D-14

Example D.7 Pin-Connected Tension Member

Given:

An ASTM A36 pin connected tension member with the dimensions shown below carries a dead load of 12 kips and a live load of 4 kips in tension. The diameter of the pin is 1 inch, in a Q-in. oversized hole. Assume that the pin itself is adequate. Verify the strength by both LRFD and ASD.

Solution:

Material Properties: Plate ASTM A36 Fy = 36 ksi Fu = 58 ksi Manual

Table 2-4 Geometric Properties:

w = 4.25 in. t = 0.500 in. d = 1.00 in. a = 2.25 in. c = 2.50 in. Check dimensional requirements: 1) beff = 2t + 0.63 in. = 2(0.500 in.) + 0.63 in. = 1.63 in. 2) a > 1.33beff 2.25 in. > (1.33)(1.63 in.) = 2.17 in. o.k. 3) w > 2beff + d 4.25 in. > 2(1.63 in.) + 1.00 in. = 4.26in. 4.25in.≅ o.k. 4) c > a 2.50 in. > 2.25 in. o.k.

Section D5.2

Calculate the required tensile strength

LRFD ASD Pu = 1.2(12.0 kips) + 1.6(4.00 kips)

= 20.8 kips Pa = 12.0 kips + 4.00 kips = 16.0 kips

D-15

Calculate the available tensile rupture strength on the net effective area

Pn = 2tbeffFu = (2)(0.500 in.)(1.63 in.)(58 ksi) = 94.5 kips Eqn. D5-1

LRFD ASD φt = 0.75

φtPn = 0.75(94.5 kips) = 70.9 kips

Ωt = 2.00 Pn/Ωt = (94.5 kips) / 2.00 = 47.3 kips

Section D5.1

Calculate the available shear rupture strength

Asf = 2t(a + d/2) = 2(0.500 in.)[2.25 in. + (1.00 in. /2)] = 2.75 in.2

Pn = 0.6FuAsf = (0.6)(58 ksi)(2.75 in.2) = 95.7 kips

Section D5.1 Eqn. D5-2

LRFD ASD φsf = 0.75

φsfPn = 0.75(95.7 kips) = 71.8 kips

Ωsf = 2.00 Pn/Ωsf = (95.7 kips) / 2.00 = 47.9 kips

Section D5.1

Calculate the available bearing strength Apb = 0.500 in.(1.00 in.) = 0.500 in.2 Rn = 1.8FyApb = 1.8(36 ksi)(0.500 in.2) = 32.4 kips

Eqn. J7.1

LRFD ASD φt = 0.75

φtPn = 0.75(32.4 kips) = 24.3 kips

Ωt = 2.00 Pn/Ωt = (32.4 kips) / 2.00 = 16.2 kips

Section J7

Calculate the available tensile yielding strength Ag = 4.25 in. (0.500 in.) = 2.13 in.2

Pn = FyAg = 36 ksi (2.13 in.2) = 76.5 kips

Section D2 Eqn. D2.1

LRFD ASD φt = 0.90

φtPn = 0.90(76.5 kips) = 68.9 kips

Ωt = 1.67 Pn/Ωt = (76.5 kips) / 1.67 = 45.8 kips

Section D2

The available tensile strength is governed by the bearing strength limit state

LRFD ASD φtPn = 24.3 kips

24.3 kips > 20.8 kips o.k.

Pn/Ωt = 16.2 kips 16.2 kips > 16.0 kips o.k.