Drive shafts (Figure 5.3) in cars are generally made of steel. An automobile manufacturer is seriously thinking of changing the material to a composite material. The reasons for changing the material to composite materials are that composites 1. Reduce the weight of the drive shaft and thus reduce

Question

Drive shafts (Figure 5.3) in cars are generally made of steel. An automobile manufacturer is seriously thinking of changing the material to a composite material. The reasons for changing the material to composite materials are that composites

1. Reduce the weight of the drive shaft and thus reduce energy consumption
2. Are fatigue resistant and thus have a long life
3. Are noncorrosive and thus reduce maintenance costs and increase life of the drive shaft
4. Allow single piece manufacturing and thus reduce manufacturing cost

The design constraints are as follows:

1. Based on the engine overload torque of 140 N-m, the drive shaft needs to withstand a torque of 550 N-m.
2. The shaft needs to withstand torsional buckling.
3. The shaft has a minimum bending natural frequency of at least 80 Hz.
4. Outside radius of drive shaft = 50 mm.
5. Length of drive shaft = 148 cm.
6. Factor of safety = 3.
7. Only 0, +45, –45, +60, –60, and 90° plies can be used.

For steel, use the following properties:

Young’s modulus E = 210 GPa,
Poisson’s ratio ν = 0.3,
Density of steel ρ = 7800 kg/m³
Ultimate shear strength [latex]τ_{ult}[/latex] = 80 MPa.

For the composite, use properties of glass/epoxy from Table 2.1 and Table 3.1 and assume that ply thickness is 0.125 mm. Design the drive shaft using

1. Steel
2. Glass/epoxy
TABLE 2.1
Typical Mechanical Properties of a Unidirectional Lamina (SI System of Units)

Property	Symbol	Units	Glass/ epoxy	Boron/ epoxy	Graphite/ epoxy
Fiber volume fraction	[latex]V_f[/latex]		0.45	0.50	0.70
Longitudinal elastic modulus	[latex]E_1[/latex]	GPa	38.6	204	181
Transverse elastic modulus	[latex]E_2[/latex]	GPa	8.27	18.50	10.30
Major Poisson’s ratio	[latex]ν_{12}[/latex]		0.26	0.23	0.28
Shear modulus	[latex]G_{12}[/latex]	GPa	4.14	5.59	7.17
Ultimate longitudinal tensile strength	[latex](\sigma_1^T)_{ult}[/latex]	MPa	1062	1260	1500
Ultimate longitudinal compressive strength	[latex](\sigma_1^C)_{ult}[/latex]	MPa	610	2500	1500
Ultimate transverse tensile strength	[latex](\sigma_2^T)_{ult}[/latex]	MPa	31	61	40
Ultimate transverse compressive strength	[latex](\sigma_2^C)_{ult}[/latex]	MPa	118	202	246
Ultimate in-plane shear strength	[latex]( au_{12})_{ult}[/latex]	MPa	72	67	68
Longitudinal coefficient of thermal expansion	[latex]\alpha_1[/latex]	μm/m/°C	8.6	6.1	0.02
Transverse coefficient of thermal expansion	[latex]\alpha_2[/latex]	μm/m/°C	22.1	30.3	22.5
Longitudinal coefficient of moisture expansion	[latex]\beta_1[/latex]	m/m/kg/kg	0.00	0.00	0.00
Transverse coefficient of moisture expansion	[latex]\beta_2[/latex]	m/m/kg/kg	0.60	0.60	0.60

Source: Tsai, S.W. and Hahn, H.T., Introduction to Composite Materials, CRC Press, Boca Raton, FL, Table 1.7, p. 19; Table 7.1, p. 292; Table 8.3, p. 344. Reprinted with permission.

