Examveda
Examveda

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


This Question Belongs to Civil Engineering >> Theory Of Structures

Join The Discussion

Comments ( 5 )

  1. Vivek Vaishnav
    Vivek Vaishnav :
    3 years ago

    2¢/(3es/L)

  2. Pronab Kumar
    Pronab Kumar :
    4 years ago

    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

  3. Sayantan Das
    Sayantan Das :
    4 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.

  4. Wanbok Kharjana
    Wanbok Kharjana :
    5 years ago

    Solution

  5. Jogarao Karni
    Jogarao Karni :
    5 years ago

    JJ

Related Questions on Theory of Structures

Y are the bending moment, moment of inertia, radius of curvature, modulus of If M, I, R, E, F and elasticity stress and the depth of the neutral axis at section, then

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. $$\frac{{\text{M}}}{{\text{I}}} = \frac{{\text{E}}}{{\text{R}}} = \frac{{\text{Y}}}{{\text{F}}}$$