Download - Welding Design PPT
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