Question

1.
A bar 0.3 m long is 50 mm square in section for 120 mm of its length, 25 mm diameter for 80 mm and
of 40 mm diameter for the remaining length. If a tensile force of 100 KN is applied to the bar, calculate
the maximum and minimum stresses produced in it, and the total elongation. Take E= 200 GN/m2 and
assume uniform distribution of load over the cross-sections.
[Ans. 204 MN/m², 40 MN/m .0.1453 mm)​

Answer 1

Answer:

Maximum stress occur at the region of minimum cross section and the minimum stress occur in the region of maximum cross sectional area. Also the total elongation is the sum of the elongations of each section. Square cross section area= 2500mm2. Corresponding stress = 100kN/ 2500mm2=40MPa.( minimum stress). circular cross section diameter is 25mm corresponding stress =100kN/(3.14*252)=203Mpa.( maximum stress)....