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Beam Deflection Review (4.3-4.5)
MAE 316 – Strength of Mechanical ComponentsNC State University Department of Mechanical and Aerospace Engineering
Beam Deflection Review1
Beam Deflection Review2
Deflection Due to Bending (4.3)
ydx
dy
ds
θ
tandx
dy
cosds
dx
sinds
dydsd
1dcurvature of beam
ds
Slope of the deflection curve
Beam Deflection Review3
Deflection Due to Bending (4.3)
Assumption 1: θ is small. 1. 2.
Assumption 2: Beam is linearly elastic. Thus, the differential equation for the deflection curve is:
dxdsdx
d
ds
d
1
tandx
dy
2
2
dx
yd
dx
d2
21
dx
yd
EI
M
1
EI
M
dx
yd
2
2
Beam Deflection Review4
Deflection Due to Bending (4.3)
Recall: So we can write: Deflection curve can be found by integrating
Bending moment equation (2 constants of integration) Shear-force equation (3 constants of integration) Load equation (4 constants of integration)
Chosen method depends on which is more convenient.
wdx
dV V
dx
dM
wdx
ydEI
4
4
Vdx
ydEI
3
3
Mdx
ydEI
2
2
Beam Deflection Review5
Method of Superposition (4.5)
Deflection and slope of a beam produced by multiple loads acting simultaneously can be found by superposing the deflections produced by the same loads acting separately. Reference Appendix A-9 (Beam Deflections and Slopes) Method of superposition can be applied to statically determinate and statically indeterminate beams.
Consider the following example: Find reactions at A and C. Method 1: Choose MC and RC asredundant.
Method 2: Choose MC and MA as redundant.Beam Deflection Review6
Method of Superposition (4.5)
Beam Deflection: Method of Superposition
7
Example Problem
For the beam and loading shown, determine (a) the deflection at C, and (b) the slope at end A.
8
Example Problem
Loading I: Case 6 in Appendix D.
4 35
384 24C AwL wL
yEI EI
Loading II: Case 7 in Appendix D.
3 2
2 1
6 2 2 16 3A A A
C AM M L M LL L
y LEIL EI EI
with 2 4 31 1
12 192 36A C AwL wL wL
M yEI EI
Loading III: Case 7 in Appendix D.
31(using Loading II result)
16
6
BC
BA
M Ly
EIM L
EI
with 2 4 31 1
12 192 72B C AwL wL wL
M yEI EI
(a) Deflection at C.
4 4 4 45 1 1 1
384 192 192 384CwL wL wL wL
yEI EI EI EI
41
384CwL
yEI
(b) Slope at A.
3 3 31 1 1
024 36 72A
wL wL wL
EI EI EI
0A
Beam Deflection: Method of Superposition
9
Example Problem
For the beam shown, determine the reaction at B.
10
Example Problem
Beam is second degree indeterminate. Choose BR and BM as redundant reactions.
Loading I: Case 1 of Appendix D.
3 2
I I( ) , ( )3 2B B
B BR L R L
yEI EI
Loading II: Case 3 of Appendix D.
2
II II( ) , ( )2B B
B BM L M L
yEI EI
Loading III: Case 3 applied to Portion AC.
2 20 0
III
0 0III
( /2)( )
2 8( /2)
( )2
C
C
M L M Ly
EI EIM L M L
EI EI
Portion CB remains straight.
20
III III III
20
III III
3( ) ( ) ( )
2 8
1( ) ( )
2
B C C
B C
M LLy y
EI
M L
EI
Super position and constraint:
I II III
23 20
( ) ( ) ( ) 0
30
3 2 8
B B B B
B B
y y y y
M LL LR M
EI EI EI
(1)
I II III
20
( ) ( ) ( ) 0
10
2 2
B B B B
B B
M LL LR M
EI EI EI
(2)
Solving (1) and (2) simultaneously,
03
2B
MR
L
01
4BM M
Castigliano’s Theorem(4.7-4.10)
MAE 316 – Strength of Mechanical ComponentsNC State University Department of Mechanical and Aerospace Engineering
Castigliano’s Theorem11
Castigliano’s Theorem (4.8)
Castigliano’s Theorem12
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. 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 (4.8)
Castigliano’s Theorem13
“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.”
ii F
U
Where:Fi = Force at i-th application pointδi = Displacement at i-th point in the direction of Fi
U = Total strain energy
Castigliano’s Theorem (4.8)
Castigliano’s Theorem14
We can also use this method to find the angle of rotation (θ).
