mielke
Mechanical
- Aug 24, 2009
- 181
I am using Austenitic SS 304 annealed bolts (3/8"-16) and hypothetically say i am torquing it to 290 in-lb. using the following relationships i determine the tensile stress developed in the bolt...
F = T/[KxD] = (290)/(0.2x0.375) = 3875 lb
P = F/A = (3875)/(0.0775) = 50,000 psi
where
F is force in pounds
T is torque in inch pounds
D is nom diameter of bolt
P is stress
A is tensile stress area
I get a stress of 50,000 psi and I am a little confused on how to use this value. the yield strength for this bolt is 30ksi while the tensile strength is 75ksi. For such an application would i use the yield strength or tensile strength?
F = T/[KxD] = (290)/(0.2x0.375) = 3875 lb
P = F/A = (3875)/(0.0775) = 50,000 psi
where
F is force in pounds
T is torque in inch pounds
D is nom diameter of bolt
P is stress
A is tensile stress area
I get a stress of 50,000 psi and I am a little confused on how to use this value. the yield strength for this bolt is 30ksi while the tensile strength is 75ksi. For such an application would i use the yield strength or tensile strength?