Welding DesignWelding Design

1 This topics will be discussed in one session of the lecture

2 However an extensive discussion about welding will NOT be covered for mechanical program such as1048708Stresses in welded joint in torsion and bending1048708The strength of welded joints1048708Static and fatigue loading applied on welded joints1048708Resistance welding

Permanent JointsPermanent Joints

Permanent joint covers joint such as

Welding Brazing Soldering Cementing Gluing

The advantages of permanent joint are it will eliminate fasteners holes and assembly cost

Welding SymbolsWelding Symbols

Basic Weld SymbolsBasic Weld Symbols

Fillet WeldsFillet Welds

Butt and Fillet WeldsButt and Fillet Welds

A Transverse Fillet WeldA Transverse Fillet Weld

This stress can be divided into two components a shear stress and This stress can be divided into two components a shear stress and a normal stress There area normal stress There are

In fig below these are entered into a Morrrsquos circle diagram The In fig below these are entered into a Morrrsquos circle diagram The largest principal stress is seen to belargest principal stress is seen to be

Stress Distribution in Fillet Stress Distribution in Fillet WeldsWelds

Thus the equation for average stress isThus the equation for average stress is

and is normally used in designing joints having and is normally used in designing joints having fillet weldsfillet welds

A double-filleted lap jointA double-filleted lap joint

Since there are two welds the throat area for both isSince there are two welds the throat area for both is

The average shear stress is thereforeThe average shear stress is therefore

(9-3)(9-3)

Torsion in Welded JointsTorsion in Welded Joints

The reaction at the support cantilever always consists of a The reaction at the support cantilever always consists of a shear force shear force VV and a moment and a moment MM The shear force produces a The shear force produces a primary shear in the welds of magnitudeprimary shear in the welds of magnitude

(9-4)(9-4)

Where A is the throat area of all the weldsWhere A is the throat area of all the weldsThe moment at the support produces secondary shear or The moment at the support produces secondary shear or torsion of the welds and this stress istorsion of the welds and this stress is

(9-5)(9-5)

r = distance from the centroidr = distance from the centroid J = polar moment inertiaJ = polar moment inertia

Figure 9-11 A moment connectionFigure 9-11 A moment connection

Figure 9-12 shows two welds in a group The rectangle represent the throat areas of Figure 9-12 shows two welds in a group The rectangle represent the throat areas of the welds Weld 1 has a throat width bthe welds Weld 1 has a throat width b1 1 ndash 070hl and weld 2 has a throat width dndash 070hl and weld 2 has a throat width d22 = =

0707h0707h22 Note that h1 and h2 are the respective weld sizes The throat area of both Note that h1 and h2 are the respective weld sizes The throat area of both

welds iswelds is

A = AA = A11 + A + A22 = B = B11DD11 + B + B22DD22 (a)(a)

Since the throat width of a fillet weld is 0707h the relationship Since the throat width of a fillet weld is 0707h the relationship between the unit polar moment of inertia and the polar moment of between the unit polar moment of inertia and the polar moment of inertia of a fillet weld isinertia of a fillet weld is

J = 0707hJJ = 0707hJuu

Table 9-1 Torsional Table 9-1 Torsional Properties of Fillet Properties of Fillet

WeldsWelds

EXAMPLE 9-1EXAMPLE 9-1A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated in figure 9-13 Compute the maximum stress in the weldin figure 9-13 Compute the maximum stress in the weld

SolutionSolutiona)a) Label the end and corners of each weld by letter Sometimes it is desirable to label Label the end and corners of each weld by letter Sometimes it is desirable to label

each weld of a set by number See figure 9-14each weld of a set by number See figure 9-14b)b) Compute the primary shear stress As shown in figure 9-13 each plate is welded to Compute the primary shear stress As shown in figure 9-13 each plate is welded to

the channel using three 6mm fillet welds Figure 9-14 show that we have divided the channel using three 6mm fillet welds Figure 9-14 show that we have divided the load in half and are considering only a single plate From case 4 of table 9-1 the load in half and are considering only a single plate From case 4 of table 9-1 we find the throat area aswe find the throat area as

A = 0707(6)[2(56) + 190] = 1280 mmA = 0707(6)[2(56) + 190] = 1280 mm2 2

Then the primary shear stress ishellipThen the primary shear stress ishellip

Permanent JointsPermanent Joints

Permanent joint covers joint such as

Welding Brazing Soldering Cementing Gluing

The advantages of permanent joint are it will eliminate fasteners holes and assembly cost

Welding SymbolsWelding Symbols

Basic Weld SymbolsBasic Weld Symbols

Fillet WeldsFillet Welds

Butt and Fillet WeldsButt and Fillet Welds

A Transverse Fillet WeldA Transverse Fillet Weld

This stress can be divided into two components a shear stress and This stress can be divided into two components a shear stress and a normal stress There area normal stress There are

In fig below these are entered into a Morrrsquos circle diagram The In fig below these are entered into a Morrrsquos circle diagram The largest principal stress is seen to belargest principal stress is seen to be

Stress Distribution in Fillet Stress Distribution in Fillet WeldsWelds

Thus the equation for average stress isThus the equation for average stress is

and is normally used in designing joints having and is normally used in designing joints having fillet weldsfillet welds

A double-filleted lap jointA double-filleted lap joint

Since there are two welds the throat area for both isSince there are two welds the throat area for both is

The average shear stress is thereforeThe average shear stress is therefore

(9-3)(9-3)

Torsion in Welded JointsTorsion in Welded Joints

The reaction at the support cantilever always consists of a The reaction at the support cantilever always consists of a shear force shear force VV and a moment and a moment MM The shear force produces a The shear force produces a primary shear in the welds of magnitudeprimary shear in the welds of magnitude

(9-4)(9-4)

Where A is the throat area of all the weldsWhere A is the throat area of all the weldsThe moment at the support produces secondary shear or The moment at the support produces secondary shear or torsion of the welds and this stress istorsion of the welds and this stress is

(9-5)(9-5)

r = distance from the centroidr = distance from the centroid J = polar moment inertiaJ = polar moment inertia

