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![Page 1: Chapter 5 Design and Analysis of a Laminate The Drive Shaft Problem Dr. Autar Kaw Department of Mechanical Engineering University of South Florida, Tampa,](https://reader035.fdocuments.in/reader035/viewer/2022062515/56649cf75503460f949c6d58/html5/thumbnails/1.jpg)
EML 4230 Introduction to Composite Materials
Chapter 5 Design and Analysis of a Laminate
The Drive Shaft Problem
Dr. Autar KawDepartment of Mechanical Engineering
University of South Florida, Tampa, FL 33620
Courtesy of the TextbookMechanics of Composite Materials by Kaw
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Problem Statement
A drive shaft for a Chevy Pickup truck is made of steel. Check whether replacing it with a drive shaft made of composite materials will save weight?
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Design of a Composite Drive Shaft
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Why composite materials? Light weight ___ reduces energy
consumption; increases amount of power transmitted to the wheels. About 17-22% of engine power is lost to rotating the mass of the drive train.
Fatigue resistant ___ durable life. Non-corrosive ___ reduced maintenance
cost and increased life. Single piece ___ reduces manufacturing
cost.
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Why composite materials? Prevent injuries – the composite
drive shaft “broom” on failure
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Problem Description
Design Constraints1. Maximum horsepower= 175 HP @4200
rpm2. Maximum torque = 265 lb-ft @2800
rpm3. Factor of safety = 33. Outside radius = 1.75 in4. Length = 43.5 in
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Torque in Drive Shaft
In first gear- the speed is 2800 rpm (46.67 rev/s)- assume ground speed of 23 mph (405 in/sec)
Diameter of tire = 27 inRevolutions of tire =405/[π(27)] = 4.78 rev/sDifferential ratio = 3.42Drive shaft speed = 4.78 x 3.42 = 16.35 rev/sTorque in drive shaft = (265x46.7)/16.35 =
755 lb-ft
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Maximum Frequency of Shaft
Maximum Speed = 100 mph (1760 in/sec)
Diameter of tire = 27 inRevolutions of tire =1760/[π(27)] =
20.74 rev/sDifferential ratio = 3.42Drive shaft speed = 20.74x3.42 = 71Hz
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Design Parameters
Torque Resistance. Should carry load without failure
Not rotate close to natural frequency. Need high natural frequency otherwise
whirling may take place
Buckling Resistance. May buckle before failing
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Steel Shaft – Torque Resistance
Shear Strength, max = 25 Ksi
Torque, T = 755 lb-ftFactor of Safety, FS = 3Outer Radius, c = 1.75 inPolar moment of area, J =cin = 1.69 int = 1.75-1.69 = 0.06 in = 1/16 in
J
Tc
FS
τmax
4475.12 inc
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Steel Shaft - Natural Frequency
fn =
Acceleration due to gravity, g= 32.2 ft/s2
Young’s modulus, E = 30 MsiWeight per unit length, W = 0.19011 lbf/inLength, L = 43.5 inSecond Moment of Area, I =
fn = 204 Hz
Meets minimum of 71.1Hz
42 WL
gEI
.in9973.016
175.175.1
44
44
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Steel Shaft - Torsional Buckling
T=
Mean radius, rm = 1.6875 inThickness, t = 1/16 inYoung’s modulus, E = 30 MsiCritical Buckling Load, T = 5519 lb-ft
Meets minimum of 755 lb-ft
23
22272.0
mm r
ttEr
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Designing with a composite
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Load calculations for PROMAL
Nxy = (Why)
T = 755 lb-ftr = 1.75 in
Nxy = 470.8 lb / in
Neglecting centrifugal force contribution
22 mr
T
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Composite Shaft-Torque Resistance
Inputs to PROMAL:Glass/Epoxy from Table 2.1Lamina Thickness = 0.0049213 inStacking Sequence: (45/-45/45/-45/45)s Load Nxy = 470.8 lb / in
Outputs of PROMAL:Smallest Strength Ratio = 1.52 (not safe)
Thickness of Laminate:h = 0.0049213*10 = 0.04921 in
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Composite Shaft - Natural Frequency
fn =
g = 32.2 ft/s2
Ex = 1.814 MsiI = 0.7942 in4
W = 0.03438 lbf/inL = 43.5 in
Hencefn = 105.6 Hz (meets minimum 71.1 Hz)
42 WL
IgEx
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Composite Shaft - Torsional Buckling
T =
rm = 1.75 -0.04921/2= 1.72539 in
t = 0.04921 inEx = 1.814 MsiEy = 1.814 Msi
T = 183 lb-ft (does not meet 755 lb-ft torque)
23
41
322272.0
mxm r
tEEtr
y
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Comparison of Mass
Steel Glass/ Epoxy (not acceptable
design) Specific Grav 7.850 1.785
Inside radius 1.6875" 1.7008"
Outside radius 1.75" 1.75"
Length 43.5" 43.5"
Mass 8.322 lbm 1.496 lbm
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END