What are the horizontal structures beneath a slab that help transfer the load from the slab to the columns?
2 answers:
You might be interested in
Answer:
a) 1.16 m/s
b) 1/216000
c) (√15)/6480000
Explanation:
The parameters given are;
Length of boat prototype, lp = 30 m
Speed of boat prototype = 9 m/s
Length of boat model, lm= 0.5 m
a) lm/lp = 0.5/30 = 1/60 = ∝
(vm/vp) = ∝^(1/2) = √∝ = (1/60)^(1/2)
vm = 9 × (1/60)^(1/2) = 1.16 m/s
b) The ratio of the model to prototype drag, Fm/Fp, is given as follows;
Fm/Fp = (vm/vp)²×(lm/lp)² = ∝³
Fm/Fp = (1/60)³ = 1/216000
c) The ratio of the model to prototype power pm/p = (Fm/Fp) × (vm/vp) = ∝³×√∝
The ratio of the model to prototype power pm/p = √(1/60) × (1/60)³
pm/p = √(1/60) × (1/60)³ = (√15)/6480000
Answer:
3.115× meter
Explanation:
hall-petch constant for copper is given by
=25 MPa
k=0.12 for copper
now according to hall-petch equation
=
+
240=25+
D=3.115× meter
so the grain diameter using the hall-petch equation=3.115× meter