Engineering Tripos Part A Supervisor Version Module 3D4 Structural Analysis and Stability Handout 1 Elastic Analysis (8 Lectures) Fehmi Cirak (fc86@) Stability (8 Lectures) Allan McRobie (fam @eng)
January 008 ELASTC ANALYSS Handout by C. Burgoyne and F. Cirak Bending of asymmetric beams Biaxial bending St Venant torsion Restrained warping torsion Grillage analysis Macaulay s Method Reciprocal theorem nfluence lines Displacement method
BENDNG OF ASYMMETRC BEAMS 3 Consider channel section as cantilever Self-weight loading f beam split along middle the two halves displace more and move sideways. Failure to understand this leads to many problems.
4 Should be able to analyse asymmetric beam bending about arbitrary axes. Some practical examples: Angle Z-Purlin Crane beam Z-Purlin used as supporting beam Many beams designed to carry asymmetric loads or to fit in asymmetric spaces. New forms of construction made from folded plates often asymmetric.
Consider general cross-section: 5 Define arbitrary axes (x, y) through centroid (G). z-axis is line of centroids (out of page)
Displacements of centroid in x and y directions are u and v. 6 Define curvature about x-axis. Look at y-z plane. dv d v Gradient decreasing.! is ve dz dz Curvature " x! d u Similarly, for bending about y-axis, " y! dz (other sign conventions possible be consistent) d dz Axial strain at centroid! a v
Plane sections remain plane 7 Axial strain at any point is function of x and y! a + y." x + x." y Axial stress at point (x, y) (" +! y! x) # E + a x y. Axial force N!" da! da + " x! yda + E# xda a " y But! yda! xda 0 if bending related to centroid! and! da A so N E! a A
Bending about x-axis 8 M x "!.y da ( ) " ye # a +$ x y +$ y x da A E# a " yda + E$ x " y da + E $ y " xyda
Similarly 9 M y! xyda + E" y! E" x da x Define (Units L 4 ) xx yy xy!!! y x da da xyda ( + ve) ( + ve) ( ± ve) ' & M $ % M x y #! " & E$ % xx xy xy yy #&(! $ "%( x y #! " can easily be inverted &( $ %( x y #! " E & 1 yy xy & x # ( ) $! ' ' $ M! " xx yy xy % xy ' xx # "% M y xy term means that M x causes κ y and vice-versa. Need to be able to calculate all values of for any cross-section.
10 WARNNG!! xx & yy are always positive xy can be +ve or ve. The difference matters!! A wing or a propeller blade is an asymmetric beam under biaxial bending. Module 3D4 Handout 1B Supervisor Version
To find 4 methods: 11 1. DOUBLE NTEGRATON Used for basic shapes. BULT-UP SECTON Combines basic shapes Commonest method 3. LNE NTEGRALS Used for thin sections (folded plates bent sheet, etc.) 4. PERMETER METHOD Best for complex shapes Needs computer.
DOUBLE NTEGRATON Revision 1! f da!!. " f. dx. dy. choose to break area into strips running in either x or y direction. ntegrate along strips. Then integrate all strips. b ' $! f. da!! f. dx ". dy %& a # nner integral is along the strip with appropriate limits Outer integral is to sum the strips, over whole section.
For circular sections, use polar coordinates. 13 ) / ' $! f. da!! f. rd( " # dr %& 0 Greatly simplifies limits on integration
Parallel-axis theorem (revision) 14 xx " y da "( ys + y! ) y y s " da + ys " y. da + " s A +! xx! da y!. da Similarly yy xy x s x s y A + s! A + yy! xy mportant that! related to own centroid. xx
DOUBLE NTEGRATON 15 Usually used to determine properties of standard components or section shapes; e.g. xy for triangle (about its own centroid). Much more convenient to work in terms of origin at corner (O) and local axes s, t. First find st Equation of hypoteneuse s b(1 - t/d) ( 1( t d ) d b ' $ st! %!. st ds ". 0 & 0 # d ( 0 tb & $ 1' % t d t + d #! ". dt dt b d & 1 $ % ' 3 + 1 4 b d 4 #! "
16 But we know that the centroid G is at s b/3, t d/3 Apply Parallel-axis theorem. Remember you go from the centroid of the section to another axis, so st xy + b d A.. 3 3 " xy st! bd b d.. 3 3! b d 7 This is where the negative sign comes from.
ALTERNATVELY 17 Built-up sections 1. Find about centroid of each region.. Use parallel-axis theorem to find of combined section. Basic building blocks usually rectangles, triangles, circles or elements taken from data sheets.
