When yielding of steel is not required!!

[Yousef ZAA]

Hi all,
Sometimes you see equations of minimum reinforcement of different types of structural system and you see for instance the tensile force is divided by (0.5*fy)to obtain that amount of reinforcement. My question why do we want to put steel that do not yield?!

 

[hokie66]

I thought that would be obvious. If reinforcement yields, the ultimate capacity of the member has been reached, and lots of bad things happen, like excessive deflection, followed by collapse.

 

[Geoff13]

There needs to be additional strength in our designs to reflect our lack of perfect knowledge. Are the design loads actually the maximum that will happen? Are the materials actually as strong as we predict? Did everything get built exactly as drawn? Is our calculation only an approximate model of reality? Etc, etc, etc.

Allowable stress design does this limiting the design strength to a smaller percentage of failure, such as you reference. Ultimate strength design factors up the loads to something larger than the expected loads, but then allows stresses much closer to a failure (perhaps 0.85 or even 0.9 of yield).

Both methods result in structures where, if all things were perfect (i.e. loads, materials strength, dimensions, etc) there would be a reserve strength before failure, commonly refered to as the factor of safety (FoS). Some codes are very conservative, such as ASME Section VIII Division 1 for pressure vessels which aims for a FoS of 3.5, whereas structural codes typically allow FoS to be in the 1.6 to 1.8 range.

 

[Yousef ZAA]

Thank you all.

 

[rapt]

It could also be a service stress condition. I think there is or previously was a condition in prestress service stress calculations in ACI318 where .5fsy was the steel stress used to determine the reinforcement area required.