This document discusses the fundamentals of mechanics of elastic materials. It defines key terms like stress, strain, and modulus of elasticity used to characterize how materials deform under force. Stress represents the force applied over an area, strain is the deformation normalized by length. Modulus of elasticity relates stress and strain and indicates a material's stiffness. Examples show how to calculate stress, strain, and modulus of elasticity for an elastic rod under tension. The stress-strain curve is also explained, distinguishing elastic deformation, permanent plastic deformation, yield point, ultimate strength, and failure point.
Page 6 of 8Engineering Materials ScienceMetals LabLEEDS .docxbunyansaturnina
Page 6 of 8Engineering Materials Science
Metals Lab
LEEDS BECKETT UNIVERSITY
SCHOOL OF THE BUILT ENVIRONMENT & ENGINEERING
Course: BSc (Hons) Civil Engineering BEng (Hons) Civil Engineering
HND Civil Engineering
Laboratory Experiment:
Stress-Strain Behaviour of Mild Steel and High Yield Steel bars.
Associated Module(s)
Level 4 Engineering Materials Science
Object of Experiment
To investigate the stress-strain behaviour of the above materials.
Theory/Analysis
A knowledge of the behaviour of structural steel under load is essential to ensure structural collapse does not occur and that serviceability requirements are achieved. In these respects the following mechanical properties of a material are required:-
1. The yield stress, σy (or 0.2% proof stress)
2. The Elastic (or Young’s) Modulus, E
3. The maximum tensile strength, σmax
4. The stress at failure, ie the fracture stress, σf
5. The % elongation at failure
Apparatus
1. 500kN Denison Testing Machine
2. Extensometer and Denison extension gauge (measures cross head movement)
3. Grade 250 plain round mild steel bar, 20mm diameter
Characteristic strength = 250 N/mm²
Conforms to BS 4449.
4. Grade 460 deformed high yield steel.
Reinforcing bar, T16, 16mm diameter.
Characteristic strength = 460 N/mm²
Conforms to BS 4449.
Method
Each of the bars in turn is placed in the jaws of the testing machine.
The 50mm extensometer is attached to the bar and zeroed.
Load is applied and recorded in increments up to failure. For each load increment, extension readings from the extensometer and the Denison extension gauge are noted.
At the yield point, the extensometer is removed to prevent damage to it and readings continue on the Denison extension gauge.
The load at failure and the manner of failure are noted.
See the Figure below showing the Test Setup.
(
L
G
values; L
G
= 100 mm for the plain
round
bar, and L
G
= 80 mm for the deformed
high yield
bar
) (
L
G
,
gauge length of the samples
) (
P = the tensile force applied to bars from Dennison testing machine
) (
P
) (
Extension of the sample bars is measured by:
the
Dennison (on-board) extension gauge which monitors cross-head
movement
. This effectively gives sample extension readings from the start of the test (P = 0) through to failure.
An extensometer gauge. This is accurate only over the initial linear-elastic phase of the test.
) (
P
)
Each student should prepare and submit a laboratory report, the results and discussion sections are outlined below:a) Results and Calculations
Readings of load (P), against extension (e), have been recorded for each specimen tested and provided to you (appended at the end of this laboratory briefing document).
Knowing the original bar diameters (d), load data can converted to stress (σ) by dividing each load reading by the appropriate cross sectional area.
Strain values are determined by dividing the extension (e) data by the appropriate gauge length for each bar (LG); the g.
Page 6 of 8Engineering Materials ScienceMetals LabLEEDS .docxbunyansaturnina
Page 6 of 8Engineering Materials Science
Metals Lab
LEEDS BECKETT UNIVERSITY
SCHOOL OF THE BUILT ENVIRONMENT & ENGINEERING
Course: BSc (Hons) Civil Engineering BEng (Hons) Civil Engineering
HND Civil Engineering
Laboratory Experiment:
Stress-Strain Behaviour of Mild Steel and High Yield Steel bars.
Associated Module(s)
Level 4 Engineering Materials Science
Object of Experiment
To investigate the stress-strain behaviour of the above materials.
