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![Page 1: FE Review Mechanics of Materials - Auburn · PDF fileFE Review Mechanics of Materials. FE Mechanics of Materials Review Stress N V M F N = internal normal force (or P) V = internal](https://reader030.vdocuments.site/reader030/viewer/2022021508/5a72da047f8b9aa7538e0688/html5/thumbnails/1.jpg)
FE ReviewMechanics of Materials
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FE Mechanics of Materials Review
Stress
N
VM
F
N = internal normal force (or P)V = internal shear forceM = internal moment Double Shear
N PA A
σ = = F/2
F/2
F
Normal Stress = F
VA
τ =Average Shear Stress =
V = F/2
2F
Aτ =
V = F/2
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FE Mechanics of Materials Review
Strain
0
0 0 0
L L LL L L
δε ∆ −= = =Normal Strain Units of length/length
ε = normal strain∆L = change in length = δL0 = original lengthL = length after deformation (after axial load is applied)
Percent Elongation = 0
100LL∆
×
Percent Reduction in Area = 100i f
i
A AA−
×
Ai = initial cross- sectional areaAf = final cross-sectional area
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FE Mechanics of Materials Review
Strain
Shear Strain = change in angle , usually expressed in radians
θ
xyγ
y
B B'
x
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FE Mechanics of Materials Review
Stress-Strain Diagram for Normal Stress-Strain
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FE Mechanics of Materials Review
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FE Mechanics of Materials Review
Hooke's Law (one-dimension)
Eσ ε=σ = normal stress, force/length^2E = modulus of elasticity, force/length^2
= normal strain, length/length
Gτ γ=ε
= shear stress, force/length^2G = shear modulus of rigidity, force/length^2
= shear strain, radiansγτ
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FE Mechanics of Materials Review
2(1 )EGν
=+
ν = Poisson's ratio = -(lateral strain)/(longitudinal strain)
lat
long
ενε
= −
'lat r
δε = change in radius over original radius
long Lδε = change in length over original length
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FE Mechanics of Materials Review
Axial Load
If A (cross-sectional area), E (modulus of elasticity), and P (load) are constant in a member (and L is its length):
PLAE
δ =
If A, E, or P change from one region to the next:
P AEL
σε δ
= = ⇒ Change in length
PLAE
δ = ∑ Apply to each section where A, E, & P are constant
/A Bδ = displacement of pt A relative to pt B
Aδ = displacement of pt A relative to fixed end
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FE Mechanics of Materials Review-Remember principle of superposition used for indeterminate structures
- equilibrium/compatibility
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FE Mechanics of Materials Review
Thermal Deformations
0( ) ( )t T L T T Lδ α α= ∆ = −
tδ = change in length due to temperature change, units of length
α = coefficient of thermal expansion, units of 1/°
T = final temperature, degrees
0T = initial temperature, degrees
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FE Mechanics of Materials Review
Torsion
Torque – a moment that tends to twist a member about its longitudinal axis
Shear stress, , and shear strain, , vary linearly from 0 at center to maximum at outside of shaft
τ γ
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FE Mechanics of Materials Review
r
T
TrJ
τ = τ = shear stress, force/length^2
T = applied torque, force·length
r = distance from center to point of interest in cross-section(maximum is the total radius dimension)
J = polar moment of inertia (see table at end of STATICSsection in FE review manual), length^4
TLJGφ = φ = angle of twist, radians
L = length of shaft
G = shear modulus of rigidity, force/length^2
( / )z zG Gr d dzφ φτ γ φ= =( / )d dzφ = twist per unit length, or rate of twist
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FE Mechanics of Materials Review
Bending
Positive Bending
Makes compression in top fibers and tension in bottom fibers
Negative Bending
Makes tension in top fibers and compression in bottom fibers
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FE Mechanics of Materials Review
Slope of shear diagram = negative of distributed loading value ( )dV q xdx
− =
Slope of moment diagram = shear value dM Vdx
=
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FE Mechanics of Materials Review
Change in shear between two points = neg. of area under distributed loading diagram between those two points
2
1
2 1 [ ( )]x
xV V q x dx− = −∫
2
1
2 1 [ ( )]x
xM M V x dx− = ∫
Change in moment between two points = area under shear diagram between those two points
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FE Mechanics of Materials Review
Stresses in Beams
MyI
σ = − = normal stress due to bending moment, force/length^2σy = distance from neutral axis to the longitudinal fiber in
question, length (y positive above NA, neg below)
I = moment of inertia of cross-section, length^4
maxMcI
σ = ± c = maximum value of y;distance from neutral axis toextreme fiber
xyε ρ= − ρ = radius of curvature of deflected
axis of the beam
yE Eσ ε ρ= = − 1 MEIρ
=From and
MyI
σ = −
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FE Mechanics of Materials Review
IS c= S = elastic section modulus of beam
Then maxMc MI S
σ = ± = ±
VQIt
τ =Transverse Shear Stress:
Transverse Shear Flow: VQqI
=
' 'Q y A=t = thickness of cross-section at point of interest
t = b here
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FE Mechanics of Materials Review
Thin-Walled Pressure Vessels (r/t >= 10)
Cylindrical Vessels
1tprt
σ σ= =
= hoop stress in circumferential direction= gage pressure, force/length^2= inner radius= wall thicknesst
rp1σ
22aprt
σ σ= = = axial stress in longitudinal direction
See FE review manual for thick-walled pressurevessel formulas.
