lecture slides - philadelphia university · 2016. 10. 26. · utilizing a fictitious force...
TRANSCRIPT
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Chapter 4
Deflection and Stiffness
Lecture Slides
The McGraw-Hill Companies © 2012
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Chapter Outline
Shigley’s Mechanical Engineering Design
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Force vs Deflection
Elasticity – property of a material that enables it to regain its
original configuration after deformation
Spring – a mechanical element that exerts a force when
deformed
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Linear spring
Nonlinear
stiffening spring Nonlinear
softening spring
Fig. 4–1
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Spring Rate
Relation between force and deflection, F = F(y)
Spring rate
For linear springs, k is constant, called spring constant
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Axially-Loaded Stiffness
Total extension or contraction of a uniform bar in tension or
compression
Spring constant, with k = F/d
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Torsionally-Loaded Stiffness
Angular deflection (in radians) of a uniform solid or hollow
round bar subjected to a twisting moment T
Converting to degrees, and including J = pd4/32 for round solid
Torsional spring constant for round bar
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Deflection Due to Bending
Curvature of beam subjected to bending moment M
From mathematics, curvature of plane curve
Slope of beam at any point x along the length
If the slope is very small, the denominator of Eq. (4-9)
approaches unity.
Combining Eqs. (4-8) and (4-9), for beams with small slopes,
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Deflection Due to Bending
Recall Eqs. (3-3) and (3-4)
Successively differentiating
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Deflection Due to Bending
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(4-10)
(4-11)
(4-12)
(4-13)
(4-14)
Fig. 4–2
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Example 4-1
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Fig. 4–2
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Example 4-1
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Example 4-1
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Example 4-1
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Beam Deflection Methods
Some of the more common methods for solving the integration
problem for beam deflection
◦ Superposition
◦ Moment-area method
◦ Singularity functions
◦ Numerical integration
Other methods that use alternate approaches
◦ Castigliano energy method
◦ Finite element software
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Beam Deflection by Superposition
Superposition determines the effects of each load separately, then adds the results.
Separate parts are solved using any method for simple load cases.
Many load cases and boundary conditions are solved and available in Table A-9, or in references such as Roark’s Formulas for Stress and Strain.
Conditions
◦ Each effect is linearly related to the load that produces it.
◦ A load does not create a condition that affects the result of another load.
◦ The deformations resulting from any specific load are not large enough to appreciably alter the geometric relations of the parts of the structural system.
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Example 4-2
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Fig. 4–3
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Example 4-2
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Example 4-2
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Example 4-3
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Fig. 4–4
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Example 4-3
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Example 4-3
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Fig. 4–4
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Example 4-3
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Example 4-4
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Fig. 4–5
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Example 4-4
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Example 4-4
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Beam Deflection by Singularity Functions
A notation useful
for integrating
across
discontinuities
Angle brackets
indicate special
function to
determine whether
forces and moments
are active
Shigley’s Mechanical Engineering Design
Table 3–1
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Example 4-5
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Example 4-5
Shigley’s Mechanical Engineering Design
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Example 4-5
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Example 4-5
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Example 4-6
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Fig. 4–6
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Example 4-6
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Example 4-6
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Example 4-6
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Example 4-7
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Fig. 4–7
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Example 4-7
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Fig. 4–7
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Example 4-7
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Example 4-7
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Example 4-7
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Example 4-7
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Example 4-7
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Example 4-7
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Strain Energy
External work done on elastic member in deforming it is
transformed into strain energy, or potential energy.
Strain energy equals product of average force and deflection.
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Some Common Strain Energy Formulas
For axial loading, applying k = AE/l from Eq. (4-4),
For torsional loading, applying k = GJ/l from Eq. (4-7),
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Some Common Strain Energy Formulas
For direct shear loading,
For bending loading,
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Some Common Strain Energy Formulas
For transverse shear loading,
where C is a modifier dependent on the cross sectional shape.
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Summary of Common Strain Energy Formulas
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Example 4-8
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Fig. 4–9
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Example 4-8
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Fig. 4–9
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Example 4-8
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Castigliano’s Theorem
When forces act on elastic systems subject to small
displacements, the displacement corresponding to any force, in
the direction of the force, is equal to the partial derivative of the
total strain energy with respect to that force.
