Question

A steel bar has a rectangular cross section measuring 1.000cm by 2.500cm and is 8.100m in length. A load of 150000.000N causes the bar to extend 2.000mm. Calculate the stress and strain.

Area = l*w
= 1.000cm*2.500cm
= 2.500cm = 0.025m
= 0.025m*8.100m
= 0.2025m

Stress = Load/ Area
150000/.02025
= 740740.7407Pa
= 740.740kPa

Strain
= 0.002m/8.100m
= 0.00024691

Is this correct?

Answer 1

σ= F/A
A=1•2.5=2.5 cm²=2.5•10⁻⁴m²
σ= 150000/2.5•10⁻⁴=6•10⁸ N/m².

Strain is defined as "deformation of a solid due to stress" and can be expressed as
ε= σ/E,
where E= Young’s modulus (Modulus of elasticity)
For steel E=200 GPa =>
ε= σ/E = 6•10⁸/2•10¹¹=3•10⁻³ m = 3 mm

Answer 2

Area = lw (correct)
=1cm x 2.5 cm
=0.0100 cm x 0.025 cm
=0.00025 m²
(remember, length 8.1 does not come into the calculation of cross section area)

Stress
=force (in N)/area (in m²)
=150000/0.00025 Pa
=6×10^8 Pa
=6×10^5 mPa
=600,000 mPa

Strain
=deformation (in metres) /original length (in metres)
=2 mm/ 8.1 m
=0.002 m / 8.1 m
= 2.47×10^-4
=0.000247

Note Elena's formula for strain
ε=σ/E
is also correct, except that E is not directly given in the question.
Also, since both σ and E are in pressure units, strain is unitless.