Can anyone explain how gravity beam loads get distributed into a fully fixed end or a zero tolerance type fitting. I have seen it assumed as an triangular distribution with the higher ends of the triangle opposing one another on each end. Any help would be appreciated.
Tek-Tips is the largest IT community on the Internet today!
Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!
-
Congratulations JStephen on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!
Fixed End or Zero Tolerance Load Distribution 1
- Thread starter cap4000
- Start date
- Thread starter
- #3
Please check the embedment load calcs on page 3 of this link.
-
1
- #5
cap4000: The triangular load distribution you mentioned would be applicable if all material is linear and the stress doesn't exceed the yield strength. If this is a cantilever, and we neglect the bond horizontal shear stress (which is equivalent to assuming zero friction), and assume the stress doesn't exceed the yield strength, I'm currently getting the following rigid-body mechanics solution, where a = distance from wall surface to where triangular load distribution crosses x axis, L = beam length (measured from wall surface), Le = beam embedment length, w1 = beam self weight per unit length (positive downward), w2 = triangular load distribution ordinate at wall surface (force per unit length, positive upward), and w3 = triangular load distribution ordinate at embedment length Le (positive upward).
a = (Le/6)(3L+4Le)/(L+Le).
w2 = (w1)(L)(3L+4Le)/(Le^2).
w3 = (w2)[1-(Le/a)].
a = (Le/6)(3L+4Le)/(L+Le).
w2 = (w1)(L)(3L+4Le)/(Le^2).
w3 = (w2)[1-(Le/a)].
- Thread starter
- #6
- Thread starter
- #7
cap4000: My w3 equation is written positive upward, which means if it gives a negative result (which it does), then the force is downward.
By the way, using a rigid-body mechanics solution, such as in my previous post, is reasonable only if the beam is relatively stiff compared to the embedment material. If, instead, the beam is relatively flexible compared to the embedment stiffness, the load will transfer more rapidly to the embedment material, such that embedment beyond a certain distance becomes ineffective. In other words, if the beam is relatively flexible, then one can't ignore elasticity and compliance, and the problem becomes more complicated, in which case you could use, e.g., finite element and surface contact to solve the problem.
By the way, using a rigid-body mechanics solution, such as in my previous post, is reasonable only if the beam is relatively stiff compared to the embedment material. If, instead, the beam is relatively flexible compared to the embedment stiffness, the load will transfer more rapidly to the embedment material, such that embedment beyond a certain distance becomes ineffective. In other words, if the beam is relatively flexible, then one can't ignore elasticity and compliance, and the problem becomes more complicated, in which case you could use, e.g., finite element and surface contact to solve the problem.
- Thread starter
- #9
vonlueke
If you have the PCI Design Handbook 3rd edition pgs 6-26 and 6-68 have a lot of conditions already calculated in a nice tabular form. If you do any precast or prestressed concrete this book is a must. Thanks Again for the Tips.
If you have the PCI Design Handbook 3rd edition pgs 6-26 and 6-68 have a lot of conditions already calculated in a nice tabular form. If you do any precast or prestressed concrete this book is a must. Thanks Again for the Tips.
- Status
Similar threads
- Question
- Question
- Question
