Prepared by: KDK

1
Prepared by: Kerem Doruk Katlıoğlu
Room: G-156
Phone: 5246
E-mail: [email protected]
Date: 10.12.2014
ME 307 – MACHINE ELEMENTS I
TUTORIAL 10
“DESIGN OF PERMANENT JOINTS”
METU – ME 307 Machine Elements I – Fall 2014
2
Problem 2
3 plates are connected by 6 rivets as shown in the figure. The first row has 3 rivets
each with 15mm diameter, the second row has a single rivet with 20mm diameter
and the third row has 2 rivets each with 18mm diameter. Find the forces and stresses
acting on the rivets, and determine the safety factor of the joint. (For the rivets 𝑆𝑦 =
250𝑀𝑃𝑎 , 𝑆𝑠𝑦 = 80𝑀𝑃𝑎 and for the plates 𝑆𝑦 = 200𝑀𝑃𝑎 )
15kN
120
8
Plate 1
10
Main Plate
15kN
Plate 2
6
Row 1
Row 2
Row 3
Solution
The shear stresses on the shear areas of the rivets will be the same. It can be seen
clearly from the free body diagram of the main plate that for the rivets in rows 1
and 2 there are 2 shear areas (double shear) for each rivet. In the third row there is
only a single shear area for each rivet. There are totally 10 shear areas.
METU – ME 307 Machine Elements I – Fall 2014
3
Analysis of the rivets in the second row:
𝐹1′
𝐹1′
𝐹1′
𝐹2′
Bearing stress on plate 1: 𝜎𝑏 =
15kN
𝐹1′
𝐹1′
𝐹1′
Then, 𝜏 ′ =
𝐹
∑𝐴
Bearing stress on plate 2: 𝜎𝑏 =
and
𝐹𝑖′
=
𝐹∙𝐴𝑖
∑𝐴
𝜋𝑑12
Area of the rivet in the second row: 𝐴2 =
Area of the rivets in the third row: 𝐴3 =
=
4
𝜋𝑑22
4
𝜋𝑑32
4
𝜋152
=
=
= 176.72𝑚𝑚2
4
𝜋202
4
𝜋182
4
= 314.16𝑚𝑚2
= 254.47𝑚𝑚
2
Then the total shear area: ∑ 𝐴 = 2(3𝐴1 + 𝐴2 ) + 2𝐴3 = 2198𝑚𝑚2
′
Shear stress: 𝜏 =
=
=
𝐹2′
𝑑2 ∙ℎ2
2144
20∙8
2∙𝐹2′
𝑑2 ∙ℎ𝑚
=
2144
20∙6
= 13.402𝑀𝑃𝑎
=
2∙2144
20∙10
= 21.444𝑀𝑃𝑎
= 17.87𝑀𝑃𝑎
where 𝐹 = 15𝑘𝑁 .
Area of the rivets in the first row: 𝐴1 =
𝐹
∑𝐴
𝑑2 ∙ℎ1
Bearing stress on the main plate: 𝜎𝑏 =
𝐹3′
𝐹3′
𝐹2′
𝐹2′
15000
2198
= 6.824𝑀𝑃𝑎
𝐹 ∙ 𝐴1 15 ∙ 176.72
=
=
= 1206𝑁
∑𝐴
2198
𝐹 ∙ 𝐴2 15 ∙ 314.16
𝐹2′ =
=
= 2144𝑁
∑𝐴
2198
𝐹 ∙ 𝐴3 15 ∙ 254.47
𝐹3′ =
=
= 1737𝑁
∑𝐴
2198
=
Bearing stress on the main plate: 𝜎𝑏 =
Bearing stress on plate 2: 𝜎𝑏 =
𝐹1′
𝑑1 ∙ℎ2
1206
15∙8
2∙𝐹1′
𝑑1 ∙ℎ𝑚
=
1206
15∙6
𝑑3 ∙ℎ2
𝑑3 ∙ℎ𝑚
=
1737
18∙6
=
1737
18∙10
= 9.650𝑀𝑃𝑎
= 16.083𝑀𝑃𝑎
𝑆𝑠𝑦
𝜏
80
= 6.826 = 11.72
2∙1206
15∙10
𝑆𝑦
𝜎𝑏
=
𝑆𝑦
𝜎𝑏
200
21.444
250
= 21.444 = 11.66, the safety factor for
= 9.33
Analysis of the main plate:
Thickness of the main plate: ℎ𝑚 = 10𝑚𝑚
Row 3:
Cross-sectional area:
= 10.052𝑀𝑃𝑎
=
𝐹3′
𝐹3′
The safety factor for the shear failure of the rivets: 𝑛 =
plate bearing stress is 𝑛𝑏 =
Analysis of the rivets in the first row:
𝑑1 ∙ℎ1
Bearing stress on plate 2: 𝜎𝑏 =
factor for rivet bearing stress is 𝑛𝑏 =
𝐹1′
𝐹1′
Bearing stress on the main plate: 𝜎𝑏 =
The maximum bearing stress is on the main plate (second row). Then, the safety
Note that all rivets have the same shear stress value.
