CombinedLoading

CombinedLoading - 8.5 COMBINED LOADINGS Pressure vessel (b)...

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Unformatted text preview: 8.5 COMBINED LOADINGS Pressure vessel (b) (C) FIG. 8-20 Examples of structures subjected to combined loadings: (a) wide—flange beam supported by a cable (combined bending and axial load), (b) cylindrical pressure vessel Supported as a beam, and (c) shaft in Combined torsion and bending In previous chapters we analyzed structural members subjected to a single type of loading. For instance, we analyzed axially loaded bars in Chapters 1 and 2, shafts in torsion in Chapter 3, and beams in bending in Chapters 4, 5, and 6. We also analyzed pressure vessels earlier in this chapter. For each type of loading, we developed methods for finding stresses, strains, and deformations. However, in many structures the members are required to resist more than one kind of loading. For example, a beam may be subjected to the simultaneous action of bending moments and axial forces (Fig. 842021), a pressure vessel may be supported so that it also func— tions as a beam (Fig. 8‘20b). or a shaft in torsion may carry a bending load (Fig. 8720c). Known as combined loadings, situations similar to those shown in Fig. 820 occur in a great variety of machines, build- ings. vehicles, tools, equipment, and many other kinds of structures. A structural member subjected to combined loadings can often be analyzed by superimposing the stresses and strains caused by each load acting separately. However, superposition of both stresses and strains is permissible only under certain conditions, as explained in earlier chap: ters. One requirement is that the stresses and strains must be linear functions of the applied loads. which in turn requires that the material follow Hooke‘s law and the displacements remain small. A second requirement is that there must be no interaction between the various loads, that is, the stresses and strains due to one load must not be affected by the presence of the other loads. Most ordinary struc— tures satisfy these two conditions, and therefore the use of superposition is very common in engineering work Method of Analysis While there are many ways to analyze a structure subjected to more than one type of load, the procedure usually includes the following steps: 1. Select a point in the structure where the stresses and strains are to be determined. (The point is usually selected at a cross section where the stresses are large, such as at a cross section where the bending moment has its maximum value.) 2.« For each load on the structure, determine the stress resultants at the cross section containing the selected point. (The possible stress resultants are an axial force, a twisting moment, a bending moment, and a shear force.) Calculate the normal and shear stresses at the selected point due to each of the stress resultants. Also, if the structure is a pressure ves— sel, detemiine the stresses due to the internal pressure. (The stresses are found from the stress fomrulas derived previously; for instance, 0" : P/A, 7' : Tye/1,50 : My/l, 1' : VQ/Ib, and (I : pr/t.) 4. Combine the individual stresses to obtain the resultant stresses at the selected point. In other words, obtain the stresses (TX, 0),, and TX), 0) ta l 1'1: [2/7 FIG. 3-21 Cantilever bar subjected to combined torsion and bending: (a) loads acting on the bar, (b) stress resultants at a cross section. and (c) stresses at points A and B acting on a stress element at the point. (Note that in this chapter we, are dealing only with elements in plane stress.) 5. Determine the principal stresses and maximum shear stresses at the selected point, using either the stress-transformation equations or Mohr’s circle. If required: determine the stresses acting on other inclined planes. 