MECHANICS OF MATERIALS
UNIAXIAL STRESS-STRAIN
Uniaxial Loading and Deformation
σ = P/A, where
Stress-Strain Curve for Mild Steel
♦
σ = stress on the cross section,
P = loading, and
A = cross-sectional area.
ε = δ/L, where
δ = elastic longitudinal deformation and
L = length of member.
E=σ ε=
δ=
P A
δL
PL
AE
THERMAL DEFORMATIONS
δt = αL (Τ – Τo), where
The slope of the linear portion of the curve equals the
modulus of elasticity.
δt =
deformation caused by a change in temperature,
α =
temperature coefficient of expansion,
DEFINITIONS
L =
length of member,
Engineering Strain
ε = ∆L / L0, where
Τ =
Τo =
final temperature, and
initial temperature.
ε =
∆L =
engineering strain (units per unit),
change in length (units) of member,
CYLINDRICAL PRESSURE VESSEL
L0 =
original length (units) of member.
Cylindrical Pressure Vessel
For internal pressure only, the stresses at the inside wall are:
Percent Elongation
% Elongation
σ t = Pi
⎛ ∆L ⎞
= ⎜ ⎟ × 100
⎝L ⎠
Percent Reduction in Area (RA)
The % reduction in area from initial area, Ai, to final area,
Af, is:
σ t = − Po
ro2 + ri2
ro2 − ri2
σr = radial stress,
Pi = internal pressure,
Shear Stress-Strain
γ = τ/G, where
Po = external pressure,
γ
=
shear strain,
ro = outside radius.
τ
=
shear stress, and
ri
σ a = Pi
E
, where
2(1 + ν )
E
=
modulus of elasticity
v
=
Poisson's ratio, and
=
– (lateral strain)/(longitudinal strain).
= inside radius, and
For vessels with end caps, the axial stress is:
shear modulus (constant in linear force-deformation
relationship).
G=
and 0 > σ r > − Po , where
σt = tangential (hoop) stress,
⎛ Ai − A f ⎞
=⎜
× 100
⎝ Ai ⎟⎠
True Stress is load divided by actual cross-sectional area.
G =
and 0 > σ r > − Pi
For external pressure only, the stresses at the outside wall
are:
o
%RA
ro2 + ri2
ro2 − ri2
ri2
ro2 − ri2
These are principal stresses.
♦ Flinn, Richard A. & Paul K. Trojan, Engineering Materials & Their Applications,
4th ed., Houghton Mifflin Co., 1990.
38
MECHANICS OF MATERIALS (continued)
The circle drawn with the center on the normal stress
(horizontal) axis with center, C, and radius, R, where
When the thickness of the cylinder wall is about one-tenth
or less, of inside radius, the cylinder can be considered as
thin-walled. In which case, the internal pressure is resisted
by the hoop stress and the axial stress.
Pr
σt = i
t
and
Pr
σa = i
2t
C=
where t = wall thickness.
♦
Principal Stresses
For the special case of a two-dimensional stress state, the
equations for principal stress reduce to
σ a ,σ b =
σc = 0
2
2
⎛ σx − σ y
, R = ⎜⎜
2
⎝
2
⎞
⎟ + τ 2xy
⎟
⎠
The two nonzero principal stresses are then:
STRESS AND STRAIN
σx + σ y
σx + σ y
⎛ σx − σ y
± ⎜⎜
2
⎝
σa = C + R
σb = C − R
τin = R
(σy, τxy)
2
⎞
⎟ + τ 2xy
⎟
⎠
(σx, τxy)
The two nonzero values calculated from this equation are
temporarily labeled σa and σb and the third value σc is
always zero in this case. Depending on their values, the
three roots are then labeled according to the convention:
algebraically largest = σ1, algebraically smallest = σ3, other
= σ2. A typical 2D stress element is shown below with all
indicated components shown in their positive sense.
The maximum inplane shear stress is τin = R. However, the
maximum shear stress considering three dimensions is
always
♦
τ max =
To construct a Mohr's circle, the following sign conventions
are used.
σ1 − σ 3
.
