Review Lecture. AE1108-II: Aerospace Mechanics of Materials. Dr. Calvin Rans Dr. Sofia Teixeira De Freitas

Review Lecture AE1108-II: Aerospace Mechanics of Materials Dr. Calvin Rans Dr. Sofia Teixeira De Freitas Aerospace Structures & Materials Faculty of Aerospace Engineering

Analysis of an Engineering System Overview of Mechanics Treat as rigid body M a F W R 1 R 2 Statics: Dynamics: R 1 + R 2 = W F = M a External analysis provides performance requirements of the system

Analysis of an Engineering System Overview of Mechanics External viewpoint Speed, acceleration, trajectory Weight, power, drag Internal viewpoint Part interaction, deformation, failures Combustion, fuel flow, and many more 3

Engineering Systems are Not Rigid lobs! Overview of Mechanics 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 Solid Mechanics (Mechanics of Materials) Is the structure strong enough and stiff enough?

Overview of Engineering Mechanics External Analysis: Particle Rigid ody Static Analysis (ΣF = 0) Statics (AE1130-I) Dynamic Analysis (ΣF = ma) Dynamics (AE1130-II) Internal Analysis: Solids Solid Mechanics (AE1108-II) Structural Analysis (AE2135-I ) Vibrations (AE2106) Fluids Aerodynamics (AE1110, AE2130) Energy Physics I, Power & Propulsion (AE1240, AE2203)

Anatomy of a Solid Mechanics Problem Compatibility Applied Loads Force-Displacement Relationship FD & Equilibrium Reaction & Internal Loads Stress Strain (Hooke s Law) Deformation Geometry Material Properties Statically Determinate Statically Indeterminate Performance Requirements

Stress, Strain, & Hooke s Law Definition of Stress Stress is defined as the intensity of a force Normal Stress: F lim z z A0 A Shear Stress: zx zy lim A0 lim A0 F x A F y A

Stress, Strain, & Hooke s Law Definition of Strain Strain is defined as the intensity of a deformation Normal Strain: n lim A along n A A A t C A π 2 Undeformed body θ n Shear Strain: nt 2 A CA lim along n along t C A Deformed body

Stress, Strain, & Hooke s Law Generalized Hooke s Law T E E E T E E E T E E E xy xy G xz xz G yz yz G x y z x y x z y z x y z

Developing a Force-Displacement Relation z Displacement y Visualize Deformation Compatible Strain Hooke s Law σ z Force y

Force-Stress-Displacement Relations P P Axial Stress P A Deformation PL EA T M M T Torsion (circular) (thin-walled) ending Moment TR J My I EI TL GJ T TL 2tA 2 m 4A G 2 dv 2 dz m L m 0 M 1 ds t V V Transverse Shear VQ It Negligible for long beams (moment deformation dominates)

Sign Convention P P Axial P + - P P P T T Torsion T + - T T T M M ending Moment M + M - M M V V Transverse Shear V + V V - V

Sign Convention eam Coordinate System The y-axis points downwards such that a positive moment results in positive (tensile) stress for positive y values eam deflections are positive in the positive y-direction Positive distributed loads result in positive deflections +w N.A. z x +M + +M y +v y +V +V NOTE: The text book defines y and +v as upwards!

Section Properties Solid Rectangular Section Solid Circular Section Thin-walled Circular Section b d d x h/2 O h x O x O t b/2 y y y I x bh 12 3 I x d 64 4 J d 32 4 I x td 8 3 J td 4 3 2 I I Ad Parallel axis theorem x x d A x x'

eam Deflections Need to Integrate the Moment-Curvature Relationship EI 2 dv dz 2 M Direct Integration M EI Express M using Macaulay Step Functions n n f x x a M EI Divide Problem using Superposition Use standard/handbook solutions

Cantilever eam Standard Solutions xz q Deflection (at ) Slope (at ) A L EI v 4 ql 8EI 3 ql 6EI A xz L P EI v 3 PL 3EI 2 PL 2EI A xz L EI M v 2 ML 2EI ML EI

A q L EI x A L/2 EI x P A EI x M Deflection Slope 4 5 384 C ql v EI 3, 24 A ql EI 3 48 C PL v EI 2, 16 A PL EI 2 8 C ML v EI Simply Supported eam Standard Solutions 0 C C C C L/2 L/2 L/2 M z z z

Compatibility Compatibility is needed for statically indeterminate problems Too many supports/reaction forces to be determined by equilibrium Statically Determinate Statically Indeterminate P 2 P 1 P 1 FD FD R R P 2 P 1 R A P 1 We remove support in FD, but we need to maintain its constraint!

