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The document outlines various aspects of steel structure design, including its advantages, disadvantages, and methods of design such as Working Stress Method (WSM) and Limit State Method (LSM). It covers the analysis and design of joints, tension and compression members, beams, plate girders, and roof trusses. Additionally, it discusses the structural behavior of steel, emphasizing its strength, ductility, and the challenges of corrosion and fire resistance.
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0% found this document useful (0 votes)
82 views209 pagesSteel
The document outlines various aspects of steel structure design, including its advantages, disadvantages, and methods of design such as Working Stress Method (WSM) and Limit State Method (LSM). It covers the analysis and design of joints, tension and compression members, beams, plate girders, and roof trusses. Additionally, it discusses the structural behavior of steel, emphasizing its strength, ductility, and the challenges of corrosion and fire resistance.
Uploaded by
mp8315199Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF or read online on Scribd
/ 209
me)
bi “p
er tot yan"
US tO CO) ayn
| POKHARA UNIVERSITY
|
Er. Kamlesh Ray
Er. Rajendra Air
TPT Chapter 1
Introduction
11
1.2
13
14
Contents
Structural Behavior of Steel ..
‘Advantages and Disadvantages of Steel Structure
Methods of Design of Steel Structures (WSM, LSM).
Plastic Method of Design of Steel Structure (Ultimate
Load Method)...
Chapter 2
Analysis and Design of Joints
21
2.2
23
24
25
Types of Riveted Joints.
Types of Bolted Joints..
Types of Welded Joints .
Axially and Eccentrically Loaded Bolted Joints (Bracket
Connection)
‘Axially and Eccenttically Loaded Welded Joints .
Chapter 3
Design of Tension Members (LSM)
3.1
3.2
3.3
Net Cross Sectional Area of Tension Member ...
Design of Structural Members in Tension (I, Angle and
Channel).
Design of Lug Angles
Chapter 4
Design of Compression Members (LSM) 165-222
41
42
43
44
45,
4.6
Computation of Permissible Stress for Compression
Members.
Design of Standard Steel Sections for Compressive Load...... 166
Design of Built up Members....
Design of Lacing and Battens..
Design of Eccentrically Loaded Column ..
Design of Column Spices... Design of Col 223-244
yn of Column Bases (LSM) -
for Axially Loaded Columns
51 Design of Base for Ax a
Loaded Columns,
52 Design of Base for Excentrically
Chapter 6 sin om
Design of Stee! Beam (LSM) 8
61 Design of Latrally Restrained Beams 285
62 Web Crippling fect and ingot Ba. 368
65 Dg Uri eam (Case 822) 8
64 Design of Built-up Beams.
Chapter 7
De’ign of Pate Girders (LSM) 299-316
7 Hlmentsof Pte Gide ~ 28
72 Curtailment of Plate. 300
73. Design Procedure for Plat Gieder. 300
7A Optimus Depthof Pate Gide. on
Chapter
Design of Roof Trusses (LSM) 317-366
‘1 Typesof Roof Truss and Ther Selection 20
82. Load Calton in Ro! Tress onc 322
83. Designo Pati nn 38
BA Designo Becngynnnnn 331
Chapter 9
Tinbertrvctores es - 10
81 Allowable Sirsa in Tiber soe eames
92. Design ool Built-up and Spaced Columna.. gp
93. Designof Timber Beams es
94 Types of Joints and Their Detailing es
95 sol and Des f Comes Beso Sel and
Annex A 7 a
oe is
Irvadction et Chapter 1
Chapter 1
Introduction
Tal Structural behavior gf sted ae
12. Advantages and disadvantages of ste! structures cn 3)
13. Method sof design of ste! structure (WSM and LSM) wn1-3)
14 _Plasticmethod of design of steel structure een 5
A steel structure is a metal structure which is made of stractral stl
‘components connected to each other to cary leads and provide rigid
‘Because of the high strength of ste these structures ae reliable and requite
fewer raw materials than some other typeof structure such concrete,
In stool structures, structural ste i the main load carrying material to
‘transfer the load within them and to transfer load tothe ground.
‘Common steel structures
8) Roof truss in factories, cinema hall, allways ee
) Crane girders, columns, beams
Plate girders, bridges
<4) Transmission towers, water tank, chimney ete.
1.1 Structural Behavior of Steel
“The behavior of structure depends on the load transferring ation of its
‘members and joints, This may be almost entirely by axial tension oF
‘compression, as inthe triangulated structures wit joint loading as shown
inthe figure below:
£
(0) Axial force (0) Bending 1p Design of tel & Tiber Sractres
MM
{
cntetntoee
wy bending and shear actions. Usually the bending, action
ae Re a ra
Sy
Alternatively,
G) Geometic nom insrty
XQ Mattia nontinerity
Material and geometric nonlinearity
Deformation
ig Site aor
Purely flexural structures, and purely axial structures with lightly loaded
‘compression members behave as linear. However, structures with both
flexural and aval actions behave non inary, even nea the service loads,
This isa result ofthe geometrically nonlnea behavior of its members
Most steel structares behave nonlinearly near thee ultimate loads, unless
they fil prematurely duc to brite fracture fatigue or local buckling This
‘non-linear behavior is due either to.
Duckling or both.
Inaxial structures failure may
bucking either of some compression members or ofthe rams Lot,
material yielding or member or frame
Inodtion ea Chapte | 3
1.2 Advantages and Disadvantages of Steel Structures
‘The advantages of stel over other materials for construction are:
4) thas high strength per unit mass, Hence, even fr large structures,
the size of steel structural element is small, in
sina, saving. space
‘onstruction and improving aesthetic view.
by Ithasassured quality and high duabiiy
2) Speed of construction is another important advantage of steel
structure, Snee, standard sections of steel are available which can be
prefabricated inthe workshop/site, they may be kept ready by the
time the site is ready and the structure erected as soon a the
rwady. Hence there is lot of saving in construction ti
4) Setstatrrcn be eget ar ery:
©) By sing ttl crm net eases cn Be aly
hand vpn ste guy
6) ite ae nan wii he Bot wer and pa
data Bee on ton fe tag arin
Mite raaie
Tedd of te tee
1) nase coms
5) Nanna ot gh ce ds unig ope
iy Sectmemtes arco
13 Methods of Desig of Sie! trates SM ISH)
Desi a sel suc cot fd sl ments air
Ccvetan ota opiate dre
‘comes
Doge atl races on loving en
2) Winger eto HS)
B} intone deen)
3 Uinta eign) oe
S) Waking or metho (VSN); Thi i he let yma
7 ey Sent hs the aes ae os
SOT E ST ue ego etl ot ie ne a
ar emt detping donee ey Hes,
See tinct roost cmbraionot wong | Design of Stel & Timber races
F allowable or
toad and the members are proportioned such that allows
permissible streses never exceeded
Working stress < Permissible tres
vel tress
or Working stress Fair of saftyFO5)
vantages of working stress method
2) Remi ree
stress method
Disadvantages of working tess ethod
j) Tis oes an uneconomial design ofthe structures,
i) edo apply ther val of the safety factor.
ii)_tesa time consuming method.
