Standard vs Custom Battery Packs - Decoding the Power Play
Timber and steel flexure
1.
2. 2
Table of Content
Table of Content
1) Introduction………………………………………………….3
2) Scope ………………………………………………………....6
3) Experimental Apparatus & Procedure………………….….8
4) Graphs and Tables ………………………………………...10
5) Sample Calculations………………………………………..22
6) Summary and Comparison of Results……….…………....26
7) Discussion and Conclusion………………………………...27
8) Questions ……………………………………………………28
9) References ………………….…………………………..….29
3. 3
INTRODUCTION
Engineering is the branch of science and technology concerned with the design, building, and use
of engines, machines, and structures. Always the engineer is looking for parameters that lead to
the best decisions while referring to the use of a material. As we know, Strength of Materials or
mechanics of materials, deals with the behavior of objects subjected to stresses and strains. It
appears the couple stress-strain (fig 1) is the result of the previous loads that were applied on the
element made of a specific material since the loading is caused by a tensile or compressive force.
Fig 1: Stress-Strain Diagram, source Google.com
In fact, an element under a load always is in tension and in compression at the same time (fig 2)
Saying an element is in tension or in compression is about the kind of element that is under
analysis. As an illustration, columns are known to be studied for compression most of the time and
beams for tension. However, tensile and compressive or shear stresses cannot be separated so
Flexure is the term to express the outcome from the effect of the three stresses previously cited
since it is about the geometry and the applied load on a structural element such beams that have
bent under the loading. In sum, Flexure, also called bending stresses is the stresses caused by the
bending moment (Force multiplied by arm of lever) and the formula for flexure or bending stress
(fb) according to the assumptions that stress is proportional to strain ,a plane section before
bending remains plane after the bending:
4. 4
fb = Mc/I =M/S (1)
M = moment max, k-in(beam simply supported M=PL/4);c=distance farthest fiber to neutral
axis(in),I=moment of Inertia,in4
;S= section modulus ,in3
; (S=I / c)
Fig 2: Beam under Load, source mathalino.com
As we said earlier, Engineers have to know the behavior of materials in tension, compression
and flexure meaning the determination of their ultimate or breaking strength. Therefore, it is
useful to test a material that undergoes bending forces via a flexure test. (Fig 3a,Fig 3b).
Fig 3a: set up for a flexural test, source google.com
F:force, d:thickness, L:span
5. 5
Fig 3b:Flexure test, source CMCE 2306, EO3O Fig 4: stress vs deformation, source CMCE
2306, EO3O
Mainly, the flexure test allows to measure the flexural strength and flexural modulus. Also, the
flexural test leads to determine the modulus of elasticity, E (ksi) and the proportional limit, ksi
(elastic limit) (Fig 4)
The modulus of elasticity is given by the following:
E = ΔPL3
/Δe48I (2)
ΔP: differential load, Δe: differential deformation, L: Span, I: Inertia,in4
, (I=bh3
/12 :rectangular
section, I=Πd2
/64 :circular section)
As a definition, Flexural strength (fb) , known as modulus of rupture, the bending stress at the onset
of rupture(ksi),or bend strength, or transverse rupture strength is a stress in a material just before
it yields. It is the maximum stress at the uttermost fiber on either the compression or tension side
of the specimen (fig 2).
Likewise, the Flexural modulus is obtained from the slope stress vs. strain diagram.
Once those two stresses are determined, they become tools to evaluate the capacity of the material
to adapt the flexure effect.
As an example, steel, wood and timber are products that will experience these forces or properties
in various forms. So, they are subjected to mechanical testing such as the flexure test before they
are used in the construction, furniture and common goods manufacturing industries depending
upon their application. As a result, the measured strength of the steel, or wood and timber material
will determine if they are acceptable candidates, and if they correspond to the values standard for
a particular application.
6. 6
Properties Steel Timber
AISI 1080 Yellow Pine Douglas
Modulus of Elasticity, ksi 29000 1460 1765
Tensil strength ultimate,
ksi
63.8 0.4 0.35
Modulus of rupture, ksi 120-130 6.29 3.6
Flexural modulus, ksi 1.26-1.31 1.18
Rupture, ksi 9.14 6.4
Density, lb/in3 0.284 0.145 0.145
Stress at the proportional
limit (elastic limit), 𝜎𝜎𝜎𝜎𝜎𝜎𝜎𝜎
(𝑘𝑘𝑘𝑘𝑘𝑘)
4-6 6.525-10.585 5.95-12.035
Table 1: Values Standard for steel, Timber in bending, source Matweb.com
Scope
Throughout history, wood has found favor as a building material due to its strength,
economy, workability, beauty and durability. Wood-frame buildings are economical to build,
heat and cool, and provide maximum comfort to occupants. Wood construction is readily
adaptable to traditional, contemporary and the most cutting-edge building styles. Its architectural
possibilities are limitless and its durability spans the centuries.