TABLE 1.7
Chemical Composition of E-Glass and S-Glass Fibers

Material	% Weight
Material	E-Glass	S-Glass
Silicon oxide	54	64
Aluminum oxide	15	25
Calcium oxide	17	0.01
Magnesium oxide	4.5	10
Boron oxide	8	0.01
Others	1.5	0.8

TABLE 3.1
Typical Properties of Fibers (SI System of Units)

Property	Units	Graphite	Glass	Aramid
Axial modulus	Gpa	230	85	124
Transverse modulus	GPa	22	85	8
Axial Poisson’s ratio	—	0.30	0.20	0.36
Transverse Poisson’s ratio	—	0.35	0.20	0.37
Axial shear modulus	GPa	22	35.42	3
Axial coefficient of thermal expansion	μm/m/°C	-1.3	5	-5.0
Transverse coefficient of thermal expansion	μm/m/°C	7.0	5	4.1
Axial tensile strength	MPa	2067	1550	1379
Axial compressive strength	MPa	1999	1550	276
Transverse tensile strength	MPa	77	1550	7
Transverse compressive strength	MPa	42	1550	7
Shear strength	MPa	36	35	21
Specific gravity	—	1.8	2.5	1.4

Accepted Answer

1. STEEL DESIGN.
Torsional strength: The primary load in the drive shaft is torsion. The maximum shear stress, [latex]τ_{\max}[/latex], in the drive shaft is at the outer radius, [latex]r_o[/latex], and is given as

[latex] au_{\max }=\frac{T r_{o}}{J} [/latex] (5.27)

where

T = maximum torque applied in drive shaft (N-m)
[latex]r_o[/latex] = outer radius of shaft (m)
J = polar moment of area ([latex]m^4[/latex])

Because the ultimate shear strength of steel is 80 MPa and the safety factor used is 3, using Equation (5.27) gives

[latex] \frac{80 imes 10^{6}}{3}=\frac{(550)(0.050)}{\frac{\pi}{2}\left(0.050^{4}-r_{i}^{4} ight)} \ \space \ r_i=0.04863 m.[/latex]

Therefore, the thickness of the steel shaft is

[latex]t = r_o – r_i[/latex]
= 0.050 – 0.04863
= 1.368 mm.

Torsional buckling: This requirement asks that the applied torsion be less than the critical torsional buckling moment. For a thin, hollow cylinder made of isotropic materials, the critical buckling torsion, [latex]T_b[/latex], is given by[latex]^4[/latex]

[latex] T_{b}=\left(2 \pi r_{m}^{2} t ight)(0.272)(E)\left(\frac{t}{r_{m}} ight)^{2 / 3} [/latex], (5.28)

where

[latex]r_m[/latex] = mean radius of the shaft (m)
t = wall thickness of the drive shaft (m)
E = Young’s modulus (Pa)

Using the thickness t = 1.368 mm calculated in criterion (1) and the mean radius

[latex] r_{m}=\frac{r_{o}+r_{i}}{2} \ =\frac{0.050+0.04863}{2} \ =0.049315 m, \ T_{b}=2(0.049315)^{2}(0.001368)(0.272)\left(210 imes 10^{9} ight)\left(\frac{0.001368}{0.049315} ight)^{3 / 2} \ = 109442 N-m[/latex].

The value of critical torsional buckling moment is larger than the applied torque of 550 N-m.
Natural frequency: The lowest natural frequency for a rotating shaft is given by[latex]^5[/latex]

[latex] f_{n}=\frac{\pi}{2} \sqrt{\frac{E I}{m L^{4}}} [/latex] (5.29)

where
g = acceleration due to gravity (m/s²)
E = Young’s modulus of elasticity (Pa)
I = second moment of area ([latex]m^4[/latex])
m = mass per unit length (kg/m)
L = length of drive shaft (m)

Now the second moment of area, I, is

[latex] I=\frac{\pi}{4}\left(r_{o}^{4}-r_{i}^{4} ight) \ =\frac{\pi}{4}\left(0.050^{4}-0.04863^{4} ight) \ -5.162 imes 10^{-7} m^4[/latex].

The mass per unit length of the shaft is

[latex] m=\pi\left(r_{o}^{2}-r_{i}^{2} ight) ho \ = \pi\left(0.050^{2}-0.04863^{2} ight)(7800)\=3.307kg/m.[/latex]

Therefore,

[latex] f_{n}=\frac{\pi}{2} \sqrt{\frac{\left(210 imes 10^{9} ight)\left(5.162 imes 10^{-7} ight)}{(3.307)(1.48)^{4}}} \ =129.8 Hz.[/latex]

This value is greater than the minimum desired natural frequency of 80 Hz.
Thus, the steel design of a hollow shaft of outer radius 50 mm and thickness t = 1.368 mm is an acceptable design.