Lint sate method: The lint ste method, considers the ultimate
stent of material which is ignored in the working stress method
nd aso asures thatthe siructue i serviceable for it intended
petod of design.
“The various iit tates tobe considered indesign may’ be grouped
inthe aon wo major categories:
4) Limit stat of strengths The limi state of strength inclades
+ Los of ql of whole or part ofthe structure.
+ Lasof stability of structure asa whole or pat fit
+ Fall by excessive deformation,
+ Frecture dt fatigue
+ Breet.
4) Limit sat of servcebliy: Te limit state of servieabity
»)
‘Vibrations in structures or any past of its component
limiting its functional effectiveness
‘+ Repairable damage or rack due o fatigue,
+ Corson.
© Bre,
‘Advantages of limit state method:
+ Ttuses a modem way of designing a structure that
‘Wide range of logical and technical considerations, :
Intvodtion ca Chapter 1} 5
It includes separate consideration
typeof material and type of oad,
‘This provides an economical design ofthe structure.
Disadvantages of limit state method:
* It wses possible approach for design so it cannot provide
‘oper an savy dae nea
‘of loads and peak conditions. “ee =
of diferent kinds of failure,
1.4 Plastic Method of Design of Steel Structure (Ultimate Load
Method)
‘The limitation of working stress method to asses actual load carrying
capacity, made researchers to-develop ultimate load method, which is
also known as loa factor method (LEM), When applied to steel strctures
it is referred as plastic design method.
In this method a section is said to have forme plastic hinge when all the
fibers yield. After that it continues to resist load which as caused plastic
Fringe but will not resist any more load. But structure continues to resist
further load til sufficient plastic hinges ae formed to developed collapse
mechanism.
In this method safety measures are introduced by suggesting a load
factor, which is defined as the ratio of design load to working load, The
suggested load factors as per 15 8001984 were shown below:
SI No. Workingiond (Minted
1 [Dead oa 7
2_[Dead ade imposed oad 7
2, [Dead lad wind ora sd 7
[ead oad + imposed ond racimictond [13
Advantages of ULD
j) Redistribution of internal forces is accounted.
fi) _Itallows varied selection of load factors
Disadvantages of ULD
It doesn't guarantee serviceability performance. To acount for this 1S
800-1984 suggested limitations on deflection. However, it didnt guarantee
‘other serviceability limits ike instability and fatigues Crna
‘Write short notes on
antage of steel
[p03 Spring, 2018 Fal]
age and dis
Solution: Advantage of steel a
tvengts tetas gh sen io
eb mater or consiaction FO
damage from natural disasters such
1g it a popul
and other large
strongand d
Durability Ste a
a eathguakes and hurrcanes, a
aia
es re ni
cost, Ey aera conse
ee ct an ele 1 6 leet
come revo acomplh ir
‘ison forthe structure,
‘Quick fabian time: Sel stractres can be fabric
aehing tem an economia choice fo constraton projets
bier lar choice for
creative fred
ted quickly,
Susana Sto! i 2 sustainable materia that can be recycled
nd reused reducing wasteand environmental impact
Disadvantages of tel
than other
Initial cost: Steel is typically more expensive
construction materials, suchas concrete or wood, making ita costly
‘option for constrtion projects.
“Maintenance: Solis susceptible to rust and corrosion, requiring
‘regular inspection and maintenance to prevent structural damage
‘Weight Sto eaviee than other construction material, which can
make it more dificult to transport and handle on construction sites.
Valneability to corosion: Ste is vulnerable to coreosion when
exposed to airand water, requiring regular painting and special care.
Fire resistance: Stel is not fire-resistant and requires additional
fireproofing measures to meet building codes,
2. Wie shor notes on tnt sate method as Become more pops
than working ses mend tats Spring
Solan Uni ste method as tse more pope than working
seve mtn So ir apo ay ae
mae stenghandsieblyrgeements
Hee are somehydiferences betwen te to eee
Irrodution Chapter 1) 7.
Working stress method
+ Based on the elastic theory which assumes that concrete and
steel are clastic and the stress-strain curve linear for both,
+ Design load is kased on uniqueness theorem, desig strength
fof material is based on lower bound theorem,
4+ Serviceabilty isnot considered
+ Thasic method and its hnowledge i essential for understanding,
the concepts of design.
Limit state method
4+ Based on predictions unlike working stress method which i
deterministic in nature, assumes that the loads factors of safety
and material stresses are hnown accurately
Takes into account the ultimate strength of the structure and
also the serviceabilit
‘+ A judicious combination of working stress and ultimate load
‘methods of design.
+The acceptable limits of safety and serviceability requirements
before failure occurs is called a limit tate
+ Divided into two parts: Limit state of Strength and Limit state
of Serviceability
+ Overal, the limit state method is based on physical parameters
‘and partial factors that are based on statistical and probabilistic
rounds, which can be controlled. This makes it a more
rational and reliable method for designing structures
Write short notes on structural behavior of steel. [2015 Spring]
Solution: Structural behavior of steel
Duetilty: One ofthe key advantages of ste! as a structural material
is its high ductility, Ductilty refers to the ability of a material to
‘undergo significant deformation befor failure, allowing it to absorb
during loading, This property makes steel structures more
resilient andl better able to withstand dynamic loads such as seismic
events or impact loads
Strength: Steel is known for its high strength-toweight ratio,
‘meaning it can carry substantial loads without being excessively
heavy, ‘This characteristic allows for the design of slender and
efficient structural members, reducing the overall weight of the
structure and facilitating ease of construction. ——————
1 oso Sel Tnber SIUC
cast behavior, which means
excellent
Basic be aes Searmatons under fd and return ti
cn ene pe aires Tis ope er ha
Seer sutures can withstand
iene arin ei
‘or faling under compression :
sr rettance: Whi tet ass some of siengih when exposed
Fe al mc nai
1h ent al se cts an
ee re tf ny
Cention cet andeo
SO a onrcion pr
ca cane encom me
Seen enh! pope fs canbe aoe
ote
‘ns cg) erent ond pectese Bw mises
Sonar ntact sy.
Sense Mee rely da nfs Bl
msl mo el
fractue in eel structures ean
Comer Wie
= cal oper i
ssp cotson wien expos to mest nd agresive
‘ctenial to enhanee the er comes and maintenance are
‘Settee
ral tonalite mah
wide range of construct poplar choise Sar;
lig nds ache an anh
Scone ies
‘minor deflections and vibrations
Irvoduction ot Chapter 1] 9
Write short notes on limit state and stress method of
Can {2015 Fat, 2017 Fall, 2018 Fall]
Solution: Limit state method and working stess method are two diferent
approaches tothe design of structures inckuding’ steel structures, that
aim to ensure their safety and serviceability. Both methods are
commonly used in structural engineering, bu the Limit State Method
{snow the more widely accepted and adopted approach.
