A revisionary list of formulae for Mechanics of Solids
A revisionary list of formulae for Mechanics of Solids
Formula
Formula symbols
Units
Stress =
applied force cross-sectionalarea
σ=
F A
Strain =
change in length original length
ε=
x L
E=
σ ε
Pa
k=
F δ
N/m
G=
τ γ
Pa
Young’s modulus of elasticity = Stiffness =
stress strain
force extension
Modulus of rigidity =
shear stress shear strain
Thermal strain = coefficient of linear expansion × temperature rise
ε = αT
Thermal stress in compound bar
σ1 =
Ultimate tensile strength =
(α1 − α 2 ) E1 E 2 A2 T ( A1 E1 + A2 E 2 )
maximum load original cross-sectional area
Moment = force × perpendicular distance
Pa
Pa Pa
M = Fd
stress stress bendingbending momentmoment σ M E = = = = distancedistance from neutral of area of areay fromaxis neutralsecond axis moment second moment I R ,s modulus Young ,sYoung modulus = = radius ofradius curvature of curvature
Nm N/m3
Torque = force × perpendicular distance
T = Fd
Nm
Power = torque × angular velocity
P = Tω = 2πnT
W
Horsepower
1 hp = 745.7 W
Torque = moment of inertia × angular acceleration
T = Iα
Nm
shear stress torque = radius polar second moment of area (rigidity)(angle of twist) = length
τ T Gθ = = r J L
N/m3
Average velocity = Acceleration =
distance travelled time taken
change in velocity time taken
v=
s t
m/s
a=
v−u t
m/s2
Mechanics of Solids, Second Edition © 2016 C. Ross, J. Bird and A. Little. Published by Taylor & Francis. All rights reserved.
Formula
Formula symbols
Units
Linear velocity
v = ω r
m/s
Angular velocity
ω=
Linear acceleration
a = rα
θ = 2πn t
rad/s m/s2 m/s
Relationships between initial velocity u, final velocity v, displacement s, time t and constant acceleration a
v 2 = v1 + at 1 s = ut + at 2 2 2 2 v = u + 2as
rad/s
Relationships between initial angular velocity ω1, final angular velocity ω2, angle θ, time t and angular acceleration α
ω 2 = ω1 + αt 1 2 θ = ω 1t + α t 2 ω 12 = ω 22 + 2αθ
m (m/s)2
rad (rad/s)2
Momentum = mass × velocity
kg m/s
Impulse = applied force × time = change in momentum
kg m/s
Force = mass × acceleration
F = ma
N
Weight = mass × gravitational field
W = mg
N
Centripetal acceleration
a=
v2 r
m/s2
Centripetal force
F=
mv 2 r
N
mass Density = volume
ρ=
m V
kg/m3
Work done = force × distance moved
W = Fs
Efficiency = Power =
J
useful output energy input energy
energy used (or work done) = force × velocity time taken
P=
E = Fv t
W
Potential energy = weight × change in height 1 kinetic energy = × mass × (speed)2 2
Ep = mgh 1 Ek = mv2 2
J
kinetic energy of rotation 1 = × moment of inertia × (angular velocity)2 2
Ek =
1 2 Iω 2
J
Mechanics of Solids, Second Edition © 2016 C. Ross, J. Bird and A. Little. Published by Taylor & Francis. All rights reserved.
J
Formula
Formula symbols
Units
Frictional force = coefficient of friction × normal force
F = μN
N
Angle of repose, θ, on an inclined plane
tan θ = μ
Efficiency of screw jack
η=
tan θ tan(λ + θ )
displacement acceleration
T = 2π
y a
s
mass stiffness
T= 2π
m k
s
simple pendulum
T = 2π
L g
s
compound pendulum
T = 2π
(k G2 + h 2 ) gh
s
SHM periodic time T = 2 π T = 2π
Force ratio =
load effort
Movement ratio = Efficiency =
distance moved by effort distance moved by load
force ratio movement ratio
Kelvin temperature = degrees Celsius + 273 Quantity of heat energy = mass × specific heat capacity × change in temperature New length = original length + expansion
Q = mc(t2 – t1)
J
L2 = L1 [1 + α (t2 – t1)]
m
New surface area = o riginal surface area + increase in area
A2 = A1 [1 + β (t2 – t1)]
m2
New volume = original volume + increase in volume
V2 = V1 [1 + γ (t2 – t1)]
m3
force area = density × gravitational acceleration × height
Pressure =
F A p = ρgh 1 bar = 105 Pa
p=
Absolute pressure = gauge pressure + atmospheric pressure
Mechanics of Solids, Second Edition © 2016 C. Ross, J. Bird and A. Little. Published by Taylor & Francis. All rights reserved.
