Welding Design Analysis An Introduction
MOHD ZAIDI BIN HAMZAH
POLITEKNIK SULTAN HAJI AHMAD SHAH
Welding Design Analysis An Introduction
Published by POLITEKNIK SULTAN HAJI AHMAD SHAH SEMAMBU 25350 KUANTAN
Copyright ©2021, by Politeknik Sultan Haji Ahmad Shah Materials published in this book under the copyright of Politeknik Sultan Haji Ahmad Shah. All rights reserved. No part of this publication may be reproduced or distributed in any form or by means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers.
Preface
This book aims to help readers understand the basics of basic welded joint analysis and explore more knowledge in this subject. From my teaching experience, I found that the reference books on basic welded joint analysis are difficult to understand by students, hence this book is written in details with steps and guidelines to facilitate students to understand the basics of analysis in welding design. I sincerely hope that you will gain a lot from this book. Hopefully you will get new knowledge and experience from reading this. Thanks Mohd Zaidi Bin Hamzah
Table of Contents
INTRODUCTION
1
Advantages dan Disadvantages of Welded Joint
1
Weld symbols in annotated drawing
1
Fillet Weld
3
Butt – Groove Weld
4
Types of fillet weld
6
MATHEMATICAL ANALYSIS FOR WELDED JOINT
7
Lap Joint Stress Analysis
7
Throat Thickness
7
Minimum Area Of The Weld
8
Tensile stress
8
Shear stress
9
Butt Joint Stress Analysis
11
Throat Thickness
11
Minimum area of the weld
11
Tensile and shear stress
12
Cantilever Lap Joint Stress Analysis
13
Tensile and Shear Load or Combination
13
Eccentric load – Torsion
14
SAFE LOAD AND STANDARD CODE
17
PROBLEM AND SOLUTION
18
BIBLIOGRAPHY
32
WELDING DESIGN ANALYSIS
INTRODUCTION Welding is one of important joint methods in mechanical engineering design, categorized as permanent joint. Welding is a fabrication or sculptural process that joins materials usually metal or thermoplastics, by causing coalescence. This is often done by melting the work pieces and adding a filler materials to form a pool of molten materials (weld pool) that cools to become a strong and permanent joint, with pressure sometimes used in conjunction with heat, or by itself, to produce the weld bead. This is in contrast with soldering and brazing, which involve melting a lower melting point of filler materials between the work pieces to form a bond between them, without melting the work pieces.
Advantages and Disadvantages of Welded Joint Advantages: Produces permanent and strong joints, better in finishing and painting, greater joint strength due to coalescence of the same materials, suitable for unlimited joints than riveting and faster than riveting. The processes can be performed manually, semi-automatically, or completely automatically. Continuous welds provide fluid tightness therefore become a choice for fabricating pressure vessels. Disadvantages: Due to permanent joints, disassemble without destroying detail parts is impossible, parts that need to be in regular maintenance for example, should not be welded. The welding is a hot work process leads to metallurgical changes in the parent metal in the vicinity of the weld. Residual stresses may be introduced due to restrained member of clamping or holding, while the cooling process may cause extra stress. It is requires skilled welders, and welded joint inspection requires skilled and certified inspector.
Weld Symbols in Annotated Drawing The welds must be precisely specified on working drawings, and this is done by using the welding symbol, shown in Figure 1, as standardized by the American Welding Society
1
WELDING DESIGN ANALYSIS (AWS). The arrow of this symbol points to the joint that to be welded. The body of the symbol contains as many of the following elements as are deemed necessary: • Reference line • Arrow • Basic weld symbols • Dimensions and other data • Supplementary symbols • Finish symbols • Tail • Specification or process
Figure 1 The AWS standard welding symbol showing the location of the symbol elements. The arrow used in welding symbols is the first need to be understand. The welding part indicated by the arrow is called the arrow side, while the side opposite of the arrow side is called the other side. Referring to the Figure 2, if a welds need to be performed on the 2
WELDING DESIGN ANALYSIS arrow side the basic weld symbol or detail reference must be put below the reference line, vice versa the basic weld symbol will be drawn above the reference line.
Fillet Welds
(a) The work pieces is in preparation holds in T-joint position to be fillet welded.
(b)
The number indicates the leg size of the welds. The arrow points to the arrow side and the basic weld symbols below the reference line show that the fillet weld need to be executed on the arrow side with the leg size of 5 mm.
(c)
The arrow points to the arrow side and the basic welds symbols above the reference line show that the fillet weld need to be executed on the other side with leg size of 5 mm.
(d)
The arrow should point only to one weld when both sides are the same. The arrow points to the arrow side and the basic welds symbols on the reference line show that the fillet weld need to be executed on both; arrow side and other side with leg size of 5 mm.
Figure 2 Weld symbols on fillet welds 3
WELDING DESIGN ANALYSIS
Figure 3 All around welds symbol
The arrow point on the arrow side with additional circle symbols shown in Figure 3 indicates that the welding is to go all around. The weld symbols below the reference line indicate that the fillet welds must be performed on the arrow side which is at the outside of the pipe with leg size of 5 mm.
Figure 4 Types of weld and groove symbols in arc and gas welding
Butt – Groove Weld Figure 5 shows typical butt weld joints with common groove types used in arc and gas welding. Of course this groove should be prepared by means of machining before the 4
WELDING DESIGN ANALYSIS welding takes place. Be caution to understand the welding drawing before to cut some materials otherwise the company will suffer a loss. (a)
The welding symbol indicates that a Vgroove of 60° must be prepared with a gap of 2 mm between the two parent materials. The gap also called as root opening. The arrow points to the arrow side and the weld symbols below the reference line indicates that the welds must be executed on the arrow side. The V symbol means V-groove and the number 2 and 60° mean gap value and angle of V-groove, respectively.
(b) The weld symbols indicates that the butt welds must be done on both sides with square groove preparation. (c) The weld symbols indicates that the butt welds must be done on both sides with double 60° V-groove preparation. (d) The weld symbols indicates that the butt welds must be done on the arrow side with 45° bevel-groove preparation
Figure 5 Butt weld joint and groove types.
