FE Review – Mechanics of
Materials
1
Resources
„
„
You can get the sample reference book:
www.ncees.org – main site
http://www.ncees.org/exams/study_ma
terials/fe_handbook
Multimedia learning material web site:
http://web.umr.edu/~mecmovie/index.
html
FE Review
Mechanics of Materials
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1
First Concept – Stress
„
Normal Stress (normal to surface)
„
Shear Stress (along surface)
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Mechanics of Materials
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Second Concept – Strain
„
„
Normal Strain – length change
„
Mechanical
„
Thermal
Shear Strain – angle change
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Mechanics of Materials
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2
Material Properties
„
Hooke’s Law
„
Normal (1D)
„
Normal (3D)
„
Shear
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Material Properties
„
Poisson’s ratio
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3
Axial Loading
F
F
σx
F
„
Stress σ x = P
A
„
Deformation δ = ∑ PL
AE
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Mechanics of Materials
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Torsional Loading
T
„
„
T
Stress τ =T ρ
J
τ max =Tc
J
ρ
τmax
τ
Deformation θ = ∑ TL
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JG
Mechanics of Materials
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4
Bending Stress
M
M
„
„
„
Stress σ x =− M r y
I
σ max = M rc
I
σx
Find centroid of cross-section
Calculate I about the Neutral Axis
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Mechanics of Materials
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Transverse Shear Equation
τ ave =V
Average over entire cross-section
A
τ ave =VQ
Ib
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Average over line
V = internal shear force
b = thickness
I = 2nd moment of area
Q = 1st moment of area of partial section
Mechanics of Materials
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5
Partial 1st Moment of Area (Q)
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Mechanics of Materials
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Max. Shear Stresses on Specific
Cross-Sectional Shapes
Rectangular Cross-Section
τ max = 3V
2A
τ
Circular Cross-Section
τ max = 4V
3A
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τ Mechanics of Materials
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6
Max. Shear Stresses on Specific
Cross-Sectional Shapes
Wide-Flange Beam
τ max ≈ V
Aweb
τ
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V & M Diagrams
V
w = dV
dx
M
V = dM
dx
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Six Rules for Drawing V & M
Diagrams
1.
2.
3.
4.
5.
6.
w = dV/dx
The value of the distributed load at any point in the beam is equal to the
slope of the shear force curve.
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.
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).
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).
The shear force will be zero at each end of the beam unless a point force is
applied at the end.
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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Mechanics of Materials
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11
Deflection Equation
d2y = M
dx2 EI
y = deflection of midplane
M = internal bending moment
E = elastic modulus
I = 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 twice
3. Apply boundary conditions
4. Apply matching conditions (if applicable)
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Mechanics of Materials
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Method of Superposition
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Stress Transformation
Plane Stress Transformation Equations:
σy
σn =
τxy
σ x +σ y σ x −σ y
+
2
τ nt =−
2
σ x −σ y ⎞⎟⎠
⎛
⎜
⎝
2
cos2θ +τ xy sin2θ
sin2θ +τ xy cos2θ
σx
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Stress Transformation
Principal Stresses:
σ p1, p2 =
σ x +σ y
2
tan ( 2θ p ) =
+
σ x −σ y ⎞⎟2
⎟
⎟
⎠
2
2
+τ xy
τ xy
⎛σ x −σ y ⎞
⎜
⎝
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⎛
⎜
⎜
⎜
⎝
2
Mechanics of Materials
⎟
⎠
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Stress Transformation
Max Shear Stress:
τ max =
σ p1 − σ p 2
τ max =
2
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σ p1
τ max =
2
σ p2
2
Mechanics of Materials
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Stress Transformation
τ
Mohr’s Circle
(σ
σy
y
,τ xy )
τxy
σx
σ
C
R
(σ
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Mechanics of Materials
x
, −τ xy )
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Combined Loading
We have derived stress equations for four
different loading types:
σx =
P
A
τ max = k
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Mechanics of Materials
τ=
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V
A
29
Tc
J
σx = −
Mc
I
σx = +
Mc
I
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Method for Solving Combined
Loading Problems
1. 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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Mechanics of Materials
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Thin-Walled Pressure Vessels
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Column Buckling
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Maximum Shear Stress Theory
σp2
σY
−σY
σY
−σY
σp1
Failure occurs when:
σ p1 > σ Y
σ p1 − σ p 2 > σ Y
if σp1 and σp2 have the same sign
if σp1 and σp2 have different signs
where σp1 is the largest principal stress.
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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.
σp2
σY
−σY
σY
σp1
Failure occurs when:
σ p21 − σ p1σ p 2 + σ p2 2 > σ Y2
−σY
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