DMHSS: digital material
          hierarchical space structures 
         
         
         
          current hierarchical prototype
         
         
         
          1km x 1km x 10m space truss
          100 x 100 x 1 cells
          100,000 struts @ 10m length
          1,000,000m total length
         
         
         
         
          TBD:
          -packing volume
          -build time
          -structural requirements
         
         
         
          300m diameter dish "arecibo in orbit", source: www.tethers.com
         
          
          
          applications:
          -high bandwidth communications with Mars & deep space probes
          -detection and characterization of NEOs
          -solar thermal power
          -space habitation
         
        
        
         
         analysis
         
        
        
          -the standard procedure for
              minimum-mass, high strength slender structures is to equate
                global and local buckling modes
             -NOTE: this is a linear elastic
              analysis. non linear behavior will likely worsen performance, and
              will be investigated soon.
             
             
             
             
             -local buckling of injection molded
              parts:
                         
                  
              
             
                           
                           
                           
                           
                           
                           
                           
                           
                           
                           
                           
                           
                           
                           
                        F = n π2 E
                I / L2   
                (1) 
              
          
             
             E = 1.35 x 10^6 psi
              / 9300 MPa
             
              assuming n = 1 for pinned connections
             
              F = π2
                  (1.35x10^6 psi) (0.00005in^4) / (2in)^2 =
                  
                  16.65 lb / 75 N
                      
                      
                      for 1P applied load, the struts take 0.2P in compression
                      (ignoring boundary conditions)
                      
                      if 0.2P = 16.65 lb, P = 83.25 lb
                      
                      I of the column is determined by taking the cross section
                      as shown:
                      
                      

                      
                      Iy is found to be 0.092 in^4
                      
                      Now we determine the global length that will result in
                      buckling at P
                      
                      L = (π2EI/P)^(1/2)
                      
                      L = (π2(1.35 x 10^6
                      psi)(0.092in^4) / 83.25 lb)^(1/2)
                      
                      L = 121 in or 10.11 ft
                      
                      
                      So for the given design, a ~10' (~50 cell) column will
                      fail at a load of 83 lbs, and the struts will fail with
                      their corresponding load of 16 lbs
                      
                      
                      zytel density = 1.39 g/cm3 = 0.05 lb/in3
                      
                      1 part = 0.1335 in3 = 0.006675 lb
                      
                      1 cell = 3 parts = 0.02 lb
                      
                      50 cell column = 1 lb
                      
                      comparison with different cross sections with same I:
                    
            
                                
                                truss wins for mass!
                                
                              
                      
                      
                      BUT
                      
                      improvement can be made by replacing the solid struts of
                      the part with tubes. this increasing buckling capacity and
                      potentially reduces mass
                      (the manufacturing challenges will be considered at a
                      later time)
                      
                      
                      QUICK MASS/VOLUME CALC FOR LAUNCH SHROUD
                      
                      MASS
                      
                      1lb/10ft= .15 kg/m
                      
                      10^6m = 1.5 * 10^5 kg
                      
                      Ariane 5 launch shroud: 
                      LEO: @21,000 kg = x7 launch shrouds
                      GTO: @10,500 kg = x14 launch shrouds
                      
                      VOLUME
                        
                      150 parts/10ft = 50 parts / m
                      
                      1,000,000m = 50,000,000 parts
                      
                      Ariane 5 launch shroud:
                      7.7m x 4.5m x 4.5m = 155 m3
                      
                      = 20,000,000 parts
                      
                      = x2.5 launch shrouds ----> MASS LIMITED!