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Unformatted text preview: Applications of Stress/ Strain Transformations 1. Spherical Pressure Vessels Cut through the sphere on a vertical diametral plane and isolate half of the shell and its
ﬂuid contents as a single free body. Acting on this free body are the tensile stress 0' in the wall of the vessel and the ﬂuid
pressure p. The pressure acts horizontally against the plane circular area of ﬂuid
remaining inside the hemisphere. Since pressure is uniform, the resultant pressure force P=p(7rr2) (1) The resultant of the tensile stresses 0' in the wall is a horizontal force equal to the stress
times the area over which it acts, that is 0' (27rrmt) (2) where t is the thickness of the wall and rm is its mean radius:
I r = r + — 3 m 2 0
Consequently, equilibrium of forces in the horizontal direction gives 6(27rrmt)—p(7rr2)=0 (4) Applications 1  10/07/2008 Solving for the tensile stresses in the wall of the vessel: 2
$512 <5) For thin shells (r>>t), disregarding the difference between r and rm (i.e. replace rm by r),
we obtain the following formula for calculating the tensile stresses in the wall of a sphere shell. pr 0' — 2t (6)
As is evident from the symmetry of spherical shell, we obtain the same equation for the
tensile stresses when we cut through the center of the sphere in any direction whatsoever.
Thus, we reach the following conclusion: A pressurized sphere is subjected to uniform tensile stress 0' in all directions. Stresses at surfaces , (Ty = 0'
(3)654 mum iurfuu (b) {*4 fume/r Surfau’.
Outer surface think about 3D Mohr’s circle
pr
01:02:}? 03:0 (7) Therefore
01 '— 03 _ p_r r = 8
max 2 ( )
Outer surface gthink about 3D Mohr’s circle)
pr
0‘ = 0' = — 0' 2 — 9
 2 2t 3 P ( )
Therefore
7 =0'_03=ﬂ+£=ﬂ(1+_2_t_) (10) max 2 4t 2 4t r
When the vessel is thinwalled, the ratio 2t/r is very small. we can disregard the term 2t/r in Eq (10). Consequently, the maximum shear stresses at the inner and outer surfaces are
the same. In other words, we can disregard the existence of the principal stress 0'3 in the z
direction because it is small when compared with the principal stresses 0'] and 02. Applications —2— 10/07/2008 FIG. 8—5 Example 81. Spherical pressure
vessel. (Attachments and supports are
not shown.) A compressed—air tank having an inner diameter of 18 inches and a wall thickness
of 1/4 inch is formed by welding two steel hemispheres (Fig. 85). (a) If the allowable tensile stress in the steel is 14,000 psi, what is the
maximum permissible air pressure p,, in the tank? (b) If the allowable shear stress in the steel is 5,700 psi, what is the
maximum permissible pressure p,,? (c) If the normal strain at the outer surface of the tank is not to exceed
0.0003, what is the maximum permissible pressure pf? (Assume that Hooke’s
law is valid and that the modulus of elasticity for the steel is 29 X 106 psi and
Poisson’s ratio is 0.28.) (d) Tests on the welded seam show that failure occurs when the tensile
load on the welds exceeds 8.1 kips per inch of weld. If the required factor of
safety against failure of the weld is 2.5, what is the maximum permissible
pressure pd? (e) Considering the four preceding factors, what is the allowable pressure
pallow in the tank? Solution
(21) Allowable pressure based upon the tensile stress in the steel. The maximum tensile stress in the wall of the tank is given by the formula 0 : pr/Zt
Solving this equation for the pressure in terms of the allowable stress, we get I, s : r 9.0 in. : 777.8 psi T bus, the maximum allowable pressure based upon tension in the wall of the tank
is p” : 777 psi. (Note that in a calculation of this kind, we round downward, not upward.)
(b) Allowable pressure based upon the shear stress in the steel. The max— imum shear stress in the wall of the tank is given by g from which we
get the following equation for the pressure: 41m“: : 4(025 in.)(5,700 psi)
9.0m. ~ 633.3 psi pl) 7’: Therefore, the allowable pressure based upon shear is p}, : 633 psi.
(c) Allowable pressure based upon the normal strain in the steel. The normal strain is obtained from Hooke’s law for biaxial stress continued 626 CHAPTER 8 Applications of Plane Stress 1
6x : EOE — My) (h)
Substituting (TX : (Ty = (T : pr/Zt , we obtain
0" [)r
: _ 1 A : a 1 _ _
6x E( v) ZtE ( V) (8 4) This equation can be solved for the pressure pC: ageing“, 2(0.25 in.)(29 x 106 psi)(0.0003) .
.* i — 671.3 s
p‘ 1(1 A u) (9.0 in.)(l u 0.28) p“ Thus. the allowable pressure based upon the nomial strain in the wall is
pc 2 671 psi. ((1) Allowable pressure based upon the tension in the welded seam. The
allowable tensile load on the welded seam is equal to the failure load divided by
the factor of satiety: Ti'ii re ‘ i . . .
Trim. : e" '” : ELIE}:le : 3.24 k/m. : 3240 lb/1n.
n .r The corresponding allowable tensile stress is equal to the allowable load on a
oneeinch length of weld divided by the cross—sectional area of a one—inch length
of weld: M TmOWUU in.) . (32401b/in.)(l.0 in) . x, . . —12,960 U“”‘" (10 mo) (1.0111.)(0.25 m.) p“ Finally, we solve for the internal pressure by using Eq. (8»1): 2t , w . '  ' r '
01110 7 2(0 25 in )(129601381) V 7200 ps1
r 9.0111. pd T This result gives the allowable pressure based upon tension in the welded seam.
(e) Allowable pressure. Comparing the preceding results for pa, p1,, p6, and pd,
we see that shear stress in the wall governs and the allowable pressure in the tank is pallow : This example illustrates how various stresses and strains enter into the design of
a spherical pressure vessel. Note: When the internal pressure is at its maximum allowable value
(633 psi), the tensile stresses in the shell are a : B'— = ——ﬁ(633 950930 1”) : 11,400 psi 2: 2(0.25 1n.)
Thus, at the inner surface of the shell / the ratio of the principal stress
in the z direction (633 psi) tie the in—plane principal stresses (12,000 psi) is only
0.0561 Therefore, our earlier assumption that we can disregard the principal stress
03 in the z direction and consider the entire shell to be in biaxial stress is justiﬁed. 2. Cylindrical Pressure Vessels A , where 0'] is called the circumferential stress or the hoop stress; and 02 is called the
longitudinal stress or the axial stress. As demonstrated in the following derivation: 01>0'2. For circumferential stress, based on the free body in (b), we have the following equation
of equilibrium. 01(2bt)—2pbr=0 (ll) Solve for 0'1 Application 5 10/07/2008 0—1:? (12) For longitudinal stress, based on the free body in (c), we have the following equation of
equilibrium. 02(2727‘!) — par2 2 0 (l3)
Solve for 0'2
pr
0 2— l4
2 2t ( ) Comparing Eqs. (12) and (14), we notice that the circumferential stress in a cylindrical
vessel is equal to twice the longitudinal stress: a, = 20—2 (15) In addition, the longitudinal stress of a cylindrical vessel is equal to the membrane
stress in a spherical vessel. [see Eqs. (6) and (14)] Outer surface tthink about 3D Mohr’s circle) pr pr
0' =— 0' =— 0' =0 16
l t 2 3 ( )
Therefore
7 :0‘1—0'3_pr (17) max 2 — Outer surface gthink about 3D Mohr’s circle) 19’” P’”
0 =— 0‘ =— 0' =— 18
1 t 2 21 3 p ( )
Therefore
0'1—0'3 pr p pr t
r = =—— ——=—— 1+— 19 2 zt 2 2t( r) ( ) When the vessel is thin—walled, the ratio t/r is very small, we can disregard the term t/r in
Eq. (19). Consequently, the maximum shear stresses at the inner and outer surfaces are the
same. In other words, we can disregard the existence of the principal stress 03 in the z
direction because it is small when compared with the principal stresses 0'; and 02. Note: >The maximum shear stress in cylindrical pressure vessels is twice that of
spherical pressure vessels. [see Eqs. (8) and (17)] > An obvious discontinuity exists at the ends of the cylinder where the heads are
attached, because the geometry of the structure changes abruptly. Stress
concentrations occur there. The stresses at such points can not be determined
solely from the derived equations; instead, more advanced methods (such as shell
theory and ﬁnite element analysis) must be used. Application 6 10/07/2008 Helical weld FIG. 8—9 Example 8—2. Cylindrical
pressure vessel with a helical weld FlG.8—1D Solution to Example 82 A cylindrical pressure vessel is constructed from a long, narrow steel plate by
wrapping the plate around a mandrel and then welding along the edges of the
plate to make a helical joint (Fig. 8—9). The helical weld makes an angle
a : 55° with the longitudinal axis. The vessel has inner radius r = 1.8 m and
wall thickness t = 20 mm. The material is steel with modulus E = 200 GPa and
Poisson’s ratio 1/ = 0.30. The internal pressure p is 800 kPa. Calculate the following quantities for the cylindrical part of the vessel:
(21) the circumferential and longitudinal stresses 0, and 02, respectively; (b) the
maximum in—plane and out—of—plane shear stresses; (0) the circumferential and
longitudinal strains 61 and 62, respectively; and (d) the normal stress (rw and shear
stress 'rw acting perpendicular and parallel, respectively, to the welded seam. Solution (a) Circumferential and longitudinal stresses. The circumferential and
longitudinal stresses 0] and 02, respectively, are pictured in Fig. 8«10a, where
they are shown acting on a stress element at point A on the wall of the vessel. The
magnitudes of the stresses can be calculated from Eqsff and {filll: a, = B: = HMO kpa)(1'8’“)— : 72 MPa 02 : B: : 51 : 36 MPa
1 20 mm 2t 2 (b) (C) continued 632 CHAPTER 8 Applications of Plane Stress The stress element at point A is shown again in Fig. 8—10b, where the x axis is in
the longitudinal direction of the cylinder and the y axis is in the circumferential
direction. Since there is no stress in the z direction (03 Z 0), the element is in
biaxial stress. Note that the ratio of the internal pressure (800 kPa) to the smaller lH—plan€
principal stress (36 MPa) is 0.022. Therefore, our assumption that we may disret
gard any stresses in the z direction and consider all elements in the cylindrical
shell, even those at the inner surface, to be in biaaial stress is justified. , r. 7 ~ s. a .
(b) Maximum shear stresses. The largesttLinplane shear stress"‘~18 Obtained
M..._....« w ._ . 24' (Tmax)z : 01— 02 2 4 4! Because we are disregarding the normal stress in the z direction, the largest
outrof~plane shear stress is obtained from Eq. (8—911): : 3i 2 35 : 36 MPa
2 2: This last stress is the absolute maximum shear stress in the wall of the vessel (0) Circumferential and longitudinal strains. Since the largest stresses are
well below the yield stress of steel . we may assume
that Hooke’s law applies to the wall of the vessel. Then we can obtain the strains
in the x and y directions (Fig. 8—10b) , 65150;: m) e : éray A m) (gm
We note that the strain ex is the same as the principal strain 62 in the longitudinal
direction and that the strain 6), is the same as the principal strain 6, in the cir»
cumferential direction. Also, the stress 0; is the same as the stress 02, and the
stress (T). is the same as the stress (fl. Therefore, the preceding two equations can
be written in the following forms: e2: ?(1—2u)=%(1*2w (8—1121)
5, =%(2‘ V)='2%(2—2V) (811b)
Substituting numerical values, we ﬁnd
52 e UEZU 2y) — (36 MpggglGPim'wn “ 72 x 10"6
61 *— (2 u) * (72 giggégag'w) a 306 >< 10’6 These are the longitudinal and circumferential strains in the cylinder. (L y FIG. 8406 (Repeated) 47.8 MPa x1 /\ SECTION 8.3 Cylindrical Pressure Vessels 633 (d) Normal and shmr Stresses acting on the welded seam. The stress element
at point B in the wall of the cylinder (Fig. 8—10a) is oriented so that its sides are
parallel and perpendicular to the weld. The angle 6 for the element is 0:90°*a:35° as shown in Fig. 8—10c. Either the stress—transformation equations or Mohr’s
circle may be used to obtain the normal and shear stresses acting on the side faces
