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Unit 5
Threaded & Welded Joint
Prepared By
Prof. M.C. Shinde [9970160753]
Mech. Engg. Dept., JSCOE, Hadapsar
SPPU Syllabus Content: (10 hrs)
Threaded Joint: Basic types of screw fasteners, Bolts of
uniform strength, I.S.O. Metric screw threads, Bolts under
tension, eccentrically loaded bolted joint in shear, Eccentric
load perpendicular and parallel to axis of bolt, Eccentric load
on circular base, design of Turn Buckle.
Welded Joint : Welding symbols, Stresses in butt and fillet
welds, Strength of butt, parallel and transverse fillet welds,
Axially loaded unsymmetrical welded joints, Eccentric load in
plane of welds, Welded joints subjected to bending and
torsional moments
Part A
Threaded Joint: Basic types of screw fasteners, Bolts of
uniform strength, I.S.O. Metric screw threads, Bolts under
tension, eccentrically loaded bolted joint in shear,
Eccentric load perpendicular and parallel to axis of bolt,
Eccentric load on circular base, design of Turn Buckle.
Threaded Joint
Need to Use Threaded Joints
.
•To connect the components together which should be
readily disassembled. E.g. Screwed fasteners
•To transmit a power or energy e.g. power screws
•For obtaining accurate movements in measuring
instruments.
Advantages of Threaded Joints
.
•They are convenient to assemble and disassemble.
•The assembly and disassembly require only spanners
and do not require any special tooling's;
•Threaded joints are relatively cheap;
•They give high clamping force;
•They are highly reliable.
Note:- Tooling is the process of acquiring the manufacturing
components and machines needed for production
Disadvantages of Threaded Joints
.
•The component becomes weak because of the holes;
•There is a possibility of loosening of joint due to
excusive vibrations;
•There is stress concentration in threaded portion,
which may be vulnerable under fatigue load.
I.S.O. Metric Screw Thread
.
•fig. shows profile of an ISO metric screw threads which
are commonly used in India.
•It is not easy to get mathematical relation between
d,p,dc ,hence it is necessary to refer the standard tables
to find the dimensions of screw threads.
Type of I.S.O. Metric Screw Thread
.
•Coarse threads
•Fine threads
Coarse Thread
.
•For given nominal diameter, pitch and hence lead angle
are larger compare to fine threads
Advantages
•They are stronger and hence less possibility of thread
crushing and shearing.
•Due to large lead angle threads offer less resistance to
unscrewing.
Designation of ISO Metric
Coarse Thread
.
•It is designated by
letter M followed by
nominal diameter. E.g.
M30
Fine Thread
.
•For given nominal diameter, pitch and hence lead angle
are smaller compare to coarse threads
Advantages
•Fine adjustments is possible with fine threads.
•Though, fine threads are weaker its screw body is
stronger.
•Fine threads offer greater resistance to unscrewing.
Designation of ISO
Metric Fine Thread
.
It is designated by letter
M followed by nominal
diameter and pitch, two
being separated by sign
x. E.g. M30x3
Basic types of screw fasteners
.
1) Through bolt
2) Tap bolt
3) Studs
4) Cap screws
5) Set screws
Through Bolt
.
•It is cylindrical bar with a head at
one end the threads at other end.
•It is passed through drilled holes in
the parts to be fastened together
and nut is screwed tightened on
the threaded end.
•Nut and bolt heads are hexagonal.
Tap Bolt
.
•It is the bolt which passes through a
hole in one part and screwed into a
tapped hole in the second part which
acts as nut thus holding parts together.
• It does not need nut.
•It is used when arrangement does not
provide any space to accommodate the
nut.
• Frequent insertion or removal of tap
bolt is likely to damage threads
Studs
.
•It is a cylindrical bar threaded at both
ends.
•One end is screwed into tapped hole of
parts to be fastened while nut is
screwed into other end .
•Stud always remains in position when
two parts are disconnected
•Studs are mainly used for connecting
covers like cylinder head of engine.
Cap screws
.
•Cap screws are very similar to tap bolts,
but smaller in size as compared to tap
bolts.
• In addition variety of shapes of heads
are available for cap screws.
Set screws
.
•Set screws are very similar to cap screws but
are threaded practically throughout length
and are still smaller in size. Theses are used to
prevent the relative motion between two
parts.
•They are used instead of key in light power
transmission
Materials used for screwed fasteners
.
•Materials used for lightly loaded nuts, bolts and studs
are medium carbon steels like: 35C8,40C8,45C8
•Materials used for highly loaded high strength nuts,
bolts and studs are high carbon steels(60C4) and alloy
steels(40Ni4,35Mn6Mo4)
•Materials used for corrosion resistance fasteners are
chromium alloy steels like: 40Cr4Mo2,
40Ni6Cr4Mo2,31Ni100Cr3Mo6
Bolts of uniform strength
.
•If a bolt is subjected to shock or impact loading as in
case of connecting rod bolts or fasteners of power
hammers, it should be designed to absorb impact
energy
Drawback of ordinary bolt:
.
•In ordinary bolts ,cross-sectional area is minimum in
threaded portion. Hence, stress in threaded portion of
the bolt will be higher than that in shank or body.
•Impact energy absorbed at any point in a body is
directly proportional to square of stress at that point.
•Therefore large portion of impact energy will be
absorbed in the threaded portion and relatively small
portion of energy is absorbed by a shank.
•This uneven distribution of impact energy may lead to
fracture of bolt in threaded portion.
Methods of achieving bolts of uniform strength
.
•Reduction of shank diameter
•Drilling axial hole
Reduction of shank diameter
.
•If shank diameter is reduced to core diameter as shown
in fig. stress become same throughout length of bolt.
•Hence impact energy is distributed uniformly
throughout the bolt length.
•The bolt in this way becomes stronger and lighter.
•This type of bolt is known as bolt of uniform strength.
Drilling axial hole
.
•In this method an axial hole is drilled through the head
down to the threaded portion such that cross sectional
area of threaded portion.
•For bolts of uniform strength
Stresses in screw fastener body
.
1) Direct tensile stress
2) Direct shear stress
3) Maximum shear stress
Direct tensile stress
.
Direct shear stress
.
Maximum shear stress
.
Eccentrically loaded bolted joints
.
1) Eccentric load in plane of bolts
2) Eccentric load perpendicular to axes of bolts
3) Eccentric load parallel to axes of bolts
Eccentric load in plane of bolts
.
Eccentric load in plane of bolts
.
Effect of load W at a distance e from the centre of gravity of the bolt system is
equal to;
1) Direct parallel force (W): the direct parallel force W through CG of bolt system
results in primary shear force on each bolt.
2) Turning Moment(W.e): the turning moment W.e about C.G. of bolt system
results in secondary shear force on each bolt.
Procedure for design of bolted joints with load in Plane of
Bolts
.
Step1: Find the C.G. of bolt system
.
•Let A1,A2,A3,A4……… cross sectional areas of the
bolts,mm2
•x1,x2,x3,x4………distances of bolt centres from Y-axis,
mm
•y1, y2,y3,y4………distances of bolt centres from X-axis,
mm
Step1: Find the C.G. of bolt system
.
•Distance of C.G. of bolt system from Y-axis is,
Step1: Find the C.G. of bolt system
.
•Distance of C.G. of bolt system from X-axis is,
Step2: Find the primary shear force on each bolt(Fp)
.
•The direct parallel force W through C.G. of the bolt
system results in primary shear force Fp on each bolt
and is given by
Fp1=Fp2=Fp3=Fp4=Fp=W/n
Where
n=number of bolts
Fp1,Fp2,Fp3,Fp4 =primary shear forces on bolts,N
Step3: Find the secondary shear force on each bolt(Fs)
.
