This preview shows pages 1–14. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Problem 121 An element in pmne stress is subjected to stresses 1,. or: = 6500 psi, 0'}, = 1300 psi, and 1'1}, = 1750 psi, as shown in the figure. as, = 11'00 psi
Determine the stresses acting on an element oriented at Eltl angle 9 = 60° from the .1' axis, where the angle 5' is positive when counterclockwise. Show these stresses on a sketch of an element _
0",. = 6500 p31
0 J.“ oriented tit the single 0. I“. = 27'50 psi 1,. or: = 6500 psi o", = 1300 psi 1'“, = 2750 psi
' s=sw ' '
292B P31 “.1: + [T's ".1: _ [rs _
or, = —2 ' +—ﬁ ' cos 2” + r_,._,. sin 2“ = 5180 psi 1'— ‘TJ; _ [T's _
7_,.l_,.l — —T' sin 2H + 7,}. cos 2“ = —3=‘150 psi 1— rr_.Fl = rr_,. + tr}. — is“ = 2910 p51 1— Problem 1214 Solve the preceding prohlent for ax = 31 MPa _~.,~ and o", = —5£I Iii[Pa [see figure].
' SCI MPa (3
/§
32 MP4: ——
0 .T
a
Solutlun 1214 Biaitial stress
a STEEss ELEMENT
er n.“ =1] e = sass”
/ﬂ( [1H =:r_,.. +tr}. —r:'_1.. = — [8 MPa 4—
”" . — rr .
a1 = 32 MPH '_.'.'L't'l = ——J ,3 J 51:12“ + s“. cosEIi
o", = —51} MPa u '
a 15.: [I = NIPa '—
Find angles {1" for 0' = ELI .‘t‘
or = normal stress on plane aa 15 NH
(I: + “rs “'1'— “'1' ,
rrll = + cos 2“ + a}. sin 2“
= —‘} + :1] coslliI [MPa]
. 9
For rrl.l = I}, we ohtarn cos 2“ = 2H = 7132” and H = 33.6150 1—
Problem 1215 An element in plane stress from the frame of a racing
car is oriented at a known angle 5' [see ﬁgure]. On this inclined element,
the normal and shear stresses have the magnitudes and directions shown 4139 P51 2.360 psi ,r’i
in the figure. I} = 30" Determine the nornta] and shear stresses acting on an element whose
sides are parallel to the I}? axes; that is, determine o'x, 0",, and 1'“. Show the results on a sketch of an elentent oriented at H = ﬂ”; Problem 1219 At a point in a structure subjected to plane stress,
the stresses are or}, = —4ﬂﬂﬂ psi, 0". = 25oz: psi, and 1'“. = 2342"} psi
[the sign convention for these stresses is shown in Fingl }. A stress
element located at the same point in the structure, but oriented at a counterclockwise angle H] with respect to the I axis, is subjected
to the stresses shown in the figure iota, Tb, and EllﬂEII psi). Assuming that the angle 9] is between zero and 913°, calculate
the normal stress 0.5, the shear stress Tb, and the angle El]. Solution 1219 Platte stress
cl'I = —4ﬂDEI psi o", = 15GB psi Tm, = EEGD psi ANGLE El, rr.+rr. tT.—U'.
.1. J. .1. 1 FOR 9 = '5']: rr_1.= T + cos 1“ + 7,}. sin 1“
"x. = 20”“ Psi “3. = “in Ta, = 1's soon psi = —?so — ssso cos so, + soon grass,
Find ohmb, and ﬁll or —65 cosEIEilJ + 56 sinErEi'1 — 55 = {I
Solve numerically:
STRESS Us se,= 89.12” e,= 44.5450 — o',, = or + o". — EDDIE psi = —35l2ﬂ psi 4—
' SHEAR STRESS Tb
I'FI. — (I 1. y .
= —T sinEHl + 7,}. coslrl1 ' b ' .‘L'L'r'l 3'39"} psi 4— Prehlsm ?.31 An element in plane stress is subjected te
stresses ax = esee psi, a", = lTDrIJ psi, and 1'”, = ETSD psi
{see the figure fer F'rehlem "LE—l }. Determine the principal stresses and sheer them en a
sketch ef a preperl'j.r erienterl element. Salutlen 131 Princile stresses cs, = 650'} Psi a, = tree psi en, = ease psi Therefere, er, = arse psi and a, = 24.44“
PRMCIPAL smsssss I 0'2 = 451} psi and “P, = ] 14.44“ “ft—
... I 1}. tan En = = 1.1453 F' _
(II (1,. 7,. =48.E~9° and EP=E=L=14° as .
q
as} erase“ and sp= 114.44“ 4'GP“ TIE!) psi rrl. + {1'}. rr_,.. — rrv re“ — ,3 + ,3 ' ces EH + 7,}. sin 2H Fer 1HP= 48.89”: er_,.l = Ti‘ﬁﬂ psi
Fer EHP= 223.39”: rrxl = 451] psi Prehlsm 1.32 An element in plane stress is subjected te stresses
0', = SI} MPa, e". = 52 MPa, and 1'“, = 48 MPa [see the figure fer
Prehlent "LE—2}. ' ' Determine the principal stresses and shew them en a sketch ef
a preperly eriented element. 1.} Problems 1311 through 1315 An e1emettt in plane stress {see ﬁgure}
is subjected to stresses err, 0",, and 1'“.
