It is important not to confuse a force with its representation.Mohmmad Maaitah said:
I didn't do any work because expect saying "it's zero".berkeman said:
Stress in a hollow pipe when a force is uniformly applied refers to the internal forces per unit area that develop within the pipe material to resist deformation. This stress can be axial (along the length of the pipe), radial (perpendicular to the surface), or circumferential (around the pipe's circumference), depending on how the force is applied.
Axial stress in a hollow pipe under a uniformly applied force is calculated using the formula: σ = F / A, where σ is the axial stress, F is the applied force, and A is the cross-sectional area of the pipe. For a hollow pipe, the cross-sectional area is the difference between the area of the outer circle and the inner circle.
The stress distribution in a hollow pipe is influenced by several factors, including the magnitude and direction of the applied force, the pipe's material properties (such as Young's modulus), the pipe's geometry (inner and outer diameters), and any existing imperfections or residual stresses within the pipe material.
The wall thickness of a hollow pipe significantly affects its ability to withstand stress. Thicker walls generally mean the pipe can handle higher stress levels because the material can distribute the applied force over a larger area, reducing the stress intensity. Conversely, thinner walls may lead to higher stress concentrations and a greater likelihood of material failure.
Common failure modes of a hollow pipe under uniform stress include yielding (permanent deformation), buckling (sudden lateral deflection), and fracture (cracking or breaking). The specific failure mode depends on the material properties, the magnitude of the applied stress, and the pipe's geometric characteristics.