Design of Steel structures

1. Introduction
2. Riveted connection**
3. Bolted connection
4. Welded connection**
5. Compression member**
6. Tension member
7. Flexural members
8. Plastic theory
9. Girders
Introduction
Design: Assign the size of members so that it can withstand the external loads safely and economically (Cost +
Safety)
Criteria for design
Primary criteria Secondary criteria
1. Strength Criteria 1. Stability Criteria
2. Stiffness Criteria
Strength: Resistance against gradually applied load.
Toughness: Resistance against impact (load having velocity) loading.
Hardness: Resistance against surface scratches (abrasion).
Stiffness: Resistance against deformation (Rigid bodies has no deformation).
Durability: Resistance against natural pressure force.
Elasticity: Regain its original position.
Plasticity: Showing permanent deformation at constant value of load.
Fatigue: Resistance against reversible forces.
Endurance Limit: Maximum stress up to which material can withstand infinite number (106) of stresses
For Mild steel Endurance limit is 183 MPa
Ductile material (Steel) Brittle material (Concrete)
Large deformation before fracture (gives warning) Showing sudden fracture
σ y comp (250 MPa)> σ y tens (240 MPa)> τ (100 Mpa) σ y comp (50 Mpa)> τ (15 MPa)>σ y tens (2.5 MPa)
Stress strain Curve for mild steel (Constitute experimental curve)
 To eliminate size affect in Load deflection we use stress strain curve
 Strain controlled test
 Extensiometer is use to measure deflection or Strain
 A = Proportionality limit (Stress ∝ Strain)
B = Elastic limit
C = Upper yield point
D = Lower yield point
DE = Yielding zone/Plastic zone (Strain
increasing at constant value of stress)
(Carbon and Iron atom slipping)
EF = Strain Hardening zone
F = Ultimate strength, maximum stress.
G = Strain softening zone (necking)

Maximum Permissible stresses in structural steel as per WSM

σ at =Maximum permissible axial tension=0.6 f y For Mild Steel
σ ac=Maximum permissible axial compression=0.6 f y Ultimate strength = 410 MPa
σ bt =Maximum permissible bendingtension=0.66 f y Yield strength = 250 MPa
σ bc=Maximum permissible bending compression=0.66 f y Shear Strength = 100 MPa
τ avg= Average Shear=0.4 f y
τ max=Maximum shear stress=0.45 f y
σ P=Maximum permissible bearing stress=0.75 f y
 In case of Slab Base strength is always 185 MPa
 Endurance limit is 183 MPa
Bearing Stress is the compressive stress where it is transferred from one cross section to another cross section

Factor of Safety
Ultimate strength
FOS= Brittle material
Working stress
Yield strength
FOS= Ductile material
Working stress
Yield strength fy
FOS ∈axial tension∧compression= = =1.67
Working stress 0.6 f y
Yield strength fy
FOS ∈Bendin g tension∧compression= = =1.5
Working stress 0.66 f y
Yield strength fy
FOS ∈Shear = = =2.5
Working stress 0.4 f y
Maximum Permissible deflection in structural steel as per WSM
Span
Maximum Permissible deflection as per WSM =
325
Maximum Permissible stresses in structural steel as per LSM (Strength Criteria)
fy
σ at , σ ac , σ bt , σ bc , σ P=
1.1
Where 1.1 is Partial FOS
fy
τ max=Maximum permissible shear stress=
√ 3 ×1.1
fy
Note: isthe max shear stress that a ductile material can withstand at the time of failure as per Distortion energy
√3
theory of failure and FOS 1.1 is applied over the value
fU
Maximum permissible strength for bolts∧welds=
√3 ×1.25
fU
Note: Bolts and welds can be loaded to ultimate strength without the danger of large deformation, so is the
√3
maximum permissible shear stress for bolts and welds and FOS of 1.25 is applied over this value to get the
value of maximum permissible shear.
Maximum Permissible deflection in structural steel as per LSM (Stiffness Criteria)
For Simply Supported beams For Cantilever beams
1. Maximum Permissible deflections are twice to that
1. If the beams are supporting such elements which
Span Span
are not susceptible to cracking than of simply supported beam i.e. and
Span 150 180
Maximum Permissible deflection = respectively.
300
Gantry Graders
2. If the beams are supporting such elements which 1. For manually operated cranes
are susceptible to cracking (glass panel) than Span
Maximum Permissible deflection =
Span 500
Maximum Permissible deflection =
360 2. For electrically operated cranes having capacity
50t
Span
Maximum Permissible deflection =
750
3. For electrically operated cranes having capacity
greater than 50t
Span
Maximum Permissible deflection =
1000

