My goal is to find a standard product CF tube which could reach the E value and the compressive strength of mild steel.
I can find plenty of tubes that you make, but none have the needed information to make a decision, so I have attached a design document showing the requirement.
Hopefully help is close.
Cheers
John
For front push rod-find Ixx for worst case load, 15660 N cornering,with SF = 2
S.F of 2 to reach YP of MS = 250 MPa t & c, 150 MPa shear. For pure compression τ max = ½ axial stress
A = F/σ = 15660/125= 125.3 mm2
Tube dia = 25.4 mm , circ = 79.8 mm find thickness.
A = Pi*r2t = A/circ = 125.3 /79.8= 1.57 mm(1x1/16” 25.4 x 1.57mm) Ixx = 8350 mm4 (stress limit)
Pcr = π2EI / l2(pin – pin ends)
Steel = π2 *207000*8350 /11002
= 14100 N (= S.F 0.9)not ok2.3 Kg (bow & ball joint eccentricity of less than (0.035”) 0.1%)
(1x3/16” 25.4 x 4.78mm) Ixx = 17300 mm4
= π2 *207000*17300 /11002
= 29210 N (= S.F 1.86)just ok2.67 Kg
(1 ¼x1/16” 31.75 x 1.65 mm)Ixx = 17700 mm4
= π2 *207000*17700 /11002
= 29855 N (= S.F 1.9)just ok1.32 Kghigher drag
(1 ¼x0.083” 31.75 x 2.11mm)Ixx = 21600 mm4
= π2 *207000*21600 /11002
= 36470 N (= S.F 2.3)ok1.73 Kg higher drag
Ti 1.31x0.065”E=114 GPaYP 1100 MPa
= π2 *114000*36200 /11002
= 33660 N (= S.F 2.15)ok1.73 Kg higher drag,($300/tube)no benifit
Al (1.25x0.125” 31.75 x 3.175 mm, max wall) Ixx = 29400 mm4E=69 GPaYP 275 MPa
= π2 *69000*29400 /11002(6061 T6 tube)
= 16540 lbf (= S.F 1.06)not ok(0.8 Kg) higher drag
CF ???
My goal is to find a standard product CF tube which could reach the E value and the compressive strength of mild steel. Then the 31.75 x 1.65 mm size would be ok and the weight would be much less.
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JohnM
Hi All,
My goal is to find a standard product CF tube which could reach the E value and the compressive strength of mild steel.
I can find plenty of tubes that you make, but none have the needed information to make a decision, so I have attached a design document showing the requirement.
Hopefully help is close.
Cheers
John
For front push rod -find Ixx for worst case load, 15660 N cornering, with SF = 2
S.F of 2 to reach YP of MS = 250 MPa t & c, 150 MPa shear. For pure compression τ max = ½ axial stress
A = F/σ = 15660/125 = 125.3 mm2
Tube dia = 25.4 mm , circ = 79.8 mm find thickness.
A = Pi*r2 t = A/circ = 125.3 /79.8 = 1.57 mm (1x1/16” 25.4 x 1.57mm) Ixx = 8350 mm4 (stress limit)
Pcr = π2EI / l2 (pin – pin ends)
Steel = π2 *207000*8350 /11002
= 14100 N (= S.F 0.9) not ok 2.3 Kg (bow & ball joint eccentricity of less than (0.035”) 0.1%)
(1x3/16” 25.4 x 4.78mm) Ixx = 17300 mm4
= π2 *207000*17300 /11002
= 29210 N (= S.F 1.86) just ok 2.67 Kg
(1 ¼x1/16” 31.75 x 1.65 mm) Ixx = 17700 mm4
= π2 *207000*17700 /11002
= 29855 N (= S.F 1.9) just ok 1.32 Kg higher drag
(1 ¼x0.083” 31.75 x 2.11mm) Ixx = 21600 mm4
= π2 *207000*21600 /11002
= 36470 N (= S.F 2.3) ok 1.73 Kg higher drag
Ti 1.31x0.065” E=114 GPa YP 1100 MPa
= π2 *114000*36200 /11002
= 33660 N (= S.F 2.15) ok 1.73 Kg higher drag, ($300/tube) no benifit
Al (1.25x0.125” 31.75 x 3.175 mm, max wall) Ixx = 29400 mm4 E=69 GPa YP 275 MPa
= π2 *69000*29400 /11002 (6061 T6 tube)
= 16540 lbf (= S.F 1.06) not ok (0.8 Kg) higher drag
CF ???
My goal is to find a standard product CF tube which could reach the E value and the compressive strength of mild steel. Then the 31.75 x 1.65 mm size would be ok and the weight would be much less.
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