The deflection of a fixed-fixed pier is Δ = Δbending + Δshear = Ph^3/12EI + 1.2Ph/EA. If you assume that E = 1,000,000 psi, t = 1 inch, and P = 100,000 lbs, this can be rewritten as Δ = 0.1(h/d)^3 + 0.3(h/d). If you assume that P = 1,000,000 lbs, the equation becomes Δ = (h/d)^3 + 3(h/d). Since your actual walls won't match any of these assumptions, the latter two equations are only useful for finding the relative rigidities of the walls to be used in distributing forces. If you want to find the actual displacements, you will need to use the the first equation.

Not to hijack the thread, but...

I just used a first principles approach to try to find a corresponding shear deformation value for a concrete wall.
Long story short, I used a value of 0.15 for poisson's ratio (v) and got 2.3Ph/EA. Why would the masonry code give a smaller shear deformation all things remaining equal?

Some background, I used (delta) = Ph/GA, where G = E/2(1+v)

Furthermore, in order to achieve the masonry equation above (1.2Ph/EA), you would need a negative value for 'v.'...

What am I missing?

Could it be that the flexural connection is at each end and the distance from the inflection point to the end is 1/2 assuming a point of inflection at mid height?

Dik

That may be true for the flexural component, but I'm dealing with the shear component which should be constant through the wall section regardless of the wall fixity conditions.

The shear deflection is a function of the effective shear area, which accounts for the non-uniform shear stress across the depth of the section. For rectangular sections, the effective shear area is 5/6 the gross area. Invert 5/6 to get the 1.2 shown in the shear deflection term in the above equation. Also, poisson's ratio is commonly assumed to be 0.25 which results in G = 0.4E. Some engineers reduce this further to account for cracking.

Also, I should have written Ev (or G) in the shear deflection term of the equation I wrote above. 1.2Ph/EvA or 1.2Ph/GA.

That makes sense. So in equivalent terms a masonry wall would allow for 30% more shear deformation.

Thanks for the clear up. I was hoping I wasn't going crazy there.

No problem. I don't follow your statement about 30% more shear deformation. What are you comparing it to?

atrzzy,
Also, the masonry code, ACI 530-11 section 1.8.2.2.2 defines E_v = 0.4 E_m. Similarly for concrete, ACI 318-11 commentary section R8.8.2 states ". . . the shear modulus may be taken as 0.4 E_c."

Deker,

I'm working in Canada with CSA A23.3 and as far as I can tell there's no clause stating that the shear modulus should be taken as 0.4Ec. I could be mistaken about this so if anyone's aware of this clause please let me know.

Anyhow, for that reason I revert to first principles: G = Ev = Ec/2(1+poisson), for poisson = 0.15, G = Ev = 0.43Ec. So far so good.

So then I compared the shear deformation formulas for masonry and concrete separately. The reason being is that I would expect more deformation to come out of the masonry formula, all things equal. (I realize that all things aren't equal, but bear with me).

For masonry, as per previous posts, = 1.2Ph/0.4EA = 3Ph/EA

For concrete, as per my derivation = Ph/0.43EA = 2.32Ph/EA

Hence the 30% more shear deformation in a masonry wall compared to a concrete wall with identical P,H,E,A (should such a thing exist).

The 1.2 factor is common to both concrete and masonry walls since it is based on the section shape and not the material. Concrete walls are stiffer, but it's because Ec is higher than Em, not because the deflection equations are different.

Does ACI specify the 1.2 factor for concrete walls? I haven't seen such a reference in CSA. Straightforward isotropic material mechanics theory doesn't suggest that a 1.2 factor is innate to a rectangular wall shape as far as I can tell.

Thanks for your time.

ACI doesn't specify the 1.2 factor that I'm aware of. I've only seen it in textbooks. Here are two links that discuss the concept: Link 1, Link 2.

Very helpful. Thank you.

Deker, could you clarify how you determined 'P=100,000 lbs'?
I used to use P=1000KN based on my senior notes but I have never seen any books discussing why P is assumed to be 1000KN or other values if any.

Thank you.

gotlboys, if you're only determining relative rigidities, P is just taken as any convenient value.

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The name is a long story -- just call me Lo.