lab #2 report due today
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Lab #2 Report due TODAY. 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 - PowerPoint PPT PresentationTRANSCRIPT
Lab #2 Report due TODAY 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 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
Perfectly linear-elastic behavior Circular x-section (solid or hollow) Small rotations Constant length Constant diameter
No deformation A highly
deformable member
A circular x-section
A grid of parallel circles and longitudinal lines on the outer surface
After Deformation
Two torques are applied
Circles remain circles
Longitudinal lines become twisted
All angles are equal
End x-section remains flat
Angle of Twist 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
9
Shear Strain Distribution
max = c• max - occurs on the
outer surface• = (/c) max
• Result valid also for circular tubes within the material
• Valid for plastic deformation also
Shear Strain Distribution
Shear Strain & Angle of Twist
= * / L
max = * D / 2 / L
12
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
Shear Stress & Torque- Linear Elastic Deformation Only
= T * / J
max = T * D / 2 / J
Shear Stress-Strain Curve
max, rad
max, psi
Modulusof Rupture
Slope = = Shear Modulus, G
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
n
P
NV
t
2sin)2/(sincos
)2cos1)(2/(cos
: )/ that note (and plane inclined on the
ly,respectivestresses,shear and normal the
obtain to with V and N Normalize
)cos(
:is plane inclined theof Area
)sin(:0
)cos(:0
xxnt
x2
xn
x
AP
A
AA
PVF
PNF
n
n
t
n
x
Maximum Stress – Uniaxial Tension
Comparing Tensile vs. Torsion Results
Measuring Angle of TwistT=(R-R0)-(L-L0)
Torsion Data Sheet
Computing the Slip Angle
Manually Recorded Data
R0
L0
T=(R-R0)-(L-L0)
slip= H-(LH/LT)T
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 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, & 3Experimental Results
Lab #2 Report due TODAY 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