Answer: 0.95 inches
Explanation:
A direct load on a column is considered or referred to as an axial compressive load. A direct concentric load is considered axial. If the load is off center it is termed eccentric and is no longer axially applied.
The length= 64 inches
Ends are fixed Le= 64/2 = 32 inches
Factor Of Safety (FOS) = 3. 0
E= 10.6× 10^6 ps
σy= 4000ps
The square cross-section= ia^4/12
PE= π^2EI/Le^2
6500= 3.142^2 × 10^6 × a^4/12×32^2
a^4= 0.81 => a=0.81 inches => a=0.95 inches
Given σy= 4000ps
σallowable= σy/3= 40000/3= 13333. 33psi
Load acting= 6500
Area= a^2= 0.95 ×0.95= 0.9025
σactual=6500/0.9025
σ actual < σallowable
The dimension a= 0.95 inches
Answer:
Time period = 41654.08 s
Explanation:
Given data:
Internal volume is 210 m^3
Rate of air infiltration 
length of cracks 62 m
air density = 1.186 kg/m^3
Total rate of air infiltration 
total volume of air infiltration
Time period 
Answer:
slenderness ratio = 147.8
buckling load = 13.62 kips
Explanation:
Given data:
outside diameter is 3.50 inc
wall thickness 0.30 inc
length of column is 14 ft
E = 10,000 ksi
moment of inertia 
Area 
r = 1.136 in
slenderness ratio 
buckling load 
Answer:
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Explanation:
