machine design me-317

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

• Weight.

• 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

• It is a joint between two components lying approximately in the same plane.

• It connects the ends of two plates.

Types of Butt Joints

• Square Butt

• V-Butt

• U-Butt

• Double V-Butt

• 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:

𝜎𝑡 =𝑃

ℎ𝑙

The throat of the weld does not include the bulge.

Tensile force on plates are given by:

𝑃 = 𝜎𝑡 ∗ 𝑡 ∗ 𝑙

QUESTION

Strength of Parallel Fillet Joint

• Parallel fillet weld is subjected to tensile force

P as shown in figure.

• ‘h’ is the leg

• ‘t’ is the throat

• Cross section of the weld is a right angle triangle

having two equal sides.

• Length of each side is called leg.

• As a rule, the leg h is equal to the plate thickness.

• Throat ‘t’ is the minimum cross section of the weld.

• 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:

𝑡 = ℎcos45𝑜 𝑜𝑟 𝑡 = 0.707ℎ

𝜏 =𝑃

0.707ℎ𝑙

𝑷 = 𝟎. 𝟕𝟎𝟕𝒉𝒍𝝉

For two welds of equal length on both sides of the plate:

OR

𝑷 = (𝟐 ∗ 𝟎. 𝟕𝟎𝟕)𝒉𝒍𝝉

𝑷 = 𝟏. 𝟒𝟏𝟒𝒉𝒍𝝉

Strength of Transverse Fillet Joint

• Transverse fillet weld is subjected to tensile force

P as shown in figure.

• ‘h’ is the leg

• ‘t’ is the throat

• Cross section of the weld is a right angle triangle

having two equal sides.

• Length of each side is called leg.

• As a rule, the leg h is equal to the plate thickness.

• Throat ‘t’ is the minimum cross section of the weld.

• 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 ‘t’ in the above equation we have:

Rearranging the terms:

Where,

𝜎𝑡= permissible tensile stress for the weld (N/mm2)

𝜎𝑡 =𝑃

𝑡𝑙

𝜎𝑡 =𝑃

0.707ℎ𝑙

𝑷 = 𝟎. 𝟕𝟎𝟕𝒉𝒍𝝈𝒕

For two welds of equal length on both sides of the plate:

OR

𝑷 = (𝟐 ∗ 𝟎. 𝟕𝟎𝟕)𝒉𝒍𝝈𝒕

𝑷 = 𝟏. 𝟒𝟏𝟒𝒉𝒍𝝈𝒕

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

• G is the centre of gravity of the angle section.

• The external force acting on joint passes through G.

• 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:

𝑷𝟏 = 𝟎. 𝟕𝟎𝟕𝒉𝒍𝟏𝝉

𝑷𝟐 = 𝟎. 𝟕𝟎𝟕𝒉𝒍𝟐𝝉

𝑷 = 𝑷𝟏 + 𝑷𝟐

𝑷𝟏𝒚𝟏 = 𝑷𝟐𝒚𝟐

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

𝒍𝟏𝒚𝟏 = 𝒍𝟐𝒚𝟐

𝒍𝟏 + 𝒍𝟐 = 𝒍

QUESTION

Eccentrically Loaded

Welded Joints

• Primary Shear Stress

• Due to the load acting on the G.

• Uniformly distributed over the throat of the weld.• The direct or primary shear stress is calculated as:

• Secondary Shear Stress• Due to bending moment (P*e)• Directly proportional to the distance from G.• The secondary shear stress is calculated as:

𝝉𝟏 =𝑷

𝑨

𝝉𝟐 =𝑴𝒓

𝑱

𝝉𝟐 =𝑴𝒓

𝑱

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 ‘G’.