can you expleain repititive member for a joist. Cr =1.15 joist spacing is less than 24",...

can you expleain repititive member for a joist. Cr =1.15 joist spacing is less than 24", there are more than three. how to determine if there are more than three?

Expert Solution

?? ? As per ASTM the bending stress for each member is calculated by Formula
x = Fb/(1-(KV/N)
where Fb is the 5% exclusion limit of the allowable bending stress of an individual member, K is the distance from the mean to the lower percentile in terms of standard deviates, V is the coefficient of variation (COV), N is the number of members in the assembly, and ?? is the allowable bending stress of a member as a result of load-sharing.
Generally K value is 1.645 from Normal distribution , N = no of members. V = 0.25 for wood lumbers .
Explaination - The repetitive member factor is used to adjust the allowable design stress for the member you are designing. The allowable design stresses for wood are based on the 95th percentile of actual load capacity. So for every 100 joists loaded, 95 will have an actual strength greater than the design strength. When designing a single member, there is a probability that the member will fall into the lower 5th percentile. When you have 3 or more members that can share load, what is the likelihood that all three will fall into the lower 5th percentile. Through physical testing, a 15% strength increase was calculated to account for this probability.

Ally Wells answered 1 year ago

for joist spaced less than 24 inches OC there are 3 commonly used adjustment factors name...

for joist spaced less than 24 inches OC there are 3 commonly used adjustment factors name them

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