Question

A 6x50 mm flat alloy bar clongates 1.22 mm in a length of 1.5 m under a total axial load of 35.0 NThe proportional limit of this alloy is 300 MPa Determine: '.50" 35 kN 35 kN 6 x 50 A. The axial stress in the bar. B. The modulus of elasticity of the material. C. The change in the two lateral dimensions if Poisson's ratio for the material is 0.32

Answer 1

Answer:

A. Axial stress is 116.66 MPa

B. Modulus of elasticity is 143.5 GPa

C. The changes in two lateral dimensions are 1.3 μm and 0.156 μm

Explanation:

The following information is given:

Axial Load = P = 35 KN = 35000 N

Original Length = L = 1.5 m

Change in length = ΔL = 1.22 mm = 1.22 x 10^-3 m

width = w = 50 mm = 0.05 m

breadth = b = 6 mm = 0.006 m

A.

The axial stress is given as:

σ = P/A

σ = P/wb

σ = (35000 N)/(0.05 m)(0.006 m)

σ = 116.66 MPa

B.

The modulus of elasticity is given in the proportional limit as:

Modulus of Elasticity = E = Stress/Strain = σ/∈

Strain = ∈ = ΔL/L

∈ = 0.00122 m/1.5 m

∈ = 8.13 x 10^-4 mm/mm

Therefore,

E = (116.66 MPa)/8.13 x 10^-4)

E = 143.5 GPa

C.

Poisson's Ratio = n = -(Lateral Strain)/(Axial Strain)

n = 0.32 = - (Lateral Strain)/8.13 x 10^-4

Lateral Strain = - 2.6 x 10^-4 mm/mm

Here, the negative sign indicates the decrease in dimensions.

For change in width (ΔW):

Lateral Strain = - 2.6 x 10^-4 mm/mm = ΔW/W

ΔW = (- 2.6 x 10^-4 mm/mm)(50 mm)

ΔW = - 130 x 10^-4 mm

ΔW = - 1.3 μm

Here, the negative sign indicates the decrease in width.

For change in breadth (Δb):

Lateral Strain = - 2.6 x 10^-4 mm/mm = Δb/b

ΔW = (- 2.6 x 10^-4 mm/mm)(6 mm)

ΔW = - 15.6 x 10^-4 mm

ΔW = - 0.156 μm

Here, the negative sign indicates the decrease in breadth.