Evaluate secure axial load, Mechanical Engineering

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, rxx = 64 mm, ryy = 26.4 mm.

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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


1783_Evaluate secure axial load1.png

Here,   rxx = 64 mm  and   ryy = 26.4 mm fmax = 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.

Posted Date: 1/21/2013 7:22:21 AM | Location : United States

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