In Fig 12.5, alpha is the angle between T and the Y axis, not between T and the X axis.Mechaman said:
haruspex said:
The torque is around the X axis, so the vector is along the X axis, i.e. where alpha is zero. So yes, it is cos because cos(0)=1.Mechaman said:
Thanks for response sorry for late reply. I see now.haruspex said:
The spring force due to axial and torque load is the force exerted by a spring when it is stretched or compressed due to an applied axial or torque load. This force is directly proportional to the displacement of the spring from its equilibrium position.
The formula for calculating the spring force due to axial and torque load is F = kx, where F is the spring force, k is the spring constant, and x is the displacement of the spring from its equilibrium position. This formula applies to both axial and torque loads.
The spring force due to axial and torque load is affected by several factors including the spring's material, diameter, and length, as well as the magnitude and direction of the applied load. These factors can impact the spring's stiffness and its ability to resist deformation under a load.
The spring force due to axial and torque load increases proportionally with the applied load. This means that as the load increases, the spring will stretch or compress further, resulting in a greater force exerted by the spring. However, if the load exceeds the spring's elastic limit, the force will no longer increase proportionally and the spring may become permanently deformed.
Axial load is a force applied along the axis of a spring, while torque load is a rotational force applied to the spring. Both types of loads cause the spring to compress or stretch, but the direction of the force and the resulting displacement of the spring will differ. Additionally, the formula for calculating the spring force will also differ depending on whether it is an axial or torque load.