2. COMPOSITE MATERIALS DESIGN.
Torsional strength: Assuming that the drive shaft is a thin, hollow cylinder, an element in the cylinder can be assumed to be a flat laminate. The only nonzero load on this element is the shear force, [latex]N_{xy}[/latex]. If the average shear stress is [latex](τ_{xy})_{average}[/latex], the applied torque then is

T = (shear stress) (area) (moment arm)
[latex] T=\left( au_{x y} ight)_{ ext {average }} \pi\left(r_{o}^{2}-r_{i}^{2} ight) r_{m} . [/latex] (5.30)

The shear force per unit width is given by

[latex] N_{x y}=\left( au_{x y} ight)_{ ext {average }} t [/latex].

Because

[latex]t = r_o – r_i \ r_{m}=\frac{r_{o}+r_{i}}{2},[/latex]

then

[latex] N_{x y}=\frac{T}{2 \pi r_{m}^{2}} \ \space \ =\frac{550}{2 \pi(0.050)^{2}} [/latex] (5.31)

= 35,014 N/m.

To find approximately how many layers of glass/epoxy may be needed to resist the shear load, choose a four-ply [latex][±45]_s[/latex] laminate. Inputting a value of [latex]N_{xy} = 35,014 N/m[/latex] into the PROMAL program, the minimum strength ratio obtained using Tsai–Wu theory is 1.261. A strength ratio of at least 3 is needed, so the number of plies is increased proportionately as [latex]\frac{3}{1.264} imes 4\cong 10.[/latex] The next laminate chosen is [latex][±45_2/45]_s[/latex] laminate. A minimum strength ratio of 3.58 is obtained, so it is an acceptable design based on torsional strength criterion.
Torsional buckling: An orthotropic thin hollow cylinder will buckle torsionally if the applied torque is greater than the critical torsional buckling load given by[latex]^4[/latex]

[latex] T_{c}=\left(2 \pi r_{m}^{2} t ight)(0.272)\left(E_{x} E_{y}^{3} ight)^{1 / 4}\left(\frac{t}{r_{m}} ight)^{3 / 2} [/latex] . (5.32)

From PROMAL, the longitudinal Young’s moduli [latex]E_x[/latex] and the transverse Young’s moduli [latex]E_y of the [±45_2/45]_s[/latex] glass/epoxy laminate based on properties from Table 2.1 are

[latex]E_x = 12.51 GPa \ E_y = 12.51 GPa[/latex]

Because lamina thickness is 0.125 mm, the thickness of the ten-ply [latex][±45_2/45]_s[/latex] laminate, t, is

t = 10 × 0.125 = 1.25 mm.

The mean radius, [latex]r_m[/latex], is

[latex] r_{m}=r_{o}-\frac{t}{2} = 50 -\frac{1.25}{2} \ =49.375 mm [/latex].

Therefore,

[latex] T_{c}=2 \pi(0.049375)^{2}(0.00125)(0.272) imes \ \space \ \left[\left(12.51 imes 10^{9} ight)\left(12.51 imes 10^{9} ight)^{3} ight]^{1 / 4}\left(\frac{0.00125}{0.049375} ight)^{2 / 3} \ \space \ =262 N-m.[/latex]

This is less than the applied torque of 550 N-m. Thus, the [latex][±45_2/45]_s[/latex] laminate would torsionally buckle. Per the formula, the torsional buckling is proportional to [latex]E_y^{3/4} and E_x^{1/4}[/latex]. Because the modulus in the y-direction is more effective in increasing the critical torsional buckling load, it will be necessary to substitute by or add 90° plies.
Natural frequency: Although the [latex][±45_2/45]_s[/latex] laminate is inadequate, per the torsional buckling criterion, let us still find the minimum natural frequency of the drive shaft, which is given by[latex]^5[/latex]

[latex] f_{n}=\frac{\pi}{2} \sqrt{\frac{E_{x} I}{m L^{4}}} . [/latex] (5.33)

