In the box section beam shown in Fig. 24.18 the covers 1 and 3 are identical laminates each comprising four orthotropic plies, 0.15 mm thick, and for which the ply angle is 45°. The ply elastic constants are El=140 000 N/mm^2, Et=10 000 N/mm^2, Glt=5000 N/mm^2 and νlt=0.3. The webs 2 and 4 are also

Question

In the box section beam shown in Fig. 24.18 the covers 1 and 3 are identical laminates each comprising four orthotropic plies, 0.15 mm thick, and for which the ply angle is [latex]45^{\circ}[/latex]. The ply elastic constants are [latex]E_{l}=140000 N / mm ^{2}[/latex], [latex]E_{t}=10000 N / mm ^{2}, G_{l t}=5000 N / mm ^{2} ext { and } v_{l t}=0.3 .[/latex] The webs 2 and 4 are also identical laminates each comprising four plies 0.13 mm thick but in which the outer ply angle is [latex]30^{\circ}[/latex] while the angle of the two inner plies is [latex]45^{\circ}[/latex] For the outer plies [latex]E_{l}=200000 N / mm ^{2}, E_{t}=15000 N / mm ^{2}, G_{l t}=10000 N / mm ^{2} ext { and } v_{l t}=0.3[/latex]. and for the two inner plies [latex]E_{l}=140000 N / mm ^{2}, E_{t}=10000 N / mm ^{2}, G_{l t}=5000 N / mm ^{2} ext { and } v_{l t}=0.3 .[/latex] If the beam is subjected to an axial tensile load of 50 kN determine the share of the load carried by the covers and the webs.

Accepted Answer

The equivalent elastic constants for the covers have been calculated in Ex. 24.10 and are

[latex]E_{Z}=E_{x}=10893 N / mm ^{2}[/latex]

For the webs, from Example 24.11

[latex]E_{Z}=E_{x}=29908 N / mm ^{2}[/latex]

Then, for each cover

[latex]b_{i} t_{i} E_{Z, i}=150 imes 4 imes 0.15 imes 10893=9.8 imes 10^{5} N[/latex]

For each web

[latex]b_{i} t_{i} E_{Z, i}=50 imes 4 imes 0.13 imes 29908=7.8 imes 10^{5} N[/latex]

Then

[latex]\sum\limits_{i=1}^{n} b_{i} t_{i} E_{Z, i}=2 imes 9.8 imes 10^{5}+2 imes 7.8 imes 10^{5}=35.2 imes 10^{5} N[/latex]

From Eq. (24.66)

[latex]\varepsilon_{Z}=\frac{P}{\sum\limits_{i=1}^{n} b_{i} t_{i} E_{Z, i}}[/latex] (24.66)

[latex]\varepsilon_{Z}=\frac{50 imes 10^{3}}{35.2 imes 10^{5}}=1.42 imes 10^{-2}[/latex]

Therefore, from Eq. (24.65)

[latex]P=\varepsilon_{Z} \sum\limits_{i=1}^{n} b_{i} t_{i} E_{Z, i}[/latex] (24.65)

[latex]P( ext { covers })=1.42 imes 10^{-2} imes 9.8 imes 10^{5}=13916 N[/latex]

[latex]P( ext { webs })=1.42 imes 10^{-2} imes 7.8 imes 10^{5}=11076 N[/latex]

Check: Total load=2(13 916+11 076)=49 984 N which is approximately 50 kN the discrepancy being due to rounding off errors. Having obtained the axial force in each cover and web the axial force intensities can be calculated. The strength of the box beam can then be investigated as illustrated in Example 24.12.

Question 24.14: In the box section beam shown in Fig. 24.18 the covers 1 and...

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