6. Determine the strains at the point with the aid of Hooke’s law for plane stress. ‘ 7. Select additional points and repeat the process. Continue until enough stress and strain information is available to satisfy the purposes of the analysis. illustration of the mattress To illustrate the procedure tor analyzing a member subjected to corn— bined loadings. we will discuss in general terms the stresses in the cantilever bar of circular cross section shown in Fig. 8—21a. This bar is subjected to lW() types ot‘ loadha torque T and a vertical load P, both acting at the tree end of the bar. Let us begin by arbitraril) selecting two points A and B for investi— gation (Fig. 8—2 la). Point rl is located at the top of the bar and point B is located on the side. Both points are located at the same cross section. The stress resultants acting at the cross section (Fig. 8-21b) are a twisting moment equal to the torque a bending moment M equal to the load P times the distance I) from the free end of the bar to the cross section. and a shear t‘orce V equal to the load P. The stresses aeting at points A and B are shown in Fig. 8‘2lc. The twisting moment T produces torsional shear stresses _ Tr 2T T i h t . (a) I [p 77"”; in which r is the radius of the bar and [p : 7TF4/2 is the polar moment of inertia of the cross—sectional area. The stress 7'1 acts horizontally to the left at point A and vertically downward at point B, as shown in the figure. The bending moment M produces a tensile stress at point A: Mr 4M —~« : a. (b) I 7Tr (TA : in which [ = 772‘4/4 is the moment of inertia about the neutral axis. How— ever, the bending moment produces no stress at point B, because B is located on the neutral axis. The shear force V produces no shear stress at the top of the bar (point A), but at point B the shear stress is as follows t._ 57.. ' u sflfajfl; T2 T 3/1 T 377% . . 2 . . in whicer : 77F is the cross—sectional area. 2. FIG. 8—21 (Repeated) FIG. 8-22 Stress element at point A The stresses (IA and r, acting at point A (Fig. 8—2lc) are shown acting on a stress element in Fig. 8—2221. This element is cut from the top of the bar at point A. A twoidimensional View of the element, obtained by looking vertically downward on the element, is shown in Fig. 8—22b. For the purpose of determining the principal stresses and maximum shear stresses. we construct X and y axes through the element. The x axis is parallel to the longitudinalaxis of the circular bar (Fig. 8—21a) and the y axis is horizontal. Note that the element is in plane stress with (IX : ([4, (IV : 0, and TM, : i 7,. A stress element at point B (also in plane stress) is shown in Fig. 8—23a. The only stresses acting on this element are the shear stresses, equal to 71 + 72 (see Fig. 8~2lc). A two—dimensional View of the stress element is shown in Fig. 8—23b, with the X axis parallel to the longitudinal axis of the bar and the y axis in the vertical direction. The stresses acting on the ele ment are (5 : or). i: (l and TH. : Wt r1 + 73). Now that we have determined the stresses acting at points A and B and constructed the corresponding stress elements, we can use the transformation equations of plane stress » ' ' 7, or Mohr’s circle i". .) to determine principal stresses, maximum shear stresses and stresses acting in inclined directions. We can also use Hooke‘s law - A and B. The procedure described preyiousl) for analyzing the stresses at points A and 13’ (Fig. Rillai can be used at other points in the bar. Of particular interest are the points where the stresses calculated from the flexure and shear formulas have maximum or minimum values, called critical points. For instance. the normal stresses due to bending are largest at the cross section oi maximum bending moment, which is at the support. Therefore. ‘points (7' and D at the' top and bottom of the beam at the fixed end (Fig. 8721a) are critical points where the stresses should be calculated. Another critical point is point B itself, because the shear stresses are a maximum at this point. (Note that in this example the shear stresses do not change if point B is moved along the bar in the longitudinal