Please show solutions on how to get the 7.85 value. The circle green in the figure.

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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Please show solutions on how to get the 7.85 value. The circle green in the figure.

Civil engineering- steel design

3.5-3
A WT8 x 13 of A992 steel is used as a tension member. The connection is with 7/8-in.
diameter bolts as shown in Figure P3.5-3. Compute the nominal block shear strength.
T
"-3
1/2".
4½"
FIGURE P3.5-3
O
the shear areas are
= 0.250(13.5) = 3.375 in.²
and, since there are 4.5 hole diameters,
Anv = 0.250[13.5-4.5(7/8 + 1/8)] = 2.25 in.²
The tension areas are
Agt = 0.25 (7.85-4.5) = 0.8375 in.²
Ant = 0.25[7.85-4.5-0.5(1.0)] = 0.7125 in.²
Fy= 50 ksi, Fu = 65 ksi
Rn = 0.6FuAny + UbsFuAnt = 0.6(65)(2.25) + 1.0(65) (0.7125) = 134 kips
Check upper limit:
0.6FyAgy + UbsFuAnt = 0.6(50) (3.375) + 1.0(65)(0.7125)
148 kips > 134 kips
=
R₁
=
134 kips
Transcribed Image Text:3.5-3 A WT8 x 13 of A992 steel is used as a tension member. The connection is with 7/8-in. diameter bolts as shown in Figure P3.5-3. Compute the nominal block shear strength. T "-3 1/2". 4½" FIGURE P3.5-3 O the shear areas are = 0.250(13.5) = 3.375 in.² and, since there are 4.5 hole diameters, Anv = 0.250[13.5-4.5(7/8 + 1/8)] = 2.25 in.² The tension areas are Agt = 0.25 (7.85-4.5) = 0.8375 in.² Ant = 0.25[7.85-4.5-0.5(1.0)] = 0.7125 in.² Fy= 50 ksi, Fu = 65 ksi Rn = 0.6FuAny + UbsFuAnt = 0.6(65)(2.25) + 1.0(65) (0.7125) = 134 kips Check upper limit: 0.6FyAgy + UbsFuAnt = 0.6(50) (3.375) + 1.0(65)(0.7125) 148 kips > 134 kips = R₁ = 134 kips
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