•The twisting moment W.e about C.G. of the bolt system
results in secondary shear forces which are not same
on all the bolts.
•secondary shear force on any bolt is proportional to its
distance from C.G.
Let, Fs1,Fs2,Fs3,Fs4 =secondary shear forces on bolts, N
l1,l2,l3,l4=distances of the bolt centres from C.G.
Step3: Find the secondary shear force on each bolt(Fs)
.
•w= secondary shear force on bolt per unit distance
,N/mm
Therefore , Fs1=w*l1, Fs2=w*l2
Fs3=w*l3, Fs4=w*l4
•Taking moment about the C.G.
W*e= Fs1*l1+ Fs2*l2+ Fs3*l3+Fs4*l4
Step3: Find the secondary shear force on each bolt(Fs)
Step3: Find the secondary shear force on each bolt(Fs)
By putting w in Fs=w.l equations the secondary shear force on
each bolt(Fs) is given by
Note :-the direction of secondary shear force on any bolt is perpendicular to
the line joining the bolt centre and the C.G. of the bolt system
Step4: Find the resultant shear force on most heavily
loaded bolt(FR)
Note :-the direction of secondary shear force on any bolt is perpendicular to
the line joining the bolt centre and the C.G. of the bolt system
Step5: Find the bolt Size
Where,
FR=Resultant shear force
Ac=Cross-sectional area of bolt
Summary of Steps:
design of bolted joints
with load in Plane of Bolts
Ex.5.1 a steel plate subjected to a force of 8kN is fixed to a channel by
means of three identical bolts as shown in fig. the bolts are made of
45C8 (Syt=380N/mm2). If the required factor of safety is 2.5 determine
size of the bolts.
Given
Step1: Find the C.G. of bolt system
.
The C.G. of the bolt system is at the geometric centre of
the bolt system.
Step2: Find the primary shear force on each bolt(Fp)
.
Step3: Find the secondary shear force on loaded (1&3)
bolt(Fs)
Step4: Find the resultant shear force on most heavily
loaded bolt(FR)
Step5: Find the bolt Size
Ex.5.2 a steel plate subjected to a force of 3kN is fixed to a vertical
channel by means of four identical bolts as shown in fig. the bolts are
made of plain carbon steel 45C8 (Syt=380N/mm2). If the required factor
of safety is 2. determine diameter of the bolts.
Given
Step1: Find the C.G. of bolt system
.
The C.G. of the bolt system is at the geometric centre of
the bolt system.
Step2: Find the primary shear force on each bolt(Fp)
.
Step3: Find the secondary shear force on bolt(Fs)
Step4: Find the resultant shear force on most heavily
loaded bolt(FR)
Step5: Find the bolt Size
Ex.5.3 a bracket made of steel subjected to a force of 8kN is fixed to a
vertical channel by means of four identical bolts as shown in fig. the
bolts are made of plain carbon steel 45C8 (Syt=254N/mm2). If the
required factor of safety is 2. determine diameter of the bolts.
Given
Step1: Find the C.G. of bolt system
.
The C.G. of the bolt system is at the geometric centre of
the bolt system.
Step2: Find the primary shear force on each bolt(Fp)
.
Step3: Find the secondary shear force on bolt(Fs)
Step4: Find the resultant shear force on most heavily
loaded bolt(FR)
Step5: Find the bolt Size
Ex.5.4 a bracket is bolted to a column by 6 bolts of equal size as shown
in fig. it carries a load of 50kN at the distance of 150mm from the
centre of column. If the maximum stress in the bolt is to be limited to
150N/mm2. determine diameter of the bolts.
Given
Step1: Find the C.G. of bolt system
.
The C.G. of the bolt system is at the geometric centre of
the bolt system.
Step2: Find the primary shear force on each bolt(Fp)
.
Step3: Find the secondary shear force on bolt(Fs)
Step4: Find the resultant shear force on most heavily
loaded bolt(FR)
Hence most heavily loaded
bolt are 2 & 6
Step5: Find the bolt Size
Eccentric load perpendicular to axis of bolts
.
Eccentric load perpendicular to axis of bolts
.
Effect of load W at a distance e from the centre of gravity of the bolt system is
equal to;
1) Force in a plane between wall & bracket: force W between the wall & bracket
results in the direct shear force on each bolt.
2) Turning Moment(W.e): the turning moment W.e tends to cause tilting of
bracket about an edge O-O. this results in tensile force on each bolt.
Procedure for design of bolted joints with load
perpendicular to axes of Bolts
.
Flowchart
design of bolted joints
with load perpendicular to
axes of Bolts
Step1: Find the direct shear force in each bolt
.
Step2: Find the direct shear stress in each bolt
.
Step3: Find the tensile stress in most heavily loaded bolt
.
•w= tensile force on bolt per unit distance ,N/mm
Therefore , Ft1=w*l1, Ft2=w*l2
Ft3=w*l3, Ft4=w*l4
•Taking moment about the C.G.
W*e= Ft1*l1+ Ft2*l2+ Ft3*l3+Ft4*l4
Step3: Find the tensile stress in most heavily loaded bolt
.
Step3: Find the tensile stress in most heavily loaded bolt
By putting w in Ft=w.l equations the tensile force on each bolt(Ft)
is given by
Step3: Find the tensile stress in most heavily loaded bolt
Note :-in this case most heavily loaded bolts are upper bolts 2 & 3
Step4: Find the maximum shear stress in most heavily
loaded bolt
Step5: Find the bolt Size
By knowing value of permissible shear stress for bolt material the
size of bolt can be determined by using equation
Ex.5.5 fig. shows a pulley bracket supported to vertical wall by 4 bolts
,two each at locations A & B. the pull W on each side of the wire rope
over the pulley is 22kN. determine the size of the coarse threaded
metric bolts using allowable shear stress of 30MPa for bolt material.
Bolts may be selected from following table.
Given
Step1: Find the direct shear force in each bolt
.
Step2: Find the direct shear stress in each bolt
.
Step3: Find the tensile stress in most heavily loaded bolt
Step4: Find the maximum shear stress in most heavily
loaded bolt
Step5: Find the bolt Size
M30 bolt with a stress area of 561 mm2 is selected
Ex.5.6 a bracket shown in fig. is fixed to the support by means of 3 bolts.
The dimensions given in figures are in mm. the bolts are maded of plain
carbon steel 45C8 (Syt=380N/mm2) and factor of safety is 2.5 .specify
the size of bolts.
Given
Step1: Find the direct shear force in each bolt
.
Step2: Find the direct shear stress in each bolt
.
Step3: Find the tensile stress in most heavily loaded bolt
Step4: Find the maximum shear stress in most heavily
loaded bolt
Step5: Find the bolt Size
Step5: Find the bolt Size
Ex.5.7 a steel bracket is fixed to vertical support by 3 bolts of size M20,
two at the top and one at the bottom, as shown in fig. if The permissible
tensile stress for the bolt is 60N/mm2. determine the maximum load
that can be supported by bracket at 350mm from the vertical support.
Given
Step1: Find the direct shear force in each bolt
.
Step2: Find the direct shear stress in each bolt
.
Step3: Find the tensile stress in most heavily loaded bolt
Step4: Find the maximum shear stress in most heavily
loaded bolt
Step4: Find the maximum load
Eccentric load parallel to axis of bolts
.
Effect of load W is equal to;
1) Parallel force through CG of bolt system(W): Parallel force through CG of bolt
system result in primary tensile force on each bolt.
2) Turning Moment(W.e): the turning moment W.e tends to cause tilting of
bracket about an edge O-O. this results in secondary tensile force on each
bolt.
Procedure for design of bolted joints with load parallel to
axes of Bolts
.
Flowchart
design of bolted joints
with load parallel to axes
of Bolts
Step1: Find the primary tensile force in each bolt
.
Step2: Find the secondary tensile forces in most heavily
loaded bolt
.
•w= tensile force on bolt per unit distance ,N/mm
Therefore , Ft1=Ft4=w*l1, Ft3=Ft2=w*l2,
•Taking moment about the C.G.
W*e= Ft1*l1+ Ft2*l2+ Ft3*l3+Ft4*l4
Step2:Find the secondary tensile forces in most heavily
loaded bolt
.
Step2: Find the secondary tensile forces in most heavily
loaded bolt
By putting w in Ft=w.l equations the tensile force on each bolt(Ft)
is given by
Step3: Find the resultant tensile force in most heavily
loaded bolt
Step4: Find the bolt Size
Ex.5.8 a cast iron bracket fixed to the steel structure, as shown in fig.
supports a load W of 25kN. There are two bolts each at A & B. if
permissible tensile stress for bolts is 50N/mm2,determine the size of the
bolts.
Given
Step1: Find the primary tensile force in each bolt
.
Step2:Find the secondary tensile forces in most heavily
loaded bolt
.
Step2: Find the secondary tensile forces in most heavily
loaded bolt
Step3: Find the resultant tensile force in most heavily
loaded bolt
Step4: Find the bolt Size
Eccentric load on circular base
.
In some applications, the base of a bracket is made circular as in case of
a flanged of a heavy machine tool or a pillar crane.
Eccentric load on circular base
.
Let,
R=outside radius of the flange, mm
r=radius of the bolt pitch circle, mm
l1,l2,l3,l4=distances of bolt centres from the tilting edge, mm
W=load per unit distance from tilting edge, N/mm
• l1=R-r cosƟ
• l2=R+r sinƟ
• l3=R+r cosƟ
• l4=R-r sinƟ
Ex.5.9 the bracket is bolted to the support and subjected to the load of
20kN as shown in fig. the bolts are made of 45C8(Syt=360N/m2). If the
required safety of margin is 1.8. find the size of bolts.
Given
Step1: Find the direct shear force in each bolt
.
Step1: Find the direct shear stress in each bolt
.
Step2:Find tensile stress in each bolt
.
Step2: Find the tensile forces in most heavily loaded bolt
Step2: Find the tensile stress in most heavily loaded bolt
Step3: Find the maximum shear stress in bolt
Step4: Find the bolt Size
Turn Buckle
Turn Buckle
A turnbuckle, stretching screw or bottle screw is a device for adjusting the tension
or length of ropes, cables, tie rods, and other tensioning systems.
A turnbuckle or coupler is mechanical joint used for connecting two members
which are subjected to tensile loading and which require slight adjustment of
length or tension under loaded condition.
Components of Turn Buckle
i. Threaded tie rod with right hand thread
ii. Threaded tie rod with left hand thread
iii. coupler
Threaded tie rod
• Two threaded rods are used.
• One of the tie rods has right hand threads
• Other has left hand threads.
Coupler
• It connects two tie rods.
• It is of hexagonal or
rectangular cross
section in the centre in
order to facilitate the
turning with help of
spanner and round at
both ends.
• Sometimes it is also
known as tommy bar
Materials for Turn Buckle
Normally tie rods are made of steel, while coupler is made of steel or
cast iron
Applications for Turn Buckle
Tie rod of wall crane, tension members of bridges, steel structures
,electric poles
Design of Turn Buckle
Design of Turn Buckle
• Diameter of tie rod(d)
• Stresses in tie rod
• Length of coupler nut(Ln)
• Outside diameter of coupler nut(D)
• Outside diameter of coupler(D2)
• Length of coupler between nuts(L)
Diameter of tie rod(d)
Tensile Stresses in tie rod
Torsional Shear Stresses in tie rod
• dm=pitch diameter of tie rod
• l=lead
• μ1=virtual coefficient of friction= (μ/ cos β)
• Ф1=virtual friction angle=tan-1 μ1
• β=Semi thread angle=30 for ISO Metric Screw Threads
• λ=lead angle=tan-1[ λ/ π.dm]
Maximum Shear Stresses in tie rod
Length of coupler nut(Ln)
Take largest value of 3 for length of coupler nut(Ln)
Outside diameter of coupler nut(D)
Take largest value of 2 for Outside diameter of coupler nut(D)
Outside diameter of coupler (D2)
Take largest value of 2 for Outside diameter of coupler (D2)
Length of coupler between nuts(L)
Ex.5.10 the pull in tie rod of an iron truss is 50kN. If the permissible
stresses are 75MPa in tension,45MPa in shear and 90MPa in crushing,
design a suitable adjustable steel screwed joint.
Given
Diameter of tie rod(d)
Tensile Stresses in tie rod
To find lead angle
To find friction angle
To find torque required
Torsional Shear Stresses in tie rod
Maximum Shear Stresses in tie rod
Hence rod is safe
Length of coupler nut(Ln)
Take largest value of 2 for length of coupler nut(Ln)
Outside diameter of coupler nut(D)
Take largest value of 2 for Outside diameter of coupler nut(D)
Outside diameter of coupler (D2)
Take largest value of 2 for Outside diameter of coupler (D2)
Length of coupler between nuts(L)
Problems for Practice
a bracket shown in fig. is fixed to steel column by using 4 bolts of size M14. A load
of W acts on the bracket at a distance of 400mm from the face of the column. The
permissible tensile stress for the bolt and bracket material is 84N/mm2. if the b/t
ratio for the cross-section of the arm of the bracket is 45,determine;
i)The maximum load that can be supported by bracket
ii) Cross section of the arm of bracket
Part B
Welded Joint : Welding symbols, Stresses in butt and fillet
welds, Strength of butt, parallel and transverse fillet welds,
Axially loaded unsymmetrical welded joints, Eccentric load in
plane of welds, Welded joints subjected to bending and
torsional moments
Advantages of Welded Joints over riveted or threaded
joints
.
• fluid tight
• give light weight construction.
• Economical
• Can be produced at much faster rate and with automation.
• The complicated components easily welded.
• Do not weakens the parts to be connected.
Disadvantages of Welded Joints
.
• Welded components are much poor in vibration.
• Welding cannot be used to join dissimilar materials.
• Quality and strength of welded joint depend upon the skill of
operator.
• For different materials, different welding processes ad hence
different machines are required.
Types of Welded Joints
.
1. Butt weld
2. Fillet or lap weld
3. Other types of weld
Butt Weld
.
Butt Weld
.
• Butt weld is obtained by placing the plates to be joined side
by side with their edges nearly touching each other.
• Small gap is maintained between the edges.
Types of Butt Weld
.
1. Square butt weld
2. Single V butt weld
3. Single U butt weld
4. Double V butt weld
5. Double U butt weld
Square Butt Weld
.
• If thickness of the plates is less than 5mm,edges of the plates
do not required bevelling, and hence joint used is known as
square butt weld.
Single V-butt or single U-butt Weld
.
• If thickness of the plates is between 5mm and 12.5mm,edges
of the plates bevelled to V or U groove, and accordingly
single V-butt or single U-butt weld may be used.
Double V-butt or double U-butt Weld
.
•If thickness of the plates is between more than 12.5mm,it is
necessary to bevel and weld plates from both sides. In such
cases double V-butt or double U-butt weld may be used.
Fillet or Lap Weld
.
Fillet or Lap Weld
.
• It is obtained by overlapping two plates and then welding
edges of plates.
Types of Fillet or Lap Weld
.
1. Parallel fillet weld
2. Transverse fillet weld
Parallel fillet Weld
.
