Respuesta :
Answer:
Part a
The amount by which the specimen will elongate [tex](\bigtriangleup L)[/tex] in the direction of the applied stress is 0.5912 mm
.
Part b
The change in diameter [tex](\bigtriangleup d)[/tex] of the specimen is -0.0184 mm
. Explanation:
When a material is loaded with a force, internal stress is created inside the material and material gets deformed due to the applied load. The deformation occurs in both lateral and longitudinal.
Force:
Force is a push or pull applied on an object, which can resultant changes in the physical properties of the object or causes a certain amount of work done.
Tensile force:
It is the force that acts along the length of the material in the opposite direction that pulls the material apart. The tensile force is positive in magnitude.
Compressive force:
It is the force that acts along the length of the material in the opposite direction that squeezes the material together. The compressive force is negative in magnitude.
Stress:
The internal resistance force acting on the surface of the material per unit area is known as stress.
Longitudinal Strain:
The ratio of change in length to the original length of the specimen is known as strain.
Young’s modulus:
The ratio of true stress to the true strain of the given specimen is known as Young’s modulus.
Poisson ratio:
The ratio of lateral strain to the longitudinal strain is known as Poisson ratio.
Fundamentals
The relation for area (A) of the circular section is given as follows:
[tex]A=\frac{\pi}{4}d^2[/tex]
Here, the diameter of the rod is d.
The relation for stress is given as follows:
[tex]\sigma=\frac{P}{A}[/tex]
Here, stress is [tex]\sigma[/tex] , load applied is Р , and cross sectional area is A
. The relation for longitudinal strain is given as follows:
[tex]\epsilon=\frac{\bigtriangleup L}{L}[/tex]
Here, longitudinal strain is , change in length is [tex]\bigtriangleup L[/tex], and original length is L.
The relation for Young’s modulus is given as follows:
[tex]E=\frac{\sigma}{\epsilon}[/tex]
Here, young’s modulus is E
. The relation for Poisson ratio is given as follows:
[tex]\mu=\frac{lateral\, strain}{longitudinal\, strain}\\\\=-\frac{\bigtriangleup d/d}{\bigtriableup l/l}[/tex]
Here, Poisson ratio is [tex]\mu[/tex] , change in dimension is [tex]\bigtriangleup d[/tex], and original diameter is d.
The negative sign is due the decrease in the diameter of the specimen due to the tensile force.
(a)
Find the area of cross section of the specimen.
[tex]A=\frac{\pi}{4}d^2\\\\=\frac{\pi}{4}(18.4mm)^2\\\\A=265.9mm^2[/tex]
Find the change in length of the specimen using young’s modulus formula.
[tex]E=\frac{\sigma}{\epsilon}\\\\=\frac{P}{A}\times \frac{L}{\bigtriangleup L}\\\\\bigtriangleup L=\frac{PL}{AE}=\frac{52400N\times 192mm}{265.9mm^2\times 64\times 10^9\frac{N}{m^2}\times(\frac{1m}{10^3mm})^2}\\\\\bigtriangleup L=0.5912mm[/tex]
(b)
Write the relation for Poisson ratio.
[tex]\mu=\frac{lateral\, strain}{longitudinal\, strain}\\\\=-\frac{\bigtriangleup d/d}{\bigtriableup l/l}\\\\0.32=-\frac{\bigtriangleup d/18.4}{0.591/192}\\\\=-0.0184mm[/tex]
