machine element I

1. Chapter Three
Strength Calculation and Dimensioning of
Joints:
Bolted Joints; Riveted Joints; Welding
Joints
1

2. Chapter outline
How to calculate the strength and dimensioning of joints ?
• Bolted
• Riveted
• Welding Joints
2

3. Introduction
• The design of joints is one of fascinating aspect.
• The success or failure of a design can be hinge on proper selection use of
its fastener.
• It is also a very big business and is a significant part of our economy
• Thousands to millions of fasteners are used in a single complex assembly.
E.g. the Boeing 747 uses about 2.5 million fasteners, some of which costs
several dollars.
• There are several types of commercially available fasteners.
3

4. Cont’d…
• Methods of joining parts are extremely important in the engineering of a
quality design, and it is necessary to have a thorough understanding of the
performance of fasteners and joints under all conditions of use and design.
Thread Standards and Definitions
 The pitch is the distance between
adjacent thread parallel to thread axis.
 The major diameter d is the largest
diameter of a screw thread.
 The minor (or root) diameter 𝑑 𝑟 is the
smallest diameter
 The pitch diameter 𝑑 𝑝 is a theoretical
diameter between the major and minor
diameters.
 The lead l is the distance the nut moves when the nut is given one turn.
4

5. Cont’d…
• The American National (Unified) thread standard has been approved
for use on all standard threaded products.
5

6. Threaded Fasteners
• Points of stress concentration are at the fillet, at the start of the threads (run
out), and at the thread-root fillet in the plane of the nut when it is present.
• The thread length of inch-series bolts, where d is the nominal diameter, is
where the dimensions are in millimeters.
6

7. Cont’d…
7

8. Bolted and Riveted Joints Loaded in Shear
• Riveted and bolted joints loaded in shear are treated exactly alike in design and
analysis.
• Figure 8–23a shows a riveted connection loaded in shear. Let us now study the
various means by which this connection might fail. Figure 8–23b shows a failure
by bending of the rivet or of the riveted members. The bending moment is
approximately M = Ft/2, where F is the shearing force and t is the grip of the
rivet, that is, the total thickness of the connected parts. The bending stress in
the members or in the rivet is, neglecting stress concentration,
8

9. Bolted and Riveted Joints Loaded in Shear
• Riveted and bolted joints loaded in shear are
treated exactly alike in design and analysis.
• Figure 8–23b shows a failure by
bending of the rivet or of the riveted
members. approximately 𝑀 =
𝐹𝑡
2
• where F is the shearing force and
• t is the grip of the rivet, that is, the total
thickness of the connected parts
• The bending stress 𝜎 =
𝑀
𝐼/𝐶
• In Fig. 8–23c failure of the rivet by
pure shear is shown; the shear stress in the
rivet is 𝜏 =
𝐹
𝐴
• Rupture of one of the connected
members or plates by pure tension is
illustrated in Fig. 8–23d. The tensile stress is
𝜎 =
𝐹
𝐴
9

10. Cont.
• Figure 8–23e illustrates a failure by crushing of the rivet or plate.
• Calculation of this stress, which is usually called a bearing stress, is
complicated by the distribution of the load on the cylindrical surface of
the rivet.
• The exact values of the forces acting upon the rivet are unknown, and so
it is customary to assume that the components of these forces are
uniformly distributed over the projected contact area of the rivet.
• This gives for the stress
σ = −
𝐹
𝐴
• where the projected area for a single rivet is A = td. Here, t is the
thickness of the thinnest plate and d is the rivet or bolt diameter.
10

11. Cont.
• Edge shearing, or tearing, of the margin is shown in Fig. f and g, respectively.
• In structural practice this failure is avoided by spacing the rivets at least 1
1
2
diameters away from the edge.
• Bolted connections usually are spaced an even greater distance than this for
satisfactory appearance, and hence this type of failure may usually be
neglected.
• In a rivet joint, the rivets all share the load in shear, bearing in the rivet,
bearing in the member, and shear in the rivet.
• In a bolted joint, shear is taken by clamping friction, and bearing does not
exist.
11

