CE-204 Final Exam Formula Sheet
Prof. D. Jansen
CE 204 Final Exam Formula Sheet
Centroids and Moments of Inertia
Q x = ∫ y d A = yA
Q y = ∫ x d A = xA
J =∫r2 dA
J = Ix + Iy
y
h
I y = ∫ x2 d A
I = I + Ad 2
1 3
bh
12
1 3
I y' =
b h
12
x′
1
I x = bh 3
3
x
1
I y = b3h
3
1
bh(b 2 + h 2 )
JC =
12
y′
Ix = ∫ y2 dA
I x' =
C
b
h
C
h/3
y
C
4r/3π
O
x′
x
b
y
r
1
bh 3
36
1 3
Ix =
bh
12
I x' =
x′
Ix = Iy =
r
O
x
1
π r4
4
JO =
1
π r4
2
τ =
Tr
J
x
8 4
1
1 4
π
4
I x' = −
r I x = I y = π r J O = π r
8
4
8 9π
1-D Stress Components
σ =
τ max =
P
A
σ =−
3V
rectangle
2A
τ max =
My
I
τ =
4V
solid circle
3A
τ max =
VQ
It
2V
thin tube
A
q=
∆H VQ
=
∆x
I
2
3V
y
Shear stress distribution in a rectangle: τ =
1 −
2 A
c
where y is measured from the neutral axis and c is the half-height of the beam (c = h/2)
1 of 2
CE-204 Final Exam Formula Sheet
Prof. D. Jansen
1-D Deformations
Axial Deformation: δ =
PL
AE
Torsional Deformation: φ =
L
P( x)
dx
AE
0
δ =∫
TL
JG
L
T ( x)
dx
JG
0
φ=∫
Power in a circular shaft: P = 2π f T
Thermal Strain: ε T = α (∆T )
Unrestrained Change in Length of Rod due to Temperature Change: δ T = α (∆T )L
α is the coefficient of thermal expansion and ∆T is the change in temperature from a
reference temperature
Relationships Between Stress and Strain
1
ν
ν
σx − σy − σz
E
E
E
ν
ν
1
εy = − σx + σy − σz
E
E
E
ν
ν
1
εz = − σx − σ y + σz
E
E
E
εx =
ν =−
εy
ε
=− z
εx
εx
γ yz =
γ xz =
τ xy
G
τ yz
G
τ xz
G
σ x = (λ + 2G )ε x + λε y + λε z
σ y = λε x + (λ + 2G )ε y + λε z
σ z = λε x + λε y + (λ + 2G )ε z
Plane stress (σz = τyz = τzx= 0) :
(Defined for uniaxial
stress in the x-direction)
G=µ=
λ=
γ xy =
νE
E
εx +
εy
2
1 −ν
1 −ν 2
νE
E
σy =
εx +
εy
2
1 −ν
1 −ν 2
σx =
E
2(1 + ν )
νE
εz = −
(1 + ν )(1 − 2ν )
2 of 2
ν
1 −ν
(ε + ε )
x
y