Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 2 (Q.No. 6)
6.
The length of a column, having a uniform circular cross-section of 7.5 cm diameter and whose ends are hinged, is 5 m. If the value of E for the material is 2100 tonnes/cm2, the permissible maximum crippling load will be
Discussion:
9 comments Page 1 of 1.
Mounika reddy said:
1 decade ago
P = (n^2*pi^2*EI)/L^2.
I = pi*r^4/4, n=1.
P = 128800*10^-4.
P = 12.88 tonnes.
I = pi*r^4/4, n=1.
P = 128800*10^-4.
P = 12.88 tonnes.
Abhay said:
10 years ago
P = pi^2EI/L^2.
Mohan krishna said:
8 years ago
permissible crippling load (P) = π^2EI/L^2,
where I = πD^4/64,
therefore P = (π^2*2100*7.5^4/64)/500^2,
= 12.8796 tonnes.
where I = πD^4/64,
therefore P = (π^2*2100*7.5^4/64)/500^2,
= 12.8796 tonnes.
Tailma mahesh said:
8 years ago
Superb thanks @ Mohan Krishna.
Akshay kumar said:
8 years ago
But the ends are hinged so l=L so why do we used 2π?
(1)
Shobha khare said:
7 years ago
Thank you @ Mohan Krishna.
Protap said:
7 years ago
Thanks @Mohan.
Rosy said:
6 years ago
(P) = π^2EI/L^2,
I = πD^4/64,
therefore P = (π^2*2100*7.5^4/64)/500^2,
So, B = 12.8796 tonnes.
I = πD^4/64,
therefore P = (π^2*2100*7.5^4/64)/500^2,
So, B = 12.8796 tonnes.
(2)
Mun said:
5 years ago
Crippling load (p) = π^2EI/L^2.
I= πD^4/64.
Max crippling load (P)= (π^2*2100* 7.5^4)/64)/500^2= 12.85 tonnes.
I= πD^4/64.
Max crippling load (P)= (π^2*2100* 7.5^4)/64)/500^2= 12.85 tonnes.
(2)
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