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Transcript of 1 Load and Stress Analysis Section III. 2 Introduction about stresses Shearing force and bending...
1
Load and Stress Analysis
Section III
2
Introduction about stresses Shearing force and bending moment
diagrams Bending, Transverse, & Torsional
stresses Compound stresses and Mohr’s circle Stress concentration Stresses in pressurized cylinders,
rotating rings, curved beams, & contact
Talking Points
3
Assume downward force as negative and upward force as positive; and counterclockwise moment as positive and clockwise as negative.
Loads may act on multiple planes.
Introduction about stresses
i. Static Equilibrium and Free-Body Diagram
0 F 0M
0T
4
The load is applied along the axis of the bar (perpendicular to the cross-sectional area) and it is uniformly distributed across the cross-sectional area of the bar.
The normal stress can be tensile (+) or compressive (-) depending on the direction of the applied load P.
The stress unit in N/m2 or Pa or multiple of this unit, i.e. MPa, GPa.
Introduction about stresses – Cont.
ii. Direct Normal Stress & Strain
A
P
E
Assuming elasticity
oL
L
A
P
E
E
A
LPL o
oL
L
Hooke’s Law
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Sometimes, a body is subjected to a number of forces acting on its outer edges as well as at some other sections, along the length of the body. In such case, the forces are split up, and their effects are considered on individual sections. The resulting deformation of the body is equal to the algebraic sum of the deformation of the individual sections. Such a principle of finding out the resultant deformation is called the principle of superposition.
Introduction about stresses – Cont.
n
E1
o
A
LPL
Principle of Superposition:
L1 L2 L3
d1 d2
d3
L3 L2L1
d3
d2
d1
P1 P2P3
P4
6
Introduction about stresses – Cont.
Example on Principle of Superposition: A brass bar, having cross sectional area of 10 cm2 is subjected to axial forces as shown in the figure. Find the total elongation of the bar (L). Take E = 80 GPa.
L = -150 m
7
For engineering materials, = 0.25 to 0.33. For a rounded bar, the lateral strain is equal to the
reduction in the bar diameter divided by the original diameter.
Introduction about stresses – Cont.
iii. Poisson’s Ratio
Strain Axial
Strain Lateral Ratio sPoisson'
x
z
x
y
or
From Hooke’s Law:
Ex
x
Ex
zy
For 1D stress system ( ) 1D
stress system
0 zy
For 2D stress system ( )
0,0 zy
yxx E
1 xyy E
1and
8
Introduction about stresses – Cont.
Example on Poisson’s Ratio: A 500 mm long, 16 mm diameter rod made of a homogenous, isotropic
material is observed to increase in length by 300 m, and to decrease in diameter by 2.4 m when subjected to an axial 12 kN load. Determine the modulus of elasticity and Poisson’s ratio of the material.
E = 99.5 GPa = 0.25
9
Introduction about stresses – Cont.
iii. Direct Shear Stress & Strain Assuming
elasticity The load,
here, is applied in a direction parallel to the cross-sectional area of the bar.
A
Q
G
StrainShear
G is known as modulus of rigidity
Single & Double Shear
The rivet is subjected to single shear
The rivet is subjected to double shear
A
Q
2
12G
E
Relation between E, G, and
Q
Q
10
Shearing Force (S.F.) and Bending Moment (B.M.) Diagrams
Simply supported beam
Cantilever beam
Sign Convention
Relationship between shear force and bending moment
dx
dMQ
QdxM Or
diagram forceshear under the area The M
11
Shearing Force (S.F.) and Bending Moment (B.M.) Diagrams - Examples
i. Concentrated Load Only:
12
Shearing Force (S.F.) and Bending Moment (B.M.) Diagrams - Examples
ii. Distributed Load Only:
13
Shearing Force (S.F.) and Bending Moment (B.M.) Diagrams - Examples
iii. Combination of Concentrated and Distributed Load:
14
Shearing Force (S.F.) and Bending Moment (B.M.) Diagrams - Examples
iv. If Couple or Moment is Applied:
15
Bending, Transverse, & Torsional stresses
I
yM
i. Bending Stress
where, M is the applied bending moment (B.M.) at a transverse
section, I is the second moment of area of the beam cross-section
about the neutral axis (N.A.), i.e. , is the stress
at distance y from the N.A. of the beam cross-section.
dAyI 2
16
ii. Transverse Stress
Bending, Transverse, & Torsional stresses – Cont.