Figure 9-11 A moment connectionFigure 9-11 A moment connection

Figure 9-12 shows two welds in a group The rectangle represent the throat areas of Figure 9-12 shows two welds in a group The rectangle represent the throat areas of the welds Weld 1 has a throat width bthe welds Weld 1 has a throat width b1 1 ndash 070hl and weld 2 has a throat width dndash 070hl and weld 2 has a throat width d22 = =

0707h0707h22 Note that h1 and h2 are the respective weld sizes The throat area of both Note that h1 and h2 are the respective weld sizes The throat area of both

welds iswelds is

A = AA = A11 + A + A22 = B = B11DD11 + B + B22DD22 (a)(a)

Since the throat width of a fillet weld is 0707h the relationship Since the throat width of a fillet weld is 0707h the relationship between the unit polar moment of inertia and the polar moment of between the unit polar moment of inertia and the polar moment of inertia of a fillet weld isinertia of a fillet weld is

J = 0707hJJ = 0707hJuu

Table 9-1 Torsional Table 9-1 Torsional Properties of Fillet Properties of Fillet

WeldsWelds

EXAMPLE 9-1EXAMPLE 9-1A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated in figure 9-13 Compute the maximum stress in the weldin figure 9-13 Compute the maximum stress in the weld

SolutionSolutiona)a) Label the end and corners of each weld by letter Sometimes it is desirable to label Label the end and corners of each weld by letter Sometimes it is desirable to label

each weld of a set by number See figure 9-14each weld of a set by number See figure 9-14b)b) Compute the primary shear stress As shown in figure 9-13 each plate is welded to Compute the primary shear stress As shown in figure 9-13 each plate is welded to

the channel using three 6mm fillet welds Figure 9-14 show that we have divided the channel using three 6mm fillet welds Figure 9-14 show that we have divided the load in half and are considering only a single plate From case 4 of table 9-1 the load in half and are considering only a single plate From case 4 of table 9-1 we find the throat area aswe find the throat area as

A = 0707(6)[2(56) + 190] = 1280 mmA = 0707(6)[2(56) + 190] = 1280 mm2 2

Then the primary shear stress ishellipThen the primary shear stress ishellip

Welding SymbolsWelding Symbols

Basic Weld SymbolsBasic Weld Symbols

Fillet WeldsFillet Welds

Butt and Fillet WeldsButt and Fillet Welds

A Transverse Fillet WeldA Transverse Fillet Weld

This stress can be divided into two components a shear stress and This stress can be divided into two components a shear stress and a normal stress There area normal stress There are

In fig below these are entered into a Morrrsquos circle diagram The In fig below these are entered into a Morrrsquos circle diagram The largest principal stress is seen to belargest principal stress is seen to be

Stress Distribution in Fillet Stress Distribution in Fillet WeldsWelds

Thus the equation for average stress isThus the equation for average stress is

and is normally used in designing joints having and is normally used in designing joints having fillet weldsfillet welds

A double-filleted lap jointA double-filleted lap joint

Since there are two welds the throat area for both isSince there are two welds the throat area for both is

The average shear stress is thereforeThe average shear stress is therefore

(9-3)(9-3)

Torsion in Welded JointsTorsion in Welded Joints

The reaction at the support cantilever always consists of a The reaction at the support cantilever always consists of a shear force shear force VV and a moment and a moment MM The shear force produces a The shear force produces a primary shear in the welds of magnitudeprimary shear in the welds of magnitude

(9-4)(9-4)

Where A is the throat area of all the weldsWhere A is the throat area of all the weldsThe moment at the support produces secondary shear or The moment at the support produces secondary shear or torsion of the welds and this stress istorsion of the welds and this stress is

(9-5)(9-5)

r = distance from the centroidr = distance from the centroid J = polar moment inertiaJ = polar moment inertia

Figure 9-11 A moment connectionFigure 9-11 A moment connection

Figure 9-12 shows two welds in a group The rectangle represent the throat areas of Figure 9-12 shows two welds in a group The rectangle represent the throat areas of the welds Weld 1 has a throat width bthe welds Weld 1 has a throat width b1 1 ndash 070hl and weld 2 has a throat width dndash 070hl and weld 2 has a throat width d22 = =

0707h0707h22 Note that h1 and h2 are the respective weld sizes The throat area of both Note that h1 and h2 are the respective weld sizes The throat area of both

welds iswelds is

A = AA = A11 + A + A22 = B = B11DD11 + B + B22DD22 (a)(a)

Since the throat width of a fillet weld is 0707h the relationship Since the throat width of a fillet weld is 0707h the relationship between the unit polar moment of inertia and the polar moment of between the unit polar moment of inertia and the polar moment of inertia of a fillet weld isinertia of a fillet weld is

J = 0707hJJ = 0707hJuu

Table 9-1 Torsional Table 9-1 Torsional Properties of Fillet Properties of Fillet

WeldsWelds

EXAMPLE 9-1EXAMPLE 9-1A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated in figure 9-13 Compute the maximum stress in the weldin figure 9-13 Compute the maximum stress in the weld

SolutionSolutiona)a) Label the end and corners of each weld by letter Sometimes it is desirable to label Label the end and corners of each weld by letter Sometimes it is desirable to label

each weld of a set by number See figure 9-14each weld of a set by number See figure 9-14b)b) Compute the primary shear stress As shown in figure 9-13 each plate is welded to Compute the primary shear stress As shown in figure 9-13 each plate is welded to

the channel using three 6mm fillet welds Figure 9-14 show that we have divided the channel using three 6mm fillet welds Figure 9-14 show that we have divided the load in half and are considering only a single plate From case 4 of table 9-1 the load in half and are considering only a single plate From case 4 of table 9-1 we find the throat area aswe find the throat area as

A = 0707(6)[2(56) + 190] = 1280 mmA = 0707(6)[2(56) + 190] = 1280 mm2 2

Then the primary shear stress ishellipThen the primary shear stress ishellip

Basic Weld SymbolsBasic Weld Symbols

Fillet WeldsFillet Welds

Butt and Fillet WeldsButt and Fillet Welds

A Transverse Fillet WeldA Transverse Fillet Weld

This stress can be divided into two components a shear stress and This stress can be divided into two components a shear stress and a normal stress There area normal stress There are

