Question

Q.23 A mild steel bar of square cross-section 25 mm
x 25 mm is 1 m long and it is subjected to biaxial
stress as shown below:
2
0,= 400 N/mm
(Compression)
2
o = 480 N/mm
(Tension)
2x70
- Pulang menent's
E = 2 x 105 N/mm2, u = 0.3, what is the
elongation of the bar in mm in x-direction
(a) 1.0 mm
(b) 0.5 mm
(c) 0.05 mm
(d) 0.075 mm
​

Answer 1

It has given that, A mild steel bar of square cross sectional area, A = 25 mm² = 6.25 × 10¯⁴ m² is L = 1m long and it is subjected to biaxial stress σx = 480 × 10⁶ N/m² and σy = 400 × 10⁶ N/m² in perpendicular directions , E = 2 × 10¹¹ N/m², Poisson's ratio, μ = 0.3

we have to find the elongation of the bar in mm in x - direction ( longitudinal)

solution : the elongation of the bar in the longitudinal direction = strain in longitudinal direction × length of bar

= [(σx - μσy)/E ] × L

= [(480 - 0.3 × 400) × 10⁶/2 × 10¹¹] × 1

= 180 × 10¯⁵ m

= 1.8 mm

Therefore the elongation of the bar in the x direction is 1.8 mm