of this element. Stress—transtrmalirm equations. The normal stress ax] and the shear stress
Txm actingy on the X1 face of the element (Fig. 8~lOc) are obtained a, + (TV (IX ‘7 0V 0“‘ :4 e + e 2 ’ cos 26 + 7x). sin 2 6 (8—123)
(7r I a\' ~
73m ’~  if 2 ' S11] 20 + 7:“, cos 26 (8—12b)
Substituting (rA : (r2 : [)F/zl, (n, i 01 : pr/r, and 7n, : 0, we obtain
W pf I \ ‘  f pr 4
(le i (3 A cos 20) Arm A ~47 5m 26 (8—l3zi,b) These equations give the normal and shear stresses acting on an inclined plane
oriented at an angle () with the longitudinal axis of the cylinder. Substituting pr/4r : 18.MPa and f} : 350 into Eqs. (8—l3a) and (8—13b),
we obtain (51 : 47.8 MPa rxm : 169 MPa These stresses are shown on the stress element of Fig. 8— We
To complete the stress element, we can calculate the normal stress 03,] acting
on the y] face of the element from the sum of the normal stresses on perpendicular faces (8—14) 01+ (72: 0x] + (Ty, Substituting numerical values? we get 03'] : 01+ 02 i a“ : 72 MP3 + 36 MPa — 47.8 MPa : 60.2 MP3 as shown in Fig. 8—l()c. continued 534 CHAPTER 8 Applications of Plane Stress FIG. 811 Mohr's circle for the biaxial
stress element of Fig. 8—10b. (Note: All
stresses on the circle have units of MP3.) '_ B(6=90°) 0
X1 TKD‘I From the ﬁgure, we see that the normal and shear stresses acting perpendicular
and parallel, respectively, to the welded seam are (J'w = 47.8 MP3 73, = 169 MPa Mollr’s circle. The Mohr’s circle construction for the biaxial stress element
of Fig. 8—10b is shown in Fig. 8~l 1. Point A represents the stress 02 z 36 MP3 on
the x face (6 = 0) of the element, and point 8 represents the stress 01 2 72 MPa
on the y face (6 : 90°). The center C of the circle is at a stress of 54 MPa. and the
radius of the circle is : 72 MPa * 36 MPa R
2 = 18 MPa A counterclockwise angle 20 I 700 (measured on the circle from point A)
locates point D, which corresponds to the stresses on the x, face (6 = 35°) of the
element. The coordinates of point D (from the geometry of the circle) are axl = 54 MPa # Rcos 70O : 54 MP3 A (18 MPa)(cos 70°) = 47.8 MPa
7' : R sin 700 : (18 MPa)(sin 70°) 2 16.9 MP3 le1 These results are the same as those found earlier from the stress—transforrnation
equations. [0 3. Maximum Stresses in Beams M VQ
:i— d =— 20
0' I y an r 1b ( ) Direction and magnitude of the principal stress vary continuously from top to bottom of
the beam. (b) (C) ((1) lb) NorMM— &$lse.~r stresses
m Ward saw“,
(at) Mame slseu guesses Application 1 l  10/07/2008 Throughout the beam, we can construct two systems of orthogonal curves, called stress
trajectories that give the direction of the principal stresses. Solid lines are used for tensile principal stresses and dashed lines for compressive
principal stresses. The curves for tensile and compressive principal stresses always intersect at right angles,
and every trajectory crosses the longitudinal axis at 45°. At the top and bottom surfaces of the beam, where shear stress is zero, the trajectories are
either horizontal or vertical. 00 Application 12 10/07/2008 FIG. EH7 Example 8—3. Beam of
rectangular cross section FlG. 8—18 Plane—stress element at cross
section mn of the beam of Fig. 8—17
(Example 8—3) A simple beam AB with span length L : o it supports a concentrated load P :
10,800 lb acting at distance C = 2 ft from the righchand support (Fig. 817). The
beam is made of steel and has a rectangular cross section of width b 2 2 in. and
height h : 6 in. Investigate the principal stresses and maximum shear stresses at cross
section mn, located at distance x z 9 in, from end A of the beam. (Consider
only the inplane stresses.) y  ‘ P: 10,800 lb /l c=2ft—> 1f 7 risen
P RA : : 3.600 lb Solution
We begin by using the flexure and shear formulas to calculate the stresses acting on cross section (rm. Once those stresses are known, We can determine the
principal stresses and maximum shear stresses from the equations of plane stress.