In fig below these are entered into a Morrrsquos circle diagram The In fig below these are entered into a Morrrsquos circle diagram The largest principal stress is seen to belargest principal stress is seen to be

Stress Distribution in Fillet Stress Distribution in Fillet WeldsWelds

Thus the equation for average stress isThus the equation for average stress is

and is normally used in designing joints having and is normally used in designing joints having fillet weldsfillet welds

A double-filleted lap jointA double-filleted lap joint

Since there are two welds the throat area for both isSince there are two welds the throat area for both is

The average shear stress is thereforeThe average shear stress is therefore

(9-3)(9-3)

Torsion in Welded JointsTorsion in Welded Joints

The reaction at the support cantilever always consists of a The reaction at the support cantilever always consists of a shear force shear force VV and a moment and a moment MM The shear force produces a The shear force produces a primary shear in the welds of magnitudeprimary shear in the welds of magnitude

(9-4)(9-4)

Where A is the throat area of all the weldsWhere A is the throat area of all the weldsThe moment at the support produces secondary shear or The moment at the support produces secondary shear or torsion of the welds and this stress istorsion of the welds and this stress is

(9-5)(9-5)

r = distance from the centroidr = distance from the centroid J = polar moment inertiaJ = polar moment inertia

Figure 9-11 A moment connectionFigure 9-11 A moment connection

Figure 9-12 shows two welds in a group The rectangle represent the throat areas of Figure 9-12 shows two welds in a group The rectangle represent the throat areas of the welds Weld 1 has a throat width bthe welds Weld 1 has a throat width b1 1 ndash 070hl and weld 2 has a throat width dndash 070hl and weld 2 has a throat width d22 = =

0707h0707h22 Note that h1 and h2 are the respective weld sizes The throat area of both Note that h1 and h2 are the respective weld sizes The throat area of both

welds iswelds is

A = AA = A11 + A + A22 = B = B11DD11 + B + B22DD22 (a)(a)

Since the throat width of a fillet weld is 0707h the relationship Since the throat width of a fillet weld is 0707h the relationship between the unit polar moment of inertia and the polar moment of between the unit polar moment of inertia and the polar moment of inertia of a fillet weld isinertia of a fillet weld is

J = 0707hJJ = 0707hJuu

Table 9-1 Torsional Table 9-1 Torsional Properties of Fillet Properties of Fillet

WeldsWelds

EXAMPLE 9-1EXAMPLE 9-1A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated in figure 9-13 Compute the maximum stress in the weldin figure 9-13 Compute the maximum stress in the weld

SolutionSolutiona)a) Label the end and corners of each weld by letter Sometimes it is desirable to label Label the end and corners of each weld by letter Sometimes it is desirable to label

each weld of a set by number See figure 9-14each weld of a set by number See figure 9-14b)b) Compute the primary shear stress As shown in figure 9-13 each plate is welded to Compute the primary shear stress As shown in figure 9-13 each plate is welded to

the channel using three 6mm fillet welds Figure 9-14 show that we have divided the channel using three 6mm fillet welds Figure 9-14 show that we have divided the load in half and are considering only a single plate From case 4 of table 9-1 the load in half and are considering only a single plate From case 4 of table 9-1 we find the throat area aswe find the throat area as

A = 0707(6)[2(56) + 190] = 1280 mmA = 0707(6)[2(56) + 190] = 1280 mm2 2

Then the primary shear stress ishellipThen the primary shear stress ishellip

Fillet WeldsFillet Welds

Butt and Fillet WeldsButt and Fillet Welds

A Transverse Fillet WeldA Transverse Fillet Weld

This stress can be divided into two components a shear stress and This stress can be divided into two components a shear stress and a normal stress There area normal stress There are

In fig below these are entered into a Morrrsquos circle diagram The In fig below these are entered into a Morrrsquos circle diagram The largest principal stress is seen to belargest principal stress is seen to be

Stress Distribution in Fillet Stress Distribution in Fillet WeldsWelds

Thus the equation for average stress isThus the equation for average stress is

and is normally used in designing joints having and is normally used in designing joints having fillet weldsfillet welds

A double-filleted lap jointA double-filleted lap joint

Since there are two welds the throat area for both isSince there are two welds the throat area for both is

The average shear stress is thereforeThe average shear stress is therefore

(9-3)(9-3)

Torsion in Welded JointsTorsion in Welded Joints

The reaction at the support cantilever always consists of a The reaction at the support cantilever always consists of a shear force shear force VV and a moment and a moment MM The shear force produces a The shear force produces a primary shear in the welds of magnitudeprimary shear in the welds of magnitude

(9-4)(9-4)

Where A is the throat area of all the weldsWhere A is the throat area of all the weldsThe moment at the support produces secondary shear or The moment at the support produces secondary shear or torsion of the welds and this stress istorsion of the welds and this stress is

(9-5)(9-5)

r = distance from the centroidr = distance from the centroid J = polar moment inertiaJ = polar moment inertia

Figure 9-11 A moment connectionFigure 9-11 A moment connection

Figure 9-12 shows two welds in a group The rectangle represent the throat areas of Figure 9-12 shows two welds in a group The rectangle represent the throat areas of the welds Weld 1 has a throat width bthe welds Weld 1 has a throat width b1 1 ndash 070hl and weld 2 has a throat width dndash 070hl and weld 2 has a throat width d22 = =

0707h0707h22 Note that h1 and h2 are the respective weld sizes The throat area of both Note that h1 and h2 are the respective weld sizes The throat area of both

welds iswelds is

A = AA = A11 + A + A22 = B = B11DD11 + B + B22DD22 (a)(a)

Since the throat width of a fillet weld is 0707h the relationship Since the throat width of a fillet weld is 0707h the relationship between the unit polar moment of inertia and the polar moment of between the unit polar moment of inertia and the polar moment of inertia of a fillet weld isinertia of a fillet weld is

