design for static load - universiti teknologi malaysiajamalt/compdesign/topic1 static.pdf · static...
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Static load design 1
Fx
FY
FZ
Design for Static Load
Objective Strengthen the knowledge: static and strength of material. Know how to do analysis on component under static load Find the relationship between these topics Know 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 head Eg: Fx = 10 kN, Fy = 20 kN and Fz = 30 kN
F = (10i + 20j + 30k) kN
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
Static load design 3
Static:
000 zyx FFF
000 zyx MMM
Discuss the following problem
Static load design 4
Based on T, which one has higher force value (F1 or F2). If pulley diameter is D, find F1 in function of D and T.
Find F1 in function of D (sprocket diameter) and d (pulley diameter)
Static load design 5
Pulley
Chain
Sprocket
Ball Bearing
V-Belt
F=600 N
F1=1.75F2
F2
Shaft rotated at
350 rpm (ccw)
F=600 N
Isometric View
1. 3-80 pp 140
2. 3-84 pp 140
Static load design 6
3. 3-68 pp 137
FBD Figure 3-72, 3-73
Static load design 7
INDETERMINATE PROBLEM
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 to bending or torsion
Transverse shear stress A
V
A: cross sectional area
P
V
Static load design 9
Moment
Bending stress Z
M
I
Mc
c: distance from the center axis I: 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
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 well as the loads themselves. Note We should identify the most critical element in order to proceed with our analysis. We should calculate the maximum stresses resulted from stress on the element of the most the critical position. How?
Shear Stress J
Tr
R: distance from center axis J: second polar of moment inertia
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
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
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
Static load design 14
Mohr Circle Mohr Circle is used to represent graphically the state of stress on the entire structure.
Element
Refer to Eq 3-13, 3-14 pp 77, 78 to calculate the principal stress Read Mohr Circle background theory: Section 3-6 (pp 70 – 84)
τxy
sxy
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 element equals or exceeds the maximum shear stress in a tension test specimen of the same material when that specimen begins to yield.
LIMIT DUE YIELDING
LIM
IT D
UE T
O S
HEAR
SytSyc
Static load design 16
Therefore, any Mohr circle generated from the most critical element must lie inside this limit
Failure occurs when the Mohr circle lies outside the limits When the limits are plot
against 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
Static load design 17
Case 1
Case 2
Case 3
QA
QB
Sy
Sy
Syc
Syc
Static load design 18
Distortion Energy Theory (DE) The effective stress is called Von Misses
222
22
3'
'
xyyyxx
BBAA
If sy = 0
22 3' xyx
Therefore failure happens when s’ > Sy.
To plot the graph sa against sb for DE Theory, s’ = Sy.
QA
QB
Sy
Sy
Syc
Syc
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 across
the (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 these intersecting 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 Line When the element is subject to shear only. The intersection of this line with MSS and DE curve will remark the relationship between yield stress and shear yield stress. There MSS : Ssy = 0.5 Sy DE: Ssy = 0.577 Sy
sA
sB
Sy
Sy
Syc
Syc
Pure Shear Ln
Load Line
Static load design 20
Static failure using mathematical approach Three-dimensional Stresses (Section 3-7 pp 82)
Principal Stresses s3 < s2 < s1 Maximum Shear Stress
When there is no pressure, sz is equal to zero, therefore one of
principal stress must equal to zero.
τxy
sxy
s1 s2 s3
Static load design 21
What is the value of s3, s2, s1 and τmax for a, b and c?
MSST theory
Maximum shear 2
31max
Allowable shear stress n
SS
y
sy2
Safety factor 31
yS
n
DET
Static load design 22
Total stress 22 3' xyx
22' BBAA
Safety factor '
ySn
*NOTE Please differentiate between
stresses on element xyx ,
2D Principal stresses BA ,
3D Principal stresses 321 ,,
Brittle Material
Static load design 23
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
Static load design 25
Table A-18
Table A-20 to A-22
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