Finally, we can plot graphs of these stresses to show how they vary over the height of the beam.
As a preliminary matter. we note that the reaction of the beam at support A is R14 : P/3 : 3600 lb. and therefore the bending moment and shear force at section mn Lift: M : RAX : (3600 lb)('§ in.) 1 32,400 lb~in. V : RA : 3600 lb Normal stresses on cross section mn. These stresses are found from the flexure formula , as follows: 1202.400 than.»
(2 in.)(6 in.)3 My A lZMy ' (5:) I "F—biﬁ F : e 900y (a) in which y has units of inches (in) and (IA has units of pounds per square inch
(psi). The stresses calculated from Eq. (3) are positive when in tension and
negative when in compression. For instance. note that a positive value of y
(upper half of the beam) gives a negative stress, as expected. A stress element cut from the side of the beam at cross section mm (Fig 8—17)
is shown in Fig. 818. For reference purposes. a set of xv axes is associated with
the element. The nomial stress (TX and the shear stress TX}. are shown acting on the
element in their positive directions. (Note that in this example there is no normal
stress or), acting on the element.) continued 15 CHAPTER 8 Applications of Plane Stress Shear stresses on cross section mn. The shear stresses are given by the Shear
formula in which the ﬁrst moment Q for a rectangular cross section is h h/2~ t b if
Q:b(2—y>(y+ 2 )>=§(7‘y2> (8~18) Thus, the shear formula becomes VQ 12V, 17 I12 ’2 _ﬂ(h2 2)
T711) T(bh3)(b)<2>(4 )) bh3 4 y (8‘19) The shear stresses TI). acting on the x face of the stress element (Fig. 8—18) are
positive upward, whereas the actual shear stresses 7 (Eq. 819) act downward.
Therefore, the shear stresses TX), are given by the following formula: 6V r13 Z n 3;?er “ (8‘20) Substituting numerical values into this equation gives  2
6(36()() lb) ((6 :1.) Vs) # _50(9 _ yz) (b) TM 2 M . . 2
' (2 in.)(6 in.)
in which y has units of inches (in) and 7,0, has units of pounds per square inch (p31).
Calculation of stresses. For the purpose of calculating the stresses at
cross section ma, let us divide the height of the beam into six equal intervals
and label the corresponding points from A to G, as shown in the side view of
the beam (Fig. 81921). The y coordinates of these points are listed in column
2 of Table 8] and the corresponding stresses ()‘x and TX), (calculated from EqS.
a and b, respectively) are listed in columns 3 and 4. These stresses are plotted
in Figs. 8—19b and 8~19c The normal stresses vary linearly from a
compressive stress of ~2700 psi at the top of the beam (point A) to a tensile
stress of 2700 psi at the bottom of the beam (point G). The shear stresses have
a parabolic distribution with the maximum stress at the neutral axis (point D).