BASC ELEMENTS 18 Rectangle xx bd 1 3 yy b 3 d 1 xy 0 Circle xx yy! R 4 4 xy 0
Triangle 19 xx bd 36 3 yy 3 b d 36 xy! b d 7 But, for :- xy + b d 7 (More examples from published data books Roark or Steel Designers Manual)
EXAMPLE 0 Centroid at t s 10.5 + 10.1 3 0 10. 1 + 10! 3.5 0
1! xx! yy! xy x y s s $! # offsets! " For the two regions (1) 1000 1 10 1 0! 1.5 + () 40 1 50 1 0 +1.5 + xx (1) () 1000 ( ) 40 +. 10 + +!.10 1 1 ( ) 166. 7 yy 10 + 1 (! 1.5). 10 + 50 + 1 ( 1.5).10 66. 7 xy 0 + (!1.5.)10. ( 1.5.! ).10! 60 +
xx OF SYMMETRCAL SECTON Consider single axis of symmetry For every area da with +ve x There is another area da with same y and ve x. " " yda ( + x). yda + (! x) 0 xy xy 0 about any axis of symmetry, but not about all axes. Can sometimes simplify xy calculation.
3 xy of shaded regions is zero since they are symmetric about G From unshaded regions xy 4.(-1.5).5 + 6. (.5). (-) -60 as before
LNE NTEGRAL 4 f section is thin, then we can turn integral over area to integral along centreline.! da "! tds xx + " 9 "( 7! s1) 0 5.5 "(! ) 0 y 161 da t t 1 ds ds 1 Compare with 166.7 (exact)
5 Discrepancy caused by overlaps f these areas large then error will be large use judgement.
PERMETER CALCULATON 6 Not for hand calculation Use for complex shapes Easy to program. Define arbitrary axes (O-s & O-t) Specify perimeter as sequence of segments. Move around section leaving material on right. Perimeter must close.
For holes 7 Cross from outside to inside and go round hole still leaving material on right. Go back to outside along same line.
ADD areas below segments moving to right. 8 SUBTRACT areas below segments moving to left. Net effect. nclude all material inside perimeter.
9 FORMAL METHOD b h h t r s i+ 1 i BE t i+ 1 t i! s! t i RGOROUS WTH SGNS! ' tda bh r hr. + bh t & $ h % r + ht 3 #! " Similarly for all other!... da Automatically accounts for direction of movement and for arbitrary position of 0.
30 Example To find of rim section of a bicycle wheel. Polished section photographed and perimeter digitised.
BENDNG ABOUT OTHER AXES s, t 31 Suppose we know,, xx xy yy s xcos! " y t xsin! + y sin! cos! ss # t sin da! #( xsin! + y cos! ) yy + sin! cos! xy da + cos! xx
SMLARLY 3 tt # s da cos! yy " sin! cos! xy + sin! xx st cos! sin! ( ) ( ) yy " xx + cos! " sin! xy Complicated, but relate to sinθ and cosθ. These are equations of Mohr s circle of nd moment of area
For our example 33 xx 166.7 yy 66.7 xy Centre of Radius of circle circle 1 60! 60 ( 166.7 + 66.7) + 116.7 ( 166.7! 116.7) 78. 1 st 0 when tan(θ) 60/(166.7-116.7) θ 5.1 These positions of s, t are known as the principal axes of the section.
Loading in one principal axis causes no curvature about other principal axis. 34
STRESS CALCULATON 35 Stress caused by strain Strain caused by curvature! y E R M this works for any axis this only works for principal axis calculate nd moment of area about both principal axes. (s,t) Then! M s t. + s. ss M tt t
DEFLECTON CALCULATON 36 Data book tip deflection PL 3 3E Resolve load into principal directions (Superposition principle) Calculate deflections for each component separately.
37 N.B. minor-axis deflection can be large because is small. Principal cause of buckling problems.
SHEAR - Revision 38 Beam subject to variable moment
39 Average stress! M y at At R.H. end average stress M y dm y +..! z dz L.H. end d" " +.! z dz Stress difference Force difference F y.! z F y. A.! z
Must be balanced by shear stress along longitudinal cut. 40 " #. t. dz F ya. $ z! #. t F ya q This is shear flow along longitudinal cut. Complementary shear then applied across the transverse cut.
41 Shear flow is continuous at junctions in order to equilibrate longitudinal shear. Therefore, q 1 + q q 3
Check vertical shear 4 Vertical shear flow q FAy q & F$ % Bt 1 D + st & $ % D ' s ##!! " "
43 Total vertical shear force in web! D q ds Q 0. small shear carried in flanges) ( 1 1 But 1 3 1 3 1 3 1 0 1 +! " + + # $ % & ' ( # # $ % & & ' ( + # # $ % & & ' ( ) + * + F Q Bt D t D Bt D t D Bt F ds t s D st D Bt F Q D So Shear Force is almost all carried in web.
SHEAR CENTRE 44 f section not doubly symmetric then shear force has to be applied away from centroid to induce pure bending. This location called SHEAR CENTRE. e.g. consider channel bent about x-axis only. Not in torsional equilibrium about centroid.
45 n top flange q FsDt So total horizontal shear force in top flange Q B! q ds 0 FDB 4 for whole section ' FB! Q D t & D # Bt $! % "
46 Take moments about centre of web.! QD FB Load must be acting at B outside section This point known as the Shear Centre (S)
Beam must be loaded through shear centre (S) to prevent torsion. 47 Shear centre lies on any axis of symmetry. Calculation of position of S requires knowledge of stresses throughout section. Perform calculation relative to principal axes. Not trivial calculation but not usually quoted in data books.