1 load and stress analysis section iii. 2 introduction about stresses shearing force and bending...

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Load and Stress Analysis

Section III

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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

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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

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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.


n

E1

o

A

LPL

Principle of Superposition:

L1 L2 L3

d1 d2

d3

L3 L2L1

d3

d2

d1

P1 P2P3

P4

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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

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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.


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

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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

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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

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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

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Shearing Force (S.F.) and Bending Moment (B.M.) Diagrams - Examples

i. Concentrated Load Only:

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ii. Distributed Load Only:

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iii. Combination of Concentrated and Distributed Load:

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iv. If Couple or Moment is Applied:

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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

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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

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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

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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.

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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)

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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:

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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

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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

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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)

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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

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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

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Case 5: Bending & Torsion:

Case 6: Torsion & Axial Load :

(Shear) MPa 9.502


(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


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

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Case 7: Bending, Axial Load, and Torsion:

(Shear) MPa 7.482


(Tension), MPa 9.21

MPa 7.40

MPa 3.531.6165.7

:Bpoint At

(Shear) MPa 3.532


8.68

2

8.68


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

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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.


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

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Stresses in pressurized cylinders, rotating rings, curved beams, & contact

1 load and stress analysis section iii. 2 introduction about stresses shearing force and bending...

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