J = 0707hJJ = 0707hJuu

Table 9-1 Torsional Table 9-1 Torsional Properties of Fillet Properties of Fillet

WeldsWelds

EXAMPLE 9-1EXAMPLE 9-1A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated in figure 9-13 Compute the maximum stress in the weldin figure 9-13 Compute the maximum stress in the weld

SolutionSolutiona)a) Label the end and corners of each weld by letter Sometimes it is desirable to label Label the end and corners of each weld by letter Sometimes it is desirable to label

each weld of a set by number See figure 9-14each weld of a set by number See figure 9-14b)b) Compute the primary shear stress As shown in figure 9-13 each plate is welded to Compute the primary shear stress As shown in figure 9-13 each plate is welded to

the channel using three 6mm fillet welds Figure 9-14 show that we have divided the channel using three 6mm fillet welds Figure 9-14 show that we have divided the load in half and are considering only a single plate From case 4 of table 9-1 the load in half and are considering only a single plate From case 4 of table 9-1 we find the throat area aswe find the throat area as

A = 0707(6)[2(56) + 190] = 1280 mmA = 0707(6)[2(56) + 190] = 1280 mm2 2

Then the primary shear stress ishellipThen the primary shear stress ishellip

Butt and Fillet WeldsButt and Fillet Welds

A Transverse Fillet WeldA Transverse Fillet Weld

This stress can be divided into two components a shear stress and This stress can be divided into two components a shear stress and a normal stress There area normal stress There are

In fig below these are entered into a Morrrsquos circle diagram The In fig below these are entered into a Morrrsquos circle diagram The largest principal stress is seen to belargest principal stress is seen to be

Stress Distribution in Fillet Stress Distribution in Fillet WeldsWelds

Thus the equation for average stress isThus the equation for average stress is

and is normally used in designing joints having and is normally used in designing joints having fillet weldsfillet welds

A double-filleted lap jointA double-filleted lap joint

Since there are two welds the throat area for both isSince there are two welds the throat area for both is

The average shear stress is thereforeThe average shear stress is therefore

(9-3)(9-3)

Torsion in Welded JointsTorsion in Welded Joints

The reaction at the support cantilever always consists of a The reaction at the support cantilever always consists of a shear force shear force VV and a moment and a moment MM The shear force produces a The shear force produces a primary shear in the welds of magnitudeprimary shear in the welds of magnitude

(9-4)(9-4)

Where A is the throat area of all the weldsWhere A is the throat area of all the weldsThe moment at the support produces secondary shear or The moment at the support produces secondary shear or torsion of the welds and this stress istorsion of the welds and this stress is

(9-5)(9-5)

r = distance from the centroidr = distance from the centroid J = polar moment inertiaJ = polar moment inertia

Figure 9-11 A moment connectionFigure 9-11 A moment connection

Figure 9-12 shows two welds in a group The rectangle represent the throat areas of Figure 9-12 shows two welds in a group The rectangle represent the throat areas of the welds Weld 1 has a throat width bthe welds Weld 1 has a throat width b1 1 ndash 070hl and weld 2 has a throat width dndash 070hl and weld 2 has a throat width d22 = =

0707h0707h22 Note that h1 and h2 are the respective weld sizes The throat area of both Note that h1 and h2 are the respective weld sizes The throat area of both

welds iswelds is

A = AA = A11 + A + A22 = B = B11DD11 + B + B22DD22 (a)(a)

Since the throat width of a fillet weld is 0707h the relationship Since the throat width of a fillet weld is 0707h the relationship between the unit polar moment of inertia and the polar moment of between the unit polar moment of inertia and the polar moment of inertia of a fillet weld isinertia of a fillet weld is

J = 0707hJJ = 0707hJuu

Table 9-1 Torsional Table 9-1 Torsional Properties of Fillet Properties of Fillet

WeldsWelds

EXAMPLE 9-1EXAMPLE 9-1A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated in figure 9-13 Compute the maximum stress in the weldin figure 9-13 Compute the maximum stress in the weld

SolutionSolutiona)a) Label the end and corners of each weld by letter Sometimes it is desirable to label Label the end and corners of each weld by letter Sometimes it is desirable to label

each weld of a set by number See figure 9-14each weld of a set by number See figure 9-14b)b) Compute the primary shear stress As shown in figure 9-13 each plate is welded to Compute the primary shear stress As shown in figure 9-13 each plate is welded to

the channel using three 6mm fillet welds Figure 9-14 show that we have divided the channel using three 6mm fillet welds Figure 9-14 show that we have divided the load in half and are considering only a single plate From case 4 of table 9-1 the load in half and are considering only a single plate From case 4 of table 9-1 we find the throat area aswe find the throat area as

A = 0707(6)[2(56) + 190] = 1280 mmA = 0707(6)[2(56) + 190] = 1280 mm2 2

Then the primary shear stress ishellipThen the primary shear stress ishellip

A Transverse Fillet WeldA Transverse Fillet Weld

This stress can be divided into two components a shear stress and This stress can be divided into two components a shear stress and a normal stress There area normal stress There are

In fig below these are entered into a Morrrsquos circle diagram The In fig below these are entered into a Morrrsquos circle diagram The largest principal stress is seen to belargest principal stress is seen to be

Stress Distribution in Fillet Stress Distribution in Fillet WeldsWelds

Thus the equation for average stress isThus the equation for average stress is

and is normally used in designing joints having and is normally used in designing joints having fillet weldsfillet welds

A double-filleted lap jointA double-filleted lap joint

Since there are two welds the throat area for both isSince there are two welds the throat area for both is

The average shear stress is thereforeThe average shear stress is therefore

(9-3)(9-3)

Torsion in Welded JointsTorsion in Welded Joints

The reaction at the support cantilever always consists of a The reaction at the support cantilever always consists of a shear force shear force VV and a moment and a moment MM The shear force produces a The shear force produces a primary shear in the welds of magnitudeprimary shear in the welds of magnitude

(9-4)(9-4)

