design course notes

7/25/2019 Design Course Notes

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MecE 360: Engineering Design II

Section 4: Loading & Stress AnalysisDay 2


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Class Notifications

Pick up bearing catalogue in machine shop von Mises next day with failure discussion

Office Hours: Extended 5:00 to 6:00 meeting

times for Feb meetings May change further depending on sign-up

trends

Project: How to convey information on design

1 page template with key information?

Currently have final report posted for one, in

process of getting reports from Chen


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Objectives

1. Stress concentration for fatigue2. Deflections and slopes for loaded beams

Many long examples


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SCF for Cyclic Loading (p.4-25)

Brittle materials are more susceptible to stress

concentration (SC) than ductile materials

Ductile materials deform if local equivalent stresses

exceed yield strength

Brittle materials fail if stress reaches Sut(no yield)

Components under cyclic loading often fail

in a brittle manner

Ductile materials that plastically deform are work

hardened and become brittle

Cracks propagate with each cycle in a brittle material


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SCF for Cyclic Loading

Fatigue SCF (Kf) Static (theoretical) SCF (Kt)

If the material being cycled is rittle

Kf=Kt and Kfs =Kts(tensile) (shear)

Brittle materials are more sensitive to

notches and stress concentrations

Need to define a characteristic that relates the

level of sensitivity to change in geometry and SCF for

cyclic loading


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Notch sensitivity q and qs provide relationship between

fatigue SCF and theoretical SCF

For very brittle materials q = 1 and qs = 1

1

1

t

f

K

K

q 1

1

ts

fs

s K

K

q


Known/given/from last day Known/given

unknown unknown

MaterialProperty, in

charts/graphs

MaterialProperty, in

charts/graphs


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Graphically find q or qs based on Sutand notch radius

For qs , add 20 kpsi to Sut, then find q

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.05 0.1 0.15 0.2notch radius, r (in)

q

240/1656

200/1379

160/1103

140/965

120/827

100/689

80/552

70/483

60/41450/345

Dd

M Mr

TT


Sut[kpsi/MPa]


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0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.05 0.1 0.15 0.2

notch radius, r (in)

q

240/1656

200/1379

160/1103

140/965

120/827

100/689

80/552

70/483

60/414

50/345

Example: Sut = 120 kpsi, r= 0.1 in

q 0.86


Sut[kpsi/MPa]


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0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.05 0.1 0.15 0.2

notch radius, r (in)

q

240/1656

200/1379

160/1103

140/965

120/827

100/689

80/552

70/483

60/414

50/345


Sut[kpsi/MPa]

Example: Sut = 120 kpsi, r= 0.1 in

For qs use

Sut+ 20 kpsi

= 140 kpsi

qs 0.89


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nafa K

1

1

t

f

K

Kq


For alternating stress

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.05 0.1 0.15 0.2notch radius, r (in)

q

240/1656

200/1379

160/1103

140/965

120/827

100/689

80/552

70/483

60/414

50/345

Shaft with groove in bending

1

2

3

4

5

0 0.1 0.2 0.3

r/d

Kt

D/d= D/d=2 D/d=1.5 D/d=1.3

D/d=1.15 D/d=1.1 D/d=1.05 D/d=1.01

DdM M

r

Nominal stress


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For mean stressnmfmm

K

Approach to solve for Kfm:


Nominal stress

If ynf SK max then: ffm KK

If yf SK n max then:

n

n

m

afy

fm

KSK

Ifyf

SKnn

2minmax then: 0

fmK

Most common case

Ductile to brittle

transition at notch

Compression and tension;

yielding occurs;

mean stress has little effect


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Example 4-3 (p.4-30)

Dd

MM

r

TT

The 50 mm diameter shaft shown has fully reversed

bending (100 Nm) and fluctuating torsion (50 25 Nm).

The small diameter is dis 40 mm and the radius of

curvature of the groove is 5 mm. The shaft is made of

AISI 1018 cold-drawn steel with Sut= 440 MPa, Sy = 370MPa.