‘Working Stress Method (WSM): Inthe Working Stress Method, the
design is based on permissible stresses or working stresses. The
permissible stresses are typically determined based on the yield
Strength ofthe material used in the structure, The idea behind this
‘method is to ensure that the applied loads do not cause the
‘material’ stress to exceed its allowable working stress.
“The design process involves calculating the stresses and deflections
caused by the applied loads and comparing them with the
permissible stresses, If the calculated stresses are within the
permissible limits, the design is considered safe. However, this
fnethod does not explicitly account for the probability of loads
‘exceeding design loads or the actual behavior ofthe material
‘Working stress method has several limitations including the lack of
‘a clear safety margin against failure and the inability to account for
‘Variations in material properties and load uncertainties. Because of
these limitations, the use of the Working Stress Method has been
largely replaced by the Limit State Method in modem structural
design codes.
Limit State Method (LSM): The Limit State Method is a more
‘advanced and rational approach to structural design. Its based on &
thorough consideration of various limit states that the structure
right encounter during its lifespan. A limit state is a condition
beyond which the structure may no longer fulfill its intended
function or where it becomes unsafe.
In the Limit State Method, two main limit states are considered:
‘Ultimate Limit State (ULS): This limit state deals withthe safety of
the structure under extreme loads, such as maximum dead Toads,
tive loads, wind loads, or seismic forces. The design ensures thatthe
‘structure can safely carry these loads without experiencing failure oF
collapse.
Serviceability Limit State (SS)
functionality and comfort ofthe strict
considers criteria such as deflection, vi
fensure that the structure remains
performance requirements,
“This limit state addresses the
ure during is service life. I
‘bration, and durability to
serviceable and meets 10 | Desgn of tel & Timber Sructres
te method include its rational approach,
ofthe iit stl :
Advantages ofthe I “seempsibiity with modem analysis
better safety assessment, a
techniques.
‘Te design proces in the Limit State Method involves calculating
me Sp ht
structure satisfies ll elev :
mit State Method provides a more comprehensive ond
Paces ‘ taking into account safety,
te url desk
Le ey uceran as eoe the peers
rmethod in modern engineering practice and is adopted in most
Trerational design codes and standards for various types of
stractars, nla tel traces,
5. Write short notes on comparision between WSM and LSM of
design of tee structure
[pms Fal 206 Fl 2017 Spring, 2000 Fal, 021 Fall, 2022 Fa}
Salton The working stress method (WSM) andthe Timi state method
(LSM) ae approaches used inthe design of tel structures
Here acomparzon betwen the io methods:
Working ses method
+ WSN is an caste design method that assumes that the
mater bchaveselastally and the relationship between loads
ane suestsi ier.
‘The permissible stesso design stress calculated by dividing
the Garces eg af the moter bythe factor of
‘safety. a iH
The design based on man
manning tess ina strctre
‘below a certain level, a
WM is simple and real, bi
le ut dows not consider the
iat sen eral estate
Limit stte method
+ LSM is pase design method ae
et considers not only the
‘rng ofthe race bua issbiliy sed een
Tedsgn sts on theo oen
acceptable limits for safety and servi sea cee ace
=e leabilty before failure °
Inoduction Chapter 11
+ In LSM, ultimate toads sidered for safety and working
fare considered for safety an
loads are considered for serviceability, :
+ The streses in an element are obtained from the design loads
and compared with the design strengths.
‘The partial safety factors are derived using probability and
statistics and are different for dlferent lood combinations,
hhence giving a more rational and scientific design procedure
‘Comparison:
WSM is an older method, while LSM sa modem method.
WSM is based on the elastic theory, while LSM is based on the
plastic theory.
+ WSM assumes thatthe loads, factors of safety, and material
streises are known accurately, while LSM_is based on
predictions and considers partial safety factors derived using
probability and statistics,
+ WSM is simple and reliable, while LSM is more comprehensive
and rational
‘+ WSM does not consider the ultimate strength or failure ofthe
structure, while LSM considers ultimate loads for safety.
In surnmary, the working stress method isan elastic design method
that is simple and reliable but does not consider the ultimate
strength or failure ofthe structure. The limit state method isa plastic
design method that is more comprehensive and rational,
considering the ultimate strength, stability, and serviceability ofthe
structure. The LSM is the preferred method of design for steel
structures in modern times.
6 Explain about safety and serviceability requirement. of steel
structure. T2019 spring]
Solution: Safety and servicebily are two fundamental requireménts
that must be carefully considered in the design, construction, and
‘maintenance of tel structures
‘ofthese requirements are essential to ensure thatthe structre
an nel agate yi lig 28
Feliable environment for occupants and uses.
Safty Safety isthe primary concern inthe design a constrution
of any strctre Forse! structures safety conser inde
the ability ofthe structure to withstand al anticipated loads
‘potential hazards without experiencing failure o collapse. a
12 sg of tel Tb SEAS
taken into account: q
quirement pe stable under all expected |
ast
srt veight), live Toads
rods (cl-WeH8
aang ded fs and any other
an pds, seismic Tonds and any oth
et he strate remaINS UPrght
ae
and can support
‘buckling o overturning
fe tel members ane connections
‘drength and load-carrying
Properly designed sections,
sc: 7
ore eet
ets ret dss
sn ra pri st
setae ow ching cal
Strength and load
‘must be designed
cops
fore acre cn Ba
‘failure points.
rndaey and robustness: 5
anny th eign pro
“Fa aes Ts relandany
Sa ening the ik of alate clase due toa
Sg re
Suey factors: Salety factors or factors of safety are used in design
eerie n mag of salty agai unexpected oF uncertain
Conlon ‘Thse ator count for varisons in material
oper costacton kanes and Id uncertain,
Fie protection Si scares may requie fie protection to
tata ther lod ering capacity during fe. ieesitant
cotngso slo anh pli to protct the sel fom high
temperatures and mati atengh,
Servicii: In ation to sty, sevicaity is another
Import spect el seat dein Serie relers othe
Fevlormance of the ste in terme of xeupent como
functional and appearance
Ke sesieliyequtemens nade
Deleon cnt Bxsive deen ander seve lous an
seeese oc ct acta
limi toenureacomiorableand sableenvemene SPO
‘Vibration contro: Sie strututes can be nen
ca el sts can fe ee ovis
nme factors. Excessive
fol structures often incorporate
ie alternative load paths in ease
jhances the overall safety of
Iniroduction ot Chapter 1] _ 13
vibrations may cause discomfort or damage sensitive equipment
within the structure. Proper design and damping mechanisms are
employed to control vibrations.
Crack control: For welded steel structures, special attention is given
tocontrol potential crack formations due to cyclic loading or fatigue.
Proper detailing and material selection help minimize the risk of
«crack propagation,
‘Aesthetics: Stel structures are often exposed and visible, especially
in architectural applications. Therefore, aesthetics and appearance
play a significant role in the design to ensure the structure blends
harmoniously with its surroundings and meets the client's aesthetic
expectations
Mecting both safety and serviceability requirements demands
careful analysis, accurate calculations, and adherence to relevant
building codes and standards, Additionally, regular inspections and
maintenance throughout the structure's Hfe are essential to ensure
continued safety and performance.
curve for mild steel. Describe in brief.
