TAM 251 E QUATION S HEET Main Equations Stress σ avg = F A τ avg = V A Strain eng = δ L 0 true = ln L f L 0 γ = δ x L y + δ y L x Constitutive Relations σ = E τ = Gγ Material Properties (Isotropic) ν = - lat long G = E 2(1 + ν ) Thermal Expansion th = α ΔT δ th = αL 0 ΔT Axial Loading δ = FL 0 EA σ = F A Torsion φ = TL 0 GJ τ = Tρ J γ = φρ L 0 Stiffness and Flexibility k axial = EA L 0 = 1 f axial k torsion = GJ L 0 = 1 f torsion Bending σ = Mc I Thin-Walled Pressure Vessels σ h = pr t σ a = pr 2 t Transverse Shear τ = VQ It q = VQ I Miscellaneous Distributed Loads, Shear, & Bending Moments dV dx = -w dM dx = V Inclined Plane: Normal Stress σ n = σ x cos 2 θ +2 τ xy sin θ cos θ + σ y sin 2 θ Inclined Plane: Shear Stress τ n,s =(σ y - σ x ) sin θ cos θ + τ xy ( cos 2 θ - sin 2 θ ) Tresca Criterion |σ 1 | = σ yield , |σ 2 | = σ yield when σ 1 ,σ 2 have the same sign |σ 1 - σ 2 | = σ yield when σ 1 ,σ 2 have the opposite sign Von-Mises Criterion σ 2 1 - σ 1 σ 2 + σ 2 2 = σ 2 yield TAM 251 Equation Sheet Page 1 Apr. 2019 TAM 251 Equation Sheet Page 1 Apr. 2019 TAM 251 Equation Sheet Page 1 Apr. 2019
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TAM 251 EQUATION SHEET
Main Equations
Stress σavg =F
Aτavg =
V
A
Strain εeng =δ
L0εtrue = ln
Lf
L0γ =
δxLy
+δyLx
Constitutive Relations σ = Eε τ = Gγ
Material Properties (Isotropic) ν = − εlatεlong
G =E
2(1 + ν)
Thermal Expansion εth = α∆T δth = αL0∆T
Axial Loading δ =F L0
EAσ =
F
A
Torsion φ =T L0
GJτ =
T ρ
Jγ =
φρ
L0
Stiffness and Flexibility kaxial =EA
L0=
1
faxialktorsion =
GJ
L0=
1
ftorsion
Bending σ =Mc
I
Thin-Walled Pressure Vessels σh =pr
tσa =
pr
2t
Transverse Shear τ =V Q
I tq =
V Q
I
Miscellaneous
Distributed Loads, Shear,& Bending Moments
dV
dx= −w dM
dx= V
Inclined Plane: Normal Stress σn = σx cos2 θ + 2τxy sin θ cos θ + σy sin2 θ
Inclined Plane: Shear Stress τn,s = (σy − σx) sin θ cos θ + τxy(cos2 θ − sin2 θ
)
Tresca Criterion
|σ1| = σyield, |σ2| = σyieldwhen σ1, σ2 have thesame sign
|σ1 − σ2| = σyieldwhen σ1, σ2 have theopposite sign
Von-Mises Criterion σ21 − σ1σ2 + σ2
2 = σ2yield
TAM 251 Equation Sheet Page 1 Apr. 2019TAM 251 Equation Sheet Page 1 Apr. 2019TAM 251 Equation Sheet Page 1 Apr. 2019
Stress Transformations
Plane Stress
σx′ =σx + σy
2+σx − σy
2cos(2θ) + τxy sin(2θ)
σy′ =σx + σy
2− σx − σy
2cos(2θ) − τxy sin(2θ)
τx′y′ = −σx − σy2
sin(2θ) + τxy cos(2θ)
Mohr’s Circle σavg =σx + σy
2R =
√(σx − σy
2
)2
+ τ 2xy
Principal Stresses σ1 = σavg + R σ2 = σavg − R τmax = R
Plane Orientations tan 2θp =2τxy
σx − σytan 2θs = −σx − σy
2τxy
Moments and Geometric Centroids
Q = yA Ix =
∫A
y2 dA Jo =
∫A
ρ2dA y =1
A
∫A
ydA
Rectangle x
y
b
h Ix =1
12bh3
Circle
ry
x Ix =π
4r4 Jz =
π
2r4
Semicircle x
y
x`
ryIx =
π
8r4 Ix′ =
(π
8− 8
9π
)r4 y =
4r
3π
Parallel Axis Theorem Ic = Ic′ + Ad2cc′
Buckling
Critical Load Pcr =π2EI
(Le)2
pinned-pinned Le = L
pinned-fixed Le = 0.7L
fixed-fixed Le = 0.5L
fixed-free Le = 2L
TAM 251 Equation Sheet Page 2 Apr. 2019TAM 251 Equation Sheet Page 2 Apr. 2019TAM 251 Equation Sheet Page 2 Apr. 2019
Beam Deflection
DiagramMax. Deflection Slope at End Elastic Curve
ymax θ y(x)
y P
xymax
L
−P L3
3EI−P L
2
2EIy(x) =
P
6EI
(x3 − 3Lx2
)
y
x
w
ymaxL
−wL4
8EI−wL
3
6EIy(x) = − w
24EI
(x4 − 4Lx3 + 6L2x2
)
y
ymax
x
L
M
−ML2
2EI−ML
EIy(x) = − M
2EIx2
yw
ymax
x
L
− 5wL4
384EI± wL3
24EIy(x) = − w
24EI
(x4 − 2Lx3 + L3x
)
yL2
PL2
ymax
x − P L3
48EI± P L2
16EI
For 0 ≤ x ≤ L2
y(x) =P
48EI
(4x3 − 3L2x
)
by
x
P
ymax
a
A B
Lxm
For a > b
−P b(L2 − b2
)3/29√
3EIL
at xm =
√L2 − b2
3
θA = −P b(L2 − b2
)6EIL
θB = +P a
(L2 − a2
)6EIL
For x < a :
y(x) =P b
6EIL
[x3 − x
(L2 − b2
)]For x = a :
y = −P a2 b2
3EIL
yM
ymax
A B
Lxm
x
− ML2
9√
3EI
at xm =L√3
θA = −ML
6EI
θB = +ML
3EI
y(x) =M
6EIL
(x3 − L2x
)
TAM 251 Equation Sheet Page 3 Apr. 2019TAM 251 Equation Sheet Page 3 Apr. 2019TAM 251 Equation Sheet Page 3 Apr. 2019
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