12. exercise
The two 1𝑖𝑛 𝑥 4𝑖𝑛 1018 cold-rolled steel bars are butt-spliced with two
0.5𝑖𝑛 𝑥 6𝑖𝑛 1018 cold-rolled splice plates using four 0.75in diameter 16 UNF
grade 5 bolts as depicted in bellow fig. For a design factor of 𝑛 𝑑 = 1.5
estimate the static load F that can be carried if the bolts lose preload.
12

13. solution
Given data: minimum strengths of 𝑆 𝑦 = 54 𝑘𝑝𝑠𝑖 and 𝑆 𝑢𝑡 = 64kpsi are found for
the members, and minimum strengths of 𝑆 𝑝 = 85kpsi and 𝑆 𝑢𝑡 = 120kpsi for the
bolts are found.
𝐹/2 is transmitted by each of the splice plates, but since the areas of the splice
plates are half those of the center bars, the stresses associated with the plates
are the same. So for stresses associated with the plates, the force and areas
used will be those of the center plates.
13

14. Cont.
Bearing in bolts, all bolts loaded:
𝜎 =
𝐹
2𝑡𝑑
=
𝑆 𝑝
2𝑡𝑑
𝑆 𝑝 = proof stress
𝐹 =
2𝑡𝑑𝑆 𝑝
𝑛 𝑑
=
2 1
3
4
85
1.5
= 85 kip
Bearing in members, all bolts active:
𝜎 =
𝐹
2𝑡𝑑
=
𝑆 𝑦 𝑚𝑒𝑚
𝑛 𝑑
𝐹 =
2𝑡𝑑 𝑆 𝑦 𝑚𝑒𝑚
𝑛 𝑑
=
2 1 0.75 54
1.5
= 54 kip
14

15. Cont.
• Shear of bolt, all bolts active: If the bolt threads do not extend into the shear
planes for four shanks:
𝜏 =
𝐹
4𝜋𝑑2 4
= 0.577
𝑆 𝑝
𝑛 𝑑
𝐹 = 0.577𝜋𝑑2
𝑆 𝑝
𝑛 𝑑
= 0.577π 0.75 2
85
1.5
= 57.8 kip
If bolt threads extend into a shear plane:
𝜏 =
𝐹
4𝐴 𝑟
= 0.577
𝑆 𝑝
𝑛 𝑑
𝐹 =
0.577 4 𝐴 𝑟 𝑆 𝑝
𝑛 𝑑
=
0.577 4 0.551 85
1.5
= 45.9 kip
Edge shearing of member at two margin bolts: from above fig.
𝜏 =
𝐹
4𝑎𝑡
=
0.577 𝑆 𝑦 𝑚𝑒𝑚
𝑛 𝑑
𝐹 =
4𝑎𝑡0.577 𝑆 𝑦 𝑚𝑒𝑛
𝑛 𝑑
=
4 1.125 1 0.577 54
1.5
= 93.5 kip
15

16. Cont.
• Tensile yielding of members across bolt holes:
𝜎 =
𝐹
𝑤 𝑚𝑒𝑚𝑏 − 2𝑥𝑑 𝑏 𝑡
=
𝐹
4 − 2𝑥0.75 𝑡
=
𝑆 𝑦 𝑚𝑒𝑚
𝑛 𝑑
𝐹 =
4 − 1.5 𝑡 𝑆 𝑦 𝑚𝑒𝑚
𝑛 𝑑
=
2.5 1 54
1.5
= 90 kip
• Member yield:
𝐹 =
𝑤𝑡 𝑆 𝑦 𝑚𝑒𝑚
𝑛 𝑑
=
4 1 54
1.5
= 144 kip
• On the basis of bolt shear, the limiting value of the force is 45.9 kip,
assuming the threads extend into a shear plane. However, it would be poor
design to allow the threads to extend into a shear plane. So, assuming a good
design based on bolt shear, the limiting value of the force is 57.8 kip. For
the members, the bearing stress limits the load to 54 kip.
16

