Strength of materials
STRENGTH OF MATERIALS deals with the relations
between external applied loads and their internal
effects on bodies.
STRENGTH OF
MATERIALS
Moreover, the bodies are no longer assumed to be
ideally rigid; the deformations however small, are the
major interest.
The properties of the materials of which a structure
or machine is made affect both its choice and the
dimensions that will satisfy the requirements of
strength and rigidity.
Mechanics of
Deformable Bodies
It includes the study of the strength capabilities and
characteristics of selected materials.
Strength of materials
Strength of materials
The subject matter includes discussions of the
fundamental concepts of stresses and strains
experienced and/or developed by different materials
in their loaded state and subjected to different
conditions of constraint.
Understanding of how bodies or materials respond to
applied load is the main area of emphasis.
Study of the relationship between externally applied
loads and their internal effects on rigid bodies.
RIGID BODY – bodies which neither change in
shape and size after the application of forces.
FREE BODY DIAGRAM – Sketch of the isolated body
showing all the forces on it.
THREE MAJOR DIVISIONS OF MECHANICS
1. Mechanics of Rigid Bodies – Engineering Mechanics
2. Mechanics of Deformable Bodies – Strength of
Materials
3. Mechanics of Fluids - Hydraulics
Strength of materials
Strength of materials
The strength of a material is its ability to withstand
an applied stress without failure
Two categories -> Yield Strength and Ultimate
Strength
Yield Strength - the stress
level at which a material
begins to deform plastically.
Yield strength refers to the point on the engineering
stress-strain curve beyond which the material begins
deformation that cannot be reversed upon removal of
the loading
Ultimate Strength - It is the
maxima of the stress-strain
curve. It is the point at
which necking will start.
Ultimate strength refers to the point on the
engineering stress-strain curve corresponding to the
maximum stress.
A material's strength
microstructure.
is
dependent
on
its
Fracture Strength - The
stress
calculated
immediately
before
the
fracture.
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Analysis of internal forces
Analysis of internal forces
By setting up the equilibrium conditions, the
inner forces of a member subjected to an external
load situation can be determined. So far neither
the material nor the type of cross section applied
for the member are being taken into account.
But both material and type of cross section
obviously have an impact on the behavior of the
member subjected to load.
y
Mxy
Pxy
Pxx
Mxz
z
Analysis of internal forces
Pxx (Axial Force) – The
component
measures
the pulling (or pushing)
action over the section.
A pull represents a
tensile
force
which
tends to elongate the
member
whereas
a
push is a compressive
force which tends to
shorten it. It is often
denoted by P.
Analysis of internal forces
Mxx (Torque) – This
component measures the
resistance to twisting the
member and is commonly
given the symbol T.
Mxy, Mxz (Bending
Movements) – These
components measure the
resistance to bending the
member about the Y or Z
axes and are often
denoted by My or Mz.
x
Pxz
To design the member therefore a closer look on
how the internal forces act along its cross section
needs to be taken.
Analysis of internal forces
Mxx
Pxy,Pxz (Shear Force) –
These are components
of the total resistance
to sliding the portion to
one
side
of
the
selection
pass
the
other. The resultant
shear force is usually
designated by V and its
components by Vy and
Vz to identify their
directions.
Fundamental concept of stress
When a force is transmitted through a body, the
body tends to change its shape or deform. The
body is said to be strained.
STRESS – is defined as the strength of material
per unit area. It is the force on a member divided
by the area which carries the force. In symbol;
𝜎=
Where:
σ = stress
P = force
A = area
𝑃
𝐴
2
Fundamental concept of stress
Units
σ, Stress
P, Force
A, Area
Note:
1
1
1
1
English
psi (lbs/in2),
ksi (kips/in2)
pounds, kips
sq. in. (m2)
Metric
Pa (N/m2)
N, kN
(mm2), m2
MPa = 1x106 Pa = 1x106 N/m2
kip = 1000 pounds (lbs)
ksi = 1000 psi
N/mm2 = 1 MN/m2 = 1 MPa
Kinds of stresses
Axial Stress – the type of stress wherein the force
applied is perpendicular or normal to the area. It
can be tensile or compressive stress.
Shearing Stress – the type of stress wherein the
force applied is parallel to the area
Bearing Stress – is the constant pressure between
separate bodies. It differs from the compressive
stress as it is an internal stress caused by the
compressive force.
Torsional Stress – stress produced due to torque
Bending Stress – stress developed due to bending
of the member
STRENGTH OF
MATERIALS
Mechanics of
Deformable Bodies
AXIAL Stress
AXIAL Stress problem 1
Axial stress may be tensile, 𝜎𝑡 or compressive, 𝜎𝑐
and result from forces acting perpendicular to the
plane of the cross-section
Determine the axial stress on members BD, CE &
CD of the truss shown. Assume cross-sectional
area of each member is 900mm².
