Design a tension member of length 800 mm and outer circular cross section to carry a tensile load of 750 kN. The material available is carbon/epoxy composite.
Unidirectional carbon/epoxy composite properties at $V_f = 0.6$ are

$E_1=130 \ GPa, X^T=2400 \ MPa, \text{ and }\rho_c=1.53 \ g/cm^3$

Question

Design a tension member of length 800 mm and outer circular cross section to carry a tensile load of 750 kN. The material available is carbon/epoxy composite.
Unidirectional carbon/epoxy composite properties at [latex]V_f = 0.6[/latex] are

[latex]E_1=130 \ GPa, X^T=2400 \ MPa, ext{ and } ho_c=1.53 \ g/cm^3[/latex]

Accepted Answer

Let us consider a tubular cross section and choose filament winding as the manufacturing process. We shall explore two ways to do filament winding of the tension member—axial winding and helical winding.

Axial Winding
In axial winding, the winding angle is zero. Unlike helical winding, it does not involve cross-overs and a fiber volume fraction of 0.6 can be expected. Then, the given unidirectional carbon/epoxy strength can be directly used, and taking a factor of safety of 1.25, the total area of cross section is readily obtained as

[latex]A_c=\frac{750 imes 1.25 imes 1000}{2400}=390.6 \ mm^2 [/latex]

Taking a nominal shell thickness as 4 mm, the mean diameter is obtained as

[latex]D_m=\frac{390.6}{\pi imes 4} =31.1 \ mm[/latex]

We take the inner diameter as 30 mm and readily compute the required outer diameter as

[latex]D_0=\sqrt{30^2+\frac{4 imes 390.6}{\pi} }=37.4 \ mm [/latex]

which implies a shell thickness of 3.7 mm. The thickness provided, however, has to be an integer multiple of ply thickness. Accordingly, considering a ply thickness of 0.5 mm, we provide eight plies. From Example 14.1 , we know that the required bandwidth for a ply thickness of 0.5 mm with two spools is 3.08 mm.
However, it must also satisfy the following condition:

[latex]\frac{\pi imes D}{B_w}=n [/latex]

in which n is an integer, which is nothing but the number of circuits per ply, and D and [latex]B_w[/latex] are the inner diameter and bandwidth, respectively, for that ply. Note that diameter D increases by twice the ply thickness after the completion of each ply. Then, the parameters n and [latex]B_w[/latex] for each ply can be tabulated as in Table 14.4.
In Table 14.4, [latex]D_i[/latex], n, [latex]B_w[/latex], t, and [latex]D_o[/latex] are the inner diameter, number of circuits, bandwidth along the circumference, ply thickness, and outer diameter,

TABLE 14.4
Axial Winding Parameters (Example 14.3)

Ply	[latex]D_i[/latex]	n	[latex]B_w[/latex]	t	[latex]D_o[/latex]
Ply-1 (hoop)	30.00	-	7.7	0.2	30.40
Ply-2 (axial)	30.40	31	3.08	0.50	31.40
Ply-3 (hoop)	31.40	-	7.7	0.2	31.80
Ply-4 (axial)	31.80	32	3.12	0.49	32.78
Ply-5 (hoop)	32.78	-	7.7	0.2	33.18
Ply-6 (axial)	33.18	33	3.16	0.49	34.16
Ply-7 (hoop)	34.16	-	7.7	0.2	34.56
Ply-8 (axial)	34.56	35	3.10	0.50	35.56
Ply-9 (hoop)	35.56	-	7.7	0.2	35.96
Ply-10 (axial)	35.96	37	3.05	0.50	36.96
Ply-11 (hoop)	36.96	-	7.7	0.2	37.36
Ply-12 (axial)	37.36	38	3.09	0.50	38.36
Ply-13 (hoop)	38.36	-	7.7	0.2	38.76
Ply-14 (axial)	38.76	40	3.04	0.51	39.78
Ply-15 (hoop)	39.78	-	7.7	0.2	40.18
Ply-16 (axial)	40.18	41	3.08	0.50	41.18
Ply-17 (hoop)	41.18	-	7.7	0.2	41.58

respectively, for the ply. Axial filament winding needs consolidation during winding and each of the above plies is consolidated by providing a hoop ply of 0.2 mm thickness. Thus, the total thickness turns out to be 5.8 mm, the cross section OD:
41.6 mm × ID: 30 mm, and the total mass 797 g.

Helical Winding
In helical winding, the winding angle is nonzero. It involves cross-overs and a lower fiber volume fraction is likely to result. [latex]V_f[/latex] depends on a number of factors, and in general, [latex]V_f = 0.5[/latex] can be expected. The longitudinal modulus is linearly influenced by fiber volume fraction but not strength. However, for simplicity, we can consider a linear dependence of longitudinal tensile strength on fiber volume fraction.
A very important aspect in helical winding is the angle of winding. For a tension member, low angle helical winding is preferable. However, very small angle may put added requirement of consolidating hoop plies. Let us choose for the present design example 15° as the helical winding. Then, the total area of cross section is obtained as

[latex]A_c=\frac{750 imes 1.25 imes 1000}{(2400 imes 0.5/0.6) imes cos^2 15^\circ } =502.4 \ mm^2[/latex]

Taking the inner diameter as 30 mm, the required outer diameter is readily computed as 39.2 mm. We choose to provide six helical plies of 0.8 mm thickness each at ±15°. Note that each helical ply is actually a compound ply of two subplies at +15° and −15° of thickness 0.4 mm.
We have seen in Example 14.1 that the bandwidth normal to the winding direction for a ply thickness of 0.5 mm for n spools is 1.54n. Accordingly, for two spools, the bandwidth normal to the meridian, that is, along the circumference, path is worked out as follows:

[latex]B_w=(1.54 imes 2) imes \left\lgroup\frac{0.5}{0.4} ight group imes \frac{1}{cos15^\circ } =4.0 \ mm[/latex]

TABLE 14.5
Helical Winding Parameters (Example 14.3)

Ply Description	[latex]D_i[/latex]	n	[latex]B_w[/latex]	t	[latex]D_o[/latex]
Ply-1 (hoop)	30	-	7.7	0.2	30.4
Ply-2 (helical)	30.4	24	3.98	0.80	32.0
Ply-3 (helical)	32.0	25	4.02	0.80	33.6
Ply-4 (helical)	33.6	26	4.06	0.79	35.2
Ply-5 (helical)	35.2	28	3.95	0.81	36.8
Ply-6 (helical)	36.8	29	3.99	0.80	38.4
Ply-7 (helical)	38.4	30	4.02	0.80	40.0
Ply-8 (hoop)	40.0	-	7.7	0.20	40.4

which implies that the required number of circuits per ply is 23.6. We provide an integer number of circuits, say, 24, which in turn, implies a reduced bandwidth and increased ply thickness. The number of circuit for each ply is worked out and tabulated in Table 14.5.
In Table 14.5, [latex]D_i[/latex] , n, [latex]B_w[/latex], t, and Do are the inner diameter, number of circuits, bandwidth along the circumference, ply thickness, and outer diameter, respectively, for the ply. Note that we have provided two hoop plies.
The total thickness turns out to be 5.2 mm, the cross section OD: 40.4 mm × ID: 30 mm, and the total mass 672 g. (A reduced density of 1.46 is considered for a reduced fiber volume fraction.)

Question 14.3: Design a tension member of length 800 mm and outer circular ...

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