direction.) ' , ‘ to determine the strains at points \« (a) (b) FIG. 8»23 Stress element at point B As a final step, the principal stresses and maximum shear stresses at the critical points can be compared with one another in order to deter: mine the absolute maximum normal and shear stresses in the bar. This example illustrates the general procedure for determining the stresses produced by combined loadings. Note that no new theories are inonvedieonly applications of previously derived formulas and con— cepts Since the variety of practical situations seems to be endless, we will not derive general formulas for calculating the maximum stresses. Instead. we will treat each structure as a special case. Seteetiea at firétteal Points If the objective of the analysis is to determine the largest stresses (now/rm! in the structure. then the critical points should be selected at cross sections where the stress resultants have their largest values Furthermore, within those cross sections, the points should be selected where either the normal stresses or the shear stresses have their largest values. By using good judgment in the selection of the points, we often can be reasonably certain of obtaining the absolute maximum stresses in the structure. However, it is sometimes difficult to recognize in advance where the maximum stresses in the member are to be found. Then it may be necessary to investigate the stresses at a large number of points, perhaps even using trial‘anderror in the selection of points. Other strategies may also prove fruitful~such as deriving equations specific to the problem at hand or making simplifying assumptions to facilitate an otherwise difficult analysis. The following examples illustrate the methods used to calculate stresses in structures subjected to combined loadings. ( a) tic) FIG. 8-24 Example 874. Rotor shaft of a helicopter (combined torsion and axial force) The rotor shaft of a helicopter drives the rotor blades that provide the lifting force to support the helicopter in the air (Fig. 8-2421) As a consequence, the shaft is subjected to a combination of torsion and axial loading (Fig. 8724b). For a 50mm diameter shaft transmitting a torque T : 2.4 ani and a tensile force P :4 l25 kN, determine the maximum tensile stress, maximum compressive stress and maximum shearstress in the shaft. Solution The stresses in the rotor shaft are produced by the combined action ol‘ the axial t‘orce P and the torque T (Fig. 8724b), Therefore, the stresses at any point on the surface of the shaft consist of a tensile stress (IO and shear stresses T”. as shown on the stress element ot Fig. 8724c. Note that the y axis is parallel to the longitudinal axis ot the shaft. The tensile stress on, equals the axial l‘orce divided bV the crossvsectional area: P 4P7 (r0 7 (v 7 ref 3 i A 7rd ‘ 4 25 . “ UK 3 kN); r 63.66 Ml’a ’rrtDO mm) The shear stress T0 is obtained from the torsion formula (3.4 - L"; $.97 t) : T V V t , [,1 7rd" 7130 tntnl‘ ’7 , v . WA W T?) : 97.7s MPa The stresses m) and TH act directly on cross sections of the shalt. .Knowing the stresses (To and TH. m: can now obtain the principal stresses and maximum shear stresse: . The principal stresses are S L (rx t m. / ’ (IY 7 (r\ i 3 , 0'] ~ 1‘ e i t i We + 7“} (d) ‘“ 2 \ 2 Substituting tr. 5 0. (I3 : (a) 5 mm MPa. and TU : r 4 Ti, '779778 MPa. we get all : 32 MPa 1' 103 MPa or (f1 2 135 MPa (73 : *7] MPa These are the maximum tensile and compressive stresses in the rotor shaft. The maximum inrplane shear stresses are P “i or1 - 0) ~ , , Tmax 1T d TX) This term was evaluated previously, so we see immediately that Tinax : MP3 Because the principal stresses m and (f2 have opposite signs, the maximum iniplane shear stresses are larger than the maximum out—of—plane shear stresses " ' "t. 'l‘herefore, the max ’imum shear stress in the shaft is 103 MPa. EXample 8-5 (a) (b) HG. 8—25 Example 8—5. Pressure vessel subjected to combined internal pressure and axial force A thin-walled cylindrical pressure vessel with a circular cross section is sub jected to internal gas pressure p and simultaneously compressed by an axial load P : 12 k (Fig. 8'25a). 