Bearing stress on plate 1: 𝜎𝑏 =
Analysis of the rivets in the third row:
= 16.083𝑀𝑃𝑎
𝐴𝑚,3 = ℎ𝑚 (120 − 2𝑑3 ) = 10(120 − 2 ∙ 18)
= 840𝑚𝑚2
From the figure, 𝐹 = 15𝑘𝑁
15kN
F
Then, the tensile stress on the plate at row 3:
𝜎𝑚,3 =
𝐹
15000
=
= 17.857𝑀𝑃𝑎
𝐴𝑚,3
840
= 13.402𝑀𝑃𝑎
METU – ME 307 Machine Elements I – Fall 2014
Row 3
4
Row 2:
Cross-sectional area:
Row 2:
Cross-sectional area:
𝐹3′
𝐴𝑚,2 = ℎ𝑚 (120 − 𝑑2 ) = 10(120 − 20)
= 1000𝑚𝑚2
15kN
From the figure,
𝐹 = 15𝑘𝑁 − 2 ∙
𝐹3′
= 11526𝑁
F
𝐹3′
Then, the tensile stress on the plate at row 2:
𝜎𝑚,2
𝐹
11526
=
=
= 11.526𝑀𝑃𝑎
𝐴𝑚,2
1000
𝜎1,2 =
15kN
From the figure,
𝐹 = 15𝑘𝑁 − 2 ∙ 𝐹2′ − 2 ∙ 𝐹3′ = 7238𝑁
F
2 ∙ 𝐹2′
Then, the tensile stress on the plate at
row 1:
𝜎𝑚,1 =
𝐹
7238
=
= 9.651𝑀𝑃𝑎
𝐴𝑚,1
750
F
𝐹1′
P1
𝐹1′
𝐹
2144
=
= 2.680𝑀𝑃𝑎
𝐴1,2
800
Analysis of plate 2:
Thickness of plate 2: ℎ2 = 6𝑚𝑚
𝐹3′
𝐴𝑚,1 = ℎ𝑚 (120 − 3𝑑1 ) = 10(120 − 3 ∙ 15)
= 750𝑚𝑚2
𝐹1′
Then, the tensile stress on the plate at row
2:
Row 2
Row 1:
Cross-sectional area:
𝐴1,2 = ℎ1 (120 − 𝑑2 ) = 8(120 − 20)
= 800𝑚𝑚2
𝐹 = 𝑃1 − 3 ∙ 𝐹1′ = 2144𝑁
Row 1:
Cross-sectional area:
𝐴2,1 = ℎ2 (120 − 3𝑑1 ) = 6(120 − 3 ∙ 15)
= 450𝑚𝑚2
′
𝑃2 = 3 ∙ 𝐹1 + 𝐹2′ + 2 ∙ 𝐹3′ = 9237𝑁
𝐹 = 𝑃2 = 9237𝑁
𝐹3′
Row 1
P2
F
Then, the tensile stress on the plate at row
Analysis of plate 1:
Thickness of plate 1: ℎ1 = 8𝑚𝑚
𝜎2,1
Row 1:
Cross-sectional area:
𝐴1,1 = ℎ1 (120 − 3𝑑1 ) = 8(120 − 3 ∙ 15)
= 600𝑚𝑚2
𝑃1 = 3 ∙ 𝐹1′ + 𝐹2′ = 5763𝑁
𝐹 = 𝑃1 = 5763𝑁
Row 2:
Cross-sectional area:
P1
F
𝐹1′
𝐴2,2 = ℎ2 (120 − 𝑑2 ) = 6(120 − 20) = 600𝑚𝑚2
𝐹 = 𝑃2 − 3 ∙ 𝐹1′ = 5618𝑁
P2
Then, the tensile stress on the plate at row 1:
𝜎1,1
1:
𝐹
9237
=
=
= 20.527𝑀𝑃𝑎
𝐴2,1
450
𝐹1′
𝐹1′
𝐹
5763
=
