The beam size is slightly deficient with re-
spect to balanced design, and the concrete will
therefore be stressed to capacity under the stip-
ulated load. In Fig. 17, let AOB represent the
stress line associated with balanced design and
A'O'B represent the stress line in the present
instance. (The magnitude QiAA' is exaggerat-
ed for clarity.)
- Develop suitable equations for
the beam
Refer to Fig. 17:
T=Tb+^d (35)
blfm™ 1?' StrCSS dlaSram f°r T where T and T» = tensile force in Present in"
stance and at balanced design, respectively.
And
bt\3d-2f)x
M = M,+ (36)
6a
- Apply the equations from step 4
Thus, M-Mb = 68,000 in-lb (7682.6 N-m). By Eq. 36, Jt = 68,000(6)(16.5)/[66(25)(49.5 -
1O)] = 103 lb/in^2 (710.2 kPa);/ 5 = 20,000 - 10(103) = 18,970 lb/in^2 (130,798.2 kPa). By
Eq. 35, T= 215,000 + 66(25)( 103)733 = 220,200 Ib (979,449.6 N).
- Design the reinforcement; establish the web width
Thus A 8 -220,200/18,970= 11.61 in^2 (74.908 cm^2 ). Use five no. 11 and three no. 10 bars,
placed in two rows. Then A 3 = 11.61 in^2 (74.908 cm^2 ); 6'min = 14.0 in (355.6 mm). It is
therefore necessary to maintain the 14-in (355.6-mm) width.
- Summarize the design
Width of web: 14 in (355.6 mm); reinforcement: five no. 11 and three no. 10 bars.
- Verify the design by computing the capacity of the member
Thus nAs = 116.1 in
2
(749.08 cm
2
); kd= [330(2.5) + 116.1(16.5)]/(330 + 116.1) = 6.14 in
(155.956 mm); Jt= 6.14/16.5 = 0.372> kb\ therefore, concrete is stressed to capacity. Then
fs = 10(1125X10.36)76.14 = 18,980 lb/in^2 (130,867.1 kPa); z = (^5 / 3 )(6.14 + 2 x 1.14)/
(6.14 + 1.14) = 1.93 in (49.022 mm); jd = 14.57 in (370.078 mm); Mallow =
11.61(18,980)(14.57) = 3,210,000 in-lb (362,665.8 N-m). This is acceptable.
DESIGN OFAT BEAM HA VING STEEL
STRESSED TO CAPACITY
Assume that the girder in the previous calculation procedure carries a total load, including
the weight of the web, of 4100 Ib/lin ft (59,835.0 N/m). Compute the area of reinforce-
ment.