Wood building materials are good for the environment, too. Wood is a renewable,
sustainable resource that is manufactured in energy-efficient processes that optimize use of
renewable energy sources. In fact, in a comparison of fossil fuel consumption associated with the
materials for three floor systems – wood, concrete and steel – the wood joist floor required the
7. 7
least amount of fossil fuel energy.
ASTM Standard D3957 (ASTM 2006) provides procedures for establishing stress grades
for structural members used in log homes (Burke 2004). The standard addresses grading and
property assignment for two types of structural members used in log buildings: wall logs and
round timber beams. Only the standard applied to round timber beams are evaluated in this
study.
There is a good correlation between modulus of rupture and modulus of elasticity in static
bending and between MOE by static bending and MOE by transverse vibration for both Douglas
fir and pine.
The wood content of the pine and the content of the Douglas fir contribute to the large
difference in properties of the logs for the two species. he technical basis for a mechanical
grading system has historically been tile relationship between bending strength and stifles, and
relationships between bending strength and strength in compression and tension parallel to tile
grain (ASTM D6570). For ' usually graded lumber, relationships between bending strength and
tensile strength parallel to tile grain, and between bending strength and compression strength
parallel to the grain, are used to reduce sampling and testing costs (ASTM D1990).
8. 8
SAE-AISI 1080 steel is plain carbon steel formulated for primary forming into wrought
products. 1080 is the designation in both the SAE and AISI systems for this material. G10800 is
the UNS number. It has a moderately low ductility and can have a moderately high tensile
strength relative to other wrought plain carbon steels. The properties of SAE-AISI 1080 steel
include two common variations. The property values on this page show ranges across all of
them. For more specific values, follow the links immediately below. The graph bars on the
material properties cards further below compare SAE-AISI 1080 steel to: wrought plain carbon
steels, all iron alloys, and the entire database.
Procedures
1. Record the dimensions of the materials to the nearest 0.001" then scribe a line on the ends of a
28” inch span and the center point of the span.
2. Activate the hydraulic pump by depressing the Pump button on the console.
3. Zero Out the load using the 7, press the "display" key to stabilize the value.
4. Insert the material between the support rollers. Lower the fixed head to barely engage the
center of the material. Use the load knob to control the rate of loading.
5. Record the deflection readings (digital readings in inches) at every100 pound increment,
plotting a running laboratory graph of load vs. reading. This plotting of the points is continued
until the graph shows an appreciable bend indicating that the proportional limit (near the elastic
limit has been exceeded.
9. 9
6. Continue applying load until the ultimate (maximum) load is achieved and there is a visible
mode of failure. Record the ultimate load.
7. Depress The unload button. Use the unload knob to control the rate of unloading.
8. When all testing
Is complete depress the emergency stop button and turn off the main power switch.
9. For each specimen, note and record the type of failure as per Figure 8,Method D143 Of the
Selected ASTM Engineering Materials Standards.
APPARATUS
36-inch scale.
Machine equipment:
Timber transverse bending table
10. 10
Graphs & Tables:
Table #1: AISI 1080 Steel Data Sheet
Load, Kips Dial Deformation, In Formulas
0 0 0
.dialxdial factorδ =
3
48
P L
E
e I
∆
=
∆
4
64
d
I
π
=
M
S
σ =
I
S
C
=
3
32
d
S
π
=
4
PL
M =
Ultimate Strength: 2.2 Kips
0.1 3 0.015
0.2 4 0.02
0.3 6 0.03
0.4 7 0.035
0.5 9 0.045
0.6 10 0.05
0.7 12 0.06
0.8 14 0.07
0.9 15.5 0.0775
1 17.5 0.875
1.1 19.5 0.0972
1.2 22 0.11
1.3 24 0.12
1.4 26.5 0.1325
1.5 29.5 0.1475
1.6 34 0.17
1.7 39.5 0.1975
1.8 48.5 0.2425
22. 22
Sample Calculations:
Deformation for steel
Typical calculation
Deformation = Dial ∗ 0.0005 = 24 ∗ 0.005 = 0.120 in.
