By: Prof Dr. Akhtar Naeem Khan [email protected]

N.W.F.P. University of Engineering and Technology

Peshawar

1

By: Prof Dr. Akhtar Naeem [email protected]

Lecture 07: Miscellaneous topics

CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 2

Residual Stresses

Factors effecting Residual Stresses

Remedial measures against Residual Stresses

Effect of Residual stresses on tension members

Topics to be Addressed


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


Residual Stresses

Residual stresses result principally from non-uniform 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


Residual Stresses

• 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.

1. Thermal Residual Stresses: W /I shape


Residual Stresses1. 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.



Distribution of Residual-Stress in W section



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.



• 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


Residual Stresses2. 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


Residual StressesResidual 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.


Factors affecting Residual Stresses

1. Geometry

2. Method of preparation

3. Fabricating operation



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




UM Plate Flame-cut Plate




H from UM Plates H from Flame-cut Plates




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.




Box figures






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.


Remedial Measures against R S

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.




• 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.




• Because they are quenched and tempered, A514 rolled steel shapes are partially stress-relieved, so residual stresses are small


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 connectionsNon-linear behavior of connectionsResidual stresses in member



The section is an idealized (web less) H

Residual-stress distribution is considered as linear





For Fig (d): P = 2(24x12x1) = 576kips

favg = 576/24 = 24ksi

→ This gives point A on the stress-strain curve

A BC





A BC

For fig (f): P = 2(36x6x1 + 30x6x1) = 792kips

favg = 792/24 = 33ksi

→ This gives point B on stress-strain curve



A BC

For fig (h):

favg = 36ksi

→ This gives point C in on stress-strain curve



No effect on the yield strength of the memberLowering 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


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 x-section in neighborhood of the hole


Net SectionNet 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)


Net Section: Staggered bolts


Net Section: Staggered boltsFailure 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)


Net Section: Staggered boltsIf g > s : failure is expected along 1-2-3-4If g < s : failure is expected along 1-2-5For fixed values of g and s…. failure is expected

along the zigzag section as the size of holes increases.


Net SectionEmpirical 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.


Net Section

The net width concept is useful when elements of uniform thickness are being evaluated

→ It is called s2t/4g rule.

g

sdww gn 4

2

gtsdtAA gn 4

2

If the plate thickness is uniform, we can divide each term by ‘t’ to get:

Thus the net area for the failure path is given as:


Net Section

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.

Staggered bolts in angles.


Net SectionStaggered 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.


Net SectionExample 01

3 in.

5 in.

5 in.

3 in.

3 in. 3 in. 3 in. 3 in. 3 in. 3 in.

b

a

c

d

e

i

j

f

h

3 in.

5 in.

5 in.

3 in.

3 in. 3 in. 3 in. 3 in. 3 in.3 in. 3 in. 3 in. 3 in. 3 in. 3 in.

b

a

c

d

e

i

j

f

h

Compute the smallest net area for the plate shown below: The holes are for 1 in. diameter bolts.



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-ewn = 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



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




Net SectionExample 02 (a)

The fastener gages shown are the usual values for 7x4 angle.

g

sdww gn 4

2

For section abde: Wn = (7+4-0.75)-2(.75+1/8) = 12



The fastener gages shown are the usual values for 7x4 angle.

g

sdww gn 4

2

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

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)


Net SectionExample 02 (b)

The net area is given as:

An = {10.25 - 3x0.875 + 22/(4x3) + 22/(4x4.25)}x0.75

= 6.15in2

gtsdtAA gn 4

2


Net Section

Example 3-6-2 and Example 3-6-3

( From Gaylord)

Due Date:

Assignment


Stresses on Net SectionConsider 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.


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.


Stresses on Net SectionAccording 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 holex = distance from center of hole to any point on

the transverse 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.


Stresses on Net SectionExample: 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.



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.


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.


Net Section efficiency1.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


Net Section efficiency1.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


Net Section efficiency2. Method of making holesPunched 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 = 1 for drilled, sub-punched and reamed holes


Net Section efficiency3. 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)


Net Section efficiency3. g/d ratio


Net Section efficiency4. 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:

1)(

dg

dttdg

AA

g

n


Net Section efficiency4. 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.


Net Section efficiency5. Shear Lag effectThe 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.


Net Section efficiency5. Shear Lag effect

Lxk_

4 1

L: L: length of the connection i.e. distance from the first fastener to the last one


Net Section efficiencyEffective Net area

The effective area is obtained by multiplying the net area with all the above mentioned coefficients:

Ae = K1 K2 K3 K4 An


Thanks

By: Prof Dr. Akhtar Naeem Khan [email protected]

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