Why WSM modified as LSM

Limit State method of design
 A limit state is condition of structure beyond which it no longer fulfilled the relevant design criteria
Condition may refer to a degree of loading on the structure while criteria refer to structural integrity fitness fir
use, durability or other design considerations.
Limit State of Strength Limit state of serviceability
1. Load carrying capacity of structure Yielding, 1. Actual performance of structure under actual application
Cracking, Buckling, Fatigue (Cyclic stress), of service load like deflection, pounding, corrosion,
Fracture vibration
 Assume the variation in both load and material properties
Design load and design strength in LSM
Design load =Workingload × ∂ FOS
Characteristic strength
Design Strength=
Suitable ∂ FOS
Note: When wind load and Earthquake are considered permissible stress in structural steel and connections is
increased by 33.33% and 25% respectively.
The most economical structures correspond to minimum weight. In steel design about 30% of the cost is of
material only.
 Design life of steel structure and RCC structure is 50 years.
 Design life of Dam is 50 years 70 years.
Physical properties of Structural steel
 Steel is an alloy of iron and carbon (carbon content approximately 0.23%)
Low Carbon Steel (MS) Medium carbon steel High carbon steel (Hard Steel)
0.15% - 0.25% 0.31% - 0.55% 0.61% - 1.5%
 As carbon content increases strength hardness durability increases and ductility reduce

EA = Axial Rigidity
EA
= Axial stiffnes
As per IS 800 2000 L
EI = Flexural rigidity
EI
=Flexural stiffness
L
 Ultimate deformation
Ductility=
Yield deformation
fU
1. ≮ 1.2 (Ultimate strength is 20% more than the yielding strength)
fy
2. % elongation (Strain) ≮ 10%
3. fy ≯ 450 MPa (Otherwise size components will be very very small)
Modulus of Elasticity
Longitudian stress 5
E S= =2.1× 10 MPa=210 GPa
Longitudinal strain
 Less E means more strain or more deformation and less stiff material.
 Aluminum has E = 70 GPa almost 1/3 of steel. So less stiff than Steel i.e.
More deformation in same load but Al has more durability that’s why
used in wires.
Shear Modulus
Shear stress 5
G= =0.77 ×10 MPa=77 GPa
Shear strain
Poisson’s Ratio (µ or 1/m)
−Lateral strain
µ= =0.5 Plastic range
Longitudinal strain
¿ 0.3 Plastic range
∫ ydA =First moment of area(helps∈ finding C .G of cross section)
Verignon’s Theorem
A y=∑ ay

CG of the section , y=
∑ ay
A
Moment of Inertia
∫ y 2 dA=Second moment of area
Note: MOI is a mathematical term but when it is combined to a material property then it tells about resistance to
rotation , resistance to bending and resistance to buckling .
EI
=Flexural stiffness
L
So if the MOI is more of a cross section about any is more than it is tough to bend that cross section to that axis

Mass Density
 3
ρ=7850 kg/m
Coefficient of expansion
−6 mm
α =12 ×10 /oC
mm
 Standard Structural Steel Rolled Section
Made by the process of hot (above crystalline temperature) rolling
1. Forging
2. Rolling
3. Molding
4. Casting
1. Beam Section
ISLB ISMB ISHB
Roof beam where Floor beam Columns
load is relatively This section has high MOI about xx axis compare to yy
less axis so the lateral buckling strength is less. To increase
this compression flange is restrained by casting it into
concrete slab or by providing lateral restrains.

ISWB (Wide) ISJB (Junior)

Columns

2. Channel Section
It is used as Purlins (beams supporting roof covering material) and built up
ISSC
3. T Section
Used in the design of inclined member of trusses, members subjected to reversal of stresses, tension member etc
ISHT ISST ISLT
Indian standard wide flange T bar Indian standard long stem T bar Indian standard light T bars
4. Angles
Used as Purlins and Girts (Beam supporting wall covering material)
Equal Angles Unequal Angle Bulb angle
Both legs are equal Used in ship design where MOI is
high
Calculation of Cross sectional area of angles
5. Flats
60 ISF 8
60 Width and 8 thickness
Used as stiffeners lacing members and battens
6. Plates (thickness ≥ 3mm)
ISPL 2000×1000×8
7. Sheets (thickness < 3mm)
ISSH
8. Bars
ISSQ (Indian standard square bars) ISRO (Indian standard round bars)
9. Strips
ISST 200×2
Indian standard Strips
Note:
 IS 800 2007 recommends no minimum thickness requirements for steel sections but a minimum
thickness of 6 mm for main members and 5 mm for secondary members must be use.
 Steel in contact with soil and water and those subjected to alternate wetting and drying then 1.5 mm of
additional thickness must be provided.
 Riveted, Bolted, pinned Connection
Joints
Lap Joint

Lg

Inefficient because of load lines don’t coincide and hence these two form undesirable couple. And bending
stress developed. This couple can be minimized be increasing the overlap length and decreasing the Grip length
(Lg)
Overlap length ≮ 4t or 40 mm (whichever is more)
Lg ≯ 5d (Where d = diameter of bolts)
FBD of bolts in lap joint
Single shear one plane and bearing

Butt Joint

t Cover Plate ≥ t Main Plate Most efficient Joint

FBD of bolts in double cover butt joint
Line of action of force is ok but non symetrical about Double Shear two plane and bearing
x axis. So efficient more than lap joint.
Single shear and bearing.

Shear Capacity is Twice to that of Lap joint and single

cover butt joint
Types of Joints design

Failure Criteria
Failure of Bolts Failure of Plates
Shear failure of bolts Shearing Failure of plate
a. Bolts may get cut into two pieces a. Plate may get shear out and cracks may developed
b. Factored shear stress exceeds the shear capacity parallel to the applied load
of bolts.