Bending Stress AISI 1080:
σ =
M
𝑆𝑆
=
6.6 𝑘𝑘. 𝑖𝑖𝑖𝑖
0.0397𝑖𝑖𝑖𝑖3
= 166.25𝑘𝑘𝑘𝑘𝑖𝑖
𝑀𝑀 = Moment (k-in), 𝑀𝑀 =
𝑃𝑃𝑃𝑃
4
=
2.2𝑘𝑘∗12𝑖𝑖 𝑖𝑖
4
= 6.6 𝑘𝑘 𝑖𝑖𝑖𝑖
S = Section Modulus 𝑆𝑆 =
πd
3
32
=
3.14∗0.743
32
= 0.0397 𝑖𝑖𝑛𝑛3
Modulus of Elasticity (AISI 1080):
𝐸𝐸 =
∆P
∆e
∗
𝐿𝐿
𝐼𝐼
=
(1.3𝑘𝑘 − 0.5𝑘𝑘)
(0.12𝑖𝑖𝑖𝑖 − 0.045𝑖𝑖𝑖𝑖)
∗
123
𝑖𝑖𝑖𝑖
48(0.0147𝑖𝑖𝑛𝑛4)
= 26,122 𝑘𝑘𝑘𝑘𝑘𝑘
𝐼𝐼 =
𝜋𝜋𝑑𝑑4
64
=
3.14 ∗ 0.744
64
= 0.0147 𝑖𝑖𝑛𝑛4
∆𝑃𝑃=differential load (k) from graph shown above
∆𝑒𝑒=differential deflection (in.) from graph shown above
𝐿𝐿 =span (in.)
𝐼𝐼 = Second moment of Inertia (in4),
Stress at the proportional limit (elastic limit), 𝜎𝜎𝜎𝜎𝜎𝜎𝜎𝜎 (𝑘𝑘𝑘𝑘𝑘𝑘) for steel
The Pu (kips) is obtained from graph (an estimate)
23. 23
2 2
2
Pr 2
1.5
0.74
0.430
4 4
1.5
3.48
0.430
u
op
Pu kips
d x
A in
P kips
ksi
A in
π π
σ
=
= = =
= = =
Deformation: For pine
Typical calculation
Deformation = Dial ∗ 0.0005 = 119 ∗ 0.0005 = 0.0595 in.
Bending Stress Timber (Pine):
σ =
M
𝑆𝑆
=
7.35 𝑘𝑘 𝑖𝑖𝑖𝑖
0.503𝑖𝑖𝑖𝑖3
= 14.61
𝑘𝑘
𝑖𝑖𝑛𝑛2
𝑀𝑀 = Moment (k-in), 𝑀𝑀 =
𝑃𝑃𝑃𝑃
4
=
1.05𝑘𝑘∗28𝑖𝑖 𝑖𝑖
4
= 7.35 𝑘𝑘 𝑖𝑖𝑖𝑖
S = Section Modulus 𝑆𝑆 =
bh
3
12
=
1.56∗1.573
12
= 0.503 𝑖𝑖𝑛𝑛3
Modulus of Elasticity (Pine):
𝐸𝐸 =
∆P
∆e
∗
𝐿𝐿
𝐼𝐼
=
(0.6𝑘𝑘 − 0.3𝑘𝑘)
(0.036𝑖𝑖𝑖𝑖 − 0.018𝑖𝑖𝑖𝑖)
∗
283
𝑖𝑖𝑖𝑖
48(0.503𝑖𝑖𝑛𝑛4)
= 14,206 𝑘𝑘𝑘𝑘𝑖𝑖
𝐼𝐼 =
𝑏𝑏ℎ3
12
=
1.56𝑖𝑖𝑖𝑖 ∗ 1.573
𝑖𝑖𝑖𝑖
12
= 0.503 𝑖𝑖𝑛𝑛4
∆𝑃𝑃=differential load (k) from graph shown above
∆𝑒𝑒=differential deflection (in.) from graph shown above
𝐿𝐿 =span (in.)
𝐼𝐼 = Second moment of Inertia (in4),
Stress at the proportional limit (elastic limit), 𝜎𝜎𝜎𝜎𝜎𝜎𝜎𝜎 (𝑘𝑘𝑘𝑘𝑘𝑘) for pine
The Pu (kips) is obtained from graph (an estimate)
24. 24
2
Pr 2
1.56 1.57 2.434
0.7
0.288
2.434
u
op
A bd x in
P kips
ksi
A in
σ
= = =
= = =
Deformation for fir:
Typical calculation
Deformation = Dial ∗ 0.0005 = 115 ∗ 0.0005 = 0.575 in.
