During the operation of the sliding spiral, it mainly bears the action of torque and axial tension (or pressure).
At the same time, there is significant relative slippage between the screw and nut threads when they rotate relative to each other. The main form of failure is thread wear.
Therefore, the basic size of the sliding spiral is generally determined based on the wear resistance conditions.
Calculation of screw wear resistance
Data: F=41048 (N), according to “Mechanical Project”:
For trapezoidal thread, if h=0.5P, then according to “Mechanical Project”, (p)=22 MP.
Therefore,
where Φ = 4 can be obtained for high transmission accuracy, heavy load and long service life.
(p) is the allowable pressure of the material, according to “Mechanical Project”, (p)=22 MPa.
d² can be considered 50.5 mm to meet the d² requirement. Therefore, the nominal diameter d=55mm.
Thread outer diameter d 3 =45mm.
Internal thread diameter D 4 =56mm and D 1 =46mm.
Thread pitch p=9mm.
Nut height H=202mm.
Calculation of screw strength
Given the torque.
According to “Mechanical Design”, for general threads with M10~M16, tanφ v =0.17.
Given D = 100 mm and t = 25 mm,
Therefore,
Critical cross-sectional area,
Where A is the minimum cross-sectional area of the screw.
According to “Mechanical Design”, σ is =320~360 MPa. Thus, σ ca ≤ (σ). Therefore, it is concluded that the previously obtained screw data meets the strength requirements.
Calculation of nut resistance
As the strength of the nut material is lower than that of the screw, simply check the strength of the nut thread. Take the nut height H = 202 mm. The flexural strength condition of the dangerous cross-section a—a of the thread is:
where b is the thickness of the wire root, mm. For trapezoidal threads, b=0.65P=5.85mm.
l is the bending moment arm, mm.
(τ) is the allowable shear stress of the nut material.
(σ b ) is the allowable bending stress of the nut material.
Therefore,
The left and right nuts are 101 mm tall.
Calculation of screw stability
The critical load Fcr can be calculated using Euler's formula.
where E is the tensile and compressive modulus of elasticity of the screw material, Mpa, E=2.06×10 5 MPa.
I is the moment of inertia of the dangerous section of the screw, mm 4 .
Therefore,
Therefore, the conclusion is that the stability of the screw is reliable.