Bearing failure of plate

Bearing Failure of Bolts a. If the material of plate is weaker than that of the bolt
a. Bolts may get out of shaped around the semi material
circumference

Splitting failure of plate

a. Occurs due to the diagonal tension in the plate at
plate in bolt level.

Tensile (Tearing) failure of plate

a. Plate may get cracked when the bolt material is
stronger than plate material and cracks may get
developed perpendicular to applied load

Note: Shearing failure, bearing failure and splitting failure of plates are due to insufficient end distance, so by
providing proper end distance we can prevent these failures.
Now for the design bolted connection for remaining threes
1. Shear failure of bolts
2. Bearing Failure of Bolts
 3. Tensile (Tearing) failure of plate
In the of bolted joint we ensure that the shearing and bearing strength of bolts is always greater than tensile
strength of plate (as we never want the failure of connection because it is more dangerous)
Strength of bolted joints

{
Shearing strength of all the bolts(P S )
Strengthof Bolted joints=Minimum Bearing strength of all the bolts∈ joints (P B )
Tensile strengthof plate(P t )
Lap Joints

πd
2
P =Bearing strength of all thebolts ∈ joint =n ( d ×t ) × f b
PS =Shearing strength of all the bolts∈ joint =n ×f S B
4 2.5 K B f U
f fUb=Maximum Bearing strength of bolt material= 1.25 =30
f S= Maximum permissible strength of bolt material= =100 MPa(WSM )(IS 800)
√3 × 1.25 K B=1(Assume)
n=9 t = thickness of thinner plate
d ×t = Projected area n = 9

{
fy
Ag × (Based on gross area yielding)
Pt =Tensile strength of plate=Minimum of 1.1
0.9 f U
A net × (Based on net area cracking)
1.25
A g=B ×t A net=( B−nd )× t
n = 3 here d= Diameter of bolt hole
Note: 0.9fU is same as 0.67fck in RCC (if plate is subjected to direct tension 0.9fU is taken as its ultimate
strength)
Note: There is no concept of Gross area yielding in WSM and the plate is assumed to be cracked along net
section at σ at =0.6 f y
Pt =A net × σ at =(B−nd)×t × 0.6 f y
 Double Cover Butt joint

Critical Section for main plate is 1 -1

Critical Section for cover plate 2 - 2

( )
π d PB =Bearing strength of all thebolts ∈ joint =n ( d ×t ) × f b
2
PS =Shearing strength of all the bolts∈ joint =n 2× ×fS 2.5 K B f U
4
f b=Maximum Bearing strength of bolt material= ( LSM
fU 1.25
f S= Maximum permissible strength of bolt material= ( LSM )=0.4 f yK=100 MPa (WSM )(IS 800)
√3 × 1.25 B=1(Assume)

n = 9 here not 18 d = diameter of bolts t = thickness of thinner plate

d = diameter of bolts
d ×t = Projected area n = 9 here, not 18
t=minimum of { thickness of thinner main plate
Sumesion of thickness of cover plates

{
fy
Ag × (Based on gross area yielding)
Pt =Tensile strength of plate=Minimum of 1.1
0.9 f U
A net × (Based on net area cracking)
1.25
A g=B ×t A net=( B−nd )× t
n (no of holes in critical section) = 3 here d= Diameter of bolt hole
t = thickness of thinner main plate
Note: 0.9 fU is same as 0.67fck in RCC (if plate is subjected to direct tension 0.9fU is taken as its ultimate
strength)
Note: There is no concept of Gross area yielding in WSM and the plate is assumed to be cracked along net
section at σ at =0.6 f y
Pt =A net × σ at =(B−nd)×t × 0.6 f y
Arrangement of bolts to get the maximum value of Pt
Chain Riveting Diamond Riveting

From the above analysis we get that Diamond riveting is more efficient than chain riveting because very less
area is reduced from the most critical section of main plate i.e. section 1-1. So Anet is maximum and hence Pt is
maximum.
Note; In the diamond riveting at section 3 3. Anet is very less because of more number of bolt holes so
possibility of failure of cover plate is increased. To increase the strength of cover plates also a special type of
Diamond riveting can be provided as shown in the figure below…

Single bolted joint means there are one bolt line on each side
******For Mild Steel Bolt and Weld******
Ultimate strength (fu) = 410 MPa
Yield strength (fy) = 250 MPa
Shear Strength (τ)= 100 MPa

Single bolted joint Double riveted or double bolted joint

Gusset Plate: is rectangular plate to accommodate hr There are two bolt lines.
connection of different members meeting at a joint
As per IS 800 2000 thickness of gusset plate t ≮ 12 mm

Rivet Value or Bolt Value: It is the strength of a single bolt i.e. how much load a single bolt can transfer.
It is taken as minimum values of PS or Pb for a single bolt
Bolts in single shear Bolts in double shear

{ {
2
πd π d2
RV =Minimum of P S = × f S RV =Minimum of S P =2× ×fS
4 4
P b=( d ×t ) × f b P b= ( d × t ) × f b