[st] Determine the principal stresses and show them on a sketch of EL properly.r oriented element.
[11 J Determine the moximutii shear stresses and associated normal Tn
a). stresses and show them on a sketeh of El properl],r oriented eleirtent. 0 2:
Data tor 1311 or}: = 3512"] psi, o3. = 1121} psi, i'M = —12llJ psi
SDIIJIIIJII 1311 Platte. stress
or: = 351er psi o", = ] 121] psi Tn, = —12IllII psi Therefore,
' ' titJ = 41301] psi and “pl = —22.61°
h
ta] PRINCIPAL STRESSES I.‘J'_._. = 62!] psi” and tipa = 45133”
27A}.
taitﬂtiP = = — ] .ﬂﬂSﬁ
(TI — rt"). 23? = —45.24° and HP = —12.62°
29? = 134.7%” and Hp = 6133” “rs; + “T's fr.1.'_ [rs
2 ' + ' cos 2“ + .3}. stnEH For 23F = —45.E4°: re“ =4£Iﬂll psi
For 23? = 1351.16”: :rxl =ﬁ2ll psi rrl. = Salutlun 1319 Plane stress
HI = 65m psi 11}, = —EEI}EI psi 0'}, = ?
lCline principal stress = T301} psi (tension) [a] STRESS 0", Because crI is smaller than the given principal stress, we knew that the given stress is the larger principal stress.
0'] = T301} psi rr_1..+ tr}. til". — :il' I r 2 2
rr1 = 1 + 2 + TI}.
Substitute numerical values and solve for a"; a", = —25EHII psi 4— 1'} {b} PRINCIPAL STRESSES 27A}.
tan EN}, = = —ﬂ.62222
rrl. — If}.
23 = —3l.391° and HP = —15.945°
seﬁ = 143.1139“ and tip = “H.053”
rr.+r,1'. :r.—rr.
(r11 = A '1 + A J cos 1“ + Tl}. sin 1“ “1 ‘4 For 23? = —31.39]°: [rm = TEDEl psi
For 23? = 143.109“: crxl = —33{IE psi Therefore,
:JI'J = TECH} psi anti“? = —15.95°
tr2 = —33ﬂﬂ psi and upI = 7435” 3300 psi Problem 141 An element in mimic? stress is subjected to tensile
stresses a1 = 14,513] psi, as shown in the figure. Using Mohr's circle, determine {a} the stresses acting on all 145m] P51
eleiitent oriented at a counterclockwise angle H = 24° from the .‘r
.‘r axis and [h] the maximum shear stresses and associated normal
stresses. Show all results on sketches of properlyr oriented elements. Solutlnn 1.41 Ultimtisl stress
err = 14,501} psi at,“ = {I 1'”. = {II Point 5': rrIl = R + R cos2t1= 12,100 psi
rm = R sihEIr = —53‘;HJ psi [rt] ELEMENT AT 5' = “4° [All stresses in psi] point 5: rrl.l = R — R cos EN = Alll] psi
33 = 43° = “L ° R = T251] psi 7W = 539G psi
Point ('3': ct“ = T251] psi {h} MAKE:IUM SHEAR STRESSES Point .5]: EHEI = —+§.Iﬂ° MEI = —45°
Tmﬂ = R = T251] psi Point .52: ENE: = 911° us: = 45°
ijn = —R = —?15ﬂpsi ﬂ'mr = R = 7251] psi TE 50 psi Problem "L445 An elernertt in bi'oxr'at‘ stress is subjected to stresses y err = EDEN} psi and o", = —15Eﬂ psi, as shown in the ﬁgure.
