Handbook of Civil Engineering Calculations

3. Select the stirrup size
   Equate the spacing near the support to the minimum practical value, which is generally
   considered to be 4 in (101.6 mm). The equation for stirrup spacing is

<K/y

>-&

(17)

Then Av = sv'ubl($fy = 4(122)(15)/[0.85(40,000)] = 0.215 in^2 (1.3871 cm^2 ). Since each
stirrup is bent into the form of a U, the total cross-sectional area is twice that of a straight
bar. Use no. 3 stirrups for which Av = 2(0.11) = 0.22 in^2 (1.419 cm^2 ).

4. Establish the maximum allowable stirrup spacing
   Apply the criteria of the Code, or smax = d/4 ifv> 6(fc')^0 -^5. The right-hand member of
   this inequality has the value 279 lb/in^2 (1923.70 kPa), and this limit therefore does not ap-
   ply. Then smax = d/2 = 11.25 in (285.75 mm), or ,smax = Av /(0.00156) = 0.22/[0.OO 15(15)]
   = 9.8 in (248.92 mm). The latter limit applies, and the stirrup spacing will therefore be re-
   stricted to 9 in (228.6 mm).


5. Locate the beam sections at which the required stirrup spacing
   is 6 in (152.4 mm) and 9 in (228.6 mm)
   Use Eq. 17. Then Avfy/b = 0.85(0.22)(40,000)/15 = 499 lb/in (87.38 kN/m). At C: vj =
   499/6 = 83 lb/in^2 (572.3 kPa); vu = 83 + 93 = 176 lb/in^2 (1213.52 kPa); AC = (272 -
   176)72.52 = 38 in (965.2 mm). At D: v^ = 499/9 = 55 lb/in^2 (379.2 kPa); vu = 55 + 93 =
   148 lb/in^2 (1020.46 kPa); AD = (272 - 148)72.52 = 49 in (1244.6 mm).


6. Devise a stirrup spacing conforming to the computed results
   The following spacing, which requires 17 stirrups for each half of the span, is satisfactory
   and conforms with the foregoing results:

Distance from last

stirrup to face of

Quantity Spacing, in (mm) Total, in (mm) support, in (mm)

1 2 (50.8) 2 (50.8) 2 (50.8)

9 4(101.6) 36 (914.4) 38 (965.2)

2 6 (152.4) 12 (304.8) 50 (1270)

5 9(228.6) 45(1143) 95(2413)

DETERMINATION OF BOND STRESS

A beam of 4000-lb/in^2 (27,580-kPa) concrete has an effective depth of 15 in (381 mm)
and is reinforced with four no. 7 bars. Determine the ultimate bond stress at a section
where the ultimate shear is 72 kips (320.3 kN). Compare this with the allowable stress.

Calculation Procedure:

1. Determine the ultimate shear flow hu
   The adhesion of the concrete and steel must be sufficiently strong to resist the horizontal
   shear flow. Let uu = ultimate bond stress, lb/in^2 (kPa); Vu = ultimate vertical shear, Ib (N);