Beam Deflection Review
(4.3-4.5)
MAE 316 – Strength of Mechanical Components
Y. Zhu
1
Beam Deflection Review
Deflection Due to Bending (4.3)
ds
dy
θ
dx
y
d  ds
Slopeofthedeflectioncurve
d 1
  curvature of beam
ds 
2
Beam Deflection Review
dy
 tan 
dx
dx
 cos 
ds
dy
 sin 
ds
Deflection Due to Bending (4.3)
Assumption1:θ issmall.

ds  dx  d  1  d
ds  dx

1.

dy
1 d2y
d d 2 y
 tan    
 2
 2 
2.
dx
 dx
dx dx
Assumption2:Beamislinearlyelastic.

1


M
EI
Thus,thedifferentialequationforthedeflectioncurveis:

d2y M

2
dx
EI
3
Beam Deflection Review
Deflection Due to Bending (4.3)

Recall:

Sowecanwrite:

Deflectioncurvecanbefoundbyintegrating



dV
 w
dx
dM
V
dx
d2y
EI 2  M
dx
d3y
EI 3  V
dx
d4y
EI 4   w
dx
Bendingmomentequation 2 constantsofintegration
Shear‐forceequation 3 constantsofintegration
Loadequation 4 constantsofintegration
Chosenmethoddependsonwhichismoreconvenient.

4
Beam Deflection Review
Method of Superposition (4.5)

Deflectionandslopeofabeamproducedbymultipleloadsacting
simultaneouslycanbefoundbysuperposing thedeflections
producedbythesameloadsactingseparately.

ReferenceAppendixA‐9 BeamDeflectionsandSlopes

Methodofsuperpositioncanbeappliedtostaticallydeterminate
andstaticallyindeterminatebeams.
5
Beam Deflection Review
Method of Superposition (4.5)

Considerthefollowingexample:

FindreactionsatAandC.

Method1:ChooseMC andRC as
redundant.

Method2:ChooseMC andMA asredundant.
6
Beam Deflection Review
Castigliano’s Theorem
(4.7-4.10)
MAE 316 – Strength of Mechanical Components
NC State University Department of Mechanical and Aerospace Engineering
7
Castigliano’s Theorem
Castigliano’s Theorem (4.8)

This method is a powerful new way to determine
deflections in many types of structures – bars, beams,
frames, trusses, curved beams, etc.

We can calculate both horizontal and vertical
displacements and rotations (slopes).

There are actually two Castigliano’s Theorems.


8
The first can be used for structures made of both linear and
non-linear elastic materials.
The second is restricted to structures made of linear elastic
materials only. This is the one we will use.
Castigliano’s Theorem
Castigliano’s Theorem (4.8)
“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 w.r.t. that force.”

U
i 
Fi
Where:
Fi = Force at i-th application point
δi = Displacement at i-th point in the direction of Fi
U = Total strain energy
9
Castigliano’s Theorem
Castigliano’s Theorem (4.8)

We can also use this method to find the angle of rotation (θ).
U
i 
M i
Where:
Mi = Moment at i-th application point
θi = Slope at i-th point resulting from Mi
U = Total strain energy
10
Castigliano’s Theorem
Castigliano’s Theorem (4.8)

General case
U
1 
F1
δ1
F1
F2 δ2
F3 δ3
U
3 
F3
U stored in
structure
Fn
δn
11
U
2 
F2
U
n 
Fn
Castigliano’s Theorem
Strain Energy in Common Members (4.7)

Spring
k
F
δ
1 2
U  k
2
Note:
F  k    F / k
1 F
1 F 2    U  F

Check: U  k   
F k
2 k
2 k
2
12
Castigliano’s Theorem
Strain Energy in Common Members (4.7)

Bar subject to axial load
A,E
F
F
L
L
1 F 2L
F ( x)2
U
or U  
dx
2 AE
2 A( x) E ( x)
0
13
Castigliano’s Theorem
Strain Energy in Common Members (4.7)

Shaft subject to torque
J,G
T
T
L
L
1 T 2L
T ( x)2
U
or U  
dx
2 GJ
2G ( x) J ( x)
0
14
Castigliano’s Theorem
Strain Energy in Common Members (4.7)

Beam subject to bending
I,E
M
M
L
L
1 M 2L
M 2 ( x)
U
or U  
dx
2 EI
2 E ( x) I ( x)
0
15
Castigliano’s Theorem
Strain Energy in Common Members (4.7)

Beam in direct shear
2
L
F L
F 2 ( x)
U
or U  
dx
2 AG
2 A( x)G ( x)
0
16
Castigliano’s Theorem
Strain Energy in Common Members (4.7)

Beam in transverse shear
I,E
V
V
L
L
CV 2 ( x)
U 
dx
2 A( x)G ( x)
0
Correction factor for transverse shear
(see table 4-1 in textbook)
17
Castigliano’s Theorem
Strain Energy in Common Members (4.7)

For structures with combined loading (or multicomponent structures) add up contributions to U.
F 2 ( x)
T 2 ( x)
M 2 ( x)
U 
dx  
dx  
dx 
2 A( x) E ( x)
2G ( x) J ( x)
2 E ( x) I ( x)
L
F 2 ( x)
CV 2 ( x)
0 2 A( x)G( x) dx   2 A( x)G( x) dx
18
Castigliano’s Theorem
Castigliano’s Theorem - Frame (4.8)

For the structure and loading shown below, determine the vertical
deflection at point B. Neglect axial force in the column.
P
L2
B
E, I
L1
A
19
Castigliano’s Theorem
Castigliano’s Theorem - Frame (4.8)

For the structure and loading shown below, determine the vertical
and horizontal deflection at point B. Neglect axial force in the
column.
L2
w
B
E, I
L1
A
20
Castigliano’s Theorem
Castigliano’s Theorem – Curved Beam (4.9)

For the structure and loading shown below, determine the vertical
and horizontal deflection at point B. Consider the effects of bending
only.
Fv
E, I
B
R
A
21
Castigliano’s Theorem
Fh
Castigliano’s Theorem - Trusses

For the structure and loading shown below, determine the vertical
deflection at D and horizontal deflection at C. Let L = 16 ft, h = 6 ft,
E = 30 x 103 ksi, P = 18 kips,Atens = 2.5 in2, and Acomp = 5 in2.
B
h
A
C
D
L/2
22
P
L/2
Castigliano’s Theorem
Statically Indeterminate Problems (4.10)

For the structure and loading shown below, find the
fixed end reactions.
w
A
B
L
23
Castigliano’s Theorem
Statically Indeterminate Problems (4.10)

A curved frame ABC is fixed at one end, hinged at
another, and subjected to a concentrated load P, as
shown in the figure below. What are the horizontal H
and vertical F reactions? Consider bending only.
24
Castigliano’s Theorem
Special Cases: Hollow Tapered Beam

Find the tip deflection for the structure and loading
shown below.
P
t
dA
dB= 2dA
x
L
25
Castigliano’s Theorem
Special Cases: Beam With Spring

For the beam-spring system below, find the deflection
at end C.
P
E, I
B
A
C
k
x
L
26
a
Castigliano’s Theorem