N.W.F.P. University of Engineering and
Technology Peshawar
Lecture 07: Miscellaneous topics
By: Prof Dr. Akhtar Naeem Khan
chairciv@nwfpuet.edu.pk
1
Topics to be Addressed
 Residual Stresses
 Factors effecting Residual Stresses
 Remedial measures against Residual
Stresses
 Effect of Residual stresses on tension
members
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Prof. Dr Akhtar Naeem Khan
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Residual Stresses
• Tension members response to load is much
•
similar to that of tensile-test coupon .
Member behavior may differ from coupon
behavior because of:
• Slip in bolted & riveted connections
• Non-linear behavior of connections
• Residual stresses in member
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Residual Stresses
 Residual stresses result principally from nonuniform cooling of hot rolled or welded shapes
and from cold straightening of bent members.
1. Thermal Residual Stresses
2. Residual Stresses caused by cold
straightening
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Residual Stresses
1. Thermal Residual Stresses: W /I shape
• As they have more surface exposure per unit of
volume, flange tips and central parts of the webs
tends to cool faster than juncture of flange-web
of a section (w-f) i.e. rate of cooling of juncture is
slower than rate of cooling of tips
 As a result, metal at junctures continues to
contract as it cools after flange tips and web
interior have cooled to temperature of
surroundings.
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Residual Stresses
1. Thermal Residual Stresses: W /I shape
• This contraction is partially restrained by cooler
metal which causes:
• Tensile stresses to develop at the juncture
of flange-web
• Compressive stresses in the remainder of
the cross-section
 These are called residual Stresses.
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Residual Stresses
1. Thermal Residual Stresses: W /I shape
Distribution of Residual-Stress in W section
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Residual Stresses
1. Thermal Residual Stresses: W /I shape
20 W’s shapes were investigated:
 It Revealed that flange-tip stress frc varied from 4.1 to
18.7 Ksi, the average being 12.8 ksi
 Residual stresses in web center varied from 41Ksi
compression to 18.2Ksi tension.
 Showing some W’s develop residual tension over entire
web, instead the pattern shown.
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Residual Stresses
1. Thermal Residual Stresses: W /I shape
• Only one out of 20 sections was thicker than 1 in.
• Therefore, above values are not representative
of W’s with thick flanges and webs
 Residual stresses tend to increase in magnitude with
increase in thickness
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Residual Stresses
2. Residual Stresses caused by cold straightening
Two straightening procedures:
1. Rotorizing: continuous straightening procedure
•
Residual stress distribution changes along
entire length of member.
2. Gagging: Concentrated straightening at few
points
•
Almost no change in thermal R.S
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Residual Stresses
Residual Stresses in welded connection
 Because of high concentration of heat,
tensile residual stresses at the weld in
welded members usually equal the yield
strength of the weld metal itself which may
be as much as 50% higher than that of the
parent metal.
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Factors affecting Residual
Stresses
1. Geometry
2. Method of preparation
3. Fabricating operation
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Factors affecting Residual
Stresses
1. Geometry
 Magnitude and distribution of thermal residual
stresses are influenced to considerable degree by
geometry of x-section
 Residual stresses tend to increase in magnitude with
increase in thickness
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Factors affecting Residual
Stresses
2. Method of preparation
UM Plate
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Flame-cut Plate
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Factors affecting Residual
Stresses
2. Method of preparation
H from UM Plates
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H from Flame-cut Plates
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Factors affecting Residual
Stresses
2. Method of preparation
 Large residual stresses develop at the
corners of the welded box
 On the other hand, residual stresses in the
hot-rolled square box are very low and in
one investigation averaged less than 5 ksi.
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Factors affecting Residual
Stresses
2. Method of preparation
 Box figures
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Factors affecting Residual
Stresses
2. Method of preparation
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Factors affecting Residual
Stresses
3. Fabricating operation
• Fabricating operations such as cambering and
straightening by cold bending also induce
residual stresses.
• These are of about the same magnitude but
differ in distribution
• These stresses are superimposed on the
thermal residual stresses.
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Remedial Measures against
RS
Quenching and Tempering
• Quenching is the act of rapidly cooling the hot
steel to harden the steel.
• Quenched steel is hard and brittle.
• Often it is just too brittle and must be made
more malleable, This is achieved by a process
known as tempering.
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Remedial Measures against
RS
Quenching and Tempering
• The quenched steel is heated again but this
time to a temperature between 200 °C and
300 °C.
• When the metal reaches the tempering
temperature, it is quenched again in cold water
or oil. The result is a steel that is still hard but
is more malleable and ductile.
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Remedial Measures against
RS
Quenching and Tempering
• Because they are quenched and tempered,
A514 rolled steel shapes are partially stressrelieved, so residual stresses are small
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Effect of RS on Tension
Members
 Generally tension members response to
load is much similar to that of tensile-test
coupon but not identical.
 However member behavior may differ
from coupon behavior because of:

