Chapter 6: Design of

Mechanical Joint 2: WELD

DR. AMIR PUTRA BIN MD SAAD

C24-322

amirputra@utm.my | amirputra@mail.fkm.utm.my

mech.utm.my/amirputra

6.1 INTRODUCTION

The strongest and most common method of permanently joining steel

components together is by welding. Of the many welding techniques available,

arc welding is the most important since it is adaptable to various manufacturing

environments and is relatively cheap.

Economic ?Safe ?

Which ???

6.1 INTRODUCTION

This chapter considers on the arc fillet weld type only.

6.1 INTRODUCTION

The welding symbol is standardized by the American Welding Society (AWS).

6.2 WELDING SYMBOL

6.2 WELDING SYMBOL: Example

6.2 WELDING SYMBOL: Example

Throat Length = 0.707h

450

Shear

Stress

h

h

450

0.707h

Throat

plane

toe

toe root

6.3 WELDING MODEL

Effective Area = Smallest Area = Throat Area

a. Primary Stress:

i. Axial Stress

ii. Shear Stress

b. Secondary Stress:

i. Bending Stress

ii. Torsion Stress

𝝉𝒂𝒙𝒊𝒂𝒍′ =

𝑷

𝑨𝒘

𝝉𝒔𝒉𝒆𝒂𝒓′ =

𝑽

𝑨𝒘

𝝉𝒃𝒆𝒏𝒅𝒊𝒏𝒈′′ =

𝑴𝒄

𝑰𝒘

𝝉𝒕𝒐𝒓𝒔𝒊𝒐𝒏′′ =

𝑻𝒓

𝑱𝒘

6.4 TYPES OF STRESS

A

B

C

xy

z

F

Primary Load:

ForcesSecondary Load:

Moments

x

y

zG

r

Mz

In-Plane: Torsion

Mx

Out-of-Plane: BendingAxial or/and Shear

My

6.5 STRESS DISTRIBUTION

Direct Shear Stress:

Bending Shear Stress:

𝜏𝑠ℎ𝑒𝑎𝑟′ =

𝑉

𝐴𝑤

𝜏𝑏𝑒𝑛𝑑𝑖𝑛𝑔′′ =

𝑀𝑐

𝐼𝑤

M

V

𝜏𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 = 𝜏𝑠ℎ𝑒𝑎𝑟′ 2

+ 𝜏𝑏𝑒𝑛𝑑𝑖𝑛𝑔′′ 2

𝑰𝒘 = 𝟎. 𝟕𝟎𝟕 𝒉 𝑰𝒖

Front View

Side View

weld

Combine Stress: Theorem Pythagoras

6.6 SECONDARY STRESS: BENDING STRESS

𝜏𝑡𝑜𝑟𝑠𝑖𝑜𝑛′′ =

𝑇𝑟

𝐽𝑤

Direct Shear Stress:

Torsion Stress:

𝜏𝑠ℎ𝑒𝑎𝑟′ =

𝑉

𝐴𝑤

𝑱𝒘 = 𝟎. 𝟕𝟎𝟕 𝒉 𝑱𝒖

Combine Stress: Cosine Rule (triangle shape)

𝜏𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 = 𝜏𝑠ℎ𝑒𝑎𝑟′ 2

+ 𝜏𝑡𝑜𝑟𝑠𝑖𝑜𝑛′′ 2

− 2𝜏𝑠ℎ𝑒𝑎𝑟′ 𝜏𝑡𝑜𝑟𝑠𝑖𝑜𝑛

′′ 𝑐𝑜𝑠(𝜃)

𝜃 is the angle formed between vector 𝜏𝑠ℎ𝑒𝑎𝑟′ and 𝜏𝑡𝑜𝑟𝑠𝑖𝑜𝑛

′′

M

V

6.7 SECONDARY STRESS: TORSION STRESS

6.8 SECOND MOMENT OF AREA

Moments of inertia of linear weld segments.

h

For simplicity it is assumed that the effective weld width in the plane of the paper isthe same as throat length ℎ. Rectangular moment of inertia about axes of symmetryof the weld segment, 𝑋′ and 𝑌′.

