Question

A symmetric I-section of size 180 mm × 40 mm, 8 mm thick is strengthened with 240 mm × 10 mm rectangular plate on top flange as shown is Fig If permissible stress in the material is 150 N/mm² , determine how much concentrated load the beam of this section can carry at centre of 4 m span. Given ends of beam are simply supported.

Answer 1

Area of section A 

= 240 × 10 + 180 × 8 + 384 × 8 + 180 × 8 = 8352 mm²

Let centroid of the section be at a distance y from the bottom most fibre. Then 

A \(\bar y\)= 240 × 10 × 405 + 180 × 8 × (400 – 4) + 384 × 8 × 200 + 180 × 8 × 4

8352 \(\bar y\) = 2162400 

∴ \(\bar y\) = 258.9 mm

Moment carrying capacity of the section 

= f_per Z = 150 × 853588.3 

= 128038238.7 N-mm 

= 128.038 kN-m.

Let P kN be the central concentrated load the simply supported beam can carry. Then max bending movement in the beam

= \(\frac{P\times4}4\) = p kN-m

Equating maximum moment to moment carrying capacity, we get 

P = 128.038 kN.