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Unformatted text preview: FIG.7—18 Example 7—4. (a) Element in
plane stress, and (b) the corresponding
MOhI’s circle. (Note: All stresses on the
Circle have units of MPa.) At a point on the surface of a pressurized cylinder, the material is subjected to
biaxial stresses (TX I 90 MPa and cry = 20 MPa, as shown on the stress element
of Fig. 7—183. Using Mohr’s Circle, determine the stresses acting on an element inclined at
an angle 6 = 300. (Consider only the in—plane stresses, and show the results on a
sketch of a properly oriented element.) Solution Construction of Mohr’s circle. We begin by setting up the axes for
the normal and shear stresses, with ax] positive to the right and rm] positive
downward, as shown in Fig. 7—18b. Then we place the center C of the circle
on the an axis at the point where the stress equals the average normal stress (Eq. 731a): +
(raver—0X20} Point A, representing the stresses on the x face of the element (6 = O), has
coordinates (7X! 2 90 MP3 TX r 0 M # 4 . 72.5» W
l I) ((9 Z 300)
«I 20 i“ l
mt m . . N . . 30.3 N
A l UXI
0 B (9 = 0)
(0 : 90°)
D'
(8 = 120°)
<——~—— 55 i'—‘
h 90
TXU’I
(b)
continued 558 CHAPTER 7 Analysis of Stress and Strain Similarly, the coordinates of point 8, representing the stresses on the y face
(6 = 90°), are ‘ (III I 20 MPa Txm = 0
Now we draw the circle through points A and B with center at C and radius R
(see Eq. 7—31b) equal to _ 2 # 2 5
R:\/(0X20y)+75_\/(90MP3220MPa> +0_35Mpa Stresses On an element inclined at 6 : 30°, The stresses acting on a plane
oriented at an angle 9 I 300 are given by the coordinates of point D, which is at
an angle 26 Z 600 from point A (Fig. 718b). By inspection of the circle, we see
that the coordinates of point D are ll (Point D) 0Xl (r3Vtr + R cos 600 I 55 MPa + (35 MPa)(cos 600) = 72.5 Ml’a r : *R sin 60" 2 ~ (35 MPa)(sin 600) : —303 MPa le1 ln a‘similar manner, we can find the stresses represented by point D’,'which cor—
responds to an angle 9 = 1200 (or 29 = 240°): (Point D’) (le : (ram — R cos 60O : 55 MPa ~ (35 MPa)(cos 600) : 37.5 MPa 71m : R sin 60° : (35 MPa)(sin 60°) : 30.3 MPa These results are shown in Fig 7—19 on a sketch of an element oriented at an
angle 6 = 300, with all stresses shown in their true directions. Note that the
sum of the normal stresses on the inclined element is equal to UK + 0y, or
110 MPa FIG. 7—19 Example 7—4 (continued).
Stresses acting on an element oriented at
an angle 6 = 30° Example??? I 5,000 psi \.——» \i
l (a)
FIG. 720 Example 775. (21) Element in
plane stress, and (b) the corresponding
Moth circlerr(Noa': Allrstresses onrtheie r
circle have units of psi.) 4000 psi 15,000psi
0— . —> An element in plane stress at the surface of a large machine is subjected to stresses
(7, = 15,000 psi, (fy = 5,000 psi, and T“, = 4,000 psi, as shown in Fig 7720a. Using Mohr’s circle, determine the following quantities: (a) the stresses
acting on an element inclined at an angle 0 : 400, (b) the principal stresses, and
(c) the maximum shear stresses. (Consider only the in—plane stresses, and show
all results on sketches of properly oriented elements.) 82 (053 : 64.30) /T\\\
\
0(9:40°)
a 805
59’ 41340
W L \ 81(6p121930) C X A (a ; 0)?
// t5: i
'(951 ~25?) :
M» 10000 ——~—— 5000 A
l
<~ 15,000 ieeieeﬁl
"""i" ' "71.)" ’7 if 7’ ’ ’
0)) Solution Construction, ofMohr’s Circle. The first step in the solution is to set up the
axes for Mohr’s circle, with (I,] positive to the right and Txm positive downward
(Fig. 7—20b). The center C of the circle is located on the (IX.I axis at the point
where {IX} equals the average normal stress (Eq. 7~3la‘). 70“, + a) _5,000 psi + 5000 psi 2 2 0‘aver : l0,000 psi Point A, representing the stresSes on the x face of the element ((9 : 0), has
coordinates all : 15,000 psi 71m : 4,000 psi Similarly, the coordinates of point 8, representing the stresses on the y face ((9 = 900) are (le : 5,000 psi TX”! : 74,000 psi conﬁrmed I 10.400 psi 7?.
