14
CE 2210 Quiz Name: _______________________
CE 2210 Quiz Name: _______________________
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
828
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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= − − + −
829
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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= =
830
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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
BMappendixE.indd Page 831 13/10/16 9:15 PM f-389 /208/WB01929/9781119227489/bmmatter/App_E/text_s
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=
−−
+
BMappendixC.indd Page 822 12/10/16 3:44 PM f-389 /208/WB01929/9781119227489/bmmatter/App_C/text_s
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
=−
−+
−
BMappendixC.indd Page 823 12/10/16 3:44 PM f-389 /208/WB01929/9781119227489/bmmatter/App_C/text_s