'l‘he cylinder has inner radius r : 2.1 in. and wall thickness t ’4: 0.15 in. Determine the maximum allowable internal pressure palm“, based upon an allowable shear stress 016500 psi in the wall of the vessel. Solution The stresses in the wall ot‘ the pressure vessel are caused by the combined action of the internal pressure and the axial l'orce. Since both actions produce uniform normal stresses thrtinighout the wall. oe can select any point on the sur- tace for investigation. At a typical point. such as point A (Fig. 8—2521), we isolate Fig 8—1511 The r axis is parallel to the longitudinal axis of the pressure vessel and the t. axrs is circuinterential. Note that there are no shear stresses acting on the element. Primripal stresses. The longitudinal stress (I, is equal to the tensile stress a stress element as shown in produced by the internal pressure ’ ' minus the comv pressiye stress produced by the axral liorce; thus m 2 ,7- ._ ; : A. , (t) in which A : 27711 is the cross-sectional area 01 the cylinder. (Note that for con~ venience we are using the inner radius r in all calculations.) The circumferential stress (I. is equal to the tensile stress or} produced by the internal pressure (rigs, cw (r‘ r (E) ' 1‘ Note that (I), is algebraically larger than 0‘. Since no shear stresses act on the eletnent (Fig. 825‘), the normal stresses (I. and (r‘. are also the principal stresses: .r r , . (71 :(TV:L (f;t:(f)t:£"" P (ill) ’ t 2t 27TH Now substituting numerical values. we obtain ' 7 2.1 ' . (r1 : p' : Kw 1”) : 140p t 015 in. pr P [)(2.1 in.) 12 k a — e it w m r 2 2t 2771'! 2(().15 in.) 277(2.l in.)(0.15 in.) : 7.0;] e 6063 psi in which 01. (f2. and p have units of pounds per square inch (psi). [nplant’ shear strerscs. The maximum invplane shear stress (1,. t m is f, T M iiii "4' : 317114.01) * 70p + 6063 psi) : 3.5;) + 3032 psi [flux 7 Since rum is limited to (1500 psi, the preceding equation becomes 6500 psi :r— 3,5!) l 3032 psi from which we get 34(18 psi , t - I) W 900.91)“ or warmth 7’ 090 p81 33 because we round dmvnwaid. ()itl-ofipluue ,\/‘l(‘(tl' stresses: The maximum out»ol'~plane shear stress (see Eqs. 728:1 and 7728b) is either (T: (T, THIJ\ 7 if Or Tinax l m ‘ From the lirst ol these two equations we get _o5t)t) psi 5 35/) * l0}: psi or ([l‘wmil ’7 .7720 psi From the second equation \x e get moo psi : 70p a (MUM); : 928 psi Alltllt/‘(lb/C’ internal pressure. Comparing the three ealeulated values for the allowable pressure. we see that (WNW); governs, and therefore the allowable internal pressure is ' pallou : At this pressure the prineipal stresses are al : 13.000 psi and (r3 : 430 psi These stresses haVe the same signs, thus confirming that one of the outiofrplane shear stresses must be the largest shear stress ’f Note: in this example. we determined the allowable pressure in the vessel assuming that the axial load was equal to 12 k. A more complete analysis would include the possibility that the axial force may not be present (As it turns out? the allowable pressure does not Change if the axial force is remOVed from this example) 3 t "t l t t 1‘77 -r i l l ilt'llmmtt 4 ~ >t Ill) tttltl‘ HEB—26 Example so Wind pressure against a sign teombined bending. torsion and shear ot~ the pole) A sign of dimensions 2.0 m X’ 1.2 m is supported by a hollow Circular p019 havingy outer diameter 120 min and inner diameter 180 mm (Fig. 8—26) The sign is oltset 0.5 m trom the renterline of the pole and its lower edge is 6.0 m above the ground Determine the prineipal stresses and maximum shear stresses at points A and [I at the base of the pole due to a wind pressure ot‘ 20 kPa against the sign. Solution Stress resultant»; The Wind pressure against the sign produees a resultant toree W that aets at the midpoint of the sign (Figs 8»Z7a) and is equal to the pres, sure [I times the area .t o‘uer whieh it acts: ll” 7 psi 1 (2.0 kl’aXZt) m >< 1.2 m) : 478 kN The line oti aetion ot‘ this t‘oree is at height h : be m above the ground and at distanee t) l5 m trom the eenterline of the pole. The \\ ind toree aeting on the. sign is statieally equivalent to a lateral loree W and a torque T acting. on the pole tliig. 