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Unformatted text preview: PR M 4
Use Eq. (5 25) to determine the principal stresses for an arbitrary state of plane stress Gt. oy. rm, and conﬁrm Equ' I I
(5 15). l . §QLuTlgNz From equations (526):
1, = a, + 0'). i— + + 22 2 +0+0~z'2 o'I:I"r1
2 — 0x0} JyJ: 0201 fry r}: — 7:1 _ Uxay xy _ x r xv 13 =arrcr),(0)=0
From Equation (525):
3—[10‘2+120'13=0
2  (0'; + 03):)“ + [010“. — Ti]: 0 Using the quadratic equation to solve for the values of o: At a point p a material iS subjected to the state of triaxial stress I .'2 AN
cI  240 MPa cy — 420 MPa oz: 240 MPa. Determine the ‘ = :
principal stresses and the abso!ute maximum shear stress. ‘ P_B
1 r ; A [
LUTI N: "v "I I I cs,
From equations (526): ; / \‘ ‘ I,I 5”
I, = a" + 0,, + 0': = 240mg ~120MPa + 24mm: = 360MPa j i 3C r2 = axay + dryer: + age, — r; — ri — 1;: (240MPa)(—120MPa) + (~120MP0)(240MP¢:) + (240MP0)(240MP0) 0; 1’
[3 = mayor: = (240)(120)(240) = —6,912,000 ;‘ _ 3 From Equation (5—25): 0'3 — 1,03 + 12a» [3 :0 I73 — 360a2 + (0)0 + 6,912,000 = 0
Using a graphing calculator to solve for the values of o:
ANS g1 32 = 240MP§ g. = —120MPa Usig Equations (527) to ﬁnd the maximum shear stress: a, — Ulzazl 4240041251; 240MPa _ 0 «73 =l240MPﬂ_2(_120MPa)~180MPa
02:03_ 240MPa—2(—120MPa) 180MPa— Maximum shear stress is: ANS 1% = 1§QMP§ 280 a paint p a material is subjected to a state of stress (in ksi) rI . rm 300 150 100
n q, r“ = 150 200 100 Determine the principal
3 r3, :7: —~ 100 100 — 200 ' asses and the absolute maximum shear stress. Conﬁrm the
solute maximum shear stress by drawing the superimposed
3hr's circle as shown in Fig. 5.36. )LuTIQN:
3m equations (526):
II = a" + 0'7“ + 0': = 300+ 200— 200: 300 12 = 0'le + oyaz + max — rj‘.  r; — r; = (300x200) + (200x400) + (—200)(300) — (150)2 — (—100)l — (100)1 = 432,500
I} = Ottoyo“: _ (7.: r5: _ arr; _ 0—: 73} + 27.95712: I; = {300)(200)(—200} — (300)(—]00)2 (200)(100)2  (—2Ci0)(150)2 + 2(150)(—100)(100) = —15,500.000 3m Equation (525):
a3 who} +Izcr13 =0
a3 — ziooa2 — 82.5000 + 15,500,000 = 0
.lng a graphing calculator to solve for the values of 0': 43 g = 499 k§i g2 =14§ ksi g; = ~2§7 I<§i .ing Equations (527) to determine the maximum shear stress: ‘01 —az : 4091m—148kn =1305m
2 2 0‘1 «:3 =’409k§1—(257k5i) =333k5i
2 2 CT: SJ] 3 148k31_;—257ksl) = 202’s“: a maximum shear stress is: is In, = gag k§i PBQBLEM §z.11 ..
At point p of the housing of the bearing shown in Problem 62.10 ‘ is subjected to the state of plane strain a. = 0.0032. 0.. = 00026.
and 7,... = 00044. If 5’, = 0.0037. s, = 00031. determine the angle 0 and the strain if... at p. ' gg. “1.9". " 2 .5“. Equation (67):
a, + 5.. a, — a}, ' a" = (cos it?) + Lin—Lain 219) ‘ 2 2
. —. . — —. . 44 .
03037:0003244 0002:0{00032 £00026)](C0526)+[00§ ){swlﬁ} Using a graphing calculator to solve for 0: ANS 0 = 201"
Using this value for 0 in equation (6—9): 0.0044 y“. a. — 6..
2 ’7 .. {00558.30} [0.0032 — (—0.0020] (sin 26') + {if—(cos 2€)(00526) = — (510 53.3“) +[ ._ ‘ZPonis Pan are one millimeter apart in the reference state
of a material. It the material is subjected to the homogeneous y state of plane strain 5. = 0.003. a, = 0002. and y... = 0.006.
what is the distance between points P and Q in the deformed material? Q9 60°
f i. .f l
SOLUTION: p "
We use equation (6?) with 0 = 60°: : 0', — 5..
2 . g 0.003 — (—0.002) + [0.003 — 90002)] " 2 2 :3. =  0.00335 + (cos 29) + film 20)
— 0.006 («151200) +l
(sin 120°)
\ The deformed distance from P to Q is: (POT = a" (PO) + (P0) = (P0) (1 + 5', ) = (1 mm) (0.00335 + t) ANS [PQI' = Oﬁgﬁﬁ mm 312 ...
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 Spring '06
 Keil

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