to tensile ratio, max. --- --- 0.85

Elongation in 8 in., min. 20% 18% 18%

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Design Philosophies

The design of structural members involves the selection of a cross section that will safely and
economically resist the applied loads.

The fundamental requirement of structural design is that the required strength not exceeds the
available strength.

Required strength ≤ available strength

In allowable strength design (ASD) (NSCP 2015 Section 502.3.4), a member is selected that has
cross-sectional properties such as area and moment of inertia that are large enough to prevent
the maximum applied axial force, shear or bending moment from exceeding an allowable, or
permissible value. This value is obtained by dividing the nominal, or theoretical, strength by a
factor of safety. This can be expressed as

Required strength ≤ allowable strength Eq. 1

where
𝑛𝑜𝑚𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ
𝐴𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ =
𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟

If stresses are used instead of forces or moments the relationship of Equation 1 becomes

Maximum applied stress ≤ allowable stress Eq. 2

This approach is called allowable stress design or elastic design or working stress design. The
allowable stress will be in elastic range of material. Working stresses are those resulting from the
working or service loads.

Plastic design is based on a consideration of failure conditions rather than working load
conditions. A member is selected by using the criterion that the structure will fail at a load
substantially higher than the working load. The term plastic is used because, at failure, parts of
the member will be subjected to very large strains – large enough to put the member into the
plastic range. When the entire cross section becomes plastic at enough locations, “plastic hinges”
will form at those locations, creating collapse mechanism. As the actual loads will be less than the
failure loads by a factor of safety known as the load factor, member designed this way are safe
despite being designed based on what happens at failure. This procedure is roughly as follows:

1. Multiply the working loads (service loads) by the load factor to obtain the failure load.
2. Determine the cross-sectional properties needed to resist failure under these loads.
3. Select the lightest cross-sectional shape that has these properties.

Load and resistance factor design (LRFD) (NSCP 2015 Section 502.3.3) is similar to plastic
design in that strength, or the failure condition, is considered. Load factors are applied to the
service loads, and a member is selected that will have enough strength to resist the factored
loads. In addition, the theoretical strength of the member is reduced by the application of a
resistance factor. The criterion that must be satisfied in the selection of a member is

Factored load ≤ factored strength Eq. 3

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Why LRFD?
By Lynn S. Beedle
Modern Steel Construction, AISC, 26, 4(4th Quarter 1986), 30-31

 LRFD is another “tool” for structural engineers to use in steel design. Why not
have the same tools (variable overload factors and resistance factors) available for
steel design as are available for concrete design?
 Adoption of LRFD is not mandatory but provides a flexibility of options to the
designer. The marketplace will dictate whether or not LRFD will become the sole
method.
 ASD is an approximate way to account for what LRFD does in a more rational way.
The use of plastic design concepts in ASD has made ASD such that it no longer may
be called an "elastic design" method.
 The rationality of LRFD has always been attractive, and becomes an incentive
permitting the better and more economical use of material for some load
combinations and structural configurations. It will also likely lead to having safer
structures in view of the arbitrary practice under ASD of combining dead and live
loads and treating them the same.
 Using multiple load factor combinations should lead to economy.
 LRFD will facilitate the input of new information on loads and load variations as
such information becomes available. Considerable knowledge of the resistance of
steel structures is available. On the other hand, our knowledge of loads and their
variation is much less. Separating the loading from the resistance allows one to be
changed without the other if that should be desired.
 Changes in overload factors and resistance factors φ are much easier to make than
to change the allowable stress in ASD.
 LRFD makes design in all materials more compatible. The variability of loads is
actually unrelated to the material used in the design. Future specifications not in
the limit states format for any material will put that material at a disadvantage in
design.
 LRFD provides the framework to handle unusual loads that may not be covered by
the Specification. The design may have uncertainty relating to the resistance of the
structure, in which case the resistance factors may be modified. On the other hand,
the uncertainty may relate to the loads and different overload factors may be used.
 Future adjustments in the calibration of the method can be made without much
complication. Calibration for LRFD was done for an average situation but might be
adjusted in the future.
 Economy is likely to result for low live load to dead load ratios. For high live load to
dead load ratios there will be slightly greater costs.
 Safer structures may result under LRFD because the method should lead to a better
awareness of structural behavior.
 Design practice is still at the beginning with regard to serviceability limit states;
however, at least LRFD provides the approach.

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Design of Tension Members

Tension Members
 Structural members subjected to axial tensile forces.

