Steel Tension Member: Yielding and Rupture

ϕ P_{n} = min (0.90 F_{y} A_{g}, 0.75 F_{u} A_{e})

Two limit states: yielding on the gross section (φ=0.90) and rupture on the effective net section (φ=0.75). Ae = An U with U the shear-lag factor. Holes: An = Ag − ndb t (plus 1/16 in on hole diameter for punching).

$A_{g}$: Gross area
$A_{e}$: Effective net area
$F_{y}, F_{u}$: Yield and tensile strength

Step-by-step solved example

Ag=5.00 in², Fy=36 ksi, Fu=58 ksi, Ae=4.50 in². φPn?

Gross yielding vs net rupture — two φPn checks.

1. Yield
0.90×36×5=162 kip.
2. Rupture
0.75×58×4.50=196 kip. Controls? min=162 kip (yielding).

Answer: 162 kip (yielding controls)

10 practice questions

0/10 correct

1.φ for tensile rupture is

2.φ for tensile yielding is

3.U accounts for

4.Bolt holes reduce

5.Block shear is

6.If Ae is much smaller than Ag, rupture may

7.A36 Fu is at least about

8.An for a chain of holes uses

9.Slenderness of tension members is

10.φPn is compared with