Pa
Circular segment In Figure F1, shaded area =
R2 (α − sin α ) 2
Figure F1
Summary of standard results of the second moments of areas of regular sections Shape
Position of axis
Second moment of area, I
(1) Coinciding with B
BD 3 3 DB 3 3
Rectangle length D breadth B
(2) Coinciding with D (3) Through centroid, parallel to B
BD 3 12
(4) Through centroid, parallel to D
DB 3 12
Triangle Perpendicular height H base B
(1) Coinciding with B (2) Through centroid, parallel to base (3) Through vertex, parallel to base
BH 3 4
(1) Through centre perpendicular to plane (i.e. polar axis)
π D4 π R4 or 32 2
Circle radius R diameter D
(2) Coinciding with diameter
π D4 π R4 or 64 4
(3) About a tangent
5π D 4 5π R 4 or 64 4
Coinciding with diameter
π R4 8
Semicircle radius R
BH 3 12 BH 3 36
Mechanics of Solids, Second Edition © 2016 C. Ross, J. Bird and A. Little. Published by Taylor & Francis. All rights reserved.
Bending stresses in beams σ M E = = y I R
I
M σ = Z y
Z=
Beam deflections due to bending M = E I
d2 y dx 2
Torsion τ T Gθ = = r J l
) (
(
)
)
σθ =
1 1 σ + σ y + σ x − σ y cos 2θ + τ xy sin 2θ 2 x 2
τθ =
1 σ − σ y sin 2θ – τ xy cos 2θ 2 x tan 2θ =
(σ
2τ xy x
−σy
)
=
)
(
(σ
1 1 σx + σ y + 2 2
x
−σy
)
−σy
)
2
+ 4 τ xy 2
σ 2 = minimum principal stress =
)
(
1 1 σx + σ y − 2 2
(σ
y
− σx
(σ
tan 2θ =
1 σ − σx τ = ± 4 y
2 τ xy
(
x
2
)
)
τ =
σ1 =
σ2 =
τ = ±
16 M− πd 3 16 πd 3
)
ε2 =
1 1 ε + εy − 2 x 2
σx =
σy =
σx =
σy =
τ xy = G γ xy
E
(1 − ν 2 E
(1 − ν 2
(ε
x
− εy
)
− εy
)
(ε ε + vε ) )( ε + vε ) )(
x
x
y
y
x
2
2
+ γ xy 2 + γ xy 2
E 1 − ν ε + νε x y 1 + ν 1 − 2ν
(
)(
)
(
)
E 1 − ν ε + νε y x 1 + ν 1 − 2ν
(
σH =
)(
)
(
)
+ τ xy
2
( M 2 + T 2 ) ( M 2 + T 2 )
PR ηL t
σL =
PR 2 ηc t
σ =
RP 2η t
Energy methods + 4 τ xy 2
∂U e
∂P
=u
( )
UT =
∂U e ∂R
=λ
σ2 × volume of rod 2E
WD = d U b =
M 2 +T 2
(
2 16 M+ πd 3
)
1 1 ε + εy + 2 x 2
(σ1 − σ 2 )
(
)
ε1 =
Ue = 2
− εy
x
Membrane theory for thin-walled circular cylinders and spheres
σ 1 = maximum principal stress
(ε
γ xy
P=Tω
Complex stress and strain
(
tan 2θ =
M2 × dx 2E I
τ2 T 2 × dx Us = × volume 2G 2G × J
Theories of elastic failure
σ 1 = σ yp
σ 3 = σ ypc
) σ 3 − v (σ 1 + σ 2 ) = σ ypc (
σ 1 − v σ 2 + σ 3 = σ yp
Mechanics of Solids, Second Edition © 2016 C. Ross, J. Bird and A. Little. Published by Taylor & Francis. All rights reserved.
