Engineering Tripos Part A Module 3D4 Elasticity Analysis and Stability Supervisor Version Elastic Analysis Handout 1 8 lectures Fehmi Cirak (fc86@) Stability 8 lectures Sergio Pellegrino January 007 Handout courtesy of Chris Burgoyne
ELASTC ANALYSS Bending of asymmetric beams Biaxial bending St Venant torsion Restrained warping torsion Grillage analysis Macaulay s Method Reciprocal theorem nfluence lines
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 Must be able to analyse asymmetric beam bending about arbitrary axes. Some practical examples: Angle Z-Purlin Crane beam Angles and purlins cheaper to make than -beams. 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 EXAM MARKS WLL BE DEDUCTED F SGNS ARE NOT TREATED PROPERLY
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.
Typical Data book symmetric section 1 Gives values for bending about horizontal axis only. Values given as elastic section modulus (/y)
But same data book for asymmetric section does not give xy. 13 To analyse properly engineer will need to calculate y and xy
DOUBLE NTEGRATON Revision 14 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. 15 ) / ' $ f. da f. rd( " # dr %& 0 Greatly simplifies limits on integration
Parallel-axis theorem (revision) 16 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 17 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 # "
18 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 19 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 0 Rectangle xx bd 1 3 yy b 3 d 1 xy 0 Circle xx yy R 4 4 xy 0
Triangle 1 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 Centroid at t s 10.5 + 10.1 3 0 10. 1 + 10 3.5 0
3 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 +
4 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.
5 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 6 f section 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)
7 Discrepancy caused by overlaps f these areas large then error will be large use judgement.
PERMETER CALCULATON 8 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 9 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. 30 SUBTRACT areas below segments moving to left. Net effect. nclude all material inside perimeter.
31 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.
3 Example To find of rim section of a bicycle wheel. Polished section photographed and perimeter digitised.
BENDNG ABOUT OTHER AXES s, t 33 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 34 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 35 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. 36
STRESS CALCULATON 37 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 CALCULATONS 38 Two methods 1. Relate to principal axes. Allow for xy term use x-y axes. Data book tip deflection PL 3 3E
First method 39 Resolve load into principal directions Calculate deflections for each component separately.
40 N.B. minor-axis deflection can be large because is small. Principal cause of buckling problems. ALTERNATVELY &( $ $ $ %( x y # " E 1 ( ' ) xx yy xy & yy $ $ $ %' xy ' xy xx #& M $ $ " $ % M x y # " 1 7510 E & 66.7 60 #& M $ $ $ $ $ % 60 166.7 " $ % 0 x # " M x " x " 113E y M x 15E These values apply to this loading only since they depend on M y being zero.
41 " x PL 3 3E PL E 3. 1 3.113 PL.009 E 3 " y PL E 3. 1 3.15 PL.006 E 3 Result the same Principal axis method better. xy method gets more awkward when loading becomes complex.
SHEAR - Revision 4 Beam subject to variable moment
43 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. 44 " #. t. dz F ya. $ z #. t F ya q This is shear flow along longitudinal cut. Complementary shear then applied across the transverse cut.
45 Shear flow is continuous at junctions in order to equilibrate longitudinal shear. Therefore, q 1 + q q 3
Check vertical shear 46 Vertical shear flow q FAy q & F$ % Bt 1 D + st & $ % D ' s ## " "
47 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 48 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.
49 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 $ % "
50 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. 51 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.
TORSON 5 Revision - Thin-walled tube Assume tube rotates about arbitrary position 0. Shear flow q continuous round tube. Consider length Torque due to ds Total torque ds dt ds. q. p T dt q. A
Also know 53 4A T G.. " where ds # t " d dz Under pure torque section does not distort. we can nest tubes inside one another and add effects.
Consider rectangle 54 Consider thin tube dz thick, then sum effects of all the thin tubes to get total effect. Enclosed area for thin tube z( b t + z) % ds t ( b $ t + z) ) dz + 4z dz # T G 4A ds % t. " GJ "
Now integrate for all the strips 55 t / ( J ) 0 z ( b ' t + z) ( b ' t + z) + 4.4. dz z & bt $ % 3 3 4 t # ' 1 " bt 3 3 ( if b >> t) (This analysis not strictly accurate since tubes do interact but error is in the t 4 term so the overall result is correct) A better analysis makes use of the Prandtl Stress Function and is described in Timoshenko & Goodier, Chapter 8.
Shear stress maximum on outer surface τ 56 Consider outer layer dz this carries torque. dt 4A G' " % ds t qa A q &. dz G' ". ds % t bt G' " dz $ & # G' ". t ( b + t) But so T bt G' ". 3 3 & 3T bt These values apply for b>>t. See Timoshenko for more accurate values.
N.B. 1. This analysis approximate since end effects not treated correctly. Works for thin rectangles but not for chunky sections. 57 Proper analysis requires Prandtl stress-function.. Assumed rotation about centre of strip. But consider: This is equivalent to rotation plus rigid-body translation. Stiffness and stresses valid for small rotations about any axis.
