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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] ...
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 Spring '08
 Fosdick
 Force, Shear Stress

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