eml3500 ch 4 slides
TRANSCRIPT
-
7/28/2019 EML3500 Ch 4 Slides
1/134
Chapter 4
Deflection and Stiffness
Lecture Slides
The McGraw-Hill Companies 2012
-
7/28/2019 EML3500 Ch 4 Slides
2/134
Chapter Outline
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
3/134
Force vs Deflection
Elasticityproperty of a material that enables it to regain its
original configuration after deformation Springa mechanical element that exerts a force when
deformed
Shigleys Mechanical Engineering Design
Linear spring
Nonlinear
stiffening spring
Nonlinear
softening spring
Fig. 41
-
7/28/2019 EML3500 Ch 4 Slides
4/134
Spring Rate
Relation between force and deflection,F=F(y)
Spring rate
For linear springs, k is constant, calledspring constant
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
5/134
Axially-Loaded Stiffness
Total extension or contraction of a uniform bar in tension or
compression
Spring constant, with k=F/d
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
6/134
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 includingJ= pd4/32 for round solid
Torsional spring constant for round bar
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
7/134
Deflection Due to Bending
Curvature of beam subjected to bending momentM
From mathematics, curvature of plane curve
Slope of beam at any pointx 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,
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
8/134
Deflection Due to Bending
Recall Eqs. (3-3) and (3-4)
Successively differentiating
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
9/134
Deflection Due to Bending
Shigleys Mechanical Engineering Design
(4-10)
(4-11)
(4-12)
(4-13)
(4-14)
Fig. 42
-
7/28/2019 EML3500 Ch 4 Slides
10/134
Example 4-1
Shigleys Mechanical Engineering Design
Fig. 42
-
7/28/2019 EML3500 Ch 4 Slides
11/134
Example 4-1
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
12/134
Example 4-1
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
13/134
Example 4-1
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
14/134
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
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
15/134
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 loadcases.
Many load cases and boundary conditions are solved andavailable in Table A-9, or in references such asRoarks
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 ofthe parts of the structural system.
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
16/134
Example 4-2
Shigleys Mechanical Engineering Design
Fig. 43
-
7/28/2019 EML3500 Ch 4 Slides
17/134
Example 4-2
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
18/134
Example 4-2
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
19/134
Example 4-3
Shigleys Mechanical Engineering Design
Fig. 44
-
7/28/2019 EML3500 Ch 4 Slides
20/134
Example 4-3
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
21/134
Example 4-3
Shigleys Mechanical Engineering Design
Fig. 44
-
7/28/2019 EML3500 Ch 4 Slides
22/134
Example 4-3
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
23/134
Example 4-4
Shigleys Mechanical Engineering Design
Fig. 45
-
7/28/2019 EML3500 Ch 4 Slides
24/134
Example 4-4
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
25/134
Example 4-4
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
26/134
Beam Deflection by Singularity Functions
A notation useful
for integrating
across
discontinuities
Angle brackets
indicate special
function todetermine whether
forces and moments
are active
Shigleys Mechanical Engineering Design
Table 31
-
7/28/2019 EML3500 Ch 4 Slides
27/134
Example 4-5
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
28/134
Example 4-5
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
29/134
Example 4-5
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
30/134
Example 4-5
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
31/134
Example 4-6
Shigleys Mechanical Engineering Design
Fig. 46
-
7/28/2019 EML3500 Ch 4 Slides
32/134
Example 4-6
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
33/134
Example 4-6
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
34/134
Example 4-6
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
35/134
Example 4-7
Shigleys Mechanical Engineering Design
Fig. 47
-
7/28/2019 EML3500 Ch 4 Slides
36/134
Example 4-7
Shigleys Mechanical Engineering DesignFig. 47
-
7/28/2019 EML3500 Ch 4 Slides
37/134
Example 4-7
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
38/134
Example 4-7
Shigleys Mechanical Engineering Design
-
7/28/2019 EML3500 Ch 4 Slides
39/134
Example 4-7
Shigleys Mechanical Engineering Design
4
-
7/28/2019 EML3500 Ch 4 Slides
40/134
Example 4-7
Shigleys Mechanical Engineering Design
E l 4 7
-
7/28/2019 EML3500 Ch 4 Slides
41/134
Example 4-7
Shigleys Mechanical Engineering Design
E l 4 7
-
7/28/2019 EML3500 Ch 4 Slides
42/134
Example 4-7
Shigleys Mechanical Engineering Design
St i E
-
7/28/2019 EML3500 Ch 4 Slides
43/134
Strain Energy
External work done on elastic member in deforming it is
transformed intostrain energy, orpotential energy.
