534 BASIC KNOWLEDGE IN STRENGTH OF MATERIALS APPLIED IN MARINE ENGINEERING FOR MARITIME OFFICERS Application 11.2 EQUIVALENT STRESSES IN A RECTANGULAR CROSS SECTION We consider that the crankshaft web can be modeled as a bar having a rectangular cross section. Tasks: h H |ட E N B D Y F b Figure 11.13 - Rectangular section and the designation of the points where the equivalent stresses must be computed 1. plot the diagrams of the normal stress; 2. determine the values of the equivalent stresses in the contour points: A, B, C, D, E, F, G, H. Input data: 1. geometric data: h = 18.a, α = 0.267, b = 6.a, 2. internal forces and moments in the cross section where the calculus is performed: -160. pa MX = +800. pa² My = -1050. pa² M₂ =-750-pa² Mz Coviedol M v2 3BN 17806630 53 Reminder of the theoretical notions Solution Normal stress is produced by axial forces, N, and bending moments, My and Mz. σ(y, z) = Mz Iz computed using Navier's relation, (10.21), Mz ··y=σª +σ¾¢ +σ¼z, where, according to (8.1) and Y = ' My (y, z) = σr (y,z)= My.z, Mz (y, z) = == A Mz Iz ∙y. Iy The neutral axis is the locus² of the points in the section where the normal stress is null and it has the equation, (10.26): σ(y, z) =0, namely, (10.27), N MY + A Iy Mz y= 0. The coordinates of the neutral axis intersection points Iz with the two axes OY, and OZ, respectively are obtained by equaling with zero each of the coordinates, one at a time, which leads to the set of points 2 the set of all entities (points, lines or surfaces) that satisfy a given requirement FAILURE THEORIES 535 Iz N . YI₁ = A Mz and y₁₁ = 0 Drawing the neutral axis is important in order Iy N Z12 = A MY to easily identify the most remote point in the section with respect to this axis, points where the normal stresses have extreme values. Tangential or shear stresses are produced by the torque (MX) and by the shear forces (Ty,Tz). To compute the tangential stresses produced by torsion, there are specific calculus relations for each shape of the cross section. In this case, having a rectangular cross section, and considering that h>b, according to (9.61), use the following calculus relations: Mx max max Mx max M X = a.h.b² M X a'.h.b² The overall maximum tangential stress □ Mx 2 max max found at the middle max of the long side (of h dimension), and a local ‘maximum' tangential stress TMX is located at the middle of the shorter side of the rectangle (of b dimension). Coefficients a and a' are taken from the tables where are listed in respect with h the ratio, or are determined using some alculus relations resulted from the b book Son0-630 53 Theory of Elasticity. In this case coefficients a and a' are given in the problem hypothesis, being considered as associated information to the primary geometric data, i. e. h and b. Tangential stresses generated by the shear forces Ty and T₂ are computed using Juravski' Tz Sy by relation, (10.68), Ty ST TTY bz A series of geometrical characteristics appear in these relations and they have to be computed: 1. Area of the cross section is computed by A=h·b 3 b.h³ h.b³ 3 moments of area are computed using: Iy = ' Iz = These 12 12 relations can be better remembered by using the mnemonic rule I = axis 12 where || is the side parallel to the reference axis and _ is the side perpendicular on the reference axis. 3. Maximum values of the first moments of area Sy and S₂ can be calculated for half of the cross section, which has a rectangular shape. Thus, they are computed as a product between the area of a half of section and distance from the centroid of that given area to the axis in respect with which the first moment of area is being computed. 536 BASIC KNOWLEDGE IN STRENGTH OF MATERIALS APPLIED IN MARINE ENGINEERING FOR MARITIME OFFICERS 0.25-b A=h.0.5b 0.5.h Y 0.25-h Y Z A=0.5h-b Sy=(0.5h-b)-0.25h 0.5-b Z Sz=(h-0.5b)-0.25b Figure 11.14 - Calculus scheme of the first moment with respect to axis Y and Z rea, Therefore, for the rectangular cross section the first moment of area has the maximum value in its centroid, namely for half of the section. Its corresponding area is A = h h 2 ··b. The marked area its centroid at distance 4 in respect with the horizontal axis OY. Hence, the maximum first moment of area is Sy = 2 hed ISBN10-00539.4 h 4 distance = h².b 8 Area Similarly, for the rectangular cross section, the S₂ first moment of area has its maximum value in its centroid, namely for a half cross section. Its corresponding area is 4 = h. The marked area has the centroid at b 4 distance in respect to the vertical axis OZ. Thus, the according first moment of area is b