Answer: the allowable load P is 242.7877 kips
Explanation:
Given that;
diameter bolts d = 1.83 in
ultimate shear strength of the bolts = 60 ksi
we know that
shear area = 2×(π/4)d²
= 2×(π/4)×(1.83)² = 5.2604 in²
so
p/3(5.2604) = 60000/3.9
p/15.7812 = 15384.6153
p = 15.7812 × 15384.6153
p = 242787.691 lb
p = 242.7877 kips
therefore the allowable load P is 242.7877 kips
Explanation:
1) x= y/(1+y)
2)x= -y/(1-y)
3) x=y/(1-y)
4)square root (2E/m-v^2)
5) (( x^2b^2- a^2b^2)/a^2) square root
6) square root( x^3/a)
Answer:
hello your question has some missing information attached to the answer is the missing component
Answer : αaxial,p = -6.034 ksi ( compressive )
αbend,p = 19.648 ksi ( tensile )
Explanation:
αaxial, p =
equation 1
αbend, p =
equation 2
P = load = 35 kips
A = area of column = 5.8 
d = column cross section depth = 9.5 in
= 55.0 
Hence equation 1 becomes
αaxial,p = -35 / 5.8 = - 6.034 ksi ( compressive )
equation 2 becomes
αbend, p =
= + 19.648 ksi ( tensile )
Sorry man i don’t know sorry