Week_9_Lecture_16 Power Transmission Thin Walled Vessels and Strain Gauges

5/17/2018 Week_9_Lecture_16 Power Transmission Thin Walled Vessels and Strain Gauges

1/23

13MMA100 Mechanics of Materials

Thinned Walled Vessels, Strain Gauges

and Power Transmission.

What stresses should we consider for a thinwalled vessel?

What can strain gauges tell us?

How do we calculate shaft power?

How do we calculate torque in geared shafts?


2/23


Thin Walled Pressure

Vessels

The radius to thic!ness ration is large, i.e., r"t # $%Stress distri&ution through thic!ness is constant

We often need to predict the stress in &oilers andtan!s.

To calculate the stresses when the' are (thin walled),we assume that


3/23


Gauge Pressure

What is gauge pressure?

*t is the pressure a&ove normal atmospheric

pressure.

We are interested in the pressure differential

&etween the inside and the outside of the vessel.

This means that we generall' use the gaugepressure inside the vessel.


4/23


+'lindrical Vessels

$

2

The skin is stressed due to the circumference trying to expand,

assuming positive internal pressure, and due to the ends being

pushed outards by the internal pressure!

"ircumferential

#ongitudinal

$

2


5/23


+ircumferential Hoop-

Stresses

d'

( )[ ] ( ) 0d!2d!2 1 = yrPyt

t

$r1 =Therefore

1

1

r

/quili&rium of forces

t 0 Vessel wall thic!nessP 0 Vessel pressure

$


6/23


1ongitudinal Stresses.

.

P r/quili&rium of forces

t 0 Vessel wall thic!ness

( ) ( ) % .

. = rprt

Thereforet

pr

=


7/23


+'lindrical Vessels

Hoop stress is twice as large as longitudinal stress

1ongitudinal 2oints must carr' twice asmuch stress as circumferential 2oints

t

pr=

$

t

pr

.. =


8/23


Spherical Vessels

/quili&rium of forces

t2

$r

2 =

( ) 02 22 = rPrt

Therefore

3ut this will &e the same in an' direction, &ecause

of spherical s'mmetr'.

.

.

P r


9/23


Strain Gauges


10/23


Axial %train


11/23


How do the' wor!?


12/23


The strain is proportional to the voltage measured at &!


13/23



14/23


Strain 4osettes

5or each spo!e, we can use

sincossincos 22 XYYXgauge ++=

6nd we have three equations for three un!nowns


15/23


1et7s loo! at the e8ample of a 9:; or


16/23


>sing 9.$ we o&tain

XA =

XYYXB

$

$

$

++=

YC =

sincossincos 22 XYYXgauge ++=

00sin

10cos

=

=

2

1'(sin

2

1'(cos

=

=

1)0sin

0)0cos

=

=

*+n '!1


17/23


Therefore we now have the components of planar stress

AX = CY = ( )CABXY +=

XYYXB

$

$

$++=4e arrange


18/23



19/23


Power Transmissionthrough Shafts

$oer -atts. / Tor+ue m. x Angular %peed rads%ec.

The angular speed /

$oer in / $oer ut #osses

4022 RPM

f =

h i f i l


20/23


J

Tc=max

A pump is rated at 135HP (1HP = 746 W) at a speed of 1500 rpm.

The drive shaft to the pump sits i t!o "eari#s ad eds i a 60 tooth #ear !hee$ e%$osed i a #ear "o&.

The #ear !hee$ meshes !ith a se%od #ear of '0 teeth.

The #ear "o& is drive via a e$e%tri% motor !hose output shaft is 30mm i diameter.

Assumi# 10 po!er $osses i the #ear"o& (pump side) !hat tor*ue must the motor "e a"$e to provide at rated speed+

Pump po!er %osumed (W) = 746 & 135 = 100710 W

,otor po!er provided (W) = 100710 - 0. = 11100 W

Pump speed (H/) = 1500 - 60 Pump speed (rads-) = ' & 3.14 & 1500 - 60 = 157 ads-

,otor speed (H/) = (1500 - 60) & (60-'0) ,otor peed (rads-) = 471 ads-

,otor tor*ue & ,otor speed = ,otor po!er

,otor tor*ue (2m) = 11100 - 471 = '37.5 2m

f the ma&imum shear stress a$$o!a"$e is 55 ,Pa !ou$d a 30mm diameter so$id shaft "e suffi%iet for the motor+

==

Motor

5earbox $ump

13MMA100 M h i f M i l


21/23


,otor tor*ue (2m) = '37.5 2m

,a& a$$o!a"$e shear stress 55,Pa

haft diameter = 30mm J

Tc

=max

'' 01(!022

== CJ

MPa61!''01(!0

(6!2372

3max =

=

f the mai "od of the motor armature had a shaft si/e of

60mm ad redu%ed via a 6mm radius fi$$et to the

sma$$er shaft !ou$d the shaft "e suita"$e+

13MMA100 M h i f M i l


22/23


f the mai "od of the armature had a shaft si/e of 60mm ad redu%ed via a 6mm radius fi$$et to the sma$$er shaft

!ou$d the shaft "e suita"$e+

2!030

4

230

40

==

==

d

r

d

D

8/ 1!3 from chart.

"orrected stress!!!!

MPaMPa

K

((24!(6

61!''max

>=

=

13MMA100 M h i f M t i l


23/23


Week_9_Lecture_16 Power Transmission Thin Walled Vessels and Strain Gauges

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