 Proportioning of tension members is among the simpler problems that face the structural
engineer but requires great care in the design and detailing of their connections.
 It does not have the inherent stability problems of beams and columns. Thus, do not
generally require bracing.
 The resulting tension member structures are less redundant, and the potential for sudden
failure exists if any inadequacy is present, such as weakness in a connection.

Examples of Tension Members:

 Hangers for catwalks and storage bins,
 Truss web and chord members,
 Cables for direct support of roofs and canopies,
 Sag rods,
 Tie rods and
 Various type of braces

In the determination of the available strength of a tension member the following are needed:

Gross Area Ag – needed for tensile yielding limit state. It is the original, unaltered cross-
sectional area of the member.

Effective Net Area, Ae – needed for tensile rupture limit state. It is the cross-sectional area
available to be stressed in tension (An) multiplied by reduction coefficient U.

Slenderness Limitation

There is no maximum slenderness limit for design of member in tension.

For members designed on the basis of tension, the slenderness ratio

L/r preferably should not exceed 300. This suggestion does not apply
to rods or hangers in tension.

Tension Member Analysis

 The direct stress formula is the basis for tension members analysis (and design).

𝑃
𝑓 =
𝐴

Or for tensile capacity

𝑃 = 𝐹𝐴

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Where
ft computed tensile stress
P applied axial load
Pt axial tensile load capacity (or maximum allowable axial tensile load)
Ft allowable axial tensile stress
A cross-sectional area of axially loaded tension members

Tensile Strength

Mode of Failure of Tension Members:

Prevention

 Excessive deformation (initiated by Load on GROSS SECTION must be small enough

yielding) that the stress in gross section is less than the
yield strength Fy.

Pn = FyAg

 Fracture The stress on the NET SECTION must be less

than the tensile strength Fu.

Pn = FuAe

The design tensile strength, ΦtPn, and the allowable tensile strength, Pn/Ωt tension members, shall
be the lower value obtained according to the limit states of tensile yielding in the gross section and
tensile rupture in the net section.

LRFD ASD
(Load and Resistance Factor Design) (Allowable Strength Design)

Factored Tensile strength is compared to Total service load is compared to the

the design strength allowable strength

For Tension
Members 𝑃𝑢 = ∅ 𝑃 𝑃
𝑃 ≤
Ω
Where:
For yielding ∅ = 0.9 Ω = 1.67
For fracture ∅ = 0.75 Ω = 2.0

Thus:
𝑃𝑢 = 0.9𝐹 𝐴 𝑃𝑛 = 0.6𝐹 𝐴
𝑃𝑢 = 0.75𝐹 𝐴 𝑃𝑛 = 0.5𝐹 𝐴

Sect 510.4.1 NSCP C101-15, 7th Ed.

where

Ae = effective net area, mm2

Ag = gross section area of member, mm2
Fy = specified minimum yield stress of the type of steel being used, MPa
Fu = specified minimum tensile strength of the type of steel being used, MPa

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When members without holes are fully connected by welds, the effective net area used in
equation 1-2 shall be as defined in section 504.3. When holes are present in a member, with
welded end connections, or at the welded connection in the case of plug or slot welds, the
effective net area through the holes shall be used in equation 1-2

Area Determination

Gross Area, Ag, of a member is the total cross sectional area.

Net Area, An, of a member is the sum of the products of the thickness and the net width of each
element computed as follows:

In computing net area for tension and shear, the width of a bolt hole shall be taken as 1/16 in.
(1.6 mm) greater than the nominal dimension of the hole.

Comment:
The exact amount of area to be deducted from the gross area to
account for presence of bolt holes depends on the fabrication
procedure. The usual practice is to drill or punch standard holes (not
oversized) with a diameter 1/16 inch (1.6 mm) larger than the fastener
diameter. To account for possible roughness around the edges of the
hole, Section B4.3 of the AISC Specification requires the addition of
1/16 inch (1.6 mm) to the actual hole diameter. This amounts to using
an effective hole diameter 1/8 inch (3.2 mm) larger than the fastener
diameter.

For a chain of holes extending across a part in any diagonal or zigzag line, the net width of
the part shall be obtained by deducting from the gross width the sum of the diameters or slot
dimensions as provided in Section 510.3.2 (Section J3.2 of AISC), of all holes in the chain,
and adding, for each gage space in the chain, quantity s2/4g

where

s longitudinal center-to-center spacing (pitch) of any consecutive holes, mm.

g transverse center-to-center spacing (gage) between fastener gage lines, mm.

