Strength of Materials

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CTC / MTC 222

Strength of Materials


Final Review

Final Exam


Tuesday, December 13,
3:00


5:00


30% of grade


Graded on the basis of 30 points in increments of ½ point


Open book


May use notes from first two tests plus two additional sheets
of notes


Equations, definitions, procedures, no worked examples


Also may use any photocopied material handed out in class


Work problems on separate sheets of engineering
paper


Hand in test paper, answer sheets and notes stapled to back
of answer sheets

Course Objectives


To provide students with the necessary
tools and knowledge to analyze forces,
stresses, strains, and deformations in
mechanical and structural components.


To help students understand how the
properties of materials relate the
applied loads to the corresponding
strains and deformations.

Chapter One


Basic Concepts


SI metric unit system and U.S. Customary
unit system


Unit conversions


Basic definitions


Mass and weight


Stress, direct normal stress, direct shear stress
and bearing stress


Single shear and double shear


Strain, normal strain and shearing strain


Poisson’s ratio, modulus of elasticity in tension
and modulus of elasticity in shear

Direct Stresses


Direct Normal Stress ,




σ

= Applied Force/Cross
-
sectional Area = F/A


Direct Shear Stress,



Shear force is resisted uniformly by the area of the part in
shear




= Applied Force/Shear Area = F/A
s


Single shear


applied shear force is resisted by a single cross
-
section of the member


Double shear


applied shear force is resisted by two cross
-
sections of the member

Direct Stresses


Bearing Stress,
σ
b



σ
b

= Applied Load/Bearing Area = F/A
b


Area A
b
is the area over which the load is transferred


For flat surfaces in contact, A
b

is the area of the smaller of the
two surfaces


For a pin in a close fitting hole, A
b

is the projected area,
A
b
= Diameter of pin x material thickness


Chapter Two


Design Properties


Basic Definitions


Yield point, ultimate strength, proportional
limit, and elastic limit


Modulus of elasticity and how it relates
strain to stress


Hooke’s Law


Ductility
-

ductile material, brittle material

Chapter Three


Direct Stress


Basic Definitions


Design stress and design factor


Understand the relationship between design stress,
allowable stress and working stress


Understand the relationship between design factor, factor of
safety and margin of safety


Design / analyze members subject to direct stress


Normal stress


tension or compression


Shear stress


shear stress on a surface, single shear and
double shear on fasteners


Bearing stress


bearing stress between two surfaces,
bearing stress on a fastener

Chapter Three


Axial
Deformation and Thermal Stress


Axial strain
ε
,


ε

=
δ

/ L , where
δ

= total deformation, and L = original
length


Axial deformation,
δ


δ

= F

L
/
A E


If unrestrained, thermal expansion will occur due to
temperature change


δ

=
α

x L x ∆T


If
restrained
, deformation due to temperature change
will be prevented, and stress will be developed


σ

= E

α

(∆T)

Chapter Four


Torsional
Shear Stress and Deformation


For a circular member,
τ
max
= Tc / J


T = applied torque, c = radius of cross section, J = polar moment
of inertia


Polar moment of Inertia, J


Solid circular section, J =
π

D
4
/ 32


Hollow circular section, J =
π

(D
o
4
-

D
i
4
)

/ 32


Expression can be simplified by defining the
polar
section modulus
, Z
p
= J / c, where c = r = D/2


Solid circular section,
Z
p

=
π

D
3
/ 16


Hollow circularsection,
Z
p

=
π

(D
o
4
-

D
i
4
)

/ (16D
o
)


Then,
τ
max
= T / Z
p

Chapter Five


Shear Forces and
Bending Moments in Beams


Sign Convention


Positive Moment M


Bends segment concave upward


compression on top



Relationships Between Load,
Shear and Moment


Shear Diagram


Application

of a downward concentrated load
causes a downward jump in the shear diagram.
An upward load causes an upward jump.


The slope of the shear diagram at a point
(dV/dx) is equal to the (negative) intensity of
the distributed load w(x) at the point.


The change in shear between any two points on
a beam equals the (negative) area under the
distributed loading diagram between the points.


Relationships Between Load,
Shear and Moment


Moment Diagram


Application of a clockwise concentrated moment
causes an upward jump in the moment diagram.
A counter
-
clockwise moment causes a
downward jump.


