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FE Review – Mechanics of Materials
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ResourcesYou can get the sample reference book:www.ncees.org – main sitehttp://www.ncees.org/exams/study_materials/fe_handbookMultimedia learning material web site:http://web.umr.edu/~mecmovie/index.html
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Normal Stress (normal to surface)
Shear Stress (along surface)
First Concept – Stress
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Normal Strain – length change
Mechanical
Thermal
Shear Strain – angle change
Second Concept – Strain
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Material PropertiesHooke’s Law
Normal (1D)
Normal (3D)
Shear
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Material PropertiesPoisson’s ratio
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Axial Loading
Stress
Deformation
FF
PLAEδ =∑
xPAσ = F σx
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Torsional Loading
Stress
Deformation TLJGθ =∑
TJρτ =
TT
maxTcJτ =
ρ
τ
τmax
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Bending Stress
Stress
Find centroid of cross-sectionCalculate I about the Neutral Axis
rxM y
Iσ =− max rM cIσ =
MM
σx
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Transverse Shear Equation
ave VAτ = Average over entire cross-section
aveVQIbτ = Average over line
V = internal shear forceb = thicknessI = 2nd moment of areaQ = 1st moment of area of partial section
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Partial 1st Moment of Area (Q)
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Max. Shear Stresses on Specific Cross-Sectional Shapes
Rectangular Cross-Section
max32VAτ =
τCircular Cross-Section
max43VAτ =
τ
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Max. Shear Stresses on Specific Cross-Sectional Shapes
Wide-Flange Beam
maxweb
VAτ ≈
τ
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V & M Diagrams
dVwdx
=V
M dMVdx
=
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Six Rules for Drawing V & M Diagrams
1. w = dV/dxThe value of the distributed load at any point in the beam is equal to the
slope of the shear force curve.2. V = dM/dx
The value of the shear force at any point in the beam is equal to the slope of the bending moment curve.
3. The shear force curve is continuous unless there is a point force on the beam. The curve then “jumps” by the magnitude of the point force (+ for upward force).
4. The bending moment curve is continuous unless there is a point moment on the beam. The curve then “jumps” by the magnitude of the point moment (+ for CW moment).
5. The shear force will be zero at each end of the beam unless a point force is applied at the end.
6. The bending moment will be zero at each end of the beam unless a point moment is applied at the end.
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Deflection Equation
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d y MEIdx
=y = deflection of midplaneM = internal bending momentE = elastic modulusI = 2nd moment of area with
respect to neutral axis
To solve bending deflection problems (find y):1. Write the moment equation(s) M(x)2. Integrate it twice3. Apply boundary conditions4. Apply matching conditions (if applicable)
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Method of Superposition
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Stress TransformationPlane Stress Transformation Equations:
cos2 sin22 2x y x y
n xyσ σ σ σσ θ τ θ+ −
= + +
sin2 cos22x y
xyntσ σ
τ θ τ θ⎛ ⎞⎜ ⎟⎝ ⎠
−=− +
τxy
σx
σy
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Stress TransformationPrincipal Stresses:
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1, 2 2 2 xyx y x y
p pσ σ σ σσ τ
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
+ −= + +
( )tan 2
2
xyp
x y
τθ σ σ=
−⎛ ⎞⎜ ⎟⎝ ⎠
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Stress TransformationMax Shear Stress:
1 2max 2
p pσ στ
−= 1
max 2pσ
τ =2
max 2pσ
τ =
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Stress Transformation
Mohr’s Circle
σ
τ
C
( ),x xyσ τ−
( ),y xyσ τ
R
τxy
σx
σy
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Combined LoadingWe have derived stress equations for four different loading types:
xPA
σ =
maxVkA
τ =
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xMcI
σ = −
xMcI
σ = +
TcJ
τ =
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Method for Solving Combined Loading Problems1. Find internal forces and moments at
cross-section of concern.2. Find stress caused by each individual
force and moment at the point in question.
3. Add them up.
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Thin-Walled Pressure Vessels
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Column Buckling
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σY
σY
−σY
−σY
Failure occurs when:
1p Yσ σ>
where σp1 is the largest principal stress.
if σp1 and σp2 have the same sign
1 2p p Yσ σ σ− > if σp1 and σp2 have different signs
σp1
σp2
Maximum Shear Stress Theory
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σY
σY
−σY
−σY
Failure occurs when:2 2 21 1 2 2p p p p Yσ σ σ σ σ− + >
σp1
σp2
Maximum Distortion Energy Theory
This theory assumes that failure occurs when the distortion energy of the material is greater than that which causes yielding in a tension test.
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