=
= 9.605𝑀𝑃𝑎
𝐴1,1
600
METU – ME 307 Machine Elements I – Fall 2014
F
5
Then, the tensile stress on the plate at row 2:
𝜎2,2
Row 3:
Cross-sectional area:
60
h
30
𝐴2,3 = ℎ2 (120 − 2𝑑3 ) = 6(120 − 2 ∙ 18)
= 504𝑚𝑚2
𝐹 = 𝑃2 − 3 ∙ 𝐹1′ − 𝐹2′ = 3474𝑁
P2
𝐹1′
h
30
𝐹2′
F
𝐹1′
60
h
𝐹1′
Then, the tensile stress on the plate at
row 2:
𝜎2,3
Problem 3:
𝐹
5618
=
=
= 9.364𝑀𝑃𝑎
𝐴2,2
600
h
60
h
h
h
30
h
h
𝐹
3474
=
=
= 6.893𝑀𝑃𝑎
𝐴2,3
504
30
70
30
60
5 kN
The maximum tensile stress occurs in plate 2 at row 1: 𝜎2,1 = 20.527𝑀𝑃𝑎
The safety factor for plate tension:
𝑛𝑡 =
𝑆𝑦
200
=
= 9.74
𝜎2,1 20.527
Therefore, the plate bearing stress is the most critical.
Then, the safety factor of the joints 𝑛 = 9.33
The two beams are connected to each other and one of them is connected to support
by welded joints as shown in figure below. Determine the weld leg size h such that
welded joints can carry 5kN load. All the dimensions are in mm, Ssy=80MPa, and
take the factor of safety, n=3.
Solution: There are two joints in the problem.
Y’
B
y
y
𝑥
B
x
x
𝑦
rA
𝑦
A
β
METU – ME 307 Machine Elements I – Fall 2014
𝑥
rB
rA
X’
α
A
6
The welded joint on the left: Let a=60mm, b=30mm
Weld throat area:
Since t=0.707h,
Then the resultant shear stress at point A is,
𝜏𝐴 = √(𝜏𝐴𝚤 )2 + (𝜏𝐴𝚤𝚤 )2 + 2𝜏𝐴𝚤 𝜏𝐴𝚤𝚤 𝑐𝑜𝑠(𝛽)
𝐴 = [2𝑎 + 5𝑏] ∙ 𝑡 = [2 ∙ 60𝑚𝑚 + 5 ∙ 30𝑚𝑚] ∙ 0.707ℎ = 191ℎ 𝑚𝑚2
Centroid of weld group:
𝑥=
[2𝑎 ∙
𝑎
𝑏
+ 2𝑏 ∙ + 𝑏 ∙ 𝑏] 𝑡
2
2
= 20𝑚𝑚
𝐴
𝜏=
𝑦 = 45 𝑚𝑚
Polar moment of area, J=Ix+Iy
1
1
1
𝑏 2
1 3
𝐼𝑥 = 2 [ 𝑎𝑡 3 + 𝑎𝑡𝑦 2 ] + 2 [ 𝑏 3 𝑡 + 𝑏𝑡𝑏 2 ] + 2 [ 𝑏𝑡 3 + 𝑏𝑡 ( ) ] +
𝑏 𝑡
12
12
12
2
12
𝐼𝑥 = 5.3ℎ3 + 224300ℎ 𝑚𝑚4
Note that first term (5.3h3) in the Ix can be neglected.