Strain energy equals product of average force and deflection.
Shigleys Mechanical Engineering Design
S C St i E F l
-
7/28/2019 EML3500 Ch 4 Slides
44/134
Some Common Strain Energy Formulas
For axial loading, applying k=AE/lfrom Eq. (4-4),
For torsional loading, applying k= GJ/lfrom Eq. (4-7),
Shigleys Mechanical Engineering Design
S C St i E F l
-
7/28/2019 EML3500 Ch 4 Slides
45/134
Some Common Strain Energy Formulas
For direct shear loading,
For bending loading,
Shigleys Mechanical Engineering Design
S C St i E F l
-
7/28/2019 EML3500 Ch 4 Slides
46/134
Some Common Strain Energy Formulas
For transverse shear loading,
where Cis a modifier dependent on the cross sectional shape.
Shigleys Mechanical Engineering Design
S mmar of Common Strain Energ Form las
-
7/28/2019 EML3500 Ch 4 Slides
47/134
Summary of Common Strain Energy Formulas
Shigleys Mechanical Engineering Design
Example 4 8
-
7/28/2019 EML3500 Ch 4 Slides
48/134
Example 4-8
Shigleys Mechanical Engineering Design
Fig. 49
Example 4 8
-
7/28/2019 EML3500 Ch 4 Slides
49/134
Example 4-8
Shigleys Mechanical Engineering Design
Fig. 49
Example 4 8
-
7/28/2019 EML3500 Ch 4 Slides
50/134
Example 4-8
Shigleys Mechanical Engineering Design
Castiglianos Theorem
-
7/28/2019 EML3500 Ch 4 Slides
51/134
Castiglianos 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,
Shigleys Mechanical Engineering Design
Example 4 9
-
7/28/2019 EML3500 Ch 4 Slides
52/134
Example 4-9
Shigleys Mechanical Engineering Design
Fig. 49
Example 4 9
-
7/28/2019 EML3500 Ch 4 Slides
53/134
Example 4-9
Shigleys Mechanical Engineering Design
Utilizing a Fictitious Force
-
7/28/2019 EML3500 Ch 4 Slides
54/134
Utilizing a Fictitious Force
Castiglianos 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.
Shigleys Mechanical Engineering Design
Finding Deflection Without Finding Energy
-
7/28/2019 EML3500 Ch 4 Slides
55/134
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.
Shigleys Mechanical Engineering Design
Common Deflection Equations
-
7/28/2019 EML3500 Ch 4 Slides
56/134
Common Deflection Equations
Shigleys Mechanical Engineering Design
Example 4-10
-
7/28/2019 EML3500 Ch 4 Slides
57/134
Example 4-10
Shigleys Mechanical Engineering Design
Fig. 410
Example 4-10
-
7/28/2019 EML3500 Ch 4 Slides
58/134
Example 4 10
Shigleys Mechanical Engineering Design
Example 4-10
-
7/28/2019 EML3500 Ch 4 Slides
59/134
Example 4 10
Shigleys Mechanical Engineering Design
Example 4-10
-
7/28/2019 EML3500 Ch 4 Slides
60/134
Example 4 10
Shigleys Mechanical Engineering Design
Example 4-11
-
7/28/2019 EML3500 Ch 4 Slides
61/134
Example 4 11
Shigleys Mechanical Engineering Design
Fig. 411
Example 4-11
-
7/28/2019 EML3500 Ch 4 Slides
62/134
Example 4 11
Shigleys Mechanical Engineering DesignFig. 411
Example 4-11
-
7/28/2019 EML3500 Ch 4 Slides
63/134
Example 4 11
Shigleys Mechanical Engineering Design
Fig. 411
Example 4-11
-
7/28/2019 EML3500 Ch 4 Slides
64/134
p
Shigleys Mechanical Engineering DesignFig. 411
Example 4-11
-
7/28/2019 EML3500 Ch 4 Slides
65/134
p
Shigleys Mechanical Engineering DesignFig. 411
Example 4-11
-
7/28/2019 EML3500 Ch 4 Slides
66/134
p
Shigleys Mechanical Engineering DesignFig. 411
Example 4-11
-
7/28/2019 EML3500 Ch 4 Slides
67/134
p
Shigleys Mechanical Engineering DesignFig. 411
Deflection of Curved Members
-
7/28/2019 EML3500 Ch 4 Slides
68/134
Consider case of thick curved member in bending (See Sec. 3-18)
Four strain energy terms due to
Bending momentM
Axial forceFq