No of loads required at a joint,

Factored load (Pu )
n=
Rivet Value( R¿¿ V ) LSM ¿
Working load (Pu )
n=
Rivet Value( R¿¿ V )WSM ¿
Efficiency is the measure of strength of bolted joint in comparison to the solid main plate.
Strength of bolted joint Least value of Ps , Pb , Pt
η= ×100= ×100
Strength of solid main plate Strength of solid main plate
 In Case of WSM
Least value of Ps , Pb , Pt (B−nd )× t × σ at (B−nd)
η= ×100= =
Strength of solid main plate B ×t × σ at B

Specifications of Bolted or Riveted joint

1. Rivets are expressed by its Shank dia or Nominal dia
(𝞍n)
So according to Unwin’s formula
ϕ n=6.04 √ t
t = minimum thickness of plate
2. After adding clearence it becames Gross dia (𝞍d)
ϕ d =ϕ n +1.5 mm for ϕ ≤ 25 mm
ϕ d =ϕ n +2.0 mm for ϕ >25 mm
3. Diameter of bolt hole (for the ease of fitting, extra
provide)
Nominal diameter of bolt Diameter of bolt hole
12mm – 14mm Provide 1mm extra
16mm – 24mm Provide 2mm extra
>24mm Provide 3mm extra
4. Fe 4.6 means fU=400 MPa and fy = 0.6× fU = 0.6×400 = 240MPa
Pitch (p) Gauge(g)
Pitch ≮ 2.5𝞍n (lesser distance causes difficulties in the installation)
Pitch (maximum) = Min of(12t or 200mm) Compression (Buckling occur between bolts)
Pitch (maximum) = Min of(16t or 200mm) Tension (Splitting of plates between plates)
In any case the centre to centre distance should not be more than 32t or 300mm
Maximum pitch of Tack bolts
Pitch (maximum) = Min of(32t or 300mm)
If Tack bolts are exposed to weather Pitch (maximum) = Min of(16t or 200mm)
 Maximum pitch of Tack bolts when steel section like angels section, channel section, and T section are used
as tension member is taken as 1000 mm (shear leg affect)
 Maximum pitch of Tack bolts when steel section angels section, channel section, and T section are used as
compression member is taken as 600 mm (buckling)
Tack Bolts: Not transfer any load just used as stitching
Gusset Plate: is rectangular plate to accommodate hr connection of different members meeting at a joint
As per IS 800 2000 thickness of gusset plate t ≮ 12 mm

End distance Edge distance

 {
Minimum= 1.5 d for machine cut element
1.7 d for hand cut element
(to prevent shearing bearing and tearing failure)
d = diameter of bolt hole
Maximum=12 tɛ=12 t where ɛ ¿
√ √250
fy
(To prevent segregation of plate)
=
250
250
=1

Staggered bolting or Zigzag bolting means two consecutive bolts in the direction of applied load must not be I
the same line.

Maximum gauge is 75 mm
For calculating maximum pitch above pitch values are increased by 50%
Pitch (maximum) = Min of (18t or 300mm) Compression (Buckling occur between bolts)
Pitch (maximum) = Min of (24t or 300mm) Tension (Splitting of plates between plates
Permissible Value for Shear stress and Bear stress for rivets
Rivets fS fb
Shop Rivets 100 300
Power driven field rivets (10% reduce) 90 270
Hand driven field rivets 80 250
 Minimum number of bolts provided to any joint must be two.
 Welded connection
1. Welded connections are more economical
2. In case of reversal of stress bolt connection is better than welded connection.
Fillet Weld
When two plates are to be jointed in different planes
then weld is known as Fillet weld.
Note: Weld is strong in tension and weak in shear that
why it fails along plane 1-1 i.e. plane of maximum
shear that is the parallel to the load applied.
 Fillet weld always gives resistance in Shear

Size of Fillet weld (S) is smaller leg length (root to toe length)
 The maximum size of fillet weld is a function of thinner plate
 The minimum size of fillet weld is a function of thicker plate. Thicker plate will reduce the residual
tensile stresses on cooling
Butt weld Square edge Rounded edge
Maximum size of weld = Plate The maximum size of fillet weld
thickness – 1.5 mm should not exceed 75% or 3/4th of
the thickness of the plate to be joint

As per IS 800 2007 the minimum size is specified as follows

Thickness of thicker plate (mm) Min size of weld (mm)
Up to 10 mm 3 mm (any case)
10 – 20mm 5 mm
20 – 32mm 6 mm
>32mm 8 mm
Note: So in any case size of weld should not be less than 3 mm and not greater than thickness of thinner plate.
As far as possible smallest size weld should be provided because with the increase in size of weld throat
thickness is increased but not in great amount but the volume of weld material increases without increasing the
strength
3. Perfect triangle angle of the weld is 45O
4. For a standard 45o fillet weld the ratio of size of the fillet to throat thickness is √ 2:1
Throat thickness (t t ¿
1. Shortest dimension of fillet weld
2. Welds are classified by its throat thickness and the throat thickness is the weakest section in the throat.
3. It is part of fillet which is assumed to be effective in transferring stresses. The area of throat plane is taken
as effective in design of fillet weld.
4. Since the throat area is minimum it is the weakest in fillet weld.
5. As per IS 800 2007
Throat thickness (tt) = K S
K = 0.7 for angle of infusion 60O – 90O
Angle 60 – 90 91 – 100 101 – 106 107 – 113 114 - 120
K 0.70 0.65 0.60 0.55 0.50

Note: Fillet weld should not be provided if the angel between fusion faces is less than 60o or greater than 120o

Effective length is the length of weld which is assumed to be effective in transferring load.
1. Effective length of Fillet weld = (Actual length (L) - 2S)

As per IS 800 2007 actual length must not be less than 4times the size of weld
Effective area is the area of weld which is assumed to be effective. This is taken as the throat area.

