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Steel Design Plastic Analysis Designing means to fabricate structures which will be able to support the given load conditions.

These may be either steel structures or RCC structures.

There are three methods of steel design:

a) Working Stress method: The worst combination of working (service) loads is ascertained and

the members are proportioned on the basis of working stress. The worst combination should

not exceed the permissible stresses. The permissible stresses are generally some fraction

(factor of safety) of the yield stress.

b) Plastic Method: We find that steel can withstand higher loads than the elastic limit. In this

method the reserve strength of the structure is used. This method is vased on failure

conditions and not working stresses. This method extends the structural usefulness of the

material upto the ultimate strength. <explain that steel is ductile and therefore can

withstand a bit more of stress than elastic. Bends a bit until it collapses>. In this method the

loads are primarily supported by resistance to bending as deformation is small in case of

bending. So continuous beams and rigid frames are analysed using this method. Gen not

used for statically determinate frames or pin connected beams.

c) Limit State method: this is based on limit strength and serviceability. That is the acceptable

limit for the safety and serviceability before failure. The objective is to achieve a structure

which will not become unfit for the period design like say 50 years.

Stress Strain curve of steel

Fig 2.1 shows the stress strain curve of annealed(i.e it has been given metallurgical treatment) mild

steel specimen in tension.

A is the limit of proportionality.

B is the elastic limit.

C represents the upper yield point at which there is a definite increase in strain without any further

increase in stress.

C’ is the lower yield point as the stress drops abruptly, then strain increases on constant stress up to

D. Beyond D strain increases further and the material is said to be strain hardening at E. F is failure

point.

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The strain scale is enlarged considerably in Fig 2.1 (b), in order to study the yield range. fu and fL are

the upper and lower yield point in fig 2.1(b). For the purpose of plastic design the elastic limit and

the lower yield point may be assumed to be numerically identical. The yield strength is fixed at fy

and the strain hardening region is ignored. Ignoring the strain hardening adds to the safety.

In fig 2.1 (c) the stress strain behaviour is idealized and it is assumed that extension is unlimited at

constant stress fy

So during Bending of beams the following picture illustrates what happens assuming the idealised

stress-strain curve:

Collapse takes place when plastic moment is reached at the highly stressed points.

Shape factor:

Shape factor is the ratio of plastic moment to yield moment of the section. It depends on the cross

section of the shape. It is a measure of the reserve strength available in the section after bending.

S=Mp/My =𝜎𝑦𝑍𝑝

𝜎𝑦𝑍𝑒=

𝑍𝑝

𝑍𝑒

Shape Moment of inertia about Cg Moment of inertia about base

Rectangle 𝑏𝑑3

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𝑏𝑑3

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Triangle 𝑏ℎ3

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𝑏ℎ3

12

Circle 𝜋𝑑4

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NA

Shape factor of rectangle

Shape factor 𝑆 =𝑀𝑝

𝑀𝑦=

𝜎𝑦𝑍𝑃

𝜎𝑦𝑍𝐸=

𝑍𝑃

𝑍𝐸=

𝑏𝑑2

4𝑏𝑑2

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= 1.5

Zp =𝐴

2(𝑦1̅̅ ̅ + 𝑦2̅̅ ̅); A is the area of the whole cross-section and 𝑦1̅̅ ̅ and 𝑦2̅̅ ̅ are the distances from the

centroidal axis to the centroids of the area in compression and tension respectively. In fig 2.5 C is

compression and T is tension.

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Some shape factors are as givrn below

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Load factor:

Ratio of collapse load to the working load.

F=Pu/Pw=Mp/Mw=fyZp/(f x Ze)=fy/f x S= FOS x S(Shape factor)

What is the load factor that is generally assumed in case of steel design?

1.7 to 2

What is the reduction in load factor when the structures are subject to winds?

25 %

<Explain that the maximum stressed points attain plastic state>

<Pn 70: Fig:2.12>

For any structure to fail a number of hinges need to formed:

There is no relation between the number of hinges and no. of redundancies. You have to find by

experience.

Redundancy is the static indeterminacy of the structure.

What is partial collapse of structure?

No of plastic hinges in the collapse mechanism are less than r+1

What is complete collapse of structure?

