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Review LectureAE1108-II: Aerospace Mechanics of Materials
Aerospace Structures& Materials
Faculty of Aerospace Engineering
Dr. Calvin RansDr. Sofia Teixeira De Freitas
Analysis of an Engineering SystemOverview of Mechanics
W
M∙a
R1 R2
F
Treat as rigid body
Statics:
Dynamics:
R1 + R2 = W
F = M∙a
External analysis provides performance requirements of the system
Analysis of an Engineering System
External viewpoint• Speed, acceleration, trajectory• Weight, power, drag
3
Overview of Mechanics
Internal viewpoint• Part interaction, deformation, failures• Combustion, fuel flow, and many more
Engineering Systems are Not Rigid Blobs!
• What can happen to the car?• Suspension bottoms out• Chassis deforms and contact ground• Drive-shaft fails due to overload• Interference of moving parts• Vibration issues• Engine overheats• Lubricant issues
Overview of Mechanics
Solid Mechanics(Mechanics of Materials)
Is the structure strong enough and stiff enough?
Overview of Engineering MechanicsStatic Analysis
(ΣF = 0)Dynamic Analysis
(ΣF = ma)
External Analysis:• Particle
• Rigid BodyStatics (AE1130-I)
Internal Analysis:
• Solids
• Fluids
• Energy
Dynamics(AE1130-II)
Solid Mechanics(AE1108-II)
Physics I, Power & Propulsion(AE1240, AE2203)
Aerodynamics(AE1110, AE2130)
Vibrations(AE2106)
Structural Analysis(AE2135-I )
Anatomy of a Solid Mechanics Problem
Force-Displacement Relationship
Applied Loads
Geometry
Reaction & Internal Loads
Strain(Hooke’s Law)Stress Deformation
MaterialProperties
Compatibility
Performance Requirements
FBD & Equilibrium
Statically Determinate
Statically Indeterminate
Stress, Strain, & Hooke’s Law
• Stress is defined as the intensity of a force
Definition of Stress
0lim z
z A
FA
0
0
lim
lim
xzx A
yzy A
FAFA
Shear Stress:
Normal Stress:
Stress, Strain, & Hooke’s Law
• Strain is defined as the intensity of a deformation
Definition of Strain
B
A
n
π 2
Undeformed body
C
t
A’
B’
Deformed body
C’
θ’
along n along t
lim2nt B A
C A
along nlimn B A
B A BABA
Shear Strain:
Normal Strain:
Stress, Strain, & Hooke’s LawGeneralized Hooke’s Law
yx zx
y x zy
yxzz
TE E E
TE E E
TE E E
xyxy
xzxz
yzyz
G
G
G
Anatomy of a Solid Mechanics Problem
Force-Displacement Relationship
Applied Loads
Geometry
Reaction & Internal Loads
Strain(Hooke’s Law)Stress Deformation
MaterialProperties
Compatibility
Performance Requirements
FBD & Equilibrium
Statically Determinate
Statically Indeterminate
Developing a Force-Displacement Relation
Visualize Deformation
z
y
z
y
Compatible Strain
Hooke’s Law σ
Displacement
Force
Force-Stress-Displacement RelationsAxial
Torsion
Transverse Shear
Bending Moment
PA
PP
T
T
MM
VV
TRJ
MyI
VQIt
PLEA
TLGJ
2
2
d vEI Mdz
Stress Deformation
Negligible for long beams
(moment deformation dominates)
T2tAm
20
14
mL
m
TL dsA G t
(thin-walled)
(circular)
Sign Convention
PP
T
T
MM
VV
Axial
Torsion
Transverse Shear
Bending Moment
- + V
V
V
V
- +
M M
M M
+ - TT
TT
+ -PP
PP
Sign Convention
• The y-axis points downwards such that a positive moment results in positive (tensile) stress for positive y values
• Beam deflections are positive in the positive y-direction• Positive distributed loads result in positive deflections
Beam Coordinate System
N.A.
y
x
y
z++M +M
+V +V
+w
+v
NOTE: The text book defines y and +v as upwards!
Section Properties
d
Ox
y
b
h
b/2
h/2
O
y
x
Solid Rectangular Section Solid Circular Section Thin-walled Circular Section
3
12xbhI
4
64xdI
4
32dJ
3
8xtdI
3
4tdJ
x •
x'
A
d 2
x xI I A d Parallel axis theorem
d
Ox
y
t
Beam Deflections
2
2
d vEI Mdz
Need to Integrate the Moment-Curvature Relationship
nnf x x a
MEI
MEI
Direct Integration Express M using Macaulay Step Functions
Divide Problem using Superposition
Use standard/handbook solutions
A B
q
L
EIx
A BL
EIx P
A BL
EIx
M
Deflection (at B) Slope (at B)
4
8BqLvEI
3
6BqLEI
3
3BPLvEI
2
2BPLEI
2
2BMLv
EI
Cantilever Beam Standard Solutions
BMLEI
z
z
z
A B
q
L
EIx
A B
L/2
EIx P
A B
EIx
M
Deflection Slope
45384C
qLvEI
3
, 24A BqL
EI
3
48CPLv
EI
2
, 16A BPL
EI
2
8CMLv
EI
Simply Supported Beam Standard Solutions
0C
C
C
C
L/2
L/2L/2
M
z
z
z
Anatomy of a Solid Mechanics Problem
Force-Displacement Relationship
Applied Loads
Geometry
Reaction & Internal Loads
Strain(Hooke’s Law)Stress Deformation
MaterialProperties
Compatibility
Performance Requirements
FBD & Equilibrium
Statically Determinate
Statically Indeterminate
• Compatibility is needed for statically indeterminate problems• Too many supports/reaction forces to be determined by equilibrium
Compatibility
P1P2
P1
Statically Determinate Statically Indeterminate
P1P2
RP1RA
RB
FBD FBD
We remove support in FBD, but we need to maintain its constraint!
CompatibilitySome Examples
P1A BC
A B
C
P
P
w
A B
0AB 0AC CB 0Bv
A
P
B
VB
HBMB
B
C
P
VB HB
MB
= +
0B BCABv
0B ABBCv
B BAB BC
CompatibilityReal World Examples
CompatibilityReal World Examples
Superposition
P
w(x)
w(x)
P
Difficult problem Multiple simple problems
1 2
Anatomy of a Solid Mechanics Problem
Force-Displacement Relationship
Applied Loads
Geometry
Reaction & Internal Loads
Strain(Hooke’s Law)Stress Deformation
MaterialProperties
Compatibility
Performance Requirements
FBD & Equilibrium
Statically Determinate
Statically Indeterminate
Stress Transformations
• Complex stress state including many components of shear and normal stress possible
• Stress state at a point depends on orientation
Towards Failure Analysis
σx
τxy
σy
σx′
τx′y′
x′y′
θ
Stress Transformations
• Orthogonal set of planes exist at every point where shear stress is zero
• Principle stress planes• Max and min normal stress• Known as Principle Stresses
• Plane of maximum shear stress
• Inclined 45° from principle stress plane
• Normal stresses can be non-zero
Towards Failure Analysis
Principle stresses Maximum shear stress
Stress Transformations
• Stress transformations can be described by equation of a circle
• Mohr Circle
Towards Failure Analysis
Oσ
τ
C
σave
τxyσx x
τxy
σy
y
τyx
σx
A
σy
B
R
2 2 2ave R
2x y
ave
AB 2 2
x ave xyR
2θ
θ
(ccw)
(cw)
Stress Transformations
Why do we care about stress transformations?
Brittle failureDuctile failure
- Dictated by maximum shear stresses
- Dictated by maximum normal stresses
Towards Failure Analysis
Tresca Yield Criterion
σ1
σ2+σyield
+σyield
-σyield
-σyield
O σ
τ
σ1
σ2
max 2yield
Oσ
τ
σ1
σ2
O
σ
τ
σ1
σ2
Oσ
τ
σ1
σ2
Anatomy of a Solid Mechanics Problem
Force-Displacement Relationship
Applied Loads
Geometry
Reaction & Internal Loads
Strain(Hooke’s Law)Stress Deformation
MaterialProperties
Compatibility
Performance Requirements
FBD & Equilibrium
Statically Determinate
Statically Indeterminate
Solving Solid Mechanics Problems
1. Draw FBD! • Establish sign convention
2. Equilibrium Equations• Determine if statically determinate or indeterminate
3. Compatibility Conditions4. Force-Displacement Relations
• If force-displacement relation is unknown, use Hooke’s Law to relate stress and strain
5. Solve• Desired reaction forces• Desired internal stresses• Desired displacement
General Procedure
Solving Solid Mechanics Problems
• Problem may have a design element• Maximum load structure can carry• Minimum span, height, or other geometrical parameter
• Design element centres around structural requirement• Maximum allowable stress• Maximum allowable deflection• Principle stresses from Mohr Circle• Failure criteria
General Procedure
Solving Solid Mechanics Problems
• Describe your understanding of the problem in words first
• Is the problem statically determinate or indeterminate?• What is the compatibility condition?• Is there any trick to the problem?• How does the design constraint affect the problem?
• Always look at your final numerical answer and reiterate the meaning of the sign
• Elongation or contraction?• Tension or compression?• Clockwise or counter clockwise rotation?
• Does the answer make sense, and if not, describe where you think you went wrong
Recommendations
Exam Hints
• There is always a Mohr Circle question!• Difficult questions have a twist on a concept or
condition you have analysed before: identify it and describe it
Past Exam Questions
Two identical thin walled beams (AB & BC) are connected at B via a hinge. At A and at C the beams are clamped. Beam AB is loaded in bending by the load qAB and beam BC is loaded in bending by the load qBC.
Assume:(EI)AB=(EI)BC=EILAB=LBC=L
a) What is the deflection of point B when qAB=qBC=q?b) What is the deflection of point B when qAB=2qBC=2q?
A B C
qAB qBC
exam question (1)
What is the twist? Hinge mid-span. Hinges can carry shear, but not moments, therefore the moment is zero midspan.Problem is both statically determinate and indeterminate
a) What is the deflection of point B when qAB=qBC=q?
(I) FBD:
A B
FB
q
CB
FB
q
(Reactions at fixed boundary conditions not shown)
Problem is symmetric!
0BF
(since FBD must be symmetric, and FB must be equal and opposite on
opposing faces)
Since we only want displacement, we can recognize the standard case and skip directly to solving for displacement:
Standard Solution:4
3
8
6
B
B
qLvEI
qLEI
4
8BqLvEI
Deflection is downwards, which is logical for a distributed load acting downwards
(IV) Force-Displacement:
z
b) What is the deflection of point B when qAB=2qBC=2q?
(I) FBD:
A B
FB
2q
CB
FB
q
(Reactions a fixed boundary conditions not shown)
Problem is not symmetric!
0BF
Standard Solutions: 4
3
8
6
B
B
qLvEI
qLEI
3 34 428 3 8 3
B BF L F LqL qLEI EI EI EI
Deflection is downwards, which is logical for a distributed load acting downwards
Problem is actually statically indeterminate!
(III) Compatibility: B BAB BCv v Beam deflection at B must be the same for both beams
(IV) Force-Displacement:
3
2
3
2
B
B
PLvEI
PLEI
316BF qL
4
34 4 43
2 2 3168 3 8 3 16
BB AB
qLF LqL qL qLvEI EI EI EI EI
z
z
exam question(2)Consider the Airbus A380 aircraft:• The engines of the A380 are suspended
below the wing, having a distance of 2.5 m between the centre of the wing box and the engine thrust line. This 2.5 m distance results in the engines imparting a torsional moment into the wingbox. Each of the A380 engines can generate approximately 350kN of thrust. The inboard engine is located 8m away from the wing root and the outboard engine is located 16m from the wing root.
• The wingbox is approximated as a constant cross section (figure above) with a fixed support condition at the wing-to-fuselage connection.
a. Determine the maximum shear stress in the wingbox at the wing root (A) and the angle of twist relative to point A at each engine location in the wing (B and C) DUE TO THE TORQUE OF THE ENGINES ONLY. (G = 26 GPa)
b. Part b is later (too much text for here!)c. Part c is later (too much text for here!)
A
B
8m16m
30m
C
D2.5m1m
3m
t = 1cm
y
What is the twist? Twist is in part b, will get to it later
a. Determine the maximum shear stress in the wingbox at the wing root (A) and the angle of twist relative to point A at each engine location in the wing (B and C) DUE TO THE TORQUE OF THE ENGINES ONLY. (G = 26 GPa)
(I) FBD:
A B C D
TA875kNm
Engine torque = 350kN x 2.5m = 875kNm875kNm
(II) Equilibrium: 875 875 1750AT T kNm Statically determinate!
(IV) Solve: Shear Stress
Thin walled torque box:2 m
TtA
23 1 3
1 0.011750
mA m m mt cm mT kNm
@A:
2
1750 29.22 0.01 3A
kNm MPam m
a. Determine the maximum shear stress in the wingbox at the wing root (A) and the angle of twist relative to point A at each engine location in the wing (B and C) DUE TO THE TORQUE OF THE ENGINES ONLY. (G = 26 GPa)
(IV) Solve: Angle of Twist
Thin walled torque box:
(I) FBD:
A B C D
TA875kNm 875kNm
T-line
0CDT kNm
875BCT kNm
1750ABT kNm
24 m
TL dsGA t
23mA m
)/(80001.032
01.012 mm
tds
3
9
875 10 8 800 0.01795 1.02834 26 10 9C B rad degrees
3
9
1750 10 8 800 0.01197 0.68564 26 10 9B rad degrees
b. The expression for the deflection of the wing due to distributed lift over the wing and engine weight is given below. Based on that expression (where x is distance from the wing root in meters, and force units in the expression are given in kN, and positive v is upwards in the direction of lift), plot the shear force diagram for the wing and the distributed load diagram (do not plot the bending moment diagram!)
4 3 2
4 3 2
5
55 1205for 0 8: 85660 12 355 1235for 8 16: 87100 1920 512012 3110 3868444for 16 30: 30 2224451680 3
x EIv x x x
x EIv x x x x
x EIv x x
A
B
8m16m
30m
C
D2.5m1m
3m
t = 1cm
y
( )( )
EIv w xEIv V xEIv M x
for v for v
( )( )
EIv w xEIv V xEIv M x
What is the twist? You can differentiate deflection to get loading
A
B
8m16m
30m
C
D2.5m1m
3m
t = 1cmy
4 3 255 1205for 0 8: 85660 12 3
x EIv x x x
3 2
2
55' 1205 1713203
'' 55 2410 171320''' 110 2410 ( )'''' 110 ( )
EIv x x x
EIv x xEIv x V xEIv w x
V -linex kN‐1530
‐2410
w -linex
‐110
kN/m
b. Plot the shear force diagram for the wing and the distributed load diagram
A
B
8m16m
30m
C
D2.5m1m
3m
t = 1cmy
b. Plot the shear force diagram for the wing and the distributed load diagram
4 3 255 1235for 8 16: 87100 1920 512012 3
x EIv x x x x
3 2
2
55' 1235 174200 19203
'' 55 2470 174200''' 110 2470 ( )'''' 110 ( )
EIv x x x
EIv x xEIv x V xEIv w x
V -linex ‐710
kN
‐1590
‐1530
‐2410
w -linex
‐110
kN/m
A
B
8m16m
30m
C
D2.5m1m
3m
t = 1cmy
b. Plot the shear force diagram for the wing and the distributed load diagram
V -linex ‐710
kN
-770‐1590
‐1530
‐2410
w -linex
‐110
kN/m
5110 3868444for 16 30: 30 2224451680 3
x EIv x x
4
3
2
110' 1 (30 ) 222445336
110'' (30 )84
55''' 1 (30 ) ( )14
55'''' (30 ) ( )7
EIv x
EIv x
EIv x V x
EIv x w x
c. Based on your previous analysis, determine what the weight of each engine is (Point loads at B and C in kN) and determine the total lift produced by the wing (resultant of distributed loads)
A
B
8m16m
30m
C
D2.5m1m
3m
t = 1cmy
Engine weight : -1530 - (-1590) = 60kN
Lift : (-110)(8) + (-110)(8) + (0.5)(-110)(14) = -2530 kN
V -linex ‐710
kN
-770‐1590
‐1530
‐2410
w -linex
‐110
kN/m
(positive is downwards, makes sense!)
(negative is upwards, makes sense!)
(total thrust for one wing is (2)(350kN) = 700 kN, thus T/W < 1, which makes sense!)
exam question(3)
A steel I-beam is to be reinforced by bonding platesof the same material to the top and bottom flanges as shown in the figure to the right. Given the above allowables:a) Determine the maximum bending moment that can be carried by the
unreinforced beam.
b) Determine the maximum shear force that can be carried by the unreinforcedbeam.
c) Determine the maximum bending moment that can be carried by the reinforced beam.
d) Determine the maximum shear force that can be carried by the reinforcedbeam.
What is the twist?For reinforced beam, two conditions need to be checked for max shear force (part d): Transverse shear in steel I-beam and shear along adhesive bondline
max
max
max
150
80
5
steel
steel
adhesive
MPa
MPa
MPa
A steel I-beam is to be reinforced by bonding plates of the same material to the top and bottom flanges as shown in the figure.a) Determine the maximum bending moment that can be carried by the unreinforced beam.
3 32 4 420 160 (2101 12· 160 20 15 400 1012 1
) 3.6252
mm mm mI mm mm m mm mm m
4 4150 3.625 10, 247.2
0.220My Iso kN
MPa mM m
I y m
max
max
max
150
80
5
steel
steel
adhesive
MPa
MPa
MPa
A steel I-beam is to be reinforced by bonding plates of the same material to the top and bottom flanges as shown in the figure.
b) Determine the maximum shear force that can be carried by the unreinforced beam.
max
max
max
150
80
5
steel
steel
adhesive
MPa
MPa
MPa
VQ IbsoVIb Q
4 3(200 10 )·(160 20 ) (100 )·(200 15 ) 9.72 10Q mm mm mm mm mm mm mm m
4 4
4 3
·3.625 10 ·0.0159.72
81
0 4470
.5maxMPa m k
mm NV
4 43.625 10I mMaximal shear is at the middle.
A steel I-beam is to be reinforced by bonding plates of the same material to the top and bottom flanges as shown in the figure.
c) Determine the maximum bending moment that can be carried by the reinforced beam.
max
max
max
150
80
5
steel
steel
adhesive
MPa
MPa
MPa
24 3 4 4153.625 10 2 ·(15 ) (150 )(1 150 5.9615 ) 22 0
120 1
2reinfmmI mm mm mm mm mmm
4 4150 ·5.96 38100.235
0.4maxreinf
MPa mm
M kNm
A steel I-beam is to be reinforced by bonding plates of the same material to the top and bottom flanges as shown in the figure.
d) Determine the maximum shear force that can be carried by the reinforcedbeam.
max
max
max
150
80
5
steel
steel
adhesive
MPa
MPa
MPa
4
3
4
3
80 ·5.96 10 ·0.015 481.91.484 10maxreinf
MPa m mV kNm
4 3 30.015(0.220 ) 0.150 0.0152
9.72 10 1.484 10maxreinfQ m
4 45.96 10reinf mI IbVQ
AND THE BONDLINE?
A steel I-beam is to be reinforced by bonding plates of the same material to the top and bottom flanges as shown in the figure.
d) Determine the maximum shear force that can be carried by the reinforcedbeam.
max
max
max
150
80
5
steel
steel
adhesive
MPa
MPa
MPa
4 45.96 10reinf mI IbVQ
4 315220 · 150 5.1215 102adhmmmm mm mQ m m
4 4
4 4
5 ·5.96 10 ·5.1
0.15 802
721maxadh
MPa m mV kNm
BONDLINE!
481.9maxreinfV kN