Compatibility Some Examples w P 1 A A C 0 0 v 0 A AC C A P C P = A P v C 0 A v A 0 C A C H M V + M V C H P

Compatibility Real World Examples

Superposition Difficult problem Multiple simple problems w(x) w(x) P 1 2 P

Stress Transformations Towards Failure Analysis Complex stress state including many components of shear and normal stress possible Stress state at a point depends on orientation y x σ x σ x θ τ x y τ xy σ y

Stress Transformations Towards Failure Analysis 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 Towards Failure Analysis Stress transformations can be described by equation of a circle (cw) τ Mohr Circle 2 2 2 R ave τ yx θ y σ y A τ xy σ x x ave x 2 R 2 2 y x ave xy O τ xy σ y σ ave C R 2θ σ x A σ (ccw)

Stress Transformations Towards Failure Analysis Why do we care about stress transformations? Ductile failure rittle failure - Dictated by maximum shear stresses - Dictated by maximum normal stresses

Tresca Yield Criterion max yield 2 σ 2 σ 2 σ 1 O σ σ 2 +σ yield O σ 1 σ τ +σ yield τ σ 1 σ 1 -σ yield σ 1 σ 2 O σ -σ yield σ 2 O σ τ τ

Solving Solid Mechanics Problems General Procedure 1. Draw FD! Establish sign convention 2. Equilibrium Equations Determine if statically determinate or indeterminate 3. Compatibility Conditions 4. 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

Solving Solid Mechanics Problems 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

Solving Solid Mechanics Problems Recommendations 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

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

exam question (1) Two identical thin walled beams (A & C) are connected at via a hinge. At A and at C the beams are clamped. eam A is loaded in bending by the load q A and beam C is loaded in bending by the load q C. Assume: (EI) A =(EI) C =EI L A =L C =L A C q A q C a) What is the deflection of point when q A =q C =q? b) What is the deflection of point when q A =2q C =2q? 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 when q A =q C =q? (I) FD: (Reactions at fixed boundary conditions not shown) A C Problem is symmetric! F F F 0 q q (since FD must be symmetric, and F 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: (IV) Force-Displacement: z Standard Solution: v 4 ql 8EI 3 ql 6EI v 4 ql 8EI Deflection is downwards, which is logical for a distributed load acting downwards

b) What is the deflection of point when q A =2q C =2q? (I) FD: (Reactions a fixed boundary conditions not shown) A C Problem is not symmetric! F F F 0 2q q Problem is actually statically indeterminate! (III) Compatibility: eam deflection at must be the same for both beams v v A C (IV) Force-Displacement: 4 Standard Solutions: ql v z 8EI 3 ql 6EI z 3 PL v 3EI 2 PL 2EI v A 4 3 4 3 2qL FL ql FL 8EI 3EI 8EI 3EI 3 F ql 16 3 4 4 3 4 ql 4 2qL FL 2qL 16 3qL 8EI 3EI 8EI 3EI 16EI Deflection is downwards, which is logical for a distributed load acting downwards

exam question(2) Consider the Airbus A380 aircraft: The engines of the A380 are suspended below the wing, having a distance of A 2.5m D 2.5 m between the centre of the wing box and the engine thrust line. This y C 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 ( 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!) 8m 16m 30m 1m 3m t = 1cm 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 ( and C) DUE TO THE TORQUE OF THE ENGINES ONLY. (G = 26 GPa) (I) FD: 875kNm T A 875kNm Engine torque = 350kN x 2.5m = 875kNm A C D (II) Equilibrium: T T 875 875 1750kNm A Statically determinate! (IV) Solve: Shear Stress Thin walled torque box: T ta 2 m @A: A 3m1m3m m t 1cm0.01m T 1750kNm 2 1750kNm A 29.2MPa 20.01 3 2 m 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 ( and C) DUE TO THE TORQUE OF THE ENGINES ONLY. (G = 26 GPa) (I) FD: T A 875kNm 875kNm A C D T-line TCD TC TA 0kNm 875kNm 1750kNm (IV) Solve: Angle of Twist Thin walled torque box: TL GA 2 4 m ds t Am ds t 3m 2 1 3 2 2 800( m / m) 0.01 0.01 3 175010 8 9 800 0.01197 rad 0.6856 degrees 42610 9 3 87510 8 C 9 800 0.01795 rad 1.0283 degrees 42610 9

