proppastie
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To calculate the buckling axial load of a tube, we use the Euler's formula for buckling of a column:
Pcr=π2EI(KL)2P_{cr} = \frac{\pi^2 E I}{(K L)^2}Pcr=(KL)2π2EI
Where:
I=π64(D4−d4)I = \frac{\pi}{64} \left( D^4 - d^4 \right)I=64π(D4−d4)
Where:
d=D−2t=2−2(0.065)=1.87 inchesd = D - 2t = 2 - 2(0.065) = 1.87 \, \text{inches}d=D−2t=2−2(0.065)=1.87inches
Now, we can calculate the moment of inertia:
I=π64(24−1.874)I = \frac{\pi}{64} \left( 2^4 - 1.87^4 \right) I=64π(24−1.874)I=π64(16−12.423)I = \frac{\pi}{64} \left( 16 - 12.423 \right) I=64π(16−12.423)I=π64×3.577≈0.1756 in4I = \frac{\pi}{64} \times 3.577 \approx 0.1756 \, \text{in}^4I=64π×3.577≈0.1756in4
Pcr=π2(10,000,000)(0.1756)(1.0×10.5)2P_{cr} = \frac{\pi^2 (10,000,000) (0.1756)}{(1.0 \times 10.5)^2}Pcr=(1.0×10.5)2π2(10,000,000)(0.1756)Pcr=9.8696×107×0.1756110.25P_{cr} = \frac{9.8696 \times 10^7 \times 0.1756}{110.25}Pcr=110.259.8696×107×0.1756Pcr=1.734×107110.25≈157,000 lbsP_{cr} = \frac{1.734 \times 10^7}{110.25} \approx 157,000 \, \text{lbs}Pcr=110.251.734×107≈157,000lbs
If the boundary conditions are different (such as fixed ends or free ends), the effective length factor KKK would need to be adjusted accordingly.
4o mini
To calculate the buckling length of the tube for a given axial load of P=157,000 lbsP = 157,000 \, \text{lbs}P=157,000lbs, we can rearrange the Euler's buckling formula to solve for the length LLL:
Pcr=π2EI(KL)2P_{cr} = \frac{\pi^2 E I}{(K L)^2}Pcr=(KL)2π2EI
Rearranging for LLL:
L=π2EIPcr(K)2L = \sqrt{\frac{\pi^2 E I}{P_{cr} (K)^2}}L=Pcr(K)2π2EI
Where:
L=π2(10,000,000)(0.1756)157,000(1.0)2L = \sqrt{\frac{\pi^2 (10,000,000) (0.1756)}{157,000 (1.0)^2}}L=157,000(1.0)2π2(10,000,000)(0.1756)
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You said:
WHAT IS THE BUCKLING AXIAL LOAD OF A 10.5 INCH LONG 2 X .065 6061-T6 TUBEChatGPT said:
To calculate the buckling axial load of a tube, we use the Euler's formula for buckling of a column:
Pcr=π2EI(KL)2P_{cr} = \frac{\pi^2 E I}{(K L)^2}Pcr=(KL)2π2EI
Where:
- PcrP_{cr}Pcr is the critical buckling load,
- EEE is the modulus of elasticity for the material (for 6061-T6, E=10,000,000E = 10,000,000E=10,000,000 psi),
- III is the minimum moment of inertia of the cross-section,
- LLL is the length of the tube (in inches),
- KKK is the effective length factor, which depends on the boundary conditions (typically K=1.0K = 1.0K=1.0 for a column with both ends pinned).
Step 1: Moment of Inertia for a Hollow Tube
The moment of inertia for a hollow circular tube is given by:I=π64(D4−d4)I = \frac{\pi}{64} \left( D^4 - d^4 \right)I=64π(D4−d4)
Where:
- DDD is the outer diameter of the tube,
- ddd is the inner diameter of the tube.
