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/mm², 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 mm²

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/mm²

∴          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.