Now,

[latex]E_x = 12.51 GPa \ I=\frac{\pi}{4}\left(r_{o}^{4}-r_{i}^{4} ight) \ =\frac{\pi}{4}\left(0.050^{4}-0.04875^{4} ight) \ = 4.728 imes 10^{-7} m^4.[/latex]

The mass per unit length of the beam is

[latex] m=\frac{\pi\left(r_{o}^{2}-r_{i}^{2} ight) L ho}{L} \ =\pi\left(r_{o}^{2}-r_{i}^{2} ight) ho \ =\left(0.05^{2}-0.04875^{2} ight)(1785) \ =0.6922 \frac{ kg }{m} [/latex]

Thus,

[latex] f_{n}=\frac{\pi}{2} \sqrt{\frac{\left(12.51 imes 10^{9} ight)\left(4.728 imes 10^{-7} ight)}{0.6922 imes 1.48^{4}}} \ =66.3 Hz .[/latex]

Because the minimum bending natural frequency is required to be 80 Hz, this requirement is also not met by the [latex][±45_2/45]_s[/latex] laminate. The minimum natural frequency can be increased by increasing the value of [latex]E_x[/latex] because the natural frequency [latex]f_n[/latex] is proportional to [latex]\sqrt{E_x}[/latex]. To achieve this, 0° plies can be added or substituted.

TABLE 5.12
Acceptable and Nonacceptable Designs of Drive Shaft Based on Three Criteria of Torsional Strength, Critical Torsional Buckling Load, and Minimum Natural Frequency

Laminate stacking sequence	No. plies	Minimum strength ratio	Critical torsional buckling load (N-m)	[latex]E_x[/latex] (GPa)	[latex]E_y[/latex] (GPa)	Minimum natural frequency (Hz)	Acceptable design
[latex][0/±45_2/45/90]_s[/latex]	14	3.982	797.8	16.44	16.44	75.6	No
[latex][0_2/±45_2/90]_s[/latex]	14	3.248	828.8	20.16	16.16	83.7	Yes
[latex][0/±45_2/90]_s[/latex]	12	3.006	564.1	17.07	17.07	77.2	No
[latex][0/±45_2]_s[/latex]	10	2.764	291.2	17.86	12.76	79.2	No
[latex][45/90_3/0_2]_s[/latex]	12	4.127	763.5	19.44	24.47	82.4	Yes
Design constraints		>3	>550			>80

Note: Numbers given in bold italics to show the reason for unacceptable designs.

From the three criteria, we see that ±45° plies increase the torsional strength, 90° plies increase the critical torsional buckling load, and the 0° plies increase the natural frequency of the drive shaft. Therefore, having ±45°, 90°, and 0° plies may be the key to an optimum design.
In Table 5.12, several other combinations have been evaluated to find an acceptable design.
The last stacking sequence [latex][45/90_3/0_2]_s[/latex] is a 12-ply laminate and meets the three requirements of torsional strength, critical torsional buckling load, and minimum natural frequency.
MASS SAVINGS. The savings in the mass of the drive shaft are calculated as follows:

Mass of steel drive shaft = [latex]π (r_o^2 – r_i^2) L ρ \ = π (0.050^2 – 0.04863^2) (1.48) (7800) \ = 4.894 kg.[/latex]

The thickness, t, of the [latex][45/90_3/0_2]_s[/latex] glass/epoxy shaft is

t = 12 × 0.125
=1.5 mm

The inner radius of the [latex][45/90_3/0_2]_s[/latex] glass/epoxy shaft then is

[latex]r_i = r_o – t [/latex]
= 0.05 – 0.0015
= 0.0485 m.

Mass of [latex][45/90_3/0_2]_s[/latex] glass/epoxy shaft is

[latex]= π (r_o^2 – r_i^2) L ρ \ = π (0.05^2 – 0.0485^2) (1.48) (1758) \ = 1.226 kg.[/latex]

Percentage mass saving over steel is

[latex] =\frac{4.894-1.226}{4.894} imes 100 \ 75 \%[/latex].

Would an 11-ply, [latex][45/90_4/\overline{90}]_s[/latex] glass/epoxy laminate meet all the requirements?

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