5
WELDING DESIGN ANALYSIS
Types of Fillet Weld The fillet welds are widely used in many types of welded joint such as lap joint, T-joint and corner joint. The types of fillet weld can be summarize as: • Single transverse fillet • Double transverse fillets • Single parallel fillet • Double parallel fillets These types of fillet weld give influences on strength of welded joints that will be discussed further in analysis section.
(a) Single transverse fillet
(b) Double transverse fillets
(c) Double parallel fillets
(d) Double parallel and single transverse fillets
Figure 6 Types of fillet weld
6
WELDING DESIGN ANALYSIS
MATHEMATICAL ANALYSIS FOR WELDED JOINT There are three basic strength analyses of welded joint that will be discussed in this section namely: • Lap joint • Butt joint • Cantilever lap joint
Lap Joint Stress Analysis
Figure 7 Terminology of transverse fillet welded joint Consider a single transverse fillet welded joint as shown in Figure 7. It is assumed that the section of the fillet weld is a right-angled tringle ABC with 45°angle. Length of the fillet weld, ℓ is length of CE or can be measured from the root of the weld BG where the parts
of parent metal are bound. The length of each side BA and BC is known as leg size or weld
size, 𝑠𝑠. The perpendicular distance, BD of the hypotenuse from the intersection of the legs,
is the thickness of the fillet weld and is known as throat thickness, 𝑡𝑡.
Throat Thickness
Referring to the right-triangle ABD as shown in Figure 7 on the right figure, the throat thickness, 𝑡𝑡 which is the length of BD is calculated by using SOH CAH TOA trigonometry. Knowing that angle of BAD is 45°, Hypotenuse is 𝑠𝑠 and Opposite is 𝑡𝑡, therefore; 7
WELDING DESIGN ANALYSIS sin 45 =
𝑡𝑡 𝑠𝑠
𝑡𝑡 = sin 45 × 𝑠𝑠 𝑡𝑡 = 0.707𝑠𝑠
Minimum Area of The Weld The minimum area of the weld is also known as throat area. In Figure 7 the minimum area of the weld is obtained by calculating the rectangular BDGH’s area, that is; 𝐴𝐴 = 𝑡𝑡ℎ𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑡𝑡ℎ𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 × ℓ𝑒𝑒𝑒𝑒𝑒𝑒ℎ𝑡𝑡 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤. 𝐴𝐴 = 𝑡𝑡 × ℓ
𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑡𝑡 = 0.707𝑠𝑠 ∴ 𝐴𝐴 = 0.707𝑠𝑠ℓ
Tensile stress
(a)
(b)
Figure 8 (a) Single and (b) Double transverse fillet welded joints under tensile load Considering the a single and a double transverse fillet welded joints are under tensile load, 𝐹𝐹 as shown in Figure 8, if 𝜎𝜎 is the allowable tensile stress for the fillet welded joint, then the tensile stress for the single transverse joint is; 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝜎𝜎 =
𝐹𝐹 𝐴𝐴
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑇𝑇ℎ𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 8
WELDING DESIGN ANALYSIS 𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝐴𝐴 = 0.707𝑠𝑠ℓ
𝜎𝜎 =
𝐹𝐹 0.707𝑠𝑠ℓ
As the double transverse fillet welded joint has double throat areas that resist the tensile load 𝐹𝐹, then the tensile stress for the double transverse joints is; 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =
𝜎𝜎 =
𝐹𝐹 2𝐴𝐴
𝜎𝜎 =
𝐹𝐹 1.414𝑠𝑠ℓ
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 2 × 𝑇𝑇ℎ𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 2𝐴𝐴 = 2 × 0.707𝑠𝑠ℓ
Shear stress
(a)
(b)
Figure 9 (a) Single and (b) Double transverse fillet welded joints under shear load Considering a single and a double transverse fillet welded joints are under shear load, 𝐹𝐹 as
shown in Figure 9, if 𝜏𝜏 is the allowable shear stress for the fillet welded joint, then the shear stress for the single transverse joint is; 𝑆𝑆ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝜏𝜏 =
𝐹𝐹 𝐴𝐴
𝑆𝑆ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑇𝑇ℎ𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
9
WELDING DESIGN ANALYSIS 𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝐴𝐴 = 0.707𝑠𝑠ℓ
𝜏𝜏 =
𝐹𝐹 0.707𝑠𝑠ℓ
As the double transverse fillet welded joint has double throat areas that resist the shear load 𝐹𝐹, then the tensile stress for the double transverse joints is; 𝑆𝑆ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =
𝜏𝜏 =
𝐹𝐹 2𝐴𝐴
𝜏𝜏 =
𝐹𝐹 1.414𝑠𝑠ℓ
𝑆𝑆ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 2 × 𝑇𝑇ℎ𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 2𝐴𝐴 = 2 × 0.707𝑠𝑠ℓ
Notice that the equation of shear stress is seems to be the same as tensile stress equation but it is not. The load in shear stress is applied parallel to the transverse fillet weld as oppose to the direction of tensile load. Therefore the letter 𝑉𝑉 is sometimes used to represent the shear stress to avoid confusion.
10
WELDING DESIGN ANALYSIS
Butt Joint Stress Analysis Consider a single complete penetration V-groove butt joint as shown in Figure 10. Length of the weld, ℓ is the length of CD. Throat thickness, 𝑡𝑡 is the length of AB measured from
root of the weld to face of the weld. Note that the measurement of throat thickness does not include the reinforcement bead. The reinforcement can be beneficial however leads to
stress concentration at the weld toe, and it is good practice to grind off the reinforcement in a presence of fatigue load condition.