Welded joint on the right
Let a=60mm
Weld throat area:
𝐴 = (2𝑎)0.707ℎ = (2 ∙ 60𝑚𝑚)0.707ℎ = 85ℎ 𝑚𝑚2
224300h mm 4
The centroid of the weld group:
2
1
𝑎
1
1
𝑏 2
𝐼𝑦 = 2 [ 𝑎3 𝑡 + 𝑎𝑡 ( − 𝑥 ) ] + 2 [ 𝑏𝑡 3 + 𝑏𝑡𝑥 2 ] + 2 [ 𝑏 3 𝑡 + 𝑏𝑡 (𝑥 − ) ]
12
2
12
12
2
1 3
+
𝑏𝑡 + 𝑏𝑡(𝑥 − 𝑏)2 = 2.65ℎ3 + 57270ℎ 𝑚𝑚4
12
𝑥 = 30𝑚𝑚
𝑦 = 30𝑚𝑚
Polar moment of area, J=Ix+Iy
1 3
1
𝑡𝑎 ] = 2 [ 603 ∙ 0.707ℎ] = 25452ℎ 𝑚𝑚4
12
12
1
1
𝐼𝑦 = 2 [ 𝑎𝑡 3 + 𝑎𝑡𝑥 2 ] = 2 [ 60 ∙ (0.707ℎ)3 + 60(0.707ℎ)302 ]
12
12
= 3.534ℎ3 + 76356ℎ 𝑚𝑚4
𝐼𝑥 = 2 [
Similarly, first term (2.65h3) in the Iy can be neglected.
Iy
57270h mm 4
J= Iy+ Iy=281600h mm4
The torsion of 5 kN force about the centroid of the first weld group is:
𝑀 = 5𝑘𝑁 ∙ 200𝑚𝑚 = 1000𝑁𝑚
𝜏𝚤 =
A
(45mm)2
76356h mm 4
J= Iy+ Iy=101808h mm4
𝐹
5𝑘𝑁
=
𝐴 191ℎ 𝑚𝑚2
As can be seen from the figure above critical points are A (or B).
The secondary shear stress at point A,
1000 Nm (40mm)2
First term (3.534h3) in the Iy can be neglected.
Iy
The primary shear stress is,
MrA
J
𝑆𝑠𝑦 80
=
≈ 27𝑀𝑃𝑎
𝑛
3
Then, by equating the resultant shear stress at point A to the allowable shear
stress, the weld led size h for the weld group on the left can be determined as
8.6 mm.
Due to symmetry,
Ix
where the angle between the primary and the secondary shear stress at point A is,
β=48o. The allowable shear stress in the welds is;
The torsion of 5 kN force about the centroid of the second weld group is,
𝑀 = 5𝑘𝑁 ∙ 60𝑚𝑚 = 300 𝑁𝑚
The primary shear stress is,
𝜏𝚤 =
281600h mm4
METU – ME 307 Machine Elements I – Fall 2014
𝐹
5𝑘𝑁
=
𝐴 85ℎ 𝑚𝑚2
7
The secondary shear stress is maximum at point A(or at point B),
A
MrA
J
300 Nm (30mm)2
(30mm) 2
101808h mm4
Then, the resultant shear stress at point x is,
𝜏𝐴 = √(𝜏𝐴𝚤 )2 + (𝜏𝐴𝚤𝚤 )2 + 2𝜏𝐴𝚤 𝜏𝐴𝚤𝚤 𝑐𝑜𝑠(𝛽)
where the angle between primary and secondary shear stress at point A is α=45o
The allowable shear stress in the welds is;
𝜏=
Problem 4:
An I-shaped beam made of AISI 1040 HR steel is welded to the fixed support as shown in
the figure. The beam is subjected to the load F which is varying between 0 and 5 kN. The
load F is applied on the end of the beam at the mid-plane of the section. Find the factor of
safety of the welded joint. Take 𝑆𝑠𝑢 = 0.5 ∙ 𝑆𝑢𝑡 .