Principal stresses and maximum shear stresses. The principal stresses at
each of the seven points A through G may be determined a; + 0). (IX * 0y 2 Z
(le2 = T i 2 + T”. (8—21) Since there is no normal stress in the y direction (Fig. 8—18), this equation simplifies to
0x 0X 2 2
01,2 —— 7 + + TXy Also. the maximum shear stresses ' ' , i Z T
’Tmax : ‘ (0X 2 0y) + TXZ)’ 2 ,k
7inax : szy which simplifies to By
(C)
1350
FIG. 8—19 Stresses in the beam of
Fig. 817 (Example 83). (:1) Points A,
B, C, D, E, F, and G at cross section
mn; (b) normal stresses aY acting on
cross section mn; (c) shear stresses TX).
acting on cross section rm; ((1) principal . 934
tensile stresses (7]; (e) principal com~ 1350 pressive stresses a2; and (t) maximum
shear stresses Tum. (Note: All stresses
have units of psi.) (d) (e) (f) (I l 02 7in ax TABLE 81 STRESSES AT CROSS SECTION mn IN THE BEAM OF FIG. 817  W (1) (2) (3) p (4) (5) (6) (7) Point .y 0*. Txy. ‘7 1. 02. Tm} (m) (p81) (p51) (p81) (p81) (p51) New" A 3 v2700 0 0 —2700 1350 T A I B 2 ~ 1800 —250 34 ~ 1834 934
M _ _ _ ( 1.1) l y Y MM: C 1 900 400 152 1052 602 D 0 0 — 450 450 — 450 450 Me outHM physrt owl. E 7 1 900 e 400 1052 e 152 602 ‘ s F —2 1800 A250 1834 —34 934 " '0 m G ~3 2700 0 2700 0 1350 continued [5 344 CHAPTER 8 Applications of Plane Stress Thus, by substituting the values of 0x and T”. (from Table 8—1) into Eqs. (822)
and (8—24), we can calculate the principal stresses 0'1 and (1'2 and the maximum
shear stress Tm“. These quantities are listed in the last three columns of
Table 8—1 and are plotted in Figs. 8—19d, e. and f. The tensile principal stresses (71 increase from zero at the top of the beam to
a maximum of 2700 psi at the bottom (Fig. 8—l9d). The directions of the stresses
also change, varying from vertical at the top to horizontal at the bottom. At mid
height. the stress 01 acts on a 45° plane. Similar comments apply to the com—
pressive principal stress 02, except in reverse. For instance, the stress is largest
at the top of the beam and zero at the bottom (Fig. 8—19e). The maximum shear stresses at cross section mn occur on 450 planes at
the top and bottom of the beam. These stresses are equal to one—half of the
normal stresses 0. at the same points. At the neutral axis, where the normal
stress (1. is zero. the maximum shear stresses occur on the horizontal and Ver—
tical planes. Note 1: If we consider other cross sections of the beam, the maximum
normal and shear stresses will be different from those shown in Fig. 8—19. For
instance. at a cross section between section mn and the concentrated load
(Fig. 8—17), the normal stresses ax are larger than shown in Fig. 8—l9b because
the bending moment is larger. However. the shear stresses 7'”. are the same as
those shown in Fig. 819c because the shear force doesn’t change in that region
of the beam. Consequently, the principal stresses (71 and 02 and maximum shear
stresses 7,me will vary in the same general manner as shown in Figs. 8—l9d. e,
and f but with different numerical values. The largest tensile stress anywhere in the beam is the normal stress at the
bottom of the beam at the cross section of maximum bending moment. This stress is
(atetis)rnax : The largest compressive stress has the same numerical value and occurs at the
top of the beam at the same cross section. The largest shear stress rxy acting on a cross section of the beam occurs to
the right of the load P (Fig. 8—17) because the shear force is larger in that region
of the beam (V : R3 = 7200 lb). Therefore, the largest value of T“, which
occurs at the neutral axis, is (Txy)max I The largest shear stress anywhere in the beam occurs on 45° planes at either
the top or bottom of the beam at the cross section of maximum bending moment: 14,400 .‘ .
Z #2 pg] 2 7200 psr Tmax Note 2: In the practical design of ordinary beams, the principal stresses and
maximum shear stresses are rarely calculated. Instead. the tensile and compressive
stresses to be used in design are calculated from the flexure formula at the cross section of maximum bending moment, and the shear stress to be used in design is
calculated from the shear formula at the cross section of maximum shear force. 1L. ...
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This note was uploaded on 11/04/2008 for the course CE 207 taught by Professor Elghandour during the Spring '06 term at Cal Poly.
 Spring '06
 Elghandour

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