machine design me-317
TRANSCRIPT
MACHINE DESIGN ME-317
- ATEEB AHMAD KHAN
MACHINE DESIGN SECTION
DEPARTMENT OF MECHANICAL ENGINEERING
ZAKIR HUSAIN COLLEGE OF ENGINEERING & TECHNOLOGY
A.M.U. ALIGARH
2017-2018
MA
CH
INE
DES
IGN
Unit 1: Introduction and Welded Joints
Unit 2: Bearing and Lubrication
Unit 3: Clutches and Brakes
Unit 4: Spring
Unit 5: Gears
Unit 1: Welded Joints
JOINT
TemporarySemi
Permanent Permanent
Welded Joint
Welding can be defined as the process of joining metallic parts by
heating to a suitable temperature with or without the application
of pressure.
Its an economical and efficient methods for obtaining a
permanent joint of metallic parts.
Advantages of Welded Joints over Riveted Joints
1. Fish Plate hence adding
β’ Weight.
β’ Cost.
2. Tight and Leak Proof
3. Less production time
4. Stress concentration in riveted joints.
5. Strength is high.
6. Difficulty in riveting certain geometry such as circular steel pipe.
Advantages of Welded Joints over Cast iron Structures
1. Lighter in weight.
2. Easy to machine.
3. Capital investment of welding shop is less than that of foundry shop.
Dis
adva
nta
ges
of
We
lde
d J
oin
t
Poor Damping Property.
Thermal Distortion resulting in residual stress.
Strength depends upon the skill of the worker.
Inspection is costly.
Stress Relieving of Welded Joints
Steps to reduce Residual Stress.
PreheatingHeat Treatment
(Annealing)
WELDED JOINTS
BUTT FILLET
BUTT JOINT
β’ It is a joint between two components lying approximately in the same plane.
β’ It connects the ends of two plates.
Types of Butt Joints
β’ Square Butt
β’ V-Butt
β’ U-Butt
β’ Double V-Butt
β’ V-Joint with Backing Strip
Fillet Joint (Lap Joint)
It is a joint between two overlapping plates or components.
It consists of an approximately triangular cross section joining two surfaces at right angle.
FILLET JOINT
TRANSVERSE PARALLEL
TRANSVERSE FILLET JOINT
PARALLEL FILLET JOINT
Strength of Butt Weld
The average tensile stress in the weld is given by:
ππ‘ =π
βπ
The throat of the weld does not include the bulge.
Tensile force on plates are given by:
π = ππ‘ β π‘ β π
QUESTION
Strength of Parallel Fillet Joint
β’ Parallel fillet weld is subjected to tensile force
P as shown in figure.
β’ βhβ is the leg
β’ βtβ is the throat
β’ Cross section of the weld is a right angle triangle
having two equal sides.
β’ Length of each side is called leg.
β’ As a rule, the leg h is equal to the plate thickness.
β’ Throat βtβ is the minimum cross section of the weld.
β’ Failure - due to shear across the minimum cross section.
From fig. (b) it can be seen that :
The cross section area at the throat is (tl) or (0.707 hl)
The shear stress in the fillet weld is given by:
Rearranging the terms in the above equation we get the strength equation of the parallel fillet weld:
π‘ = βcos45π ππ π‘ = 0.707β
π =π
0.707βπ
π· = π. ππππππ
For two welds of equal length on both sides of the plate:
OR
π· = (π β π. πππ)πππ
π· = π. ππππππ
Strength of Transverse Fillet Joint
β’ Transverse fillet weld is subjected to tensile force
P as shown in figure.
β’ βhβ is the leg
β’ βtβ is the throat
β’ Cross section of the weld is a right angle triangle
having two equal sides.
β’ Length of each side is called leg.
β’ As a rule, the leg h is equal to the plate thickness.
β’ Throat βtβ is the minimum cross section of the weld.
β’ Failure - due to tensile stress across the minimum cross section
The tensile stress in the transverse fillet weld is given by:
Substituting the expression for βtβ in the above equation we have:
Rearranging the terms:
Where,
ππ‘= permissible tensile stress for the weld (N/mm2)
ππ‘ =π
π‘π
ππ‘ =π
0.707βπ
π· = π. πππππππ
For two welds of equal length on both sides of the plate:
OR
π· = (π β π. πππ)ππππ
π· = π. πππππππ
QUESTION
QUESTION
Two steel plates, 120 mm wide and 12.5 mm thick, are joined together by means of double transverse fillet welds as shown in figure. The maximum tensile stressFor the plates and the welding material should not exceed 110 N/mm2. Find the required length of the weld if the strength of the weld is equal to the strength of the plate.
QUESTION
Axially loaded unsymmetrical welded Joints
β’ G is the centre of gravity of the angle section.
β’ The external force acting on joint passes through G.
β’ P1 and P2 are the resisting forces set up in welds 1 and 2 respectively.
Where,
From fig. (b), the forces acting on the body are:
Taking moment about centre of gravity:
π·π = π. πππππππ
π·π = π. πππππππ
π· = π·π + π·π
π·πππ = π·πππ
Substituting the value of P1 and P2 in the previous expression we have:
Assuming total length of the weld as l,
The above two equations are used to find out the required lengths l1 and l2
ππππ = ππππ
ππ + ππ = π
QUESTION
Eccentrically Loaded
Welded Joints
β’ Primary Shear Stress
β’ Due to the load acting on the G.
β’ Uniformly distributed over the throat of the weld.β’ The direct or primary shear stress is calculated as:
β’ Secondary Shear Stressβ’ Due to bending moment (P*e)β’ Directly proportional to the distance from G.β’ The secondary shear stress is calculated as:
ππ =π·
π¨
ππ =π΄π
π±
ππ =π΄π
π±
Where,r = distance of a point in weld from G.J = polar moment of inertia of all welds about G.
The secondary shear stress at any point in the weld is proportional to its distance from the center of gravity.
Therefore, its maximum at the farthest point.
Hence in the above equation r is the maximum distance of a point on weld from G.
1
3
1
3 2
1
12
12 12
G xx yy
xx
G xx yy yy
G
J I I
ltI
J I I I
l t AlJ
3
12yy
l tI
Where,A is the throat area of the weld.JG1 is the polar moment of inertia of the weld about its center of gravity.
The polar moment of inertia about an axis passing through G is determined by the parallel axis theorem:
Where, r1 is the distance between G and G1.
2
1 1G GJ J Ar
22
112
G
lJ A r
When there are number of welds, then in that case:
The above value of J is to be used to calculate secondary stress.
1 2 3.......... nJ J J J J
Welded Joint Subjected to Bending Moment
POLAR MOMENT
OF INERTIA OF THE THROAT
ABOUT βGβ.