Answer:
The appropriate answer will be "2353.8 K". A further solution is given below.
Explanation:
The given values are:
Ultimate capacity,
= 1700kN
Resistance factor,
F.S = 0.65
According to the the LRFD, the allowable bearing pressure will be:
⇒ [tex]Pallo w = \phi_c\times \frac{Pn}{F.S}[/tex]
On substituting the values, we get
⇒ [tex]=0.9\times \frac{1700}{0.65}[/tex]
⇒ [tex]=2353.8 \ K[/tex]
According to the the ASD, the allowable bearing pressure will be:
⇒ [tex]Pallow=\frac{Pn}{\Omega \times F.S}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{1700}{1.67\times 0.65}[/tex]
⇒ [tex]= \frac{1700}{1.0855}[/tex]
⇒ [tex]=1845.35 \ K[/tex]
