Determine the elongation of the tension member designed in Example 14.3. Consider the following data:

$E_1=130 \ GPa, \ E_2=6.5 \ GPa, \ G_{12}=5.0 \ GPa, \ \text{ and } \ ν _{12}=0.25$

If the axial elongation is to be limited to 0.02 mm, what should be the design modifications, if any?

Question

Determine the elongation of the tension member designed in Example 14.3. Consider the following data:

[latex]E_1=130 \ GPa, \ E_2=6.5 \ GPa, \ G_{12}=5.0 \ GPa, \ ext{ and } \ ν _{12}=0.25[/latex]

If the axial elongation is to be limited to 0.02 mm, what should be the design modifications, if any?

Accepted Answer

Axial Winding
Using Equation 14.56, the axial moduli for the 0° and 90° plies are determined as follows:

[latex]E_x=\left[\frac{\cos^4 heta}{E_1}+\frac{\sin^4 heta}{E_2}+\left\lgroup\frac{1}{G_{12} }-\frac{2\upsilon _{12}}{E_1} ight group \sin^2 heta \cos^2 heta ight] ^{-1} \quad\quad\quad\quad\quad (14.56) \ E_{x,0^\circ }=\left[\frac{1}{130}+0+0 ight]^{-1}=130 \ GPa \ E_{x,90^\circ }=\left[0+\frac{1}{6.5}+0 ight] ^{-1}=6.5 \ GPa[/latex]

The areas of cross section of the 0° and 90° plies are calculated from the design data from Example 14.3, as follows:

[latex]A_{0^\circ }=448.7 \ mm^2 \ ext{ and } \ A_{90^\circ }=202.3 \ mm^2[/latex]

Then, using Equation 14.57, the axial elongation is computed as follows:

[latex]\sum\limits_{i=1}^{n}{A_iE_{xi}}=AE_x=\frac{PL}{\Delta} \quad\quad\quad\quad\quad (14.57) \ \Delta=\frac{750 imes 1.25 imes 800}{448.7 imes 130,000+202.3 imes 6500} =0.013 \ mm[/latex]

Helical Winding
Using Equation 14.56, the axial modulus for the 15° plies is determined as follows:

[latex]E_{x,15^\circ }=\left[\frac{\cos^415^\circ }{130}+\frac{\sin^4 15}{6.5}+\left\lgroup\frac{1}{5}-\frac{2 imes 0.25}{130} ight group imes \sin^2 15^\circ \cos^2 15^\circ ight] ^{-1}=50.9 \ GPa[/latex]

We have already found that the axial modulus of the 90° plies is 6.5 GPa.
The areas of cross section of the 15° and 90° plies are calculated from the design data from Example 14.3, as follows:

[latex]A_{15^\circ }=530.8 \ mm^2 \ ext{ and } \ A_{90^\circ }=44.2 \ mm^2[/latex]

Then, using Equation 14.57, the axial elongation is computed as follows:

[latex]\Delta=\frac{750 imes 1.25 imes 800}{530.8 imes 50,900+44.2 imes 6500}=0.028 \ mm [/latex]

The axial elongation can be reduced most conveniently by reducing the angle of winding. Let us consider a winding angle of 10°, which, it can be calculated to show, results in an axial modulus of 76.3 GPa. Then, without changing any other design parameters, it can be seen that the axial elongation comes down to 0.018 mm. Note, however, that with a reduced helical winding angle, it may become necessary to increase the consolidation hoop plies.

Question 14.4: Determine the elongation of the tension member designed in E...

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