Bending Stress Timber (Douglas Fir):
σ =
M
𝑆𝑆
=
6.65 𝑘𝑘 𝑖𝑖𝑖𝑖
0.371𝑖𝑖𝑖𝑖3
= 17.92
𝑘𝑘
𝑖𝑖𝑛𝑛2
𝑀𝑀 = Moment (k-in), 𝑀𝑀 =
𝑃𝑃𝑃𝑃
4
=
0.95𝑘𝑘∗28𝑖𝑖 𝑖𝑖
4
= 6.65 𝑘𝑘 𝑖𝑖𝑖𝑖
S = Section Modulus 𝑆𝑆 =
bh3
12
=
1.43∗1.463
12
= 0.371 𝑖𝑖𝑛𝑛3
Modulus of Elasticity (Douglas Fir):
𝐸𝐸 =
∆P
∆e
∗
𝐿𝐿
𝐼𝐼
=
(0.6𝑘𝑘 − 0.3𝑘𝑘)
(0.0575𝑖𝑖𝑖𝑖 − 0.021𝑖𝑖𝑖𝑖)
∗
283
𝑖𝑖𝑖𝑖
48(0.371𝑖𝑖𝑛𝑛4)
= 16,886 𝑘𝑘𝑘𝑘𝑖𝑖
𝐼𝐼 =
𝑏𝑏ℎ3
12
=
1.43𝑖𝑖𝑖𝑖 ∗ 1.463
𝑖𝑖𝑖𝑖
12
= 0.371 𝑖𝑖𝑛𝑛4
∆𝑃𝑃=differential load (k) from graph shown above
∆𝑒𝑒=differential deflection (in.) from graph shown above
𝐿𝐿 =span (in.)
𝐼𝐼 = Second moment of Inertia (in4),
25. 25
Stress at the proportional limit (elastic limit), 𝜎𝜎𝜎𝜎𝜎𝜎𝜎𝜎 (𝑘𝑘𝑘𝑘𝑘𝑘) for fir
2
Pr 2
1.43 1.46 2.1
0.7
0.33
2.1
u
op
A bd x in
P kips
ksi
A in
σ
= = =
= = =
Summary and comparison of results
The obtained results from samples calculation, graphs and standard values
Table #
Properties Steel Timber
AISI
1080
Standard
Value
AISI
1080
Yellow Pine
Standard
Value
Yellow
Pine
Douglas Fir
Standard
Value
Douglas
Fir
Modulus of
Elasticity, ksi
29000 26,122 1460 14,206 1765 16,886
Tensil strength
ultimate, ksi
63.8 …….. 0.4 …… 0.35 ………..
Modulus of
rupture, ksi
120-130 166.25 9.70 14.61 6.40 17.92
Flexural
modulus, ksi
…….. ……. 1.26-1.31 ………. 1.18 ……….
Density, lb/in3 0.284 ……… 0.145 …… 0.145 ……..
Stress at the
proportional
limit (elastic
limit), 𝜎𝜎𝜎𝜎𝜎𝜎𝜎𝜎
(𝑘𝑘𝑘𝑘𝑘𝑘)
4-6 3.48 6.525-
10.585
0.22 5.95-12.035 0.33
Discussion and conclusion.
The overall experiment was performed fallowing the presented methodology. There were three
experiments as: timber flexure pine, fir and steel flexure (AISI 1080).
26. 26
A. Modulus of Elasticity (ksi): this property was calculated for timber (pine and fir) and
steel (AISI 1080).
• The standard value for pine is 1460ksi and the actual result is 14,206ksi. This modulus of
elasticity is close to the standard value, which means that the experiment was done
properly. The timber presented efficient behavior when the force was applied.
• The Douglas fir presents 1765ksi as standard value, and the values obtained from lab is
16,886𝑘𝑘𝑘𝑘𝑘𝑘. These values are a slightly different since the material, used during the
experiment, did not respond efficiently to external applied force.
• The standard value of steel (AISI 1080) is 29000ksi and the actual value is 26,122ksi.
The actual value is acceptable since it is close to the standard value. The actual specimen
behaves differently since the exerted force was applied uneven at some point, but overall
the specimen reacted efficiently to vertical applied force.
B. Modulus of rupture, ksi
The actual value of the pine is 14.60ksi and the standard value is 9.70ksi.
apparently, these values are differently. the actual value is higher than standard
value, which means that the material responds differently than standard materials’
behavior.
The obtained actual value of the fir is 17.92ksi and the standard value is 6.40ksi.
Clearly the specimen performed has high value since it has different mechanical
properties than the standard values’ behaviors.