Note: Weld area is treated as line area i.e. the higher power terms of thickness like t2, t3 are neglected in
calculations.
Overlap of the plates to be joined by the fillet weld should not be less than 4 times the thickness of thinner
member to be joined or 40 mm whichever is more.

Pitch of weld (same as bolting)

For weld in compression zone For weld in tension zone
Pitch (maximum) = Min of(12t or 200mm) Pitch (maximum) = Min of(16t or 200mm) Tension
Compression (buckling) (Splitting)

Note: Weld at the junction of flange plate or web plate is designed for the horizontal shear stress (direct shear
FA y
stress) i.e. τ = I b

Slot Welding
If the overlap length is limited then slot welding is done by making slots in connecting plates
The slot thickness should not be less than 3t or 25 mm (whichever is less)
Strength of Weld (Always weak in shear that’s why shear strength)
Shear strength of Fillet weld =PS =( Leff ×t t ) × f S
fU
f S= Max permissible shear strength of fillet material= ( LSM ) =100 MPa (WSM )
√ 3 ×1.25
Questions
 Analysis of moment free connection
It means we have to ensure that there is no moment in the welds in the welded connection.
End fillet is not provided End fillet is provided

Eccentric Connections
If the cg of bolt group or weld group does not lie on the line of action of load, then the connection is referred as
eccentric connection.

Two types
In plane Eccentricity Out of plane eccentricity
 Load is in the same plane i.e. in the plane of bolt group  Load is in the different plane i.e. bolt group or
or weld group. weld group and loads are in different plane
 Twisting moment on the connection  Bending moment on the connection
Case 1: Bolted Connection Case 1: Bolted Connection
Bolt group and load is in the same plane but line of action Direct load P and Bending moment M = Pe
of load does not pass through the cg of bolt group due to Due to the direct load P all the bolts are subjected to
which connection is subjected to Twisting moment. shear and bearing. Due to BM bolts above NA are
1. Bracket is subjected to moment but Bolt group is subjected to tension but the bolts below NA are useless
subjected to twisting moment. in resisting compression because the compressive
 2. Due to the direct load P direct shear stresses stress below NA is resisted by flange of I section and
developed in the bolts and due to the twisting Bracket below the Na by pressing against each other.
moment torsion shear stresses are developed in the The bolts above NA are therefore subjected to
bolts. shearing, bearing and tension and bolts below NA are
subjected to shearing bearing only.

For Bolts above NA is analysed by the interaction

equation given the IS 800 2007

( )( )
2 2
PS Cal P t Cal
+ ≤1
PS Pt
PS cal = Calculated Shear force in each bolt
PS = Shear capacity of in each bolt
Pt cal = Calculated tensile force in each bolt
Pt = Tensile strength in each bolt

Direct shear force in each bolt due to the direct load P

P
F 1= ×A
∑ Ai i
Direct load P is shared by the bolts depends on their area.
If all the bolts are same (equal area)
P P
F 1= × A i= … … … … ..(1)
n× A i n
Direct shear force always acts on the direction of external Case 2: Welded Connection
applied load. The effect of eccentric load on the CG of weld group is
the direct load P and the BM M =Pe. Due to the direct
Torsional shear force due to the twisting moment (T=Pe) at load P direct shear stress f1 is developed in the weld at
the connection point A.
( )
τ
F 2=T i × A i= ×r × Ai
j
f1 is uniform throughout the weld.
Due to the BM Pe bending stresses f2 are developed in
Where J=∫ r dA=∑ r i Ai
2 2
weld.
(As bolts are discrete function that’s why we are using f2 is maximum at extreme fibers
Summation not integration) In this case compression is also occurring because
below NA the gap is filled by weld material and the
So putting the value of j, F 2= ( τ
∑ r i2 A i )
× r i × Ai compressive force is transferred through weld only.
P
Direct Shear stress , f 1=

( )
Pe Pe ( D ×t ) × 2
F 2= × r i × Ai = ×r i … … …(2) M
∑ ri A i
2
∑ r i2 Bending compressive stress at point A due ¿the BM , f 2= ×
I xx
F 2 ∝ A i × ri
Torsional shear force always acts on the direction of Resultant stress , f r =√ f 21 + f 22 ≯ f s
external applied torque or perpendicular to the radius of
bolt. As per IS 800 resultant stress can be taken as
Resultant shear fortce ( F R ) =√ F + F +2 F1 F2 Cosθ
2
1
2
2
follows
 FR is dependent upon F1 and F2 both and the angle Resultant shear stress , f r= √ 3 f 21+ f 22 ≯ f s
between them (cosθ). If all the bolts are same then
value of F1 is same for all and F1 is maximum for the
bolt which is furthest from the CG.
 FR is more significantly dependent on ri compared to θ
as maximum value of cosθ is only 1.
 For bolts to be safe FR should not be more than
permissible shear capacity (Rivet Value=RV)
Case 2: Welded Connection
The effect of eccentric load on the CG of weld group

Question Which of the following bolts are more critical?

