Masonry Magazine July 1976 Page. 15
TIN TERMS OF w2/ (SINGLE SPANS)
TIN TERMS OF w/ ICONTINUOUS SPANS)
0.30
SUPPORT
1/20
0.25
0.20
0.15
0.10
1/10
TABLE 1. Moment stress coefficients for simply supported single-span wall beams (uniform loading);
½
MIDSPAN & SUPPORT,
1/2
0.06
h
β 21
Top of Beam
Bottom of Beam
1/2
+0.75
-1.2
02
04
06
08
1.0
2/3
+0.33
-1.05
1
+0.06
-0.97
HEIGHT TO SPAN RATIO-B
FIGURE 2. Resultant, T, of Tensile Stresses in Girders Having Uniform Loading.
EXAMPLE 2. Design an interior span of a continuous deep beam for which the following data are given.
fm 2000 psi
Length of span, 30 ft.
Width of beam, b 11.625 in.
Length of support, c = 3 ft.
Height of beam, h= 15 ft.
Uniform load, w 5000 lb/ft. 416.67 lb./in.
(a) Check compressive stress at support.
Allowable comp. stress, Fa 0.20 fm [1-48]
Fa 0.2(2000) 1-0.033)
387 psi
33 (5000)
Actual comp. stress= (11.625) [36+2(11.625)]
239 psi; OK
(b) Compute flexural stresses and required reinforcement.
3
30
Flexural stresses
1
h
15
30
(coefficient from Table 2)()
416.67
Midspan: Top flexural stress = (1.05) (11.625) = 37.6 psi
Bottom flexural stress = (-1.31) (11.625) = -46.9 psi
Support: Top flexural stress = (-1.25) (35.8)-44.7 psi
Bottom flexural stress = (9.32)(35.8) = 333.6 psi
Allowable flexural com stress 03366660
Based on data contained in "Design of Deep Girders," Portland Cement Association, 1951. Positive values indicate compression; negative values indicate tension,
From Figure 2, lind coefficients for T and multiply by w/.
Midspan: T (0.12) (5000)(30) 18,000 lb.
18,000
As 20,000 =0.9 sq.in. (locate near bottom of beam)
Support: T (0.23)(5000)(30) 34,500 lb,
34,500
As = 20,000 1.725 sq.in. (locate near top of beam)
(c) Compute shear stresses and required reinforcement.
As in Example 1, assume critical section for shear is located 0.15 In from face of support
0.15 0.15 (27) 4.05 ft.
V= 133-(4.05 +1.5)) (5000) = 137,250 lb.
V
137,250
bd
(11.625) (1804)
= 67 psi
(Note: Horiz, steel assumed to be located 4" from bottom of beam)
Allowable V 1.1m 49.2 psi
(maximum of 50 psi without shear reinforcement)
Allowable V3m 134 psi
(maximum of 150 psi with shear reinforcement taking all of shear)
Calculate required shear reinforcement.
Assuming spacing of shear reinf., s=16",
Αν V (137,250) (16)
fyd (20,000)(176)
0.62 sq.in.
Use No. 8 vertical bars @ 16" o.c." (0.59 sq.in./ft.)
Minimum wall steel 0.002(11.625)(12)
0.279 sq.in./ft.
Use minimum of No. 4 bars 48" o.c. in horiz. direction
(A.049 sq.in./ft.)
*Note: ACI Code requires that shear reinforcement required ut critical section be used throughout the soon