Determine the maximum stress intensities:
A rolled steel I-section, flanges 150 mm wide & 25 mm thick, web 200 mm long & 10 mm thick is utilized to carry & axial load of 800 kN. The load line is eccentric, 50 mm above XX & 30 mm to the left of yy. Determine the maximum & minimum stress intensities in the section.
Figure
Solution
Area of the cross-section, A = 2 × (150 × 25) + (200 × 10) = 9500 mm^{2}
Moment of inertia about XX-axis,
I xx = 150 × 250/12 - 140 (20)^{3}/12
= 8565 × 10^{4} mm^{4}
Moment of inertia about YY-axis,
I yy =2 × (25 × 150^{3} )/ 12 -200 (10)^{3}/12 = 1407 × 10^{4} mm^{4}
Eccentricity,
ex = 50 mm
ey =- 30 mm
Vertical load, W = 800 kN
Direct stress at any point, f_{0} = P/A =800 × 103/9500
= 84.2 N/mm^{2}
Maximum bending compressive stress shall take place at edge 4 of the section in Figure
= ( fb )_{4} = ((P × ex /I_{xx}) × y )+(( P × ey/ I _{yy })× x)
= (800 × 103 × 50/ 8565 × 10^{4}) × (- 125) +( 800 × 10^{3} × (- 30) /1407 × 10^{4}) × 75
= 186.4 N/mm2
Maximum bending tensile stress shall take place at edge 2 at the section.
( f_{ b} )2 =( 800 × 10 ^{3} × 50)/( 800 × 10^{4} ))× (- 125)
+( 8565 × 10^{3} (- 30)/ (1407 × 10^{4} ))× 75
= - 18.64 N/mm^{2}
In the section Resultant stress would be as follows:
Maximum at corner 4 = 84.2 + 186.4 = 270.6 N/mm^{2}
Minimum at corner 5 = 84.2 - 186.4 = - 102.2 N/mm^{2} (tensile)