Where:Mi = Moment at i-th application pointθi = Slope at i-th point resulting from Mi
U = Total strain energy
ii M
U
Castigliano’s Theorem (4.8)
Castigliano’s Theorem15
General case
U stored in structure
F1
F2
F3
Fn
δ1
δ2
δ3
δn
11 F
U
22 F
U
33 F
U
nn F
U
Strain Energy in Common Members (4.7)
Castigliano’s Theorem16
Spring
kFδ
2
2
1 kU
kFkF /
k
F
k
FkU
22
2
1
2
1
k
F
F
U
Note:
Check:
Strain Energy in Common Members (4.7)
Castigliano’s Theorem17
Bar subject to axial load
2 2
0
1 ( )
2 2 ( ) ( )
LF L F xU or U dx
AE A x E x
F F
L
A,E
Strain Energy in Common Members (4.7)
Castigliano’s Theorem18
Shaft subject to torque
2 2
0
1 ( )
2 2 ( ) ( )
LT L T xU or U dx
GJ G x J x
L
J,G
T T
Beam subject to bending
Strain Energy in Common Members (4.7)
Castigliano’s Theorem19
2 2
0
1 ( )
2 2 ( ) ( )
LM L M xU or U dx
EI E x I x
L
I,E
M M
Beam in direct shear
Strain Energy in Common Members (4.7)
Castigliano’s Theorem20
2 2
0
( )
2 2 ( ) ( )
LF L F xU or U dx
AG A x G x
Beam in transverse shear
Strain Energy in Common Members (4.7)
Castigliano’s Theorem21
2
0
( )
2 ( ) ( )
L CV xU dx
A x G x
Correction factor for transverse shear (see table 4-1 in textbook)
L
I,E
V V
For structures with combined loading (or multi-component structures) add up contributions to U.
Strain Energy in Common Members (4.7)
Castigliano’s Theorem22
2 2 2
2 2
0
( ) ( ) ( )
2 ( ) ( ) 2 ( ) ( ) 2 ( ) ( )
( ) ( )
2 ( ) ( ) 2 ( ) ( )
L
F x T x M xU dx dx dx
A x E x G x J x E x I x
F x CV xdx dx
A x G x A x G x
For the structure and loading shown below, determine the vertical deflection at point B. Neglect axial force in the column.
Castigliano’s Theorem - Frame (4.8)
Castigliano’s Theorem23
P
B
A
L1
L2
E, I
For the structure and loading shown below, determine the vertical and horizontal deflection at point B. Neglect axial force in the column.
Castigliano’s Theorem - Frame (4.8)
Castigliano’s Theorem24
w
B
A
L1
L2
E, I
For the structure and loading shown below, determine the vertical and horizontal deflection at point B. Consider the effects of bending only.
Castigliano’s Theorem – Curved Beam (4.9)
Castigliano’s Theorem25
B
A
E, I
Fv
Fh
R
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.
Castigliano’s Theorem - Trusses
Castigliano’s Theorem26
P
A C
B
D
L/2L/2
h
For the structure and loading shown below, find the fixed end reactions.
Statically Indeterminate Problems (4.10)
Castigliano’s Theorem27
L
A B
w
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.
Statically Indeterminate Problems (4.10)
Castigliano’s Theorem28
Find the tip deflection for the structure and loading shown below.
Special Cases: Hollow Tapered Beam
Castigliano’s Theorem29
P
dA dB= 2dA
t
L
x
For the beam-spring system below, find the deflection at end C.
Special Cases: Beam With Spring
Castigliano’s Theorem30
L
P
AC
B
k
a
x
E, I