The behavior of a fixed-end A992 steel shaft when subjected to torque can be complex. Given that the shaft has a diameter of 65 mm, it is essential to understand how the shear modulus of rigidity, which is 75 GPa, influences its response to applied torques.
When a torque is applied to the shaft, it induces shear stress and causes the shaft to twist. The amount of twist depends on the magnitude of the torque, the length of the shaft, and the shear modulus of the material. In this case, the fixed ends at points A and B impose boundary conditions that affect the distribution of shear stress along the shaft.
To calculate the shear stress (τ), we can use the formula:
τ = T * r / Jwhere:
- τ is the shear stress,
- T is the applied torque,
- r is the radius of the shaft (which is half of the diameter), and
- J is the polar moment of inertia.
The polar moment of inertia for a circular shaft is given by:
J = (π/32) * d^4Inserting the diameter (d = 65 mm or 0.065 m), we can calculate J and subsequently the shear stress for various values of T.
Moreover, the angle of twist (θ) can be estimated using the formula:
θ = (T * L) / (J * G)Here, L represents the length of the shaft and G is the shear modulus (75 GPa). The fixed ends prevent any rotation at points A and B, thus the angle of twist will provide insight into how the shaft behaves under specific loading conditions.
In summary, the interaction between applied torque and the mechanical properties of A992 steel results in shear stresses and angular deflections, describing the shaft’s performance under torsional loads.