17. Shear Joints with Eccentric Loading
• In the previous example, the load distributed equally to the
bolts since the load acted along a line of symmetry of the
fasteners.
• The analysis of a shear joint undergoing eccentric loading
requires locating the center of relative motion between the
two members.
• In bellow fig. let 𝐴1to 𝐴5 be the respective cross-sectional
areas of a group of five pins, rivets, or bolts.
• Under this assumption the rotational pivot point lies at the
centroid of the cross-sectional area pattern of the pins,
rivets, or bolts.
• Using statics, we learn that the centroid G is located by the
coordinates 𝑥 and 𝑦,
𝑥 =
𝐴1 𝑥1 + 𝐴2 𝑥2 + 𝐴3 𝑥3 + 𝐴4 𝑥4 + 𝐴5 𝑥5
𝐴1 + 𝐴2 + 𝐴3 + 𝐴4 + 𝐴5
𝑦 =
𝐴1 𝑦 + 𝐴2 𝑦2 + 𝐴3 𝑦 + 𝐴4 𝑦4 + 𝐴5 𝑦5
𝐴1 + 𝐴2 + 𝐴3 + 𝐴4 + 𝐴5
17

18. example
• Introduce two forces P1 and P2
at the center of gravity ‘G’ of
the rivet system.
• Assuming that all the rivets are
of the same size, the effect of
P1 = P is to produce direct shear
load on each rivet of equal
magnitude. Therefore, direct
shear load on each rivet
18

19. Cont.
• The effect of 𝑝2 = 𝑃 is to produce a turning moment of magnitude
𝑷 × 𝒆 which tends to rotate the joint about the center of gravity ‘G’ of the rivet
system in a clockwise direction.
• Due to the turning moment, secondary shear load on each rivet is produced.
• In order to find the secondary shear load, the following two assumptions are made :
a) The secondary shear load is proportional to the radial distance of the rivet under
consideration from the center of gravity of the rivet system.
b) The direction of secondary shear load is perpendicular to the line joining the center
of the rivet to the center of gravity of the rivet system.
Let 𝐹1, 𝐹2, 𝐹3 ... = Secondary shear loads on the rivets 1, 2, 3...etc. 𝑙1, 𝑙2, 𝑙3... = Radial
distance of the rivets 1, 2, 3 …etc. from the center of gravity ‘G’ of the rivet system.
19

20. Cont.
• The force taken by each bolt depends upon its radial distance from the
centroid; that is, the bolt farthest from the centroid takes the greatest load,
while the nearest bolt takes the smallest. We can therefore write
𝐹1
𝑙1
=
𝐹2
𝑙2
=
𝐹3
𝑙3
= …
𝐹2 = 𝐹1 𝑥
𝑙2
𝑙1
, and 𝐹3 = 𝐹1 𝑥
𝑙3
𝑙1
We know that the sum of the external turning moment due to the eccentric load
and of internal resisting moment of the rivets must be equal to zero.
𝑃 𝑥 𝑒 = 𝐹1 𝑙1 + 𝐹2 𝑙2 + 𝐹3 𝑙3 + …
= 𝐹1 𝑙1 + 𝐹1 𝑥
𝑙2
𝑙1
𝑥 𝑙2 + 𝐹1 𝑥
𝑙3
𝑙1
𝑥 𝑙3 + …
P 𝑥 𝑒 =
𝐹1
𝑙1
𝑙1
2 + 𝑙2
2 + 𝑙3
2 + …
20

21. Cont.
• From the above expression, the value of F1 may be calculated and hence F2
and F3 etc. are known.
• The primary and secondary shear load vector addition to determine the
resultant shear load (R) on each rivet by using the relation
𝑅 = 𝑃𝑠
2 + 𝐹2 + 2𝑃𝑠 𝑥 𝐹 𝑥 𝑐𝑜𝑠𝜃
Where 𝜃 = 𝐴𝑛𝑔𝑙𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡ℎ𝑒 𝑝𝑟𝑖𝑚𝑎𝑟𝑦 𝑠ℎ𝑒𝑎𝑟 𝑙𝑜𝑎𝑑 𝑃𝑆 and secondary shear load F.
where F is the largest force among rivets
• When the secondary shear load on each rivet is equal, then the heavily
loaded rivet will be one in which the included angle between the direct shear
load and secondary shear load is minimum. The maximum loaded rivet
becomes the critical one for determining the strength of the riveted joint.
Knowing the permissible shear stress (𝜏), the diameter of the rivet hole may
be obtained by using the relation,
Maximum resultant shear load 𝑅 =
𝜋
4
𝑥 𝑑2 𝑥 𝜏
21