Where A is the throat area of all the weldsWhere A is the throat area of all the weldsThe moment at the support produces secondary shear or The moment at the support produces secondary shear or torsion of the welds and this stress istorsion of the welds and this stress is

(9-5)(9-5)

r = distance from the centroidr = distance from the centroid J = polar moment inertiaJ = polar moment inertia

Figure 9-11 A moment connectionFigure 9-11 A moment connection

Figure 9-12 shows two welds in a group The rectangle represent the throat areas of Figure 9-12 shows two welds in a group The rectangle represent the throat areas of the welds Weld 1 has a throat width bthe welds Weld 1 has a throat width b1 1 ndash 070hl and weld 2 has a throat width dndash 070hl and weld 2 has a throat width d22 = =

0707h0707h22 Note that h1 and h2 are the respective weld sizes The throat area of both Note that h1 and h2 are the respective weld sizes The throat area of both

welds iswelds is

A = AA = A11 + A + A22 = B = B11DD11 + B + B22DD22 (a)(a)

Since the throat width of a fillet weld is 0707h the relationship Since the throat width of a fillet weld is 0707h the relationship between the unit polar moment of inertia and the polar moment of between the unit polar moment of inertia and the polar moment of inertia of a fillet weld isinertia of a fillet weld is

J = 0707hJJ = 0707hJuu

Table 9-1 Torsional Table 9-1 Torsional Properties of Fillet Properties of Fillet

WeldsWelds

EXAMPLE 9-1EXAMPLE 9-1A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated in figure 9-13 Compute the maximum stress in the weldin figure 9-13 Compute the maximum stress in the weld

SolutionSolutiona)a) Label the end and corners of each weld by letter Sometimes it is desirable to label Label the end and corners of each weld by letter Sometimes it is desirable to label

each weld of a set by number See figure 9-14each weld of a set by number See figure 9-14b)b) Compute the primary shear stress As shown in figure 9-13 each plate is welded to Compute the primary shear stress As shown in figure 9-13 each plate is welded to

the channel using three 6mm fillet welds Figure 9-14 show that we have divided the channel using three 6mm fillet welds Figure 9-14 show that we have divided the load in half and are considering only a single plate From case 4 of table 9-1 the load in half and are considering only a single plate From case 4 of table 9-1 we find the throat area aswe find the throat area as

A = 0707(6)[2(56) + 190] = 1280 mmA = 0707(6)[2(56) + 190] = 1280 mm2 2

Then the primary shear stress ishellipThen the primary shear stress ishellip

This stress can be divided into two components a shear stress and This stress can be divided into two components a shear stress and a normal stress There area normal stress There are

In fig below these are entered into a Morrrsquos circle diagram The In fig below these are entered into a Morrrsquos circle diagram The largest principal stress is seen to belargest principal stress is seen to be

Stress Distribution in Fillet Stress Distribution in Fillet WeldsWelds

Thus the equation for average stress isThus the equation for average stress is

and is normally used in designing joints having and is normally used in designing joints having fillet weldsfillet welds

A double-filleted lap jointA double-filleted lap joint

Since there are two welds the throat area for both isSince there are two welds the throat area for both is

The average shear stress is thereforeThe average shear stress is therefore

(9-3)(9-3)

Torsion in Welded JointsTorsion in Welded Joints

The reaction at the support cantilever always consists of a The reaction at the support cantilever always consists of a shear force shear force VV and a moment and a moment MM The shear force produces a The shear force produces a primary shear in the welds of magnitudeprimary shear in the welds of magnitude

(9-4)(9-4)

Where A is the throat area of all the weldsWhere A is the throat area of all the weldsThe moment at the support produces secondary shear or The moment at the support produces secondary shear or torsion of the welds and this stress istorsion of the welds and this stress is

(9-5)(9-5)

r = distance from the centroidr = distance from the centroid J = polar moment inertiaJ = polar moment inertia

Figure 9-11 A moment connectionFigure 9-11 A moment connection

Figure 9-12 shows two welds in a group The rectangle represent the throat areas of Figure 9-12 shows two welds in a group The rectangle represent the throat areas of the welds Weld 1 has a throat width bthe welds Weld 1 has a throat width b1 1 ndash 070hl and weld 2 has a throat width dndash 070hl and weld 2 has a throat width d22 = =

0707h0707h22 Note that h1 and h2 are the respective weld sizes The throat area of both Note that h1 and h2 are the respective weld sizes The throat area of both

welds iswelds is

A = AA = A11 + A + A22 = B = B11DD11 + B + B22DD22 (a)(a)

Since the throat width of a fillet weld is 0707h the relationship Since the throat width of a fillet weld is 0707h the relationship between the unit polar moment of inertia and the polar moment of between the unit polar moment of inertia and the polar moment of inertia of a fillet weld isinertia of a fillet weld is

J = 0707hJJ = 0707hJuu

Table 9-1 Torsional Table 9-1 Torsional Properties of Fillet Properties of Fillet

WeldsWelds

EXAMPLE 9-1EXAMPLE 9-1A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated in figure 9-13 Compute the maximum stress in the weldin figure 9-13 Compute the maximum stress in the weld

SolutionSolutiona)a) Label the end and corners of each weld by letter Sometimes it is desirable to label Label the end and corners of each weld by letter Sometimes it is desirable to label

each weld of a set by number See figure 9-14each weld of a set by number See figure 9-14b)b) Compute the primary shear stress As shown in figure 9-13 each plate is welded to Compute the primary shear stress As shown in figure 9-13 each plate is welded to

the channel using three 6mm fillet welds Figure 9-14 show that we have divided the channel using three 6mm fillet welds Figure 9-14 show that we have divided the load in half and are considering only a single plate From case 4 of table 9-1 the load in half and are considering only a single plate From case 4 of table 9-1 we find the throat area aswe find the throat area as

A = 0707(6)[2(56) + 190] = 1280 mmA = 0707(6)[2(56) + 190] = 1280 mm2 2

Then the primary shear stress ishellipThen the primary shear stress ishellip

Stress Distribution in Fillet Stress Distribution in Fillet WeldsWelds

Thus the equation for average stress isThus the equation for average stress is

and is normally used in designing joints having and is normally used in designing joints having fillet weldsfillet welds