Ib
yAQ
where Q is the applied vertical shear force at that section; A
is the area of cross-section “above” y, i.e. the area between y
and the outside of the section, which may be above or below
the neutral axis (N.A.); y is the distance of the centroid of
area A from the N.A.; I is the second moment of area of the
complete cross-section; and b is the breadth of the section at
position y.
or dAyIb
Q
d
b
R
17
iii. Torsional Stress
Bending, Transverse, & Torsional stresses – Cont.
J
T
where T is the applied external torque; is the radial direction
from the shaft center; J is the polar second moment of area of
shaft cross-section; r is the shaft radius; and is the shear
stress at radius .
J
rT max
4 2
1rJ 44
2
1io rrJ
Solid section
Hollow shaft
Note: when torsion is present
Ductile materials tends to break in a plane perpendicular to its longitudinal axis; while brittle material breaks along surfaces perpendicular to direction where tension is maximum; i.e. along surfaces forming 45o angle with longitudinal axis.
Ductile material Brittle material
18
Compound stresses and Mohr’s circle
Machine Design involves among other considerations, the proper sizing of a machine member to safely withstand the maximum stress which is induced within the member when it is subjected separately or to any combination of bending, torsion, axial, or transverse load.
19
Compound stresses and Mohr’s circle – Cont.
Maximum & Minimum Normal Stresses
2
2
(min)
2
2
(max)
22
22
xyyxyx
n
xyyxyx
n
Stress State
3D General Stress State
2D Stress State
For 2D Case:
Where: x is a stress at a critical point in tension or compression normal
to the cross section under consideration, and may be either bending or axial load, or a combination of the two.
y is a stress at the same critical point and in direction normal to the x stress.
xy is the shear stress at the same critical point acting in the plane normal to the Y axis (which is the XZ plane) and in a plane normal to the X axis (which is the YZ plane). This shear stress may be due to a torsional moment, transverse load, or a combination of the two.
n(max) and n(min) are called principal stresses and occurs on planes that are at 90° to each other, called principle planes also planes of zero shear.
Note: x, y, z all +ve, xy, yx, zy, yz, xz, zx all +ve. Due to static balance, xy = yx, zy = yz, and xz = zx.
Counterclockwise (CCW)
Clockwise (CW)
20
Compound stresses and Mohr’s circle – Cont.
yx
xy
2
2tan
max at the critical point being investigated is equal to half of the greatest difference of any of the three principal stresses. For the case of two-dimensional loading on a particle causing a two-dimensional stresses; The planes of maximum shear are inclined at 45° with the principal planes.
2
1
2 minmax2
2
max nnxyyx
Maximum Shear Stresses (max)
The planes of maximum shear are inclined at 45° with the principal planes.
The angle between the principal plane and the X-Y plane is defined by:
21
Compound stresses and Mohr’s circle – Cont.
Mohr’s Circle It is a graphical method to find the maximum and minimum normal
stresses and maximum shear stress of any member.From the diagram:x = OA, xy = AB, y =OC, and yx = CD. The line
BEDis the diameter of Mohr's circle with center at E on
the axis. Point B represents the stress coordinates x, xy
on the X faces and point D the stress coordinates y, yx
on the Y faces. Thus EB corresponds to the X-axis and
ED to the Y-axis. The maximum principal normal stress
max
occurs at F, and the minimum principal normal stress
min at G. The two extreme-value shear stresses one
clockwise and one counterclockwise, occurs at H and I,
respectively. We can construct this diagram with compass and scale and find the required
information with the aid of scales. A semi-graphical approach is easier and quicker and offer fewer opportunities forerror.
2-D
22
Compound stresses and Mohr’s circle – Cont.
Principal Element
True views on the various faces of the principal element
MaxMin
max is equal to half of the greatest difference of any of the three principal stresses. In the case of the below figure:
3-D
2
13113max
where, 32232112 2
1 ,
2
1
23
Example: A machine member 50 mm diameter by 250 mm long is supported at one
end as a cantilever. In this example note that y = 0 at the critical point.
Compound stresses and Mohr’s circle – Examples.
Case 1: Axial load only:
Case 2: Bending only:
In this case all points in the member are subjected to
the same stress.