Figure 10 V-groove butt joint
Throat Thickness Throat thickness, 𝑡𝑡 of the butt welded joint is equal to the length of leg size or weld size, 𝑠𝑠
which is equal to plate’s thickness of the parent materials. 𝑇𝑇ℎ𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑡𝑡ℎ𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 = 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡 = 𝑠𝑠
Minimum Area of The Weld Minimum area of the weld, 𝐴𝐴 is the throat area;
𝐴𝐴 = 𝑡𝑡ℎ𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑡𝑡ℎ𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 × 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙ℎ𝑡𝑡 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝐴𝐴 = 𝑡𝑡 × ℓ
𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑡𝑡 = 𝑠𝑠 ∴ 𝐴𝐴 = 𝑠𝑠𝑠𝑠
11
WELDING DESIGN ANALYSIS Tensile Stress and Shear Stress
(a)
(b)
Figure 11 V-groove butt joint is (a) under tensile load, and (b) under shear load Figure 11 (a) shows V-groove butt joint under tensile load, 𝐹𝐹. If 𝜎𝜎 is the allowable tensile stress for the welded joint, then the tensile stress is; 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝜎𝜎 =
𝐹𝐹 𝐴𝐴
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑇𝑇ℎ𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝐴𝐴 = 𝑠𝑠ℓ ∴ 𝜎𝜎 =
𝐹𝐹 𝑠𝑠ℓ
While Figure 11 (b) shows the same weld joint but under shear load. If 𝜏𝜏 is the allowable shear stress for the welded joint, then the shear stress is; 𝑆𝑆ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝜏𝜏 =
𝐹𝐹 𝐴𝐴
𝑆𝑆ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑇𝑇ℎ𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝐴𝐴 = 𝑠𝑠ℓ ∴ 𝜏𝜏 =
𝐹𝐹 𝑠𝑠ℓ
12
WELDING DESIGN ANALYSIS
Cantilever Lap Joint Stress Analysis Terminology of the fillet welded lap joint have been discussed before. As shown in Figure 12 the cantilever plate can be connected to the support base plate either with transverse fillet weld or parallel fillet or combination of both. For instant, the cantilever weld joints in the figure is a combination of single transverse and double parallel fillet welds. Length of the weld, ℓ𝑝𝑝 is the length of parallel fillet, while length of weld, ℓ𝑡𝑡 is the length of transverse fillet as shown in the figure below.
Figure 12 Cantilever fillet welds under shear and tensile load
Tensile and Shear Loads or Combination. Consider a load, 𝐹𝐹 acts on the cantilever as shown in Figure 12. Since the load is in parallel with the parallel fillet joints, both of the double parallel fillet are then under shear load, while the single transverse fillet is under tensile load. Therefore; 𝑆𝑆ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠, 𝜏𝜏 =
𝐹𝐹 2𝐴𝐴
𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝐴𝐴 = 0.707𝑠𝑠ℓ𝑝𝑝 ∴ 𝜏𝜏 =
𝐹𝐹 1.414𝑠𝑠ℓ𝑝𝑝
𝐹𝐹 = 1.414𝜏𝜏𝜏𝜏ℓ𝑝𝑝
13
WELDING DESIGN ANALYSIS 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠, 𝜎𝜎 = ∴ 𝜎𝜎 =
𝐹𝐹 0.707𝑠𝑠𝑙𝑙𝑡𝑡
𝐹𝐹 𝐴𝐴
𝐹𝐹 = 0.707𝜎𝜎𝜎𝜎ℓ𝑡𝑡
The total strength of the weld joints is the sum of the strength of the double parallel fillet and the single transverse fillet, that is; ∑𝐹𝐹 = 1.414𝜏𝜏𝜏𝜏ℓ𝑝𝑝 + 0.707𝜎𝜎𝜎𝜎ℓ𝑡𝑡
Eccentric Load – Torsion
Figure 13 Cantilever fillet welds under eccentric load Consider a cantilever welded to a support base with a double parallel fillet welded joint as shown in Figure 13. The eccentric load, 𝐹𝐹 causes two reactions at the support there are a shear force, 𝑉𝑉 and a moment, 𝑀𝑀. The shear force produces primary shear stress, denoted as 𝜏𝜏1 in the figure.
𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 =
𝜏𝜏1 =
𝐹𝐹 𝐴𝐴
𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡ℎ𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑜𝑜𝑜𝑜 𝑎𝑎𝑎𝑎𝑎𝑎 𝑡𝑡ℎ𝑒𝑒 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
14
WELDING DESIGN ANALYSIS While the moment, 𝑀𝑀 at the support produces secondary shear stress or torsion, denoted as 𝜏𝜏2 in the figure.
𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 =
𝜏𝜏2 =
𝑀𝑀 × 𝑟𝑟 𝐽𝐽
𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 × 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑜𝑜𝑜𝑜 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑡𝑡ℎ𝑒𝑒 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚, 𝑀𝑀 = 𝐹𝐹 × 𝑒𝑒 ∴ 𝜏𝜏2 =
𝐹𝐹 × 𝑒𝑒 × 𝑟𝑟 𝐽𝐽
The radial distance, 𝑟𝑟 must be measured from location of centroid, 𝐺𝐺 and the moment M
computed about G. Note that 𝑟𝑟 is usually the farthest distance from the centroid, 𝐺𝐺 of the
weld group. 𝐹𝐹 is the eccentric load and 𝑒𝑒 is eccentricity distance that is measured from eccentric load to the location of centroid, along X-axis. The location of centroid, 𝐺𝐺 is calculated based on the weld group and has been simplified in Table 1 based on case. While 𝐽𝐽 is the second polar moment of area of the weld group and is calculated as: 𝐽𝐽 = 𝑡𝑡ℎ𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑡𝑡ℎ𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 × 𝐽𝐽𝑢𝑢
𝐽𝐽 = 𝑡𝑡 × 𝐽𝐽𝑢𝑢
𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑡𝑡 𝑓𝑓𝑓𝑓𝑓𝑓 𝑡𝑡ℎ𝑒𝑒 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = 0.707𝑠𝑠 ∴ 𝐽𝐽 = 0.707𝑠𝑠𝐽𝐽𝑢𝑢
The weld size is treating as a line so that the value of the 𝐽𝐽𝑢𝑢 is the same regardless of the weld size. The equation of 𝐽𝐽𝑢𝑢 can be obtained from Table 1 based on the case of weld
group.
Finally the resultant of the shear stress is summed up from primary shear and secondary shear stress mathematically as:
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠, 𝜏𝜏 = �𝜏𝜏1 2 + 𝜏𝜏2 2 + 2𝜏𝜏1 𝜏𝜏2 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 The angle, 𝜃𝜃 is the angle between primary and secondary shear stress, and it is equal to
the angle of 𝑟𝑟 from the horizontal line as indicated in the Figure 13 above.