A double-filleted lap jointA double-filleted lap joint

Since there are two welds the throat area for both isSince there are two welds the throat area for both is

The average shear stress is thereforeThe average shear stress is therefore

(9-3)(9-3)

Torsion in Welded JointsTorsion in Welded Joints

The reaction at the support cantilever always consists of a The reaction at the support cantilever always consists of a shear force shear force VV and a moment and a moment MM The shear force produces a The shear force produces a primary shear in the welds of magnitudeprimary shear in the welds of magnitude

(9-4)(9-4)

Where A is the throat area of all the weldsWhere A is the throat area of all the weldsThe moment at the support produces secondary shear or The moment at the support produces secondary shear or torsion of the welds and this stress istorsion of the welds and this stress is

(9-5)(9-5)

r = distance from the centroidr = distance from the centroid J = polar moment inertiaJ = polar moment inertia

Figure 9-11 A moment connectionFigure 9-11 A moment connection

Figure 9-12 shows two welds in a group The rectangle represent the throat areas of Figure 9-12 shows two welds in a group The rectangle represent the throat areas of the welds Weld 1 has a throat width bthe welds Weld 1 has a throat width b1 1 ndash 070hl and weld 2 has a throat width dndash 070hl and weld 2 has a throat width d22 = =

0707h0707h22 Note that h1 and h2 are the respective weld sizes The throat area of both Note that h1 and h2 are the respective weld sizes The throat area of both

welds iswelds is

A = AA = A11 + A + A22 = B = B11DD11 + B + B22DD22 (a)(a)

Since the throat width of a fillet weld is 0707h the relationship Since the throat width of a fillet weld is 0707h the relationship between the unit polar moment of inertia and the polar moment of between the unit polar moment of inertia and the polar moment of inertia of a fillet weld isinertia of a fillet weld is

J = 0707hJJ = 0707hJuu

Table 9-1 Torsional Table 9-1 Torsional Properties of Fillet Properties of Fillet

WeldsWelds

EXAMPLE 9-1EXAMPLE 9-1A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated in figure 9-13 Compute the maximum stress in the weldin figure 9-13 Compute the maximum stress in the weld

SolutionSolutiona)a) Label the end and corners of each weld by letter Sometimes it is desirable to label Label the end and corners of each weld by letter Sometimes it is desirable to label

each weld of a set by number See figure 9-14each weld of a set by number See figure 9-14b)b) Compute the primary shear stress As shown in figure 9-13 each plate is welded to Compute the primary shear stress As shown in figure 9-13 each plate is welded to

the channel using three 6mm fillet welds Figure 9-14 show that we have divided the channel using three 6mm fillet welds Figure 9-14 show that we have divided the load in half and are considering only a single plate From case 4 of table 9-1 the load in half and are considering only a single plate From case 4 of table 9-1 we find the throat area aswe find the throat area as

A = 0707(6)[2(56) + 190] = 1280 mmA = 0707(6)[2(56) + 190] = 1280 mm2 2

Then the primary shear stress ishellipThen the primary shear stress ishellip

Thus the equation for average stress isThus the equation for average stress is

and is normally used in designing joints having and is normally used in designing joints having fillet weldsfillet welds

A double-filleted lap jointA double-filleted lap joint

Since there are two welds the throat area for both isSince there are two welds the throat area for both is

The average shear stress is thereforeThe average shear stress is therefore

(9-3)(9-3)

Torsion in Welded JointsTorsion in Welded Joints

The reaction at the support cantilever always consists of a The reaction at the support cantilever always consists of a shear force shear force VV and a moment and a moment MM The shear force produces a The shear force produces a primary shear in the welds of magnitudeprimary shear in the welds of magnitude

(9-4)(9-4)

Where A is the throat area of all the weldsWhere A is the throat area of all the weldsThe moment at the support produces secondary shear or The moment at the support produces secondary shear or torsion of the welds and this stress istorsion of the welds and this stress is

(9-5)(9-5)

r = distance from the centroidr = distance from the centroid J = polar moment inertiaJ = polar moment inertia

Figure 9-11 A moment connectionFigure 9-11 A moment connection

Figure 9-12 shows two welds in a group The rectangle represent the throat areas of Figure 9-12 shows two welds in a group The rectangle represent the throat areas of the welds Weld 1 has a throat width bthe welds Weld 1 has a throat width b1 1 ndash 070hl and weld 2 has a throat width dndash 070hl and weld 2 has a throat width d22 = =

0707h0707h22 Note that h1 and h2 are the respective weld sizes The throat area of both Note that h1 and h2 are the respective weld sizes The throat area of both

welds iswelds is

A = AA = A11 + A + A22 = B = B11DD11 + B + B22DD22 (a)(a)

Since the throat width of a fillet weld is 0707h the relationship Since the throat width of a fillet weld is 0707h the relationship between the unit polar moment of inertia and the polar moment of between the unit polar moment of inertia and the polar moment of inertia of a fillet weld isinertia of a fillet weld is

J = 0707hJJ = 0707hJuu

Table 9-1 Torsional Table 9-1 Torsional Properties of Fillet Properties of Fillet

WeldsWelds

EXAMPLE 9-1EXAMPLE 9-1A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated in figure 9-13 Compute the maximum stress in the weldin figure 9-13 Compute the maximum stress in the weld

SolutionSolutiona)a) Label the end and corners of each weld by letter Sometimes it is desirable to label Label the end and corners of each weld by letter Sometimes it is desirable to label

each weld of a set by number See figure 9-14each weld of a set by number See figure 9-14b)b) Compute the primary shear stress As shown in figure 9-13 each plate is welded to Compute the primary shear stress As shown in figure 9-13 each plate is welded to

the channel using three 6mm fillet welds Figure 9-14 show that we have divided the channel using three 6mm fillet welds Figure 9-14 show that we have divided the load in half and are considering only a single plate From case 4 of table 9-1 the load in half and are considering only a single plate From case 4 of table 9-1 we find the throat area aswe find the throat area as

