Question

A tie bar $50\text{mm}\times 8\text{mm}$ is to carry a load of 80 kN. A specimen of the same quality steel of cross -sectional area $250m{m}^2$ was tested in the laboratory. The maximum load carried by the specimen was 125 kN. The factor of safety in the design is.

• A. 2.5

Answer 1

The correct option is A 2.5

Yield strength for the steel $=\frac{\text{Maximum load}}{\text{Cross - sectional area}}$
$=\frac{125\times 1000}{250}=500N/m{m}^2$
Working stress $=\frac{80\times 1000}{50\times 8}=200N/m{m}^2$
$\therefore FOS=\frac{\text{Yield Strength}}{\text{Working stress}}=\frac{500}{200}=2.5$