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Beams And Columns
1. Subject:Subject: Machine Design and Industrial Drafting (2141907)Machine Design and Industrial Drafting (2141907)
Chapter: Design Of Beams And ColumnsChapter: Design Of Beams And Columns
Department Mechanical Engineering
Name of Subject Teacher
Dr. Divyang Patel
2. Team MembersTeam Members
Name Enrollment Number
Vinay PatelVinay Patel 170990119014170990119014
Dhananjay PatelDhananjay Patel 170990119015170990119015
Dhyey ShuklaDhyey Shukla 170990119016170990119016
22. Deflection by Bending MomentDeflection by Bending Moment
EquationEquation
22
)(
2
'' qxqLx
xMEI −==υ
1
32
'
64
C
qxqLx
EI +−=υ
21
43
2412
CxC
qxqLx
EI ++−=υ
0,
24
2
3
1 =−= C
qL
C
23. Deflection by Loading EquationDeflection by Loading Equation
L
xLq
qEI
)(0'''' −
−==υ
0,
2
)(
11
2
0'''
=+
−
= CC
L
xLq
EIυ
0,
6
)(
22
3
0''
=+
−
−= CC
L
xLq
EIυ
,
24
)(
3
4
0'
C
L
xLq
EI +
−
=υ
,
120
)(
43
5
0 CxC
L
xLq
EI ++
−
−=υ
25. Strain Energy of Pure BendingStrain Energy of Pure Bending
EI
ML
L
L
=== κ
ρ
θ
L
EI
EI
LM
U
22
22
θ
==
26. Strain Energy of BendingStrain Energy of Bending
L
EI
EI
LM
U
22
22
θ
==
∫∫ == dx
EI
EI
dxM
U 2''
2
)(
22
υ
27. Strain Energy of a Beam in ShearStrain Energy of a Beam in Shear
∫=
AG
dxCV
U
2
2
Rectangular: 1.2
Circular: 1.11
Thin-walled tubular, round: 2.00
Box section: 1.00
Structural section: 1.00
28. Strain Energy of BendingStrain Energy of Bending
22
)(
2
qxqLx
xM −=
EI
Lq
EI
dxM
U
2402
522
== ∫
qx
qL
xV −=
2
)(
AG
Lq
AG
dxV
U
202
2.1 322
== ∫
29. Castigliano’s TheoremCastigliano’s Theorem
i
i
P
U
∂
∂
=δ
When forces act on a elastic system
subject to small displacements, the
displacement corresponding to any
force, collinear with the force, is equal
to the partial derivative of the total
strain energy with respect to that
force. It can also be used to find
the displacement when no force is
applied at that point.
i
i
M
U
∂
∂
=θ
0| =
∂
∂
= iQ
i
i
Q
U
δ
30. Modified Castigliano’s TheoremModified Castigliano’s Theorem
∫∫ ∂
∂
=
∂
∂
=
∂
∂
= dx
P
M
EI
M
EI
dxM
PP
U
iii
i
2
2
δ
0MPxM −−=
1,
0
−=
∂
∂
−=
∂
∂
M
M
x
P
M
EI
LM
EI
PL
dx
P
M
EI
M
i
A
23
2
0
3
+=
∂
∂
= ∫δ
EI
LM
EI
PL
dx
M
M
EI
M
A
0
2
0 2
+=
∂
∂
= ∫θ
39. Design ConsiderationsDesign Considerations
• Stress – Yield Failure or Code Compliance
• Deflection
• Strain
• Stiffness
• Stability – Important in compressive members
• Stress and strain relationships can be studied with Mohr’s circle
Often the controlling factor for
functionality
40. Deflection [Everything’s a Spring]Deflection [Everything’s a Spring]
• When loads are applied, we have deflection
• Depends on
• Type of loading
• Tension
• Compression
• Bending
• Torsion
• Cross-section of member
• Comparable to pushing on a spring
• We can calculate the amount of beam deflection by
various methods
41. SuperpositionSuperposition
• Determine effects of individual loads separately and
add the results [see examples 4-2,3,4]
• Tables are useful – see A-9
• May be applied if
• 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
• Deformations resulting from any specific load are not large
enough to appreciably alter the geometric relations of the
parts of the structural system
42. Deflection --- Energy MethodDeflection --- Energy Method
• There are situations where the tables are insufficient
• We can use energy-methods in these circumstances
• Define strain energy
•
• Define strain energy density**
• V – volume
• Put in terms of σ, ε
∫=
1
0
x
FdxU
dV
dU
=µ
∫
=
=
=
==
=
dV
E
U
dUdV
dV
dU
E
E
x
x
xx
xx
2
2
2
1
2
1
2
1
σ
µ
µ
σ
εσµ
εσ
43. Example – beam in bendingExample – beam in bending
)(
2
2
2
2
2
2
22
2
xf
EI
M
dAdxdV
dV
EI
yM
U
dV
E
U
I
My
x
=
=
=
=
=
∫
∫
σ
σ
( )
dx
EI
M
U
dx
EI
dAyM
dAdx
EI
yM
dV
EI
yM
U
dAyI
∫
∫
∫
∫∫
∫
=
===
=
2
2
)(
22
2
2
22
2
22
2
22
2
44. Castigliano’s TheoremCastigliano’s Theorem
• Deflection at any point along a beam subjected to n loads may be
expressed as the partial derivative of the strain energy of the
structure WRT the load at that point
• We can derive the strain energy equations as we did for bending
• Then we take the partial derivative to determine the deflection
equation
• Plug in load and solve!