No of plastic hinges in the collapse mechanism are equal to r+1

What is over complete collapse of structure?

No of plastic hinges in the collapse mechanism are more than r+1

For calculating the failure load there are three theorems:

Static or lower bound theorem; Kinematic or upper bound theorem; Uniquness theorem.

What is the Static or lower bound theorem?

For a given frame and loading if there exists a set of loads which is both safe and statically admissible

with a set of load P, then the value of the load P must be less than or equal to the collapse load Pu.

Lower limit to the true load

What is Kinematic or upper bound theorem?

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For a given frame subjected to a set of loads P, the value of P which is found to correspond to any

assumed mechanism must be either greater than or equal to the collapse load Pu. Upper limit to the

true load

What is Uniqueness theorem?

For a given frame if one load is found which is both statically admissible and safe and if in this dist

the bending moment is such as to cause fully plastic moment at sufficient cross section to cause

failure as a mechanism due to rotation of plastic hinges then the load will be equal to the collapse

load.

Conditions to be satisfied in the plastic moment of analysis:

Equilibrium condition: Summation of all moments and forces are zero

Mechanism condition or Continuity condition: Ultimate load is reached when a mechanism forms

Yield Condition: Bending moment at any section must not exceed full plastic moment at section.

Note: How the area is calc to find

ext. work done.

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For static method you have to know the BMD well. And for kinematic method the principle of virtual

work is used. The main aim is to find the maximum Mp and minimum Ultimate load Wu which

causes this maximum Mp in the beam. Thus we don’t equate 𝑀𝑃 =𝑊𝐿

12 in the above Fig 2.16 as this

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is not the maximum Mp but we equate 2𝑀𝑝 =𝑊𝐿

8. This is because the middle hinge has to form to

complete the mechanism and it requires 2𝑀𝑝.

What is the principle of virtual work?

Work done by the external forces equal the work done by the internal forces.

We=Wi

Find the reaction at A in the following figure. Ans =5wL/12. Hint ∑ 𝑀𝐵 = 0

Hinge length for a point load on a simple beam with rectangular cross-section.

Hinge length for uniformly distributed load with simple beam is L/√3;

/m

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The independent mechanisms are

first formed, using only single

loads and then these mechanisms

are combined. They are combined

such that the geometry remains

the same, if you can’t keep the

geometry same you do not need

to do those mechanisms. The

geometry remains the same

meaning that the mechanisms

should be combined such that the

geometry of the independent

mechs remain (i.e the angles

should be same)

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What are the number of independent mechanisms?

n=N-r; n = possible no. of mechanisms, N= number of plastic hinges; r= number of redundancies

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A gable is the generally triangular portion of a wall between the edges of a sloping roof.

What is the max yield stress of grade of steel for plastic design?

450 Mpa

What are the classification of cross Sections and their char?

Plastic(class 1): Cross Section which can develop plastic hinges. Only these section are used for

plastic analysys

Compact(class 2): Plastic moment developed but does not form plastic hinges

Semi – Compact(class 3): Extreme fibre in compression can reach yield stress, but due to local

buckling doesn’t develop plastic hinges

Slender(class 4): Buckle locally even before attainment of yield stress.

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𝑀𝑝 =5𝑤𝐿

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Connections The three different types of connections are: Riveted, bolted and pinned connections: These

connections are alike while the welded connection is different.

There are certain terms associated with riveted connections: These are as explained below:

What is the grip of the rivet?

The distance between the undersides of the two heads of rivet.

What is the nominal diameter?

Diameter of shank.

What are hot driven field rivets?

When the rivets are heated before driving

What is the gross diameter of rivets?

The diameter of the rivets when it is hot is equal to the diameter of hole is called gross diameter.

Where are cold driven rivets used?

These are inserted into the rivet hole at room temperature, and very high pressures are required to

form the heads. Only limited to applications where there is high pressure

Is riveted connection practiced nowadays?

No

The main advantages of bolted connection over riveted connections are:

a) Erection of structure can be speeded up.

b) Less skilled persons are required.

c) More economical due to lesser equipment and labour costs.