Tension
Compression
conclusion: the normal stress acting along a
section of a member only depends on the external
load applied (e.g. a normal force F) and the
geometry of its cross section A (true for statically
determinant systems).
3
AXIAL Stress problem 1
AXIAL Stress problem 1
FBD
Av
To get the force AV:
𝜮ME = 0
AV(6) – 200(3) – 100(1.5) + 50(3) = 0
AV(6) = 200(3) + 100(1.5) – 50(3)
Ev
Av
AV = 100kN
AXIAL Stress problem 1
AXIAL Stress problem 1
To get the stress of member BD, use the formula:
PBD
A
100kN
=
900mm2
σBD =
B
BD
CD
CE
3m
C
A
200kN
Av=100kN
σBD
Av
To get force BD,
𝜮MC = 0
AV(3) – BD(3) = 0
100(3) – BD(3) = 0
BD = 100kNcompression
σBD = 111.11MPa compression
AXIAL Stress problem 1
AXIAL Stress problem 1
To get the stress of member CE, use the formula:
PCE
A
50kN
=
900mm2
σCE =
B
BD=100kN D
Av
To get force CE,
𝜮MD = 0
BD(4.5) – 200(1.5) – CE(3) = 0
100(4.5) – 300 = CE(3)
CE
BD= =50kN
1 tension
A
3m
Av=100kN
C
CE
100kN
σCE
σCE = 55.55MPa tension
200kN
4
AXIAL Stress problem 1
B
To get the stress of member CD, use the formula:
BD=100kN
CE=50kN
3m
σCD
C
200kN
Av
Av=100kN
θ = tan−1
3m
θ
1.5m
θ = 63.43°
3
1.5
PCD
A
111.81kN
=
900mm2
= 124.23MPa tension
σCD =
CD
A
AXIAL Stress problem 1
To get force CD,
𝜮MA = 0
200(3) – BD(3) – CDV(3) = 0
200(3) – 100(3) – CD(sin63.43)(3) = 0
CD = 111.81kNtension
AXIAL Stress problem 2
σCD
AXIAL Stress problem 2
An aluminum tube is rigidly fastened between a
bronze rod and a steel rod as shown. Axial loads
are applied at the positions indicated. Determine
the stress in each material.
For Bronze,
𝜮FX = 0
PB = 20kN
PB 20,000N
σB = =
AB 700mm2
σB = 28.57MPa compression
AXIAL Stress problem 2
For Aluminum,
𝜮FX = 0
20kN − 15kN = PA
PA = 5kN
5,000N
PA
=
σA =
AA 1,000mm2
σA = 5MPa compression
AXIAL Stress problem 2
For Steel,
𝜮FX = 0
20kN − 15kN − 15kN + PS = 0
PS = 10kN
PS 10,000N
σS = =
AS 800mm2
σS = 12.5MPa tension
5
AXIAL Stress problem 3
A 12in. square steel bearing plate lies between an
8in. diameter wooden post and a concrete footing.
Determine the maximum value of the axial load P
if the stress in wood is limited to 1800psi and in
concrete is limited to 650psi.
AXIAL Stress problem 3
For concrete,
PC
σC =
AC
For wood,
PW
σW =
AW
1800
lbs
=
in2
PW
8in
𝜋
4
650
2
PC = 93,600lbs
PW = 90,477.87lbs
AXIAL Stress problem 4
lbs
PC
=
in2
12in 12in
Pmax = 90,477.87lbs
AXIAL Stress problem 4
A homogenous 150kg bar AB carries a 2kN force
as shown. The bar is supported by a pin at B and
a 10mm diameter cable CD. Determine the stress
in the cable.
𝜮MB = 0
4
150 ∗ 9.81 3
PCD
3 −2 6 −
=0
5
1000
PCD = 6.84kN tension
AXIAL Stress problem 4
A=
SCD
SCD
SCD
π
10 2 = 25πmm2
4
PCD 6.84kN 103 N/kN
=
=
A
25πmm2
= 87.09N/mm2 tension
= 87.09MPa tension
AXIAL Stress problem 5
Determine the largest weight W which can be
supported by the two wires shown. The stresses in
wires AB and AC are not to exceed 100MPa and
150MPa respectively. The cross-sectional areas of
the two wires are 400mm2 for wire AB and
200mm2 for wire
AC.