22. Exercise
Shown in Fig. 8–28 is a 15- by 200-mm rectangular steel bar cantilevered to a
250-mm steel channel using four tightly fitted bolts located at A, B, C, and D. For
F = 16 kN load find
(a) The resultant load on each bolt
(b) The maximum shear stress in each bolt
(c) The maximum bearing stress
(d) The critical bending stress in the bar
22

23. Solution
(a) Point O, the centroid of the bolt group in Fig. 8–28, is found by symmetry. If
a free-body diagram of the beam were constructed, the shear reaction V
would pass through O and the moment reactions M would be about O. These
reactions are
𝑃 = 16𝑘𝑁 , 𝑀 = 𝑃 𝑥 𝑒 = 16 425 = 6800 𝑁. 𝑚
In Fig. 8–29, the bolt group has been drawn to a larger scale and the reactions
are shown. The distance from the centroid to the center of each bolt is
𝑟 = 60 2 + 75 2 = 96.0 mm
23

24. Cont.
The primary shear load per bolt is
𝐹′ =
𝐹
𝑛𝑜. 𝑜𝑓 𝑏𝑜𝑙𝑡
=
16
4
= 4kN
Since the secondary shear forces are
equal, 𝑟𝐴 = 𝑟𝐵 = 𝑟𝐶 = 𝑟 𝐷 = r
𝑀 =
𝐹𝐴
"
𝑟𝐴
( 𝑟𝐴
2
+ 𝑟𝐵
2
+ 𝑟𝐶
2
+ 𝑟 𝐷
2
)
𝑀 =
𝐹"
𝑟
(𝑟2
+ 𝑟2
+ 𝑟2
+ 𝑟2
)
𝐹"
=
𝑀𝑟
4𝑟2
𝐹"
=
6800 𝑥 96
4 96 2
= 17.7 𝑘𝑁
𝐹"
= 𝐹𝐴
"
= 𝐹𝐵
"
= 𝐹𝐶
"
= 𝐹 𝐷
"
determine the resultant shear load (R) on
each bolt.
𝑅 𝐴 = 𝐹′2
+ 𝐹"2
+ 2𝐹′ 𝐹𝐴
"
𝐶𝑂𝑆𝜃 𝐴
24

25. Cont.
𝑅 𝐵 = 𝐹′2
+ 𝐹"2
+ 2𝐹′ 𝐹" 𝐶𝑂𝑆𝜃 𝐵
𝑅 𝐶 = 𝐹′2
+ 𝐹"2
+ 2𝐹′ 𝐹" 𝐶𝑂𝑆𝜃 𝐶
𝑅 𝐷 = 𝐹′2
+ 𝐹"2
+ 2𝐹′ 𝐹" 𝐶𝑂𝑆𝜃 𝐷
Let us now find the angles between the direct and secondary shear load for
these four bolts.
𝐶𝑜𝑠𝜃 𝐴 =
75
𝑟𝐴
=
75
96
= 0.78125
𝐶𝑜𝑠𝜃 𝐵 =
75
𝑟 𝐵
=
75
96
= 0.78125
𝐶𝑜𝑠𝜃 𝐶 =
60
𝑟 𝐶
=
60
96
= 0.625
𝐶𝑜𝑠𝜃 𝐷 =
60
𝑟 𝐷
=
60
96
= 0.625
25

