for a 5" long 3/4"-16 grade 9 bolt clamping a joint that is 4" thick what is the correct pre torque and torque turn degrees using a .2 nut factor to get get 90% of the proof load
ChatGPT said:
Thought for 52s
Nice — let’s do the math. I’ll show the assumptions, formulas, and the step-by-step numbers so you can verify or change any assumption (for example a different proof stress, different nut-factor, or a different pre-torque percentage).
Assumptions I used
- Bolt: 3/4"-16, SAE Grade 9.
I used a proof stress = 145,000 psi (common value quoted for Grade 9 / high-strength alloy inch bolts). Halsey ManufacturingEngineering ToolBox
- Desired preload: 90% of the proof load (you requested 90%).
- Nut factor (K): 0.20 (you specified).
- Grip (g) / loaded length: 4.00 in (you stated joint thickness = 4").
- Elastic modulus (E) for steel: 30,000,000 psi.
- Thread pitch: 1/16 in per turn (3/4"-16 → 16 TPI → pitch = 1/16 in).
- Pre-torque (snug) chosen: 10% of final torque (common practical choice — if you want a different % say 20% I can recalc).
- Torque–tension relation used: T=K⋅D⋅FT = K \cdot D \cdot FT=K⋅D⋅F (standard approximation). Repair Engineering
If you want different assumptions (e.g., proof stress from a different spec, or using a measured nut factor) tell me and I’ll re-run the numbers.
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Calculations (results)
- Stress area AsA_sAs for 3/4-16 (using the formula above):
As=0.37296 in2A_s = 0.37296\ \text{in}^2As=0.37296 in2.
- Proof load (full proof): Fproof=(proof stress)⋅As=145,000 psi×0.37296=54,079 lbf.F_{proof} = (\text{proof stress})\cdot A_s = 145{,}000\ \text{psi}\times0.37296 = 54{,}079\ \text{lbf}.Fproof=(proof stress)⋅As=145,000 psi×0.37296=54,079 lbf.
- Target preload (90% of proof):
Ftarget=0.9×54,079=48,671 lbf\boxed{F_{target} = 0.9\times54{,}079 = 48{,}671\ \text{lbf}}Ftarget=0.9×54,079=48,671 lbf.
- Final torque (to produce FtargetF_{target}Ftarget) using T=KDFT = KDFT=KDF with K=0.20K=0.20K=0.20, D=0.75D=0.75D=0.75 in:
T=0.20×0.75×48,671=7,300.7 inch-lb=608.4 ft-lbT = 0.20\times0.75\times48{,}671 = 7{,}300.7\ \text{inch-lb} = 608.4\ \text{ft-lb}T=0.20×0.75×48,671=7,300.7 inch-lb=608.4 ft-lb.
Tfinal≈608 ft\cdotplb (≈825 N\cdotpm)\boxed{T_{\text{final}} \approx 608\ \text{ft·lb} \ ( \approx 825\ \text{N·m} )}Tfinal≈608 ft\cdotplb (≈825 N\cdotpm). Repair Engineering
- Pre-torque (10% of final torque) (chosen snug):
Tpre≈61 ft\cdotplb (≈82 N\cdotpm)\boxed{T_{\text{pre}} \approx 61\ \text{ft·lb} \ ( \approx 82\ \text{N·m} )}Tpre≈61 ft\cdotplb (≈82 N\cdotpm).
That pre-torque produces a preload of about 4,867 lbf4{,}867\ \text{lbf}4,867 lbf.
- Degrees of turn after pre-torque to reach the target preload:
Using bolt stiffness k=AsE/L=0.37296×30×106/4=2.797×106 lbf/ink = A_s E / L = 0.37296\times30\times10^6 / 4 = 2.797\times10^6\ \text{lbf/in}k=AsE/L=0.37296×30×106/4=2.797×106 lbf/in and pitch = 1/16 = 0.0625 in/rev, the additional rotation needed is about 90.2° (≈ quarter-turn).
θ≈90∘\boxed{\theta \approx 90^\circ}θ≈90∘
Summary (short)
- Final torque to reach 90% proof: ≈ 608 ft·lb (≈ 825 N·m).
- Suggested pre-torque (10% of final): ≈ 61 ft·lb (≈ 82 N·m).
- Additional rotation after pre-torque: ≈ 90° (quarter turn).
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If you want I can:
- Recalculate with a different K (e.g., 0.14 for lightly oiled, or 0.12 for machine-finished), or
- Use 20% pre-torque instead of 10%, or
- Give turn-of-nut only procedure (no torque wrench) starting from snug, or
- Show the full step-by-step spreadsheet of the calculations.
Which would you like me to do next?
You said:
Yes recalculate with turn of nut only with 20% pre torque.
ChatGPT said:
Thought for 13s
Nice — done. I used the same assumptions as before (3/4-16 Grade 9, proof stress = 145 ksi, stress area = 0.37296 in², nut factor K = 0.20, grip = 4.00 in, pitch = 1/16 in/rev). You asked for a
turn-of-nut only procedure where the initial snug is a
20% pre-torque (i.e. pre-torque = 20% of the final torque), then finish by turning the nut the required degrees.
Results (rounded)
- Target preload (90% of proof): 48,671 lbf.
- Final torque (theoretical, from T = K·D·F): ≈ 7,300.7 in·lb = 608.4 ft·lb = 825 N·m.
- Pre-torque (20% of final): ≈ 1,460.1 in·lb = 121.7 ft·lb = 165 N·m.
That pre-torque produces a preload of ≈ 9,734 lbf (≈ 20% of the target preload).
- Additional preload required after pre-torque: ≈ 38,937 lbf.
- Turn-of-nut required after pre-torque: ≈ 80.2° (≈ 0.2227 revolutions).
What to do, step-by-step
- Apply the pre-torque: 121.7 ft·lb (165 N·m).
- From that condition, turn the nut 80° (i.e., just under a quarter turn) to reach the target preload (≈ 90% of proof).
- Optionally verify with a calibrated tension-indicating method if available.
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You said:
what is the final clamp load for that
ChatGPT said:
Thought for 5s
The final clamp (preload) is the target we used:
48,671 lbf (≈
216.5 kN).
That’s based on 90% of the Grade-9 proof load used in the earlier calc (proof load ≈ 54,079 lbf → 0.90× = 48,671 lbf).
Practical note: torque-to-tension scatter (lubrication, thread condition, nut factor uncertainty) can make the actual clamp load vary ±10–20% from this value. If you need a tighter tolerance, use a direct tension-control method (strain-gauged bolts, load-indicating washers, ultrasonic, or calibrated test data).