58 For built-up sections torsional stiffness 1 bt 3 3 J 1 3.10.1 3 + 1 3 6. 3 58 3 19.3 ( L 4 ) J 1 3. 6.1 3. + 1 3. 8.1 3 0 3 6.7 ( L 4 )
OPEN vs CLOSED SECTONS 59 Closed J 4( bh) ( b + h) t ( bh). t ( 3 0 b t) b + h
Open 60 1 1 J bt 3 3 ( ht 3 + bt 3 ) ( bt 3 ) 3 0 Closed sections much stiffer than open sections in torsion.
Box beams used to: 61 1. Distribute loads better than plate girders
. Provide torsional stiffness to resist aerodynamic flutter 6 (e.g Severn & Humber suspension bridges) 3. But difficulties with indeterminate supports. Beam ends up rocking on two supports.
Junction effects 63 Thickness significantly increased in junction region and torsional stiffness varies as t 3. A significant part of the torsional stiffness of rolled steel beams and cast concrete beams comes from this region. Ref:.Struct.E. Junction effects in St. Venant s Torsional Stiffness. March 1993. C J Burgoyne.
ST VENANT TORSON and RESTRANED WARPNG TORSON 64 Axial movement known as warping. Analysis so far assumes section is free to warp, so no axial stresses set up. Only true if torque is constant, and if ends free. d result is T GJ GJ " dz (known as St Venant torsion)
But, if torque varies or if ends offer axial restraint, adjacent sections will try to warp by different amounts, so axial strains and stresses will be set up. 65 These will have associated shear stresses which will carry some of the torque. Need to consider this Restrained warping torsion [The only section which does not warp is a circle but in -legged angle sections and box sections warping is negligible.]
Ex. -beam cantilever 66 Consider an beam mounted as a cantilever, with torque applied at the free end. The two flanges will bend in opposite directions view from above:-
Curvature of top flange in its own plane. # d u d # " dz 67 d B.M in flange E f " Shear force in flange d dm " E f # dz Q Similar force in other flange but of opposite sense. Torque carried is Qd d # E f. $ """. d # E $ """ f d
Total torque 68 T GJ " $ E# """ where f d (units 6 L ) Γ is known as the Restrained Warping Torsion Constant. To solve the differential equation for torsion:- Boundary conditions At root: no twist, so θ 0 du dz Flange built-in, so 0 " # 0 At free end no moment in flange E f d # 0 " # 0
General equations not just this example. 69 T GJ " $ E# """ Boundary conditions Warping restrained " 0 Free to warp " 0 For this example solution " T GJ # z + z ( z + c) + ae + be (particular int.) (complementary fn.) Substitute into diff. eq. 1 GJ " $ # E " # GJ E λ has dimensions of length and is a property of the cross-section
70 Boundary conditions ( ) ( ) ( ) ( ) ( ) L L z z L L L L e e e GJ T c z GJ T e e c z e GJ T a a b a GJ T z e a b L z " " " " # " # " " # # ) ( 1 1 1 0 at 0 1 0 1 0 at 0 at 0 $ $ $ $ $ $ $ + % % & ' ( ( ) * $ + + + $ $ + + + % & ' ( ) * $ $ +, $ +,,
f beam long L>>λ 71 C " b " 0 a " T GJ so ) T & $ z $ % & ' ( $ 1 ' e % ' ( GJ z ## "" Clamping one end of a beam reduces its effective E length for torsion by " even for GJ constant torque.
HOW MPORTANT S WARPNG RESTRANT? 7 Ex. 610 305 179 kg U.B. (see data book) J 1 bt 3 3 3.10 6 mm 4 f 3 307.3 1 55.10 6 mm 4 " f d 55.10 6 ( 570 + 3) 9.75.10 1 mm 6 " # E GJ 10 9.75.10. 6 81 3.10 1 900 mm Typical span/depth ratio 0 so span likely to be 1 m. L 4 " # warping restraint will be significant.
73 Warping restraint normally significant for open sections. Circular sections do not warp because of symmetries. Angle sections warp slightly but shear centre at junction of the legs so shear forces have no lever arm, so warping can be ignored. Closed sections warp, but warping displacements smaller and GJ higher. λ much lower. Calculation of shear centre and EΓ is not trivial. Most data books do not give them. Best summary of values Buckling Strength of Metal Structures Bleich. (FD10) (attached below) Theory given in Theory of Elastic Stability Timoshenko & Gere Sect. 5.3 (FD14).
SUMMARY 74 Centroid (G) Apply Calculate N, M x, M,, xx xy ss, tt, Principal axes xy y Shear centre (S) Apply Calculate F, Fy, EΓ, GJ T
Difference between S and G causes interaction problems for large deflections and buckling 75
Data for position of Shear Centre and value of Restrained Warping Torsion constant. 76 Taken from Buckling Strength of Metal Structures by F. Bleich. McGraw-Hill, 195 (CUED Library FD10).
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