• If the load axis is parallel to the axis of the fillet ,it is known
as parallel fillet weld.
Transverse fillet Weld
.
• If the load axis is perpendicular to the axis of the fillet ,it is
known as transverse fillet weld.
• It can be single or double transverse fillet weld.
Welding symbols
.
.
Stresses in Butt Welds
.
Tensile stress in single V- Butt Welds
• = average tensile stress in the weld, N/mm2
•P=tensile force on the weld ,N
•h=weld throat thickness of butt weld or plate thickness, mm
•l=length of weld, mm
Tensile stress in double V- Butt Welds
• = average tensile stress in the weld, N/mm2
•P=tensile force on the weld ,N
•h1=throat thickness at the top, mm
•h2=throat thickness at the bottom, mm
Shear stress in single butt weld
• = average shear stress in the weld, N/mm2
•Ps=shear force on the weld ,N
•l=length of the weld, mm
•h=throat thickness or plate thickness, mm
Stresses in fillet Welds
.
• Weld size or leg size(h): the length of each equal sides of isosceles
triangle is known as weld size or leg size h.
• Weld throat thickness(t): the perpendicular distance of hypotenuse
from intersection of legs is known as weld throat thickness t.
• Cross sectional area of weld is minimum at the throat which is located
at 45 to the leg.
Shear stress in fillet weld
• = average shear stress in the weld, N/mm2
• P=tensile or shear force on the weld ,N
• l=length of the fillet weld, mm
• t=weld throat thickness, mm
• h=weld size or leg size of fillet weld, mm
Strength of Welds
.
Strength of Butt Welds
.
• Tensile strength
•Compressive strength
• Shear strength
Strength of single transverse fillet Welds
.
Strength of double transverse fillet Welds
.
Strength of double parallel fillet Welds
.
Procedure for design of Weld with In-plane eccentric
.
Procedure for design of Weld with In-
plane eccentric loads
.
Welded joints with In-plane eccentric loads
.
1) Parallel force (P): Parallel force through CG of bolt system result in
primary shear stress.
2) Twisting Moment(T): the twisting moment about the CG of weld
results in secondary shear stress.
Step 1 find polar Moment of Inertia of Weld Group
.
Step 1 find polar Moment of Inertia of Weld Group
.
Weld figure Moment of Inertia about XX [Ixx]
Step 1 find polar Moment of Inertia of Weld Group
.
Weld figure Moment of Inertia about XX [Ixx]
Step 1 find polar Moment of Inertia of Weld Group
Step 2 find direct(Primary) Shear stress
• P=eccentric force on the weld ,N
• A=throat area of all weld, mm2
=A1+A2+A3=l1*t+l2*t+l3*t=(2*l1+l2)*t
• t=throat thickness of fillet weld, mm
Step 3 find Torsional(Secondary) Shear stress
Step 4 find Resultant Shear stress
Step 5 Calculate the weld size(h)
Ex.5.11 A bracket is welded to a column as shown in fig. Determine the
size of the weld, if permissible shear stress in the weld is 80 N/mm2
Given
Step 1 find polar Moment of Inertia of Weld Group
.
Step 1 find polar Moment of Inertia of Weld Group
.
Step 1 find polar Moment of Inertia of Weld Group
.
Step 1 find polar Moment of Inertia of Weld Group
.
Step 2 find direct(Primary) Shear stress
Step 3 find Torsional(Secondary) Shear stress
Step 4 find Resultant Shear stress
Step 5 Calculate the weld size(h)
Ex.5.12 fig. shows a rectangular steel plate welded as a cantilever to a
vertical column and supports a single concentrated load of 60kN.
Determine the weld size, if shear stress in the same is not to exceed
140MPa.
Given
Step 1 find polar Moment of Inertia of Weld Group
.
Step 1 find polar Moment of Inertia of Weld Group
.
Step 1 find polar Moment of Inertia of Weld Group
.
Step 1 find polar Moment of Inertia of Weld Group
.
Step 2 find direct(Primary) Shear stress
Step 3 find Torsional(Secondary) Shear stress
Step 4 find Resultant Shear stress
Step 5 Calculate the weld size(h)
Ex.5.13 A bracket is welded to the column and carries an eccentric load
P as shown in fig. the size of weld is 12mm. If the maximum shear stress
in the weld is 80MPa,determine the load P.
Given
Step 1 find polar Moment of Inertia of Weld Group
.
Step 1 find polar Moment of Inertia of Weld Group
.
Step 1 find polar Moment of Inertia of Weld Group
.
Step 1 find polar Moment of Inertia of Weld Group
.
Step 2 find direct(Primary) Shear stress
Step 3 find Torsional(Secondary) Shear stress
Step 4 find Resultant Shear stress
Step 5 Calculate the weld size(h)
Ex. fig. shows a welded joint subjected to an eccentric load of 20kN. The
welding is only on one side. If the permissible shear stress for the weld
material is 80MPa,determine the weld size.
Procedure for design of Weld with subjected to bending
.
Procedure for design of Weld with
subjected to bending moment
.
Welded joints subjected to bending moment
.
1) Parallel force (P): Parallel force through the plane of the welds
result in direct shear stress or primary shear stress.
2) Bending Moment(M): the bending moment causes the moment
induced shear stress or secondary shear stress.
Step 1 find Moment of Inertia of Weld Group about
horizontal axis through C.G.(Ixx):
.
Step 2: find the direct (primary) shear stress
.
Step 3: find moment induced (secondary) shear stress:
.
Step 4: find the resultant shear stress:
.
Step 5 Calculate the weld size(h)
Ex.5.14 a solid rectangular bar of cross-section 100mm*150mm is
welded to a support by means of fillet weld as shown in fig. it is
subjected to a load of 25kN at a distance of 500mm from the plane of
weld. If the permissible shear stress for the weld is 85 N/mm2.
determine the weld size and throat thickness.
Given
Step 1 find Moment of Inertia of Weld Group about
horizontal axis through C.G.(Ixx):
.
Step 2: find the direct (primary) shear stress
.
Step 3: find moment induced (secondary) shear stress:
.
Step 4: find the resultant shear stress:
.
Step 5 Calculate the weld size(h)
Ex.5.15 a bracket is welded to the vertical plate by two fillet welds, as
shown in fig. determine the weld size, if permissible shear stress is
limited to 70MPa.
Given
Step 1 find Moment of Inertia of Weld Group about
horizontal axis through C.G.(Ixx):
.
Step 2: find the direct (primary) shear stress
.
Step 3: find moment induced (secondary) shear stress:
.
Step 4: find the resultant shear stress:
.
Step 5 Calculate the weld size(h)
Welded joints subjected to torsional moment
Ex. 5.16 A circular bar of 50mm diameter is welded to a steel plate by
an annular fillet weld and is subjected to a twisting moment of 2kN-m.
if the allowable shear stress in the weld material is 85MPa, determine
the size of the weld.
Given
To find polar MI
Torsional shear stress
To find weld size

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Unit 5 Design of Threaded and Welded Joints

  • 1. Unit 5 Threaded & Welded Joint Prepared By Prof. M.C. Shinde [9970160753] Mech. Engg. Dept., JSCOE, Hadapsar
  • 2. SPPU Syllabus Content: (10 hrs) Threaded Joint: Basic types of screw fasteners, Bolts of uniform strength, I.S.O. Metric screw threads, Bolts under tension, eccentrically loaded bolted joint in shear, Eccentric load perpendicular and parallel to axis of bolt, Eccentric load on circular base, design of Turn Buckle. Welded Joint : Welding symbols, Stresses in butt and fillet welds, Strength of butt, parallel and transverse fillet welds, Axially loaded unsymmetrical welded joints, Eccentric load in plane of welds, Welded joints subjected to bending and torsional moments
  • 3. Part A Threaded Joint: Basic types of screw fasteners, Bolts of uniform strength, I.S.O. Metric screw threads, Bolts under tension, eccentrically loaded bolted joint in shear, Eccentric load perpendicular and parallel to axis of bolt, Eccentric load on circular base, design of Turn Buckle.