MPa 83.32 MPa, 65.7
0
MPa 65.71096.11015AP
m1096.110504A
(max)(max)(max)
33
2323
nn
xy
x
(Shear) MPa 60.302
on)(Compressi MPa 1.61 ,0
MPa 1.61
:Bpoint At
(Shear) MPa 60.302
0 (Tension), MPa 1.61
MPa 1.61641050
102510250103
:Apoint At
(max)(max)
(min)(max)
(max)(max)
(min)(max)
43
333
n
nn
x
n
nn
x
I
yM
I
yM
24
Compound stresses and Mohr’s circle – Examples.
Case 3: Torsion only: Case 4: Bending & Axial Load :
In this case the critical point occur along the outer
surface of the member.
(Shear) MPa 7.2625.53
on)(Compressi MPa 5.53 ,0
on)(Compressi MPa 5.531.6165.7
:Bpoint At
(Shear) MPa 4.3428.68
0 (Tension), MPa 8.68
(Tension) MPa 8.681.6165.7
:Apoint At
(max)
(min)(max)
(max)
(min)(max)
nn
x
nn
x
I
yM
A
P
I
yM
A
P
(Shear) MPa 7.40
on)(Compressi MPa 7.40
(Tension) MPa 7.40
MPa 7.40321050
1025101
0
(max)
(min)
(max)
43
33
n
n
xy
x
J
rT
25
Compound stresses and Mohr’s circle – Examples.
Case 5: Bending & Torsion:
Case 6: Torsion & Axial Load :
(Shear) MPa 9.502
on)(Compressi MPa 4.81
(Tension), MPa 3.20
MPa 7.40
MPa 1.61
:Bpoint At
(Shear) MPa 9.502
on)(Compressi MPa 3.207.402
1.61
2
1.61
(Tension), MPa 4.817.402
1.61
2
1.61
MPa 7.40
MPa 1.61
:Apoint At
(min)(max)(max)
(min)
(max)
(min)(max)(max)
22
(min)
22
(max)
nn
n
n
xy
x
nn
n
n
xy
x
J
rTI
yM
(Shear) MPa 9.402
on)(Compressi MPa 1.377.402
65.7
2
65.7
(Tension), MPa 7.447.402
65.7
2
65.7
MPa 7.40
MPa 65.7AP
(min)(max)(max)
22
(min)
22
(max)
nn
n
n
xy
x
J
rT
26
Compound stresses and Mohr’s circle – Examples.
Case 7: Bending, Axial Load, and Torsion:
(Shear) MPa 7.482
on)(Compressi MPa 5.75
(Tension), MPa 9.21
MPa 7.40
MPa 3.531.6165.7
:Bpoint At
(Shear) MPa 3.532
on)(Compressi MPa 197.402
8.68
2
8.68
(Tension), MPa 7.877.402
8.68
2
8.68
MPa 7.40
MPa 8.681.6165.7
:Apoint At
(min)(max)(max)
(min)
(max)
(min)(max)(max)
22
(min)
22
(max)
nn
n
n
xy
x
nn
n
n
xy
x
I
yM
A
P
J
rTI
yM
A
P
27
Example on Mohr’s circle: The stress element shown in figure has x = 80
MPa and xy, = 50 MPa (CW). Find the principal stresses and directions.
Compound stresses and Mohr’s circle – Examples.
Locate x = 80 MPa along the axis. Then from x,
locate xy = 50 MPa in the (CW) direction of the axis to
establish point A. Corresponding to y = 0, locate yx = 50
MPa in the (CCW) direction along the axis to obtain point
D. The line AD forms the diameter of the required circle
which can now be drawn. The intersection of the circle
with the axis defines max and min as shown.
3.5140
50tan2
:is CW to axis-X thefrom 2 angle The
MPa 246440 MPa, 1046440
MPa 644050
1-
max
(min)(max)
22(max)
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Stress Concentration Occurs when there is sudden changes in cross-sections of
members under consideration. Such as holes, grooves, notches of various kinds.
The regions of these sudden changes are called areas of stress concentration.
Stress-concentration factor (Kt or Kts)
The analysis of geometric shapes to determine stress-concentration factors is a difficult problem, and not many solutions can be found.
ots
ot KK
maxmax Theoretical
ly
29
Stresses in pressurized cylinders, rotating rings, curved beams, & contact