The standard values for steel is 120 -130ksi and the actual vale is 166ksi. The
actual value is over than actual value. The used specimen apparently present
higher resistance due to exerted vertical force along x axis. Also, there some
factors that may affect to get close result to the standard value as shown below.
C. Stress at the proportional limit (elastic limit), 𝜎𝜎𝜎𝜎𝜎𝜎𝜎𝜎 (𝑘𝑘𝑘𝑘𝑘𝑘)
The pine’s actual value is 0.22ksi and the standard values is 6.525-10.585ksi. The
difference the actual and standard value is not huge because the elastic limit point
along the curved was picked randomly. This could be one the reason why the
actual value is off. Also, the elastic limit has to deal with quantity of exerted load.
27. 27
The standard value of fir is 5.95-12.035ksi and the actual value is 0.33ksi. once
again, the estimate value along the curve at graph # 3 was not accurate. Also, the
specimen presented poor limit elastic, as result the elastic limit is less.
The standard value of steel is 4-6ksi and the actual value is 3.48ksi. The result is
acceptable since the obtained number is close to the standard value. In terms of
the elastic limit this specimen reacted efficiently. The located point the of elastic
limit was picked accurately along the curve (graph #1).
It is important to state, the performed experiment could have been presented some mistakes: poor
measurement of dimension of timber and poor recording data, manufacture fabrication, poor
calibration of equipment (excessive or not vertical force applied on the specimen). These
mistakes could make to specimen go over and under to the standard values. It is very important
to pay attention these details to get precise and accurate results.
The properties of materials are very essential to analyze and study when comes to design a new
structure and redesign an existing structure. Designer studies the elastic modulus of beam,
timber, or concrete. These properties are indicator to accomplish efficient structures. Also,
flexural testing is predominately used in many construction industries and others as well, where
the material is subject to some of bending force. To illustrate, the construction industry is the
most common test for structural steel, concrete beam, timber joists since it is widely used to
evaluate materials that can be difficult to test in tensile mode
A. Fir B. Pine C. Steel (AISI1080)
The bending stress of pine, fir and steel at maximum applied load.
28. 28
The rupture occurs more in fir than pine since the fir has wide parallel grains and pine has close
grains. The grain resist to the exerted force
Questions
A. for timber
1. Are values for short, clear beams applicable to beams of large size? Explain.
For the large size of beam are not applicable since the serviceability and structural
requirement for larger beam should be studied.
2. Is the modulus of elasticity for bending the same as for compression or tension parallel to
the grain? Explain.
At some point the modulus of elasticity for bending to parallel to the grain will be same
to the compression or tension, but at some point it will not be same since the wood
timber has different and mechanical behavior properties it is not unique as steel.
3. Since the modulus of rupture is calculated using the usual flexure formula for stress (σ =
𝑀𝑀𝑀𝑀/ 𝐼𝐼), why can't be called the breaking stress?
The use the flexure strength up to yield point only not up to breaking point. As the given
formula the moment applied are within limits not up to breaking point. Hence it is called
modulus of rupture or flexural strength. However, the modulus of rupture is the flexural
strength which is the stress located before yielding point. Then, it cannot be called as
breaking stress
B. For steel 1080
Why is it not possible to obtain the modulus of elasticity in transverse bending by the same
method used in the steel tension/compression test?
The strain is proportional to stress in the tension and compression test. As result, the modulus of
elasticity is found using Hook’s law up to proportional limit. However, in case of the transverse
bending stress, strain relation is not proportional and it is not possible to obtain modulus of
elasticity. In addition, to find the modulus of elasticity the difference of values is taken before
elastic limit multiplicated by distance and constant times moment of inertia.
29. 29
References
Michael Ashby (2011). Materials selection in mechanical design, Butterworth-
Heinemann. p. 40.
William D. Callister, Jr., Materials Science and Engineering, Hoken: John Wiley & Sons,
Inc., 2003.
ASTM C1161-02c(2008)e1, Standard Test Method for Flexural Strength of Advanced
Ceramics at Ambient Temperature, ASTM
Flexural Strength of Timber-Lightweight concrete composite Beam , E.C David, H.B Koh,
Timber Compression strength perpendicular to the grain,testing of glugam beams with and
without reinforcement, Daniel Ed,Fredrik Hasselqvist, Lund Intitute of technology,Lund
University,2011.
Flexure Formulas. Web: mathalino.com
Static and Dynamic Wood Test Methods.Web:Ressourcetest.net
Flexural strength, Wikipedia, free encyclopedia. Web: Google.com
American Forest and Paper Association. Inc. (AF&PA). 2005. National design
specification for wood construction...&PA. Washington. D.C.merican Society for
Testing
