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 × 106 N/mm2, the depth of the joist, is
A. 200 mm
B. 250 mm
C. 300 mm
D. 400 mm
Answer: Option D

2¢/(3es/L)
F=MC/I
F=(M/I)*C
Here F=75N/mm^2
Slope=0.005=ML/3EI
or, M/I=0.005*3E/L=0.375
so F=(M/I)*C=0.375*C
C=75/0.375=200mm
D=2*C=2*200=400mm
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.
Solution
JJ