Bending moment due toFq
(since neutral axis and centroidal
axis do not coincide)
Transverse shearFr
Shigleys Mechanical Engineering DesignFig. 412
Deflection of Curved Members
-
7/28/2019 EML3500 Ch 4 Slides
69/134
Strain energy due to bending momentM
Shigleys Mechanical Engineering Design
where rn is the radius of the neutral axis
Fig. 412
Deflection of Curved Members
-
7/28/2019 EML3500 Ch 4 Slides
70/134
Strain energy due to axial forceFq
Shigleys Mechanical Engineering DesignFig. 412
Deflection of Curved Members
-
7/28/2019 EML3500 Ch 4 Slides
71/134
Strain energy due to bending moment due toFq
Shigleys Mechanical Engineering DesignFig. 412
Deflection of Curved Members
-
7/28/2019 EML3500 Ch 4 Slides
72/134
Strain energy due to transverse shearFr
Shigleys Mechanical Engineering DesignFig. 412
Deflection of Curved Members
-
7/28/2019 EML3500 Ch 4 Slides
73/134
Combining four energy terms
Deflection by Castiglianos method
General for any thick circular curved member, with appropriate
limits of integration
Shigleys Mechanical Engineering Design
Deflection of Curved Members
-
7/28/2019 EML3500 Ch 4 Slides
74/134
For specific example in figure,
Shigleys Mechanical Engineering DesignFig. 412
Deflection of Curved Members
-
7/28/2019 EML3500 Ch 4 Slides
75/134
Substituting and factoring,
Shigleys Mechanical Engineering DesignFig. 412
Deflection of Thin Curved Members
-
7/28/2019 EML3500 Ch 4 Slides
76/134
Forthin curved members, sayR/h > 10, eccentricity is small
Strain energies can be approximated with regular energy equations
with substitution ofRdqfordx AsR increases, bending component dominates all other terms
Shigleys Mechanical Engineering Design
Example 4-12
-
7/28/2019 EML3500 Ch 4 Slides
77/134
Shigleys Mechanical Engineering Design
Fig. 413
Example 4-12
-
7/28/2019 EML3500 Ch 4 Slides
78/134
Shigleys Mechanical Engineering Design
Fig. 413
Example 4-12
-
7/28/2019 EML3500 Ch 4 Slides
79/134
Shigleys Mechanical Engineering Design
Example 4-12
-
7/28/2019 EML3500 Ch 4 Slides
80/134
Shigleys Mechanical Engineering Design
Example 4-12
-
7/28/2019 EML3500 Ch 4 Slides
81/134
Shigleys Mechanical Engineering Design
Example 4-12
-
7/28/2019 EML3500 Ch 4 Slides
82/134
Shigleys Mechanical Engineering Design
Example 4-13
-
7/28/2019 EML3500 Ch 4 Slides
83/134
Shigleys Mechanical Engineering Design
Example 4-13
-
7/28/2019 EML3500 Ch 4 Slides
84/134
Shigleys Mechanical Engineering DesignFig. 414
Example 4-13
-
7/28/2019 EML3500 Ch 4 Slides
85/134
Shigleys Mechanical Engineering Design
Example 4-13
-
7/28/2019 EML3500 Ch 4 Slides
86/134
Shigleys Mechanical Engineering Design
Example 4-13
-
7/28/2019 EML3500 Ch 4 Slides
87/134
Shigleys Mechanical Engineering DesignFig. 414 (b)
Statically Indeterminate Problems
-
7/28/2019 EML3500 Ch 4 Slides
88/134
A system is overconstrainedwhen it has more unknown support
(reaction) forces and/or moments than static equilibrium
equations.
Such a system is said to bestatically indeterminate.
The extra constraint supports are call redundant supports.
To solve, a deflection equation is required for each redundant
support.
Shigleys Mechanical Engineering Design
Statically Indeterminate Problems
-
7/28/2019 EML3500 Ch 4 Slides
89/134
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,
Shigleys Mechanical Engineering Design
Fig. 415
Procedure 1 for Statically Indeterminate Problems
-
7/28/2019 EML3500 Ch 4 Slides
90/134
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.