Reliability: 𝑅 = 0.9
𝑎 = 250𝑚𝑚, 𝑏 = 75𝑚𝑚, 𝑐 = 10𝑚𝑚, 𝑑 = 150𝑚𝑚, ℎ = 8𝑚𝑚
F
𝑆𝑠𝑦 80
=
≈ 27 𝑀𝑃𝑎
𝑛
3
h
The by equating the resultant shear stress at point A to the allowable shear
stress, the weld leg size h for the weld group on the right can be determined
as 6.4 mm.
d
c
Therefore, the weld leg size for the given loading and design criteria is to
be taken as 9 mm.
h
b
a
Solution:
The force F produces primary shear stress and
the moment M produces secondary shear stress
in the welds.
y
A
B
𝑦
F
M
x
G
C
D
The throat area is:
𝐴 = 𝑡 ∙ (4 ∙ 𝑏 − 2 ∙ 𝑐)
where 𝑡 = 0.707ℎ = 5.657𝑚𝑚
𝐴 = 5.657 ∙ (4 ∙ 75 − 2 ∙ 10) = 1584𝑚𝑚2
Due to symmetry, the distance to the edges from
the centroid of the weld pattern is:
𝑦 = 75𝑚𝑚
All the points at the corners will have the same stress during loading. Only point A will be
analyzed.
The second moment of area based on the weld throat:
METU – ME 307 Machine Elements I – Fall 2014
8
1
𝐼𝑥 = 2 ∙ ( ∙ 𝑏 ∙ 𝑡 3 + 𝑏 ∙ 𝑡 ∙ (𝑦)2 ) + 2
12
1
1
∙ ( ∙ 𝑏 ∙ 𝑡3 −
∙ 𝑐 ∙ 𝑡 3 + 𝑏 ∙ 𝑡 ∙ (𝑦 − 𝑐)2 − 𝑐 ∙ 𝑡 ∙ (𝑦 − 𝑐)2 )
12
12
1
𝐼𝑥 = 2 ∙ ( ∙ 75 ∙ 5.6573 + 75 ∙ 5.657 ∙ (𝑦 )2 ) + 2
12
1
∙ ( ∙ (75 − 10) ∙ 5.6573 + (75 − 10) ∙ 5.657 ∙ (75 − 10)2 )
12
𝐼𝑥 = 7.884 ∙ 106 𝑚𝑚4
Note that terms having t3 are negligible due to low value of weld thickness compared
to weld length.
𝑘𝑎 = 0.47 (Formula sheet, Figure 7-8: An as-forged surface should be used for welds
unless a superior finish is specified.)
𝐹𝑚𝑎𝑥 = 5000𝑁 & 𝐹𝑚𝑖𝑛 = 0
𝑘𝑏 = 0.85 (Formula sheet: for 50𝑚𝑚 ≥ 𝑑𝑒 ≥ 8𝑚𝑚)
𝑘𝑐 = 0.897 (Formula sheet: Table 7-7, for 𝑅 = 0.90)
𝑘𝑑 = 1 (Nothing specified)
Then, 𝐹𝑎 =
𝐹𝑚 =
𝐹𝑚𝑎𝑥 −𝐹𝑚𝑖𝑛
𝐹𝑚𝑎𝑥 +𝐹𝑚𝑖𝑛
2
2
= 2500𝑁 → 𝑀𝑎 = 𝐹𝑎 ∙ 𝑎 = 625000𝑁 ∙ 𝑚𝑚
= 2500𝑁 → 𝑀𝑚 = 𝐹𝑚 ∙ 𝑎 = 625000𝑁 ∙ 𝑚𝑚
The effective dimension 𝑑𝑒 will be obtained by equating the area of material stress at and
above the 95% of the maximum stress to the same area in rotating beam specimen. The
weld sections on the outer surfaces of the I-beam will have equivalent stress.