B
C
30o
A
45o
W
6
AXIAL Stress problem 5
AXIAL Stress problem 5
FBD
P = AS
𝜮Y = 0
ACY = WY
AC sin 75 = W sin 60
AC = 0.897W
𝜮H = 0
ABH = ACH
AB cos 30 = AC cos 45
AB cos 30 = 0.897W cos 45
AB = 0.732W
AXIAL Stress problem 6
The Bell Crank shown is in equilibrium.
Determine the required diameter of the connecting
rod AB if its axial stress is limited to 100MPa.
For AC:
AC = (200)(150) = 30000N
0.897W = 30,000
W = 33,444.82N
For AB:
W = 33,444.82N
AB = 400(100) = 40,000
0.732W = 40,000
W = 54644.81N
AXIAL Stress problem 6
ΣMD = 0
30kN sin 60 240mm − P 200mm = 0
P = 31.18 kN
P
A AB
N
31.18 x 103 N
100
=
mm2
πd2
4
d = 19.92mm
Say: d = 20mm
σAB =
STRENGTH OF
MATERIALS
Mechanics of
Deformable Bodies
7
shearing Stress
Shearing Stress is produced whenever the applied
load cause one section of a body to tend to slide
past its adjacent section.
τ=
Where
τ – shear stress
V – shear force
A – area in shear
V
A
shearing Stress
Shear Stresses are produced by equal and
opposite parallel forces not in line.
The forces tend to make one part of the material
slide over the other part.
Shear Stress is tangential to the area over which
it acts.
Shear Stress is a measure of the internal
resistance of a material to an externally applied
shear load.
shearing Stress
The rivet resists shear across its cross-sectional
area.
Single Shear
shearing Stress
Bolt resists shear across two cross-sectional
areas.
Double Shear
ΣFy = 0
P = 2V
P
V=
2
shearing Stress
A circular slug is about to be punched out of a
plate.
Punching Shear
ΣFy = 0
P=V
Ashear = C * t
Ashear = πdt
shearing Stress problem 1
A hole is to be punched out of a plate having an
ultimate shearing stress of 300MPa. If the
compressive stress in the punch is limited to
400MPa. Determine the maximum thickness of
the plate from which a hole of 100mm in diameter
can be punched. If the plate is 10mm thick,
compute the smallest diameter hole that can be
punched.
8
shearing Stress problem 1
shearing Stress problem 1
For punching force, P
P
σc =
Ac
N
P
400
=
π(100mm)2
mm2
4
𝐏 = 𝟑. 𝟏𝟒 𝐱 𝟏𝟎𝟔 𝐍
First Situation
τp = 300MPa
σc = 400MPa
d = 100mm
For shear force, V
ΣFy = 0
P=V
𝟑. 𝟏𝟒 𝐱 𝟏𝟎𝟔 𝐍 = 𝐕
shearing Stress problem 1
P
Ac
P
σc =
πd2
4
Solving for thickness, t
V
Av
V
τp =
πdt
N
3.14 x 106 N
300
=
2
mm
π 100 mm t
t = 33.33 mm
τp =
First Situation
τp = 300MPa
σc = 400MPa
d = 100mm
shearing Stress problem 1
ΣFy = 0
P=V
σc =
πd2
4
N πd2
P = 400
mm2 4
N πd2
P = 400
mm2 4
P = σc
Second Situation
τp = 300MPa
σc = 400MPa
t = 10mm
𝐏 = 𝟏𝟎𝟎𝛑𝐝𝟐
shearing Stress problem 2
The end chord of a timber truss is framed into the
bottom chord as shown in the figure. Neglecting
friction, compute dimension b if the allowable
shearing stress is 900kPa.
P=50KN
b
30o
c
100πd2 = V
V
τp =
Av
100πd2
τp =
πdt
N
100πd2
300
=
mm2 πd 10 mm
d = 30 mm
Second Situation
τp = 300MPa
σc = 400MPa
t = 10mm
shearing Stress problem 2
P=50KN
b
30o
c
τ=
V
Av
V = PH
V = 50cos30
V = 43.3kN
N
43.3x103 N
0.90
=
2
mm
150 mm (b)
b = 320.74mm
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shearing Stress problem 3
shearing Stress problem 3
The Bell Crank shown is in equilibrium.
Determine the shearing stress in the pin at D if its
diameter is 20mm.
ΣMD = 0
30kN sin 60 240mm − P 200mm = 0
P = 31.18 kN
ΣFH = 0
DH = P + 30kN cos 60
DH = 31.18 kN + 30kN cos 60
DH = 46.18 kN
DH
ΣFV = 0
DV
DV = 30kN sin 60
Dv = 25.98kN
𝐃 = 𝟓𝟐. 𝟗𝟖𝐤𝐍
shearing Stress problem 3
τ=
V
A
52,980N
π
2 ∗ 20mm
4
τ = 84.33MPa
τ=
2
59
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