Theory/Analysis
A knowledge of the behaviour of structural steel under load is essential to ensure structural collapse does not occur and that serviceability requirements are achieved. In these respects the following mechanical properties of a material are required:-
1. The yield stress, σy (or 0.2% proof stress)
2. The Elastic (or Young’s) Modulus, E
3. The maximum tensile strength, σmax
4. The stress at failure, ie the fracture stress, σf
5. The % elongation at failure
Apparatus
1. 500kN Denison Testing Machine
2. Extensometer and Denison extension gauge (measures cross head movement)
3. Grade 250 plain round mild steel bar, 20mm diameter
Characteristic strength = 250 N/mm²
Conforms to BS 4449.
4. Grade 460 deformed high yield steel.
Reinforcing bar, T16, 16mm diameter.
Characteristic strength = 460 N/mm²
Conforms to BS 4449.
Method
Each of the bars in turn is placed in the jaws of the testing machine.
The 50mm extensometer is attached to the bar and zeroed.
Load is applied and recorded in increments up to failure. For each load increment, extension readings from the extensometer and the Denison extension gauge are noted.
At the yield point, the extensometer is removed to prevent damage to it and readings continue on the Denison extension gauge.
The load at failure and the manner of failure are noted.
See the Figure below showing the Test Setup.
(
L
G
values; L
G
= 100 mm for the plain
round
bar, and L
G
= 80 mm for the deformed
high yield
bar
) (
L
G
,
gauge length of the samples
) (
P = the tensile force applied to bars from Dennison testing machine
) (
P
) (
Extension of the sample bars is measured by:
the
Dennison (on-board) extension gauge which monitors cross-head
movement
. This effectively gives sample extension readings from the start of the test (P = 0) through to failure.
An extensometer gauge. This is accurate only over the initial linear-elastic phase of the test.
) (
P
)
Each student should prepare and submit a laboratory report, the results and discussion sections are outlined below:a) Results and Calculations
Readings of load (P), against extension (e), have been recorded for each specimen tested and provided to you (appended at the end of this laboratory briefing document).
Knowing the original bar diameters (d), load data can converted to stress (σ) by dividing each load reading by the appropriate cross sectional area.
Strain values are determined by dividing the extension (e) data by the appropriate gauge length for each bar (LG); the g.
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also calculation of modulus of elasticity using
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Analysis of Cross-ply Laminate composite under UD load based on CLPT by Ansys...IJERA Editor
In current study the strength of composite material configuration is obtained from the properties of constituent
laminate by using classical laminate plate theory. For the purpose of analysis various configurations of 2 layered
and 4 layered cross ply laminates are used. The material of laminate is supposed to be boron/epoxy having
orthotropic properties. The loading in current study is supposed to be of uniformly distributed load type. For the
analysis purpose software working on finite element analysis logics i.e. Ansys mechanical APDL is used. By the
help of Ansys mechanical APDL the deflection and stress intensity is found out. The effect of variation of
laminate layers is also studied in current study along with the effect of variation of stacking patterns. The current
study will also help to conclude which stacking pattern is best in 2 layered and 4 layered cross ply laminate.
Experimental evaluation of strain in concrete elementsnisarg gandhi
Evaluation of strain using
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also calculation of modulus of elasticity using
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also for the procedure to use electrical strain gauge see the following link
https://drive.google.com/open?id=0Bw9bdaDxJsb8enFZOFhlRWFMYWs&authuser=1
Theory of Plasticity is a very important topic in solid mechanics & Strength of Materials. It is very very useful for Mechanical & Civil Engineering students.
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2. Why study mechanics?
Useful for the analysis and
design of load-bearing
structures, such as:
buildings
bridges
space shuttles
prosthetics
biological implants
Also used to
characterize materials
3. Stress
The force per unit area, or intensity of the forces
distributed over a given section. (units = Pascals
[Pa] or pounds per square inch [psi])
σ = F/A
• Stress is how engineers normalize the force that is
applied to a material to account for differences in
geometry.
• Useful for predicting failure conditions for materials.
4. Strain
Deformation per unit length (units: none [unitless])
ε = ΔL/L
Strain is how engineers normalize the deformation
that a material experiences to account for
differences in geometry.
Useful for determining how much a material can
deform before failure.