8727b). The torque is equal to the loree tt' times the distanee ('7: 'l' WI) ” tts kN)(l.5 m) i 7.2 kN-m The stress resultants at the base of the pole tFig. 8*Z7C) eonstst ot a bending moment M. a torque T. and a shear toree V, Their magnitudes are M " Wli I’ (4.8 kNltbo m) i 31.6% kN'm ’l' 7.2 kN'm \" r W: 4.8 KN Examination ot‘ these stress resultants shows that maximum bending stresses oeeur at point A and maximum shear stresses at point 8. Therefore. A and B are critical points where the stresses should be detemtirted. (Another eritieal point is diametrically opposite point A. as explained in the Note at the end ot‘ this example) Stresses at points A and l}. The bending moment M produces a tensile stress (rA at point A tl7ig. 827d) but no stress at point B (which is located on the neutral axis). The stress (5‘ is obtained from the flexure formula: Moll/2) (at : i'e' I in which (12 is the outer diameter (220 mm) and [ is the moment ol‘ inertia of the cross section. The moment of inertia is W i 4 4‘ 7T 7 ' * I : flit, 7- (r I) : ~‘ (220 mm)4 - (180 mm)" : 63,46 >< to ‘J m1 04 \ “ , ()4 in which cz’l is the inner diameter. Therefore, the stress (IA is ‘ Mtl’l (rt :7 / a (3 lio8'kN-m)(220 mm) - .. i; ( )r 2/ 2(6346 >< to’énfi) 54'” Ml“ W: 4.8 kN (L) (d) \ (r : (n ! \' i [ \ L x 0 iii W 7 O W¥g , “WJ , 4—— 4—— FIG. 8—27 Solution to Example 8%» (C) m The torque '1' pmdua‘m shear stresses 7, at points A and B (Fig. 827(1). We can calculate these slrcssss from [he torxmn formula: 7 'I‘(:d2/2) [I W I P in which [p is the polar moment 0f incrlizt: 1,) z (1‘: d“ : 21 :7 126.92 >< 10m“ 1) \ ~ ‘ (Onlmucd Thus‘ "- ° 5/} ; Vit’izolnrx’it)’ 'll/l, (7‘7.7 kN-niXZZO mm) : " 7:‘;1';r)’ (124 Mm Finally we calculate the shear stresses at points A and B tlue to the shear toree V. the shear stress at point A is zero‘ and the shear stress at point [3 ttlenotetl Ti in Fig. firl'hll is obtained from the shear l‘ormula tor a Cll'CUl‘At' tube ( j‘) tll \Klllt’ll I i and r} are the outer and inner radiil respectively. and A is the cross» seetional aiea’ ru ,rt 7 mr} * r31) :: l257llmm Substituting nuineiit'al Values into litre ti). we obtain 7; '5 077(3 MPa The stresses aeting on the cross section at points A antl B have now been calculated. Srrms violin-um: "the next step is to show these stresses on stress elements (Figs. «’427e and tr For both elements the y axis is parallel to the longitudinal axis ot the pole and the x axis is horizontal. At point A the stresses acting on thC element are (r\ 1: l) T“. Ti lv 72 I (7.24 MPa + 0.76 MP8 :1 7.00 Ml’a Since there are no normal stresses acting on the elemenL point B is in puffi shear, Now that all stresses acting on the stress elements (Figs. 82% and f) are kmwn_ We, can , ' ' 2 determine the princlpal stresses and maximum shear stresses. Principal stressm‘ and U'IUXlHlMHl shear stresses at point A The principal stresses are obtained here: 7 (r‘ + m, “1.3 " ,9 00 Substituting (rr ': (l in 54.91 lVlPZL and 7”. 4 6/24 Ml’a, we get in ,3 i 275 Ml’a :tf 28.2 MPa or m ’4 55.7 t/lPa (rl : *ll 7 Ml’a The maximum m-plane shear stresses may he obtained : ll (rt 7 (rt ‘7: . 7 , Tlllli\ 777 7 7 W ) l7 771V This term was evaluated previnusly so we see immediatel) that r 38.2 Ml’a '91" iti’.t\ Because the principal stresses (I, and (r, have npposite signs, the maximum in— han the maumum (tl1l70l’pltllifi‘ shear stresses t“ . plane shear stresses are larger t i i ‘ ' ) 'l‘herelorei the maximum shear stress at pmnt A\ ts ZR 2 Ml’a Pruitt/ml strum”. tl/lt/ llltl,tlllllllll slii'm' \Il't’NHC) ul [mint H. The stresses at 7.