Handbook of Civil Engineering Calculations

Afy = 11.8 in

2

x 36 ksi = 425 kips (1890 kN)

Vh = 425 kips (189OkN)

3. Find the number of shear studs required
   The nominal strength of a single shear stud [from Eq. (/5-7)] is

Qn = Q.5AscVftEc<AscFu

For a^3 /4-in-diameter stud,

A*c = *( ^p^1 ]^2 = 0.44 in^2 (2.84 cm^2 )

E 0 = w

1

• 5
  V£ = 145
  15
  VsiO = 3904 ksi (26,899 kNm)

Fn= 60ksi (413 kNm)

Qn = 0.5 x 0.44 in^2 V5.0 ksi x 3904 ksi < 0.44 in^2 x 60 ksi (413 kNm)

= 30.9 kips < 26.4 kips (117.4 kN)

= 26.4 kips per stud (117.4 kN per stud)

The number of shear connectors between the points of zero and maximum moments is

VH 425 kips ... J

W = ^ = ^^klpS/Stud

= 16.1 or 17 studs

For the beam shown in Fig. 48, the required number of shear studs is 2n = 2 x 17 = 34.

Assuming a single line of shear studs (over the beam web), stud spacing = 30.0 ft/34 =

0.88 ft = 10.6 in (26.9 cm). This is greater than the six-stud diameter [or 6 x % in = 4.5 in

(11.4 cm)] minimum spacing, and less than the eight slab thickness [or 8 x 5 in = 40 in

(101.6 cm)] maximum spacing, which is satisfactory.

Related Calculations. This procedure is the work of Abraham J. Rokach, MSCE,
Associate Director of Education, Ajmerican Institute of Steel Construction. SI values were
prepared by the handbook editor.

n studs n studs

FIGURE 48