{2021 Fal]
Daw a stress str
Solution: The stress-strain curve for mild ste sa graphical representation
ofthe relationship between stress and strain ofthe material The stress
is plotted on the y-axis, and the strain is plotted on the xaxis. The
curve goes through several stages before the specimen experiences
failure.
7 nally plastic
“Tratess
Una
strandigeam
viet
“ ‘Conventional stress-strain
guna
SEE Tan
eis the tessa fll
Stain —>
Lined range
a ence ET TVET spn eprom
me 2
Sete
Pesali epr erm |
tte Ht Bj: The ness ofste8 0 HEP 1B, the strain also
hn le it. oN ay
ewe one he anus 2 gal
ean al te mas
vend yor Tepito te c= sa
ere Tcl anne
aah mania soos req OF 8 Tat
I rman ical sup iF |
eri (9 Te posto te cave whee te ar
‘ere ead sy ata deceit
sea ah minimum se is equed fr 3 rod to |
Ie daenton cled efowe yl pint
tar pit (The pint nthe carve where the material
Chapter 2
Analysis and Design of Joints
Types of Riveted Jorn
“Types of Bolted Joints.
23. Typesof Welded Joints = St
‘Aadaly and Feventrically Loaded Bolted Joints (Bracket Connection) 78
‘Axally and Becentriealy Loaded Welded Joins. 401
2.1 Types of Riveted Joints
‘The joining of sheets or plates of any metal by means of rivets sealed a
reaches its maximum stress, riveted join. The riveted joint is a permanent joint and it is also known
-Fractare point (F} The pint on he curve where the material fractures. fs permanent fastener joints. Mainly two types of joints are used while
208 joining metal sheets by rivets:
9) LapJoints B) ButtJoints
Wi DC, Pe
BB Date Hy Dest sank dame
Tig Singer ait
eer tap joint Wen nes
ep eit hen it called the doa
2,2,
a
rete at
seen rors fet are usd in
zm ip eit cad 295 PP
32
aut
oy ai eT et oe kl
sume cee
neler fanart
smicic
focal erect tiny
i aheptmcnenege
Scouring mien
ene
FS rae
cp a
Analysis & Design ofits ca Chapter 2) 97
Double strap butt joint: In double strap butt joint, the edges of|
e t int the edges
the main plates butt against each other & two cover plates are
placed on both sides ofthe main plates
2.2 Types of Bolted Joints
“There ae two predominant types of bolted joints subjected to aval forces
{na bolted connection namely, lap joint and but joint.
4) Lap joints: In a lap joint, the main members to be connected are
placed over one another to form an overlap between the members,
fnd then the bolting is done onthe overlapped portion.
Single ted ap oint Doubt pin
'b) Butt joints: In this type of joint, a cover plate Is used to join two
members. Based on the numberof cover plates there are two types of
‘put joint namely, single cover butt joint and double cover but joint
Fig: Single cover but jit
ig: Doe oer but joint
‘Types of failure in bolted connections
Ina bolted connections either the connecting plate right fair oF the bolt
3 to consider the failure modes
‘might fail. Therefore, i becomes importan
of both bolt and the plate
Failure modes of bol
4) Shear failure of bolt: As the name suggests, this failure ocurs due
tm aheat force atthe interface of surfaces in a joint. Depending on the
bolted connection.
TMomber of shear surfaces, there are two types of shear failure that
auld ocear in a bolted connection, namely, the single shear failure
fand double shear failure. ————
o_[ enol el Tier Sra
tare: Here, the bolt & 8
Single shea far: He ge
foe wh er tan et
pile single shear Plane
ete 10 shear forces
ove shear lle: Here, The ols subjected to
nt tear planes. Ths typ of lle occurs in the
‘oublecoverbtt point
wo shea planes |
Fane oat she aire
1) Being faire of bot Ia this faire, the
Ie fain rng du to conta withthe I
pes. Tis ype of fare eccurs in cases] )
‘here a low strength bolt is used with plate i
of very high grad, which ually doesnt —*{_ }
cornne
©) Tensile failure of bolts: The —
toe gh of he to
sewn le ten th
Sot ae pepe
Sine ae ott
a ell en in pope
is oes he re aon
oft thn he a
Tee fore due to
iping/ phoning of ble
Pie Te fee oftts
Anata & Designo fins on Chapter 2119.
Failure modes of plate ina bolted connection
1) Shear failure of plate
Fig: Shar failure of plate
fi) Bearing failure of plate
Fig: Boring fur of plate
Both shear and bearing flute of the plate can be avoided by
providing sufficient centre to centre distances between the
Bolts as mentioned i section 10 of 18800: 2007
1) ‘Tensile failure of plate: Due to reduction in the net area (i.
due to bolt holes) of the plate along the bolt Tine, the tensile
Strength ofthe plate will be lesser than the actual value at this
section. Because ofthis, the plate might fail under tension
Fig: Tensile fur of plate
Basic terminologies in designing a bolted connection
Pitch distance (p): Centre to centre distance between two adjacent bolt
1nd the applied load.
ig): Centre to cenre distance between two adjacent belt
holes inthe direct
Gauge distance (
‘oes in the perpendicular direction of the applied load,
Edge distance (c): The distance between the edge of the plate to the
reaest centre line ofthe bolt Hole inthe direction ofthe applied loud
Fig: Ra Trios sing ai reeton 20 | Design of Stel & Timber Sct
of the plate to the nearest
he pe applied load.
Jrove mentioned lerms refer
End distance(e}: The distance between t
Ee lin of the ot olen the deton ott
For detailed speictions regarding all he
SSsuwaractoniDometons
aera af bole Back bats ae oe Kas! expensive bol
Pee eT cad ota ened
may we int et
trosen bang et es of the bo
mene sea aint the properties af the ol
de of a ee gh and enh. Geral
‘ior pad tse nconscton ce
Pahoa nner ar heute eae strength of ot
a4 100* AON?
an, Oba tat te id abegth of tebe
{pus Uline eng
freassa0
fg = 240 N/m?
ear imate ese strenght he af) "N/mm
cen
sept eb 0640-20
‘The design strength of bearing bolts under shear is the least of the
following
8) Shear capacity ofa bolt As pr 15 8002007, the design shear strength
of ol (Vs) is given by;
Ves
Van V2 (ce section 10 clause 1033)
where yo ~ Partial safety factor
Vas = Nominal shear capacity ofthe bolt
‘
as
Van eA tm Ae)
where,
‘fay~ Ultimate tensile strength ofa bolt
‘ne = Number of shear
planes
n= Number of shear planes withot thread intercepting the shear
‘lane (shank ares intercept the shea pane) usally taken
‘a= Net sar ara of he tts
ate, taken as the area
corespondingo ret dumetat tea
planes with treads intercepting the shear
feast expensive bolts |
posi
»
nat
‘Aap 078 x7; dcdiameter ofthe bolt and
‘w= Nominal plain shank area ofthe bolt
Aa
fs
"yi = diameter of shank
Bering capacity ofthe bolt: As por 5 600 207,
; section 19, ase
1034 the design bearing tengo the bol Vag) spe
Vase :
where,
Yoe= Patil safety factor (se table 5 of 15 800: 2007)
‘Vn = Nominal bearing strength ofthe bolt
Vn" 25 hy *
Not Her fi Tessar of (and
where,
4 Nominal diameter of bots
{= The sum of the thickness of the connected plates experienc
Tearngatesinteamedicton.