(σ12 + σ 22 + σ 32 ) − 2v (σ1σ 2 + σ1σ 3 + σ 2σ 3 ) = σ yp2
I xy = A h k
σ 1 – σ 3 = σ yp
tan 2θ =
(σ1 − σ 2 ) + (σ1 − σ 3 ) + (σ 2 − σ 3 ) 2
2
2
= 2 σ yp
2
Thick cylinders and spheres σr = A −
σ=
B
σθ = A +
r2
ρω 2 r 2 2 B B σ = A+ A+ − + 1 3 v 8 θ 3 r2 2r 3
(
)
The buckling of struts P= PR =
σ yc × A
l2
2
σc =
L a 0 +1 k
P Δ P Pe y + A P −P I e
(
Effective lengths of struts (L0) Type of strut
l
Euler
BS449
L0 = l
L0 = l
L0 = l
l
L0 = 0.5l L0 = 0.7l L0 = 2l
(
)
M cos α y I xx
+
(
) ( ) − 12 ( I
) ) sec 2θ
1 1 I + I y + I x − I y sec 2θ 2 x 2
IV
σ=
I tan θ β = tan −1 − UU IVV
)
+ Iy
x
x
− Iy
M cos θ v M sin θ u + IUU IVV
0.85l
)
M sin α x I yy
F bI
τ=
∫
y dA τ =
(
)
F B − Z y I
Composites
ε1 S11 ε 2 S12 ε 3 S13 = ε4 0 ε5 0 ε 6 0
S12
S12
0
0
S22
S23
0
0
S23
S33
0
0
0
0
S44
0
0
0
0
S55
0
0
0
0
0 0 0 0 0 S66
1/ E 0 −υ 21 / E2 1 1 / E2 0 S = −υ12 / E1 0 0 1 / G12
( )
0.7l
2l
Unsymmetrical bending of beams σ=
)
IU =
− Ix
Shear stresses in bending and shear deflections
π 2 EI
( 1 = (I 2
y
B r2
(I
2 I xy
1 1 = EE G S 1 − υ12 υ 21 1 2 12
1 0 0 R = 0 1 0 0 0 2
(
( )
)
( Q * = (T
(Q ) = ( S )
−1
Mechanics of Solids, Second Edition © 2016 C. Ross, J. Bird and A. Little. Published by Taylor & Francis. All rights reserved.
)
)−1 (Q ) ( R ) (T ) ( R )−1
N A B εo M = B D κ
σ1 σ2 σ3 σ4 σ5 σ 6
The Deviatoric strain energy (Tsai-Hill) failure criterion σ 12 F12
+
σ 22 F22
+
τ 122
−
F122
σ1 σ 2 F12
The Interactive tensor polynomial (Tsai-Wu) failure criterion f1σ 1 + f2σ 2 + f11σ 12 + f22σ 22 + f66 τ 122 + 2 f12σ 1σ 22 = 1
=1
The matrix displacement method
( ) k0
AE = l
Yi Mi Yj Mj
u10
v10
u20
C2
CS
−C 2
CS
S2
− CS
−C 2
−CS
C2
− CS
−S 2
CS
12 / l 3 2 = EI −6 / l −12 / l 3 −6 / l 2
−6 / l 2 4/l 6 / l2 2/l
v20
f1 =
1 1 − F1T F1C
f11 =
1 F1T F1C
f2 =
1 1 − F2 T F2 C
f22 =
1 F2 T F2 C
u10
− CS −S 2 CS S 2
−12 / l 3 6 / l2 12 / l 3 6 / l2
1
v10
u20
f66 =
v2 0 −6 / l 2 2/l 6 / l2 4/l
1 F122
f12
2 1 1 =− 2 F1T F1C F2 T F2 C
vi θi vj θj
The finite element method
(ko )
1 =∫ 4Δ 2
b1
0
b2
0
b3
0
0
c1
0
c2
0
c3
c1 c2 1 c3 1 E μ b1 0 b2 b3
μ 0 b1 b2 1 0 0 0 0 γ c c 1 2
b3
0
0
c1 c2
c3
b1 b2
Mechanics of Solids, Second Edition © 2016 C. Ross, J. Bird and A. Little. Published by Taylor & Francis. All rights reserved.
0
0 c3 t dA b3
Formula
Formula symbols
Units
Stress =
applied force cross-sectionalarea
σ=
F A
Strain =
change in length original length
ε=
x L
E=
σ ε
Pa
k=
F δ
N/m
G=
τ γ
Pa
Young’s modulus of elasticity = Stiffness =
stress strain
force extension
Modulus of rigidity =
shear stress shear strain
Thermal strain = coefficient of linear expansion × temperature rise
ε = αT
Thermal stress in compound bar
σ1 =
Ultimate tensile strength =
(α1 − α 2 ) E1 E 2 A2 T ( A1 E1 + A2 E 2 )
maximum load original cross-sectional area
Moment = force × perpendicular distance
Pa
Pa Pa
M = Fd
stress stress bendingbending momentmoment σ M E = = = = distancedistance from neutral of area of areay fromaxis neutralsecond axis moment second moment I R ,s modulus Young ,sYoung modulus = = radius ofradius curvature of curvature
Nm N/m3
Torque = force × perpendicular distance
T = Fd
Nm
Power = torque × angular velocity
P = Tω = 2πnT
W
Horsepower
1 hp = 745.7 W
Torque = moment of inertia × angular acceleration
T = Iα
Nm
shear stress torque = radius polar second moment of area (rigidity)(angle of twist) = length
τ T Gθ = = r J L
N/m3
Average velocity = Acceleration =
distance travelled time taken
change in velocity time taken
v=
s t
m/s
a=
v−u t
m/s2
Mechanics of Solids, Second Edition © 2016 C. Ross, J. Bird and A. Little. Published by Taylor & Francis. All rights reserved.
Formula
Formula symbols
Units
Linear velocity
v = ω r
m/s
Angular velocity
ω=
Linear acceleration
a = rα
θ = 2πn t
rad/s m/s2 m/s
Relationships between initial velocity u, final velocity v, displacement s, time t and constant acceleration a
v 2 = v1 + at 1 s = ut + at 2 2 2 2 v = u + 2as
rad/s
Relationships between initial angular velocity ω1, final angular velocity ω2, angle θ, time t and angular acceleration α
ω 2 = ω1 + αt 1 2 θ = ω 1t + α t 2 ω 12 = ω 22 + 2αθ
m (m/s)2
rad (rad/s)2
Momentum = mass × velocity
kg m/s
Impulse = applied force × time = change in momentum
kg m/s
Force = mass × acceleration
F = ma
N
Weight = mass × gravitational field
W = mg
N
Centripetal acceleration
a=
v2 r
m/s2
Centripetal force
F=
mv 2 r
N
mass Density = volume
ρ=
m V
kg/m3
Work done = force × distance moved
W = Fs
Efficiency = Power =
J
useful output energy input energy
energy used (or work done) = force × velocity time taken
P=
E = Fv t
W
Potential energy = weight × change in height 1 kinetic energy = × mass × (speed)2 2
Ep = mgh 1 Ek = mv2 2
J
kinetic energy of rotation 1 = × moment of inertia × (angular velocity)2 2
Ek =
1 2 Iω 2
J
Mechanics of Solids, Second Edition © 2016 C. Ross, J. Bird and A. Little. Published by Taylor & Francis. All rights reserved.
J
Formula
Formula symbols
Units
Frictional force = coefficient of friction × normal force
F = μN
N
Angle of repose, θ, on an inclined plane
tan θ = μ
Efficiency of screw jack
η=
tan θ tan(λ + θ )
displacement acceleration
T = 2π
y a
s
mass stiffness
T= 2π
m k
s
simple pendulum
T = 2π
L g
s
compound pendulum
T = 2π
(k G2 + h 2 ) gh
s
SHM periodic time T = 2 π T = 2π
Force ratio =
load effort
Movement ratio = Efficiency =
distance moved by effort distance moved by load
force ratio movement ratio
Kelvin temperature = degrees Celsius + 273 Quantity of heat energy = mass × specific heat capacity × change in temperature New length = original length + expansion
Q = mc(t2 – t1)
J
L2 = L1 [1 + α (t2 – t1)]
m
New surface area = o riginal surface area + increase in area
A2 = A1 [1 + β (t2 – t1)]
m2
New volume = original volume + increase in volume
V2 = V1 [1 + γ (t2 – t1)]
m3
force area = density × gravitational acceleration × height
Pressure =
F A p = ρgh 1 bar = 105 Pa
p=
Absolute pressure = gauge pressure + atmospheric pressure
Mechanics of Solids, Second Edition © 2016 C. Ross, J. Bird and A. Little. Published by Taylor & Francis. All rights reserved.
Pa
Circular segment In Figure F1, shaded area =
R2 (α − sin α ) 2
Figure F1
Summary of standard results of the second moments of areas of regular sections Shape
Position of axis
Second moment of area, I
(1) Coinciding with B
BD 3 3 DB 3 3
Rectangle length D breadth B
(2) Coinciding with D (3) Through centroid, parallel to B
BD 3 12
(4) Through centroid, parallel to D
DB 3 12
Triangle Perpendicular height H base B
(1) Coinciding with B (2) Through centroid, parallel to base (3) Through vertex, parallel to base
BH 3 4
(1) Through centre perpendicular to plane (i.e. polar axis)
π D4 π R4 or 32 2
Circle radius R diameter D
(2) Coinciding with diameter
π D4 π R4 or 64 4
(3) About a tangent
5π D 4 5π R 4 or 64 4
Coinciding with diameter
π R4 8
Semicircle radius R
BH 3 12 BH 3 36
Mechanics of Solids, Second Edition © 2016 C. Ross, J. Bird and A. Little. Published by Taylor & Francis. All rights reserved.
Bending stresses in beams σ M E = = y I R
I
M σ = Z y
Z=
Beam deflections due to bending M = E I
d2 y dx 2
Torsion τ T Gθ = = r J l
) (
(
)
)
σθ =
1 1 σ + σ y + σ x − σ y cos 2θ + τ xy sin 2θ 2 x 2
τθ =
1 σ − σ y sin 2θ – τ xy cos 2θ 2 x tan 2θ =
(σ
2τ xy x
−σy
)
=
)
(
(σ
1 1 σx + σ y + 2 2
x
−σy
)
−σy
)
2
+ 4 τ xy 2
σ 2 = minimum principal stress =
)
(
1 1 σx + σ y − 2 2
(σ
y
− σx
(σ
tan 2θ =
1 σ − σx τ = ± 4 y
2 τ xy
(
x
2
)
)
τ =
σ1 =
σ2 =
τ = ±
16 M− πd 3 16 πd 3
)
ε2 =
1 1 ε + εy − 2 x 2
σx =
σy =
σx =
σy =
τ xy = G γ xy
E
(1 − ν 2 E
(1 − ν 2
(ε
x
− εy
)
− εy
)
(ε ε + vε ) )( ε + vε ) )(
x
x
y
y
x
2
2
+ γ xy 2 + γ xy 2
E 1 − ν ε + νε x y 1 + ν 1 − 2ν
(
)(
)
(
)
E 1 − ν ε + νε y x 1 + ν 1 − 2ν
(
σH =
)(
)
(
)
+ τ xy
2
( M 2 + T 2 ) ( M 2 + T 2 )
PR ηL t
σL =
PR 2 ηc t
σ =
RP 2η t
Energy methods + 4 τ xy 2
∂U e
∂P
=u
( )
UT =
∂U e ∂R
=λ
σ2 × volume of rod 2E
WD = d U b =
M 2 +T 2
(
2 16 M+ πd 3
)
1 1 ε + εy + 2 x 2
(σ1 − σ 2 )
(
)
ε1 =
Ue = 2
− εy
x
Membrane theory for thin-walled circular cylinders and spheres
σ 1 = maximum principal stress
(ε
γ xy
P=Tω
Complex stress and strain
(
tan 2θ =
M2 × dx 2E I
τ2 T 2 × dx Us = × volume 2G 2G × J
Theories of elastic failure
σ 1 = σ yp
σ 3 = σ ypc
) σ 3 − v (σ 1 + σ 2 ) = σ ypc (
σ 1 − v σ 2 + σ 3 = σ yp
Mechanics of Solids, Second Edition © 2016 C. Ross, J. Bird and A. Little. Published by Taylor & Francis. All rights reserved.