2
Hooke's Law
Three-dimensional case:
εx = (1/E)[σx – v(σy + σz)]
γxy = τxy /G
εy = (1/E)[σy – v(σz + σx)]
γyz = τyz /G
εz = (1/E)[σz – v(σx + σy)]
γzx = τzx /G
Plane stress case (σz = 0):
εx = (1/E)(σx – vσy)
εy = (1/E)(σy – vσx)
εz = – (1/E)(vσx + vσy)
Mohr's Circle – Stress, 2D
⎤
⎡
⎢1 v
0 ⎥ ⎧ε x ⎫
⎢v 1 0 ⎥ ⎪ε ⎪
⎥⎨ y ⎬
⎢
⎢0 0 1 − v ⎥ ⎪⎩γ xy ⎪⎭
2 ⎦⎥
⎣⎢
σx = Eεx or σ = Eε, where
⎧σ x ⎫
E
⎪ ⎪
⎨σ y ⎬ =
2
⎪τ ⎪ 1 − v
⎩ xy ⎭
To construct a Mohr's circle, the following sign conventions
are used.
Uniaxial case (σy = σz = 0):
1. Tensile normal stress components are plotted on the
horizontal axis and are considered positive.
Compressive normal stress components are negative.
σx, σy, σz = normal stress,
εx, εy, εz = normal strain,
γxy, γyz, γzx = shear strain,
τxy, τyz, τzx = shear stress,
E = modulus of elasticity,
G = shear modulus, and
v = Poisson's ratio.
2. For constructing Mohr's circle only, shearing stresses
are plotted above the normal stress axis when the pair of
shearing stresses, acting on opposite and parallel faces
of an element, forms a clockwise couple. Shearing
stresses are plotted below the normal axis when the
shear stresses form a counterclockwise couple.
♦ Crandall, S.H. & N.C. Dahl, An Introduction to The Mechanics of Solids, McGraw-Hill Book Co.,
Inc., 1959.
39
MECHANICS OF MATERIALS (continued)
The term on the left side of the inequality is known as the
effective or Von Mises stress. For a biaxial stress state the
effective stress becomes
STATIC LOADING FAILURE THEORIES
Brittle Materials
(
Maximum-Normal-Stress Theory
The maximum-normal-stress theory states that failure occurs
when one of the three principal stresses equals the strength
of the material. If σ1 ≥ σ2 ≥ σ3, then the theory predicts that
failure occurs whenever σ1 ≥ Sut or σ3 ≤ – Suc where Sut and
Suc are the tensile and compressive strengths, respectively.
σ ′ = σ 2A − σ Aσ B + σ 2B
(
σ ′ = σ 2x − σ x σ y + σ 2y + 3τ 2xy
σ1
Sut
12
σ x , σ y , and τ xy are the stresses in orthogonal directions.
VARIABLE LOADING FAILURE THEORIES
Modified Goodman Theory: The modified Goodman
criterion states that a fatigue failure will occur whenever
σa σm
+
≥1
S e S ut
where
Se =
Sut =
Sy =
σa =
σm =
σmax =
σ
or
σ max
≥ 1,
Sy
σm ≥ 0 ,
fatigue strength,
ultimate strength,
yield strength,
alternating stress, and
mean stress.
σm + σa
Soderberg Theory: The Soderberg theory states that a
fatigue failure will occur whenever
If σ 1 ≥ σ2 ≥ σ3 and σ3 < 0, then the theory predicts that
yielding will occur whenever
σ1 σ3
−
≥1
Sut Suc
σa σm
+
≥1,
Se S y
σm ≥ 0
Endurance Limit for Steels: When test data is unavailable,
the endurance limit for steels may be estimated as
Ductile Materials
Maximum-Shear-Stress Theory
The maximum-shear-stress theory states that yielding begins
when the maximum shear stress equals the maximum shear
stress in a tension-test specimen of the same material when
that specimen begins to yield. If σ 1 ≥ σ2 ≥ σ3, then the
theory predicts that yielding will occur whenever τmax ≥ Sy /2
where Sy is the yield strength.
⎧⎪ 0.5 Sut , Sut ≤ 1, 400 MPa ⎫⎪
Se′ = ⎨
⎬
⎩⎪ 700 MPa, Sut > 1, 400 MPa ⎭⎪
Distortion-Energy Theory
The distortion-energy theory states that yielding begins
whenever the distortion energy in a unit volume equals the
distortion energy in the same volume when uniaxially
stressed to the yield strength. The theory predicts that
yielding will occur whenever
⎡ (σ1 − σ 2 )2 + (σ 2 − σ 3 )2 + (σ1 − σ 3 )2 ⎤
⎥
⎢
2
⎥⎦
⎢⎣
)
where σ A and σ B are the two nonzero principal stresses and
τ
σ3
12
or
Coulomb-Mohr Theory
The Coulomb-Mohr theory is based upon the results of
tensile and compression tests. On the σ, τ coordinate system,
one circle is plotted for Sut and one for Suc . As shown in the
figure, lines are then drawn tangent to these circles. The
Coulomb-Mohr theory then states that fracture will occur for
any stress situation that produces a circle that is either
tangent to or crosses the envelope defined by the lines
tangent to the Sut and Suc circles.
