Question

Q3: The engineers wants to examine the steel bar having three sections. So they have applied an axial pull
of P in kN to the steel bar of length 700 mm, 42 cm and 0.8 m which has diameters of 2 cm, 30 mm and 5
cm with a total deformation of 0.51 mm, assume E. For that purpose you are going to help the engineers
by calculating the axial pull, P and stress in all the three sections. Also do your interpretation of results
based on area of cross section, stress and deformation.

Answer 1

Assume modulus of elasticity for steel = 200 GPa

200 GPa = 200*10^3 N/mm^2

Calculate cross sectional area for each section of the steel bar.

by Hook's law we know that "Stress = E* Strain"

$\sigma = E\varepsilon = E\frac{\delta L}{L}$       therefore

$\delta L = \frac{\sigma *L}{E}$$\delta L = \frac{\sigma *L}{E} = \frac{P*L}{AE}$

Stress will be more in the section which have lesser cross section area, so stress is highest in 20mm diameter section and Lowest in 50mm Diameter section.

and change in length of each section will depend on ratio of length by cross sectional area.

P 4-30 Р &- 2cm &-5cm 08 $somm - 20 mm 42cm 700mm 0.8m = 800mm Com = 0.42 = 420mm E = 200 Gpa Let. Young modulus of steel a General formula of deflection We know anial Pull due to 8 = Px lenger Area xe Cross-section Deflection in bar having different d=deflection in section - 1 + deflection in sections + deflection section -3. PL + PL₂ P L3 P + Α, ε ALE 4. Е 2

4,- (20) mm2 = 314.15 mm? A2 = (30)?mm?? 706.85 mm2 43 = (50)2 mm = 1963.49 mm2. 0.51 mm Total de flection is limited to P + 0.51 mm - + 700 314.15 420 706.85 800 1963.49 200 x 10°% mm? 3 o. 51 x 200 X10 = P 3.2298 1 P=31580 N Answer. Now Stress in Section of 7oo mm + 20 mm dia = 100.5 N/mm?. Stress in section of 420mm & 3 omm dia=. P = 44.67 N/ A2 Stress in section of doomm length & 50 mm dia 16.08 N/mm². Az