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Unformatted text preview: 8.5 COMBINED LOADINGS Pressure vessel (b) (C) FIG. 820 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. 321 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 stresstransformation 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. 821b) 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. 822 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. 824 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. 82421) 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 85 (a) (b) HG. 8—25 Example 8—5. Pressure vessel
subjected to combined internal pressure
and axial force A thinwalled 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 crosssectional 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.
()itlofipluue ,\/‘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 kNm 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'kNm)(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 kNniXZZO 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 violinum: "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 mplane 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 outollplane 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. 828 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 lbiii. 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 crosssectional 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‘) FIGS29 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(05 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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 Spring '06
 Elghandour

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