Calculate mean and alternating bending and torsional

stresses

Given: The theoretical tensile and shear SCF are 1.75and 1.5, respectively.


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Solution

from last day.


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Solution

Sut[kpsi/MPa]


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Solution


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Solution Contd


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Solution Final


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Shaft Deflection ( p.4-32)

Besides strength considerations, it is usually

also important to examine component deflection

We will investigate shaft deflections


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Shaft Deflection

Note that a shaft is usually stepped to

accommodate gears and bearing.

It is necessary to account for changes in loading

and geometry along the length of the shaft.


20/44

21

0 00

1

00

2

2

2

2 1

CxCdxEI

Mdx

dx

dyy

CdxEI

Mdx

dx

yd

dx

dy

EI

M

dx

yd

x xx

xx

:deflection

:slope

:curvatureofradius

where

dxVM

dxqV

:moment

:shear

Shaft Deflection

Need to solve deflection equations (see CivE 270)


21/44

Some functions to

integrate/differentiate


22/44

Use of singularity functions, which are on/off

type equations (brackets: < >)

Value of a singularity function

Integration:

a-xa,xif

0x,aif

ax

axax

1nif1

0nif

1

1

n

n

n

n

ax

ax

ax

If n < 0, then x - an = 0

Shaft Deflection


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Singularity functions are applied to the different

shaft loads when defining shear flow equation

Distributed load (w), shear flow (q), [N/m]

Force (F), shear (V), [N]

Couple/moment (M), [Nm]

a is the position where the loading appears on the shaft

2

ax

0

ax 1

ax

Shaft Deflections


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Left reaction (RL

) @ a = 0m

R1 @ a = 0.025m

M@ a = 0.025m

R2 @ a = 0.075m

Right reaction (RR) @ a = 0.1m

Shaft Deflection

10

x

1025.0 x

2025.0

x

1075.0

x

11.0

x

R1

=100N

x

R2=75N

y

z

R11

2

R2

100mm

75mm

25mm

M=10Nm

RL

RR


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Shaft Deflection

11

2

21

1

1

1.0075.0

025.0025.0)(

xRxR

xMxRxRxq

R

L

Integrate shear flow q twice with respect to xto find M(x).

Use M(x) to find slope (x) and deflection y(x).

R1

=100N

x

R2=75N

100mm

75mm

25mm

M=10Nm

RL

RR


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22 ABtotalABtotal zytotal

22

zy ABtotalABtotal

Shaft Deflection

y-deflection z-deflection

Resultant deflection from all forces at critical

position


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Example 4-4

Find the equations for deflection and slope at any point on the

beam.

Example in notes package

E l P bl Si l it
http://localhost/var/www/apps/conversion/tmp/MecE360_Fall2010/Resources/Section%201%20example.jnthttp://localhost/var/www/apps/conversion/tmp/MecE360_Fall2010/Resources/Section%201%20example.jnt


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Example Problem- Singularity

Functions


29/44

Solution


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Solution


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Example Problem


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Solution

45.469 N

R2y=- R1y +P1y+ q (0.025)

=- 45.469+50 +50 N/m (0.025m)

R2y= 5.78125


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Solution


34/44

Soln contd- x-y plane


35/44

Solution- Slope and Deflection

Solution Solve Constants with


36/44

Solution- Solve Constants with

Boundary Conditions

- 0.014909

Solution Slopes and


37/44

Solution- Slopes and

Deflections in x-y plane

C1/EI


38/44

x-z plane- Fast Forward

Slope and Deflection Fast


39/44

Slope and Deflection-Fast

Forward

Solving Boundary Conditions


40/44

Solving Boundary Conditions-

Fast Forward

Boundary Conditions Fast


41/44

Boundary Conditions- Fast

Forward

Slopes and Deflections Fast


42/44

Slopes and Deflections- Fast

Forward


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Total Deflection and Slope


44/44

Whats next

Section 5: Material Failure

von Mises equivalent stress

Design criteria

Failure modes

Influential factors

design course notes

Documents