f4~ Ultimate tensile strength ofthe plate (i for fe $10, f= 10
N/m)
4y= Smaller ofthe following values
© f
Ga 3-02 =)
where,
«pend and pitch distances ofthe fasten or along bearing direction:
From clause 1022 of code 15 80: 2007
Pitch (p) «254
From clause 102.42 of code 15 80: 2007
End distance (¢) «17 dsore «15 4)
where,
ddy= dy Diameter of hoe (fer table 1, dause 1021 of 80: 2007)
f- timate tensile strength ofthe bot
Reduction factors in design shear strength ofthe bot.
Ifthe joint is too long: When the distance between the fist
land last rows of bolts in the joint, measured inthe direction of
Toad transfer exceeds 15 times the diameter ofthe bol (d:
>t the grip Iength is large: When the grip length (equa othe
thickness ofthe connected plates) exceeds 5 imes the diameter
ofthe bolt (d)
> a packing of plate more than 6 mm in thickness is provided. »
sign fie & Tiber Suc
el 18 600: 2007 10 get thy
af bot of
Refer to section 1033 shear capac Setter ‘mentioned CaS.
Fe erect acevo
ot
conve oraesnsaraneim a
si ieee ae
wader
ee icasengn
eee aan 32
sre een ine 2
erste)
Geet
ciel
i
age b tae
ee
whee,
ho= The netffective aes ofthe plate
ae afr canofls ins 0
ot :
Auefocnee RE] exsugt mgemet
t-te eatstegh ofthe st
(ole bef 180.207
strenght abtedcomecton
Te aang he ate omen he at of alt value (sign
‘Moshe ha an beds tengo he pte
Remember
2) Shearcascy ef bol For sgl shares. of shear plane = 1)
fa
Bais
Unless specified, Vay would be calculated corresponding to shear
plane intercepting the thread. Cae
For double share (of she plane =2)
Van= GEES (1% Aa) le Tand n= 0)
fos
ao egg (Aw +1 A) (eine and ne)
Under the assumption that one of
root of thread and eters intern
Bearing capacity of bolt
25¥lexd aes,
Pais ae
SONGS RETF)
the shear plane is intercepting the
pling the shank,
Voe'™
sh (Uae) and design shear
Anais & Design ofits ot Chapter 223
Fem thethicknes ofthecoverplate.
Jp Atte thickness ofthe cover pate > thickness ofthe ma $
Pee ee eer mee Cltoes ot tmelapg
‘> If two cover plates are used, t= summation ofthe thickness of both
th cover pas oi tomas an he Nek malt
piste :
2) Igri of tel not given then assume grade of tel fe 0
tate strength of sel (.)= 0 Np d
i) 1a load given i factored or designed then given lod owed oe
farther calelalon, otkervze given load © multiplied by 1 and
then should be used for further calculations. *s i
9) Tensile capacity of a bolt aI
Ta =09A, i :
oe
‘Aa =Net ese sess reo bl 2
far Ulioate tet stres of tot
n= Paral safety ator for materi of bolt en 6325.
‘d) Bolt subjected to combined shear and tension 4
‘The following equations may be used to check the safety of the
connection subjected tothe combined effect of shear and tension.
ees |
where,
\Vp= Factored shear force on bolt
‘Vas Design shear capacity ei
‘Ty Factored tensile force on bolt oa
‘T= Design tension capacity ce
‘Assumptions in the analysis of simple bolted joints
a) Friction between the plats is neglected and ood is eit by bolts,
in bearing and shearing ae
1) Incase of boli if treads occurs in the plane of shear
Ste shears ten ne he se tte ot of ee
However if treads do not occur in plane of shear, fective aes is
the crosesection area of the shank,
©) The applied load is equally resisted by al the bolts.
‘8. Distbution of stress on the portion of plate between the bol hole is
uniform, Le. stress concentration around the holes is neglected potted connection as per oon
the clearance as given below
Ts loa)
Sama Te sane
Teer shad rt Be
dane
Teemu ph
ype a0 mm whi
Fela tnt
tee forthe os neon
eat eu 10232 560-207
arn hese fr th Bln compression
eae rth lice ie pst
6) Teedprandend dees
Pera ee ad end sane om te cnte of any
Miter elgees plat should mothe ess than 17
Und Sms te ae mer rol, machine ame
‘vanadyl
i+ Thema ede dnc rom he cn of thee o he
scrote hol pote! 24 wo = 22)
‘is the thickness of the thinner outer plate, :
Design lp joint beeen the top i i
en the wo plates each of width 120 mm. if
the thickness of one plate i 16 mm and the other is 12 mm. The
Steger
Cent
soa Feh [ae ann
reen the centers of the Bolt in
than 25 times the nominal
hover is less for the boll in members
o
and t
Sot
Aral & Desi of ois on Chapter 225
Design load (P) = 200 kN
Using Ms bolts of property class 4.6;
‘Ultimate tensile strength of bolt (4) = 4 100 = 400 N/m?
Diameter of hole (&) = Nominal diameter + clearance
do= d+ 2mm. (from table 19 off code 800 2007)
Diameter of hole (d) = 20 ¢ 2 22 mm
Grade of see plate () = 410 N/mm (= Fe 10)
8) Design shear capacity of bolt: (From clause 103.3 of code 1S
800: 2007)
design shear strength of bln shear (Vs):
Since itis lap joint, boli in single shear. Assume thread lies in
the shear plane
So,1=T and.
and You * 1.25 (from table of code 15 800 2007)
0 ;
vas ig (100mm)
vases
vaea7in
oy Dapeng pact ft
Frome 84 de 80-207
“The design bring rng fa (ag) = 22S
“Note: Hote fy Will be lesser of faland fa'and twill be thickneis'of
thinner ltt).
where, y= Smaller of (36;+ 2-025 ant)
minimum agen tance
From douse 102420f coe 15500-2007
on 15 (for machine Dame)
17 orkand fame ot)
(ie ean 1500174) a
Mini pitch:
From ause 1022 of ole 8 80:2007
Yon =25d (4, £254)
Now,
End ditinee (#156 Analysis & Design fois Chapter 2) 27
cfbole = 22 “The design strength of plates given by;
cere, dee Diameter ene by
oe, e<15*B Ton 09 Ante
<8 tuget2bandd tt
Providee 4070 ag OOS
and es jr 1.25 (om able Sof 5 code S00: 2007)
Pach diane 2200 09 (120-1 20) «12 «410
dehore d= Diacte olt Tes =z
on, pe25e2 Tay = 3471552.
peso | Ta# H7A54N
Provide p= «0 | ce, ive bolts ar proved in the connection.
