Figure E8.4 shows a material chart with the two indices of Exercise E8.2 as axes. Identify and plot coupling lines for selecting materials for a column with F = 10° N and H=3m (the same conditions as above), and for a second column with F = 10³ N and H = 20m. M2 Density* Price / (Modulus^0.5), ($/m³)/(MPa^0.5) 10- 0.1- Technical ceramics CFRP Silicon carbide Zirconia Aluminium Soda-lime glass Cast iron. Carbon steel 1 Ti alloys Concrete Cement Polymers Composites GFRP Brick Granite Al alloys Bamboo Softwood Nontechnical ceramics PTFE PP Phenolics Natural materials. Hardwood Cu alloys PE Metals and alloys Pb alloys 10 10² M₁ = Density * Price/ Compressive strength, ($/m³)/(MPa) 103 104 MFA 16 108
Design Against Fluctuating Loads
Machine elements are subjected to varieties of loads, some components are subjected to static loads, while some machine components are subjected to fluctuating loads, whose load magnitude tends to fluctuate. The components of a machine, when rotating at a high speed, are subjected to a high degree of load, which fluctuates from a high value to a low value. For the machine elements under the action of static loads, static failure theories are applied to know the safe and hazardous working conditions and regions. However, most of the machine elements are subjected to variable or fluctuating stresses, due to the nature of load that fluctuates from high magnitude to low magnitude. Also, the nature of the loads is repetitive. For instance, shafts, bearings, cams and followers, and so on.
Design Against Fluctuating Load
Stress is defined as force per unit area. When there is localization of huge stresses in mechanical components, due to irregularities present in components and sudden changes in cross-section is known as stress concentration. For example, groves, keyways, screw threads, oil holes, splines etc. are irregularities.
![Problem E8.3
Figure E8.4 shows a material chart with the two indices of Exercise E8.2 as axes. Identify and plot
coupling lines for selecting materials for a column with F= 10° N and H=3m (the same
conditions as above), and for a second column with F= 10³ N and H= 20m.
M2 = Density* Price / (Modulus^0.5), ($/m³)/(MPa^0.5)
10-
0.1-
Technical ceramics
Zirconia
Aluminium
Silicon carbide
Soda-lime glass
Cast iron
Carbon steel
1
Concrete
Ti alloys
CFRP
Cement
Polymers
Composites
GFRP
Brick
Granite
Al alloys
Bamboo
Softwood
Nontechnical
ceramics
Natural materials....
Hardwood
TTTT
PTFE
PP
Phenolics
103
Cu alloys
PE Metals and alloys
TTTT
10
10²
M₁ = Density * Price/ Compressive strength, ($/m³)/(MPa)
Pb alloys
104
MFA 16
105](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe648a5cf-871e-427a-ae67-8615d084e87f%2Fe040cd96-ec34-44cf-b294-63e18052ec78%2Fpr9asf_processed.jpeg&w=3840&q=75)
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