Question

A tie bar 50 mm × 8 mm is to carry a load of 80 kN. A specimen of the same quality steel of cross-sectional area is 250 mm². For a maximum load of 125 kN carried by the specimen, the factor of safety in the design will be
1. 3.0
2. 2.5
3. 2.0
4. 1.5

Answer 1

Correct Answer - Option 2 : 2.5

Concept:

Allowable stress (or Permissible stress): It is the maximum stress allowed at which a member is expected to perform its function without failing under the given loading conditions.

Working stress: It is actual stress at which the member is subjected to under the given loading conditions and should never cross allowable stress.

Factor of Safety (FOS): It represents the required margin of safety for a structure or component and often decided according to code, law, or design requirements. In general, FOS is given by,

\({\rm{FOS}} = \frac{{{\rm{Allowable\;Stress}}}}{{{\rm{Working\;Stress}}}}\)

Calculation:

\({\rm{Allowable\;stress}} = \frac{{{\rm{Maximum\;Load\;on\;the\;specimen}}}}{{{\rm{Cross\;sectional\;Area\;of\;the\;specimen}}}} = \frac{{125}}{{250}}\frac{{{\rm{kN}}}}{{{\rm{m}}{{\rm{m}}^2}}} = 0.5{\rm{\;}}\frac{{{\rm{kN}}}}{{{\rm{m}}{{\rm{m}}^2}}}{\rm{\;\;}}\)

\({\rm{Working\;stress}} = \frac{{{\rm{Working\;Load\;on\;the\;tie}}}}{{{\rm{Cross\;sectional\;Area\;of\;the\;tie}}}} = \frac{{80}}{{50\; \times \;8}}\frac{{{\rm{kN}}}}{{{\rm{m}}{{\rm{m}}^2}}} = 0.2{\rm{\;}}\frac{{{\rm{kN}}}}{{{\rm{m}}{{\rm{m}}^2}}}{\rm{\;\;}}\)

\(\therefore {\rm{\;FOS}} = \frac{{{\rm{Allowable\;Stress}}}}{{{\rm{Working\;Stress}}}} = \frac{{0.5}}{{0.2}} = 2.5\)