# A garden hose has stopped flow. There is constant static pressure and no water hammer. Why does the hose tear in the axial direction at the end of the hose?

Hint: Mechanics of Materials

If you assume a thin wall pressure vessel, rho_long = pr/2t while rho_hoop = pr/t. Since hoop stress is twice as large as longitudinal stress, the hose will tear axially.

As @Dexter_Kenta_Yanagis mentions, this problem disguises a garden hose as a thin-walled pressure vessel, which is commonly covered in Mechanics of Materials courses.

In a cylindrical vessel with Pressure P, there are 2 stresses to consider: hoop stress trying to pull the cylinder apart radially, and longitudinal (axial) stress trying to pull the cylinder apart axially.

A visual representation of the hose as a thin-walled pressure vessel is shown here, where you can see the directions in which hoop and longitudinal stress act.

The equations for the two stresses are:

σ_h = PR/T
σ_l = PR/2T

where P = pressure, R = radius of the cylinder, & T = thickness

Since the hoop stress is twice of the longitudinal stress, the hose will tear axially. Think of a hotdog cracking along the longitudinal direction first (i.e. skin fails from hoop stress generated by internal pressure!)