Question 14.4: Determine the elongation of the tension member designed in E...
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?
Axial Winding
Using Equation 14.56, the axial moduli for the 0° and 90° plies are determined as follows:
E_x=\left[\frac{\cos^4 \theta}{E_1}+\frac{\sin^4 \theta}{E_2}+\left\lgroup\frac{1}{G_{12} }-\frac{2\upsilon _{12}}{E_1} \right\rgroup \sin^2 \theta \cos^2 \theta \right] ^{-1} \quad\quad\quad\quad\quad (14.56) \\ E_{x,0^\circ }=\left[\frac{1}{130}+0+0 \right]^{-1}=130 \ GPa \\ E_{x,90^\circ }=\left[0+\frac{1}{6.5}+0 \right] ^{-1}=6.5 \ GPa
The areas of cross section of the 0° and 90° plies are calculated from the design data from Example 14.3, as follows:
A_{0^\circ }=448.7 \ mm^2 \ \text{ and } \ A_{90^\circ }=202.3 \ mm^2
Then, using Equation 14.57, the axial elongation is computed as follows:
\sum\limits_{i=1}^{n}{A_iE_{xi}}=AE_x=\frac{PL}{\Delta} \quad\quad\quad\quad\quad (14.57) \\ \Delta=\frac{750\times 1.25\times 800}{448.7\times 130,000+202.3\times 6500} =0.013 \ mm
Helical Winding
Using Equation 14.56, the axial modulus for the 15° plies is determined as follows:
E_{x,15^\circ }=\left[\frac{\cos^415^\circ }{130}+\frac{\sin^4 15}{6.5}+\left\lgroup\frac{1}{5}-\frac{2\times 0.25}{130} \right\rgroup \times \sin^2 15^\circ \cos^2 15^\circ \right] ^{-1}=50.9 \ GPa
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:
A_{15^\circ }=530.8 \ mm^2 \ \text{ and } \ A_{90^\circ }=44.2 \ mm^2
Then, using Equation 14.57, the axial elongation is computed as follows:
\Delta=\frac{750\times 1.25\times 800}{530.8\times 50,900+44.2\times 6500}=0.028 \ mm
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.