15
WELDING DESIGN ANALYSIS Table 1 Torsional properties of fillet weld Case
1
2
3
4
Weld Group
Throat Area, 𝑨𝑨
Location of Centroid, 𝑮𝑮
Unit Second Polar Moment of Area
𝑥𝑥̅ = 0
𝑑𝑑 2
𝐴𝐴 = 0.707𝑠𝑠𝑠𝑠
𝑦𝑦� =
𝐴𝐴 = 1.414𝑠𝑠𝑠𝑠
𝑏𝑏 2 𝑑𝑑 𝑦𝑦� = 2
𝐽𝐽𝑢𝑢 =
𝑥𝑥̅ =
𝐴𝐴 = 0.707𝑠𝑠(𝑏𝑏 + 𝑑𝑑)
𝐴𝐴 = 0.707𝑠𝑠(2𝑏𝑏 + 𝑑𝑑)
5
𝐴𝐴 = 1.414𝑠𝑠(𝑏𝑏 + 𝑑𝑑)
6
𝐴𝐴 = 1.414𝜋𝜋𝜋𝜋𝜋𝜋
𝑥𝑥̅ =
𝑦𝑦� =
𝐽𝐽𝑢𝑢 =
𝑏𝑏2 2(𝑏𝑏 + 𝑑𝑑)
𝐽𝐽𝑢𝑢 =
2
𝑑𝑑 2(𝑏𝑏 + 𝑑𝑑)
𝑏𝑏2 2𝑏𝑏 + 𝑑𝑑 𝑑𝑑 𝑦𝑦� = 2
𝑥𝑥̅ =
𝑏𝑏 2 𝑑𝑑 𝑦𝑦� = 2 𝑥𝑥̅ =
𝐽𝐽𝑢𝑢 =
𝑑𝑑 3 12
𝑑𝑑(3𝑏𝑏2 + 𝑑𝑑 2 ) 6
(𝑏𝑏 + 𝑑𝑑)4 − 6𝑏𝑏2 𝑑𝑑 2 12(𝑏𝑏 + 𝑑𝑑)
8𝑏𝑏3 + 6𝑏𝑏𝑑𝑑 2 + 𝑑𝑑 3 𝑏𝑏4 − 12 2𝑏𝑏 + 𝑑𝑑
𝐽𝐽𝑢𝑢 =
(𝑏𝑏 + 𝑑𝑑)3 6
𝐽𝐽𝑢𝑢 = 2𝜋𝜋𝑟𝑟 3
*G is the centroid of weld group; s is leg size or weld size; the weld group can be turn around to be fit with related analysis
16
WELDING DESIGN ANALYSIS
SAFE LOAD AND STANDARD CODE Ensuring the appropriate strength of the weld joints is an important part of design consideration in providing a safe products for consumers. For the strength of the weld connection, the type of steel and electrode used play a role. The use of standard steel and electrode gives confidence in the safety of the connections made. It was suggested to use steels that having a UNS specification as these steels have known standard tensile and yield strength as shown for example in Table 2. This makes it easier for designer in consideration of factor of safety or permissible stress, or to replicate the values of those used by others. Table 3 shows The American Welding Society (AWS) specification code numbering system used for electrodes and Table 4 indicates American Institute of Steel Construction (AISC) code for permissible stress of weld-metal for tension and shear types of loading. These tables are only parts of full tables which can be obtained from Reference (Budynas, R. G. and Nisbett, J. K., 2015) and will be used further in problem and solution section. Table 2 ASTM Tensile and Yield Strengths for Some HotRolled (HR) and ColdDrawn (CD) Steels
UNS no.
SAE / AISI Processing no.
G10100
1010
G10150
1015
G10180
1018
HR CD HR CD HR CD
Tensile Strength (MPa) 320 370 340 390 400 440
Table 3 AWS Electrode no. Electrode code and weldmetal E60xx E70xx properties E120xx
Yield Strength (MPa)
Table 4 Weld-metal permissible stress
Permissible Stress
345 393 737
Load
Weld
Tension Shear
Butt Butt/ Fillet
0.60𝑆𝑆𝑦𝑦 0.30𝑆𝑆𝑢𝑢𝑢𝑢 *
*Shear stress on base metal should not exceed 0.40𝑆𝑆𝑦𝑦 of base metal
17
Yield Strength (MPa) 180 300 190 320 220 370 Tensile Strength (MPa) 427 482 827 Safety factor, n 1.67
WELDING DESIGN ANALYSIS
PROBLEM AND SOLUTION EXAMPLE – 1 A complete penetration V-groove butt welded joint in figure below is under tensile load of 5kN, find tensile stress.
Solution 𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺;
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓, 𝐹𝐹 = 5 𝑘𝑘𝑘𝑘 = 5000 𝑁𝑁 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠, 𝑠𝑠 = 20 𝑚𝑚𝑚𝑚 = 0.02 𝑚𝑚
ℓ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒ℎ 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤, ℓ = 100 𝑚𝑚𝑚𝑚 = 0.1 𝑚𝑚
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠;
𝜎𝜎 =
𝜎𝜎 =
𝐹𝐹 𝑠𝑠ℓ
5000 (0.02)(0.1)
𝝈𝝈 = 𝟐𝟐. 𝟓𝟓 𝑴𝑴𝑴𝑴𝑴𝑴
18
WELDING DESIGN ANALYSIS
EXAMPLE – 2 Figure below shows a cantilever connected to a support with double parallel fillet welds. Unit is in mm. Determine; (a) (b)
Total of throat area, 𝐴𝐴1 and 𝐴𝐴2 Location of centroid, 𝐺𝐺
𝐴𝐴1 𝐴𝐴2
Solution (a) The throat area of the fillet weld is calculated with formula; 𝐴𝐴 = 𝑡𝑡ℓ = 0.707𝑠𝑠ℓ
𝐴𝐴1 = 50𝑡𝑡 = 0.707(50)(𝑠𝑠) = 35.35(𝑠𝑠)
𝐴𝐴2 = 50𝑡𝑡 = 0.707(50)(𝑠𝑠) = 35.35(𝑠𝑠)
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝐴𝐴 = 𝐴𝐴1 + 𝐴𝐴2 = 2(50)𝑡𝑡 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝒎𝒎𝒎𝒎𝟐𝟐 𝑜𝑜𝑜𝑜 = 2(35.35)(𝑠𝑠) = 𝟕𝟕𝟕𝟕. 𝟕𝟕𝟕𝟕 𝒎𝒎𝒎𝒎𝟐𝟐
Total throat area of the weld group also can be solved with throat area formula from Table 1 referring Case 2. 𝐴𝐴 = 1.414𝑠𝑠𝑠𝑠
𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑑𝑑 = 50 𝑚𝑚𝑚𝑚
∴ 𝐴𝐴 = 1.414(50)(𝑠𝑠) = 𝟕𝟕𝟕𝟕. 𝟕𝟕𝟕𝟕 𝒎𝒎𝒎𝒎𝟐𝟐
19
WELDING DESIGN ANALYSIS (b) The centroid, 𝐺𝐺 of the weld group is located at the coordinate 𝑥𝑥̅ and 𝑦𝑦� from reference point.
𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑜𝑜 𝑎𝑎𝑎𝑎 𝑡𝑡ℎ𝑒𝑒 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝; 𝑥𝑥̅ =
𝐴𝐴1 𝑥𝑥1 + 𝐴𝐴2 𝑥𝑥2 𝐴𝐴1 + 𝐴𝐴2
𝐴𝐴1 = 𝐴𝐴2 = 35.35(𝑠𝑠)
𝑥𝑥1 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑥𝑥 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝐴𝐴1 𝑥𝑥1 = 25 𝑚𝑚𝑚𝑚
𝑥𝑥2 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑥𝑥 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝐴𝐴2 𝑥𝑥2 = 25 𝑚𝑚𝑚𝑚
∴ 𝑥𝑥̅ =
(35.35)(𝑠𝑠)(25) + (35.35)(𝑠𝑠)(25) (35.35)(𝑠𝑠) + (35.35)(𝑠𝑠)
𝑥𝑥̅ = 𝟐𝟐𝟐𝟐 𝒎𝒎𝒎𝒎 𝑦𝑦� =
𝐴𝐴1 𝑦𝑦1 + 𝐴𝐴2 𝑥𝑥𝑥𝑥2 𝐴𝐴1 + 𝐴𝐴2
𝑦𝑦1 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑦𝑦 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝐴𝐴1 𝑦𝑦1 = 80 𝑚𝑚𝑚𝑚
𝑦𝑦2 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑦𝑦 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝐴𝐴2 20
WELDING DESIGN ANALYSIS 𝑦𝑦2 = 0
∴ 𝑦𝑦� =
(35.35)(𝑠𝑠)(80) + (35.35)(𝑠𝑠)(0) (35.35)(𝑠𝑠) + (35.35)(𝑠𝑠)
𝑦𝑦� = 𝟒𝟒𝟒𝟒 𝒎𝒎𝒎𝒎
∴ 𝑇𝑇ℎ𝑒𝑒 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝐺𝐺 𝑖𝑖𝑖𝑖 (𝟐𝟐𝟐𝟐, 𝟒𝟒𝟒𝟒)𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑡𝑡ℎ𝑒𝑒 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝑜𝑜𝑜𝑜 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝, 𝑜𝑜.
The location of centroid 𝐺𝐺 is recommended to be calculated with simplified formula provided in Table 1 for Case 2;
𝑥𝑥̅ = 𝑦𝑦� =
𝑏𝑏 80 = = 𝟒𝟒𝟒𝟒 𝒎𝒎𝒎𝒎 ⋯ ∗ 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑡𝑡𝑡𝑡 𝑡𝑡ℎ𝑒𝑒 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝑥𝑥 𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑒𝑒 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 2 2
𝑑𝑑 50 = = 𝟐𝟐𝟐𝟐 𝒎𝒎𝒎𝒎 ⋯ ∗ 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑡𝑡𝑡𝑡 𝑡𝑡ℎ𝑒𝑒 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝑦𝑦� 𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑒𝑒 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑. 2 2
Although the value of 𝑥𝑥̅ and 𝑦𝑦� are a bit confusing due to the modification of weld group
from Table 2 Case 2 but more important is the exact same location of centroid 𝐺𝐺 obtained.
The centroid 𝐺𝐺 is the point where the eccentric load tries to twist or the centre of torsional.
It is necessary to find centroid point before the calculation proceed to find eccentricity distance, 𝑒𝑒, radial distance, 𝑟𝑟 and angle between primary and secondary shear stress, 𝜃𝜃.
21
WELDING DESIGN ANALYSIS
EXAMPLE – 3 Figure below shows a cantilever connected to a support. If the cantilever with double parallel fillet welded joints is under eccentric load of 15 𝑘𝑘𝑘𝑘, find size of the weld, 𝑠𝑠. Given
allowable shear stress is 80 𝑀𝑀𝑀𝑀𝑀𝑀.