b. The expression for the deflection of the wing due to distributed lift over the wing and engine weight is given below. ased 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!) 55 4 1205 3 2 for 0 x8: EIv x x 85660 x 12 3 55 1235 x EIv x x x x 12 3 4 3 2 for 8 16: 87100 1920 5120 110 5 3868444 for 16 x30: EIv 30 x 222445x 1680 3 What is the twist? You can differentiate deflection to get loading for v EIv w x EIv V( x) EIv M( x) EIv for v wx EIv V ( x) EIv M ( x) A y 8m 16m 30m C 2.5m D 1m 3m t = 1cm

b. Plot the shear force diagram for the wing and the distributed load diagram 8m 16m 30m 3m A y 2.5m D 1m t = 1cm C 55 1205 for 0 x8: EIv x x 85660 x 12 3 4 3 2 55 3 2 EIv ' x 1205x 171320x 3 2 EIv '' 55x 2410x 171320 EIv ''' 110x 2410 V ( x) EIv '''' 110 w( x) x 1530 V -line kn x w -line kn/m 2410 110

b. Plot the shear force diagram for the wing and the distributed load diagram 8m 16m 30m 3m A y 2.5m D 1m t = 1cm C 55 1235 12 3 55 3 2 EIv ' x 1235x 174200x 1920 3 2 EIv '' 55x 2470x 174200 4 3 2 for 8 x16: EIv x x 87100x 1920x5120 EIv ''' 110x 2470 V ( x) EIv '''' 110 w( x) x 1530 710 V -line kn x w -line kn/m 2410 1590 110

b. Plot the shear force diagram for the wing and the distributed load diagram 8m 16m 30m 3m A y 2.5m D 1m t = 1cm x 110 5 3868444 for 16 x30: EIv 30 x 222445x 1680 3 110 4 EIv ' 1 (30 x) 222445 336 110 3 EIv '' (30 x) 84 55 2 EIv ''' 1 (30 x) V ( x) 14 55 EIv '''' (30 x) w( x) 7 V -line 710 x kn 1530-770 110 1590 C w -line kn/m 2410

c. ased on your previous analysis, determine what the weight of each engine is (Point loads at and C in kn) and determine the total lift produced by the wing (resultant of distributed loads) 8m 16m 30m 3m A y 2.5m D 1m t = 1cm C x 1530 1590 710-770 V -line kn 110 x w -line kn/m 2410 Engine weight : -1530 - (-1590) = 60kN (positive is downwards, makes sense!) Lift : (-110)(8) + (-110)(8) + (0.5)(-110)(14) = -2530 kn (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) steel steel max max adhesive 150MPa 80MPa max 5MPa 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 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 unreinforced beam. 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 reinforced beam. 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

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. 1 1 I 2 160mm20mm 20mm160 mm(210mm) 15mm400mm 3.62510 m 12 12 3 2 3 4 4 4 4 150MPa 3.62510 m My I, so M 247.2kNm I y 0.220m steel steel max max adhesive 150MPa 80MPa 5MPa max

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. VQ Ib Ib sov Q steel steel max max adhesive 150MPa 80MPa 5MPa Q(200mm10 mm) (160mm20 mm) (100 mm) (200mm15 mm) 9.7210 m max Maximal shear is at the middle. I 3.62510 m 4 4 4 3 V max 80MPa 4 4 3.62510 m 0.015 9.7210 m 4 3 m 447.5kN

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. steel steel max max adhesive 150MPa 80MPa 5MPa max 1 Ireinf 3.62510 2 150 mm (15 mm) (150 mm)( 15 mm ) 22 0mm 5.96 1 0 m 12 2 2 4 3 15mm 4 4 M reinf max 150 MPa 5.9610 0.235m m 4 4 380.4kNm

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 reinforced beam. Ib V Q Qreinf max I reinf 5.9610 m 4 4 steel steel max max adhesive 150MPa 80MPa 5MPa 0.015 9.7210 (0.220 ) 0.1500.015 1.48410 2 4 3 3 max m V reinf max 4 4 80 5.9610 0.015 MPa m m 3 3 1.48410 m 481.9kN AND THE ONDLINE?

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 reinforced beam. ONDLINE! Ib V Q I reinf 5.9610 m 4 4 steel steel max max adhesive 150MPa 80MPa 5MPa 15mm Qadh 220 mm 150mm mm 5.12 m 2 max 4 3 15 10 V adh max MPa m 0.15m 4 4 5.1210 m 4 4 5 5.96 10 872kN Vreinf max 481.9kN