  • 5. Need to Use Threaded Joints . •To connect the components together which should be readily disassembled. E.g. Screwed fasteners •To transmit a power or energy e.g. power screws •For obtaining accurate movements in measuring instruments.
  • 6. Advantages of Threaded Joints . •They are convenient to assemble and disassemble. •The assembly and disassembly require only spanners and do not require any special tooling's; •Threaded joints are relatively cheap; •They give high clamping force; •They are highly reliable. Note:- Tooling is the process of acquiring the manufacturing components and machines needed for production
  • 7. Disadvantages of Threaded Joints . •The component becomes weak because of the holes; •There is a possibility of loosening of joint due to excusive vibrations; •There is stress concentration in threaded portion, which may be vulnerable under fatigue load.
  • 8. I.S.O. Metric Screw Thread . •fig. shows profile of an ISO metric screw threads which are commonly used in India. •It is not easy to get mathematical relation between d,p,dc ,hence it is necessary to refer the standard tables to find the dimensions of screw threads.
  • 9. Type of I.S.O. Metric Screw Thread . •Coarse threads •Fine threads
  • 10. Coarse Thread . •For given nominal diameter, pitch and hence lead angle are larger compare to fine threads Advantages •They are stronger and hence less possibility of thread crushing and shearing. •Due to large lead angle threads offer less resistance to unscrewing.
  • 11. Designation of ISO Metric Coarse Thread . •It is designated by letter M followed by nominal diameter. E.g. M30
  • 12. Fine Thread . •For given nominal diameter, pitch and hence lead angle are smaller compare to coarse threads Advantages •Fine adjustments is possible with fine threads. •Though, fine threads are weaker its screw body is stronger. •Fine threads offer greater resistance to unscrewing.
  • 13. Designation of ISO Metric Fine Thread . It is designated by letter M followed by nominal diameter and pitch, two being separated by sign x. E.g. M30x3
  • 14. Basic types of screw fasteners . 1) Through bolt 2) Tap bolt 3) Studs 4) Cap screws 5) Set screws
  • 15. Through Bolt . •It is cylindrical bar with a head at one end the threads at other end. •It is passed through drilled holes in the parts to be fastened together and nut is screwed tightened on the threaded end. •Nut and bolt heads are hexagonal.
  • 16. Tap Bolt . •It is the bolt which passes through a hole in one part and screwed into a tapped hole in the second part which acts as nut thus holding parts together. • It does not need nut. •It is used when arrangement does not provide any space to accommodate the nut. • Frequent insertion or removal of tap bolt is likely to damage threads
  • 17. Studs . •It is a cylindrical bar threaded at both ends. •One end is screwed into tapped hole of parts to be fastened while nut is screwed into other end . •Stud always remains in position when two parts are disconnected •Studs are mainly used for connecting covers like cylinder head of engine.
  • 18. Cap screws . •Cap screws are very similar to tap bolts, but smaller in size as compared to tap bolts. • In addition variety of shapes of heads are available for cap screws.
  • 19. Set screws . •Set screws are very similar to cap screws but are threaded practically throughout length and are still smaller in size. Theses are used to prevent the relative motion between two parts. •They are used instead of key in light power transmission
  • 20. Materials used for screwed fasteners . •Materials used for lightly loaded nuts, bolts and studs are medium carbon steels like: 35C8,40C8,45C8 •Materials used for highly loaded high strength nuts, bolts and studs are high carbon steels(60C4) and alloy steels(40Ni4,35Mn6Mo4) •Materials used for corrosion resistance fasteners are chromium alloy steels like: 40Cr4Mo2, 40Ni6Cr4Mo2,31Ni100Cr3Mo6
  • 21. Bolts of uniform strength . •If a bolt is subjected to shock or impact loading as in case of connecting rod bolts or fasteners of power hammers, it should be designed to absorb impact energy
  • 22. Drawback of ordinary bolt: . •In ordinary bolts ,cross-sectional area is minimum in threaded portion. Hence, stress in threaded portion of the bolt will be higher than that in shank or body. •Impact energy absorbed at any point in a body is directly proportional to square of stress at that point. •Therefore large portion of impact energy will be absorbed in the threaded portion and relatively small portion of energy is absorbed by a shank. •This uneven distribution of impact energy may lead to fracture of bolt in threaded portion.
  • 23. Methods of achieving bolts of uniform strength . •Reduction of shank diameter •Drilling axial hole
  • 24. Reduction of shank diameter . •If shank diameter is reduced to core diameter as shown in fig. stress become same throughout length of bolt. •Hence impact energy is distributed uniformly throughout the bolt length. •The bolt in this way becomes stronger and lighter. •This type of bolt is known as bolt of uniform strength.
  • 25. Drilling axial hole . •In this method an axial hole is drilled through the head down to the threaded portion such that cross sectional area of threaded portion. •For bolts of uniform strength
  • 26. Stresses in screw fastener body . 1) Direct tensile stress 2) Direct shear stress 3) Maximum shear stress
  • 30. Eccentrically loaded bolted joints . 1) Eccentric load in plane of bolts 2) Eccentric load perpendicular to axes of bolts 3) Eccentric load parallel to axes of bolts
  • 31. Eccentric load in plane of bolts .
  • 32. Eccentric load in plane of bolts . Effect of load W at a distance e from the centre of gravity of the bolt system is equal to; 1) Direct parallel force (W): the direct parallel force W through CG of bolt system results in primary shear force on each bolt. 2) Turning Moment(W.e): the turning moment W.e about C.G. of bolt system results in secondary shear force on each bolt.
  • 33. Procedure for design of bolted joints with load in Plane of Bolts .
  • 34. Step1: Find the C.G. of bolt system . •Let A1,A2,A3,A4……… cross sectional areas of the bolts,mm2 •x1,x2,x3,x4………distances of bolt centres from Y-axis, mm •y1, y2,y3,y4………distances of bolt centres from X-axis, mm
  • 35. Step1: Find the C.G. of bolt system . •Distance of C.G. of bolt system from Y-axis is,
  • 36. Step1: Find the C.G. of bolt system . •Distance of C.G. of bolt system from X-axis is,
  • 37. Step2: Find the primary shear force on each bolt(Fp) . •The direct parallel force W through C.G. of the bolt system results in primary shear force Fp on each bolt and is given by Fp1=Fp2=Fp3=Fp4=Fp=W/n Where n=number of bolts Fp1,Fp2,Fp3,Fp4 =primary shear forces on bolts,N
  • 38. Step3: Find the secondary shear force on each bolt(Fs) . •The twisting moment W.e about C.G. of the bolt system results in secondary shear forces which are not same on all the bolts. •secondary shear force on any bolt is proportional to its distance from C.G. Let, Fs1,Fs2,Fs3,Fs4 =secondary shear forces on bolts, N l1,l2,l3,l4=distances of the bolt centres from C.G.
