Mechanics of Deformable Bodies. Torosion Lab test Assignment | Online Assignment

Review the attached files for the assignment. Instructions how to do the assignment on the manual. I have video as well but I was unable to upload it.Steel Specimen
Fracture Surface
Aluminum Specimen
Fracture Surface
Steel Specimen
Fracture surface
Comparison of fracture surface
From left to right: Cast iron, Aluminum, Steel

University of Windsor
Faculty of Engineering
Mechanics of Deformable Bodies
GENG-2180
Summer 2020
Laboratory – 2
Torsion Test
Objective:
To study the behaviour of ductile and brittle materials subject to torsion
Equipment:
1. Torsion testing machine Tinius Olsen
2. Micrometer
3. Scale
Specimens:
Standard torsion test specimen of steel, aluminum or cast iron
Procedure: (Watch the video)
1. Using the micrometer measure the diameter of specimen to the nearest 0.02 mm at
several points along the length of the specimen and record the average diameter.
2. Check the length of the torsion specimen (200.0 mm).
3. Mount specimen in the torsion testing machine. Set the load indicator dial zero.
4. Apply the load by manually operating the machine at the following increments:
Steel 500 lb-in
Aluminum 300 lb-in
Cast Iron 250 lb-in
Take readings of angle of twist () for each load while angle of twist increases with
load proportionally in the linear range.
2
5. When the readings of angle of twist suddenly increases, the nonlinear range (yield
range) starts. Continue applying the load but take the readings of the torque at the
following increments of angle of twist:
Steel 45
Aluminum 30
Cast Iron 5
6. After the yield point has been exceeded, apply the load until failure occurs.
7. Examine the fracture surface and the type of failure that occurred in each specimen.
Report:
Prepare a lab report according to the document “Structure and Format of Laboratory
Reports” posted on Blackboard. This is a formal report that should be written using a word
processor; the only hand writing allowed is for the sample calculations. Include the following
sections and results in the report:
1. Title page
2. Objectives
3. Procedures
4. Results
a. Using the experimental data file “Torsion Data.xlsx” posted on Blackboard, calculate
the shear stress and shear strain for each recorded data point. Plot a graph of the
complete shear stress-shear strain curve for each specimen in Excel. Show the ultimate
shear stress and the shear stress at the yield point on each graph.
b. For each material, plot a graph showing the elastic portion of the shear stress – shear
strain curve (including the yield point) to an enlarged scale.
c. Provide the completed data sheet (see the next page) and sample calculations of the
mechanical properties required in the data sheet.
d. Describe the fracture and comment on the reasons for this type of fracture. Refer to the
file “Fracture Surface.pdf” poste on Blackboard.
5. Discussion
6. Conclusions
7. References (if needed)
3
Data Sheet – Torsion Test
Material Steel Aluminum Cast Iron
Type of specimen
Length
(mm)
Average diameter
(mm)
Average cross-sectional area
(mm2)
Proportional limit in shear (MPa)
Yield point or yield strength in
shear at 0.001 radians offset
(MPa)
Modulus of rigidity
(Shear modulus) (GPa)
Ultimate shear stress
(MPa)
Shear stress at rupture
(MPa)
Modulus of elasticity (calculated
using  = 0.30) (GPa)
4
Hints for filling in the data sheet:
T = torque
J = polar moment of inertia
 = shear strain
ρ = radial distance measured from the axis of the cylindrical specimen ( 0    Ro )
Ro = outside radius of a cylindrical or tubular test piece
Ri = inside radius of a tubular test piece
L = distance between collars of the cylindrical specimen
 = angle of twist (radians)
 = shear stress
G = modulus of rigidity (shear modulus)
Shear strain
L

 
Within the proportionality limit,
J
G T
L
G 


   
For solid cross sections,
4
Ro
2
J 

For tubular cross sections,
 4 
i
4
Ro R
2
J 


5
The proportional limit is estimated from the shear stress-shear strain diagram in an enlarged
graph. It is the stress at which the shear stress-shear strain relationship ceases to be linear. See
“Yield strength in shear” in the figure below.
1. Modulus of rigidity (shear modulus)
shear strain
G  shear stress
It is the slope of the shear stress-shear stain curve below the proportional limit.
2. Ultimate shear stress =
J
Tmax Ro (MPa)
(assuming the elastic torsion formula holds good up to rupture)
3. Shear stress at rupture =
J
Trupture Ro
(MPa)
(assuming the elastic torsion formula holds good up to rupture)
4. Modulus of elasticity
E = 2G (1 + )
 = Poisson’s ratio