() MPa Since the element is in pure this paint are (n r (t (f :' tt and 7H shear; the principal stresses art 7 7.0 MPa (r, 'FthPa m 7 and the maximum llt'Pl‘dntf shear stress is Tum : 7,0 Ml’a The maximum out-ollplane shear stresses are hall this value Note: ll the largest stresses anywhere in the pole are needed. then we must also determine the stresses at the eritieal paint diametrically opposite point A, because at that point the emnpressive stress due to bending has its largest value. The principal stresses at that pmrit are in 7' (l7 MPu «r; 1 "755.7 Ml’a and the maximum shear stress is 282 MPa. ’l‘herelore, the largest tensile stress in the pole is 557 MPa. the largest compressive stress is A 55.7 MPa and the largest shear stress is 28.2 MPa, (Keep in mind that only the effects ol the wind pressure are Ctmsidered in this analysis Other loads, such as the weight of the structure, alsn produce stresses at the base ol the pole.) in? “1,, ,l i r r l [7 ill FtG. 8-28 Example 8—7. Loads on (eombined axial load, bending, and shear) ti square cross section supports a horizontal platform (Fig. 828). (i in. and wall thickness t :' ()5, in. The plat- a uniformly distributed t of this distributed load A tubtilar post o The tube has outer dimension [2 r torm has ditiiensions 6.75 in X load ot It) psi acting over its upper stir 240 int and supports lace. The resultan is a vertical toice P]: [H r (20 psi‘ttb 75 in. X 24.0 in.) 3240 lb which is at distance d : ‘9 in. ie midpoint ot the platforms ’2' 800 lb acts horizon— Tliis torce acts at tl the longitudinal axis ot' the posts A second load I”: 7f 53 in. above the base aximum shear stresses at points troni tally on the post at height /1 the principal stresses and m l')eteriiiiiie hie to the loads [fl and R» A and I} at the base ot the postt Solution strains I't’,\(//[i’ltl[.\‘. The ionic i“1 ‘dtTlltlg on the plattiot'iii (Fig. 828') is statically a torte PI and a moment Mi Pal acting at the centroid “tat The load P3 is also shown in equivalent to ttoii oi the post (Fig. 87; ot the cross see this tiuui‘e, lilie stress l'QStlllllttlS ‘l/lt are shown in Fig; at the base (it the post due to the loads P1 and [’3 and the iiiotneiit . 8729b. 'l‘hese stress t‘esultaiits are the t’ollouiiig, 1. :\Il ;t\'ittl eotiipiessit'e lot‘t‘e P; i 3240 lb 2. »\ bending, moment gll, produced by the term I": Ml :7 [mi ’7 (mo tb)(9 my» :7 :wrto lb-iii. 5. :\ shear torce [’3 i “400 lb 4. A bending moment M; produced by the force PL: Mi '4 P311 :’ (800 lb)(52 in.) : 41.600 lbviri. 29b) shows that both M1 and M; t A and the shear force produces here [ixariiinatioii ot‘ these stress resultants (Figs 8- axiintini compressive stresses at poin sses at point 8. Therefore A and B are critical points w d. (Another critical point is diagonally opposite d of this example.) produce tii niaxiirium shear stre the stresses should be determine point A. as explained in the Note at the en _ Stresses at [JOHNS :l and [3‘ (l) The axial force I", (Fig. 829m produce he post. These stresses are s uniform compressive stresses throughout t in which A is the cross-sectional area of the post: A r if 7» (b I 202 : 41(1) 1 l) 4(05 mite int I 0.5 mi) 1: 11.00 in? (11‘) FIGS-29 Solution 10 fixznnplc 8A7 MH:(f/,‘+ (111‘:186()p§i , k l x 0 ()"*""1 ’ '4 vfll L7H 77W] ‘7, : [(Mps; ———> (d) (L) L mmmmd Theretore. the axial compressive stress is I) z 4 ‘ r 7 2 9T5 : 295 psi Ty :: g 7 (‘t A 11.00 in.“ town acting at points A and B in Fig 8»29e. b) produees eompressive stresses (,er are obtained from the tTexure formula: The stress (rm is sl (It The heading moment M1 (Fig 829 at points A and 1% (Fig. 829C). These stresses Ml(b/2) 7, Mil) (Wt : I 2/ in ninth I is the moment of inertia of the crossseetional area: -1 {fl lb 7 l , _ ,, . V. x— A ‘ " :V' *’ (6 in)1 * (hurt1 : 31931117 if 4 17 12 Thus the stress (rm is MI 2 ' .' ' . . 