26. Cont.
• The primary and secondary shear forces are plotted to scale in Fig. 8–29 and
the resultants obtained by using the parallelogram rule. The magnitudes are
found by analysis to be
𝑅 𝐴 = 𝑅 𝐵 = 𝐹′2
+ 𝐹𝐴
"2
+ 2𝐹′ 𝐹" 𝐶𝑂𝑆𝜃 𝐴
= 42 + 17.72 + 2 4 17.7 0.78125
= 21.0 kN
𝑅 𝐶 = 𝑅 𝐷 = 𝐹′2
+ 𝐹𝐶
"2
+ 2𝐹′ 𝐹" 𝐶𝑂𝑆𝜃 𝐶
= 42 + 17.72 + 2 4 17.7 0.78125
= 14.8 kN
26

27. Cont.
(b) Bolts A and B are critical because they carry the largest shear load.
the bolt will tend to shear across its minor diameter. Therefore the shear-
stress area is As = 144 mm2 , and so the shear stress is
𝜏 =
𝐹
𝐴 𝑠
=
21.0 10 3
144
= 146Mpa
27

28. Cont.
28
(c) The channel is thinner than the bar, and so the largest bearing stress is due
to the pressing of the bolt against the channel web. The bearing area is Ab = td
= 10(16) = 160 mm2 . Thus the bearing stress is
𝜎 = −
𝐹
𝐴 𝑏
= −
21.0 10 3
160
= −131Mpa
(d) The critical bending stress in the bar is assumed to occur in a section
parallel to the y axis and through bolts A and B. At this section the bending
moment is M = 16(300 + 50) = 5600 N · m
The second moment of area through this section is obtained by the use of the
transfer formula, as follows:

29. Cont.
𝐼 = 𝐼 𝑏𝑎𝑟 − 2 𝐼ℎ𝑜𝑙 + 𝑑2
𝐴 , 𝐴 = td
=
15 200 3
12
− 2
15 16 3
12
+ 60 2 15 16 = 8.26 10 6 𝑚𝑚4
Then 𝜎 =
𝑀𝑐
𝐼
=
5600 100
8.26 10 6 10 3 = 67.8𝑀𝑝𝑎
29

30. Welding Joints
Introduction
Study welding symbol terminology used in communicating
specification on engineering drawing
Learn how to calculate stresses in on welds and welded joint
and the strength of weld and parent materials.
30

31. Welds, Joints and Weldments• Weld
 A zone of metal that joints two or more components created by localized melting,
mixing and cooling of metal from the parent components and an expendable filler
metal of compatible composition
Melting is accomplished using a heat source such as an electric arc or oxy-acetylene
flame
• Joint
 A configuration of parent components
of defining a gap to be filled by a weld
• Weldment
A structural component constructed
by welding several components together
31

32. Welding process
• Arc welding – uses a welding power supply to create an electric arc between an electrode and
the base metals at the welding point
– Consumable electrode methods
• Shield metal arc welding (SMAW), or stick welding
• Gas metal arc welding (GMSW) is a semi-automatic or automatic or automatic welding
process that uses a continuous wire feed as an electrode and an inert or semi –inert
shield gas to protect the weld from contamination. When using an inert gas as shielding it
is known as metal inert gas (MIG)welding
– Non- consumable electrode methods
• Gas tungsten arc welding (GTAW), or tungsten inert gas(TIG) welding is a manual
welding process that uses a non consumable electrode made of tungsten, an inert or
semi inert gas mixture, and separate filler material.
32

33. Types of Welding
33

34. Types of Joint
• Describes the configuration of parent
34

35. Welding Symbol
35

36. Cont’d…
36

37. Examples
37

38. Theoretical stress in butt weld
38

39. Loading in a fillet welding
39

40. 40

41. Cont.
• Figure 9.8 illustrates a typical transverse fillet weld. In Fig. 9.9 a portion of
the welded joint has been isolated from Fig. 9.8 as a free body.
• At angle θ the forces on each weldment consist of a normal force 𝐹𝑛 and a
shear force 𝐹𝑠. Summing forces in the x and y directions gives
𝐹𝑠 = 𝐹 sin 𝜃
𝐹𝑛 = 𝐹 cos 𝜃
Using the law of sines for the triangle in Fig. 9.9 yields
𝑡
𝑠𝑖n450
=
ℎ
𝑠𝑖n(1800 − 450 − 𝜃)
=
ℎ
𝑠𝑖n(1350 − 𝜃)
=
2ℎ
cos 𝜃 + sin 𝜃
Solving for throat thickness t gives
𝑡 =
ℎ
cos 𝜃 + sin 𝜃
41

