Fundamental Mechanics of Materials Equations
Transcript of Fundamental Mechanics of Materials Equations
1. Rectangle 6. Circle
2. Right Triangle 7. Hollow Circle
3. Triangle 8. Parabola
4. Trapezoid 9. Parabolic Spandrel
5. Semicircle 10. General Spandrel
A bh=
y
h
2=
I
bh
12x
3
=
x
b
2=
I
hb
12y
3
=
Ibh
3x
3
=′
Ihb
3y =′
3
A
bh
2=
y
h
3=
I
bh
36x
3
=
x
b
3=
I
hb
36y
3
=
I
bh
12x
3
=′
Ihb
12y
3
=′
A
bh
2=
y
h
3=
I
bh
36x
3
=
x
a b( )
3=
+
I
bha ab b
36( )y
2 2= − +
I
bh
12x
3
=′
Aa b h( )
2=
+
y
a b
a bh
1
3
2=
++
Ih
a ba ab b
36( )( 4 )x
32 2=
++ +
C
b
h x
x′
x
yy′
y–
–
b
h
Cx
x′
x−yy′
y−
b
h
Cx
x
y
y′
y−
−
a
x′
b
h
Cx
y
a
−
Cx
x′
y, y′
yr −
d
Cx
y
r
d
D
C
rx
y
R
Zero slope
h Cx
bx′
x
yy′
y−
−
b
h
yCx
x′
x
yy′
Zeroslope
−
−
b
h
yCx
x′
x
yy′
Zeroslope
−
−
Table A.1 Properties of Plane Figures
A
r
2
2π=
y
r4
3π=
I r
8
8
9x
4ππ
= −
I Ir
8x y
4π= =′ ′
A r
d
42
2
π π= =
I Ir d
4 64x y
4 4π π= = =
A R r D d( )4
( )2 2 2 2π π= − = −
I I R r
4( )x y
4 4π= = −
D d
64( )4 4π
= −
yh
bx
22′ = ′
A
bh2
3=
x
b3
8=
y
h3
5=
y
h
bx
22′ = ′
Abh
3=
x
b3
4=
y
h3
10=
yh
bx
nn′ = ′
Abh
n 1=
+
xn
nb
1
2=
++
yn
nh
1
4 2=
++
791
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Common Greek letters
Alpha Mu
Beta Nu
Gamma Pi
, Delta Rho
Epsilon , Sigma
Theta TauKappa Phi
Lambda Omega
α µβ νγ π
δ ρε σθ τκ φλ ω
∆Σ
Basic definitionsAverage normal stress in an axial member
F
Aavgσ =
Average direct shear stressV
AVavgτ =
Average bearing stressF
Ab
b
σ =
Average normal strain in an axial member
L
L Ld
d
t
t
h
hor or
long
lat
ε δ
ε
=∆
=
=∆ ∆ ∆
Average normal strain caused by temperature change
TTε α= ∆
Average shear strain
change in angle from2
radγ π=
Hooke’s law (one-dimensional)
Eσ ε= and Gτ γ=
Poisson’s ratiolat
long
ν εε
= −
Relationship between E, G, and ν
GE
2(1 )ν=
+
Definition of allowable stress
FSor
FSallow
failureallow
failureσ σ τ τ= =
Factor of safety
FS or FSfailure
actual
failure
actual
σσ
ττ
= =
Axial deformationDeformation in axial members
FL
AEδ = or
F L
A Ei i
i ii∑δ =
Force-temperature-deformation relationship
FL
AET Lδ α= + ∆
TorsionMaximum torsion shear stress in a circular shaft
Tc
Jmaxτ =
where the polar moment of inertia J is defined as:
π π [ ][ ]= − = −J R r D d2 32
4 4 4 4
Angle of twist in a circular shaftTL
JGφ = or
T L
J Gi i
i ii∑φ =
Power transmission in a shaft
P Tω=Power units and conversion factors
1 W1 N m
s1 hp
550 lb ft
s
6,600 lb in.
s
1 Hz1 rev
s1 rev 2 rad
1 rpm2 rad
60 s
π
π
=⋅
=⋅
=⋅
= =
=
Fundamental Mechanics of Materials Equations
AppendixE
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Gear relationships between gears A and B
T
R
T
RR R R R
R
R
D
D
N
NGear ratio
A
A
B
BA A B B A A B B
A
B
A
B
A
B
φ φ ω ω= = − =
= = =
Six rules for constructing shear-force and bending-moment diagrams
Rule 1: V P0∆ =
Rule 2: V V V w x dx( )x
x
2 11
2∫∆ = − =
Rule 3: dV
dxw x( )=
Rule 4: M M M V dxx
x
2 11
2∫∆ = − =
Rule 5: dM
dxV=
Rule 6: M M0∆ = −
FlexureFlexural strain and stress
y1
xερ
= − E
yxσρ
= −
Flexure FormulaMy
Ix
z
σ = − or Mc
I
M
SS
I
cwheremaxσ = = =
Transformed-section method for beams of two materials [where material (2) is transformed into an equivalent amount of material (1)]
nE
E2
1
= My
In
My
Ix x1
transformed2
transformed
σ σ= − = −
Bending due to eccentric axial loadF
A
My
Ix
z
σ = −
Unsymmetric bending of arbitrary cross sections
I z I y
I I IM
I y I z
I I IMx
z yz
y z yzy
y yz
y z yzz2 2
σ =−−
+
− +−
or
M I M I y
I I I
M I M I z
I I IM I M I
M I M I
( ) ( )
tan
xz y y yz
y z yz
y z z yz
y z yz
y z z yz
z y y yz
2 2σ
β
= −+
−+
+−
=++
Unsymmetric bending of symmetric cross sections
M z
I
M y
Ix
y
y
z
z
σ = − M I
M Itan
y z
z y
β =
Bending of curved bars
∫σ
( )( )= −
−−
=M r r
r A r rr
AdA
r
wherexn
c nn
A
Horizontal shear stress associated with bending
τ = = ΣVQ
I tQ y AwhereH i i
Shear flow formula
qVQ
I=
Shear flow, fastener spacing, and fastener shear relationship
q s n V n Af f f f fτ≤ =
For circular cross sections,
Q r d2
3
1
123 3= = (solid sections)
[ ][ ]= − = −Q R r D d2
3
1
123 3 3 3 (hollow sections)
Beam deflectionsElastic curve relations between w, V, M, θ, and v for constant EI
vdv
dx
M EId v
dx
VdM
dxEI
d v
dx
wdV
dxEI
d v
dx
Deflection
Slope (for small deflections)
Moment
Shear
Load
2
2
3
3
4
4
θ
=
= =
=
= =
= =
Plane stress transformations
n
t
x
y
n
n
t
t
nt
tn
nt
tn
σ
σ σ
σθ
θ
ττ
ττ
Stresses on an arbitrary plane
cos sin 2 sin cosn x y xy2 2σ σ θ σ θ τ θ θ= + +
sin cos 2 sin cost x y xy2 2σ σ θ σ θ τ θ θ= + −
( )sin cos (cos sin )nt x y xy2 2τ σ σ θ θ τ θ θ= − − + −
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or
2 2cos 2 sin 2n
x y x yxyσ
σ σ σ σθ τ θ=
++
−+
2 2cos 2 sin 2t
x y x yxyσ
σ σ σ σθ τ θ=
+−
−−
2sin 2 cos 2nt
x yxyτ
σ σθ τ θ= −
−+
Principal stress magnitudes
2 2p p
x y x yxy1, 2
22σ
σ σ σ στ=
+±
−
+
Orientation of principal planes
tan 22
pxy
x y
θτ
σ σ=
−
Maximum in-plane shear stress magnitude
2or
2x y
xyp p
max
22
max1 2τ
σ στ τ
σ σ= ±
−
+ =−
2x y
avgσσ σ
=+
tan 22
note: 45sx y
xys pθ
σ στ
θ θ= −−
= ± °
Absolute maximum shear stress magnitude
2abs max
max minτ σ σ=
−
Normal stress invariance
x y n t p p1 2σ σ σ σ σ σ+ = + = +
Plane strain transformationsStrain in arbitrary directions
cos sin sin cosn x y xy2 2ε ε θ ε θ γ θ θ= + +
sin cos sin cost x y xy2 2ε ε θ ε θ γ θ θ= + −
2( )sin cos (cos sin )nt x y xy2 2γ ε ε θ θ γ θ θ= − − + −
or
2 2cos 2
2sin 2n
x y x y xyεε ε ε ε
θγ
θ=+
+−
+
2 2cos 2
2sin 2t
x y x y xyεε ε ε ε
θγ
θ=+
−−
−
( )sin 2 cos 2nt x y xyγ ε ε θ γ θ= − − +
Principal strain magnitudes
2 2 2p p
x y x y xy1, 2
2 2
εε ε ε ε γ
=+
±−
+
Orientation of principal strains
tan 2 pxy
x y
θγ
ε ε=
−
Maximum in-plane shear strain
22 2
orx y xy
p pmax
2 2
max 1 2γε ε γ
γ ε ε= ±−
+
= −
2x y
avgεε ε
=+
Normal strain invariance
x y n t p p1 2ε ε ε ε ε ε+ = + = +
Generalized Hooke’s lawNormal stress/normal strain relationships
E
E
E
1[ ( )]
1[ ( )]
1[ ( )]
x x y z
y y x z
z z x y
ε σ ν σ σ
ε σ ν σ σ
ε σ ν σ σ
= − +
= − +
= − +
E
E
E
(1 )(1 2 )[(1 ) ( )]
(1 )(1 2 )[(1 ) ( )]
(1 )(1 2 )[(1 ) ( )]
x x y z
y y x z
z z x y
σν ν
ν ε ν ε ε
σν ν
ν ε ν ε ε
σν ν
ν ε ν ε ε
=+ −
− + +
=+ −
− + +
=+ −
− + +
Shear stress/shear strain relationships
G G G
1;
1;
1xy xy yz yz zx zxγ τ γ τ γ τ= = =
where
GE
2(1 )ν=
+
Volumetric strain or Dilatation
eV
V E
1 2( )x y z x y zε ε ε ν σ σ σ=
∆= + + =
−+ +
Bulk modulus
KE
3(1 2 )ν=
−Normal stress/normal strain relationships for plane stress
E
E
E
1( )
1( )
( )
1( )
x x y
y y x
z x y
z x y
ε σ νσ
ε σ νσ
ε ν σ σ
ε νν
ε ε
= −
= −
= − +
= −−
+
or
E
E1
( )
1( )
x x y
y y x
2
2
σν
ε νε
σν
ε νε
=−
+
=−
+
Shear stress/shear strain relationships for plane stress
GG
1orxy xy xy xyγ τ τ γ= =
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Thin-walled pressure vesselsTangential stress and strain in spherical pressure vessel
pr
t
pd
t
pr
tE2 4 2(1 )t tσ ε ν= = = −
Longitudinal and circumferential stresses in cylindrical pressure vessels
pr
t
pd
t
pr
tE2 4 2(1 2 )long longσ ε ν= = = −
pr
t
pd
t
pr
tE2 2(2 )hoop hoopσ ε ν= = = −
Thick-walled pressure vesselsRadial stress in thick-walled cylinder
a p b p
b a
a b p p
b a r
( )
( )r
i o i o2 2
2 2
2 2
2 2 2σ =
−−
−−
−or
a p
b a
b
r
b p
b a
a
r1 1r
i o2
2 2
2
2
2
2 2
2
2σ =
−−
−−
−
Circumferential stress in thick-walled cylinder
a p b p
b a
a b p p
b a r
( )
( )i o i o
2 2
2 2
2 2
2 2 2σ =
−−
+−
−θ
ora p
b a
b
r
b p
b a
a
r1 1i o
2
2 2
2
2
2
2 2
2
2σ =
−+
−−
+
θ
Maximum shear stress
a b p p
b a r
1
2( )
( )
( )r
i omax
2 2
2 2 2τ σ σ= − =
−−θ
Longitudinal normal stress in closed cylindera p b p
b ai o
long
2 2
2 2σ =
−−
Radial displacement for internal pressure only
δ ν ν[ ]=−
− + +a p
b a rEr b
( )(1 ) (1 )r
i2
2 22 2
Radial displacement for external pressure only
δ ν ν[ ]= −−
− + +b p
b a rEr a
( )(1 ) (1 )r
o2
2 22 2
Radial displacement for external pressure on solid cylinder
p r
E
(1 )r
oδ ν= −
−
Contact pressure for interference fit connection of thick cylinder onto a thick cylinder
δ ( )( )( )=
− −−
pE c b b a
b c a2c
2 2 2 2
3 2 2
Contact pressure for interference fit connection of thick cylinder onto a solid cylinder
δ ( )=
−p
E c b
bc2c
2 2
2
Failure theoriesMises equivalent stress for plane stress
σ σ σ σ σ σ σ σ σ τ= − + = − + + 3M p p p p x x y y xy12
1 2 22 1/2 2 2 2 1/2
Column bucklingEuler buckling load
PEI
KL( )cr
2
2
π=
Euler buckling stress
E
KL r( / )cr
2
2σ π
=
Radius of gyration
rI
A2 =
Secant formula
P
A
ec
r
KL
r
P
EA1 sec
2max 2
σ = +
831
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822
Sim
ply
Su
pp
ort
ed B
eam
s Bea
mSl
ope
Defl
ecti
onE
last
ic C
urve
1
x
vP
v max
1θ2θ
L 2—L 2—
PL E
I16
12
2
θθ
=−
=−
vP
L EI
48m
ax
3
=−
vP
x EI
Lx
48(3
4)
22
=−
−
xL
for
02
≤≤
2
x
v
La
2θ1θ
P
b
Pb
Lb
LEI
()
61
22
θ=
−−
Pa
La
LEI
()
62
22
θ=
+−
vP
ab
LEI
3
22
=−
xa
at=
vP
bx LEI
Lb
x6
()
22
2=
−−
−
xa
for
0≤
≤
3
x
v
L
1θ2θ
M
ML EI
31θ
=−
ML EI
62θ
=+
vM
L EI
93
max
2
=−
xL
at1
3 3=
−
vM
x
LEI
LLx
x6
(23
)2
2=
−−
+
4
x
v
v max
1θ2θ
w
L 2—L 2—
wL E
I24
12
3
θθ
=−
=−
vw
L EI
5 384
max
4
=−
vw
x EI
LLx
x24
(2
)3
23
=−
−+
5
x
v
La
1θ2θ
ww
a LEI
La
24(2
)1
22
θ=
−−
wa LE
IL
a24
(2)
2
22
2θ
=+
−
vw
a LEI
LaL
a24
(47
3)
32
2=
−−
+
xa
at=
vw
x LEI
LxaL
xa
x
aL
aL
a24
(4
2
44
)
32
22
22
34
=−
−+
+−
+
xa
for
0≤
≤
vw
a LEI
xLx
ax L
xa
L24
(26
4)
23
22
22
=−
−+ +
−
ax
Lfo
r≤
≤
6
x
v
L
2θ1θ
w0
wL EI
7 360
10
3
θ=
−
wL EI
452
03
θ=
+
vw
L
EI
0.00
652
max
04
=−
xL
at0.
5193
=
vw
x LEI
LL
xx
360
(710
3)
04
22
4=
−−
+
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823
Can
tile
ver
Bea
ms
Bea
mSl
ope
Defl
ecti
onE
last
ic C
urve
7
L
v max
max
θ
x
vP
PL E
I2
max
2
θ=
−v
PL EI
3m
ax
3
=−
vP
x EI
Lx
6(3
)2
=−
−
8x
v
v max
max
θ
P
L 2—L 2—
PL E
I8
max
2
θ=
−v
PL E
I
5 48m
ax
3
=−
vP
x EI
Lx
xL
12(3
2)
for
02
2
=−
−≤
≤
vP
L EI
xL
Lx
L48
(6)
for
2
2
=−
−≤
≤
9
L
v max
max
θ
x
v
MM
L
EI
max
θ=
−v
ML EI
2m
ax
2
=−
vM
x EI
2
2
=−
10
L
v max
max
θ
x
vw
wL EI
6m
ax
3
θ=
−v
wL EI
8m
ax
4
=−
vw
x EI
LLx
x24
(64
)2
22
=−
−+
11
L
v max
max
θ
x
vw
0
wL E
I24
max
03
θ=
−v
wL E
I30
max
04
=−
vw
x LEI
LL
xLx
x12
0(1
010
5)
02
32
23
=−
−+
−
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