reinforcement for cantilevered and propped CMU walls 3

[EngineerofSteel]

I am designing CMU walls for dairies. These walls are typically 10 to 14 feet tall and span 50 to 100 feet. I am using the previous engineer's methods. He only differentiated the walls as to "propped" or "cantilever". He used wl^2/8 and wl^2/2 to calculate the moments.

I think these formulas are overconservative. No allowance is made for perpendicular walls at both ends (The walls form a rectangular box shape.) Additionally, there are frequently interior walls which doweled into the exterior walls or else constructed integral to the exterior wall.

How can I account for the contributing resisting force of these walls?

Also, some designs have an interior finished floor 2-4 feet above the ground elevation. I think I can show this as a "beam overhanging one support" with a uniformly distributed wind load. It seems reasonable to model a 13 foot CMU wall supported at 4' with a concrete slab and earth pressure between 0 and 4 as a 9' wall, not as a 13' wall.

Does anyone have tips for best calculating reinforcing in these situations?

Thanks, Dairy Designer

 

[DaveAtkins]

Once the wall is over twice as long as it is high, it will span in the vertical direction only. So "propped" or "cantilever" would be appropriate.
If the interior walls are indeed integral with the exterior wall (no control joint), then they can act as shear walls. You must verify they do not fail in shear or overturning, and trace the loads to the ground.
I agree that a wall supported by a slab-on-grade on one side, and soil on the other, really does not span the extra four feet to the wall footing.

DaveAtkins

 

[miecz]

You may have an isolation joint between your slab and wall, with 1/2 inch of joint filler. I wouldn't count on this to support the wall.

It's hard to carry wall forces horizontally, as you now have to provide horizontal reinforcing or count on joint reinforcment to carry the bending moment (double duty?). Also, there's the question of anisotropic behavior.

I analyze the wall as a vertical beam with axial compression, fixed at the bottom, and pinned at the top. Bending moment is wl^2/8 at the base of the wall. Axial force is at least the weight of the wall. The axial force can reduce the required reinforcing considerably. I then check the wall for the bending and axial force at 3H/8 from the top.

 

[EngineerofSteel]

Dave Atkins,

I see what you are saying. In many situations, I have situations which do not exceed the 2width:1height criterion you gave. For example, a 13' cmu wall spanning 75 feet with two interior walls evenly intersecting the wall to leave 3 short spans of 25' each. (2*13 > 25).

Now, I believe I can legitimately model the 25' spans as horizontal beams. This will produce a greater economy against using the cantilever equation w*l^2/2. AND I can avoid special inspection without using larger than #5 bar.

Pravda?

Thanks, Dairy Designer

 

[whyun]

DairyDesigner,

Horizontal bars may be designed assuming that the wall spans 25 feet horizontally. You still have to provide a minimum vertical reinforcement and design the cross walls to take the reaction as a shear wall. Design using horizontal span may not necessarily be more economical.