10-9251) | since five bls are provided in the comes
tae soaterot (sar 'Fe 3 O10 | Soy tot desig load carrie by five bolts = 5x Bolt value
= 54527
be 7
25 0406+ 202 12500 =726351N
eee “Ton (©3475 RN) > Total design Toad (= 22535 KN)
Vag 182N So, design is safe
vanes | Hence, Provide fe bls wth pte (= Om and end distance
\Weknow that rom cass 103.2 of ode 5 80: 2007 40mm
elt value= Minimum of Van and Vi 2. Designa cap joint between two plated as shown in figure so asthe
al vase 4527 kN transmit the factored load of 7O KN using Me bots of grade 46
nd plates of grade Fe 410
Now,
fmm ts
[Number of bolts required = atore lad,
Tol value a
TORN.
So provide number of olt=5
+200
Solution: See the solution of Q-No.T
3. Design a simple butt connection =
m0 ting Mu bolls of grade 56 (0 coer
transmit a service Toad of 185 &N [702 Ts
a fn the two steel plates of grade
Fe 410 witha thickness of 10mm
and 18mm.
ee Solution: Using Mis bots of grade 5
Tom lase 63.1 of code 5800 2007 fag =5 100 = 500 N/mm?
f= 500% 046 = 300 N/mm Se
Analysis & Design ooins ex Chapter 2|_ 29,
2b | Design fle & Tiber SUS
vied “15IN
Seite 1515 =20518
Ceri a0N/m?
cee (= 160
Bram rn oe Se 020)
panacea :
taco -s0mnko=
Ta ae a= Bnm> A
ee redaton in sear srength of Pat The
esa rs heb
eet ams) |
Rogan
fects
won
= te
rsh a) = E> fet A) Be
Since it isa double cover butt joint the bolts are in double shear
‘assuming umber of shear planes with threads intercepting the
shear plane r=?
So no
= :
oo Wan ite (nam Cover pate
ar
sm ckcover FE
ome H lH
Conti ie
Soltion
ow, daring analyzing and designing this type of question only half
pton iat AB taken.
Give
“Thickness ofcover pate = 6 mm
Thies of fits = 1mm
‘Using 4M bois of property clas 4.6;
Diameter of bl (@)= 16mm
Diameterof ale) =d+2:mum (fom able 190 of code I 80: 2007)
4=16+2=18mm
Ulimate tense stength of bolt, y= 4 100 = 400 N/ mun?
‘Uae tensile strength of ste! plate, f= 410 N/mm?
‘The design shear strength of bot (Va) is Given by,
Veo Bylot mAs)
Since, thore sa double cover
i Plate so total number shear plane is 2
Then,
ae BRS (ts 070
Bria (Peter rere)
Vane e624
Anais & Design ons x Chapter2) 33
‘The design bearing strength of bolt (Va) is Given by:
Vig Ell
Provide pitch (P) «80 men (Given)
End distance e) = 40:mm (Giver)
or, ke=Smatlerof (5455-3855 025-$9 1)
ken 074
25 074% 16 12% 400
on Va <1
Vago = 113.66
Now,
Bolt value = Minimum of Vas and Vas
Bolt value = 66.12 KN
Here,
‘Number of bolts half potion =4
Total strength of bolts = 4* 6612 26148 KN
Check fr plate cover plate
“The design strength of plate is given by;
Aafe
ae
(900-218) 12
r TQAEAO fs Ay (oon) *1)
“Therefore, ultimate load carying capacity of connection
“Minimum value of Bolt value and design strength of plate
264.48 KN
Jates 250 mm * 8 mm and 250 mm * 16 mm are to be
ie cover butt joint with 16 mm diameter bolts.
‘The factored tensile force on the
6. Two pl
connected in a doubl
‘The cover plate are 6 mm thick.
connection i 400 KN. Design the connection.
Solution: Given,
Diameter of bot (4) = 16mm.
Diameter of hole (de) =d +2 from table 195 code 800: 2007)
do=18mm
“Assume, bolts of grade 4.6 and Fe 410
‘Ultimate tensile strength of bolt (a) = 400 N/a?
‘Uttimate tensile strength of plate (6) =410 N/mm? cyl el &Tinbe Ss
1 enna
ected ar of ticks
Since the phates to bcos m6
‘ort sneer to prvidea poking rte
Fem
racking pte
Tra mm inst 81 Be usd
tn cmt FS Fein he th sgt offal The
‘in ceri gen
Bed 0015 4)
feedt-o0n59 09
Desnseasrenth of ls ten by
(a= Er An ted Be
Since it sa double cover butt join, the bolts are in double shear
‘assume numberof shear planes with threads intercepting shear
plane,n9*2
So a0
rit (eamabare se) 09
Ze
Ielon igre
Factored load
Vas
Number of bolt equied =
ire =i
Number of ot required = = poy
Provide 8 bots of 16 mm diameter
(Check for safety ofthe design
Design shearing strength of bolts= 85213
=A17041N>400 kN
ae eet nha 35
“The design baring strength of bl (Vay) sven bys
2S ke dt fy bak
‘
Vag
y= Sater of 6B 0.5, $8
@xtsé,
End distance 15218
477
Provide =35 mm
Pitch) £254
P<25%16
Peso
Provide? = 50 mm
= 0618 é
250648 x16%8% 40
125
or, Vy
Vape = 6635 KN
Design bearing strength of 8 bolts = 8 66.35,
"= 530.84 KN'> 400 kN
Design strength of plate from yi
From clause 6.2 of code IS 800: 2007
consideration
fe
tee
4 (From table5)
2508x250
ye T
Tug = 45154 EN > 400 kN
Design strength of plate rupture consideration
From clause 63.1 of code IS 800: 2007
(Ap"250*8 mm)
‘ya1™ 1.25 From table 5)
09 x 250-4%18) #8*410 5 nay
Ts so fe An= (= nd t)
Tay 420337 KN > 400 KN
Hence, the design i safe 36 | Design of te & Timber Sbctires
-Theaangements of itsare sown elon
i
ej
3 fe cover but joint to connect bole plates
7D nn feng. Us Mu bl of ae
‘tector plates are of 0. Find the efficiency af the joint
Sotion: Giver
“Thickness of boilerplate (= 22mm
sing Matis of ade 4.6;
‘Urinate stent fblt (a) "4% 100= 400 N/mm
Use strength of plat (6) = 410 N/mm?