Holes can be either punched or drilled. Punched holes are used in materials with yield stress less

than 360 MPa and thickness should not exceed 5660/fy mm.

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Disadvantages:

1) Cost of material is high

2) Tensile strength of bolts is reduced because of area reduction at root of threads

3) Loose fitting so less strength

4) May get loose due to vibrations

Joints may be of different kinds:

Lap joint; Butt joint Explain:

Load transfer is either by bearing and shearing or by friction.

Bearing: Bearing is the stress on the plate due to the bolt being forced onto the plate

Shearing: This is the stress on the bolt which tries to shear away or cut away the bolt at the point of

contact. This is shown in Fig.1

Fig. 1: Shearing

Explain that for bolting certain terms like diameter of bolts, spacing of bolts are reqd to be known.

Few large diameter bolts are more economical than a more number of small dia bolts.

Pitch:

It is the distance between two consecutive bolts measured parallel to the direction of stress.

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Why is a minimum amt of pitch required?

1) Prevent bearing failure of members

2) Permit efficient installation of bolts

Why is the maximum pitch required?

1) To reduce length of connection and gusset plate

2) To have uniform stress in bolts.

What is the minimum ptch?

2.5 times the nominal diameter.

What is the maximum pitch?

Tension Members: 16t or 20mm, whichever is less

Compression members: 12t or 200mm

Pitch of tacking bolts should not exceed 32t or 300mm, but when exposed to weather 16t or

200mm.

Tacking bolts are used to make members act in unison also known as stitch bolts. These bolts are

not for being subjected to stresses like tension or compression.

Gauge:

Distance between adjacent bolt lines; The gauge is such as to accommodate the bolts properly as

shown in fig 4.8.

Edge distance explain

Distance at right angle to the direction of stress from the centre of the bolt hole to the adjacent

edge of the member.

End Distance: From the centre of hole to the end of element along the direction of stress.

What is the maximum edge distance?

12t𝜖 𝑎𝑛𝑑 ∈= √250

𝑓𝑦 and t is the thickness of thinner outer plate

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N.B.: While designing generally two or three bolts are provided , even if one bolt hole is required,

this is a safety measure,as one bolt may fail to live upto its strength

The minimum of bearing, shearing, tensile strength of bolt and tensile strength of plate is taken as

strength of bolted joint, now we calculate them:

Tell we use a factor of safety of 𝛾𝑚𝑏 =1.25 for ultimate strength and for yield strength 𝛾𝑚0 = 1.1.

The best bolting pattern is “diamond”.

How is class designated?

4.6 Ultimate stress 4 x 100 =400N/mm2; Yiels dtrength =0.6 x 400 =240N/mm2

How is the cover plate thickness selected in case of butt joint?

Plate thickness not less than 5/8t

Cover plates are used in butt joints to connect the two members.

Formulae for bolt connection design Shearing strength of bolts:

𝑉𝑛𝑠𝑏 =𝑓𝑢𝑏

√3(𝑛𝑛𝐴𝑛𝑏 + 𝑛𝑠𝐴𝑠𝑏)

𝑉𝑛𝑠𝑏 Nominal shear strength of bolt

𝑓𝑢𝑏 ultimatre tensile stress of bolt

𝑛𝑛 number of shear planes with threads intercepting the shear plane

𝑛𝑠 number of shear planes without threads intercepting the shear plane

𝐴𝑠𝑏 nominal shank Area of bolt=𝜋𝑑2

4; d is dia of bolt.

𝐴𝑛𝑏 net tensile stress area. =0.78𝐴𝑠𝑏

Now for long joints and other factors we use some reduction factors:

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𝑉𝑛𝑠𝑏 =𝑓𝑢𝑏

√3(𝑛𝑛𝐴𝑛𝑏 + 𝑛𝑠𝐴𝑠𝑏)𝛽𝑙𝑗𝛽𝑙𝑔𝛽𝑝𝑘𝑔

𝛽𝑙𝑗 = reduction factor used in long bolts = 1.075−𝑙𝑗

200𝑑; 0.75 ≤ 𝛽𝑙𝑗 ≤ 1; 𝑙𝑗 is the length of joint

between the first and last row of bolts. Used when 𝒍𝒋 > 𝟏𝟓𝒅

𝛽𝑙𝑔 =reduction factor for large grip length = 8𝑑

3𝑑+𝑙𝑔; Used when total thickness of connected plates

exceed five times the nominal dia of bolt𝛽𝑙𝑔 ≯ 𝛽𝑙𝑗 𝑎𝑛𝑑 𝛽𝑙𝑔 ≯ 8𝑑

𝛽𝑝𝑘𝑔= reduction factor when packing plate thickness exceeds 6mm. =(1 − 0.0125𝑡𝑝𝑘𝑔). 𝑡𝑝𝑘𝑔 is the

thickness of thicker packing

𝑉𝑠𝑏 ≤𝑉𝑛𝑠𝑏

𝛾𝑚𝑏

𝛾𝑚𝑏 = 1.25 (𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟)

𝑉𝑠𝑏 is the shear strength of bolt used for designing. When finding bolt value use:

𝑉𝑠𝑏 =𝑉𝑛𝑠𝑏

𝛾𝑚𝑏

Remember whenever there is ultimate strength involved use partial safety factor of 1.25, when yield

strength is involved use 1.1

Bearing strength of bolt:

𝑉𝑛𝑝𝑏 = 2.5𝑘𝑏𝑑𝑡𝑓𝑢;

𝑘𝑏 =smaller of 𝑒

3𝑑0;

𝑝

3𝑑0− 0.25;

𝑓𝑢𝑏

𝑓𝑢; 𝑎𝑛𝑑 1.0;

𝑑 = nominal dia of bolt

𝑑0 = hole dia

𝑡 = aggregate thickness for connected plates (will be explained later) experiencing stress in the

same direction. If bolts are countersunk then thickness of plate minus half of the depth of

countersunk.

𝑉𝑛𝑝𝑏 =nominal bearing strength of bolt.

𝑉𝑝𝑏 ≤𝑉𝑛𝑝𝑏

𝛾𝑚𝑏

𝛾𝑚𝑏 = 1.25 (𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟)

When finding bolt value use: 𝑉𝑝𝑏 =𝑉𝑛𝑝𝑏

𝛾𝑚𝑏

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Tensile strength of bolt

𝑇𝑛𝑏 = 0.9𝑓𝑢𝑏𝐴𝑛𝑏 < 𝑓𝑦𝑏𝐴𝑠𝑏

𝛾𝑚𝑏

𝛾𝑚0

𝑇𝑛𝑏 = nominal tensile strength of bolt.

𝑇𝑑𝑏≤

𝑇𝑛𝑏

𝛾𝑚𝑏

Tensile strength of plate: (not in syllabus) 𝐴𝑛 = (𝐵 − 𝑛𝑑ℎ)𝑡; for chain bolting

𝐴𝑛 = [𝐵 − 𝑛𝑑ℎ + ∑𝑝2

4𝑔] 𝑡; for staggered bolting

𝑇𝑛𝑑 = 0.9𝐴𝑛

𝑓𝑢

𝛾𝑚1

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Efficiency of joint (not in syllabus)

𝜂 =𝑆𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑏𝑜𝑙𝑡𝑒𝑑 𝑗𝑜𝑖𝑛𝑡 𝑝𝑒𝑟 𝑝𝑖𝑡𝑐ℎ 𝑙𝑒𝑛𝑔𝑡ℎ

𝑆𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑜𝑙𝑖𝑑 𝑝𝑙𝑎𝑡𝑒 𝑝𝑒𝑟 𝑝𝑖𝑡𝑐ℎ 𝑙𝑒𝑛𝑔𝑡ℎ × 100

If pitch is staggered then take two pitch lengths and take them from the middle, such that two rivets

are included. Thus two rivets have to fail, but only one hole in a pitch (because it is (p-d) x t and no

breadth term is included).

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Which plate thickness to be used? 1) Double cover butt joint. If thickness of cover plates = tp and thickness of thinner member tm

then if

2tp>tm use tm

Else 2tp

2) If 2 angle sections or I-sections connected back to back . If thickness of gusset plates = tp and

thickness of thinner member tm then if

2tm>tp use tp else 2tm

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