Slip in bolted & riveted connections

Non-linear behavior of connections

Residual stresses in member
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Effect of RS on Tension
Members
 The section is an idealized (web less) H
 Residual-stress distribution is considered as
linear
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Effect of RS on Tension
Members
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Effect of RS on Tension
Members
 For Fig (d):
P = 2(24x12x1) = 576kips
favg = 576/24 = 24ksi
→ This gives point A on the stress-strain curve
C
A
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B
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Effect of RS on Tension
Members
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Effect of RS on Tension
Members
 For fig (f):
P = 2(36x6x1 + 30x6x1) = 792kips
favg = 792/24 = 33ksi
→ This gives point B on stress-strain curve
C
A
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Effect of RS on Tension
Members
 For fig (h):
favg = 36ksi
→ This gives point C in on stress-strain curve
C
A
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B
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Effect of RS on Tension
Members
 No effect on the yield strength of the member
 Lowering of proportional limit (P.L <36ksi)
 Increase in the strain at initiation of overall yielding
→No consequence in regard to the static strength of the
member
→Can be important if fatigue is involved
→R.S have a pronounced effect on the strength of columns
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Net Section
 Holes for bolts or rivets in tension
members effect the member in two
ways
1. Reduce area of x-section
2. Result in non uniform strain on xsection in neighborhood of the hole
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Net Section
 Net area is defined as gross section minus
area which is lost because of holes.
 Effective net area is obtained by
multiplying net area by coefficient to
account for its reduced effectiveness if not
all the member elements are connected.
(According to AISC)
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Net Section: Staggered bolts
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Net Section: Staggered bolts
 Failure paths may occur on sections normal
to axis of member (1-2-5) or may include
zigzag sections (1-2-3-4).
 Depending on the relative values of g, s and bolt
diameter d.
g: gauge (distance btw longitudinal fastener line)
s: pitch (distance btw transverse rows)
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Net Section: Staggered bolts
 If g > s : failure is expected along 1-2-3-4
 If g < s : failure is expected along 1-2-5
 For fixed values of g and s…. failure is expected
along the zigzag section as the size of holes
increases.
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Net Section
 Empirical methods have been developed
to calculate the net section fracture
strength
 Assumption: The effect of the zigzags in any
failure path can be accounted for by deducting
from the area of the section the areas lost by
the holes in the failure path and adding the
quantity (s2t/4g) for each zigzag.
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Net Section
Thus the net area for the failure path is given as:
s 2t
An  Ag   dt  
4g
If the plate thickness is uniform, we can divide each term by ‘t’ to get:
s2
wn  wg   d  
4g
The net width concept is useful when elements of uniform
thickness are being evaluated
→ It is called s2t/4g rule.
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Net Section
Staggered bolts in angles.
 If staggered lines of bolts are present in
both legs of an angle, then the net area is
found by first unfolding the angle to obtain
an equivalent plate.
 The unfolding is done at the middle
surface to obtain a plate with gross width
equal to the sum of the leg lengths minus
the angle thickness.
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Net Section
Staggered bolts in angles.
 AISC Specification B2 says that any gage line crossing
the heel of the angle should be reduced by an amount
equal to the angle thickness.
 For this situation, the distance g will be
= 3 + 2 – ½ in.
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Net Section
Example 01
Compute the smallest net area for the plate shown below: The
holes are for 1 in. diameter bolts.
i
3 in.
j
a
b
5 in.
c
5 in.
3 in.
f
h
d
e
3 in. 3 in. 3 in. 3 in. 3 in. 3 in.
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Net Section
Example 01
The effective hole diameter is 1 + 1/8 = 1.125 in.
For line a-b-d-e
wn = 16.0 – 2 (1.125) = 13.75 in.
For line a-b-c-d-e
wn = 16.0 – 3 (1.125) + 2 x 32/ (4 x 5) = 13.52 in.
The line a-b-c-d-e governs:
An = t wn = 0.75 (13.52) = 10.14 in2
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Net Section
Example 02
Example 3-6-1. A 7x4x3/4 angle is connected by two rows
of 3/4in Bolts in the 7in leg and one row in the 4in leg
as shown. Standard holes are used.
(a) Determine the pitch s so that only two holes for 3/4in
Fasteners need to be deducted in computing the net area
(b) Determine the net area of the 7x4 angle if the pitch s
is 2in
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Net Section
Example 02
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Net Section
Example 02 (a)
The fastener gages shown are the usual values for 7x4 angle.
s2
wn  wg   d  
4g
For section abde: Wn = (7+4-0.75)-2(.75+1/8) = 12
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Net Section
Example 02 (a)
The fastener gages shown are the usual values for 7x4 angle.
s2
wn  wg   d  
4g
For the section abcde: wn = 10.25-(3x0.875)+s2/(4x3) +s2/(4x4.25)
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Net Section
Example 02 (a)
For section abde: Wn = (7+4-0.75)-2(.75+1/8) = 12
For the section abcde: wn = 10.25-(3x0.875)+s2/(4x3) +s2/(4x4.25)
Equating and solving for s we get:
S = 2.48in ≈ 2.50in
An = (10.25-3x0.875)0.75
= 6.38in2
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Net Section
Example 02 (b)
The net area is given as:
s 2t
An  Ag   dt  
4g
An = {10.25 - 3x0.875 + 22/(4x3) + 22/(4x4.25)}x0.75
= 6.15in2
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Net Section
Assignment
 Example 3-6-2 and Example 3-6-3
( From Gaylord)
Due Date:
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Stresses on Net Section
Consider the non uniform strain in the vicinity of the hole in a
uniformly Stretched sheet of rubber as shown
→ The unloaded sheet upon which an orthogonal grid is drawn.
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Stresses on Net Section
The stretched sheet The strains at the edge of the elongated hole are
much larger than those elsewhere in the sheet. The disturbance is highly
localized.
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Stresses on Net Section
 According to theory of elasticity, the distribution of
stress on net section of infinitely wide plate containing
a hole at its centerline is given by
f 1 = stress that would exist if there were no hole
r = radius of hole
x = distance from center of hole to any point on
the transverse section.
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Stresses on Net Section
 As the disturbance in stress is highly
localized, this equation can be applied with
good accuracy to a plate of finite width.
 It is valid only if the stress at the edge of the
hole does not exceed the proportional limit.
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Stresses on Net Section
Example: Let the plate shown above is subjected to a uniformly
distributed tension of 12ksi as shown:
It is noted that the stress at the edge of the hole is equal to the yield
stress of A36 steel.
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Stresses on Net Section
 Tension members do not always develop a
net section average stress equal to tensile
strength.
 Reduction in strength can be expressed in
terms of efficiency of net-section.
 Net-section efficiency is the ratio of
average stress at fracture to coupon
strength.
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Net Section efficiency