Diameter oft) =16 mn
Diameter of ole (4) = d+ 2(From table 19)
Since, tis double cover but joint so the bolts are in double shear,
‘Assumeny=Tand.=1
Design shea strength of bot is given by;
fa
(a) FE Oona)
Vao= ( x x
1a BO (nomebare sta)
Vines
Assume beating stength of ots mor than
To get the maxinum eceny sen
shouldbe equated ostengtha tol
Design strength of plat pe pitch
From dus 63:1 ofS code 27,
& Wx Pays t2%04
ao ay O92 P= 19 «2x19
Aa 1535124 (0-18)
where, isthe pitch ditne of bl
SESE
sheating strength
sth of plate per pitch width
th
a i ati ct ate 21
Now,
aquating it with Vas to get maximum,
ce get maximum efficiency
35124 (P = 18) = 6.12% 108
P=3666 mm
Also, P25
P< 25x16
Peso
Hence, Provide P = 50mm
‘Check for design strength of bot in bearing
From clause 103.4 of I code 800: 2007,
25h dif
Va
Yee
Weave,
elds
e<15x18
40
wa
k= 06759
__ 25. 06759 16 12% 400
Ca 1
Vago 103.81 KN > Va (= 66:12 KN)
Heence, the assumption that bearing strength is more than design
shear is correct
Since, pitch provide is slightly more than required from strength
consideration of the plate, the strength of plate is more that the
strength of the bot.”
Design strength of joint per 40 mm width = 6612 kN
Ach
Also, design strength of slid plate per 40 mum with =“5"
40% 12250
122250 _ 09.09 kN
‘Maximum efficiency of joint
Design strength of int per 40 mm width
Design strength of solid pte per 40am wid
_ 62
177609
oo
x 100= 60.61% merce igh aang bs
nnn te bt incom ima ed
Desig roe mm wie and 2 He
ee bso att er)
ei pomited atte oie ed
ip isnot permitted atthe wl imate Ie
The con ce tt one set plne
ect denenet ett
saaion cst
ap!
service oed/ working lod "75
Breadth of plate) = 120:
“Thcknes of plate () = 20mm
Using. My high strength bots of property lass 109
‘Diameter of bolt (f) = 16 mim
iam of hole) = d+2 (fom table 19 of code 15 800: 2007)
12 mum thick
oad (P2)= 450 KN
50
= = 300 kN
=18 mm
imate tensile strength of bolt (a)
and yield strength of bolt (jy) = 09 1000
f= 500N/mn?
‘Assume, grade of sto plate () = 410N/mm?
and f= 250 N/m?
2) Slipisnot permitted atthe working load
This means tha a lip between the connections parts may occur at
the ultimate load. Therefore the joint is designed for a non-slip
condition atthe working loa but should be checked for bearing and
shearing atthe itinate load
From dlause 1043 of code 15 80: 2007
Design shear capacity a bolt as governed by si
Des she cpt al gover by ap for ton Wy
Vet
‘te
where, Vg = Nominal shear capaci
Tage, Ha Nominal shear cpacyoFblt ax governed by slp oF
Vi
Varnnte — [2 Vegetal]
Yet
ri te pr
wher ‘
tw Cocicen o ton ip a ape
oe ip factor) as specified in table 20;
1+ Number of elven orig tna et
toslip. :
fortes in cea ls
tu 110 (slp resistances eve a)
‘r= 125 sips desig t lite al)
fo= Minimum tt en (eT) ation and mn
be taken as Avs fy oe ad
tare oft at tends and
Proofs (= 0704)
70 « 1000 = 700
Vagn 0552241 #15689 * 700
on Vow i
Vou
Bolt value= 10978 KN
Working load __300
“The number of bolts needed ="Baj value" 1075
2273-3
Therefore, three bolts are needed for non-t
working load
‘The joint should be checked fo
ultimate load,
‘Check for shear at ultimate load
Design shear strength of bolt is given By;
f(g, Amt tea)
Vow
Bm
‘Assume one shea plane intercepts the threads ofthe ba
ip condition at the
1 bearing shear and block shear at
given by:
mel
So, #1 “go | Design of Sie & Tiber Sucre
A(x orse fate 416)
ce wang (ometeed
Yun 1652010
Cec for beating alia od
Desstengtht atin easing ve
stadt
‘
From dase 1022of oe 5812007
Ve2sd
pa2sxis
reson
ProndeP =m
nom nse 10242 foe 0: 2007
etd
tens
team
Prvdee = 40mm
wer,
Vas
sae ss a
sesso: 025-42 ant)
4
on, byesmaterot(55-5 95-0251)
e074
gg x BEATA 1620
= a = 1981768
The design strength of bolt = Minimum value of Vag and Vy
= 165301N
Design strength of3bolist the ultimate lod
#316530
= 485 9KN> 4501
(Check for block shear
From clause 641 of code IS 800: 2007
Avahis& Doig of one ex Chapter 2) At
i isk stg a fc tn
smaller of; :
N80 Bra 4)
From the igure;
Ae 0+ 6+) * 20 = 3200 mi? (Along 1-2-3 -4)
‘hn 4060+ 60-25 18)» 20+ 240 me (Along 1-234
‘ie 60*20= 1200 mm (Along 4-5)
w= 60] + 20-020 mn? (along-5)
Now,
yx
Tay * 720955 KN,
for, 664725 KN
Typ 664.725 KN > 450 KN
Hence, 3 bolts may be provided
bb) _Slipis not permitted at ultimate load
“This means thata sip between the connecting parts may occur at the
working, load. Therefore, the joint is designed for a. nonlip
condition a the ultimate load
Design shear capacity of bolt as governed by slip for fiction type
o (oesgaa um
"we
where,
v1.25 thes restnceis designed a ultimate ond)
QSSx2H1*ISE2*7O 1 pasty e070
QS DATAISSSEEAD |, ye Ayhyfoe OU
Ve us [o Fo= Aut fo= 0706)
Vac" 960 KN
‘imate oad
Number of bolt needed = ys,
50
= Bh eases
ee “ex [ig f Se & Tiber SIUC '
town in figure below.
Hence, five bits may be provided 3
90000] S/n
00000
“nt vo connect two plates exch 12mm
olen dete see tobe transfered fe
A etake 250 Nf} Also find the fice
Solution: Giver,
Service load?) = 200 KN
Design ad () = 15*200= 300 KN
Minimum thickness (oe) = 7250
Assume bots of grade 6
f= 40N/?
and f= 10N/mm?
