Question 14.5: A platform weighting a total of 6400 kN is supported by four...
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
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
A=\frac{1600\times 1000}{800\times \cos^215^\circ } =2144 \ mm^2
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:
E_x=\left[\frac{\cos^4 \theta}{E_1}+\frac{\sin^4 \theta}{E_2}+\left\lgroup\frac{1}{G_{12} }-\frac{2ν_{12}}{E_1} \right\rgroup \sin^2 \theta \cos^2 \theta \right] ^{-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\times 0.25}{130} \right\rgroup \times \sin^215^\circ \times \cos^215^\circ \right] ^{-1}=50.9 \ GPa
Similarly,
E_{xx,90^\circ }=6.5 \ GPa
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
(E_{xx})^{eff}=46.9 \ GPa
The moment of inertia for the chosen area of cross section is obtained as
I_{yy}=\frac{\pi\times (120^4-98^4)}{4}=90.42 \times 10^6 \ mm^4
Then, considering pin ends, using Equation 14.60, the critical buckling load is estimated as follows:
P_{cr}=k \pi^2\left\lgroup\frac{E^b_{xx}I_{yy}}{l^2} \right\rgroup \quad\quad\quad\quad\quad (14.60) \\ P_{cr}=\frac{\pi^2\times (46.9\times 1000)\times (90.42\times 10^6)}{4000^2\times 1000} =2616 \ kN
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.