design of mechanical joint 2: weld
TRANSCRIPT
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Chapter 6: Design of
Mechanical Joint 2: WELD
DR. AMIR PUTRA BIN MD SAAD
C24-322
[email protected] | [email protected]
mech.utm.my/amirputra
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6.1 INTRODUCTION
The strongest and most common method of permanently joining steel
components together is by welding. Of the many welding techniques available,
arc welding is the most important since it is adaptable to various manufacturing
environments and is relatively cheap.
Economic ?Safe ?
Which ???
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6.1 INTRODUCTION
This chapter considers on the arc fillet weld type only.
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6.1 INTRODUCTION
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The welding symbol is standardized by the American Welding Society (AWS).
6.2 WELDING SYMBOL
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6.2 WELDING SYMBOL: Example
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6.2 WELDING SYMBOL: Example
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Throat Length = 0.707h
450
Shear
Stress
h
h
450
0.707h
Throat
plane
toe
toe root
6.3 WELDING MODEL
Effective Area = Smallest Area = Throat Area
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a. Primary Stress:
i. Axial Stress
ii. Shear Stress
b. Secondary Stress:
i. Bending Stress
ii. Torsion Stress
ππππππβ² =
π·
π¨π
ππππππβ² =
π½
π¨π
πππππ πππβ²β² =
π΄π
π°π
ππππππππβ²β² =
π»π
π±π
6.4 TYPES OF STRESS
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A
B
C
xy
z
F
Primary Load:
ForcesSecondary Load:
Moments
x
y
zG
r
Mz
In-Plane: Torsion
Mx
Out-of-Plane: BendingAxial or/and Shear
My
6.5 STRESS DISTRIBUTION
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Direct Shear Stress:
Bending Shear Stress:
ππ βπππβ² =
π
π΄π€
ππππππππβ²β² =
ππ
πΌπ€
M
V
ππππ π’ππ‘πππ‘ = ππ βπππβ² 2
+ ππππππππβ²β² 2
π°π = π. πππ π π°π
Front View
Side View
weld
Combine Stress: Theorem Pythagoras
6.6 SECONDARY STRESS: BENDING STRESS
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ππ‘πππ πππβ²β² =
ππ
π½π€
Direct Shear Stress:
Torsion Stress:
ππ βπππβ² =
π
π΄π€
π±π = π. πππ π π±π
Combine Stress: Cosine Rule (triangle shape)
ππππ π’ππ‘πππ‘ = ππ βπππβ² 2
+ ππ‘πππ πππβ²β² 2
β 2ππ βπππβ² ππ‘πππ πππ
β²β² πππ (π)
π is the angle formed between vector ππ βπππβ² and ππ‘πππ πππ
β²β²
M
V
6.7 SECONDARY STRESS: TORSION STRESS
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6.8 SECOND MOMENT OF AREA
Moments of inertia of linear weld segments.
h
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For simplicity it is assumed that the effective weld width in the plane of the paper isthe same as throat length β. Rectangular moment of inertia about axes of symmetryof the weld segment, πβ² and πβ².
Rectangular moment of inertia about axes π and π through the center of gravity ofthe total weld group:
6.8 SECOND MOMENT OF AREA
πΌπ = πΌπβ² + Aπ2 =πΏ3β
12+ πΏβπ2
πΌπ = πΌπβ² + Aπ2 = πΏβπ2
πΌπβ² = ΰΆ±π¦2ππ΄ = 2ΰΆ±0
πΏ/2
π¦2 β ππ¦ =πΏ3β
12
πΌπβ² = 0
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Polar moment of inertia about an axis perpendicular to the center of gravity of the weld group:
6.8 SECOND MOMENT OF AREA
π½ = πΌπ + πΌπ =πΏ3π‘
12+ πΏπ‘ π2 + π2
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6.9 TABLE OF UNIT SECOND MOMENT OF AREA, πΌπ’
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6.9 TABLE OF UNIT SECOND MOMENT OF AREA, πΌπ’
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6.10 TABLE OF UNIT SECOND POLAR MOMENT OF AREA, π½π’
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6.10 TABLE OF UNIT SECOND POLAR MOMENT OF AREA, π½π’
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6.11 ELECTRODE STRENGTH
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πππππ‘π¦ πΉπππ‘ππ = ππ·πΈπ =π. 577ππ¦
ππππ₯
Distortion Energy Method:
ππππ₯ = π£πππ‘πππππππ¦ πππππ
6.12 STATIC FAILURE PREVENTION SCHEME
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Thank You
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3.14 EXAMPLE 3.3
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3.14 EXAMPLE 3.3
1. Establish the free-body diagram.
rA
rD
rB
rC
ππ©β²β²
ππͺβ²β²
ππ¨β²β²
ππ«β²β²
ππ¨β²
ππ«β²
ππͺβ²
ππ©β²
πΎπππ π·ππππππ
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3.14 EXAMPLE 3.3
π΄π€ = 1280 ππ2
2. Determine the primary stresses.
πβ² =π
π΄π€=
25000
1280= 19.5 πππ
3. Determine the secondary stresses.
π½ = 0.707βπ½π’ = 7.07 Γ 106 ππ4
π = 2760 ππ
ππ΄β²β² = ππ΅
β²β² =πππ΄π½
= 41 πππ
ππΆβ²β² = ππ·
β²β² =πππ·π½
= 37.3 πππ
From Table 9-1
From Table 9-1
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3.14 EXAMPLE 3.3
4. Combine the stresses for 2 and 3.
ππππ₯ = ππΆ = ππ· = 43.9 πππThus,
ππ΄ = ππ΅ = ππ΄β² + ππ΄
β²β² sin 25.6402+ ππ΄
β²β² cos 25.6402= 37 πππ
ππΆ = ππ· = ππΆβ² + ππΆ
β²β² sin 6.2502+ ππΆ
β²β² cos 6.2502= 43.9 πππ
By using Resolving Method:
5. Determine the safety factor or fillet weld size.
This question did not asking you to calculate the safety factor.