Civil Engineering - Theory of Structures - Discussion
Discussion Forum : Theory of Structures - Section 2 (Q.No. 28)
28.
A simply supported rolled steel joist 8 m long carries a uniformly distributed load over it span so that the maximum bending stress is 75 N/mm2. If the slope at the ends is 0.005 radian and the value of E = 0.2 x 106 N/mm2, the depth of the joist, is
Discussion:
8 comments Page 1 of 1.
Ranjitha said:
8 years ago
Slope = wl^3/24EI.
0.05 = wl^3/24EI.
I = wl^3/24000.
We know that,
F/Y = M/I
here F = 75N/mm2.
Y = d/2.
l = 8000mm.
So, 75/(d/2) = ((wl^2)/8)/(wl^3/24000).
75/(d/2) = 24000/(8*8000),
75/(d/2) = 0.375,
d/2= 75/0.375,
d/2 = 200,
d = 400mm.
0.05 = wl^3/24EI.
I = wl^3/24000.
We know that,
F/Y = M/I
here F = 75N/mm2.
Y = d/2.
l = 8000mm.
So, 75/(d/2) = ((wl^2)/8)/(wl^3/24000).
75/(d/2) = 24000/(8*8000),
75/(d/2) = 0.375,
d/2= 75/0.375,
d/2 = 200,
d = 400mm.
(5)
Chhaya said:
8 years ago
Slope = wl3/24EI.
Moment = wl2/8.
Stress = M/Z.
z = bd2/6.
I = bd3/12.
Slope/stress.
You got depth and it is 250.
Moment = wl2/8.
Stress = M/Z.
z = bd2/6.
I = bd3/12.
Slope/stress.
You got depth and it is 250.
Imtiaz said:
5 years ago
@Ranjitha
how I = 80000mm? please explain.
how I = 80000mm? please explain.
Sayantan said:
5 years ago
@Mtiaz.
Here, length is 8m which is 8000mm.
Here, length is 8m which is 8000mm.
Tapas Bhattacharjee said:
6 years ago
You are right, thanks @Ranjitha.
(2)
Dheeraj said:
6 years ago
@Ranjitha, very nice explanation
Hamza Shahid said:
5 years ago
Thanks for explaining @Ranjitha.
MRUDULA said:
5 years ago
Thanks @RANJITHA.
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