14 Review - Lecture - TU Delft OpenCourseWare · solutions. AB q L EI x ABL EI x P ABL EI x M ... • Compatibility is needed for statically indeterminate problems ... of a circle

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



Applied Loads

Geometry



MaterialProperties

Compatibility


FBD & Equilibrium



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



Applied Loads

Geometry



MaterialProperties

Compatibility


FBD & Equilibrium



• 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



Applied Loads

Geometry



MaterialProperties

Compatibility


FBD & Equilibrium



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


• 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


Principle stresses Maximum shear stress


• Stress transformations can be described by equation of a circle

• Mohr Circle


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)


Why do we care about stress transformations?

Brittle failureDuctile failure

- Dictated by maximum shear stresses

- Dictated by maximum normal stresses


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



Applied Loads

Geometry



MaterialProperties

Compatibility


FBD & Equilibrium



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


• 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


• 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


(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


(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


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


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.


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



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?



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

Documents

14 Review - Lecture - TU Delft OpenCourseWare · solutions. AB q L EI x ABL EI x P ABL EI x M ... • Compatibility is needed for statically indeterminate problems ... of a circle