Respuesta :
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:
The dimensions are 0.95 in
Explanation:
Given:
l = length = 64 in
lc/2 = 64/2 = 32 in
FOS = factor of safety = 3
E = 10.6x10⁶ psi
σ = 40000 psi
The square section is
I = a⁴/12
[tex]P=\frac{\pi^{2}*E*I }{lc^{2} }\\6500=\frac{\pi ^{2}*10.6x10^{6}*a^{4} }{12*32^{2} } \\a=0.95in[/tex]
σallowable = σ/3 = 40000/3 = 13333.3 psi
Area = a² = 0.95² = 0.902 in²
σactual = 6500/0.902 = 7206.2 psi
like σactual < σallowable, the dimensions are 0.95 in