  • 39. Step3: Find the secondary shear force on each bolt(Fs) . •w= secondary shear force on bolt per unit distance ,N/mm Therefore , Fs1=w*l1, Fs2=w*l2 Fs3=w*l3, Fs4=w*l4 •Taking moment about the C.G. W*e= Fs1*l1+ Fs2*l2+ Fs3*l3+Fs4*l4
  • 40. Step3: Find the secondary shear force on each bolt(Fs)
  • 41. Step3: Find the secondary shear force on each bolt(Fs) By putting w in Fs=w.l equations the secondary shear force on each bolt(Fs) is given by Note :-the direction of secondary shear force on any bolt is perpendicular to the line joining the bolt centre and the C.G. of the bolt system
  • 42. Step4: Find the resultant shear force on most heavily loaded bolt(FR) Note :-the direction of secondary shear force on any bolt is perpendicular to the line joining the bolt centre and the C.G. of the bolt system
  • 43. Step5: Find the bolt Size Where, FR=Resultant shear force Ac=Cross-sectional area of bolt
  • 44. Summary of Steps: design of bolted joints with load in Plane of Bolts
  • 45. Ex.5.1 a steel plate subjected to a force of 8kN is fixed to a channel by means of three identical bolts as shown in fig. the bolts are made of 45C8 (Syt=380N/mm2). If the required factor of safety is 2.5 determine size of the bolts.
  • 46. Given
  • 47. Step1: Find the C.G. of bolt system . The C.G. of the bolt system is at the geometric centre of the bolt system.
  • 48. Step2: Find the primary shear force on each bolt(Fp) .
  • 49. Step3: Find the secondary shear force on loaded (1&3) bolt(Fs)
  • 50. Step4: Find the resultant shear force on most heavily loaded bolt(FR)
  • 51. Step5: Find the bolt Size
  • 52. Ex.5.2 a steel plate subjected to a force of 3kN is fixed to a vertical channel by means of four identical bolts as shown in fig. the bolts are made of plain carbon steel 45C8 (Syt=380N/mm2). If the required factor of safety is 2. determine diameter of the bolts.
  • 53. Given
  • 54.
  • 55. Step1: Find the C.G. of bolt system . The C.G. of the bolt system is at the geometric centre of the bolt system.
  • 56. Step2: Find the primary shear force on each bolt(Fp) .
  • 57. Step3: Find the secondary shear force on bolt(Fs)
  • 58. Step4: Find the resultant shear force on most heavily loaded bolt(FR)
  • 59. Step5: Find the bolt Size
  • 60. Ex.5.3 a bracket made of steel subjected to a force of 8kN is fixed to a vertical channel by means of four identical bolts as shown in fig. the bolts are made of plain carbon steel 45C8 (Syt=254N/mm2). If the required factor of safety is 2. determine diameter of the bolts.
  • 61. Given
  • 62.
  • 63. Step1: Find the C.G. of bolt system . The C.G. of the bolt system is at the geometric centre of the bolt system.
  • 64. Step2: Find the primary shear force on each bolt(Fp) .
  • 65. Step3: Find the secondary shear force on bolt(Fs)
  • 66. Step4: Find the resultant shear force on most heavily loaded bolt(FR)
  • 67. Step5: Find the bolt Size
  • 68. Ex.5.4 a bracket is bolted to a column by 6 bolts of equal size as shown in fig. it carries a load of 50kN at the distance of 150mm from the centre of column. If the maximum stress in the bolt is to be limited to 150N/mm2. determine diameter of the bolts.
  • 69. Given
  • 70.
  • 71. Step1: Find the C.G. of bolt system . The C.G. of the bolt system is at the geometric centre of the bolt system.
  • 72. Step2: Find the primary shear force on each bolt(Fp) .
  • 73. Step3: Find the secondary shear force on bolt(Fs)
  • 74. Step4: Find the resultant shear force on most heavily loaded bolt(FR) Hence most heavily loaded bolt are 2 & 6
  • 75. Step5: Find the bolt Size
  • 76. Eccentric load perpendicular to axis of bolts .
  • 77. Eccentric load perpendicular to axis of bolts . Effect of load W at a distance e from the centre of gravity of the bolt system is equal to; 1) Force in a plane between wall & bracket: force W between the wall & bracket results in the direct shear force on each bolt. 2) Turning Moment(W.e): the turning moment W.e tends to cause tilting of bracket about an edge O-O. this results in tensile force on each bolt.
  • 78. Procedure for design of bolted joints with load perpendicular to axes of Bolts .
  • 79. Flowchart design of bolted joints with load perpendicular to axes of Bolts
  • 80. Step1: Find the direct shear force in each bolt .
  • 81. Step2: Find the direct shear stress in each bolt .
  • 82. Step3: Find the tensile stress in most heavily loaded bolt . •w= tensile force on bolt per unit distance ,N/mm Therefore , Ft1=w*l1, Ft2=w*l2 Ft3=w*l3, Ft4=w*l4 •Taking moment about the C.G. W*e= Ft1*l1+ Ft2*l2+ Ft3*l3+Ft4*l4
  • 83. Step3: Find the tensile stress in most heavily loaded bolt .
  • 84. Step3: Find the tensile stress in most heavily loaded bolt By putting w in Ft=w.l equations the tensile force on each bolt(Ft) is given by
  • 85. Step3: Find the tensile stress in most heavily loaded bolt Note :-in this case most heavily loaded bolts are upper bolts 2 & 3
  • 86. Step4: Find the maximum shear stress in most heavily loaded bolt
  • 87. Step5: Find the bolt Size By knowing value of permissible shear stress for bolt material the size of bolt can be determined by using equation
  • 88. Ex.5.5 fig. shows a pulley bracket supported to vertical wall by 4 bolts ,two each at locations A & B. the pull W on each side of the wire rope over the pulley is 22kN. determine the size of the coarse threaded metric bolts using allowable shear stress of 30MPa for bolt material. Bolts may be selected from following table.
  • 89. Given
  • 90. Step1: Find the direct shear force in each bolt .
  • 91. Step2: Find the direct shear stress in each bolt .
  • 92. Step3: Find the tensile stress in most heavily loaded bolt
  • 93. Step4: Find the maximum shear stress in most heavily loaded bolt
  • 94. Step5: Find the bolt Size M30 bolt with a stress area of 561 mm2 is selected
  • 95. Ex.5.6 a bracket shown in fig. is fixed to the support by means of 3 bolts. The dimensions given in figures are in mm. the bolts are maded of plain carbon steel 45C8 (Syt=380N/mm2) and factor of safety is 2.5 .specify the size of bolts.
  • 96. Given
  • 97. Step1: Find the direct shear force in each bolt .
  • 98. Step2: Find the direct shear stress in each bolt .
  • 99. Step3: Find the tensile stress in most heavily loaded bolt
  • 100. Step4: Find the maximum shear stress in most heavily loaded bolt
  • 101. Step5: Find the bolt Size
  • 102. Step5: Find the bolt Size
  • 103. Ex.5.7 a steel bracket is fixed to vertical support by 3 bolts of size M20, two at the top and one at the bottom, as shown in fig. if The permissible tensile stress for the bolt is 60N/mm2. determine the maximum load that can be supported by bracket at 350mm from the vertical support.
  • 104. Given
  • 105. Step1: Find the direct shear force in each bolt .
  • 106. Step2: Find the direct shear stress in each bolt .
  • 107. Step3: Find the tensile stress in most heavily loaded bolt
  • 108. Step4: Find the maximum shear stress in most heavily loaded bolt
  • 109. Step4: Find the maximum load
  • 110. Eccentric load parallel to axis of bolts . Effect of load W is equal to; 1) Parallel force through CG of bolt system(W): Parallel force through CG of bolt system result in primary tensile force on each bolt. 2) Turning Moment(W.e): the turning moment W.e tends to cause tilting of bracket about an edge O-O. this results in secondary tensile force on each bolt.
  • 111. Procedure for design of bolted joints with load parallel to axes of Bolts .
  • 112. Flowchart design of bolted joints with load parallel to axes of Bolts
  • 113. Step1: Find the primary tensile force in each bolt .
  • 114. Step2: Find the secondary tensile forces in most heavily loaded bolt . •w= tensile force on bolt per unit distance ,N/mm Therefore , Ft1=Ft4=w*l1, Ft3=Ft2=w*l2, •Taking moment about the C.G. W*e= Ft1*l1+ Ft2*l2+ Ft3*l3+Ft4*l4
  • 115. Step2:Find the secondary tensile forces in most heavily loaded bolt .
  • 116. Step2: Find the secondary tensile forces in most heavily loaded bolt By putting w in Ft=w.l equations the tensile force on each bolt(Ft) is given by
  • 117. Step3: Find the resultant tensile force in most heavily loaded bolt
  • 118. Step4: Find the bolt Size
  • 119. Ex.5.8 a cast iron bracket fixed to the steel structure, as shown in fig. supports a load W of 25kN. There are two bolts each at A & B. if permissible tensile stress for bolts is 50N/mm2,determine the size of the bolts.
  • 120. Given
  • 121. Step1: Find the primary tensile force in each bolt .
  • 122. Step2:Find the secondary tensile forces in most heavily loaded bolt .
  • 123. Step2: Find the secondary tensile forces in most heavily loaded bolt
  • 124. Step3: Find the resultant tensile force in most heavily loaded bolt
  • 125. Step4: Find the bolt Size
  • 126. Eccentric load on circular base . In some applications, the base of a bracket is made circular as in case of a flanged of a heavy machine tool or a pillar crane.
  • 127. Eccentric load on circular base . Let, R=outside radius of the flange, mm r=radius of the bolt pitch circle, mm l1,l2,l3,l4=distances of bolt centres from the tilting edge, mm W=load per unit distance from tilting edge, N/mm • l1=R-r cosƟ • l2=R+r sinƟ • l3=R+r cosƟ • l4=R-r sinƟ
  • 128. Ex.5.9 the bracket is bolted to the support and subjected to the load of 20kN as shown in fig. the bolts are made of 45C8(Syt=360N/m2). If the required safety of margin is 1.8. find the size of bolts.
  • 129. Given
  • 130. Step1: Find the direct shear force in each bolt .
  • 131. Step1: Find the direct shear stress in each bolt .
  • 132. Step2:Find tensile stress in each bolt .
  • 133. Step2: Find the tensile forces in most heavily loaded bolt
  • 134. Step2: Find the tensile stress in most heavily loaded bolt
  • 135. Step3: Find the maximum shear stress in bolt
  • 136. Step4: Find the bolt Size
  • 138. Turn Buckle A turnbuckle, stretching screw or bottle screw is a device for adjusting the tension or length of ropes, cables, tie rods, and other tensioning systems. A turnbuckle or coupler is mechanical joint used for connecting two members which are subjected to tensile loading and which require slight adjustment of length or tension under loaded condition.
  • 139. Components of Turn Buckle i. Threaded tie rod with right hand thread ii. Threaded tie rod with left hand thread iii. coupler
  • 140. Threaded tie rod • Two threaded rods are used. • One of the tie rods has right hand threads • Other has left hand threads.
  • 141. Coupler • It connects two tie rods. • It is of hexagonal or rectangular cross section in the centre in order to facilitate the turning with help of spanner and round at both ends. • Sometimes it is also known as tommy bar
  • 142. Materials for Turn Buckle Normally tie rods are made of steel, while coupler is made of steel or cast iron Applications for Turn Buckle Tie rod of wall crane, tension members of bridges, steel structures ,electric poles
  • 143. Design of Turn Buckle
  • 144. Design of Turn Buckle • Diameter of tie rod(d) • Stresses in tie rod • Length of coupler nut(Ln) • Outside diameter of coupler nut(D) • Outside diameter of coupler(D2) • Length of coupler between nuts(L)
  • 145. Diameter of tie rod(d)
  • 147. Torsional Shear Stresses in tie rod • dm=pitch diameter of tie rod • l=lead • μ1=virtual coefficient of friction= (μ/ cos β) • Ф1=virtual friction angle=tan-1 μ1 • β=Semi thread angle=30 for ISO Metric Screw Threads • λ=lead angle=tan-1[ λ/ π.dm]
  • 148. Maximum Shear Stresses in tie rod
  • 149. Length of coupler nut(Ln) Take largest value of 3 for length of coupler nut(Ln)
  • 150. Outside diameter of coupler nut(D) Take largest value of 2 for Outside diameter of coupler nut(D)
  • 151. Outside diameter of coupler (D2) Take largest value of 2 for Outside diameter of coupler (D2)
  • 152. Length of coupler between nuts(L)
  • 153. Ex.5.10 the pull in tie rod of an iron truss is 50kN. If the permissible stresses are 75MPa in tension,45MPa in shear and 90MPa in crushing, design a suitable adjustable steel screwed joint.
  • 154. Given
  • 155. Diameter of tie rod(d)
  • 157. To find lead angle
  • 159. To find torque required
  • 161. Maximum Shear Stresses in tie rod Hence rod is safe
  • 162. Length of coupler nut(Ln) Take largest value of 2 for length of coupler nut(Ln)
  • 163. Outside diameter of coupler nut(D) Take largest value of 2 for Outside diameter of coupler nut(D)
  • 164. Outside diameter of coupler (D2) Take largest value of 2 for Outside diameter of coupler (D2)
  • 165. Length of coupler between nuts(L)
  • 166. Problems for Practice a bracket shown in fig. is fixed to steel column by using 4 bolts of size M14. A load of W acts on the bracket at a distance of 400mm from the face of the column. The permissible tensile stress for the bolt and bracket material is 84N/mm2. if the b/t ratio for the cross-section of the arm of the bracket is 45,determine; i)The maximum load that can be supported by bracket ii) Cross section of the arm of bracket
  • 167. Part B Welded Joint : Welding symbols, Stresses in butt and fillet welds, Strength of butt, parallel and transverse fillet welds, Axially loaded unsymmetrical welded joints, Eccentric load in plane of welds, Welded joints subjected to bending and torsional moments
  • 168. Advantages of Welded Joints over riveted or threaded joints . • fluid tight • give light weight construction. • Economical • Can be produced at much faster rate and with automation. • The complicated components easily welded. • Do not weakens the parts to be connected.
  • 169. Disadvantages of Welded Joints . • Welded components are much poor in vibration. • Welding cannot be used to join dissimilar materials. • Quality and strength of welded joint depend upon the skill of operator. • For different materials, different welding processes ad hence different machines are required.
  • 170. Types of Welded Joints . 1. Butt weld 2. Fillet or lap weld 3. Other types of weld
  • 172. Butt Weld . • Butt weld is obtained by placing the plates to be joined side by side with their edges nearly touching each other. • Small gap is maintained between the edges.
  • 173. Types of Butt Weld . 1. Square butt weld 2. Single V butt weld 3. Single U butt weld 4. Double V butt weld 5. Double U butt weld
  • 174. Square Butt Weld . • If thickness of the plates is less than 5mm,edges of the plates do not required bevelling, and hence joint used is known as square butt weld.
  • 175. Single V-butt or single U-butt Weld . • If thickness of the plates is between 5mm and 12.5mm,edges of the plates bevelled to V or U groove, and accordingly single V-butt or single U-butt weld may be used.
  • 176. Double V-butt or double U-butt Weld . •If thickness of the plates is between more than 12.5mm,it is necessary to bevel and weld plates from both sides. In such cases double V-butt or double U-butt weld may be used.
  • 177. Fillet or Lap Weld .
  • 178. Fillet or Lap Weld . • It is obtained by overlapping two plates and then welding edges of plates.