Solution 𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺;
𝐹𝐹 = 15 𝑘𝑘𝑘𝑘
𝜏𝜏 = 80 𝑀𝑀𝑀𝑀𝑀𝑀
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2 𝑖𝑖𝑖𝑖 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻 𝟏𝟏
𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐, 𝐺𝐺;
𝑇𝑇ℎ𝑒𝑒 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑓𝑓𝑓𝑓𝑓𝑓 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑡𝑡𝑡𝑡 𝑏𝑏𝑏𝑏 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑡𝑡𝑡𝑡 𝑏𝑏𝑏𝑏 𝑓𝑓𝑓𝑓𝑓𝑓 𝑏𝑏 = 80 𝑚𝑚𝑚𝑚
𝑑𝑑 = 50 𝑚𝑚𝑚𝑚 𝑥𝑥̅ =
𝑦𝑦� =
𝑏𝑏 80 = = 𝟒𝟒𝟒𝟒 𝒎𝒎𝒎𝒎 2 2
𝑑𝑑 50 = = 𝟐𝟐𝟐𝟐 𝒎𝒎𝒎𝒎 2 2
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑, 𝑟𝑟 𝑎𝑎𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎, 𝜃𝜃
22
WELDING DESIGN ANALYSIS 𝑟𝑟 = �252 + 402 𝑟𝑟 = 𝟒𝟒𝟒𝟒. 𝟏𝟏𝟏𝟏 𝒎𝒎𝒎𝒎
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 =
40 25
𝜃𝜃 = 𝟓𝟓𝟓𝟓°
𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝜏𝜏1 =
𝐹𝐹 𝐴𝐴
𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝐴𝐴 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 1 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2; 𝐴𝐴 = 1.414𝑠𝑠𝑠𝑠
𝐴𝐴 = 1.414(𝑠𝑠)(50) 𝐴𝐴 = 70.7(𝑠𝑠)
15 × 103 ∴ 𝜏𝜏1 = 70.7(𝑠𝑠) 𝜏𝜏1 =
𝟐𝟐𝟐𝟐𝟐𝟐. 𝟏𝟏𝟏𝟏 𝑁𝑁/𝑚𝑚𝑚𝑚2 𝒔𝒔
𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝜏𝜏2 =
𝐹𝐹 × 𝑒𝑒 × 𝑟𝑟 𝐽𝐽
𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒, 𝑒𝑒 = 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝐹𝐹 𝑡𝑡𝑡𝑡 𝐺𝐺 𝑒𝑒 = 150 − 25 𝑒𝑒 = 𝟏𝟏𝟏𝟏𝟏𝟏 𝒎𝒎𝒎𝒎
𝑎𝑎𝑎𝑎𝑎𝑎 𝐽𝐽 = 0.707(𝑠𝑠)(𝐽𝐽𝑢𝑢 ) 23
WELDING DESIGN ANALYSIS 𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝐽𝐽𝑢𝑢 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 1 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2;
𝐽𝐽𝑢𝑢 =
𝑑𝑑(3𝑏𝑏 2 + 𝑑𝑑2 ) 6
50(3(80)2 + 502 ) 𝐽𝐽𝑢𝑢 = 6 𝐽𝐽𝑢𝑢 = 𝟏𝟏. 𝟖𝟖𝟖𝟖𝟖𝟖 × 𝟏𝟏𝟏𝟏𝟓𝟓
∴ 𝐽𝐽 = 0.707(𝑠𝑠)(1.808 × 105 ) 𝐽𝐽 = �𝟏𝟏. 𝟐𝟐𝟐𝟐𝟐𝟐 × 𝟏𝟏𝟏𝟏𝟓𝟓 �(𝒔𝒔) 𝒎𝒎𝒎𝒎𝟒𝟒
∴ 𝜏𝜏2 = 𝜏𝜏2 =
(15 × 103 )(125)(47.17) (1.278 × 105 )(𝑠𝑠)
𝟔𝟔𝟔𝟔𝟔𝟔. 𝟕𝟕𝟕𝟕 𝑁𝑁/𝑚𝑚𝑚𝑚2 𝒔𝒔
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡𝑡𝑡 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑠𝑠;
𝜏𝜏 = �𝜏𝜏1 2 + 𝜏𝜏2 2 + 2𝜏𝜏1 𝜏𝜏2 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
80 = �� 80 =
212.16 2 691.78 2 212.16 691.78 � +� � + 2� �� � 𝑐𝑐𝑐𝑐𝑐𝑐58 𝑠𝑠 𝑠𝑠 𝑠𝑠 𝑠𝑠
824.089 𝑠𝑠
∴ 𝑠𝑠 = 𝟏𝟏𝟏𝟏. 𝟑𝟑𝟑𝟑 𝒎𝒎𝒎𝒎
24
WELDING DESIGN ANALYSIS
EXAMPLE – 4 A welded connection of steel plates is shown in figure below. It is subjected to an eccentric force of 50 𝑘𝑘𝑘𝑘. Determine the size of the weld if the permissible shear stress in the weld
is not to exceed 70 𝑁𝑁/𝑚𝑚𝑚𝑚2 .
Solution 𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺;
𝐹𝐹 = 50,000 𝑁𝑁
𝜏𝜏 = 70 𝑁𝑁/𝑚𝑚𝑚𝑚2
𝑇𝑇ℎ𝑒𝑒 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑖𝑖𝑖𝑖 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 4 𝑖𝑖𝑖𝑖 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 1 𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒; 𝑏𝑏 = 100 𝑚𝑚𝑚𝑚
𝑑𝑑 = 200 𝑚𝑚𝑚𝑚 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐, 𝐺𝐺 𝑥𝑥̅ =
𝑥𝑥̅ =
𝑏𝑏 2 2𝑏𝑏 + 𝑑𝑑
1002 2(100) + 200
𝑥𝑥̅ = 𝟐𝟐𝟐𝟐 𝒎𝒎𝒎𝒎
25
WELDING DESIGN ANALYSIS 𝑦𝑦� = 𝑦𝑦� =
𝑑𝑑 2
200 2
𝑦𝑦� = 𝟏𝟏𝟏𝟏𝟏𝟏 𝒎𝒎𝒎𝒎 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑, 𝑟𝑟 𝑎𝑎𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎, 𝜃𝜃 𝑟𝑟 = �752 + 1002 𝑟𝑟 = 𝟏𝟏𝟏𝟏𝟏𝟏 𝒎𝒎𝒎𝒎 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 =
100 75
𝜃𝜃 = 𝟓𝟓𝟓𝟓. 𝟏𝟏𝟏𝟏° 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝜏𝜏1 =
𝐹𝐹 𝐴𝐴
𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝐴𝐴 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 1 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 4; 𝐴𝐴 = 0.707𝑠𝑠(2𝑏𝑏 + 𝑑𝑑)
𝐴𝐴 = 0.707𝑠𝑠(2(100) + (200)) 𝐴𝐴 = 282.8𝑠𝑠 ∴ 𝜏𝜏1 =
𝜏𝜏1 =
50000 282.8𝑠𝑠
𝟏𝟏𝟏𝟏𝟏𝟏. 𝟖𝟖𝟖𝟖 𝑁𝑁/𝑚𝑚𝑚𝑚2 𝒔𝒔
26
WELDING DESIGN ANALYSIS 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝜏𝜏2 =
𝐹𝐹 × 𝑒𝑒 × 𝑟𝑟 𝐽𝐽
𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑒𝑒 = 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝐹𝐹 𝑡𝑡𝑡𝑡 𝐺𝐺 𝑒𝑒 = 300 − 25 𝑒𝑒 = 𝟐𝟐𝟐𝟐𝟐𝟐 𝒎𝒎𝒎𝒎
𝑎𝑎𝑎𝑎𝑎𝑎 𝐽𝐽 = 0.707𝑠𝑠𝐽𝐽𝑢𝑢
𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝐽𝐽𝑢𝑢 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 1 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 4;
𝐽𝐽𝑢𝑢 = 𝐽𝐽𝑢𝑢 =
8𝑏𝑏 3 + 6𝑏𝑏𝑑𝑑2 + 𝑑𝑑3 𝑏𝑏 4 − 12 2𝑏𝑏 + 𝑑𝑑
8(100)3 + 6(100)(200)2 + (200)3 (100)4 − 12 2(100) + (200)
𝐽𝐽𝑢𝑢 = 𝟑𝟑. 𝟎𝟎𝟎𝟎 × 𝟏𝟏𝟏𝟏𝟔𝟔
∴ 𝐽𝐽 = 0.707𝑠𝑠(3.08 × 106 ) 𝐽𝐽 = (𝟐𝟐. 𝟏𝟏𝟏𝟏 × 𝟏𝟏𝟏𝟏𝟔𝟔 )𝒔𝒔 𝒎𝒎𝒎𝒎𝟒𝟒 ∴ 𝜏𝜏2 =
𝜏𝜏2 =
(50 × 103 )(275)(125) (2.18 × 106 )𝑠𝑠
𝟕𝟕𝟕𝟕𝟕𝟕. 𝟒𝟒𝟒𝟒 𝑁𝑁/𝑚𝑚𝑚𝑚2 𝒔𝒔
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡𝑡𝑡 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑠𝑠;
𝜏𝜏 = �𝜏𝜏1 2 + 𝜏𝜏2 2 + 2𝜏𝜏1 𝜏𝜏2 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
70 = �� 70 =
788.42 2 176.80 788.42 176.80 2 � +� � + 2� �� � 𝑐𝑐𝑐𝑐𝑐𝑐53.13 𝑠𝑠 𝑠𝑠 𝑠𝑠 𝑠𝑠
905.61 𝑠𝑠
∴ 𝑠𝑠 = 𝟏𝟏𝟏𝟏. 𝟗𝟗𝟗𝟗 𝒎𝒎𝒎𝒎 27
WELDING DESIGN ANALYSIS
EXAMPLE – 5 Given the allowable shear stress is 140 𝑀𝑀𝑀𝑀𝑀𝑀, determine the eccentrically load, 𝐹𝐹 shown in figure below that can cause the shearing of fillet weld connection.
Solution 𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺;
𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝜏𝜏𝑎𝑎𝑎𝑎𝑎𝑎 = 140 𝑀𝑀𝑀𝑀𝑀𝑀
𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠, 𝑠𝑠 = 6 𝑚𝑚𝑚𝑚 ⋯ ∗ 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑡𝑡ℎ𝑒𝑒 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 3 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑖𝑖𝑖𝑖 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻 𝟏𝟏 Step 1: Find Primary shear stress
𝜏𝜏1 =
𝐹𝐹 𝐴𝐴
𝐴𝐴 = 0.707(𝑠𝑠)(𝑏𝑏 + 𝑑𝑑) 𝑏𝑏 = 50 𝑚𝑚𝑚𝑚
𝑑𝑑 = 60 𝑚𝑚𝑚𝑚
∴ 𝐴𝐴 = 0.707(6)(50 + 60) 𝐴𝐴 = 𝟒𝟒𝟒𝟒𝟒𝟒. 𝟔𝟔𝟔𝟔 𝒎𝒎𝒎𝒎𝟐𝟐 ∴ 𝜏𝜏1 =
𝐹𝐹 466.62
𝜏𝜏1 = 𝟐𝟐. 𝟏𝟏𝟏𝟏𝟏𝟏 × 𝟏𝟏𝟏𝟏−𝟑𝟑 (𝑭𝑭) 𝑵𝑵/𝒎𝒎𝒎𝒎𝟐𝟐
Step 2: Find Secondary shear stress
𝜏𝜏2 =
𝐹𝐹 × 𝑒𝑒 × 𝑟𝑟 𝐽𝐽
Find 𝑒𝑒
𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐, 𝐺𝐺 𝑥𝑥̅ = 28
𝑏𝑏 2 2(𝑏𝑏 + 𝑑𝑑)
WELDING DESIGN ANALYSIS 502 𝑥𝑥̅ = 2(50 + 60)
𝑥𝑥̅ = 𝟏𝟏𝟏𝟏. 𝟑𝟑𝟑𝟑 𝒎𝒎𝒎𝒎 𝑑𝑑2 𝑦𝑦� = 2(𝑏𝑏 + 𝑑𝑑) 𝑦𝑦� =
602 2(50 + 60)
𝑦𝑦� = 𝟏𝟏𝟏𝟏. 𝟑𝟑𝟑𝟑 𝒎𝒎𝒎𝒎
∴ 𝑒𝑒 = 160 − 16.36 𝑒𝑒 = 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟔𝟔𝟔𝟔 𝒎𝒎𝒎𝒎
Find 𝑟𝑟 and 𝜃𝜃
𝑟𝑟 = �11.362 + 43.642 𝑟𝑟 = 𝟒𝟒𝟒𝟒. 𝟎𝟎𝟎𝟎𝟎𝟎 𝒎𝒎𝒎𝒎 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 =
11.36 43.64
𝜃𝜃 = 𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓° Find 𝐽𝐽
𝐽𝐽 = 0.707(𝑠𝑠)𝐽𝐽𝑢𝑢 𝐽𝐽𝑢𝑢 =
(𝑏𝑏 + 𝑑𝑑)4 − 6𝑏𝑏 2 𝑑𝑑 2 12(𝑏𝑏 + 𝑑𝑑)
(110)4 − 6(50)2 (60)2 𝐽𝐽𝑢𝑢 = 12(110) 𝐽𝐽𝑢𝑢 = 𝟕𝟕𝟕𝟕 × 𝟏𝟏𝟏𝟏𝟑𝟑
∴ 𝐽𝐽 = 0.707(6)(70 × 103 ) 𝐽𝐽 = 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 𝒎𝒎𝒎𝒎𝟒𝟒 ∴ 𝜏𝜏2 =
𝐹𝐹(143.64)(45.094) 296940
𝜏𝜏2 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎(𝑭𝑭) 𝑵𝑵/𝒎𝒎𝒎𝒎𝟐𝟐
Step 3: Find resultant shear stress 𝜏𝜏 = �𝜏𝜏1 2 + 𝜏𝜏2 2 + 2𝜏𝜏1 𝜏𝜏2 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝜏𝜏 = �(𝟐𝟐. 𝟏𝟏𝟏𝟏𝟏𝟏 × 𝟏𝟏𝟏𝟏−𝟑𝟑 (𝑭𝑭))2 + (𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎(𝑭𝑭))2 + 2(𝟐𝟐. 𝟏𝟏𝟏𝟏𝟏𝟏 × 𝟏𝟏𝟏𝟏−𝟑𝟑 (𝑭𝑭))(𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎(𝑭𝑭))𝑐𝑐𝑐𝑐𝑐𝑐𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓°