42. Cont.
 The nominal stresses at the angle 𝜃 in the weldment, 𝜏 𝑎𝑛𝑑 𝜎, 𝑎𝑟𝑒a = 𝑙𝑡
𝜏 =
𝐹𝑠
𝐴
=
𝐹𝑠𝑖𝑛𝜃(𝑐𝑜𝑠𝜃 + 𝑠𝑖𝑛𝜃)
ℎ𝑙
=
𝐹
ℎ𝑙
(sin𝜃𝑐𝑜𝑠𝜃 + 𝑠𝑖𝑛2
𝜃)
𝜎 =
𝐹 𝑛
𝐴
=
𝐹𝑐𝑜𝑠𝜃(𝑐𝑜𝑠𝜃 +𝑠𝑖𝑛𝜃)
ℎ𝑙
=
𝐹
ℎ𝑙
(𝑐𝑜𝑠2 𝜃 + 𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃)
 The von miss stress 𝜎′
at angle 𝜃 is
𝜎′ = (𝜎2 + 3𝜏2)
1
2 =
𝐹
ℎ𝑙
[(𝑐𝑜𝑠2 𝜃 + 𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃)2 +3(𝑠𝑖𝑛2 𝜃 + 𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃)2]
1
2
 The largest von misses stress occurs at 𝜃 = 62.50
with a value of 𝜎′
= 2.16𝐹 (ℎ𝑙).
The corresponding values of 𝜏 and 𝜎 are 𝜏 = 1.196𝐹 (ℎ𝑙) and 𝜎 = 0623𝐹 ℎ𝑙 .
42

43. Parallel fillet welds.
• Use distortion energy for significant stresses.
• Circumscribe typical cases by code.
• For this model, the basis for weld analysis or design employs
𝜏 =
𝐹
0.707ℎ𝑙
=
1.414𝐹
ℎ𝑙
43

44. Stresses in Welded Joints in Torsion
• Figure 9–12 illustrates a cantilever of length l welded to a column by two fillet welds.
• The reaction at the support of a cantilever always consists of a shear force V and a
moment M. The shear force produces a primary shear in the welds of magnitude
𝜏′ =
𝑉
𝐴
where A is the throat area of all the welds
• The moment at the support produces secondary shear or torsion of the welds, and
this stress is given by the equation
𝜏′′ =
𝑀𝑟
𝐽
where r is the distance from the centroid of the weld group to the point in the weld of
interest and J is the second polar moment of area of the weld group about the centroid
of the group.
When the sizes of the welds are known, these equations can be solved and the results
combined to obtain the maximum shear stress. Note that r is usually the farthest
distance from the centroid of the weld group
44

45. 45

46. Cont.
• Figure 9.13 shows two welds in a group. The rectangles represent the
throat areas of the welds.
• Weld 1 has a throat thickness 𝑡1 = 0.707ℎ1, and weld 2 has a throat
thickness 𝑡2 = 0.707ℎ2. Note that ℎ1 and ℎ2 are the respective weld sizes.
The throat area of both welds together is
𝐴 = 𝐴1 + 𝐴2 = 𝑡1 𝑑 + 𝑡2 𝑏
The x axis in fig. 9.13 passes through the centroid 𝐺1 of weld 1. the second
moment of area about this axis is
𝐼 𝑥 =
𝑡1 𝑑3
12
Similarly, the second moment of area about an axis through 𝐺1parallel to the y
axis is
𝐼 𝑦 =
𝑑𝑡1
3
12
46

