Eccentricity (e) = 0.5t, in

The moments due to lateral wind load and to axial eccentricity are additive.

f*m = 1,350 lb/in2

Fm = (0.33)f'm = 450 lb/in2

Em = 1000f'm = 1,350,000 lb/in2

Fs = 24,000 lb/in2

Es = 29,000,000 lb/in2

Es

29,000,000

n'

'

' 21.5

Em

1,350,000

(2) *Problem--*

in table B-4.

table B-47.

(3) *Solution. *Equations are from chapters 5 and 6. Flexural Check:

Horizontal reaction at the bottom of the wall is Ra:

Pe

wh

Ra '

%

12h

2

(25 lb/ft 2)(24 ft)

(1500 lb/ft)[(0.5)(11.625 in)]

'

%

(12 in/ft)(24 ft)

2

= 30.3 + 300.0 = 330.3 lb/ft of wall

Location where maximum moment occurs is "x" distance from the bottom of the wall:

wx 2

Mx ' Rax &

2

(25 lb/ft 2)(x 2)

' (330.3 lb)(x) &

2

Differentiating with respect to x;

dMx

' Ra & wx ' 330.3 & 25x ' 0

dx

Solving for x;

330.3

' 13.2 ft bottom of wall

x'

25

Maximum moment in the wall is Mmax

(25 lb/ft 2)(13.2ft)2

Mmax' (330.31b)(13.2 ft) &

2

= 4360 - 2178 = 2182 ft-lb/foot of wall

Assume the reinforcement spacing, S, is 24 inches and determine the design maximum moment, Design Mmax,

in the wall as follows:

Design Mmax = (2182 ft-lbX24 in)/(12 in/ft)

= 4364 ft-lb/S

compression area is rectangular and compare to the T-section design from table B-4.

Masonry resisting moment is Mrm:

Fmkjbd 2

Mrm '

2(12)

Where:

p = As/bd1 = (0.44 in2)/(24 in)(5.81 in) = 0.0032

np = (21.5)(0.0032) = 0.0688

k = [(np)2 + 2np] - np

k = [(0.0688)2 + (2)(0.0688)] - 0.0688

= [0.00473 + 0.1376] - 0.0688 = 0.308

j = 1 - k/3 = 1 - 0.308/3 = 0.897

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