// CODE SNIPPET
218.104.22.168 Modulus of Rupture
JUMP TO FULL CODE CHAPTER
Adobe units shall have an average modulus of rupture of 50 psi (345 kPa) when tested in accordance with the following procedure. Five samples shall be tested and individual units shall not have a modulus of rupture of less than 35 psi (241 kPa).
A cured unit shall be simply supported by 2-inch-diameter (51 mm) cylindrical supports located 2 inches (51 mm) in from each end and extending the full width of the unit.
A 2-inch-diameter (51 mm) cylinder shall be placed at midspan parallel to the supports.
A vertical load shall be applied to the cylinder at the rate of 500 pounds per minute (37 N/s) until failure occurs.
The modulus of rupture shall be determined by the equation:
f r = 3 PLs /2 Sw (St 2) (Equation 21-2)
|where, for the purposes of this section only:|
|Sw||=||Width of the test specimen measured parallel to the loading cylinder, inches (mm).|
|fr||=||Modulus of rupture, psi (MPa).|
|Ls||=||Distance between supports, inches (mm).|
|St||=||Thickness of the test specimen measured parallel to the direction of load, inches (mm).|
|P||=||The applied load at failure, pounds (N).|