Unit 2 Review - cusd80.com

Manufacturing

Product Creation Cycle

Design → Material Select
→ Manufacture → Inspe
Feedback

g Process

e

tion → Process Selection
ection →

Typical product
cost breakdown

Manufacturin

• Raw Materials undergo
processes in the produ
goods

ng Processes

o various manufacturing
uction of consumer

Material

l Testing

Material

• Engineers use a design
to solve and document

• Engineers use destruct
testing on materials fo
identifying and verifyin
various materials.

• Materials testing provi
evaluation of material

l Testing

n process and formulas
t design problems.
tive and nondestructive
or the purpose if
ng the properties of

ides reproducible
properties

Stress- Strain
from tensile

Curve: created
testing data

E = = Stress
 Strain

E is the Elastic
Modulus.
E is the slope of the
line in the elastic
region.

Using data poi
material prope
-Modulus of e
-Elastic limit
-Resilience
-Yield point
-Plastic deform
-Ultimate stre
-Failure
-Ductility

ints, you can identify and calculate
erties:
elasticity

mation
ength

Stre
average amo

exerted pe

ess:
ount of force
er unit area

Strain: a measureme
in a structure due

• Strain is calculated f

Strain = Defor
Origin

or

ε=  / L

• Strain is deformation
dimensionless quantit

ent of deformation
to applied forces.

from:
rmation
nal Length

per unit length, a
ty

Proportional Limit: greates
capable of withstanding w
straight line proportional
strain. If the force applied
the material will return to
shape.

• Yield Point: The point at w
takes place, while the loa
the same or actually drop
the material is released, t
to its original shape.

st stress a material is
without deviation from
lity between the stress and
d to a material is released,
o its original size and

which a sudden elongation
ad on the sample remains
ps. If the force applied to
the material will not return

• Ultimate Strength: The
maximum load for a sa
Beyond this point elon
continues, but the forc

e point at which a
ample is achieved.
ngation of the sample
ce exerted decreases.

• Modulus of Elasticity: A
material’s ability to reg
dimensions after the re
force. The modulus is t
line portion of the stres
the proportional limit.

• Modulus of Resilience:
material’s ability to abs
elastic limit. This modu
the area under the stre
0-force to the elastic lim

A measure of a
gain its original
emoval of a load or
the slope of the straight
ss-strain diagram up to

: A measure of a
sorb energy up to the
ulus is represented by
ess vs. strain curve from
mit.

• Modulus of Toughness
material’s ability to pla
fracturing. Work is per
absorbing energy by th
This measurement is e
the stress vs. strain cur
through the rupture po

s: A measure of a
astically deform without
rformed by the material
he blow or deformation.
equal to the area under
rve from its origin
oint.

prop
given

Calculate Ultimate stress, stress at
portional limit, and modulus of elasticity
n an initial length of 3 inches and a cross

sectional area of 0.02 in2.

U

M
3

Ultimate stress = 443 lb / .02 in2
Ultimate stress = 22,150 psi

PL stress = 340 lb / 0.02 in2
PL stress = 17,000 psi

Modulus of elasticity = P*L/(Area*
deformation)

E @ (proportional limit)=
340 lb * 3 in / (0.02 in2 * 0.01 in)

= 5,100,000 psi

A 1” diameter piece of steel is 1
tensile load in the steel is 125,00
of elasticity is 30,000,000 psi, c
engineering process:

a) The tensile stress-
b) The total elongation caused b
c) The unit elongation-

15 feet long. If the total
00 pounds and the modulus
calculate using the 5 step

by the load-

A 1” diameter piece of steel is 1
tensile load in the steel is 125,00
of elasticity is 30,000,000 psi, c
engineering process:

a) The tensile stress-
b) The total elongation caused b
c) The unit elongation-
•Stress = P/A = 125,000 lbs/ (
psi
•Elongation = P*L / (A*E) = 1
0.5 in* 0.5in* 30,000,000 psi)
•Unit Elongation is Strain, or de
=0.08 feet/15 feet = 0.00533

15 feet long. If the total
00 pounds and the modulus
calculate using the 5 step

by the load-

(pi* 0.5 in* 0.5in) = 159,155

125,000 lbs * 15 feet / (pi*
= 0.08 ft or 0.96 inches.
eformation divided by length.

A 2” by 6” rectangular stee
and supports an axial load
Calculate using the 5 step
a) The maximum unit tens
b) The maximum allowed
tensile stress must not exce
c) The total elongation if 
using the maximum allowe

el beam is 60 feet long
of 15, 000 lbs.
engineering process:
sile stress in the rod.
load (P) if the unit
eed 20,000 psi.
= 30,000,000 psi
ed load from part B.

A 2” by 6” rectangular steel
and supports an axial load o
using the 5 step engineering
a) The maximum unit tensile
b) The maximum allowed lo
stress must not exceed 20,00
c) The total elongation if =
the maximum allowed load

Area = 2” * 6” = 12 in^2
• Stress = P/A = 15000 lbs
•Stress = P/A 20,000 psi =
240,000 lbs
•Elongation is (P*L)/(A*E)
/ (12 in^2 * 30,000,000 psi)

beam is 60 feet long
of 15, 000 lbs. Calculate
g process:
e stress in the rod.
oad (P) if the unit tensile
00 psi.
= 30,000,000 psi using
from part B.

/12in^2 = 1,250 psi
= P/12 in^2 P =

= 240,000 lbs * 60 feet
= 0.04 feet or 0.48 in.

Deflection of
Axial

= P*L
A*E

Deflection is measure of the deformation in
structure.

Where: P is the applied
L is the length
A is the cross se
E is the elastic m

f Rod under
Load

na L P

A

load

ection area
modulus

Stress/ Strain

A sample of material is ¼”d
turned to a smaller diam
used in a tensile machine
point for the material is 9
strength of the material
diameter would the sam
in order to meet the spe

n Example 1

diameter and must be
meter to be able to be

e. The target breaking
925 pounds. The tensile
is 63,750 psi. What
mple have to be turned to
ecified requirements?

Stress/ Strain

Knowns:
Load = 925 lb
Stress = 63,750 psi

n Example 1

Unknowns:
Dia final = ?

Stress/ Strain

Drawing:

Equations:

• A= D 2 σ=
= .7854D2
4

n Example 1

=P
A

Stress/ Strain

Substitution:

63750 psi = 925lbs
.7854D 2

Solve:

D2 = 925lbs 

(.7854in)(63750psi)

D = 0.18454413in2 =

n Example 1

 .018454413in2

= 0.136”

Stress/ Strain

A strand of wire 1,000 ft. lo
area of 3.5 sq. inches mu
load of 2000 lb. The mod
metal is 29,000,000 psi.
deformation of this mate

n Example 2

ong with a cross-sectional
ust be stretched with a
dulus of Elasticity of this
What is the unit
erial?

Stress/ Strain

Drawing:

Equations:

= ε= 

PL

AE L

n Example 2