Civil Engineering - Theory of Structures - Discussion
Discussion Forum : Theory of Structures - Section 1 (Q.No. 28)
28.
A rolled steel joist is simply supported at its ends and carries a uniformly distributed load which causes a maximum deflection of 10 mm and slope at the ends of 0.002 radian. The length of the joist will be,
Discussion:
6 comments Page 1 of 1.
Asay said:
7 years ago
Deflection=5wl^4/384EI=10mm.
=====> EI=5wl^4/384*10 -----> (1)
Slope=wl^3/24EI=0.002mm
=====> EI=wl^3/24*0.002 -----> (2)
Comparing eq (1) & (2)
5wl^4/384*10=wl^3/24*0.002
l=384*10/5*24*0.002
l=384*10*1000/5*24*2
l=16*1000mm
l=16m.
So ,
The length of the joist will be =16m
(Option D) is correct Answer.
=====> EI=5wl^4/384*10 -----> (1)
Slope=wl^3/24EI=0.002mm
=====> EI=wl^3/24*0.002 -----> (2)
Comparing eq (1) & (2)
5wl^4/384*10=wl^3/24*0.002
l=384*10/5*24*0.002
l=384*10*1000/5*24*2
l=16*1000mm
l=16m.
So ,
The length of the joist will be =16m
(Option D) is correct Answer.
(1)
Suhasini said:
10 years ago
Max deflection = 5wL4/384EI.
Slope = wL3/24EI.
w = 24EI*slope/L3.
Max deflection = 5*(24EI*slope)*L4/(L3*384EI).
Therefore, L = max deflection*16/(5*Slope).
= 10*16/(5*0.002) = 16000 mm = 16 m.
Slope = wL3/24EI.
w = 24EI*slope/L3.
Max deflection = 5*(24EI*slope)*L4/(L3*384EI).
Therefore, L = max deflection*16/(5*Slope).
= 10*16/(5*0.002) = 16000 mm = 16 m.
(1)
Isac said:
1 year ago
Thanks for the explanation @Suhashini.
Sha said:
4 years ago
Thanks @Asay and @Suhasini.
Chhaya said:
7 years ago
Thanks @Suhasini.
Pawan said:
5 years ago
Thanks @Suhasini.
(1)
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