For angles, the gage for holes in opposite adjacent legs shall be the sum of gages from the
back of the angles less the thickness.

For slotted HSS welded to a gusset plate, the net area, An, is the gross area minus the
product of the thickness and the total width of material that is removed to form the slot. In
determining the net area across plug or slot welds, the weld metal shall not be considered as
adding to the net area.

Section 510.4.1(b) limits An to a maximum of 0.85Ag for splice plates

with holes.

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Nominal Hole Dimensions, mm (Sect 510.3.2)

Net Area, An Determination

Example 1

A ½ x 5 plate of A36 steel is used as a tension member. It is connected to a gusset plate

with four 5/8-inch-diameter bolts as shown in figure. Determine the actual net area An.

Solution:

Ag = 5(1/2) = 2.5 in2

An = Ag – Aholes = 2.5 – (1/2)(3/4)(2)

An = 1.75 in2

*** End of Example 1 ***

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Example 2

A tension member is made up of 10 mm thick plates, 225 mm wide by a lap joint made
with three columns of M20 bolts snug fit in drilled holes and arranged in the pattern
shown. Determine the least net section area, An.

Solution:

Ag = 10(225) = 2250 mm2

An = Ag – Aholes

Hole diameter for M20 bolts (oversize) = 24 mm

Along column 1 and 3

An = 2250 – 2(24) = 2202 mm2

Along column 2

An = 2250 – 3(24) = 2178 mm2

Least net section area, An = 2178 mm2

*** End of Example 2 ***

Example 3

Redo Example 2 if hole is punched out.

Solution:

Hole diameter for M20 bolts (standard) = 22 + 3 = 25 mm

Along column 1 and 3

An = 2250 – 2(25) = 2200 mm2

Along column 2

An = 2250 – 3(25) = 2175 mm2

Least net section area, An = 2175 mm2

*** End of Example 3 ***

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Effects of Staggered Holes

Whenever there is more than one hole and the holes are not lined up transverse to the
loading direction, more than one potential failure line may exist. The critical path is the path
that has the minimum net area.

CP: A-B CP: A-B / A-C

Length of Paths CP:

CP A-B = Length of (A-B) – (width of hole + Allowance)

CP A-C = Length of (A-B) – 2(width of hole + Allowance) +

Example 4

Determine the minimum net area of the plate shown, assume 24 mm-dia. bolt.

Solution:

Path A-D (2 holes):

[300 – 2(24 + 3)] = 246 mm

Path A-B-D (3 holes; 2 staggers):

[300 – 3(24 + 3) + + ] = 239.5 mm

( ) ( )

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Path A-B-C (3 holes; 2staggers)

[300 – 3(24 + 3) + + ] = 237.4 mm (controls)

( ) ( )

Therefore, CP = 237.4 mm

An = 237.4(12) = 2848.8 mm2

*** End of Example 4 ***

Gage Distance
 transverse center-to-center spacing between fastener gage lines, mm.

g = ga –t/2 + gb – t/2 = ga + gb – t

ga , gb = standard gage distance depending on the length of the leg and the
number of lines of hole

Usual Gages for Angles

in 8 7 6 5 4 3.5 3 2.5 2 1.75 1.5 1.375 1.25 1

Leg
mm 203 178 152 127 102 89 76 64 51 44 38 35 32 25
in 4.5 4 3.5 3 2.5 2 1.75 1.375 1.125 1 0.875 0.875 0.75 0.625
g
mm 114 102 89 76 64 51 44 35 29 25 22 22 19 16
in 3 2.5 2.25 2
g1
mm 76 64 57 51
in 3 3 2.5 1.75
g2
mm 76 76 64 44

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Example 5

Determine the net area An for the angle shown, assume 20 mm-dia. hole.

Solution

An = Ag - ∑Dt + ∑ t

Flattened angle

C’

Path AC:
(152 + 102 – 12) – 2(20 + 3) = 196 mm

Path ABC:

(152 + 102 – 12) – 3(20 + 3) + + = 195.3 mm

( ) ( )
Path ABC’:

(152 + 102 – 12) – 2(20 + 3) + = 210.1 mm

( )
Net Area, An:
An = 195.3(12) = 2343.6 mm2
*** End of Example 5 ***

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Effective Net Area

The net area, An gives the reduced section that resist tension but still may not correctly reflect
the strength. This is true when the tension members has a profile consisting of element not in
common plane and where the tensile load is transmitted at the end of the member by
connection to same but not all of elements. Example is an angle having connection in one leg
only.

The effective area of tension members shall be determined as follows:

Ae = AnU

where U, the shear lag factor, is determined as shown in Table 504.3.1 of NSCP C101-15.

It applies to both fastener connections (bolted and welded). For welded connections, the net
area equals the gross area Ag since there are no holes.

̅
U=1– ≤ 0.9

Where 𝑥̅distance from centroid of element being connected eccentrically to plane of

load transfer
L length of connection in the direction of loading

Member such as single angles, double angles and WT sections shall have connections
proportioned such as U is equal to or greater than 0.60. Alternatively, a lesser value of U is
permitted if these tension members are designed for the effect of eccentricity in accordance
with 508.1.2 or 508.2.

Eccentricity in Joints; determination of 𝒙 for computing U

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My Notes
Engineers Design of Steel Structures
For short tension members (connecting elements) such as splice and gusset plates, where
the elements of the cross section lie essentially in a common place, the effective net area is
taken equal to An, but may not exceed 85% of the gross area Ag.

Categories of Welded Connections

1. Loads transmitted by longitudinal welds, or by longitudinal welds in combination of

transverse welds

Ae = UAn = UAg

2. Loads transmitted only to transverse weld

Ae = UAn = Acon

Where Acon = area of directly connected elements. In this case, the shear lag effect is
approximately indirectly by using the reduced area Acon

3. Load transmitted to a plate by longitudinal welds along both sides of the plates spaced
apart such that l ≥ w

Ae = UAg

Where l = length of weld

w = distance between longitudinal welds (i.e, plate width)
U = 1.0 For l ≥ 2w
= 0.87 For 2w > l ≥ 1.5w
= 0.75 For 1.5w > l ≥ w

Example 6

Determine the reduction factor U to be applied in computing the effective net area for a
W14x82 (W360x122) section connected by plates at its two flanges as shown. There are
three bolts along each connection line.

Solution
̅ .
U=1– =1– = 0.76

Design of Steel Structures

 Structural
My Notes
Engineers Design of Steel Structures
From Table D3.1, Case 7 W, M, or S shapes having flange widths not less than two-
thirds of the depth, and structural tees cut from these shapes, U = 0.9 when there are
at least three fasteners per line in the direction of stress.

= = 0.71 > 0.67

Therefore, U = 0.9

Block Shear Strength

 Block shear – a tearing limit state

The angle tension member attached to a gusset

plate may have a tearing failure along bolt holes.
The limit state is defined by rupture along B-C plus
either yielding or rupture along plane A-B.

The four holes in the plate (Fig B and C) will

contribute to a tear out failure if the sum of the
shear strengths along A-B and C-D plus the tensile
strength along B-C is less than either of the
strengths in general yielding of the member or
rupture along E-B-C-F.

Mode of Block Shear Failure (AISC J4.3)

1. Rupture along the tensile plane (B-C) accompanied by yielding along the shear planes
(A-B and C-D).
2. Rupture along the shear planes (A-B and C-D) accompanied by yielding along the tensile
plane (B-C).
 The tensile failure is defined by rupture along the net area in both modes, while
failure along the shear planes can either be rupture along net area or yielding along
the gross area, whichever is smaller.
 From energy-of-distortion-theory

Design of Steel Structures

 Structural
My Notes
Engineers Design of Steel Structures
𝜏 = 0.6𝐹 where
𝜏 = 𝑠ℎ𝑒𝑎𝑟 𝑦𝑖𝑒𝑙𝑑 𝑠𝑡𝑟𝑒𝑠𝑠
𝜏 = 0.6𝐹 𝜏 = 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ

 Nominal shear strength

Shear yielding – tension rupture (0.6FyAgv < 0.6FuAnv)

Tn = 0.6FyAgv + FuUbsAnt

Shear fracture – tension rupture (0.6FyAgv ≥ 0.6FuAnv)

Tn = 0.6FuAnv + FuUbsAnt

Where
Agv gross area acted upon by shear
Ant net area acted upon by tension
Anv net area acted upon by shear
Fu specified minimum tensile strength
Fy specified minimum yield stress

When the tension stress is

Uniform Use Ubs = 1.0
Non-uniform Use Ubs = 0.5

Design of Tension Members

The central problem of all member design is to find a cross-section for which the required
strength does not exceed the available strength.

