TM 5-809-3/NAVFAC DM-2.9/AFM 88-3, Chap. 3
(e)
Eccentricity (e) = 0.5t, in
(f)
The moments due to lateral wind load and to axial eccentricity are additive.
f*m = 1,350 lb/in2
(g)
Fm = (0.33)f'm = 450 lb/in2
(h)
Em = 1000f'm = 1,350,000 lb/in2
(i)
Fs = 24,000 lb/in2
(j)
Es = 29,000,000 lb/in2
(k)
Es
29,000,000
(l)
n'
'
' 21.5
Em
1,350,000
(2) Problem--
(a) Determine the reinforcing bar size and spacing required to resist the given loadings.
(b) Compare the calculated resisting moment values with the values for resisting moments given
in table B-4.
(c) Compare the reinforcing results from the calculated solution with the direct solution given in
table B-47.
(3) Solution. Equations are from chapters 5 and 6. Flexural Check:
(a) First determine the maximum applied moment that must be resisted by the wall.
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
(b) Determine the resisting moments n the wall assuming 1-#6 @ 24 in. o.c. Assume the flexural
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
6-5
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