Question

For a bronze alloy, the stress at which plastic deformation begins is 275 MPa (40,000 psi), and the modulus of elasticity is 115 GPa (16.7 ×106psi).(a)What is the maximum load that may be applied to a specimen with a cross-sectional area of 325 mm2(0.5 in.2) without plastic deformation?(b)If the original specimen length is 115 mm (4.5 in.), what is the maximum length to which it may be stretched without causing plastic deformation?

Answer 1

Answer:

a) $P_{max}=89375N$

b) $L_{f}=115.275mm$

Explanation:

a) The maximum load that may be applied to a specimen of bronze alloy without plastic deformation is given by the following equation :

Pmax = (σ).(A)

Where Pmax is the maximum load

σ is the value at which plastic deformation begins

Where ''A'' is the cross-sectional area of the specimen

Let's also use the fact that :

$1 MPa=(10^{6})Pa=(10^{6})\frac{N}{m^{2}}$

We first must turn ''A'' from $(mm^{2})$ to $(m^{2})$

$325(mm^{2})=325.(mm).(mm).(\frac{1m}{1000mm}).(\frac{1m}{1000mm})$

$325(mm^{2})=(3.25).(10^{-4})m^{2}$

Using the equation :

$P_{max}=(275)(10^{6})\frac{N}{m^{2}}.(3.25).(10^{-4}).m^{2}=89375N$

The maximum load is 89375 N.

b) To calculate the maximum length we are going to use the following equation :

Lf = Li ( 1 + σ / E )

Where Lf is final length without causing plastic deformation

Li is initial length

σ is the value at which plastic deformation begins

And finally ''E'' is the modulus of elasticity from the bronze alloy

Using the equation :

$L_{f}=(115mm).(1+\frac{(275).(10^{6})Pa}{(115).(10^{9})Pa})\\L_{f}=115.275mm$