-Suc
)
12
≥ Sy
40
MECHANICS OF MATERIALS (continued)
φ = total angle (radians) of twist,
T = torque, and
Endurance Limit Modifying Factors: Endurance limit
modifying factors are used to account for the differences
between the endurance limit as determined from a rotating
beam test, S e′ , and that which would result in the real part, Se.
L = length of shaft.
T/φ gives the twisting moment per radian of twist. This is
called the torsional stiffness and is often denoted by the
symbol k or c.
Se = ka kb kc kd ke Se′
where
b
For Hollow, Thin-Walled Shafts
T
τ=
, where
2 Am t
Surface Factor, ka = aSut
Surface
Finish
Ground
Machined or
CD
Hot rolled
As forged
Factor a
kpsi
MPa
1.34
1.58
2.70
4.51
14.4
39.9
Exponent
b
57.7
272.0
–0.085
–0.265
t
–0.718
–0.995
BEAMS
Am =
kb = 1
8 mm ≤ d ≤ 250 mm;
−0.097
kb = 1.189d eff
0.6 ≤ kb ≤ 0.75
kb = 1
d > 250 mm;
For axial loading:
thickness of shaft wall and
the total mean area enclosed by the shaft measured
to the midpoint of the wall.
Shearing Force and Bending Moment Sign
Conventions
1. The bending moment is positive if it produces bending
of the beam concave upward (compression in top fibers
and tension in bottom fibers).
2. The shearing force is positive if the right portion of the
beam tends to shear downward with respect to the left.
Size Factor, kb:
For bending and torsion:
d ≤ 8 mm;
=
♦
Load Factor, kc:
kc = 0.923
axial loading, Sut ≤ 1,520 MPa
kc = 1
axial loading, Sut > 1,520 MPa
kc = 1
bending
Temperature Factor, kd:
for T ≤ 450°C, kd = 1
Miscellaneous Effects Factor, ke: Used to account for strength
reduction effects such as corrosion, plating, and residual
stresses. In the absence of known effects, use ke = 1.
The relationship between the load (q), shear (V), and
moment (M) equations are:
q ( x) = −
dV(x)
dx
TORSION
Torsion stress in circular solid or thick-walled (t > 0.1 r)
shafts:
Tr
τ=
J
V2 − V1 = ∫ x 2 ⎡−
⎣ q ( x) ⎤⎦ dx
where J = polar moment of inertia (see table at end of
DYNAMICS section).
M 2 − M1 = ∫ x 2 V ( x) dx
V =
x
1
x
1
TORSIONAL STRAIN
γ φz = limit r (∆φ ∆z ) = r (dφ dz )
Stresses in Beams
εx = – y/ρ, where
ρ = the radius of curvature of the deflected axis of the
beam, and
y = the distance from the neutral axis to the longitudinal
fiber in question.
∆z → 0
The shear strain varies in direct proportion to the radius,
from zero strain at the center to the greatest strain at the
outside of the shaft. dφ/dz is the twist per unit length or the
rate of twist.
τφz = G γφz = Gr (dφ/dz)
♦ Timoshenko, S. and Gleason H. MacCullough, Elements of Strengths of Materials, K. Van Nostrand
Co./Wadsworth Publishing Co., 1949.
2
T = G (dφ/dz) ∫A r dA = GJ(dφ/dz)
φ = ∫ oL
dM(x)
dx
T
TL
, where
dz =
GJ
GJ
41
MECHANICS OF MATERIALS (continued)
Using the stress-strain relationship σ = Eε,
Axial Stress:
σx =
M =
I =
y
=
COLUMNS
For long columns with pinned ends:
Euler's Formula
σx = –Ey/ρ, where
the normal stress of the fiber located y-distance
from the neutral axis.
Pcr =
1/ρ = M/(EI), where
the moment at the section and
the moment of inertia of the cross-section.
substitute I = r2A:
Pcr
π2 E
=
, where
A ( r )2
r = radius of gyration and
/r = slenderness ratio for the column.
For further column design theory, see the CIVIL
ENGINEERING and MECHANICAL ENGINEERING
sections.