47. 47

48. Cont.
Thus the second polar moment of weld 1 about its own centroid is
𝐽 𝐺1
= 𝐼 𝑥 + 𝐼 𝑦 =
𝑡1 𝑑3
12
+
𝑑𝑡1
3
12
In a similar manner, the second polar moment of area of weld 2 about its centroid is
𝐽 𝐺2
=
𝑏𝑡2
3
12
+
𝑡2 𝑏3
12
The centroid G of the weld group is located at
𝑥 =
𝐴1 𝑋1+ 𝐴2 𝑋2
𝐴
𝑦 =
𝐴1 𝑦1+ 𝐴2 𝑦2
𝐴
Using figure 9.13 again, we see that the distance 𝑟1 𝑎𝑛𝑑 𝑟2 from 𝐺1 𝑎𝑛𝑑 𝐺2 𝑡𝑜 𝐺,
respectively are
𝑟1 = ( 𝑥 − 𝑥1)2+ 𝑦2 𝑟2 = (𝑦2 − 𝑦)2+ (𝑥2 − 𝑥)2
48

49. Cont.
• Now, using parallel axis theorem, we find the second polar moment of area of the weld
group to be
𝐽 = (𝐽 𝐺1
+ 𝐴1 𝑟1
2
) + (𝐽 𝐺2
+ 𝐴2 𝑟2
2
)
• The distance r must be measured from G and the moment M computed about G.
• Setting the weld thicknesses 𝑡1 and 𝑡2 to unity leads to the idea of treating each fillet
weld as a line.
• The resulting second moment of area is then a unit second polar moment of area.
• The advantage of treating the weld size as a line is that the value of 𝐽 𝑢 is the same
regardless of the weld size. Since the throat width of a fillet weld is 0.707h, the
relationship between J and the unit value is 𝐽 = 0.707ℎ
49

50. 50

51. 51

52. Exercise
52

53. solution
Weld length(l) = 30mm +50mm +50mm = 130mm
𝑥 =
30 15 +50 0 +50(25)
130
= 13.08mm
𝑦 =
30 50 +50 25 +50(0)
130
= 21.15mm
𝑀 = 𝐹𝑅 = 𝐹 200 − 13.08
= 186.92 𝐹 (𝑀 𝑖𝑛 𝑁. 𝑚, 𝐹 𝑖𝑛 𝑘𝑁)
𝑟1 = (50 − 13.08)2+ (50 − 21.15)2= 28.92mm,
𝑟2 = 13.082 + (25 − 21.15)2= 13.63mm
𝑟3 = (25 − 13.08)2+ 21.152
𝐽 𝐺1
=
1
12
0.707 5 303
= 7.954(103
)𝑚𝑚4
𝐽 𝐺2
= 𝐽 𝐺3
=
1
12
0.707 5 503
= 36.82(103
)𝑚𝑚4
𝐽 = 𝑖=1
3
(𝐽𝑖 + 𝐴𝑖 𝑟𝐺 𝑖
2
)
= 7.954 103
+ 0.707 5 30 28.922
+ 36.82 103
+ 0.707 5 50 13.632
+ 36.82 103
+ 0.707 5 50 24.282
= 307.3(103
)𝑚𝑚4
53

54. Cont.
𝛼 = 𝑡𝑎𝑛−1
21.15
50 − 13.08
= 29.810
𝑟 = 21.152 + 50 − 13.08 2 = 42.55mm
𝑝𝑟𝑖𝑚𝑎𝑟𝑦 𝑠ℎ𝑎𝑒𝑟(𝜏 𝑖𝑛 𝑀𝑝𝑎, 𝐹 𝑖𝑛 𝑘𝑁)
𝜏′ =
𝐹
𝐴
=
𝐹 1𝑂3
0.707(5)(130)
= 2.176F
𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑠ℎ𝑒𝑎𝑟:
𝜏"
=
𝑀𝑟
𝐽
=
186.92𝐹(103
)(42.55)
307.3(103)
= 25.88F
𝜏 𝑚𝑎𝑥 = 25.88𝐹𝑠𝑖𝑛29.810 2 + 25.88𝐹𝑐𝑜𝑠29.810 + 2.176𝐹 2
𝜏 𝑚𝑎𝑥 = 𝜏 𝑎𝑙𝑙 → 27.79𝐹 = 140 → 𝐹 = 5.04𝑘𝑁
54

55. 55