RELATIVE EXTREMA FOR A FUNCTION 80 CHAP 3 Suppose x differs only slightly from Qk Then cj Ick for j Specialize a for this case Hint Factor out Xk and C c Use b to obtain an improved estimate for b The exact result is I 1 3 x 1 2 3 12 Using Lagrange multipliers determine the stationary values for the following constrained functions 2 f X2 g X2 2 0 b f x 2 x2 x2 4 31 a x x k Differential Geometry of a Member Element a gl xl 00 2 X3 2 0 3 13 Consider the problem of finding the stationary values of f xrax xTaTx subject to the constraint condition xx 1 Using 3 36 we write 2 X1 X2 2X 3 H f a g x ax xx 1 Show that the equations defining the stationary points of f are xTx 1 ax x b Relate this problem to the characteristic value problem for a symmetri cal matrix T 3 14 Supposef XTx and g 1 xTax 0 where a a Show that the Euler equations for Ii have the form 1 T xax ax x The geometry of a member element is defined once the curve corresponding to the reference axis and the properties of the normal cross section such as area moments of inertia etc are specified In this chapter we first discuss the differential geometry of a space curve in considerable detail and then extend the results to a member element Our primary objective is to introduce the concept of a local reference frame for a member 4 1 PARAMETRIC REPRESENTATION OF A SPACE CURVE A curve is defined as the locus of points whose position vector is a function of a single parameter We take an orthogonal cartesian reference frame having directions XI X 2 and X3 see Fig 4 1 Let be the position Vector to a point X3 We see that the Lagrange multipliers are the reciprocals of the characteristic values of a How are the multipliers related to the stationary values of f X3 y 13 X2 2 1I I x 1 0 X2 Y X1 Fig 4 1 Cartesian reference frame with position vector f y The vector directed from the origin of a fixed reference frame to a point is called the position vector A knowledge of vectors is assumed For a review see Ref 1 81 82 DIFFERENTIAL GEOMETRY OF A MEMBER ELEMENT CHAP 4 I 1 I I r 1 11 I on the curve having coordinates Xj 1J 2 J5 and let y e tne parameter we can represent the curve by y 4 1 1 X2 Fig 1 E4 1A 3 Since r xjij an alternate representation is j 1 xj xj y j 2 3 4 2 Both forms are called the parametric representation of a space curve Example 4 1 Cnnider a circ le in the XY YX nne 1 Fi 4 1A We tke v s the nolnr anole and Y let a r The coordinates are X1 a cos y Y in 12 x 3 y Y r acos yi asin and l E4 1B Y 2 2 Consider the curve Fig E4 1B defined by X1 a cos y x2 b sin y 4 3 X3 cy where a b c are constants The projection on the X1 X2 plane is an ellipse having semiaxes n sst nr h Th ntifln c Y s votlull x ort f r thiC pllrt v t t so U r a cos 4 2 thP f rlM tl t l l h Ito yl b sin y cy 3 X2 ARC LENGTH Figure 4 2 shows two neighboring points P and Q corresponding to y and y Ay The cartesian coordinates are xj and xj Axj j 1 2 3 and the r 11 1 X 1 I r n I I lengtn OI tne cnora rom r to Q is given y y Ay 3 IP Q2 As Ay 0 the chord length j 1 AXj 2 a IPQI approaches the arc length As In the limit 3 ds2 Z dxJ b j 1 Noting that dxj dx dy dv c we can express ds as dx 1 2 UO LTi dX2 2 Y 2 dx32 1 J y LdLY Y It3 a I I Fig 4 2 Differential segment of a curve 83 MEMBER ELEMENT nhcr TPY nF A f I UltItNt l I 84 V I IL CHAP 4 4 3 Finally integrating 4 4 leads to dy Y dX2 2 s y SEC 4 3 k 3 21 dy 4 5 It is customary to We have defined ds such that s increases with increasing y call the sense of increasing s the positive sense of the curve To simplify the expressions we let We consider again the neighboring points P y and Q y Ay shown in Figure 4 3 The corresponding position vectors are 7 y y Ay and Note that One can visualize a as a scale factor which converts dy into ds 1 x 0 Also if we take y s then a t lim Ay dr ds PQ IPQI ds is satisfied We suppose that b a One can always orient the axes such that this condition asC2 a express we Then 2 2 2 2 12 a b2 c 1 k sin yll 1 dr df dy dy ds da is Consider the curve defined by 4 3 Using 4 6 the scale factor 2 12 2 2 2 2 a a sin y b cos y c ady drL1 2 Ty dyy Equation 4 10 reduces to 4 6 when coordinates b2 a b r c2 2 The arc length is given by s x dy 62 1 c2 12 2 2 k sin 1 y 2 r y dy by E k y The integral for s is called an elliptic integral of the second kind and denoted Then 2 i 2 s b2 C E k y b a the curve Tables for E k y as a function of k and y are contained in Ref 3 When to reduce is called a circularhelix and the relations a a2 s ay 2 2 C 11 4 9 Fig 4 3 Unit tangent vector at P y const See Ref 1 p 401 4 10 is expressed in terms of cartesian 2 2 4 8 Since a 0 t always points in the positive direction of the curve that is in the direction of increasings or y It follows that d dy is also a tangent vector and Example 4 2 k a As Ay 0 PQ approaches the tangent to the curve at P Then the unit tangent vector at P is given by 4 7 a dy Yo where f y AP PQ y Ay Using the chain rule we can express t as Then the previous equations reduce to ds a dy s 85 UNIT TANGENT VECTOR 4 6 Tx I 22 2 3 UNIT TANGENT VECTOR 86 DIFFERENTIAL GEOMETRY OF A MEMBER ELEMENT L m r u CHAP 4 SEC …