topic1 staticstatic
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Static load design 1
Fx
FY
FZ
Design for Static Load
ObjectiveStrengthen the knowledge: static and strength of material.Know how to do analysis on component under static loadFind the relationship between these topicsKnow how to use Table in appendix A
Notation
Axis: Right Hand Rule To maintain the position of X, Y and Z axis
To show the rotation direction of Moment and Torsion
Force: arrow with single headEg: Fx = 10 kN, Fy = 20 kN and Fz = 30 kN
F = (10i + 20j + 30k) kN
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Static load design 2
Moment and torsion: arrow with two head on direction of rotation.Eg: Mx = 10 Nm, My = 20 Nm and Mz = 30 Nm
M = (10i + 20j + 30k) Nm
Are torsion and moment same and why they are named differently?
`
Mx
My
Mz
Mx
My
Mz
Mx
My
Mz
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Static load design 3
Static:
000 z y x F F F
000 z y x M M M
Discuss the following problem
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Static load design 4
Based on T, which one has higher force value (F1 or F2). If pulleydiameter is D, find F1 in function of D and T.
Find F1 in function of D (sprocket diameter) and d (pulley diameter)
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Static load design 5
Pulley
Chain
Sprocket
Ball Bearing
V-Belt
F=600 N
F1=1.75F2
F2Shaft rotated at
350 rpm (ccw)
F=600 N
Isometric View
1. 3-80 pp 140
2. 3-84 pp 140
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Static load design 6
3. 3-68 pp 137
FBD Figure 3-72, 3-73
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Static load design 7
INDETERMINATE PROBLEM
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Static load design 8
STRESS ANALYSIS
Axial Load (P)
Axial Stress A
P
A: cross sectional area
Stresses on element
Shear load (V) : can be neglected when the part is subjected tobending or torsion
Transverse shear stress A
V
A: cross sectional area
P
V
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Static load design 9
Moment
Bending stress
Z
M
I
Mc
c: distance from the center axisI: second moment of area (last page, Table
A-6 to A-8 pp1008 - 1012)Z: section modulus (Table A-6 to A-8)
Stresses on element
M
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Static load design 10
Torsion
T
Note:In design we are responsible our design is safe against failure.Therefore, parameters such A, I, Z and J play important role as wellas the loads themselves.
NoteWe should identify the most critical element in order to proceed withour analysis.
We should calculate the maximum stresses resulted from stress onthe element of the most the critical position. How?
Shear Stress J Tr
R: distance from center axisJ: second polar of momentinertia
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Static load design 11
Compounded Load
P M Compounded
Which element is most critical and why?
M
M
P
P
A
B, D
C
A
B, D
C
A
B, D
C
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Static load design 12
T M Compounded
Which element is most critical and why?
M
M T
T
A
B, D
C
A
B, D
C
A
B, D
C
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Static load design 13
M1 M2 Compounded
Which element is most critical and why?
M1
M1
M2
M2
A
B, D
C
A
B, D
C
A
B, D
C
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Static load design 14
Mohr CircleMohr Circle is used to represent graphically the state of stress on theentire structure.
Element
Refer to Eq 3-13, 3-14 pp 77, 78 to calculate the principal stressRead Mohr Circle background theory: Section 3-6 (pp 70 – 84)
τxy
sxy
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Static load design 15
STATIC FAILURE THEORIES (Chap 5)
Ductile materials (yield criteria) Maximum shear stress (MSS): Sec 5-4 pp 211 Distortion energy (DE): Sec 5-5 pp 213 Ductile Coulomb-Mohr (DCM): Sec 5-6 pp 219
Brittle materials Maximum normal stress (MNS) Sec 5-8 Brittle Coulomb-Mohr (BCM) Sec 5-9
Modified Mohr (MM) Sec 5-9
Maximum shear stress Theory (MSS)
Yielding begins whenever the maximum shear stress in any elementequals or exceeds the maximum shear stress in a tension test
specimen of the same material when that specimen begins to yield.
LIMIT DUE YIELDING
LIMIT
DUET
O
SHEAR
SytSyc
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Static load design 16
Therefore, any Mohrcircle generated fromthe most critical element
must lie inside this limit
Failure occurs when the Mohr
circle lies outside the limits
When the limits are plotagainst sa and sb.
Case1: when sa and sb are +ve
Case2: when sa is +ve and sb is –ve
Case3: when sa and sb are -ve
Syt Syc
SytSyc
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Static load design 17
Case 1
Case 2
Case 3
QA
QB
Sy
Sy
Syc
Syc
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Static load design 18
Distortion Energy Theory (DE)
The effective stress is called Von Misses
222
22
3'
'
xy y y x x
B B A A
If sy = 0
223' xy x
Therefore failure happens when s’ > Sy. To plot the graph sa against sb for DE Theory, s’ = Sy.
QA
QB
Sy
Sy
Syc
Syc
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Static load design 19
Load line and its meaning on MSS and DE.Line when it is drawn from (0, 0) to another point and this line acrossthe (sa, sb) of the Mohr Circle of the element.Intersection between load line and the MSS and DE Curve: point of
yielding according the theories defn. Any stresses exceed theseintersecting point is considered fail due to yielding ( FS < 1 ).Before the intersection point ( FS > 1)
At the intersection point (FS = 1)Beyond the intersection point ( FS < 1 ) NO GO
Pure Shear LineWhen the element is subject to shear only. The intersection of this
line with MSS and DE curve will remark the relationship between yieldstress and shear yield stress. ThereMSS : Ssy = 0.5 SyDE: Ssy = 0.577 Sy
s A
sB
y
Sy
Syc
Syc
Pure Shear Ln
Load Line
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Static load design 20
Static failure using mathematical approach
Three-dimensional Stresses (Section 3-7 pp 82)
Principal Stresses s3 < s2 < s1Maximum Shear Stress
When there is no pressure, sz is equal to zero, therefore one of
principal stress must equal to zero.
τ xy
s xy
s1s2s3
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Static load design 21
What is the value of s3, s2, s1 and τmax for a, b and c?
MSST theory
Maximum shear 2
31
max
Allowable shear stress n
S S
y
sy2
Safety factor31
yS n
DET
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Static load design 22
Total stress22
3' xy x
22' B B A A
Safety factor '
yS n
*NOTEPlease differentiate between
stresses on element xy x ,
2D Principal stresses B A ,
3D Principal stresses 321 ,,
Brittle Material
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Static load design 23
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Static load design 24
Note:Please take note of the following appendix(All the tables on the appendices should be at your finger tips)
Table A-6 to A-8
Table A-9
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Table A-18
Table A-20 to A-22