A = 0707(6)[2(56) + 190] = 1280 mmA = 0707(6)[2(56) + 190] = 1280 mm2 2

Then the primary shear stress ishellipThen the primary shear stress ishellip

A double-filleted lap jointA double-filleted lap joint

Since there are two welds the throat area for both isSince there are two welds the throat area for both is

The average shear stress is thereforeThe average shear stress is therefore

(9-3)(9-3)

Torsion in Welded JointsTorsion in Welded Joints

The reaction at the support cantilever always consists of a The reaction at the support cantilever always consists of a shear force shear force VV and a moment and a moment MM The shear force produces a The shear force produces a primary shear in the welds of magnitudeprimary shear in the welds of magnitude

(9-4)(9-4)

Where A is the throat area of all the weldsWhere A is the throat area of all the weldsThe moment at the support produces secondary shear or The moment at the support produces secondary shear or torsion of the welds and this stress istorsion of the welds and this stress is

(9-5)(9-5)

r = distance from the centroidr = distance from the centroid J = polar moment inertiaJ = polar moment inertia

Figure 9-11 A moment connectionFigure 9-11 A moment connection

Figure 9-12 shows two welds in a group The rectangle represent the throat areas of Figure 9-12 shows two welds in a group The rectangle represent the throat areas of the welds Weld 1 has a throat width bthe welds Weld 1 has a throat width b1 1 ndash 070hl and weld 2 has a throat width dndash 070hl and weld 2 has a throat width d22 = =

0707h0707h22 Note that h1 and h2 are the respective weld sizes The throat area of both Note that h1 and h2 are the respective weld sizes The throat area of both

welds iswelds is

A = AA = A11 + A + A22 = B = B11DD11 + B + B22DD22 (a)(a)

Since the throat width of a fillet weld is 0707h the relationship Since the throat width of a fillet weld is 0707h the relationship between the unit polar moment of inertia and the polar moment of between the unit polar moment of inertia and the polar moment of inertia of a fillet weld isinertia of a fillet weld is

J = 0707hJJ = 0707hJuu

Table 9-1 Torsional Table 9-1 Torsional Properties of Fillet Properties of Fillet

WeldsWelds

EXAMPLE 9-1EXAMPLE 9-1A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated in figure 9-13 Compute the maximum stress in the weldin figure 9-13 Compute the maximum stress in the weld

SolutionSolutiona)a) Label the end and corners of each weld by letter Sometimes it is desirable to label Label the end and corners of each weld by letter Sometimes it is desirable to label

each weld of a set by number See figure 9-14each weld of a set by number See figure 9-14b)b) Compute the primary shear stress As shown in figure 9-13 each plate is welded to Compute the primary shear stress As shown in figure 9-13 each plate is welded to

the channel using three 6mm fillet welds Figure 9-14 show that we have divided the channel using three 6mm fillet welds Figure 9-14 show that we have divided the load in half and are considering only a single plate From case 4 of table 9-1 the load in half and are considering only a single plate From case 4 of table 9-1 we find the throat area aswe find the throat area as

A = 0707(6)[2(56) + 190] = 1280 mmA = 0707(6)[2(56) + 190] = 1280 mm2 2

Then the primary shear stress ishellipThen the primary shear stress ishellip

Torsion in Welded JointsTorsion in Welded Joints

The reaction at the support cantilever always consists of a The reaction at the support cantilever always consists of a shear force shear force VV and a moment and a moment MM The shear force produces a The shear force produces a primary shear in the welds of magnitudeprimary shear in the welds of magnitude

(9-4)(9-4)

Where A is the throat area of all the weldsWhere A is the throat area of all the weldsThe moment at the support produces secondary shear or The moment at the support produces secondary shear or torsion of the welds and this stress istorsion of the welds and this stress is

(9-5)(9-5)

r = distance from the centroidr = distance from the centroid J = polar moment inertiaJ = polar moment inertia

Figure 9-11 A moment connectionFigure 9-11 A moment connection

Figure 9-12 shows two welds in a group The rectangle represent the throat areas of Figure 9-12 shows two welds in a group The rectangle represent the throat areas of the welds Weld 1 has a throat width bthe welds Weld 1 has a throat width b1 1 ndash 070hl and weld 2 has a throat width dndash 070hl and weld 2 has a throat width d22 = =

0707h0707h22 Note that h1 and h2 are the respective weld sizes The throat area of both Note that h1 and h2 are the respective weld sizes The throat area of both

welds iswelds is

A = AA = A11 + A + A22 = B = B11DD11 + B + B22DD22 (a)(a)

Since the throat width of a fillet weld is 0707h the relationship Since the throat width of a fillet weld is 0707h the relationship between the unit polar moment of inertia and the polar moment of between the unit polar moment of inertia and the polar moment of inertia of a fillet weld isinertia of a fillet weld is

J = 0707hJJ = 0707hJuu

Table 9-1 Torsional Table 9-1 Torsional Properties of Fillet Properties of Fillet

WeldsWelds

EXAMPLE 9-1EXAMPLE 9-1A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated in figure 9-13 Compute the maximum stress in the weldin figure 9-13 Compute the maximum stress in the weld

SolutionSolutiona)a) Label the end and corners of each weld by letter Sometimes it is desirable to label Label the end and corners of each weld by letter Sometimes it is desirable to label

each weld of a set by number See figure 9-14each weld of a set by number See figure 9-14b)b) Compute the primary shear stress As shown in figure 9-13 each plate is welded to Compute the primary shear stress As shown in figure 9-13 each plate is welded to

the channel using three 6mm fillet welds Figure 9-14 show that we have divided the channel using three 6mm fillet welds Figure 9-14 show that we have divided the load in half and are considering only a single plate From case 4 of table 9-1 the load in half and are considering only a single plate From case 4 of table 9-1 we find the throat area aswe find the throat area as