Rectangular moment of inertia about axes 𝑋 and 𝑌 through the center of gravity ofthe total weld group:

6.8 SECOND MOMENT OF AREA

𝐼𝑋 = 𝐼𝑋′ + A𝑏2 =𝐿3ℎ

12+ 𝐿ℎ𝑏2

𝐼𝑌 = 𝐼𝑌′ + A𝑎2 = 𝐿ℎ𝑎2

𝐼𝑋′ = න𝑦2𝑑𝐴 = 2න0

𝐿/2

𝑦2 ℎ 𝑑𝑦 =𝐿3ℎ

12

𝐼𝑌′ = 0

Polar moment of inertia about an axis perpendicular to the center of gravity of the weld group:

6.8 SECOND MOMENT OF AREA

𝐽 = 𝐼𝑋 + 𝐼𝑌 =𝐿3𝑡

12+ 𝐿𝑡 𝑎2 + 𝑏2

6.9 TABLE OF UNIT SECOND MOMENT OF AREA, 𝐼𝑢

6.9 TABLE OF UNIT SECOND MOMENT OF AREA, 𝐼𝑢

6.10 TABLE OF UNIT SECOND POLAR MOMENT OF AREA, 𝐽𝑢

6.10 TABLE OF UNIT SECOND POLAR MOMENT OF AREA, 𝐽𝑢

6.11 ELECTRODE STRENGTH

𝑆𝑎𝑓𝑒𝑡𝑦 𝐹𝑎𝑐𝑡𝑜𝑟 = 𝑛𝐷𝐸𝑇 =𝑂. 577𝑆𝑦

𝜏𝑚𝑎𝑥

Distortion Energy Method:

𝜏𝑚𝑎𝑥 = 𝑣𝑒𝑐𝑡𝑜𝑟𝑖𝑎𝑙𝑙𝑦 𝑎𝑑𝑑𝑒𝑑

6.12 STATIC FAILURE PREVENTION SCHEME

Thank You

3.14 EXAMPLE 3.3

3.14 EXAMPLE 3.3

1. Establish the free-body diagram.

rA

rD

rB

rC

𝝉𝑩′′

𝝉𝑪′′

𝝉𝑨′′

𝝉𝑫′′

𝝉𝑨′

𝝉𝑫′

𝝉𝑪′

𝝉𝑩′

𝑾𝒆𝒍𝒅 𝑷𝒂𝒕𝒕𝒆𝒓𝒏

3.14 EXAMPLE 3.3

𝐴𝑤 = 1280 𝑚𝑚2

2. Determine the primary stresses.

𝜏′ =𝑉

𝐴𝑤=

25000

1280= 19.5 𝑀𝑃𝑎

3. Determine the secondary stresses.

𝐽 = 0.707ℎ𝐽𝑢 = 7.07 × 106 𝑚𝑚4

𝑇 = 2760 𝑁𝑚

𝜏𝐴′′ = 𝜏𝐵

′′ =𝑇𝑟𝐴𝐽

= 41 𝑀𝑃𝑎

𝜏𝐶′′ = 𝜏𝐷

′′ =𝑇𝑟𝐷𝐽

= 37.3 𝑀𝑃𝑎

From Table 9-1

From Table 9-1

3.14 EXAMPLE 3.3

4. Combine the stresses for 2 and 3.

𝜏𝑚𝑎𝑥 = 𝜏𝐶 = 𝜏𝐷 = 43.9 𝑀𝑃𝑎Thus,

𝜏𝐴 = 𝜏𝐵 = 𝜏𝐴′ + 𝜏𝐴

′′ sin 25.6402+ 𝜏𝐴

′′ cos 25.6402= 37 𝑀𝑃𝑎

𝜏𝐶 = 𝜏𝐷 = 𝜏𝐶′ + 𝜏𝐶

′′ sin 6.2502+ 𝜏𝐶

′′ cos 6.2502= 43.9 𝑀𝑃𝑎

By using Resolving Method:

5. Determine the safety factor or fillet weld size.

This question did not asking you to calculate the safety factor.