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FE Mechanics of Materials Review2-D State of Stress
Stress Transformation
θτθσσσσ
σ 2sin2cos22' xy
yxyxx +
−+
+=
θτθσσ
τ 2cos2sin2'' xy
yxyx +
−−=
θτθσσσσ
σ 2sin2cos22' xy
yxyxy −
−−
+=
Principal Stresses
( )22
2,1 22 xyyxyx τ
σσσσσ +⎟⎟
⎠
⎞⎜⎜⎝
⎛ −±
+=
tan 2
2
xyp
x y
τθ
σ σ=
−⎛ ⎞⎜ ⎟⎝ ⎠
No shear stress actson principal planes!
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FE Mechanics of Materials Review
Maximum In-plane Shear Stress
2yx
avg
σσσ
+=( )2
2max
2 xyyx
planein τσσ
τ +⎟⎟⎠
⎞⎜⎜⎝
⎛ −=−
xyyx
s τσσ
θ /2
2tan ⎟⎟⎠
⎞⎜⎜⎝
⎛ −−=
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FE Mechanics of Materials Review
Mohr's Circle – Stress, 2D
σ, positive to the righttau, positive downward!
Center: Point C( ,0)2yx
avg
σσσ
+=
( ) ( )22xyavgxR τσσ +−=
A rotation of θ to the x’ axis on the element will correspond to a rotation of 2θ on Mohr’s circle!
1 avg aσ Rσ σ= + =
2 avg bσ Rσ σ= − =
Rplanein =−maxτ
τ+
τ−
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FE Mechanics of Materials ReviewBeam Deflections
+- Fig. 12-2
Inflection point is where the elastic curve has zero curvature = zero moment
1yε
ρ= −
MyandE Iσε σ −
= = ⇒AlsoEIM
=ρ1 ρ = radius of curvature
of deflected axis of the beam
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FE Mechanics of Materials Review
2
21 M d y
EI dxρ= = ⇒
2
2( ) d yM x EIdx
=
from calculus, for very small curvatures
( )dM xVdx
⎛ ⎞= ⇒⎜ ⎟⎝ ⎠
3
3( ) d yV x EIdx
= for EI constant
( )( ) dV xw xdx
− = ⇒ for EI constant4
4( ) d vw x EI qdx
− = = −
Double integrate moment equation to get deflection; use boundary conditionsfrom supports rollers and pins restrict displacement; fixed supports restrictdisplacements and rotations
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FE Mechanics of Materials Review
2
2( ) d yM x EIdx
= ⇒[ ( ) ]M x dx dx
yEI
= ∫ ∫
• For each integration the “constant of integration” has to be defined, based on boundary conditions
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FE Mechanics of Materials Review
Column Buckling
2
2crEIP π
= Euler Buckling Formula(for ideal column with pinned ends)
crP = critical axial loading (maximum axial load that a column can support just before it buckles)
= unbraced column length
IrA
= = radius of gyration, units of length
2Ir I r AA
= ⇒ = ⇒2
2( / )cr
crP EA r
πσ = =
= critical buckling stress/ r = slenderness ratio for the column
I = the smallest moment of inertia of the cross-section
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FE Mechanics of Materials Review
Euler’s formula is only valid when .
When , then the section will simply yield.
cr yieldσ σ≤
cr yieldσ σ>
For columns that have end conditions other than pinned-pinned:
( )22
KLEIPcr
π= K = the effective length factor (see next page)
KL = Le = the effective length
KL/r = the effective slenderness ratio
( )22
/ rKLE
crπσ =
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FE Mechanics of Materials Review
Effective LengthFactors