The slope of the moment diagram at a point
(dM/dx) is equal to the intensity of the shear at
the point.


The change in moment between any two points
on a beam equals the area under the shear
diagram between the points.

Chapter Six


Centroids and
Moments of Inertia of Areas


Centroid of complex shapes can be calculated
using:


A
T

̅Y
̅

= ∑ (A
i
y
i
) where:


A
T
= total area of composite shape


̅Y
̅

= distance to centroid of composite shape from some
reference axis


A
i
= area of one component part of shape


y
i
= distance to centroid of the component part from the
reference axis


Solve for

̅Y
̅

= ∑ (A
i
y
i
) / A
T


Perform calculation in tabular form


See Examples
6
-
1
&
6
-
2


Moment of Inertia of

Composite Shapes


Perform calculation in tabular form


Divide the shape into component parts which are simple shapes


Locate the centroid of each component part, y
i

from some
reference axis


Calculate the centroid of the composite section,

̅Y
̅

from some
reference axis


Compute the moment of inertia of each part with respect to its
own centroidal axis, I
i


Compute the distance, d
i

=

̅Y
̅

-

y
i
of the centroid of each part
from the overall centroid


Compute the transfer term A
i
d
i
2

for each part


The overall moment of inertia I
T

, is then:


I
T
=
∑ (
I
i +
A
i
d
i
2
)


See Examples
6
-
5
through
6
-
7


Chapter Seven


Stress Due to
Bending


Positive moment


compression on top, bent concave
upward


Negative moment


compression on bottom, bent
concave downward


Maximum Stress due to bending (Flexure Formula)


σ
max

= M c / I


Where M = bending moment, I = moment of inertia, and c
= distance from centroidal axis of beam to outermost fiber


For a non
-
symmetric section distance to the top fiber,
c
t
, is different than distance to bottom fiber c
b


σ
top

= M c
t

/ I


σ
bot

= M c
b

/ I


Section Modulus, S


Maximum Stress due to bending


σ
max

= M c / I


Both I and c are geometric properties of the section


Define section modulus, S = I / c


Then
σ
max

= M c / I = M / S


Units for S


in
3

, mm
3


Use consistent units


Example: if stress,
σ
, is to be in ksi (kips / in
2

), moment, M, must be
in units of kip


inches


For a non
-
symmetric section S is different for the top and the
bottom of the section


S
top

= I / c
top


S
bot

= I / c
bot

Chapter Eight


Shear Stress in
Beams


The shear stress,


, at any point within a beams cross
-
section
can be calculated from the General Shear Formula:





= VQ / I t, where


V = Vertical shear force


I = Moment of inertia of the entire cross
-
section about the centroidal
axis


t = thickness of the cross
-
section at the axis where shear stress is to
be calculated


Q = Statical moment about the neutral axis of the area of the cross
-
section between the axis where the shear stress is calculated and the
top (or bottom) of the beam


Q is also called the first moment of the area


Mathematically, Q = A
P

̅y
̅

, where:


A
P

= area of theat part of the cross
-
section between the axis where the
shear stress is calculated and the top (or bottom) of the beam



̅y
̅

= distance to the centroid of A
P

from the overall centroidal axis


Units of Q are length cubed; in
3
, mm
3
, m
3
,

Shear Stress in Common Shapes


The General Shear Formula can be used to develop
formulas for the maximum shear stress in common
shapes.


Rectangular Cross
-
section



max

= 3V / 2A


Solid Circular Cross
-
section



max

= 4V / 3A


Approximate Value for Thin
-
Walled Tubular Section



max

≈ 2V / A


Approximate Value for Thin
-
Webbed Shape



max

≈ V / t h


t = thickness of web, h = depth of beam

Chapter Twelve


Pressure
Vessels


If R
m

/ t ≥ 10, pressure vessel is considered
thin
-
walled


Stress in wall of
thin
-
walled

sphere


σ

=
p D
m

/ 4 t


Longitudinal stress

in wall of
thin
-
walled

cylinder


σ

=
p D
m

/ 4 t


Longitudinal stress is
same

as stress in a sphere


Hoop

stress in wall of cylinder


σ

=
p D
m

/ 2 t


Hoop stress is twice the magnitude of longitudinal stress


Hoop stress in the cylinder is also twice the stress in a
sphere of the same diameter carrying the same pressure