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
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Comments ( 5 )
A. $$\frac{2}{3}$$
B. $$\frac{3}{2}$$
C. $$\frac{5}{8}$$
D. $$\frac{8}{5}$$
Principal planes are subjected to
A. Normal stresses only
B. Tangential stresses only
C. Normal stresses as well as tangential stresses
D. None of these
A. $$\frac{{\text{M}}}{{\text{I}}} = \frac{{\text{R}}}{{\text{E}}} = \frac{{\text{F}}}{{\text{Y}}}$$
B. $$\frac{{\text{I}}}{{\text{M}}} = \frac{{\text{R}}}{{\text{E}}} = \frac{{\text{F}}}{{\text{Y}}}$$
C. $$\frac{{\text{M}}}{{\text{I}}} = \frac{{\text{E}}}{{\text{R}}} = \frac{{\text{F}}}{{\text{Y}}}$$
D. $$\frac{{\text{M}}}{{\text{I}}} = \frac{{\text{E}}}{{\text{R}}} = \frac{{\text{Y}}}{{\text{F}}}$$
A. $$\frac{{\text{M}}}{{\text{T}}}$$
B. $$\frac{{\text{T}}}{{\text{M}}}$$
C. $$\frac{{2{\text{M}}}}{{\text{T}}}$$
D. $$\frac{{2{\text{T}}}}{{\text{M}}}$$
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