𝜋
𝐴0.95𝜎 = 2 ∙ (0.05 ∙ 𝑑) ∙ 𝑏 = (𝑑𝑒2 − (0.95 ∙ 𝑑𝑒 )2 )
4
𝑑
8∙(0.05∙ )∙𝑏
8∙3.75∙75
2
Then, 𝑑𝑒 = √
= √𝜋(1−0.952) = 42.9𝑚𝑚
𝜋(1−0.952 )
𝑘𝑒 =
Average alternating and mean primary shear stresses at A:
𝐹𝑎
2500
𝜏𝑎′ =
=
= 1.578𝑀𝑃𝑎
𝐴
1584
𝐹𝑚
2500
′
𝜏𝑚
=
=
= 1.578𝑀𝑃𝑎
𝐴
1584
Average alternating and mean secondary shear stresses at A:
𝑀𝑎 ∙ 𝑦 625000 ∙ 75
𝜏𝑎′′ =
=
= 5.945𝑀𝑃𝑎
𝐼𝑥
7.884 ∙ 106
𝑀𝑚 ∙ 𝑦 625000 ∙ 75
′′
𝜏𝑚
=
=
= 5.945𝑀𝑃𝑎
𝐼𝑥
7.884 ∙ 106
Since primary and secondary shear stresses are perpendicular to each other:
1
𝐾𝑓
where 𝐾𝑓 = 2.7 (Shigley 8th Edition, Table 9-5 for end of parallel fillet weld)
Then, 𝑘𝑒 =
1
2.7
= 0.37
𝑘𝑓 = 1 (Nothing specified)
Corrected endurance limit: 𝑆𝑒 = 𝑘𝑎 ∙ 𝑘𝑏 ∙ 𝑘𝑐 ∙ 𝑘𝑑 ∙ 𝑘𝑒 ∙ 𝑘𝑓 ∙ 𝑆𝑒′ = 34.5𝑀𝑃𝑎
𝑆𝑠𝑒 = 0.5 ∙ 𝑆𝑒 = 17.3𝑀𝑃𝑎
𝑆𝑠𝑢 = 0.5 ∙ 𝑆𝑢𝑡 (stated in the problem)
𝑆𝑠𝑢 = 260𝑀𝑃𝑎
By using Goodman Theory of failure:
1
𝑛
𝜏
𝜏
= 𝑆 𝑎 + 𝑆𝑚
𝑠𝑒
𝑠𝑢
1 6.151 6.151
=
+
→ 𝑛 = 2.63
𝑛
17.3
260
𝜏𝑎 = √(𝜏𝑎′ )2 + (𝜏𝑎′′ )2 = √1.5782 + 5.9452 = 6.151𝑀𝑃𝑎
′ )2 + (𝜏 ′′ )2 = √1.5782 + 5.9452 = 6.151𝑀𝑃𝑎
𝜏𝑚 = √(𝜏𝑚
𝑚
The electrode weld materials are usually much stronger than the materials in the
machine elements, so the strength analysis is done for the weaker material. So only
the I-shaped beam is analyzed.
Tensile strength for AISI 1040 HR is 𝑆𝑢𝑡 = 520𝑀𝑃𝑎. (Table A-18 (Shigley 8th
Edition))
Endurance limit of the test specimen: 𝑆𝑒′ = 0.5 ∙ 𝑆𝑢𝑡 = 260𝑀𝑃𝑎
Since Goodman theory is used, a first cycle check should be done for yielding.
𝑆𝑠𝑦
𝜏𝑚𝑎𝑥 ≤
𝑛
𝜏𝑚𝑎𝑥 = 𝜏𝑎 + 𝜏𝑚 = 6.151 + 6.151 = 12.302𝑀𝑃𝑎
Yield strength for AISI 1040 HR is 𝑆𝑦 = 290𝑀𝑃𝑎. (Table A-18 (Shigley 8th Edition))
𝑆𝑠𝑦 0.5 ∙ 𝑆𝑦 0.5 ∙ 290
=
=
= 55.3𝑀𝑃𝑎
𝑛
2.63
2.63
12.302𝑀𝑃𝑎 < 55.3𝑀𝑃𝑎, so no yielding occurs at the first cycle.
METU – ME 307 Machine Elements I – Fall 2014

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