Lab #2 Report due TODAY
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Individual Data Processing and Individual
Write Up
Lab #2 Report is due TODAY by midnight
Answer Multiple Choice Questions ONLINE
TODAY by midnight
Answer Student Survey ONLINE TODAY by
midnight
Submit the Lab Report #2 in the drawer
labeled “MECE204-Strength Lab” under M. E.
Mail Folders
Torsion Testing
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Circular Cross-section Only
Determine Material Properties in Torsion
Draw Shear Stress-Strain Curve
Ductile or Brittle Failure ?
ASTM E-143 Test Method
Compare results from Tensile, Poisson’s & Torsion
Test
Become familiar with Tinius Olsen Torsion Tester
(10,000 in-lb capacity)
Sign Convention
Assumptions
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Perfectly linear-elastic behavior
Circular x-section (solid or hollow)
Small rotations
Constant length
Constant diameter
No deformation
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A highly deformable
member
A circular x-section
A grid of parallel
circles and
longitudinal lines on
the outer surface
After Deformation
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Two torques are
applied
Circles remain
circles
Longitudinal lines
become twisted
All angles are
equal
End x-section
remains flat
Angle of Twist
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A prismatic circular
member
Fixed at x = 0
Other end is FREE
Loaded by a torque
at the free end
Angle (x) - angle of
twist
It depends on x:
/x=constant
Shear Strain
Before
After
Shear Strain Distribution
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max = c
 max - occurs on the
outer surface
 = (/c)  max
Result valid also for
circular tubes within
the material
Valid for plastic
deformation also
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Shear Strain Distribution
Shear Strain & Angle of Twist
=*/L
max =  * D / 2 / L
Shear Stress Distribution
- Elastic Deformation only
 X-section: linear
distribution
 Complimentary
stresses in action
 Shear stress occur
on axial planes (normal
to the x-section
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Shear Stress & Torque
- Linear Elastic Deformation Only
=T*/J
max = T * D / 2 / J
Shear Stress-Strain Curve
Modulus
of
Rupture
max, psi
Slope = D/D = Shear Modulus, G
max, rad
Figure 3.2 Typical shear stress-strain
diagram from a Torsion Test
Linear Elastic Region
– Shear Modulus
Failure Surface – Tensile Testing
Failure Surface – Torsion Test
Brittle Failure – Torsion Test
Stresses on an Inclined Plane in
an Axially Loaded Member
Seek for equilibrium after cutting
along the oblique plane
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 F 0:
 F 0:
n
N  P cos( )
t
V  P sin( )
Area of theinclinedplaneis :
An 

n
A
cos( )
NormalizeN and V with An to obtain
the normaland shear stresses,respectively,
t
V
P
N
on theinclinedplane(and notethat x  P / A) :
 n   x cos2   ( x / 2)(1  cos 2 )
 nt   x cos sin    ( x / 2) sin 2
x
Maximum Stress – Uniaxial Tension
Comparing Tensile vs. Torsion
Results
Measuring Angle of Twist
T=(R-R0)-(L-L0)
Torsion Data Sheet
Computing the Slip Angle
Manually Recorded Data
slip= H-(LH/LT)T
T=(R-R0)-(L-L0)
R0
L0
cor = H - slip
Shear Strain and Stress
=Radians(cor)*D/2/LH
=16T//D3
Integrating Computer Data
slip= constant
cor=H - slip
Selecting Computer Data
Head Angle = 30.00 deg
When Protractors taken OFF
Max Torque = 1931 in-lb
Entire Stress Strain Curve
Elastic Curve
Proportional Limit
=G*
Proportional Limit
Plot
columns
H, I, & J
Key Ideas - Torsion
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Shear strain formula is valid in both
elastic & plastic region
Torque formula is valid in linear elastic
region ONLY
Thus, shear stress formula is valid until
proportional limit
After proportional limit, shear stress
formula overestimates stress.
Torsion Results
Comparing Tensile vs. Torsion
Results
Comparing Lab 1, 2, & 3
Experimental Results
Lab #2 Report due TODAY
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Individual Data Processing and Individual
Write Up
Lab #2 Report is due TODAY by midnight
Answer Multiple Choice Questions ONLINE
TODAY by midnight
Answer Student Survey ONLINE TODAY by
midnight
Submit the Lab Report #2 in the drawer
labeled “MECE204-Strength Lab” under M. E.
Mail Folders