Net-section efficiency depends upon
1. Ductility of metal
2. Method of making holes
3. Ratio of gage g to fastener diameter d
4. Ratio of net area in tension to area in bearing on
fastener(Called bearing ratio)
5. Distribution of x-sectional material relative to
gusset plates or other elements to which
member is connected.
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Net Section efficiency
1.Ductility of metal
 Because of non uniform stress distribution
efficiency of net section is dependent on ductility
of metal.
 Net section in highly ductile material may be 15
to 20 % stronger than same section in material
with relative low ductility (from test results)
 This effect can be expressed by net section
efficiency coefficient K1
K1=0.82 + 0.0032R <= 1
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Net Section efficiency
1.Ductility of metal
 R: percent reduction in the area of a
standard test coupon (2in gage length)
 R = 50% or more for A36 steel
→ K1 = 0.98 ≈ 1
→ A36 steel is 100% efficient as regards
ductility
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Net Section efficiency
2. Method of making holes
 Punched holes may reduce efficiency of net
section by as much as 15% compared with
drilled holes.
 This effect can be expressed by efficiency
coefficient K2.
K2 = 0.85 for punched holes
K2 =
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1
for drilled, sub-punched and
reamed holes
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Net Section efficiency
3. g/d ratio
 Net section is more efficient if ratio of
gage g to diameter d is small than if it is
large.
 Efficiency coefficient K3 is proposed for
this effect.
K3 = 1.6 – 0.7 (An / Ag)
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Net Section efficiency
3. g/d ratio
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Net Section efficiency
4. Net area in tension to area on bearing ratio
 In case of a plate with uniform gage spacing, the
bearing ratio and the fastener spacing are directly
related:
An ( g  d )t g

 1
Ag
dt
d
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Net Section efficiency
4. Net area in tension to area on bearing ratio
 Tests suggested that strength of
connection is not impaired by bearing
pressure so long as adequate spacing is
provided to produce usual failure modes.
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Net Section efficiency
5. Shear Lag effect
 The phenomenon of non-uniform straining of web.
 Ends of web are free and four forces shows
resultants of bolt shear.
 Results in shear deformation and stress in web is
said to lag because of it.
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Net Section efficiency
5. Shear Lag effect
_
k4
x
 1
L
L: length of the connection i.e. distance
from the first fastener to the last one
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Net Section efficiency
Effective Net area
The effective area is obtained by multiplying
the net area with all the above mentioned
coefficients:
Ae = K1 K2 K3 K4 An
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Thanks
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