(665psi)(0.32)(0.89)(40)(3.81) 2
=
= 4.58kipft = 54.98kipin (6.21 KNm) > 33.6kipin (3.8 KNm) , OK
M rm
2(12)
FEMA 302 Section 11.10.1 requires that the nominal flexural strength of the wall for outofplane flexure
be at least equal to 1.3 times the cracking moment of the wall. The cracking moment was calculated
previously to be 28.52 kipin. The flexural strength of the wall determined by allowable stress design was
calculated as 47.7 kipin > 37.1 kipin (= 1.3 x 28.52). The flexural strength of the wall calculated using
ultimate strength design is much greater than the strength calculated from allowable stresses. Therefore,
assume OK.
Outofplane shear strength check

Shear force demand;
The outofplane shear force demand is determined from the horizontal force on the wall face of 16.4
psf. Wall E1E2 is the most critical with an unbraced span of 20' Therefore, the shear demand for a
.
40" wide section is:
f = wL/2 = (16.4 psf)(40")(1 ft. / 12")(20 ft. / 2) = 547 lb / 40" (2.39 KN/m)
1 inch = 25.4 mm
Shear capacity;
Effective shear area, Ae = (8.3")(3.81") = 31.62 in.2 / 40"
(TM 58093 Fig. 52)
fv = Ra / bwd , where bwd = Ae and Ra = 547 lb.
(TM 58093 Eq. 617)
Fa = 547 lb / 31.62 in.2 = 17.3 psi (119 KN/m2) < 69.7 psi (481 KN/m2), OK
Outofplane bracing forces
Anchorage of walls to flexible diaphragms shall have the strength to develop the outofplane force give
by:
Fp = 1.2SDSIWp
FEMA 302 Eq. 5.2.6.3.3

Interior wall E1E2
Wp = (57psf)(20'2) = 570 plf (8.32 KN/m)
/
Fp = 1.2(0.6)(1.0)(570) = 410 plf Equivalent to 0.41 ( 40 / 12 ) = 1.37 kips / 40" (5.98 KN/m)
Minimum anchorage demand = 200 plf (2.92 KN/m) < 410 plf (5.98KN/m) (Per Sec. 72.e(2))
H145