Since A B K L has same radial distance from the cg of bolt group but B and L has small value of θ. Then B and
L bolt has felt the maximum force.
If we have to choose between B and L also then we choose B as it is nearer to the load P.

For this combination 4th number bolt is stressed more because it has least
value of θ = 0o
Question
Each bolt in the figure below resisting a shear force of 20 kN and a tensile force of 15 kN. What will be the
maximum factored that can be applied in the bracket shown below.
 The resulted compressive force below NA is assume to transferred by bracket and flange of I section and
is assumed that it acts at bottom most bolt level.

Qustion

Methods for inspection of weld

1. Magnetic Method
2. Di penetration method
 3. Ultrasonic method
4. Radiography method
Classification of weld
Shop Weld Site weld
1. Factor of Safety is 1.25 1. Factor of Safety is 1.50
 Truss Structure

Definition: Axially loaded member.

 All the members are straight need not to be prismatic and should be connected by smooth pins.
 To ensure no bending in any member the loads are placed on joints only and never on the members.
Strut is a compression member of truss. Tie is the tension member of Truss
Column is the compression member of frame that is it can have certain eccentricity i.e. bending.
Beam is structural member primarily subjected to transverse loading.
Beam Column is a structural member primarily subjected to bending but therse can be some amount of axial
forces.
Slenderness ratio (λ)
Effective Length Leff
Slenderness Ratio ( λ )= =
Corresponding Radius of gyration r corresponding
Leff
If the column has only one effective length, then λ max=
r min
 If Slenderness ratio is more than the member is easy to bend
Radius of gyration (r or k)

Radius of gyration ( r xx )=
√ I xx
A
Member Type Maximum Slenderness Ratio (λ)
Pure Tension member 400 (More value because there is no possibility of buckling)
Note: If λ is greater than 400 then tension members acts like cable or wire i.e. it will sag due to its own weight
and behaves like a wire.
Pure Compression member 180 (Lesser value because there is possibility of buckling
Pure tension members subjected to reversal of 180 (Lesser value because there is possibility of buckling)
stresses due to Live load or Dead load.
Pure tension members subjected to reversal of 350 (more value because there is lesser possibility of
stresses due to Wind load or Earthquake load. buckling) Earthquake and winds does not happen frequently.
Member subjected to compressive forces 250 (Compression member nut Secondary )
resulting only from wind and earthquake action
which does not affect other member.
For Compression flange of a beam restrained 300 (Compression flange because buckling is characteristic
against torsion and Lateral buckling. of compression only)
Hanger Bars 160 (Tension member but not in truss) (Used in Fan
psychological fear)
Lacing or Batter 145
Question
 Tension member
Shear Leg: is non uniform distribution of stress and strains.
 Shear leg is the main design issue in case of Tension member
 Shear leg exists only in connection level
 Shear leg is avoided by providing Lug angles which is a small piece of angles
Note: as per theoretical aspect tension member are most efficient structural members because there is not any
possibility o buckling and also the member can be stressed beyond the yield point.
Tension member is adversely affected in connection level
The most common modes of failure at tension member are as follows
Shear Leg
Shear leg is non uniform straining of a member deu to tension.
Shear reduces the efficiency of a tension member which is directly connected to gusset plate.
As long the resultant action and resultant reaction passes through the CG or C/s the strain and stress distribution
will be uniform throuout the c/s and there will not be any strain and stress distribution and hence there will not be
any shear leg.
Example of rubber sheet and pull it at both ends as shown in figure.

Right face of element A has been legged behind to the left face in shear so the term shear leg is derived.
Lug angle is a small piece of angel used to connect the outstand leg of a tension member with the gusset plate.
The purpose of lug angle is to reduce the shear leg effect and to reduce the length of connection to the gusset
plate.
  The loss in efficiency due to shear leg should not be more than 30% (as per IS 800 2007)
𝜼 ≯ 30%
 As per IS 800 2007 shear leg factor β should be multiplied to the outstanding leg area to take care of loss
in efficiency.
 Shear leg factor β should not be less than 0.7
β ≮ 0.7
For Single Angle For double angle placing back to back
3 A1 5 A1
β= β=
3 A 1+ A 2 5 A 1+ A 2
Effective Area, A effective = A1 +β A2
Tensile strength of plate, Pt = Aeff × σat
Long Joint
If the length of joint is more than 15d or 150tt , it is referred as long joint. It is assumed that the applied load is
shared by all the bolts equally but when the applied load is within the proportionality limit the outer bolts take
more load than inner bolts. So assumption is valid when failure condition
In long joints End bolts take more loads then inner bolts. So the failure is sequential, starting from end bolt to
centre.

 As per IS 800 2007 if the length is more than 15d or 150tt the Shear strength (PS) should be multiplied by
a reduction factor to take care of loss in efficiency.

Grip length
If the grip length increases the efficiency decreases because of additional bending stresses developed in the bolts.
To reduce the effect of unwanted bending stresses in the bolts, Is 800 2007 specifies that the grip length shoul not
be more than 5d.
 5d < Lg < 8d then Ps should be multiplied by a factor to take care of additional bending stresses.
If Lg > 8d then the section must be redesign.

