week_9_lecture_16 power transmission thin walled vessels and strain gauges
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Power TransmissionTRANSCRIPT
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5/17/2018 Week_9_Lecture_16 Power Transmission Thin Walled Vessels and Strain Gauges
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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?
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13MMA100 Mechanics of Materials
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
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13MMA100 Mechanics of Materials
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.
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13MMA100 Mechanics of Materials
+'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
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13MMA100 Mechanics of Materials
+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
$
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13MMA100 Mechanics of Materials
1ongitudinal Stresses.
.
P r/quili&rium of forces
t 0 Vessel wall thic!ness
( ) ( ) % .
. = rprt
Thereforet
pr
=
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13MMA100 Mechanics of Materials
+'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
.. =
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13MMA100 Mechanics of Materials
Spherical Vessels
/quili&rium of forces
t2
$r
2 =
( ) 02 22 = rPrt
Therefore
3ut this will &e the same in an' direction, &ecause
of spherical s'mmetr'.
.
.
P r
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13MMA100 Mechanics of Materials
Strain Gauges
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13MMA100 Mechanics of Materials
Axial %train
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13MMA100 Mechanics of Materials
How do the' wor!?
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13MMA100 Mechanics of Materials
The strain is proportional to the voltage measured at &!
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13MMA100 Mechanics of Materials
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13MMA100 Mechanics of Materials
Strain 4osettes
5or each spo!e, we can use
sincossincos 22 XYYXgauge ++=
6nd we have three equations for three un!nowns
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13MMA100 Mechanics of Materials
1et7s loo! at the e8ample of a 9:; or
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13MMA100 Mechanics of Materials
>sing 9.$ we o&tain
XA =
XYYXB
$
$
$
++=
YC =
sincossincos 22 XYYXgauge ++=
00sin
10cos
=
=
2
1'(sin
2
1'(cos
=
=
1)0sin
0)0cos
=
=
*+n '!1
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13MMA100 Mechanics of Materials
Therefore we now have the components of planar stress
AX = CY = ( )CABXY +=
XYYXB
$
$
$++=4e arrange
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13MMA100 Mechanics of Materials
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13MMA100 Mechanics of Materials
Power Transmissionthrough Shafts
$oer -atts. / Tor+ue m. x Angular %peed rads%ec.
The angular speed /
$oer in / $oer ut #osses
4022 RPM
f =
h i f i l
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13MMA100 Mechanics of Materials
J
Tc=max
A pump is rated at 135HP (1HP = 746 W) at a speed of 1500 rpm.
The drive shaft to the pump sits i t!o "eari#s ad eds i a 60 tooth #ear !hee$ 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$$o!a"$e is 55 ,Pa !ou$d a 30mm diameter so$id shaft "e suffi%iet for the motor+
==
Motor
5earbox $ump
13MMA100 M h i f M i l
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13MMA100 Mechanics of Materials
,otor tor*ue (2m) = '37.5 2m
,a& a$$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
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13MMA100 Mechanics of Materials
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
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13MMA100 Mechanics of Materials