Thus 7.6% of the total in-plane shear force on the wall will be resisted by Pier B and 92.4% of the force will

be resisted by piers C and D. The 92.4% will be distributed to piers C and D in proportion to their relative

rigidities as follows:

Therefore, 24% of the total shear force on the wall will be distributed to Pier C; 68.4% to Pier D; and 7.6%

to Pier B.

can now be accomplished. The design of each pier will begin by checking the shear and flexural stresses due

to in-plane wind loads. Axial stresses due to dead and live loads will also be checked. The flexural and axial

stresses will then be combined using the unity equation. For loading combinations that include wind loads,

the allowable stresses will be increased 33%.

reinforced and fully grouted, thus the cross section of the pier is 2*-8" by 7.62" with #5 bars in each cell.

When checking in-plane shear stresses, the assumed length of the pier, dBv, will be the actual pier length or

2*-8". When checking in-plane flexural stresses, the assumed effective depth of the beam section, dBb, will

be the actual length less the 8-inch distance from the centroid of the two end bars to the end of the pier;

therefore dBb = 2*-0".

Lateral force, VB, to Pier B = 7.6% of 10k

VB = 0.076 x 10k = 760 lbs.

Shear stress in pier B, fvB, is determined as follows:

The allowable shear stress assuming no shear reinforcement, Fvm, will be determined by equation 7-2 as

follows (assume pier fixed top and bottom):

Therefore;

Fvm = 1.0(f'm) = (1350) = 36.7 psi.

But shall not exceed 35 psi; thus Fvm = 35 psi.

fvB = 3.1 psi **< **Fvm = 35 psi x 1.33 = 46.5 psi;

Therefore, no shear reinforcement is required.

Flexural compressive stress in Pier B, fbB, is determined as follows:

Where:

Allowable flexural compressive stress in Pier B, Fb, is determined as follows:

Fb = 0.33f*m x 1.33 = 0.33(1350) X 1.33 = 600 psi

FbB = 53.9 psi < fb = 600 psi

...O.K.

Flexural tensile stress in Pier B, fsB, is determined as follows:

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