Tofind diameter of ol
4-601
a=01 yi
d=20819mm
fo eve amt i) a im abd een
Diameter of whose) =20+2
=22mm (From ble 19 of 5 code 80: 2007) |
Shear capacity of bolt
fa
Var EE (nat
Baz (mda tmAs)
Assume sear planes intercepting the thread
ne=2andiy=0
vast
Pac et
a
Vee
Analysis & Design of fins ce Chapter 2) 43
Minimum end distance, @= 15 x dy
P=25%20=50mm
15% 22°33 mm
where, ky = Smaller of #025. and?)
lo" 3do fe
3 so
ss = Smale of (5355 +5555 -025-409-1)
be 05
Now,
gg BEnOS aD IRD
a 13
Vijo = 98400 N = 98.4 KN
Bolt value = Minimum of shearing and bearing,
potalue=90541N
Design
Nembe‘of tl required = esemlaad
x00
“ast
ame4
provide nas bls
Consering wo bls in arow
Considering two bolts in a row
Noted In case, width isnot given
Her, from diagram width = e+ pis
‘Tension capacity of plate
09% Aax
et
Tua"
where,
Age (b-ndg xt
b= Width = 300 mm (Given)
do= 22mm
n= No.of bolt at eitcal section “44 [Design of tee & Timber Suc
09x ,609=2 22) x12 <0
Ta EOS
Tan sS44N
Tees sStiN
Trad Duis
Nap Ta)“ 5054
a Tectia
Tene
whee Ast
a am bis fet 022007)
1 et ego
{20 (Gre)
aint = Minimum value of
Tye 900* DTT
Tig BISISLSIN
1818 KN (Design strength of solid plate)
f + |
~Disign strength of sold plate * 10°
5054
Seis * 100 1.06% |
[Exam Question Solution
Desig lap joint between he two plates ach of width 120 mm,
the tikes of on a 6 mi the ter i 2mm te
ont has o taser a design load of 20 KN. Te plates are of Fe
‘Ho gradeand Ms bol of propery cs (201, Fl)
Solution: See example question numer 1 Same question)
2 Design lp joint betwen to plates of thickness 12 mm nd 29
fee secea ee
eaeainae "fat opr
ificieney of joint
Solution: See (2017 Fall]
Analysis & Desh of ois ea Chapter 2)
Design lap joint to join a two plates of thickness 14 mm and 16
‘nm to transmit a factored load'ef
grade and Fe410 steel, TU ale Meteo
solution: See [2017 Fall
Design a bolted butt otto connet vo pl
plate of thickness 12 mm
and 2 nm Conder Te 10 grade of ste plate and use Mi bts of
sade the jolt hast tantra servcloud of 19k [017 Spin
tion: Se [2020 Fall and [2019 Spring}
5, Design a lap joint to connect two sel plates of thickness 12 mm
and 25 mm fo transmit service lad of 80 KN, Use Mu bols of
rade 5iand Fesl0 grade of tel fan Fat
soltion
Using Mi ols of rade 6
Diameter of otf) = 16mm
Diameter of hale (i) = d+2 from tble 19 f code 8902007)
18mm
Utkiate strength of bot (4) = 5100 = 500 N/mm?
Yield strength of bol (fy) = 06500 =300'/mm?
Uitimate strength of pate) = 4101/min*
Service load = SKN
Design load =15 80 1201N
shear capacity of bot
foe
Va (mp Aw # Aa)
Bia!
Since i lap joint blt sin singe sear assume tread es inthe
180
shear plane ny:
tan a 0 x 0.78 F416)
Va" Bar" a
Veo= 3621794
[ec fl be se ol
oateot i Bro = ) ee
asso a fntontcue
‘etm revvecnan
ae rocinan
wipes Teo eel 4d
eases ooo
aan vas
cane Tate 09> Tan
30. ign is safe
west a
Hc peo ph = man ed ne
y= 055 (a ieee
cs
Design ond
Nunberofblteguied = Besa,
Provide 4s of ts
wine 0 0 0 |} Saw
Cia cone
[ ero
mang
Check for plate
Tieden stongh epithe
Prone fee 8 ay
nh Aes (end
ween "Nat eta son
SEES remeron
6 Two plates of 12 mm and 10 mm thick are joined by double bolted
double butt joint having equal cove plates of Boum thick and Mas
bolt of property class 4.6 are used at a pitch of 100 mm. Determine
the strength of joint per pitch andi’ efficiency. [2018 Spring)
Solution:
Diameter of bot (f) * 20 mm
Diameter of hole (do) 22:mm (From table 19 of code 15 8002007)
Grades,
fa
fn
Assume for Fe 10
Pitch (p) = 100 mm
“The strength of bolt in shear per pitch length
3 100 = 400 N/mm?
(06% 400 = 240 N/men?
f10 N/m
aus (rm Aso Poa)
Vau- Fr ¢ )
‘Assume numberof shear planes with threads intercepting the shear
plane, n= 2,
So, nee
400 5
. no78xEx16
Vou fuse o7ax¥x10)
Vag" 7948 KN
“The strength of bot in bearing per pitch length
Bkedtfs
Vee
where a= Smaller of (555° 357025
er 48 | Design of te! & Timber Siractres
Here,
exis
ee 5x18
e F025"
(From clause 103.4 of code 18002007)
end distance (9) £15
eex18
«27mm
Providee= 40mm
Pach distnce (9) «254
pe25x16
petdmm
Provide )= 60mm
{ and)
(0.
Nov Smaller of (55g
Minimum value of Vag and Va
= 59508 kN
Desi 300,
Bolt value “5950558
[Number of tot required =
Provide 6 nos of bot
(onsen sce bls in ow
Width of main plate*2e+ p=2 40+ 60 = 140 mm
Design of cover plate (ty)
Design strength due one section of rupture
Ta=09 Ach
Tar »
300 « 161» 99 489=2*18) 249 e410
125
by 4.88 <6mm
Provide 6 mm cover plate
Aral & Desi of fins ce Chapter 2) 51
‘Cheek
where, A= bxt
For Fe 10, 6, =250 N/m?
Now,
250
300 «100 gx 2
A,=1320 mm?
or, bet=1320mme
Wt= Hmm, b=9428mm-<140mm
let=22mm, an
Hence, provide cover plate of each 6 mm thickness 12 n08 of Mi.
bolts and two plates with 140» 14 mm and 140 «22 mm.
,
fem
© 0 0}000
9. bolted butt joint to connect two plate of thickness 16 min
and 24 mm. Consider Fe 410 grade of steel plate and My bolts of
{grade 4.6, The joint has to transer load of service load of 300 KN.
12020 Fat]
Solution:
Diameter of bolt () = 160m
Diameter of hole (és) = 18 mm
Here, the plates to be connected are of thickness 16 mm and 24 mm.
So, itis necessary to provide a packing plat of thickness = 24-16
=8mm>6mm
(ET
Ten
dima
Lo x
52 | Davignof el Tinker Stace ’
ea 6 FANS
Nove ets pacing PAE ae
facoris provided suas ag
Fb luciana ang
‘The reaon atr gen
Fromenon clase 10333 0fcoe 15 0:27
en 1-015)
pect -00035 8)
hen
Grae a= 400N/en and
Shear ep fal
=e (om hn + Aa) * Bos
verre |
ora dot rt in thle in doe sar assuming,
Fa dt etendnoeping este a, = 1
f, n=l
vane ets (0
Mae Rs
Von = 5950857 N
Vee = 59.08 KN
Bearing capcity of bolt
VasetShat
[ote Ha wile se of ad fe will Bethe nm
thickness =
where,
= #10 N/m
wore Bx16 11x90 16)
y= Smale ot (35 F-02542 and?
End distancee < 15 dy
ex 15x18
e<27mm
Provide)
Provide «25a
pe25x16
p