7 7 1) g ( 91.160 lh in )(o in ,t f 1504 PSI (T ' ~* “’1 :1 255.92 in“; t 1t'l‘he shear loree [’3 (Fig. 82%)) produces a shear stress at point H but not iseussion of shear stresses in the webs of he hear stress (an at point cl Front the d ants with llanges tSeetton 5.10). we know that an appi‘tiixiinate value of the s l by dividing the shear force by the web area lie obtainet at point B by the («tree P1 is '~ Those the shear stress produced P P7 XOl :l k ’ *re"*#rr ’*-***# : 16(7) psi '/'~ " ' 22th 7 m {2(0-5 WW” ’ Tm” \x elt The stress 7,») aets at point B in the direction shown in liig. 872%. u ‘ , (4) The bending moment M2 (Fig, 8'29b) produees a eonipressive stress at point A but no stress at point B, The stress at A is W [l 2 M7} / i J ' 1 A J/ T j 7 u) i (41,600lb11L)(:f)ln.) :7 2232 psi T ‘ “"3 1 21 2(5592 in?) This stress is also shown in Fig. 8429c Strcsx elmnents. The next step is to show the stresses elements at points A and B (Figs, 829d and e). Each element is oriente the v axis is vertical (that is, parallel to the longitudinal axis of the post) and the X axis is horizontal. At point A the only stress is a compressive stress (TA in the (v direetion (Fig. 829d): acting on stress d so that (M 1" UP} l (TMI + “Mr, T: 295 psi + 1564 psi + 2°32 psi : 4090 psi (compression) Thus, this element is in uniaxial stress [4‘ FiG.8—3O Notation for an element in plane stress At point B the compressive stress in the y direction (Fig. 8—29e‘) is (TH : (r,.1 + ter : 295 psi r 1564 psi : 1860 psi (compression) and the shear stress is T1“ : 160 psi The shear stress ziets lCllWZtt'Cl on the top [ace of the element and downward on the ,r face. of the element. Principal stresses and intuit (lard notation for an element in pl element A (Fig 82%) as follows: mun shear stresses at point A. Using the stun: tlnC stress (Fig. 8:30). we write the stresses l‘or (TY : t) (a “(f4 : 74090 psi TU : it Since the element is m lliilLtXilll stress the principal stresses tire if. r U (f2 : *4090 psi and the maximum iiirplttiie sheztr stress (Eq. 726) is m ~ in 4090 psi 701) v 7 7 Heir; ;; mew/H7 7' L 3 p51 TittitX 7’ ,1 j The maximum 0Ut4)ipl11tltf shear stress tEq. 7528M has the same magnitude Principal slit/saws (mil Hunt/Hum shear stresses (U [mint I)? Again USlt'lg the Sltmtl’drtl notation tor pltine stress tFie 873,0), we that the stresses at point B (Fig. 3*39Cl are in : (l (L " “UH ~ * 1300 psi rH * 7,)? ’3 7 lot) psi :1) obtain the prineipnl stresses 7 in 'i' U" (rx “UV 1 j if] r e :W t fl *W’ji'e 4r 7;. (in) s. \ 4.. l)y (I\ —-—————> , ~fiA»4—r¥ ‘,\’\ L (rx x <———— 0 ————> 77~ t‘oiilinuetl Substituting l‘or (rte me and 7w we, get (fig 1: " 9/30 psi 1: 9441 psi ()l‘ {I} 7;: psi (f7 7: J [)Sl The maximum in~plane shear stresses may be obtained < ' TV in) This term was evaluated previously so we see immediately that * :7 044 psi llHJ\ Because the principal stresses ir] and (r3 have opposite signs. the maximum in~ plane shear stresses are larger than the maximum outrollplane shear stresses A ' I ~: , .=;;, ,V Tlieretore‘ the maximum shear stress at point B is 94» psi. Nam ll the largest stresses anywhere at the base ol the post are needed. then we must also determine the stresses at the critical point diagonally opposite point A (Fig, 82%). because at that point each bending moment produces the max» imum tensile stress Thus; the tensile stress acting tit that point is (r; wry] + (er I" (I'M? i '7 295 psi + L364 psi + 2232 psi 7': 3500 psi The stresses acting on a stress element at that point (see Fig. 830) are 0‘ i 0 (r. : 3500 psi T“. i: 0 t and therefore the principal stresses and maximum shear stress are l l C a; 3500 psi m : 175017si Thus, the largest tensile stress anywhere at the base of the post is 3500 psi, the largest compressive stress is 4090 psi, and the largest shear stress is 2050 psi- (Keep in mind that only the effects of the loads P, and P2 are considered in this analysis. Other loads, such as the weight of the structure, also produce stresses at the base of the post) lé ...
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CombinedLoading - 8.5 COMBINED LOADINGS Pressure vessel (b)...

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