Shigleys Mechanical Engineering Design
Example 4-14
-
7/28/2019 EML3500 Ch 4 Slides
91/134
Shigleys Mechanical Engineering Design
Fig. 416
Example 4-14
-
7/28/2019 EML3500 Ch 4 Slides
92/134
Shigleys Mechanical Engineering Design
Example 4-14
-
7/28/2019 EML3500 Ch 4 Slides
93/134
Shigleys Mechanical Engineering Design
Example 4-14
-
7/28/2019 EML3500 Ch 4 Slides
94/134
Shigleys Mechanical Engineering Design
Example 4-14
-
7/28/2019 EML3500 Ch 4 Slides
95/134
Shigleys Mechanical Engineering Design
Example 4-14
-
7/28/2019 EML3500 Ch 4 Slides
96/134
Shigleys Mechanical Engineering Design
Example 4-14
-
7/28/2019 EML3500 Ch 4 Slides
97/134
Shigleys Mechanical Engineering Design
Procedure 2 for Statically Indeterminate Problems
-
7/28/2019 EML3500 Ch 4 Slides
98/134
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 andunknown restraint reactions.
3. Apply boundary conditions to the deflection equation consistent
with the restraints.
4. Solve the set of equations.
Shigleys Mechanical Engineering Design
Example 4-15
-
7/28/2019 EML3500 Ch 4 Slides
99/134
Shigleys Mechanical Engineering DesignFig. 417
Example 4-15
-
7/28/2019 EML3500 Ch 4 Slides
100/134
Shigleys Mechanical Engineering DesignFig. 417
Example 4-15
-
7/28/2019 EML3500 Ch 4 Slides
101/134
Shigleys Mechanical Engineering Design
Example 4-15
-
7/28/2019 EML3500 Ch 4 Slides
102/134
Shigleys Mechanical Engineering Design
Example 4-15
-
7/28/2019 EML3500 Ch 4 Slides
103/134
Shigleys Mechanical Engineering Design
Example 4-15
-
7/28/2019 EML3500 Ch 4 Slides
104/134
Shigleys Mechanical Engineering Design
Example 4-15
-
7/28/2019 EML3500 Ch 4 Slides
105/134
Shigleys Mechanical Engineering Design
Compression Members
-
7/28/2019 EML3500 Ch 4 Slides
106/134
ColumnA 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
Shigleys Mechanical Engineering Design
Long Columns with Central Loading
-
7/28/2019 EML3500 Ch 4 Slides
107/134
WhenPreaches critical
load, column becomes
unstable and bendingdevelops rapidly
Critical load depends on end
conditions
Shigleys Mechanical Engineering Design
Fig. 418
Euler Column Formula
-
7/28/2019 EML3500 Ch 4 Slides
108/134
For pin-ended column,
critical load is given by
Euler column formula,
Applies to other end
conditions withaddition of constant C
for each end condition
Shigleys Mechanical Engineering Design
(4-42)
(4-43) Fig. 418
Recommended Values for End Condition Constant
-
7/28/2019 EML3500 Ch 4 Slides
109/134
Fixed ends are practically
difficult to achieve
More conservative values ofCare often used, as in Table 4-2
Shigleys Mechanical Engineering Design
Table 42
Long Columns with Central Loading
-
7/28/2019 EML3500 Ch 4 Slides
110/134
UsingI = Ak2, whereA is the area and kis the radius of
gyration, Euler column formula can be expressed as
l/kis theslenderness 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 ofunstable equilibrium.