Using Mohr's circle, determine {a} the stresses acting on an element 150:] psi
oriented at a counterclockwise angle [1" = 6)" front the I axis and {b1 the
maximum shear stresses and associated normal stresses. Show all results
on sketches of properly oriented elernents. 0L 630'] P51
I Salutlun 145 Blaxial stress ax=EDEHJpsi o",=—15[HIIpsi 1' =1} 1'} [a] ELEMENTATH = 61]” [All stresses in psi} 2E3 = 121]” E = 61]” ER = 'i‘ﬁﬂﬂ psi R = STE!) psi
Point C: rrII = 2151} psi Point D: ail = 125i] — R c0561)“ = 37')" psi [11] MAKIMUM SHEAR STREESES TIM = ‘R Sinﬁﬂo = ‘3343 P5” Point 31: 2n.” = —;ui° “5' = 45'?
pm'm Dr: (Tll = 1251] + R 60560” = 4125 psi Tmﬂ = R = 3351} psi
Tim = R sinéﬂ” = 3243. psi Point 5'3: 2:13: = 91]: HS: = :159 _
rmin = _—R = —3T3ﬂ p51
(Tamer = 225D p51 3: S:
2250 psi I
4125 psi 2250 pm
600 as: = 45°
I I
3250 psi 3120 psi EDIIIHDH 2.43 Purl: shear EI=U Jv=ﬂ le=—iﬁMPa I,“ {a} ELEMENTATH = 20° 103 MP3 5.
[All stresses in MPa‘J D. / “13 MPa
2ﬂ=4ﬂ° H=2EI° R=16MPEL \ Origin 0 is at center of circle. Point D: rrxl = —R sin 2F! = —iII.2E MPa
Tm.I = —R cos 2f.I = —12.26 MPEL Point 0': = R sin 2” = 10.23 Mia;l
mm = R ms 2F.I = 12.26 I'm{Pa Problems 1445 TI'IIDIJIJII LII23 An element in plans stress is
subjected to stresses crx, or? and TI}. {see figure}. Using Mohr's circle, determine {it} the principal stresses and
lib} the maximum shear stresses and associated normal stresses.
Show all results on sketches of properly.r oriented elements. ‘1}. [11113 1D! Til11] 01. = D. a}, = —22.£1 MPa, TI? = —6.6 MP3. Eulutiun 1413 Principal slrtssts arr = 0 Cry = —22.4 MPa _‘1'
1'“ = —E.EIMPﬂ
{15111 stresses in MPa‘J P2 24.2 MP3 {b1 MAKE[UM SHEAR 5111135 3135 21’!5 = —1'J.' — 90° = — 120.51D H;I = —6I.EIS° 315: = 90° — a: = 59.49" N5: = 2934”
13"I:1111'11.11'l:1:11amI = — 11.2 MPs rm“ = R = 13.3 MPﬂ
13"I:1111'11.11'2:1:11amI = — 11.2 MPs Tm" = —13.0 MP1: R = 1’111.2J2+[6.632=13.0 MP3. 6.6
= 3121.51“
112 11.2 MB: ﬂ! = ﬂl'CtﬂTL 1:11 PRWCIPAL STRESES 211?. = —a = —30.51° 11m: —15.ss°
211“: = 130° — a: = 149.119" 1113 = MJ4°
lF"u::111'1t.F'1:::Irl = R — 11.2 = 1.3 MPﬂ 1:11:111'11.33'2:::Ir2 = —11.2— R = —24.2MPa s
1‘1 39.?4° thlem 15? Solve the preceding problem for an aluminum plate
with cl:r = 12,33!) psi {tension}, cry = —3,ﬂﬂﬂ psi fccmpressicnlt
dimensicns 2:} X 3} X 115 in., E'= 10.5 X 1135 psi, and 1' = 1133. Ecluticn 15"." Binxinl slress‘ err = 110001351 cry = —3,Elﬂﬂ psi (b; CMNGE IN TIEEC‘ICNESS
E = “35 X “351351 F = ‘133 Eq. tTB‘icl: s;= —'—.;.rx+ urn = —EE~.2.9>< 105
Dimensions of Plate: 20 in. X 30 in. X {3.5 in. E '
ShearMcclulus (Eq. 738]: hr = Pa! = — 141 X 10—5 in. 1—
E [Decrease in thickness]
6 = — = 3.941% x let psi
EH + J'l {c} CHANGE MVOLWJE
_ 2,.
fa; WWW JIMPLANE SHEAR mm me E‘l (7'43: 11" = VUC E JWx'l' ".3
Principal stresses: 0'1 = 12 DEED ps1 Vt. = [EmlmlfﬂSl = 39'] 1113
0'2 = —3.0Eﬂpsi 1—21' _6
Ir] _ “2 Also.( )trrx + rry} = 291.4 X 1'3
E . res]: .—m = = 1504:: si _
q l _ 2 P sv= tans m.3lf291.4 >< 1053
Eq‘ H351]: Tm“ = =1~gm x 1.3—6 ‘— : EDEFM in} 1— flncrease in vclu me] ...
View
Full
Document
 Spring '08
 Fosdick

Click to edit the document details