/ e $19.3“ P I FIG. 721 Example 7—5 (continued).
(a) Stresses acting on an element oriented at f) : 40“, (b) principal stresses, and (0) maximum shear stresses The circle is now drawn through points A and B with center at C. The radius of
the circle, from Eq. (7—31b), is \/( 15,000 psi — 5,000 psi
2 2
) + (4,000 psi)2 : 6,403 psi (a) Stresses on an element inclined at 6 I 400‘ The stresses acting on a
plane oriented at an angle 6 = 40° are given by the coordinates of point D, which
is at an angle 26 = 800 from point A (Fig. 7420b). To evaluate these coordinates,
we need to know the angle between line CD and the Crxl axis (that is, angle
DCPI), which in turn requires that we know the angle between line CA and the
(In axis (angle ACPI) These angles are found from the geometry of the circle,
as follows: [4 W _ 4,000 psi _
d" ‘ 5,000 psi 08 ACP, : 38.660 DCPI ¥ 809* ACP] : 800 7 38.660 : 41.34O Knowing these angles, we can determine the coordinates of point D directly from
the Figure 72121: (Ppip’iﬁi' axl = 10,000 psi’i'tpatis psi)(cos4l.34°) : 14,810 psi
rxm = —(6,403 psi)(sin 41.340) : 774,230 psi In an analogous manner, we can ﬁnd the stresses represented by point D’, which
corresponds to a plane inclined at an angle 0 : 1300 (or 20 : 2600): (Point D /) (er : 10,000 psi — (6,403 psi)(cos 41.34”) : 5,190 psi rxm = (6,403 psi)(sin 41.340) : 4,230 psi These stresses are shown in Fig. 7—21a on a sketch of an element oriented at an
angle 6 2 40° (all stresses are shown in their true directions). Also, note that the
sum of the normal stresses is equal to 0X + 0y, or 20,000 psi. (b) Principal stresses. The principal stresses are represented by points PI
and P2 on Mohr’s circle (Fig. 7420b). The algebraically larger principal stress
(point P.) is a! = 10,000 psi + 6,400 psi : 16,400 psi (a) FIG. 721 (Repeated) 16,400 psi /{
//0p‘ :\10.3“ \/
x ,
“\Pl 45.70
\ / ' 10,000 psi‘K
s,
(b) g (c) as seen by inspection of the circle. The angle 26m to point P, from point A is the
angle ACP] on the circle, that is, RF, : 20,,l : 38.600 0 : 193,0 I7] Thus, the plane of the algebraically larger principal stress is oriented at an angle
61,,‘ : 19.3“, as shown in Fig. 721b, The algebraically smaller principal stress (represented by point P2) is
obtained from the Circle in a similar manner: 02 if 10,000 psi 7 6.400 psi : 3,600 psi The angle 26),,2 to point P: on the circle is 38.66O + 18(0 : 218.660; thus, the " second "pr’i'ncrpaf planets deﬁned by the an'gle""t9i,2 = t09739’frTherprincipal stresses and principal planes are shown in Fig. 721b, and again we note that the
sum of the normal stresses is equal to 20,000 psi. (c) Maximum shear stresses. The maximum shear stresses are represented
by points 5. and $2 on Mohr‘s circle; therefore, the maximum in—plane shear
stress (equal to the radius of the circle) is Tnle : The angle ACSI from point A to point S, is 90° 7 38.660 : 51.340, and therefore
the angle 295] for point SI is 20,1: «51.340 This angle is negative because it is measured clockwise on the circle. The corre
sponding angle 63‘ to the plane of the maximum positive shear stress is one‘half
that value, or 6,] : *257", as shown in Figs. 7~20b and 7~21C. The maximum
negative shear stress (point S2 on the circle) has the same numerical value as the
maximum positive stress (6,400 psi).