Block Shear failure

If the shear strength of bolt and the bearing strength of plate is high, then a portion of plate, may fail as a block
due to shear on one side and tension on the other side as shown in figure.

Questions

Load carrying capacity of tension member:

LSM WSM
fy pt =A net × σ at =A net × 0.6 f y
Based on gross area yielding, Pt =A g ×
1.1 whichever is less There is no concept of gross area yielding
0.9 f U
Based on net area cracking , Pt =A net ×
1.25

Calculation of Anet
 Chain Riveting Diamond Riveting

Staggered Riveting

( )
2 2
S S
along failure path 1-2-6-3-4, Anet = B−nd + 1 + 2 ×t
4 g 1 4 g2
Tensile strength , pt =A net × σ at
2
S1
IS 800 2007 specifies that at the critical path corresponding
4 g1
for each inclined path.
 The factor affecting critical failure path is n, s, g.
 Critical failure path will be that path where Anet is
minimum.

Question: Which of the following is most suitable for tension?

a. ISF
b. ISST
c. ISSC
d. ISA
Answer:
Angle section has one outstanding leg. Channel has two and T has one. So Indian standard flats are most
suitable.
Since Flats may flatter when wind is there. So T section can also be used sometimes. Flat is theoretically most
suitable for no shear leg effect.
If Angel is used the unequal angles are used and longer legs must be connected to gusset plate.
Question:
 Compression members
1. Column and strut
2. Main design issue is Buckling
3. Strut are the inclined compression member
4. Euler’s Critical buckling load
2
π EI
Pcr = 2
Leff
Unsupported length is the maximum clear distance between top of floor to the bottom of beam or roof.
Effective length( Leff) is the length of column between the points of zero moment i.e. between the point of
inflection and point of contra flexure.
IS 800 2007 uses L for unsupported length and kL for effective length. Values of k as follows…
2
End Condition Diagram Effective length π EI k
Theoritical Recommende Pcr = 2
Leff
value d value IS 800
L L 2
Both ends are effectively held in π EI k =1
position but not restrained against Pcr = 2
L
rotation
One ends are effectively held in L 0.80L 2
2 π EI k =0.8
=0.707 L Pcr =
position and restrained against √2 L
2

rotation but other is not restrained

against rotation
Both ends are effectively held in L 0.65L 4 π EI
2 k =0.65
=0.5 L Pcr =
position and restrained against 2 L
2

rotation
2L 2L 2
One ends are effectively held in π EI k =2
position and restrained against P cr = 2
4L
rotation but other is neither held
in position nor restrained against
rotation
One ends are effectively held in 1.2L 1.2L k
position and restrained against =1.2L
rotation but other is only
restrained against rotation not
held in position
One end are effectively held in 2L 2L
position but not restrained against
rotation but the other end is only
restrained against rotation not
held in position
Question:
1. If 5m electrical pole or 5m chimney is given then effective length = 2×Length = 2×5m = 10m

Radius of gyration is the distance at which entire area must be kept as a strip so that it will have the same MOI
as that of original section.
 Radius of gyration is a measure of resistance to rotation, bending or buckling. If its value is more than it
is difficult to rotate or buckle that member.
 Radius of gyration (Resistance to buckling) can be increased by displacing the same area away from
axis.
  *** Compression members’ column and strut always tends to buckle about the weak axis. Weak mean
Leffective
that axis about which MOI and radius of gyration in minimum and λ ( ¿ is maximum.
r corresponding
 Wheel example
 So Euler;s critical load formula also we write Imin or rmin or λmax

K xx = xx
I
√
A
Leffective
Slenderness Ratio ( λ )=
r minimum

5. Euler’s Critical buckling load

Where r minimum=Radius of gyration= min
√ I
A

π E I min π E ( A k min ) π 2 EA π 2 EA
2 2 2

Pcr = = = 2 = 2
L2eff L2eff L eff λ max
2
k min
 The most efficient and economical section for resisting compression is thin hollow circular like well
foundation because for a given area Imin and rmin comes out to be maximum.
Euler’s Critical buckling load depends on
1. Material of column (E)
2. End conditions (Leffective)
3. Area of C/S (A)
4. MOI of C/S (Imin)
5. Axial Rigidity (EA)
6. Flexural rigidity (EI)
7. Axial stiffnes (EA/Leff)
8. Flexural stiffness (EI/Leff)
9. Slenderness Ratio (λmax)
10. Euler’s Critical buckling stress
P cr π 2 E
f cr = =
A λ2max

Rankines Formula

Design Strength of Column as per IS 800 2007

1. Perry Roburshun Approach (Earlier IS 800 1984 using the Merchant Rankine formula)
Because of the assumption involved in the derivation of Euler’s Critical stress, we get higher value of strength
of column than then actual. But columns actually buckle at lesser value of strength than fcr. So IS 800 2007
Euler critical stress and given the formula of design compressive strength of column

Buckling Class a b c d
Imperfection factor 0.21 0.34 0.43 0.76

Imperfection factor depends upon

1. Shape of column cross section
2. Direction in which buckling occurs
 3. Steel fabrication process (hot rolling etc)
4. Initial bow of column
5. Defect in material.
So the design compressive strength of column
Pcr =f cd × A g
f cd =design compressive stress of column
A g=Gross cross sectional area of column
Note: Incase of compression member we always take gross area but in case of tension member we always take
the net area of column.