A = 0707(6)[2(56) + 190] = 1280 mmA = 0707(6)[2(56) + 190] = 1280 mm2 2

Then the primary shear stress ishellipThen the primary shear stress ishellip

Figure 9-11 A moment connectionFigure 9-11 A moment connection

Figure 9-12 shows two welds in a group The rectangle represent the throat areas of Figure 9-12 shows two welds in a group The rectangle represent the throat areas of the welds Weld 1 has a throat width bthe welds Weld 1 has a throat width b1 1 ndash 070hl and weld 2 has a throat width dndash 070hl and weld 2 has a throat width d22 = =

0707h0707h22 Note that h1 and h2 are the respective weld sizes The throat area of both Note that h1 and h2 are the respective weld sizes The throat area of both

welds iswelds is

A = AA = A11 + A + A22 = B = B11DD11 + B + B22DD22 (a)(a)

Since the throat width of a fillet weld is 0707h the relationship Since the throat width of a fillet weld is 0707h the relationship between the unit polar moment of inertia and the polar moment of between the unit polar moment of inertia and the polar moment of inertia of a fillet weld isinertia of a fillet weld is

J = 0707hJJ = 0707hJuu

Table 9-1 Torsional Table 9-1 Torsional Properties of Fillet Properties of Fillet

WeldsWelds

EXAMPLE 9-1EXAMPLE 9-1A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated in figure 9-13 Compute the maximum stress in the weldin figure 9-13 Compute the maximum stress in the weld

SolutionSolutiona)a) Label the end and corners of each weld by letter Sometimes it is desirable to label Label the end and corners of each weld by letter Sometimes it is desirable to label

each weld of a set by number See figure 9-14each weld of a set by number See figure 9-14b)b) Compute the primary shear stress As shown in figure 9-13 each plate is welded to Compute the primary shear stress As shown in figure 9-13 each plate is welded to

the channel using three 6mm fillet welds Figure 9-14 show that we have divided the channel using three 6mm fillet welds Figure 9-14 show that we have divided the load in half and are considering only a single plate From case 4 of table 9-1 the load in half and are considering only a single plate From case 4 of table 9-1 we find the throat area aswe find the throat area as

A = 0707(6)[2(56) + 190] = 1280 mmA = 0707(6)[2(56) + 190] = 1280 mm2 2

Then the primary shear stress ishellipThen the primary shear stress ishellip

Since the throat width of a fillet weld is 0707h the relationship Since the throat width of a fillet weld is 0707h the relationship between the unit polar moment of inertia and the polar moment of between the unit polar moment of inertia and the polar moment of inertia of a fillet weld isinertia of a fillet weld is

J = 0707hJJ = 0707hJuu

Table 9-1 Torsional Table 9-1 Torsional Properties of Fillet Properties of Fillet

WeldsWelds

EXAMPLE 9-1EXAMPLE 9-1A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated in figure 9-13 Compute the maximum stress in the weldin figure 9-13 Compute the maximum stress in the weld

SolutionSolutiona)a) Label the end and corners of each weld by letter Sometimes it is desirable to label Label the end and corners of each weld by letter Sometimes it is desirable to label

each weld of a set by number See figure 9-14each weld of a set by number See figure 9-14b)b) Compute the primary shear stress As shown in figure 9-13 each plate is welded to Compute the primary shear stress As shown in figure 9-13 each plate is welded to

the channel using three 6mm fillet welds Figure 9-14 show that we have divided the channel using three 6mm fillet welds Figure 9-14 show that we have divided the load in half and are considering only a single plate From case 4 of table 9-1 the load in half and are considering only a single plate From case 4 of table 9-1 we find the throat area aswe find the throat area as

A = 0707(6)[2(56) + 190] = 1280 mmA = 0707(6)[2(56) + 190] = 1280 mm2 2

Then the primary shear stress ishellipThen the primary shear stress ishellip

Table 9-1 Torsional Table 9-1 Torsional Properties of Fillet Properties of Fillet

WeldsWelds

EXAMPLE 9-1EXAMPLE 9-1A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated in figure 9-13 Compute the maximum stress in the weldin figure 9-13 Compute the maximum stress in the weld

SolutionSolutiona)a) Label the end and corners of each weld by letter Sometimes it is desirable to label Label the end and corners of each weld by letter Sometimes it is desirable to label

each weld of a set by number See figure 9-14each weld of a set by number See figure 9-14b)b) Compute the primary shear stress As shown in figure 9-13 each plate is welded to Compute the primary shear stress As shown in figure 9-13 each plate is welded to

the channel using three 6mm fillet welds Figure 9-14 show that we have divided the channel using three 6mm fillet welds Figure 9-14 show that we have divided the load in half and are considering only a single plate From case 4 of table 9-1 the load in half and are considering only a single plate From case 4 of table 9-1 we find the throat area aswe find the throat area as

A = 0707(6)[2(56) + 190] = 1280 mmA = 0707(6)[2(56) + 190] = 1280 mm2 2

Then the primary shear stress ishellipThen the primary shear stress ishellip

EXAMPLE 9-1EXAMPLE 9-1A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated A 50 kN load is transferred from a welded fitting into a 200mm steel as illustrated in figure 9-13 Compute the maximum stress in the weldin figure 9-13 Compute the maximum stress in the weld

SolutionSolutiona)a) Label the end and corners of each weld by letter Sometimes it is desirable to label Label the end and corners of each weld by letter Sometimes it is desirable to label

each weld of a set by number See figure 9-14each weld of a set by number See figure 9-14b)b) Compute the primary shear stress As shown in figure 9-13 each plate is welded to Compute the primary shear stress As shown in figure 9-13 each plate is welded to

the channel using three 6mm fillet welds Figure 9-14 show that we have divided the channel using three 6mm fillet welds Figure 9-14 show that we have divided the load in half and are considering only a single plate From case 4 of table 9-1 the load in half and are considering only a single plate From case 4 of table 9-1 we find the throat area aswe find the throat area as

A = 0707(6)[2(56) + 190] = 1280 mmA = 0707(6)[2(56) + 190] = 1280 mm2 2

Then the primary shear stress ishellipThen the primary shear stress ishellip