d=D−2t=2−2(0.065)=1.87 inchesd = D - 2t = 2 - 2(0.065) = 1.87 \, \text{inches}d=D−2t=2−2(0.065)=1.87inches
Now, we can calculate the moment of inertia:
I=π64(24−1.874)I = \frac{\pi}{64} \left( 2^4 - 1.87^4 \right) I=64π(24−1.874)I=π64(16−12.423)I = \frac{\pi}{64} \left( 16 - 12.423 \right) I=64π(16−12.423)I=π64×3.577≈0.1756 in4I = \frac{\pi}{64} \times 3.577 \approx 0.1756 \, \text{in}^4I=64π×3.577≈0.1756in4
Step 2: Buckling Load Calculation
Now that we have the moment of inertia, we can plug in the values into Euler's formula. Assuming the tube is pinned at both ends, we use K=1.0K = 1.0K=1.0 and the given length L=10.5L = 10.5L=10.5 inches:Pcr=π2(10,000,000)(0.1756)(1.0×10.5)2P_{cr} = \frac{\pi^2 (10,000,000) (0.1756)}{(1.0 \times 10.5)^2}Pcr=(1.0×10.5)2π2(10,000,000)(0.1756)Pcr=9.8696×107×0.1756110.25P_{cr} = \frac{9.8696 \times 10^7 \times 0.1756}{110.25}Pcr=110.259.8696×107×0.1756Pcr=1.734×107110.25≈157,000 lbsP_{cr} = \frac{1.734 \times 10^7}{110.25} \approx 157,000 \, \text{lbs}Pcr=110.251.734×107≈157,000lbs
Final Answer:
The critical buckling load for a 10.5-inch long, 2" x 0.065" 6061-T6 tube (with both ends pinned) is approximately 157,000 lbs.If the boundary conditions are different (such as fixed ends or free ends), the effective length factor KKK would need to be adjusted accordingly.
4o mini
You said:
WHAT IS THE BUCKLING LENGTH WITH 157,000 lbs. AXIAL LOAD OF A 10.5 INCH LONG 2 X .065 6061-T6 TUBEChatGPT said:
To calculate the buckling length of the tube for a given axial load of P=157,000 lbsP = 157,000 \, \text{lbs}P=157,000lbs, we can rearrange the Euler's buckling formula to solve for the length LLL:
Pcr=π2EI(KL)2P_{cr} = \frac{\pi^2 E I}{(K L)^2}Pcr=(KL)2π2EI
Rearranging for LLL:
L=π2EIPcr(K)2L = \sqrt{\frac{\pi^2 E I}{P_{cr} (K)^2}}L=Pcr(K)2π2EI
Where:
- Pcr=157,000 lbsP_{cr} = 157,000 \, \text{lbs}Pcr=157,000lbs is the axial load (the given load),
- E=10,000,000 psiE = 10,000,000 \, \text{psi}E=10,000,000psi is the modulus of elasticity for 6061-T6 aluminum,
- I=0.1756 in4I = 0.1756 \, \text{in}^4I=0.1756in4 is the moment of inertia of the tube's cross-section,
- K=1.0K = 1.0K=1.0 is the effective length factor (assuming the tube has both ends pinned).
Step 1: Insert values into the formula
We already know that:- E=10,000,000 psiE = 10,000,000 \, \text{psi}E=10,000,000psi,
- I=0.1756 in4I = 0.1756 \, \text{in}^4I=0.1756in4,
- Pcr=157,000 lbsP_{cr} = 157,000 \, \text{lbs}Pcr=157,000lbs,
- K=1.0K = 1.0K=1.0.
L=π2(10,000,000)(0.1756)157,000(1.0)2L = \sqrt{\frac{\pi^2 (10,000,000) (0.1756)}{157,000 (1.0)^2}}L=157,000(1.0)2π2(10,000,000)(0.1756)
Step 2: Simplify the expression
L=9.8696×107×0.1756157,000L = \sqrt{\frac{9.8696 \times 10^7 \times 0.1756}{157,000}}L=157,0009.8696×107×0.1756L=1.734×107157,000L = \sqrt{\frac{1.734 \times 10^7}{157,000}} L=157,0001.734×107L=110.25≈10.5 inchesL = \sqrt{110.25} \approx 10.5 \, \text{inches}L=110.25≈10.5inchesFinal Answer:
The buckling length of the tube for an axial load of 157,000 lbs is approximately 10.5 inches, which is the same as the original length. This suggests that the given axial load is precisely the critical buckling load for this tube. If the axial load were greater than the critical load, the tube would buckle before reaching the given length.Important Notes:
- If you want to check buckling for other boundary conditions (fixed, free, or mixed ends), the value of KKK would change (e.g., K=2.0K = 2.0K=2.0 for both ends fixed).
- The calculation assumes an idealized scenario where the load is purely axial and the material behaves elastically according to the Young’s modulus.