Sz = h b Area Conventions to plot stress diagrams In case we plot free body diagrams of forces or moments, we use two conventions, namely: the sign convention and the diagrams' drawing rules. Similarly, to plot stresses there are two conventions: a sign convention and the diagrams' drawing rules. As the sign convention of forces results from the system of axes and the running sense, similarly, the sign convention for stresses results from the sign convention for internal forces. This way, we consider that the stresses applied on an elementary area which generate an internal force formally positive, may be considered formally positive. As it has been mentioned above, the problems dealing with cross section aspects (geometrical characteristics of the cross sections, stresses) are solved FAILURE THEORIES 537 in a system of axes in which the OZ axis is vertical pointing downwards, and the OY axis is horizontal pointing to the right. This system of axes results if the observer is at the left side and is looking at the cross section in the direction of the OX axis, as in the figure below. y Z P Χ y For such a position of the observer, the sign convention + Internal Force/Moment and moments is -axis ' sign = +1, internal forces was shown in detail in the chapter which presents the methods to draw the free body diagrams. Bearing in mind all the above, if for the internal forces there is the + Force + Stress then for the stresses it results the convention -axis -axis convention In what concerns the diagram drawing rules, we start from the idea that a positive stress has as effect the local elongation of the fibers in which it is 'applied'. From here, we may derive that, in respect with a zero line signifying the non-deflected section, the stress diagram is drawn 'towards the exterior of the cross section' if stresses are positive, i.e. if the fibers are elongated or ‘in tension', respectively 'towards the interior of the cross section' if the stresses are negative, i.e. if the fibers are compressed. These are general considerations and must be understood in the sense that they represent a manner of thinking, a way to logically learn the drawing rules for stress diagrams, and not to actually draw a series of stress in a more or less complex state of stresses. Calculus of values of the geometrical characteristics of the cross section the previously mentioned calculus relations, for the given cross section (h=18.a, b=6.a) we get the following values: A=108. a², Iy = 2916⋅ a², I₂ =324. aª, S₁ =243.a³, S₂ = 81. a³. 538 BASIC KNOWLEDGE IN STRENGTH OF MATERIALS APPLIED IN MARINE ENGINEERING FOR MARITIME OFFICERS Calculus of the normal stresses σ My Mz = = N A My == Iy M - 200 pa --1.8518º 108.a² Iz Z Z= a -1050. pa² •Z= -0.36008 P • Z 2916.a4 -750.pa² y 324-a4 Normal stress has the equation: P · y = +2.3148- 148 y α σ(y,z) = −1.8518º −0.36008 P· z+2.3148 P a . 2 α .y To compute normal stress in the specified points, we have to replace their coordinates in the expressions above. Therefore, we have to determine the points' coordinates. -60639.4 A Point A is placed on the OZ axis, therefore y₁ = 0. It is placed at the upper extremity of the cross section, hence in the negative range of axis OZ, therefore 18.a 2 =-9.a. Consequently, point h ZA 2 A A (y₁ = 0, ZA = = −9.a). the coordinates Point B is placed in the upper right corner, thus in the quadrant of the system of axes in which the y coordinate is positive and the z coordinate is negative. We compute the coordinates for point B and we get Ув = + b 2 = + 6.a 2 = +3.a and ZB 18.a 2 =-9.a. Hence, point B has the . coordinates B (y B = +3 · α, ZB Point C is placed on the OY axis, hence zc = 0. It is placed at the right extremity of the cross section, hence in the positive range of the OY axis, b therefore yc= =+== C(y = +3.a, zc = +3. a. Therefore, point C has the coordinates Point D is placed in the lower right corner thus in the quadrant in which both coordinates are positive. We calculate the coordinate of point D and we get YD = + 6.a 2 h 18.a = +3.a and Z₁ = + = + 2 =+9·a. Thus, point D has the 2 coordinates D (y₁ = +3⋅a, z₁ = +9⋅a). D Point E is placed on the OZ axis, therefore yĘ = 0. It is placed at the lower extremity of the cross section, hence in the positive range of the OZ axis, hence h 18.a = =+==+ 2 = +9.a. 2 ZE = +9⋅ a). E (YE = 0, Therefore, point E has the coordinates Point F is placed in the lower left corner, therefore in the quadrant in which the y coordinate is negative and the z coordinate is positive. We compute the b 6.a coordinate of point F and we get YF =-3.a and 2 2