A solid circular rod that is 600 mm long and 20 mm in diameter is subjected to an axial force of P = 50 kN. The elongation of the rod is delta = 1.40 mm and its diameter becomes df = 19.9837 mm. Determine the modulus of elasticity (E) and the modulus of elasticity in shear (G). Assume the material does not yield.

a) the modulus of elasticity (E) is 68.22 GPa

b) the modulus of elasticity in shear is 25.28 GPa

Explanation:

Given the data in the question;

First we determine the cross-sectional area A of the road;

A = π/4 × d²

given that; diameter d = 20 mm

we substitute

A = π/4 × ( 20 mm )²

A = π/4 × 400 mm

A = 314.159 mm²

Next, we find, the normal stress of the rod σ

σ = P / A

given that load p = 50 kN = 50,000 Newton

we substitute

σ = 50,000 N / 314.159 mm²

σ = 159.155 N/mm²

σ = 159.155 Mpa

Next, is the normal strain ε

ε = δ / L

given that; change in length δ = 1.40 mm and Length of rod L = 600 mm

we substitute

ε = 1.40 mm / 600 mm

ε = 0.002333

Now, we find the modulus of elasticity;

we know that;

σ = Eε

modulus of elasticity E = σ / ε

we substitute

E = 159.155 Mpa / 0.002333

E = 68219.031 MPa

E = ( 68219.031 / 1000 ) GPa

E = 68.22 GPa

Therefore, the modulus of elasticity (E) is 68.22 GPa

b)

we know that the Poisson's ratio v is;

v = -(ε / ε )

v = - (((d - d)/d) / ε )

where d  is the final diameter of the rod ( 19.9837  mm ) and ε is the lateral strain

so we substitute

v = - ((( 19.9837 - 20 ) / 20 ) / 0.002333  )

v = - (( -0.0163 / 20 ) / 0.002333  )

v = - ( - 0.000815 / 0.002333 )

v = - ( - 0.3493 )

v = 0.3493

So, our shear modulus will be;

G = E / 2( 1 + v )

we substitute

G = 68.22 GPa / 2( 1 + 0.3493  )

G = 68.22 GPa / 2( 1.3493  )

G = 68.22 GPa / 2.6986

G = 25.28 GPa

Therefore, the modulus of elasticity in shear is 25.28 GPa

nah.   i'm just sleepy.

no simpler subs form complex ones. synthesise = make