 LRFD

Pu ≤ ϕtPn or ϕtPn ≥ Pu

Where Pu is the sum of the factored load

 To prevent yielding

0.9FyAg ≥ Pu or Ag ≥
.

 To avoid fracrture

0.75FuAe ≥ Pu or Ae ≥
.

Design of Steel Structures

 Structural
My Notes
Engineers Design of Steel Structures
 ASD
 Yielding

Pa ≤ FtAg
Ag ≥ or Ag ≥
.

 Fracrture

Ae ≥ or Ae ≥
.

 Slenderness ratio limitation

𝐿
r≥
300
where r is the minimum radius of gyration of he cross-section and L is the member
length

Example 6
A tension member with a length of 2.0 m must resist a dead load of 80 kN and a service live
load of 230 kN. Select a member with rectangular cross section. Use A36 Steel (Fy = 250
MPa, Fu = 480 MPa) and assume a connection with one line of 20 mm diameter bolt.

Solution
 LRFD
Pu = 1.2D + 1.6L = 1.2(80) + 1.6(230) = 464 kN

( )
Required Ag = = = = 2062.2 mm2
∅ . . ( )

( )
Required Ae = = = = 1288.9 mm2
∅ . . ( )

Try t = 25 mm
.
Required wg = = = 82.5 mm say 90 mm

Try 25x90 cross section

Ae = An = Ag - Ahole
= 25(90) – (20 + 3)(25) = 1675 mm2 > 1288.9 mm2 Ok

Design of Steel Structures

 Structural
My Notes
Engineers Design of Steel Structures

Check slenderness ratio

( )
Imin = = 117188 mm4

A = 25(90) = 2250 mm2

rmin = = = 7.22 mm

. ( )
Maximum = = 242.4 < 300 Ok
.

Use a plate 25 x 90.

 ASD
P = D + L = 80 + 230 = 310 kN

For yielding: Ft = 0.6Fy = 0.6(250) = 150 MPa

( )
Required Ag = = = 2066.7 mm2

For fructure: Ft = 0.5Fu = 0.5(250) = 150 MPa

( )
Required Ae = = = = 1288.9 mm2
∅ . . ( )

Try t = 25 mm
.
Required wg = = = 82.5 mm say 90 mm

Try 25x90 cross section

Ae = An = Ag - Ahole
= 25(90) – (20 + 3)(25) = 1675 mm2 > 1288.9 mm2 Ok

Check slenderness ratio

Design of Steel Structures

 Structural
My Notes
Engineers Design of Steel Structures
( )
Imin = = 117188 mm4

A = 25(90) = 2250 mm2

rmin = = = 7.22 mm

. ( )
Maximum = = 242.4 < 300 Ok
.

Use a plate 25 x 90.

Built-up Members

For limitations on the longitudinal spacing of connectors between elements in continuous

contact consisting of a plate and shape or two plates, see Section 510.3.5.

Either perforated cover plates or tie plates without lacing are permitted to be used on the
open sides of built-up tension members. Tie plates shall have a length not less than two-
thirds the distance between the lines of welds or fasteners connecting them to the
components of the member. The thickness of such tie plates shall not be less than one-fiftieth
of the distance between these lines. The longitudinal spacing of intermittent welds or
fasteners at tie plates shall not exceed 150 mm.

The longitudinal spacing of connectors between components should

preferably limit the slenderness ratio in any component between the
connectors to 300.

Pin-Connected Members

Tensile Strength

The design tensile strength, ΦtPn, and the allowable tensile strength, Pn/Ωt ofpin-connected
members, shall be the lower value obtained according to the limit states of tensile rupture,
shear rupture, bearing, and yielding

1. For tensile rupture on the net effective area:

Pn = 2tbeffFu (1-1)

Φt = 0.75 (LRFD) Ωt = 2.0 (ASD)

Design of Steel Structures

 Structural
My Notes
Engineers Design of Steel Structures
2. For shear rupture on the effective area:

Pn = 0.6FuAsf (1-2)

Φsf = 0.75 (LRFD) Ωsf = 2.0 (ASD)

where

Asf = 2t(a + d/2), mm2

A = shortest distance from the edge of the pin hole to the edge of the member
measured parallel to the direction of the force, mm
beff = 2t + 16, mm but not more than the actual distance from the edge of the hole to
the edge of the part measured in the direction normal to the applied force
d = pin diameter, mm
t = thickness of plate, mm

3. For bearing on the projected area of the pin, see Section 510.7.
4. For yielding on the gross section, use the following equation

Design of Steel Structures