80 RELATIVE EXTREMA FOR A FUNCTION CHAP. 3 (b) (c) Suppose x differs only slightly from Qk. Then, cj << Ick Specialize (a) for this case. Hint: Factor out Xk and C%. Use (b) to obtain an improved estimate for ,. for j k. a=x[ 31 x {1, -3} The exact result is ) = I x = ( 1 -2) 3-12. Using Lagrange multipliers, determine the stationary values for the following constrained functions: (a) f = X2-2 g = X2 + 2= 0 (b) f = x2 + x2 + x2 gl = xl + 2 + X3-= 00 2 = X1 -X2+ 2X3+ 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 H =f + g = x'ax -(xx -1) (a) Show that the equations defining the stationary points of f are ax = )x xTx = 1 (b) Relate this problem to the characteristic value problem for a symmetri-cal matrix. 3-14. Supposef = XTx and g = 1 -xTax =0 where aT = a. Show that the Euler equations for Ii have the form 1 T ax= x xax= 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 ? 4 Differential Geometry of a Member Element 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, X2, and X3 (see Fig. 4-1). Let ? be the position Vector to a point X3 13 X3(y) X2 , 2 1I // /___ I / /' x1(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 on the curve having coordinates Xj 1J , 2, J5)and let y I* _. .1. I_ . rI 1 1 .I e tne parameter. .1 we *1 X2 . .1 Fig. E4-1A can represent the curve by 3 =(y) (4-1) 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 (1 Cnnider a circ.le in the XY.-YX,nne (Fi .4-1A We tke v s the nolnr anole and Y. let a = |r|. The coordinates are X1 = a cos y Y_ -'12 -in Y"'"' y ,x 3 and r =acos yi + asin Y)2 l E4-1B (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 -nr h Th ntifln x/ ort! f-r thiC pllrt . h-. thP f-rlM .sst c. s- [ }.'votlull v,.,~,t t*so -U*Y. Ito tl'., t-,-/l l-r = a cos yl, + b sin y, + cy' 3 X2 4-2. ARC LENGTH Figure 4-2 shows two neighboring points, P and Q, corresponding to y and y + Ay. The cartesian coordinates are , -r +X_11-1 r _-1 -I n I -lengtn OI tne cnora rom r to Q is given xj and xj + I y Axj (j = 1, 2, 3) and the 3 y +Ay) IP-Q2 j=1 (AXj)2 (a) As Ay -0, the chord length IPQI approaches the arc length, As. In the limit, 3 ds2 = Z dxJ (b) j=1 Noting that dx dxj = dv dy (c) we can express ds as [-/(dx1 \2 (dX2'2 dx32]1 2 ( a I,I UO-LTi Y J LdLY"yIt3 Y Fig. 4-2. Differential segment of a curve. 8385 ,.,,,,,f,,,I nhcr-TPY nF A MEMBER ELEMENT CHAP. 4 SEC. 4-3. UNIT TANGENT VECTOR ,V-.. ._. ... _84 UltItNt"l-I I IL Finally, integrating (4-4) leads to 4-3. UNIT TANGENT VECTOR We consider again the neighboring points, P(y) and Q(y + Ay), shown in dy ) + + (k) 21 3 dy (4-5) Figure 4-3. The corresponding position vectors are 7(y), (y + Ay), ands(y) Y=dX2)2 PQ = (y + Ay)-f:(y)= AP (a)We have defined ds such that s increases with increasingy. It is customary to call the sense of increasing s the positive sense of the curve. As Ay --0, PQ approaches the tangent to the curve at P. Then, the unit tangent To simplify the expressions, we let vector at P is given by* 3 Tx)I 22 /2 (4-6) t = lim P Q =-ds (4-8)Ay-' IPQI Using the chain rule, we can express t as Then, the previous equations reduce to ds = a dy dr df dy 1 dr (4-9) ds dy ds ady s -= a dy (4-7) Yo Since a > 0, t always points in the positive direction of the curve, that is, in the One can visualize a as a scale factor which converts dy into ds. Note that direction of increasings (or y). It follows that d:/dy is also a tangent vector and + 1.x> 0. Also, if we take y = s, then a Example 4-2 da drL1/2 Consider the curve defined by (4-3). Using (4-6), the scale factor is -Ty dyy (4-10) 2 a = [a2sin2 y + b2 cos y + c2]1/2 Equation (4-10) reduces to (4-6) when is expressed in terms'of cartesian coordinates.We suppose that b > a. One can always orient the axes such that this condition is satisfied. Then, we express a asC2 a= (b2 + c2)12 [1 --k 2 sin2 yll2 where b2 -a2 2k= 2b2 +r cThe arc length is given by 1 r(yx dy -(62 + c2)12 [1 -k2 sin2 y] /2 dys = The integral for s is called an elliptic integral of the second kind and denoted by E(k, y). Then, s = (b2 + C2)i 2 E(k, y) Tables for E(k, y) as a function of k and y are contained in Ref. 3. When b = a, the curve is called a circularhelix and the relations reduce to Fig. 4-3. Unit tangent vector at P(y). 22a = (a+ C)112 = const. s = ay * See Ref. 1,p. 401.86 DIFFERENTIAL GEOMETRY OF A MEMBER ELEMENT CHAP. 4 SEC. 4-4. PRINCIPAL NORMAL AND BINORMAL VECTORS 87 L'�m·�·r� .u-� 11 We determine the tangent vector for the curve defined by (4-3). The position vector is = a cos yt + b sin yi2 + cy 3 Differentiating with respect to y, Normal …