Evaluate secure axial load:
A small piece of ISA (200 × 100 × 15) angle carries a compressive load, the line of action of which coincides along the intersection of the middle planes of the legs. If the maximum compressive stress is not to exceed 112 N/mm^{2}, what is the secure axial load P? Given A = 4278 mm2, r_{xx} = 64 mm, r_{yy }= 26.4 mm.
Figure
Solution
Area of cross-section A = 4278 mm^{2}
Eccentricity of load with respect of xx-axis = (71.8 - 7.5) = 64.3 mm
Eccentricity of load with respect to yy-axis = 22.2 - 7.5 = 14.7 mm
Maximum compressive stress at any section will be
= (P/ A) + (M _{xx}/ I _{xx}) × y + (M _{yy}/ I _{yy}) × x
or
Here, r_{xx} = 64 mm and r_{yy} = 26.4 mm f_{max} = 112 N/mm^{2}
∴ 112 = (P/A) (1 + ((64.3 × 71.8 )/(64)^{2})+ 14.7 × 22.2 /((26.4)2)
= (P/4278) [1 + 1.127 + 0.4684]
∴ P = 184.6 kN.