• AND if we don’t have a force at the desired point:
• If there is no load acting at the point of interest, add a dummy load Q,
work out equations, then set Q = 0
i
i
F
U
∂
∂
=δ
45. Castigliano ExampleCastigliano Example
• Beam AB supports a uniformly distributed load
w. Determine the deflection at A.
• No load acting specifically at point A!
• Apply a dummy load Q
• Substitute expressions for M, M/ QA, and QA
(=0)
• We directed QA downward and found δA to be
positive
• Defection is in same direction as QA (downward)
Q
EI
wL
A
8
4
=δ
( )( )
EI
wL
dxxwx
EI
x
Q
M
wxxQxM
Q
U
L
A
A
A
A
A
8
1
)(
dx
Q
M
EI
M
4
0
2
2
1
2
2
1
A
L
0
=−−=
−=
∂
∂
−−=
∂
∂
=
∂
∂
=
∫
∫
δ
δ
Aδ
46. StabilityStability
•Up until now, 2 primary concerns
• Strength of a structure
• It’s ability to support a specified load without
experiencing excessive stress
• Ability of a structure to support a specified load
without undergoing unacceptable deformations
•Now, look at STABILITY of the structure
• It’s ability to support a load without undergoing a
sudden change in configuration
Material
failure
47. BucklingBuckling
• Buckling is a mode of failure that does not depend on
stress or strength, but rather on structural stiffness
• Examples:
49. BucklingBuckling
• The most common problem involving buckling is the design
of columns
• Compression members
• The analysis of an element in buckling involves establishing a
differential equation(s) for beam deformation and finding
the solution to the ODE, then determining which solutions
are stable
• Euler solved this problem for columns
50. Euler Column FormulaEuler Column Formula
•
• Where C is as follows:
2
2
L
EIc
Pcrit
π
=
C = ¼ ;Le=2L
Fixed-free
C = 2; Le=0.7071L
Fixed-pinned
C = 1: Le=L
Rounded-rounded
Pinned-pinned
C = 4; Le=L/2
Fixed-fixed
2
2
e
crit
L
EI
P
π
=
51. BucklingBuckling
•Geometry is crucial to correct analysis
• Euler – “long” columns
• Johnson – “intermediate” length columns
• Determine difference by slenderness ratio
•The point is that a designer must be alert to the
possibility of buckling
•A structure must not only be strong enough, but
must also be sufficiently rigid
54. Solving buckling problemsSolving buckling problems
• Find Euler-Johnson tangent point with
• For Le/ρ < tangent point (“intermediate”), use Johnson’s Equation:
• For Le/ρ > tangent point (“long”), use Euler’s equation:
• For Le/ρ < 10 (“short”), Scr = Sy
• If length is unknown, predict whether it is “long” or “intermediate”, use the
appropriate equation, then check using the Euler-Johnson tangent point once you have
a numerical solution for the critical strength
2
2
=
ρ
π
e
cr
L
E
S
y
e
S
EL 2
2π
ρ
=
2
2
2
4
−=
ρπ
ey
ycr
L
E
S
SS
55. Special Buckling CasesSpecial Buckling Cases
• Buckling in very long Pipe
2
2
L
EIc
Pcrit
π
=
Note Pcrit is inversely related to length squared
A tiny load will cause buckling
L = 10 feet vs. L = 1000 feet:
Pcrit1000/Pcrit10 = 0.0001
•Buckling under hydrostatic Pressure
56. Pipe in Horizontal Pipe BucklingPipe in Horizontal Pipe Buckling
DiagramDiagram
•
57. Far End vs. Input Load withFar End vs. Input Load with
BucklingBuckling
•
60. ImpactImpact
• Dynamic loading
• Impact – Chapter 4
• Fatigue – Chapter 6
• Shock loading = sudden loading
• Examples?
• 3 categories
• Rapidly moving loads of constant magnitude
• Driving over a bridge
• Suddenly applied loads
• Explosion, combustion
• Direct impact
• Pile driver, jack hammer, auto crash
Increasing
Severity
61. Impact, cont.Impact, cont.
• It is difficult to define the time rates of load application
• Leads to use of empirically determined stress impact factors
• If τ is time constant of the system, where
• We can define the load type by the time required to apply the load (tAL
= time required to apply the load)
• Static
• “Gray area”
• Dynamic
k
m
πτ 2=
τ3>ALt
ττ 3
2
1
<< ALt
τ
2
1
<ALt
62. Stress and deflection due to impactStress and deflection due to impact
• W – freely falling mass
• k – structure with stiffness (usually large)
• Assumptions
• Mass of structure is negligible
• Deflections within the mass are negligible
• Damping is negligible
• Equations are only a GUIDE
• h is height of freely falling mass before its release
• δ is the amount of deflection of the spring/structure
65. Energy balanceEnergy balance
• Fe is the equivalent static force necessary to
create an amount of deflection equal to δ
• Energy Balance of falling weight, W
( )
s
e
e
e
static
e
W
F
W
s
F
kF
skkW
FhW
δ
δ
δ
δ
δ
δδ
δδ
=
=
=
==
=+
2
1
++=
++=
=+
=+
s
e
s
s
s
s
h
WF
h
h
WhW
δ
δ
δδ
δ
δ
δ
δ
δ
δ
2
11
2
11
2
1
2
1
)(
2
2
66. Impact, cont.
• Sometimes we know velocity at impact rather than the
height of the fall
• An energy balance gives:
++=
++=
=
s
e
s
s
g
v
WF
g
v
ghv
δ
δ
δδ
2
2
2
11
11
2