140 = 0.02388(𝐹𝐹) ∴ 𝐹𝐹 = 𝟓𝟓. 𝟖𝟖𝟖𝟖 𝒌𝒌𝒌𝒌
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WELDING DESIGN ANALYSIS
EXAMPLE – 6 A steel bar UNS G10150 HR with rectangular cross section of 12.7 𝑚𝑚𝑚𝑚 by 50.8 𝑚𝑚𝑚𝑚 carries
a static load of 70 𝑘𝑘𝑘𝑘 as shown in figure below. It is welded to a gusset plate with fillet weld size of 9.525 𝑚𝑚𝑚𝑚 and length of weld of 50.8 𝑚𝑚𝑚𝑚 on both sides with an E70XX electrode. Given allowable force per unit length for a weld size of 9.525 𝑚𝑚𝑚𝑚 E70 electrode
weld-metal is 5.57 𝑘𝑘𝑘𝑘𝑘𝑘/𝑖𝑖𝑖𝑖𝑖𝑖ℎ of weldment. By using the welding code methods determine whether;
(a)
The weld-metal strength is satisfactory?
(b)
The attachment strength is satisfactory?
Solution (a) The weld-metal strength is satisfactory? Given; 𝐹𝐹𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 5.57 𝑘𝑘𝑘𝑘𝑘𝑘/𝑖𝑖𝑖𝑖𝑖𝑖ℎ = 975.46 𝑘𝑘𝑘𝑘/𝑚𝑚
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙ℎ 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤, ℓ = 50.8 𝑚𝑚𝑚𝑚 = 50.8 × 10−3 𝑚𝑚
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 ℓ (𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤) = 50.8 × 10−3 (2) = 0.1016 𝑚𝑚 𝐹𝐹𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 975.46(ℓ)
𝐹𝐹𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 975.46(0.1016)
𝐹𝐹𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 𝟗𝟗𝟗𝟗. 𝟏𝟏𝟏𝟏 𝒌𝒌𝒌𝒌
𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝐹𝐹𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 𝟗𝟗𝟗𝟗. 𝟏𝟏𝟏𝟏 𝒌𝒌𝒌𝒌 > 𝟕𝟕𝟕𝟕 𝒌𝒌𝒌𝒌
∴ Weld metal strength is satisfactory.
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WELDING DESIGN ANALYSIS (b) The attachment strength is satisfactory? Checking shear in attachment adjacent to the welds, from Table 4 the allowable attachment shear stress is; 𝜏𝜏𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 0.4𝑆𝑆𝑦𝑦
where 𝑆𝑆𝑦𝑦 taken from Table 2 for G10150 HR, the yield strength, 𝑆𝑆𝑦𝑦 = 190 𝑀𝑀𝑀𝑀𝑀𝑀
∴ 𝜏𝜏𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 0.4(190) = 𝟕𝟕𝟕𝟕 𝑴𝑴𝑴𝑴𝑴𝑴
while the shear stress on the base metal adjacent to the weld is; 𝜏𝜏 =
𝐹𝐹 𝐴𝐴
𝜏𝜏 =
70 𝑘𝑘𝑘𝑘 2(9.525 × 10−3 )(50.8 × 10−3 )
𝐴𝐴 = 2𝑠𝑠ℓ (𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑)
𝜏𝜏 = 𝟕𝟕𝟕𝟕. 𝟑𝟑𝟑𝟑 𝑴𝑴𝑴𝑴𝑴𝑴
𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝜏𝜏𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 𝟕𝟕𝟕𝟕 𝑴𝑴𝑴𝑴𝑴𝑴 > 𝟕𝟕𝟕𝟕. 𝟑𝟑𝟑𝟑 𝑴𝑴𝑴𝑴𝑴𝑴
∴ The attachment is satisfactory near the weld beads. Checking tensile stress of the shank, the allowable tensile stress of the shank is; 𝜎𝜎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 0.6𝑆𝑆𝑦𝑦 = 0.6(190) 𝜎𝜎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 𝟏𝟏𝟏𝟏𝟏𝟏 𝑴𝑴𝑴𝑴𝑴𝑴
while the tensile stress in the shank of the attachment is; 𝐹𝐹 70 × 103 𝜎𝜎 = = 𝑡𝑡𝑡𝑡 (12.7 × 10−3 )(50.8 × 10−3 )
𝜎𝜎 = 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓 𝑴𝑴𝑴𝑴𝑴𝑴
𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝜎𝜎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 𝟏𝟏𝟏𝟏𝟏𝟏 𝑴𝑴𝑴𝑴𝑴𝑴 ≥ 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓 𝑴𝑴𝑴𝑴𝑴𝑴 ∴ The shank tensile stress is satisfactory. 31
WELDING DESIGN ANALYSIS
Bibliography Richard G. Budynas, J. Keith Nisbett. (2015). Shigley’s Mechanical Engineering Design. 10th ed. New York: McGraw-Hill. Saleh, H. H. (2017). Fundamentals of Mechanical Components in Engineering Design. Selangor: Pelangi Professional Publishing Sdn. Bhd.
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