Transverse shear stress: τxy = VQ/(Ib), where
q = shear flow,
A′ =
y′ =
, where
= unbraced column length.
σx = – My/I, where
the distance from the neutral axis to the fiber
location above or below the axis. Let y = c, where c
= distance from the neutral axis to the outermost
fiber of a symmetrical beam section.
Let S = I/c: then, σx = ± M/S, where
S = the elastic section modulus of the beam member.
Transverse shear flow: q = VQ/I and
=
=
=
=
2
Pcr = critical axial loading,
σx = ± Mc/I
τxy
V
b
Q
π 2 EI
ELASTIC STRAIN ENERGY
If the strain remains within the elastic limit, the work done
during deflection (extension) of a member will be
transformed into potential energy and can be recovered.
If the final load is P and the corresponding elongation of a
tension member is δ, then the total energy U stored is equal
to the work W done during loading.
shear stress on the surface,
shear force at the section,
width or thickness of the cross-section, and
A′y ′ , where
area above the layer (or plane) upon which the
desired transverse shear stress acts and
distance from neutral axis to area centroid.
U = W = Pδ/2
Deflection of Beams
Using 1/ρ = M/(EI),
d2y
= M, differential equation of deflection curve
dx 2
The strain energy per unit volume is
u = U/AL = σ2/2E
3
Modulus of
Elasticity, E
Mpsi
29.0
10.0
14.5
1.6
GPa
200.0 69.0
100.0
11.0
Modulus of
Rigidity, G
Mpsi
11.5
3.8
6.0
0.6
GPa
80.0
26.0
41.4
4.1
0.30
0.33
0.21
0.33
10−6 °F
6.5
13.1
6.7
1.7
10−6 °C
11.7
23.6
12.1
3.0
Material
Determine the deflection curve equation by double
integration (apply boundary conditions applicable to the
deflection and/or slope).
EI (dy/dx) = ∫ M(x) dx
EIy = ∫ [ ∫ M(x) dx] dx
The constants of integration can be determined from the
physical geometry of the beam.
Poisson's Ratio, v
Coefficient of
Thermal
Expansion, α
42
Wood (Fir)
d4y
= dV(x)/dx = −q
dx 4
Cast Iron
EI
MATERIAL PROPERTIES
Aluminum
d y
= dM(x)/dx = V
dx 3
(for tension)
Steel
EI
Units
EI
Beam Deflection Formulas – Special Cases
(δ is positive downward)
P
y
b
a
Pa 2
(3x − a ), for x > a
6 EI
Px 2
(− x + 3a ), for x ≤ a
δ=
6 EI
δ=
δmax
x
φmax
L
y w
δmax
x
φmax
L
y
δmax
L
43
y
x
φmax
M
P
a
δ=
M o x2
2 EI
δ=
x
L
y
wo x 2 2
x + 6 L2 − 4 Lx
24 EI
Pb
6 LEI
Pb
δ=
6 LEI
b
R1 = Pb/L
(
δ=
R2 = Pa/L
w
L
R1 = w0 L/2
Ml
(
⎡L
3
3
2
2
⎢ b (x − a ) − x + L − b
⎣
[− x + (L
3
(
2
2
)]
)x⎤⎥ , for x > a
⎦
Pa 2
(3L − a )
6 EI
φ max =
Pa 2
2 EI
δ max =
wo L4
8EI
φ max =
wo L3
6 EI
δ max =
M o L2
2 EI
φ max =
MoL
EI
δ max =
(
Pb L2 − b 2
)
at x =
δ max =
5wo L4
384 EI
wo L3
24 EI
R2 = w0 L/2
Wo LOAD PER UNIT
LENGTH
Mr
δmax
Rl
Rl = Rr =
L2 − b 2
3
Pab(2 L − a )
6 LEI
Pab(2 L − b )
φ2 =
6 LEI
φ1 =
φ1 = φ 2 =
φmax
Note:
32
9 3LEI
− b x , for x ≤ a
wo x 3
L − 2 Lx 2 + x 3
24 EI
)
Rr
δ ( x) =
2
WO x
( L2 − 2 Lx + x 2 )
24 EI
2
WO L
W L
and M l = M r = O
2
12
Crandall, S.H. & N.C. Dahl, An Introduction to The Mechanics of Solids, McGraw-Hill Book Co., Inc., 1959.
δ Max =
4
WO L
l
at x =
384 EI
2
φ Max = 0.008
WO L3
24 EI
l
l
at x = ±
2
12
MECHANICS OF MATERIALS (continued)
δ=
x
)
δ max =