(σ12 + σ 22 + σ 32 ) − 2v (σ1σ 2 + σ1σ 3 + σ 2σ 3 ) = σ yp2
I xy = A h k
σ 1 – σ 3 = σ yp
tan 2θ =
(σ1 − σ 2 ) + (σ1 − σ 3 ) + (σ 2 − σ 3 ) 2
2
2
= 2 σ yp
2
Thick cylinders and spheres σr = A −
σ=
B
σθ = A +
r2
ρω 2 r 2 2 B B σ = A+ A+ − + 1 3 v 8 θ 3 r2 2r 3
(
)
The buckling of struts P= PR =
σ yc × A
l2
2
σc =
L a 0 +1 k
P Δ P Pe y + A P −P I e
(
Effective lengths of struts (L0) Type of strut
l
Euler
BS449
L0 = l
L0 = l
L0 = l
l
L0 = 0.5l L0 = 0.7l L0 = 2l
(
)
M cos α y I xx
+
(
) ( ) − 12 ( I
) ) sec 2θ
1 1 I + I y + I x − I y sec 2θ 2 x 2
IV
σ=
I tan θ β = tan −1 − UU IVV
)
+ Iy
x
x
− Iy
M cos θ v M sin θ u + IUU IVV
0.85l
)
M sin α x I yy
F bI
τ=
∫
y dA τ =
(
)
F B − Z y I
Composites
ε1 S11 ε 2 S12 ε 3 S13 = ε4 0 ε5 0 ε 6 0
S12
S12
0
0
S22
S23
0
0
S23
S33
0
0
0
0
S44
0
0
0
0
S55
0
0
0
0
0 0 0 0 0 S66
1/ E 0 −υ 21 / E2 1 1 / E2 0 S = −υ12 / E1 0 0 1 / G12
( )
0.7l
2l
Unsymmetrical bending of beams σ=
)
IU =
− Ix
Shear stresses in bending and shear deflections
π 2 EI
( 1 = (I 2
y
B r2
(I
2 I xy
1 1 = EE G S 1 − υ12 υ 21 1 2 12
1 0 0 R = 0 1 0 0 0 2
(
( )
)
( Q * = (T
(Q ) = ( S )
−1
Mechanics of Solids, Second Edition © 2016 C. Ross, J. Bird and A. Little. Published by Taylor & Francis. All rights reserved.
)
)−1 (Q ) ( R ) (T ) ( R )−1
N A B εo M = B D κ
σ1 σ2 σ3 σ4 σ5 σ 6
The Deviatoric strain energy (Tsai-Hill) failure criterion σ 12 F12
+
σ 22 F22
+
τ 122
−
F122
σ1 σ 2 F12
The Interactive tensor polynomial (Tsai-Wu) failure criterion f1σ 1 + f2σ 2 + f11σ 12 + f22σ 22 + f66 τ 122 + 2 f12σ 1σ 22 = 1
=1
The matrix displacement method
( ) k0
AE = l
Yi Mi Yj Mj
u10
v10
u20
C2
CS
−C 2
CS
S2
− CS
−C 2
−CS
C2
− CS
−S 2
CS
12 / l 3 2 = EI −6 / l −12 / l 3 −6 / l 2
−6 / l 2 4/l 6 / l2 2/l
v20
f1 =
1 1 − F1T F1C
f11 =
1 F1T F1C
f2 =
1 1 − F2 T F2 C
f22 =
1 F2 T F2 C
u10
− CS −S 2 CS S 2
−12 / l 3 6 / l2 12 / l 3 6 / l2
1
v10
u20
f66 =
v2 0 −6 / l 2 2/l 6 / l2 4/l
1 F122
f12
2 1 1 =− 2 F1T F1C F2 T F2 C
vi θi vj θj
The finite element method
(ko )
1 =∫ 4Δ 2
b1
0
b2
0
b3
0
0
c1
0
c2
0
c3
c1 c2 1 c3 1 E μ b1 0 b2 b3
μ 0 b1 b2 1 0 0 0 0 γ c c 1 2
b3
0
0
c1 c2
c3
b1 b2
Mechanics of Solids, Second Edition © 2016 C. Ross, J. Bird and A. Little. Published by Taylor & Francis. All rights reserved.
0
0 c3 t dA b3