Analysis of Truss:
1. If a strut is spanning between two gusset plates only then it is called discontinuous strut.
2. If a strut is spanning between more than two gusset plates then it is called continuous strut.
Effective length of Strut
If a single angle provided between two gusset plates as If a single angle discontinuous is connected two or
a discontinuous strut. more than two bolts or by welds to the gusset plate at
each end than
Leffective =L
Leffective =0.85 L
If two angles are placed back to back and connected to Note: If there is a possibility of buckling of the strut
gusset plate then perpendicular to the plane of strut (out of plane
Leffective =0.707 L¿ 0.85 L buckling)
Leffective =L
Depending upon the orientation of strut it can buckle it
can buckle in the plane of truss or perpendicular to the
plane of truss.
Question:

Principal Axes are the ax about which moment of inertia is either maximum or minimum and product of inertia
(Ixy) is zero.
 ydA is the 1st moment of area which is helpful in finding the CG of cross section.
 y2dA is the 2nd moment of area and measure of resistance to rotation, resistance to buckling etc. It is also
referred as Moment of Inertia.
 xydA is called product of inertia and is helpful to find out moment of inertia about any other axis.
Note: MOI is always a +ve quantity but the POI can be +ve –ve or zero depending upon the orientation about x
and y.
If a C/S is symmetrical about any axis then POI of the area w.r.t symmetrical axis and any other perpendicular
axis is always zero.
If the two axis are also passing through the CG OF C/S then are itself principal axis.

POI (Ixy) is a mathematical term. If Ixx, Iyy and Ixy are known then we can find the MOI about any other inclined
axis. The major and minor principal MOI i.e. the MOI about major and minor axis are as follows…

Note: Compression member column or strut always tens to buckle about weak axis. Weak means that axis about
which MOI and radius of gyration are minimum.
Question:

Built up columns
Lacing
Angle of lacing is in between 40o – 70o
Width of lacing
Lacing of compression member is subjected to a total transfer of Shear is 2.5%
of axial force in the column
Diameter of Rivet Width of lacing bar (almost ×3)
16 mm 50 mm
18 mm 55 mm
20 mm* 60 mm*
22 mm 65 mm
Batten
1. Effective length of battened column is increased by 10%
2. Carrying 2.5% of axial load
3.

Splices
1. These are used for increase the lengthy of column
A member acting as tie in roof truss is call tension member

A structural member subjected to tensile forces at the end are called Tension member.
1. Permissible stress in axial tension = 0.6 f y
Calculation of effective area:
Case 1
Net area= ( B−nd ) × t
B = Width of plate Tension
n = number of holes
d = gross dia****
Note : while taking compression net area =B×t
Case 2

( )
' 2
nS
Net area= B−nd+ ×t
4g
n’ = number of staggered pitch( no of zigzag)
g = gauge distance
S = Pitch

Single angle connected through one leg Pair of angles connected back to back on same side of
Net area= A 1+ k A 2 gusset plate
3 A1 Net area= A 1+ k A 2
k= 5 A1
3 A 1+ A 2 k=
5 A 1+ A 2

Double angles connected back to back on each side

of gusset plate
Net area= A g −Dudiction of rivet holes
 Minimum thickness of Gusset plate is 12 mm.
Lug Angle
 It is used to reduce the length of connection
 Beams
A structural member subjected to transverse load is called BEAM.
When provided in building it is called JOISTS.
A large beam supporting number of joists rests is called GIRDER.
Beam carrying roof load is called PURLIN
Beam which supports loads from the masonry over opening is called LINTELS
A roof beam supported by Purlins is called RAFTER.
Laterally restrained beams:
Beam which compression flange is restrained laterally
M σ E Average shear stress = 0.4fy
= =
I y R Maximum tensile stress = 0.6fy
I Bending stress = 0.66fy
M =σ × =σ × Z
y
Where Z = Section modulus Bearing stress = 0.75fy
Maximum permissible deflection in a beam
Span
As per IS 800 (WSM) δ=
325
Span
As per IS 800 (LSM) δ=
300
Purlins
WL
 Maximum bending moment in case of purlins is =
10
1
 In case of wind the bending stresses are increased by 33
3
Plate Girders
 Shear is primarily resisted by WEB
 Economical plate girder has minimum weight


Economical depth of plate girder d=1.1
M
√
σ tw
Web plate is called un stiffen if the ratio of the clear span to the depth is less
than 85
l
<85
b
Web Buckling Web Crippling or Web yielding
The web I rolled steel sections behave like a Localized buckling. To prevent Crippling
column when placed under concentrated BEARING STIFFENERS (Angle section)
load is used.
Gantry Girders

Gantry girder can carry

1. Longitudinal Load
2. Lateral Load
3. Vertical load
 Plastic Theory
Shape factor
Plastic modulus Z p
Shape factor = =
Elastic modulus Z e
Values of shape factors for different sections
Square Section Rectangular Section Circular Section Triangular Section Diamond Section
1.5 1.5 1.7 2.34 2