5. Modulus of Elasticity
A representation of the stiffness of a material that
behaves elastically (units: Pascals [Pa] or pounds per
square inch [psi])
E = σ/ε
What equation is this similar to?
k = F /Δx
Modulus of elasticity is how engineers characterize
material behavior.
Useful for knowing how materials behave, material
selection for device design, and calculating the stress in
a material since it is easier to measure deformation than
it is to determine the exact force on a material.
7. Example 1
This rod is exposed to a
tensile force of 20 N. What is
the stress in the rod?
20 N
20 N
0.5 m
0.5 m
σ =F/A
F = 20 N (given)
A = 0.5 m * 0.5 m = 0.25 m2
σ = 20 N / 0.25 m2
σ = 80 Pa
3 m
8. Example 2
The rod below is exposed to a
tensile force of 20 N and
elongates by 0.03 m. Calculate
the strain.
ε = ΔL/L
ΔL = 0.03 m (given)
L = 3 m
ε = 0.03 m / 3 m
ε = 0.01
20 N
20 N
0.5 m
0.5 m
3 m
9. Example 3
The rod below is exposed to
a tensile force of 20 N and
elongates by 0.03 m.
Calculate the modulus of
elasticity.
E = σ/ε
σ = 80 Pa (from first example)
ε = 0.01 (from second example)
E = 80 Pa / 0.01
E = 8000 Pa or 8 kPa
20 N
20 N
0.5 m
0.5 m
3 m
12. Understanding
the Stress-Strain Curve
elastic range –
The linear portion of
the stress-strain
curve. When the
force is released,
the material returns
to its original
dimensions.
plastic range –
The region of
permanent
deformation.
stress
strain
elastic
range
plastic
range
yield
stress
ultimate
tensile
strength fracture
stress
13. Understanding
the Stress-Strain Curve
yield stress –
The minimum stress
that causes
permanent
deformation.
ultimate tensile
strength –
The maximum stress
that the material can
withstand.
Also defines the
beginning of necking.
stress
strain
elastic
range
plastic
range
yield
stress
ultimate
tensile
strength fracture
stress
14. The Stress-Strain Curve
necking – A localized decrease in cross sectional
area that causes a decrease in stress with an
increase in strain.
fracture stress – Stress in which the material fails.
stress
strain
elastic
range
plastic
range
yield
stress
ultimate
tensile
strength fracture
stress
F
F
tensile load
direction
neck
steel
tensile
specime
n
Explain why we are studying tissue mechanics. First, we need to gain an understanding of mechanics in general. This ties into many different areas of engineering.
[Mechanics of Elastic Materials presentation, part of the Mechanics of Elastic Solids lesson in TeachEngineering.org]
Example: To design and build the World Trade Center, different types of steel were used, based on stress analysis of the building design. Did not use high performance steel if it was not needed. Collapse of WTC was due to too many beams being broken when the plane hit. The remaining beams could not handle the stress/weight from floors above and they buckled/failed.
Example: For biological implants, engineers analyze the stresses in bones to make sure the implant can withstand the stress that the natural joint withstood. Bone degenerates with lack of stress, so we don’t want the implant to take away from the stress borne by surrounding bone.
Explain normalizing the force by comparing two pieces of material that are the same length but different cross sectional areas. The force it takes to displace them the same amount (or break them) is different, but the stress is the same.
If you know what the yield stress is in your material and you know the dimensions, then you can calculate how much force the material can withstand before failing.
If you have a very long piece of material compared to a very short piece, then the same amount of displacement is more detrimental to the short piece, compared to the long piece. The displacement may be on the same order of magnitude as the small piece, but on a smaller order of magnitude than the large piece. Strain normalizes this and puts it on the same scale.
Yield stress corresponds to a certain strain as well. If you can only measure strain or your system depends on being strained a certain amount then you can predict failure based on the strain.
Engineers physically test materials to obtain this stress-strain curve graph. They record force and displacement data and then calculate stress and strain. Dog bone-shaped test specimens prevent sample from breaking at grips.
Explain that modulus of elasticity is the slope of the elastic region of the line.
Explain the difference between uniform deformation and necking. Even though the cross sectional area is changing, the stress is calculated using the initial area throughout the entire test because it is very difficult to measure area during testing.