The normal stresses acting on the planes of maximum shear stress are equal
to cram, which is the abscissa of the center C of the circle (10,000 psi). These
—stresses are also shown in Fig. 721c. Note that the planes of maximum shear
stress are oriented at 45° to the principal planes. (m. (a) FIG. 7—22 Example 7—6. (3) Element in
plane stress, and (b) the corresponding
Mohr’s circle. (Note: All stresses on the
circle have units of MPa.) At a point on the surface of a generator shaft the stresses are 0X : ’50 MP3,
{I}, : 10 MP3, and rxy I 410 MPa, as shown in Fig. 722a. Using Mohr’s circle, determine the following quantities: (a) the stresses
acting on an element inclined at an angle 6 : 45°, (b) the principal Stresses, and
(c) the maximum shear stresses (Consider only the inrplane stresses, and show
all results on sketches of properly oriented elements.) Solution Construction of Mohr’s circle The axes for the normal and shear stresses
are shown in Fig. 7422b, with a;l positive to the right and TX!“ positive down—
ward The center C of the circle is located on the an axis at the point Where the
stress equals the average normal stress (Eq. 7»31a): at + av ~ 50 MPa + IQ’MPa (raver : K 7 2 ’r Point A, representing the stresses on the x lace of the element (0 Z O), has
coordinates ‘ my ~50MPa 7W : 740 MPa Similarly, the coordinates of point B, representng the stresses on the y face
((9 = 90°), are ‘ (fo : 10 MPa TX!“ : 40 MP3 ta: 0)
0‘0
13° P1(6pl:116.6°)
3687" C 0T } “*1
5“ J0 40
D (a: 45°) L ‘ m0“) «4 20 10 (as. : 71.6") T 11M HWJPOint Dr)iﬁwo)l :  (50 MPa)(cos 36:87“)f 60 MPa The circle is now drawn through points A and B with center at C and radius R
(from Eq. 7~3lb) equal to —50 W3 A 10 MPa 2 2 : + (—40 MPa) 2 50 MPa (21) Stresses on an element inclined at 6 2 45°. The stresses acting on a
plane oriented at an angle 6 : 450 are given by the coordinates of point D, which
is at an angle 26 I 900 from point A (Fig 7722b). To evaluate these coordinates,
we need to know the angle between line CD and the negative axI axis (that is,
angle DCPZ), which in turn requires that we know the angle between line CA and
the negative ax] axis (angle ACPZ). These angles are found from the geometry of
the circle as follows: tan ACPn : ﬂMPa : ‘ 30 MPa 3 14cm : 53.130 [)CPZ : 90C * ACP2 : 90" ~ 53130 : 36.87C Knowing these angles, we can obtain the coordinates of point D directly from
Figure 723a: 3m = (50 MPa)(sin 36.870) 1 30 MPa In an analogous manner, we can find the stresses represented by point D’, which
corresponds to a plane inclined at an angle 6 : 135C (or 26 : 270°): (Point 0’) (le ’20 MPa + (50 MPa)(cos 36.870) : 20 MPa 71m : (750 MPa)(s1n 36.87”) : 730 MPa These stresses are shown in Fig. 72311 on a sketch of an element oriented at an
angle 0 : 45° (all stresses are shown in their true directions). Also, note that the
sum of the normal stresses is equal to (7x + 0)., or ~40 MPa. (b) Principal stresses The principal stresses are represented by points
P and P2 on Mohr’s circle. The algebraically larger principal stress (represented
by point P1) is a, Z 20 MPa + 50 MP3 2 30 MPa continued ~ (a) (b) (6)
FIG. 7—23 Example 76 (continued).
(a) Stresses acting on an element oriented at (2 = 45°, (b) principal as seen by inspection of the circle. The angle 26'pl to point P1 from point A is the
stresses. and (c) maximum shear angle ACPI measured counterclockwise on the circle, that is,
SiTCSSCS AC1;l : 26pl : 53.13” + 180“ : 233.13O 6P1 : 116.6O Thus, the plane of the algebraically larger principal stress is oriented at an angle
(9P1 : 116.60. The algebraically smaller principal stress (point P2) is obtained from the
circle in a similar manner: (I) t ’20 MPa * 50 MPa : ’70 MPa The angle 26p2 to point P2 on the circle is 53.130; thus, the second principal plane
7 7 77777 7 777777 is definedbyethe'angle (952 267697” ~ 7 r r " W W" " m’i’irrrﬂ'“
The principal stresses and principal planes are shown in Fig. 7—23b, and again
we note that the sum of the normal stresses is equal to (r, + 0y, or ~40 MPa.
(C) Maximum shear stresses. The maximum positive and negative shear
stresses are represented by points 51 and 52 on Mohr’s circle (Fig. 7—22b). Their
magnitudes, equal to the radius of the circle, are Tnm : 50 MP3 The angle ACSI from point A to point 5. is 90" + 53.130 : l43.13°, and there»
fore the angle 20Yl for point SI is 2951:1431? The corresponding angle 6.] to the plane of the maximum positive shear stress is
one—half that value, or 65' : 71.6“, as shown in Fig. 7—23c. The maximum nega tive shear stress (point 52 on the circle) has the same numerical value as the pos—
itive stress (50 MPa). The normal stresses acting on the planes of maximum shear stress are equal
to (ram, which is the coordinate of the center C of the circle (—20 MPa). These
stresses are also shown in Fig. 7~23c Note that the planes of maximum shear
stress are oriented at 45° to the principal planes. ...
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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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