  • 179. Types of Fillet or Lap Weld . 1. Parallel fillet weld 2. Transverse fillet weld
  • 180. Parallel fillet Weld . • If the load axis is parallel to the axis of the fillet ,it is known as parallel fillet weld.
  • 181. Transverse fillet Weld . • If the load axis is perpendicular to the axis of the fillet ,it is known as transverse fillet weld. • It can be single or double transverse fillet weld.
  • 183. .
  • 184. Stresses in Butt Welds .
  • 185. Tensile stress in single V- Butt Welds • = average tensile stress in the weld, N/mm2 •P=tensile force on the weld ,N •h=weld throat thickness of butt weld or plate thickness, mm •l=length of weld, mm
  • 186. Tensile stress in double V- Butt Welds • = average tensile stress in the weld, N/mm2 •P=tensile force on the weld ,N •h1=throat thickness at the top, mm •h2=throat thickness at the bottom, mm
  • 187. Shear stress in single butt weld • = average shear stress in the weld, N/mm2 •Ps=shear force on the weld ,N •l=length of the weld, mm •h=throat thickness or plate thickness, mm
  • 188. Stresses in fillet Welds .
  • 189. • Weld size or leg size(h): the length of each equal sides of isosceles triangle is known as weld size or leg size h. • Weld throat thickness(t): the perpendicular distance of hypotenuse from intersection of legs is known as weld throat thickness t. • Cross sectional area of weld is minimum at the throat which is located at 45 to the leg.
  • 190. Shear stress in fillet weld • = average shear stress in the weld, N/mm2 • P=tensile or shear force on the weld ,N • l=length of the fillet weld, mm • t=weld throat thickness, mm • h=weld size or leg size of fillet weld, mm
  • 192. Strength of Butt Welds . • Tensile strength •Compressive strength • Shear strength
  • 193. Strength of single transverse fillet Welds .
  • 194. Strength of double transverse fillet Welds .
  • 195. Strength of double parallel fillet Welds .
  • 196. Procedure for design of Weld with In-plane eccentric .
  • 197. Procedure for design of Weld with In- plane eccentric loads .
  • 198. Welded joints with In-plane eccentric loads . 1) Parallel force (P): Parallel force through CG of bolt system result in primary shear stress. 2) Twisting Moment(T): the twisting moment about the CG of weld results in secondary shear stress.
  • 199. Step 1 find polar Moment of Inertia of Weld Group .
  • 200. Step 1 find polar Moment of Inertia of Weld Group . Weld figure Moment of Inertia about XX [Ixx]
  • 201. Step 1 find polar Moment of Inertia of Weld Group . Weld figure Moment of Inertia about XX [Ixx]
  • 202. Step 1 find polar Moment of Inertia of Weld Group
  • 203. Step 2 find direct(Primary) Shear stress • P=eccentric force on the weld ,N • A=throat area of all weld, mm2 =A1+A2+A3=l1*t+l2*t+l3*t=(2*l1+l2)*t • t=throat thickness of fillet weld, mm
  • 204. Step 3 find Torsional(Secondary) Shear stress
  • 205. Step 4 find Resultant Shear stress
  • 206. Step 5 Calculate the weld size(h)
  • 207. Ex.5.11 A bracket is welded to a column as shown in fig. Determine the size of the weld, if permissible shear stress in the weld is 80 N/mm2
  • 208. Given
  • 209. Step 1 find polar Moment of Inertia of Weld Group .
  • 210. Step 1 find polar Moment of Inertia of Weld Group .
  • 211. Step 1 find polar Moment of Inertia of Weld Group .
  • 212. Step 1 find polar Moment of Inertia of Weld Group .
  • 213. Step 2 find direct(Primary) Shear stress
  • 214. Step 3 find Torsional(Secondary) Shear stress
  • 215. Step 4 find Resultant Shear stress
  • 216. Step 5 Calculate the weld size(h)
  • 217. Ex.5.12 fig. shows a rectangular steel plate welded as a cantilever to a vertical column and supports a single concentrated load of 60kN. Determine the weld size, if shear stress in the same is not to exceed 140MPa.
  • 218. Given
  • 219. Step 1 find polar Moment of Inertia of Weld Group .
  • 220. Step 1 find polar Moment of Inertia of Weld Group .
  • 221. Step 1 find polar Moment of Inertia of Weld Group .
  • 222. Step 1 find polar Moment of Inertia of Weld Group .
  • 223. Step 2 find direct(Primary) Shear stress
  • 224. Step 3 find Torsional(Secondary) Shear stress
  • 225. Step 4 find Resultant Shear stress
  • 226. Step 5 Calculate the weld size(h)
  • 227. Ex.5.13 A bracket is welded to the column and carries an eccentric load P as shown in fig. the size of weld is 12mm. If the maximum shear stress in the weld is 80MPa,determine the load P.
  • 228. Given
  • 229. Step 1 find polar Moment of Inertia of Weld Group .
  • 230. Step 1 find polar Moment of Inertia of Weld Group .
  • 231. Step 1 find polar Moment of Inertia of Weld Group .
  • 232. Step 1 find polar Moment of Inertia of Weld Group .
  • 233. Step 2 find direct(Primary) Shear stress
  • 234. Step 3 find Torsional(Secondary) Shear stress
  • 235. Step 4 find Resultant Shear stress
  • 236. Step 5 Calculate the weld size(h)
  • 237. Ex. fig. shows a welded joint subjected to an eccentric load of 20kN. The welding is only on one side. If the permissible shear stress for the weld material is 80MPa,determine the weld size.
  • 238. Procedure for design of Weld with subjected to bending .
  • 239. Procedure for design of Weld with subjected to bending moment .
  • 240. Welded joints subjected to bending moment . 1) Parallel force (P): Parallel force through the plane of the welds result in direct shear stress or primary shear stress. 2) Bending Moment(M): the bending moment causes the moment induced shear stress or secondary shear stress.
  • 241. Step 1 find Moment of Inertia of Weld Group about horizontal axis through C.G.(Ixx): .
  • 242. Step 2: find the direct (primary) shear stress .
  • 243. Step 3: find moment induced (secondary) shear stress: .
  • 244. Step 4: find the resultant shear stress: .
  • 245. Step 5 Calculate the weld size(h)
  • 246. Ex.5.14 a solid rectangular bar of cross-section 100mm*150mm is welded to a support by means of fillet weld as shown in fig. it is subjected to a load of 25kN at a distance of 500mm from the plane of weld. If the permissible shear stress for the weld is 85 N/mm2. determine the weld size and throat thickness.
  • 247. Given
  • 248. Step 1 find Moment of Inertia of Weld Group about horizontal axis through C.G.(Ixx): .
  • 249. Step 2: find the direct (primary) shear stress .
  • 250. Step 3: find moment induced (secondary) shear stress: .
  • 251. Step 4: find the resultant shear stress: .
  • 252. Step 5 Calculate the weld size(h)
  • 253. Ex.5.15 a bracket is welded to the vertical plate by two fillet welds, as shown in fig. determine the weld size, if permissible shear stress is limited to 70MPa.
  • 254. Given
  • 255. Step 1 find Moment of Inertia of Weld Group about horizontal axis through C.G.(Ixx): .
  • 256. Step 2: find the direct (primary) shear stress .
  • 257. Step 3: find moment induced (secondary) shear stress: .
  • 258. Step 4: find the resultant shear stress: .
  • 259. Step 5 Calculate the weld size(h)
  • 260. Welded joints subjected to torsional moment
  • 261. Ex. 5.16 A circular bar of 50mm diameter is welded to a steel plate by an annular fillet weld and is subjected to a twisting moment of 2kN-m. if the allowable shear stress in the weld material is 85MPa, determine the size of the weld.
  • 262. Given
  • 265. To find weld size