Shigleys Mechanical Engineering Design
Euler Curve
-
7/28/2019 EML3500 Ch 4 Slides
111/134
PlottingPcr/A vs l/k, with C = 1 gives curvePQR
Shigleys Mechanical Engineering DesignFig. 419
Long Columns with Central Loading
-
7/28/2019 EML3500 Ch 4 Slides
112/134
Tests show vulnerability to failure near point Q
Since buckling is sudden and catastrophic, a conservative
approach nearQ is desired
Point Tis usually defined such thatPcr/A = Sy/2, giving
Shigleys Mechanical Engineering DesignFig. 419
Condition for Use of Euler Equation
-
7/28/2019 EML3500 Ch 4 Slides
113/134
For (l/k) > (l/k)1, use Euler equation
For (l/k) (l/k)1, use a parabolic curve between Sy and T
Shigleys Mechanical Engineering Design
Fig. 419
Intermediate-Length Columns with Central Loading
-
7/28/2019 EML3500 Ch 4 Slides
114/134
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 inparabolic formula, also
known asJ.B. Johnson formula
Shigleys Mechanical Engineering Design
Fig. 419
Columns with Eccentric Loading
i ll l d d l
-
7/28/2019 EML3500 Ch 4 Slides
115/134
For eccentrically loaded column
with eccentricity e,
M = -P(e+y)
Substituting into d2y/dx2=M/EI,
Solving with boundary conditions
y = 0 atx = 0 and atx = l
Shigleys Mechanical Engineering Design
Fig. 420
Columns with Eccentric Loading
A id h l/2
-
7/28/2019 EML3500 Ch 4 Slides
116/134
At midspan wherex = l/2
Shigleys Mechanical Engineering DesignFig. 420
Columns with Eccentric Loading
Th i i i l d
-
7/28/2019 EML3500 Ch 4 Slides
117/134
The maximum compressive stress includes
axial and bending
SubstitutingMmax from Eq. (4-48)
Using Syc as the maximum value ofsc, and
solving forP/A, we obtain thesecant column
formula
Shigleys Mechanical Engineering Design
Secant Column Formula
Secant Column Formula
-
7/28/2019 EML3500 Ch 4 Slides
118/134
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
Shigleys Mechanical Engineering Design
Fig. 421
Example 4-16
-
7/28/2019 EML3500 Ch 4 Slides
119/134
Shigleys Mechanical Engineering Design
Example 4-16
-
7/28/2019 EML3500 Ch 4 Slides
120/134
Shigleys Mechanical Engineering Design
Example 4-17
-
7/28/2019 EML3500 Ch 4 Slides
121/134
Shigleys Mechanical Engineering Design
Example 4-17
-
7/28/2019 EML3500 Ch 4 Slides
122/134
Shigleys Mechanical Engineering Design
Example 4-18
-
7/28/2019 EML3500 Ch 4 Slides
123/134
Shigleys Mechanical Engineering Design
Example 4-19
-
7/28/2019 EML3500 Ch 4 Slides
124/134
Shigleys Mechanical Engineering Design
Example 4-19
-
7/28/2019 EML3500 Ch 4 Slides
125/134
Shigleys Mechanical Engineering Design
Struts or Short Compression Members
Strut short member loaded in compression
-
7/28/2019 EML3500 Ch 4 Slides
126/134
Strut - short member loaded in compression
If eccentricity exists, maximum stress is atB
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 ofe,
then from Eq. (4-44), the limiting slenderness
ratio for strut is
Shigleys Mechanical Engineering Design
Fig. 422
Example 4-20
-
7/28/2019 EML3500 Ch 4 Slides
127/134
Shigleys Mechanical Engineering DesignFig. 423
Example 4-20
-
7/28/2019 EML3500 Ch 4 Slides
128/134
Shigleys Mechanical Engineering Design
Elastic Stability
Be alert for buckling instability in structural members that are
-
7/28/2019 EML3500 Ch 4 Slides
129/134
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
Shigleys Mechanical Engineering Design
Fig. 424
Fig. 425
Shock and Impact
Impact collision of two masses with initial relative velocity
-
7/28/2019 EML3500 Ch 4 Slides
130/134
Impactcollision of two masses with initial relative velocity
Shocka suddenly applied force or disturbance
Shigleys Mechanical Engineering Design
Shock and Impact
Example of automobile collision
-
7/28/2019 EML3500 Ch 4 Slides
131/134
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
Shigleys Mechanical Engineering DesignFig. 426
Suddenly Applied Loading
Weight falls distance h and
-
7/28/2019 EML3500 Ch 4 Slides
132/134
Weight falls distance h and
suddenly applies a load to a
cantilever beam Find deflection and force
applied to the beam due to
impact
Shigleys Mechanical Engineering Design
Fig. 427 (a)
Suddenly Applied Loading
Abstract model considering
-
7/28/2019 EML3500 Ch 4 Slides
133/134
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 energyfrom compressing spring is
kd2/2
Conservation of energy
W(h +d) = k
d2
/2 Shigleys Mechanical Engineering Design
Fig. 427 (b)
Suddenly Applied Loading
Rearranging
-
7/28/2019 EML3500 Ch 4 Slides
134/134
Rearranging
Solving ford
Maximum deflection
Maximum force