Please show solutions on how to get the 0.250 value. The circle green in the figure.
Please show solutions on how to get the 0.250 value. The circle green in the figure.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Please show solutions on how to get the 0.250 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
1/2"-
"-3"
+3″✦✦3″✦✦3″»
FIGURE P3.5-3
O
the shear areas are
Agv 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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd26af165-a48a-4025-869b-89b03c48d51c%2Fe054ef96-8bf5-49fe-b78c-555705397cc1%2F3k8x1e1r_processed.jpeg&w=3840&q=75)
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
1/2"-
"-3"
+3″✦✦3″✦✦3″»
FIGURE P3.5-3
O
the shear areas are
Agv 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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