For rotational displacement, in radians,
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Example 4-9
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Fig. 4–9
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Example 4-9
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Utilizing a Fictitious Force
Castigliano’s method can be used to find a deflection at a point
even if there is no force applied at that point.
Apply a fictitious force Q at the point, and in the direction, of
the desired deflection.
Set up the equation for total strain energy including the energy
due to Q.
Take the derivative of the total strain energy with respect to Q.
Once the derivative is taken, Q is no longer needed and can be
set to zero.
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Finding Deflection Without Finding Energy
For cases requiring integration of strain energy equations, it is
more efficient to obtain the deflection directly without explicitly
finding the strain energy.
The partial derivative is moved inside the integral.
For example, for bending,
Derivative can be taken before integration, simplifying the math.
Especially helpful with fictitious force Q, since it can be set to
zero after the derivative is taken.
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Common Deflection Equations
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Example 4-10
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Fig. 4–10
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Example 4-10
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Example 4-10
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Example 4-10
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Example 4-11
Shigley’s Mechanical Engineering Design
Fig. 4–11
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Example 4-11
Shigley’s Mechanical Engineering Design Fig. 4–11
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Example 4-11
Shigley’s Mechanical Engineering Design
Fig. 4–11
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Example 4-11
Shigley’s Mechanical Engineering Design
Fig. 4–11
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Example 4-11
Shigley’s Mechanical Engineering Design
Fig. 4–11
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Example 4-11
Shigley’s Mechanical Engineering Design
Fig. 4–11
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Example 4-11
Shigley’s Mechanical Engineering Design
Fig. 4–11
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Deflection of Curved Members
Consider case of thick curved member in bending (See Sec. 3-18)
Four strain energy terms due to
◦ Bending moment M
◦ Axial force Fq
◦ Bending moment due to Fq (since neutral axis and centroidal
axis do not coincide)
◦ Transverse shear Fr
Shigley’s Mechanical Engineering Design
Fig. 4–12
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Deflection of Curved Members
Strain energy due to bending moment M
Shigley’s Mechanical Engineering Design
where rn is the radius of the neutral axis
Fig. 4–12
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Deflection of Curved Members
Strain energy due to axial force Fq
Shigley’s Mechanical Engineering Design
Fig. 4–12
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Deflection of Curved Members
Strain energy due to bending moment due to Fq
Shigley’s Mechanical Engineering Design
Fig. 4–12
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Deflection of Curved Members
Strain energy due to transverse shear Fr
Shigley’s Mechanical Engineering Design
Fig. 4–12
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Deflection of Curved Members
Combining four energy terms
Deflection by Castigliano’s method
General for any thick circular curved member, with appropriate
limits of integration
Shigley’s Mechanical Engineering Design
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Deflection of Curved Members
For specific example in figure,
Shigley’s Mechanical Engineering Design
Fig. 4–12
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Deflection of Curved Members
Substituting and factoring,
Shigley’s Mechanical Engineering Design
Fig. 4–12
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Deflection of Thin Curved Members
For thin curved members, say R/h > 10, eccentricity is small
Strain energies can be approximated with regular energy equations
with substitution of Rdq for dx
As R increases, bending component dominates all other terms
Shigley’s Mechanical Engineering Design
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Example 4-12
Shigley’s Mechanical Engineering Design
Fig. 4–13
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Example 4-12
Shigley’s Mechanical Engineering Design
Fig. 4–13
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Example 4-12
Shigley’s Mechanical Engineering Design
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Example 4-12
Shigley’s Mechanical Engineering Design
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Example 4-12
Shigley’s Mechanical Engineering Design
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Example 4-12
Shigley’s Mechanical Engineering Design
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Example 4-13
Shigley’s Mechanical Engineering Design
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Example 4-13
Shigley’s Mechanical Engineering Design Fig. 4–14
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Example 4-13
Shigley’s Mechanical Engineering Design
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Example 4-13
Shigley’s Mechanical Engineering Design
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Example 4-13
Shigley’s Mechanical Engineering Design
Fig. 4–14 (b)
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Statically Indeterminate Problems
A system is overconstrained when it has more unknown support
(reaction) forces and/or moments than static equilibrium
equations.
Such a system is said to be statically indeterminate.
The extra constraint supports are call redundant supports.
To solve, a deflection equation is required for each redundant
support.
Shigley’s Mechanical Engineering Design
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Statically Indeterminate Problems
Example of nested springs
One equation of static
equilibrium
Deformation equation
Use spring constant relation to
put deflection equation in terms
of force
Substituting into equilibrium
equation,
Shigley’s Mechanical Engineering Design
Fig. 4–15
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Procedure 1 for Statically Indeterminate Problems
1. Choose the redundant reaction(s)
2. Write the equations of static equilibrium for the remaining
reactions in terms of the applied loads and the redundant
reaction(s).
3. Write the deflection equation(s) for the point(s) at the locations
of the redundant reaction(s) in terms of the applied loads and
redundant reaction(s).
4. Solve equilibrium equations and deflection equations
simultaneously to determine the reactions.
Shigley’s Mechanical Engineering Design
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Example 4-14
Shigley’s Mechanical Engineering Design
Fig. 4–16
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Example 4-14
Shigley’s Mechanical Engineering Design
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Example 4-14
Shigley’s Mechanical Engineering Design
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Example 4-14
Shigley’s Mechanical Engineering Design
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Example 4-14
Shigley’s Mechanical Engineering Design
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Example 4-14
Shigley’s Mechanical Engineering Design
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Example 4-14
Shigley’s Mechanical Engineering Design
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Procedure 2 for Statically Indeterminate Problems
1. Write the equations of static equilibrium in terms of the applied
loads and unknown restraint reactions.
2. Write the deflection equation in terms of the applied loads and
unknown restraint reactions.
3. Apply boundary conditions to the deflection equation consistent
with the restraints.
4. Solve the set of equations.
Shigley’s Mechanical Engineering Design
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Example 4-15
Shigley’s Mechanical Engineering Design
Fig. 4–17
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Example 4-15
Shigley’s Mechanical Engineering Design
Fig. 4–17
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Example 4-15
Shigley’s Mechanical Engineering Design
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Example 4-15
Shigley’s Mechanical Engineering Design
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Example 4-15
Shigley’s Mechanical Engineering Design
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Example 4-15
Shigley’s Mechanical Engineering Design
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Example 4-15
Shigley’s Mechanical Engineering Design
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Compression Members
Column – A member loaded in compression such that either its
length or eccentric loading causes it to experience more than
pure compression
Four categories of columns
◦ Long columns with central loading
◦ Intermediate-length columns with central loading
◦ Columns with eccentric loading
◦ Struts or short columns with eccentric loading
Shigley’s Mechanical Engineering Design
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Long Columns with Central Loading
When P reaches critical
load, column becomes
unstable and bending
develops rapidly
Critical load depends on end
conditions
Shigley’s Mechanical Engineering Design
Fig. 4–18
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Euler Column Formula
For pin-ended column,
critical load is given by
Euler column formula,
Applies to other end
conditions with
addition of constant C
for each end condition
Shigley’s Mechanical Engineering Design
(4-42)
(4-43) Fig. 4–18
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Recommended Values for End Condition Constant
Fixed ends are practically
difficult to achieve
More conservative values of C
are often used, as in Table 4-2
Shigley’s Mechanical Engineering Design
Table 4–2
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Long Columns with Central Loading
Using I = Ak2, where A is the area and k is the radius of
gyration, Euler column formula can be expressed as
l/k is the slenderness ratio, used to classify columns according to
length categories.
Pcr/A is the critical unit load, the load per unit area necessary to
place the column in a condition of unstable equilibrium.
Shigley’s Mechanical Engineering Design
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Euler Curve
Plotting Pcr/A vs l/k, with C = 1 gives curve PQR
Shigley’s Mechanical Engineering Design
Fig. 4–19
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Long Columns with Central Loading
Tests show vulnerability to failure near point Q
Since buckling is sudden and catastrophic, a conservative
approach near Q is desired
Point T is usually defined such that Pcr/A = Sy/2, giving
Shigley’s Mechanical Engineering Design Fig. 4–19
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Condition for Use of Euler Equation
For (l/k) > (l/k)1, use Euler equation
For (l/k) ≤ (l/k)1, use a parabolic curve between Sy and T
Shigley’s Mechanical Engineering Design
Fig. 4–19
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Intermediate-Length Columns with Central Loading
For intermediate-length columns,
where (l/k) ≤ (l/k)1, use a parabolic
curve between Sy and T
General form of parabola
If parabola starts at Sy, then a = Sy
If parabola fits tangent to Euler
curve at T, then
Results in parabolic formula, also
known as J.B. Johnson formula
Shigley’s Mechanical Engineering Design
Fig. 4–19
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Columns with Eccentric Loading
For eccentrically loaded column
with eccentricity e,
M = -P(e+y)
Substituting into d2y/dx2=M/EI,
Solving with boundary conditions
y = 0 at x = 0 and at x = l
Shigley’s Mechanical Engineering Design
Fig. 4–20
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Columns with Eccentric Loading
At midspan where x = l/2
Shigley’s Mechanical Engineering Design Fig. 4–20
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Columns with Eccentric Loading
The maximum compressive stress includes
axial and bending
Substituting Mmax from Eq. (4-48)
Using Syc as the maximum value of sc, and
solving for P/A, we obtain the secant column
formula
Shigley’s Mechanical Engineering Design
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Secant Column Formula
Secant Column Formula
ec/k2 is the eccentricity ratio
Design charts of secant column formula for various eccentricity
ratio can be prepared for a given material strength
Shigley’s Mechanical Engineering Design
Fig. 4–21
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Example 4-16
Shigley’s Mechanical Engineering Design
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Example 4-16
Shigley’s Mechanical Engineering Design
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Example 4-17
Shigley’s Mechanical Engineering Design
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Example 4-17
Shigley’s Mechanical Engineering Design
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Example 4-18
Shigley’s Mechanical Engineering Design
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Example 4-19
Shigley’s Mechanical Engineering Design
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Example 4-19
Shigley’s Mechanical Engineering Design
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Struts or Short Compression Members
Strut - short member loaded in compression
If eccentricity exists, maximum stress is at B
with axial compression and bending.
Note that it is not a function of length
Differs from secant equation in that it assumes
small effect of bending deflection
If bending deflection is limited to 1 percent of e,
then from Eq. (4-44), the limiting slenderness
ratio for strut is
Shigley’s Mechanical Engineering Design
Fig. 4–22
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Example 4-20
Shigley’s Mechanical Engineering Design
Fig. 4–23
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Example 4-20
Shigley’s Mechanical Engineering Design
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Elastic Stability
Be alert for buckling instability in structural members that are
◦ Loaded in compression (directly or indirectly)
◦ Long or thin
◦ Unbraced
Instability may be
◦ Local or global
◦ Elastic or plastic
◦ Bending or torsion
Shigley’s Mechanical Engineering Design
Fig. 4–24
Fig. 4–25
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Shock and Impact
Impact – collision of two masses with initial relative velocity
Shock – a suddenly applied force or disturbance
Shigley’s Mechanical Engineering Design
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Shock and Impact
Example of automobile collision
◦ m1 is mass of engine
◦ m2 is mass of rest of vehicle
◦ Springs represent stiffnesses of various structural elements
Equations of motion, assuming linear springs
Equations can be solved for any impact velocity
Shigley’s Mechanical Engineering Design Fig. 4–26
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Suddenly Applied Loading
Weight falls distance h and
suddenly applies a load to a
cantilever beam
Find deflection and force
applied to the beam due to
impact
Shigley’s Mechanical Engineering Design
Fig. 4–27 (a)
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Suddenly Applied Loading
Abstract model considering
beam as simple spring
From Table A-9, beam 1,
k= F/y =3EI/l3
Assume the beam to be
massless, so no momentum
transfer, just energy transfer
Loss of potential energy from
change of elevation is
W(h + d)
Increase in potential energy
from compressing spring is
kd 2/2
Conservation of energy
W(h + d) = kd 2/2 Shigley’s Mechanical Engineering Design
Fig. 4–27 (b)
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Suddenly Applied Loading
Rearranging
Solving for d
Maximum deflection
Maximum force
Shigley’s Mechanical Engineering Design