A platform weighting a total of 6400 kN is supported by four struts of height 4000 mm each. Assuming the total load is equally shared, design the compression members with circular cross section. Use the carbon/epoxy composite with the following material data:

$E_1=130 \ GPa,E_2=6.5 \ GPa,G_{12}=5.0 \ GPa,ν _{12}=0.25, \ \text{ and } \ X^C=800 \ MPa$

Question

A platform weighting a total of 6400 kN is supported by four struts of height 4000 mm each. Assuming the total load is equally shared, design the compression members with circular cross section. Use the carbon/epoxy composite with the following material data:

$E_1=130 \ GPa,E_2=6.5 \ GPa,G_{12}=5.0 \ GPa,ν _{12}=0.25, \ \text{ and } \ X^C=800 \ MPa$

[latex]E_1=130 \ GPa,E_2=6.5 \ GPa,G_{12}=5.0 \ GPa,ν _{12}=0.25, \ ext{ and } \ X^C=800 \ MPa[/latex]

Accepted Answer

Let us consider a tubular compression member composed primarily of ±15° plies. From a strength consideration, the required area of cross section is worked out as

[latex]A=\frac{1600 imes 1000}{800 imes \cos^215^\circ } =2144 \ mm^2[/latex]

Let us consider a hollow tube of thickness 11 mm and outer diameter 120 mm manufactured by filament winding. The tube is to be made by 10 helical plies of 1.0 mm each and two hoop plies of 0.5 mm each. The innermost and outermost plies are provided as hoop plies. The outermost hoop helps in consolidation and better finish. The innermost hoop makes it symmetric, and in certain cases, it helps prevent fiber pull-out. Each helical is actually a combination of two plies at +15° and −15°, each of 0.5 mm thickness.
Owing to the slenderness of the tube, its design is critical from a buckling point of view. Using Equation 14.56, the axial modulus for the 15° plies is 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{2ν_{12}}{E_1} ight group \sin^2 heta \cos^2 heta ight] ^{-1} \quad\quad\quad\quad\quad (14.56) \ E_{xx,15^\circ }=\left[\frac{\cos^415^\circ }{130}+\frac{\sin^415}{6.5}+\left\lgroup\frac{1}{5}-\frac{2 imes 0.25}{130} ight group imes \sin^215^\circ imes \cos^215^\circ ight] ^{-1}=50.9 \ GPa [/latex]

Similarly,

[latex]E_{xx,90^\circ }=6.5 \ GPa[/latex]

The cross-sectional areas of the helical and hoop plies are 3424.3 and 342.4 mm², respectively. An effective axial modulus can be obtained by rule of mixture as

[latex](E_{xx})^{eff}=46.9 \ GPa[/latex]

The moment of inertia for the chosen area of cross section is obtained as

[latex]I_{yy}=\frac{\pi imes (120^4-98^4)}{4}=90.42 imes 10^6 \ mm^4 [/latex]

Then, considering pin ends, using Equation 14.60, the critical buckling load is estimated as follows:

[latex]P_{cr}=k \pi^2\left\lgroup\frac{E^b_{xx}I_{yy}}{l^2} ight group \quad\quad\quad\quad\quad (14.60) \ P_{cr}=\frac{\pi^2 imes (46.9 imes 1000) imes (90.42 imes 10^6)}{4000^2 imes 1000} =2616 \ kN[/latex]

which implies a buckling factor of 1.6.
Note: The design of a slender compression member is driven by stability. Buckling being an inherently catastrophic phenomenon, often a high buckling factor is desired. In our present case, the buckling factor can be increased by the following feasible means: (i) increase in the cross section, (ii) reduction in angle of winding, and (iii) use of fixed end connections.

Question 14.5: A platform weighting a total of 6400 kN is supported by four...

Related Answered Questions

Design a cantilever beam of length 400 mm and width 25 mm to carry a tip lateral load of 5 N. Restrict the tip deflection within 1.0 mm. Use unidirectional carbon/ epoxy prepreg with the following material data: E1 = 130 GPa, E2 = 6.5 GPa, G12 = 5.0 GPa, ν12=0.25. Each prepreg ply is of 0.25 mm ...

Verified Answer:

Design the composite shell of a cylindrical pressure vessel with an inner diameter of 200 mm and a cylindrical length of 800 mm. Internal pressure is 20 MPa. The end domes are hemispherical in shape and reinforced by polar end fittings that have equal composite pole opening diameters of 40 mm ...

Verified Answer:

Verified Answer:

Determine the elongation of the tension member designed in Example 14.3. Consider the following data: E1=130 GPa, E2=6.5 GPa, G12=5.0 GPa, and ν12=0.25 . If the axial elongation is to be limited to 0.02 mm, what should be the design modifications, if any? ...

Verified Answer:

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 Vf = 0.6 are E1 = 130 GPa, X^T = 2400 MPa, and ρc = 1.53 g/cm³ ...

Verified Answer:

Verified Answer:

Consider the data given in Example 14.1. If the axial elongation is to be limited to 0.01 mm, design the tension member adopting a micromechanics-based approach. ...

Verified Answer: