PROBLEM 7.47 Shaft cvoss section Torsion : Tt Bene ng? Cc "1 Trans verse Sheav > Resultant stresses : ty (Ă)
7.47 Solve Prob 7.25, using Mohr’s circle
7.25 A 400-tb vertical force is applied at D to a gear attached to the solid one-inch diameter shaft AB Determine the principal stresses and the maximum shearing stress at point 1 located as shown on top of the shaft SOLUTION Equivatent force -coupte system at center of shePt in sechon at peint H V= Yoo Ah, Ms (4006) = 2400 Abin T= (400)(2) = 800 4b.in, d=-lin @etd* 0.5 in T= Bet = 0,098NS mn" T= Ơ = 0.049087 in” te: Ê800M 0-52 = yopgricdd par = 4.074 kei Mc x 0.098175 (2400 (0.5 ) = q tne = ; 0.049087 24.46 xlo psi 214.d44€ ks
Stvess at peist H is zero
Trang 2TT T | mm cơ
PROBLEM 7.48 7.48 Solve Prob 7.26, using Mohr’s circle
7.26 A mechanic uses a crowfoot wrench to loosen at bolt at E Knowing that the mechanic applies a vertical 24-lb force at 4, determine the principal stresses and the maximum shearing stress at point H located as shown on top of the 3 - in diameter shaft SOLUTION Equivalent force ~ coupte system at centev of shaP} in section at point H , Ve 24 fe (A:(29(63= lý 4km T= (22)(â) > 24o Êb-in ShaPt cvoss section: d= 0.75 in c= ‡d: 0.375 in T= Ec" = 0.031063 in* T+ 4S= 0.018532 m*
Torsion: oe VY = “J * - Je _ (24o)(0.575) _ “Gosioes = 2.897 X10 pe so 2.897 ksi oe = Me _ (ua \o.378) _ 2.47 * oe, = 22
Bending € Tt > poet AI)ằ 10" ps BAT7 ks:
Transverse Shear: At pomt H_ stress dve fo tuansverse sheav is zero
Trang 3
PROBLEM 7.42 7.49 Solve Prob 7.27, using Mohr’s circle
Trang 4Cl ca coq co co ) Cay co co — PROBLEM 7.50 8 ksi =< œ —————-———ằ— im Ye >| (b) The principal stresses are (k4: 2 1 kạ:
*7,50 Solve Prob 7.28, using Mohr’s circle
7.28 For the state of plane stress shown, determine (a) the largest value of y for which the maximum in-plane shearing stress is equal to or less than 12 ksi, (6) the corresponding principal stresses SOLUTION The center of the Mohr’s cirede Ses at pomt C with coordinates ( S24 Sy â) = (234 | e ỳ = Co sr, 0) The vadius of the evete is Toman Cineplane Y 7 42 ks
The ‘stvess pont (6y,- Ty )
die atone the Bine Xi of th
Mohy cinede diagram The extveme points with Re 2k
ave Xy aud Ke
Trang 5
PROBLEM 7.51 7.51 Solve Prob 7.29, using Mohr’s circle
7.29 Determine the range of values of g, for which the maximum in-plane shearing 75 MPa stress is equal to or less than 50 MPa
40 MPa
SOLUTION
For the Mohr's civede , petut
Y Mes at (7S MPa, 40 MPa) The vad ius of chỡa
circdes ts R= 5SOMPa bet C, be the Pocation oF the Left most Rimi bing circde and C, be thet
of the night most one 6 CY = 50 MPa C,Y z $0 MP4
Noting wight twiaugdes
C,DY and C,DY Xy —:2 Gb + Dy’ = GyY* Gp = 40 =s50° B= 30 Coordinates FP potut Care (0, 75-30) = (0, 45 MPa)
Likewise, coordinates of point C, ave 1 (0,78 #80) = CO, 105 MPa)
Coordinates of point X, ( 45-30,-40) = (15 MPa, - Yo MPa) Coordinates of point % (105+ 30, - 40} = (13S MPa, - to MPs )
The permt (6, - Ty ) must Bie on the dine XX
Trang 6—1 LÍ J NN PROBLEM 7.52
7.52 Solve Prob 7.30, using Mohr’s circle
2 MPa 7.30 For the state of plane stress shown, determine (a) the value of Â, for which the in-plane shearing stress parallel to the weld is zero, (6) the corresponding principal roy stresses SOLUTION CI oo O&O om oo ^ Pomt X of Molz cirete wiost Bie on XX" Se
x" Hat Gy = IRMPa Likewise,
+} point Y Pres on dine yy"
so that sỹ z 2 MPa The
coovelinafes oF C cue
150" 222 oO = (7Mm,0)
ẻ ( 55 A Courder Dock wise rotation
Trang 7PROBLEM 7.53
7.53 Solve Prob 7.30, using Mohr’s circle and assuming that the weld forms an angle of 60° with the horizontal
Trang 8PROBLEM 7.54 25 MPa Resu #†aô† st resses = 245 + lo = = O +17,32 2
Trang 9PROBLEM 7.55 Resultant stresses G& = 44+ 4 = BSkse Gy 2-447 > 3 ks Ty * 6+0 +6ks: TY G&Ă X oO €
Trang 10
7.54 throngh 7.57 Determine the principal planes and the principal stresses for the
Trang 11PROBLEM 7.57 % * Resuftant stresses S%&: o-?tŒ - $* o+ 3đ = tỡ % + + = Ố = 466,+6,) 20 6 R= of (SS) + TH Viết Y + (Fey tan
Trang 12io C3 4 69 Co co Co ~ 1 mmo wo | —_— - r L
7.38 For the state of stress shown, determine the of values of 0 for which the
PROBLEM 7 58 Dormal stress , is equal to or less than 100 MPa i al
6% \ SOLUTION
8, = fo MPa , G= Oo
Try * ~ GO MPa
Cave * 4(Ge4 Gy) = 4S MPa
Trang 13co co IC J C35 Co E1 L1 [E Ê9 L— L_—I Cd [J C=) CC] L_—] | PROBLEM 7.59 Dorma! stress a is equal to or less than 50 MPa 7.59 For the state of stress shown, determine the range of values of 8 for which the Gy SOLUTION 6 = FOMPa, 6ÿ O Try = - 60 MPa —> 60MPa Gave = 2 (O04 5 4S MPa Ref (SgS) + 7° 4), (Pa H x = fis*s Go" = 75 MPa = Abe GEO) _ _ yg tan 2G, co ——: 3 ze 36 z - S3 \2° B o° 28 \ |A Ẳœ@ r 7 (am On = -26.5Â65°
oe Gy € SO MPa for states of stress
Y Govresponclin te the ove HBR of
Trang 14PROBLEM 7.60 magnitude 7.60 For the state of stress shown, determine the range of values of for which the of the shearing stress x., is equal to or less than 8 ksi SOLUTION Gt = 16 ksi | gy Ty = G ksi ụ oO > fears = 10 ksi 22% Œ3%&) tan 28 = : s=s~O.7S° fe Gx~ G ~Í€ : 4â, = - 36 8?O ° 6, = - 18.435°
* te) <3 ksi Por states of stress
K ee covvesponeling fo avcs HBK and
UAV & Mohv's circte The
Trang 15PROBLEM 7.61 7.641 For the clement shown, determine the range of values of &, for which the maximum tensile stress is equal to or tess than 60 MPa 120 MPa SOLUTION kỡ 6, = -20 MPa Gy = - 120 MPa Cue = 3( 8446) = -70 MPa Set Giawr GO MP2 = Gon + R R = G~- —- 6x = 130 MPa, Re (S%ŠS)' + tả La ye [đ -(#6x)" + ơlỏo°- 6ử” = 120 MPa
Range of Try ~120 MPa € Ty € 120 MPa mm
PROBLEM 7.62 mee rer ke m hung shown, Â seen ne m of ales of fy for which the 120 MPa SOLUTION con , 6, = -20 MPa G = 20 MPa, a + (6,- 6) = 50 MPa 1c Set Tratinpteny™ R= 1SO MPa But R= (SS ype zy
|#zè* ý K* (SY = 7150*- So" = 141.4 MPa
Range of Try - LY MPa € Ty S 1 MPA =
C)
Trang 16â —1 L1 7 |
7.63 For the state of stress shown it is known that the normal and shearing stresses are
PROBLEM 7.63 directed as shown and that @,= 14 ksi, ứ, = 9 ksi, and đ„„ = 5 ksi Determine (a) the
orientation of the principal planes, () the principal stress 9,,, , (c) the maximum in- plane shearing stress
SOLUTION
là z \M kei, 6= 2 kớ, - Sue = 2H VF ILS ksi
Trang 17PROBLEM cx’ = GX’ Try
7.64 The Mohr circle shown corresponds to the state of stress given in Fig xxa and b,
7.64 page yyy Noting that o,.= OC + (CX’) cos (24, - 2ỉ) and thạt 7,v.= (CX”) sin (24,
~ 2), derive the expressions for o, and 7,,, given in Eqs (7.5) and (7.6), respectively (Hint: Use sin (A + B) = sin A cos B + cos A sin B
and cos (A + B} = cos A cos B + sin A sin B.] SOLUTION OC = 4464+ 67) Cx’ = cx TX cos Zep = CK con 2p = SG Sx cx’ sin 2đ; : Cx sin 2é; = Cy ểC + cx’ Gos (Aâs- 2)
1 OE + €X” ( co 2épces2é + sỡa 28; sia 4â)
= OC + CX’ cos 2é cas2e + CX! sin 29; sỡa 2â
= Sat 6Š cà 20 4 Cy sin 20 ~ sin (2Op- 20) = TX’ (sin 20, cos 20 ~ ccs 2ấp sia 4â
sin 20, cos 28 - TX'cos 2Op sin 20
cos 20 - — sin 20 _
Trang 18
E1 F1 E ơ1 f LL E—1 E1 [E ] E1 L1 L—] Co Cc L—] C)
PROBLEM 7.65 7.65 (đ) Prove that the expression ỉ đ„ ~ Thy , where o, , 0, , and t,- are
components of stress along the rectangular axes x’ ‘and y’ , is independent of the orientation of these axes Also, show that the given expression represents the square of the tangent drawn from the origin of the coordinates to Mohr’s circle (6) Using the invariance property established in pert 2, express the shearing stress , in terms of ,,
, and the principal stresses and on, SOLUTION , el _ vt (a) From Meohr's cive be Ty: = R sin 19; Gy = Give t R cos 4p G; = Gave - R cos 2%, a G„6/- Tey! Core = Cave ~ RƠ €0s* 209 ~ R’sin* 20, lj â ——>y = 6,2 -R* 3 independent oF Đ;
a Draw Dine OK from on'gin tanges t
Trang 19
PROBLEM 7.66 9
7.66 For the state of plane stress shown, determine the maximum shearing stress when (a) a, = 20 MPa, (6) ứ = 140 MPa (Hint: Consider both in-plane and out-of- plane shearing stresses.) SOLUTION (ay & = WO MPa, Try = 3O MPa Sy = 20 MPa %2 Cape) Gave = AG + 6) = 30 MP = Go*+ 8o“ = 100 MPa 6 = Get R = 804100 = 180 MPQ Cran) 6= Sac -R = 80-100 = ~ 20 MPa (min) đ :O
(is ) đ z(& - Gx)= {00 MPa
Coan = Ơ (One - Gorn) = 100 MPa —_ +%) (mPa) (b) & = 190 MPa, G = 14o MPa Tey = 80 MPq Can = Ê(6,+G) = 140 MPa R= (S2 %)* + Ty = Jo + 8o* = 80Mf4
6% ¿z Gae t R = 220 MPa Cman)
Trang 20c ơ —— f L— f 4 = C1 Fè ] m1 | C PROBLEM 7.67 9 80 MPa 140 MPa
Trang 21a
7.68 For the state of plane stress shown, determine the maximum shearing stress PROBLEM 7.68 when (a) ứ,= 6 ksi and @,= 18 ksi, (6) a, = 14 ksi and g,=2 ksi, (Hint: Consider both
in-plane and out-of-plane shearing stresses.) SOLUTION (a) bys 6 ksi Oy = J8 ksi Tey = B ksi TQ Cksid = \Qksi R= J(S%zŠ%Ÿ + 2 = 6* + Bt = 1o ksi Gu = Gae t RE 124 10 = AQ ker (man) 6= Gx -f= 12-lo =2 kev 6.> â Comin) Tonontionplae) = R= lO kse
Toman, = (Coan Gan) = tb ksi —
Trang 22ơ
BÀ
Trang 23
PROBLEM 7.70 7.70 and 7.71 For the state of stress shown, determine the maximum shearing stress when (a) ơ, = 0, (ð) ơ,= +45 MPa, (c) 0, = -45 MPa y SOLUTION Save? 4(6,+ 6) = 60 MPa = 8S MPa Gi = Gare + R = 145 MPa Oo = Gu - R= -25 MPa (a) 6, = 0, Ge 14S MPa 5 6 = ~ 25 MPs
Gm LIS MPA | Canin = -25 MPa, Troe * 4 (Cuz - Sax) : 35 MPa =
(6b) 6,2 +45 MPa, 6,2 IWS MPa, 6, = - 25 MPa
Sine? YS MPa, Grin? -25 MPa, T > Ê (Bray - Son) = 85 MPa
(c) G6,=-4S MPa, 6, WS MPa, GL= -25 MPa
Trang 24
7.70 and 7.71 For the state of stress shown, determine the maximum shearing stress
PROBLEM 7.71 when (a) ứ,= 0, (b) 0,= +45 MPa, (c) o,=-45 MPa y SOLUTION 6, = 1SO MPa sy = 70 MPa, +% z 7Š MPa B a Cd (MPa) 3 75 MP 6+ 46,+6,) 150 MPa = NO MPa oN R= GS ay = 8S MPa 67 Cae t R= 195 MPa 6, = Gwe -R= 2&5 MPa
ta) 6° oO 2 6.2 19S MPa, OL = Z5 MPa
6G „+ 134$ HỮa , „ve O, - Tran? Ạ,„À-6„) > 97.5 MPa -
(b) 6> +ds Mfx, G2 +19 MPa, Gy 25 MPa
Gone? 19S MPa, Gai = 4S MP4, 7x" 1 „-6 )= 39 MP4 z
() 6,7 -45 MPa, G+ 195 MPa, 6,7 25 MPa
Gunn? 195 MPA, Cain = ~4SMPQ , Tom? KBoug"See)đ 120 MPa =e
Trang 25
7.7% and 7.73 For the state of stress shown, determine the maximum shearing stress PROBLEM 7.72 when (a) a, =+4 ksi, (b) 0, = ~4 ksi, (c) 0, = 0 3 SOLUTION 6= 7T kỉ, Gx 2k, Ty = -E ker @) Ces x Save? Ê(G,+ Sy) = 4S kee 8 Â A vn Ref CGS) +
2 fast eGe + 6Sbi
Gat Garr Re Il kee ` L_
6> Gx~ Re -2 ksi le 1.9 —l——s.° ——
@ 682 4ksi Sle li ks, Gy -2 kee
Sines? WMS! , Goin? = ksi, Toon đ $ (Grau Sain) = 6S Koi ~
(o) 6 =-dksỉ, ỉ.z HH kớ, GQe-2ksi
Soo? Wisi, 6xx {#ksi, Toe? EG Game) = 7S kei =
(c) & = 0, Guz Wks, Ge - 2 kee
Sang = Hk, Grin -2 ksi, Troe? % Sram Trin VF GS KSI _
Trang 26
T1 Lễ] CJ } oo oo Cd cq 3 â
PROBLEM 7.73 7.72 and 7.73 For the state of stress shown, determine the maximum shearing stress when (a) o,=+4 ksi, (6) 0, = 4 ksi, (c) o, = 0 SOLUTION GF Sk, Gy lOksi, Tt -G6 ksi Bae = Ê064 Gy) = 7S ke (ksi) RQ - (S¿& yt Ty of Casts CeF = 6.5 ksi 6 = Sie + R 6, = Gu ~ 14 ksr Ls 6.5 d | kss (@Q) GF +4 ksi, Go = 14 ksi, 6, = ] ks¿
Girne = Mksi › Guin = ] kst Corse F 4 C6 - ` = 6.5 ks ~
(bY 6,2 -4ksi, a= MY ker, Ge Ike
Sonne = TY KS Gane =~ 4 _ ` _ (œ\ỡ 6z O, Gur HM kai , 6,7 | ksi
Sra 2 1 Ms, Shin = 0, Conan đ ECCmast Crain) = 7 kes =
Trang 27
7 5 + PROBLEM 7.74 74 For the state, of ares shown, determine two values of @, for which the maximum y SOLUTION 6, = -70 MPa, Try 4o MPa Gy - Ox : Km 40 MPa Let 0= 2 Oy ? au + 6, Save = +6, +6,)= 6, +0 70 MPa ORE {TES 0= kỷ Kho Tu
Case | Tam = RF ISMPa, v= tf 75*- 4O* = & 63.44 MPa
Qe iU= + 63.494 MPa 6 = ao + 6z “ 56.83 MPa -
Cae = 1(6+6y)* -ơ 6.56 MP2
64 ~ G„+f = 68.44 MPa, GL = we RF = B56 MPa 6 =0 Ginn â C844 MPA, Guin = - BESEMPR #75 MP (ib) us -63.44 MPa Gy = 2046, = - 196.88 MPa (reject)
Gave = 4(6.+ 5) = = 133.44 MPa Gi = Cant R= -SB.44 MPa Ge = Gu -R = -208.4Â.MPa, 6.20 , Gum? O
6+ -208.44 MĐA, 2> +ÁG-„- 6a) > 104.22 MPa #75 MPa
Case (2) Assume ve â2424: 1-6 vỡ 75 HPA Saint ISO Pa = 6 6 = Ge - R= 60 - fore Te foray =.-Ốyđ+ 0 ~ & B+ Ty = (6-6) 5+ UGG ter
20 = i Ge St Gey = Cle tise" 7-160 MPa
U-= -3oMPA Gy = 20 +9, 7 +120 MPa _
R= Jur4+ ty = SOMPa
Ci: 6, + 2R = -150 +100 = - SO MPa Ok,
Trang 28
1.15 For the state ofstress shown, determine two values of ứ, for which the maximum
PROBLEM 7.75 shearing stress is 7.5 ksi
ỹ SOLUTION
6, = 10 ksi, Ty 7 kes , 2 1ẨS ksi
let us Sx Oe G7 +6,
(a) Bt +45 ksi Gy? Ro + 6y = 19 ksi veĂe<è
Gn = ‡1(6x+6š }= I46 Mỏi, 6à* Cant Re 22ksi, Ge Sue ~ R= 7 kev
Sve 22K, Sain O, Tome? ECan Gece) = MH ko Z5 hese
(by) vu =-45 ksi Sy = 4o+6x = | ksi —
Sue = 6.46, )7 SSK, CaF Gant R= 13k, 6= Gan Re ~ 2 ksi
€xx* 13 ket 5 Cun 7-2 ks: ) Cou? 4 S. %4 À - 7Ẩ5 kes aK,
Case 2 Assume Gu = O Song = ZT mn = 15 ksi = Ge
Se ee +f v*+ Tay"
SA nh
(6„-Š, -03* + Ure Ty”
(6.-G„)* - 2(-6x)u +ưé” + se + Ty”
(6.-Gat= Ty* _ (s-I!9)”- 6è 2 Lg aks:
4u 7 6.- Oy is - 10
voto bd ksi G7 20+ Gye 78 ksi =
Sue = Ê(Gut Sy ) = 8.9 Kear Re fore T= Gl kes
Ga = Gun t+ Ro= IS kee w 6+ Saw - R= 28 ksi
Gime? IS ksi, Gain = O Lg FS ksi
Trang 29
7.76 For the state of stress shown, determine the value of ô, for which the maximum
PROBLEM 7.76 shearing stress is 80 MPa 7 SOLUTION 6z !4O MP G@> 7â MPa Gave * 4(6,+& )= 45 MPa Ge G _ a = 120-70 x aS MPa
Assume Guta =O Gram = ZT = 160 MPa
Trang 30
PROBLEM 7.77 T17 For the state of: stress shown, determine the value of f,, for which the maximum
shearing stress is (a) 9 ksi, (b) 12 ksi 4 SOLUTION 6, = 1S kei Sy F 6 kes Cue = 46+ Gy) = 10.8 kee Uy S“Š cú #6 ki TQ (ksi } (a) For Z⁄2 = 3 ks¿ Center of Mobrs ccvede
Mes at pect Ce Lines
markect (a) show the
Reacts on Tae Limit On Guan 1S Grew ? 22% = 18 ksi For fhe Mohr's cirfe 6: Gv Corresponds te potst Aa R 6 * Gare 19~ 10.5 = 7.8 ksi R= fu*+ Toy Vry 2 YR - vt uot #43J75°-44.4@* + 6 ksi —_ W bh) For Ti = 12 ker
Center os Mohr's crete Des at point CQ R= ig ks:
Trang 31PROBLEM 7.78 haber Bỡ stress shown, determine two values of o, for which the maximum y SOLUTION 6, = 70 MPa 6, = 0 %> 6â MP Mohr's civede Tae stresses in Zx-plane Cave = ACR +6, ) â 1S MỸ4 (Sz) + Bt = 4st + Got = 75 MPa 6 2 Bie t R :z lẠC MPa > 6, = Can ~ Re -30 MPa Assume 6,.,2 6, = 120 MPa +) (MP2)
6 = Sain â Sum = 2 Tae ——]
Trang 32ooo ooo eo CA cỒ O) f 1 Lu
7.79 For the state of stress shown, determine the range of values of Â,, for which the
PROBLEM 7.79 maximum shearing stress is equal to or less than 60 MPa y SOLUTION oy = 100 MPa 6, = So Ma , 6,= 0 3 Gy For Mehr's circle of stresses Cave > +(G, + 6) = 30 MPa b= “Sas = 3o
Assume ve 6 = foo MPa
Sa * Oo, = Siren - 12% + CHPa^ = loo = (2@\(60) = ~ 20 MPa R 7 Gave ~ 6 = 80~(-20) = 50 MPa G = Gu + R = 30+50 = 80 MPa < 6, đ OK R s fu Te* Te = t$R* - u*
=> + 6o* - $o* = +oHPa
Trang 33
*7.80 For the state of stress of Prob 6.66, determine (a) the value of a, for which the
PROBLEM 7.80 maximum shearing stress is as small as possible, (6) the corresponding value of the shearing stress H SOLUTION Let v= SS 6y 7 6-20 Gave = z(6+ỉy } z Ốy-U ẹ = Vu°+24* G> Gw+f* 6G xU vUt + đgc S$%&=z 6x„-KRE= G-0 - ý é*+ %đ Assume Taye ts the in-plane shearing stress Voom = R
Then Ene (inptened is minimum iF yo zo
ốy: G~20 = Oe = HHO MPa, Bae đ Ge- U = HO MPa
R= Ityl = 80 MPa
Gi = Gant R= \W0+8O = 2200 Mf
Guz Gae-R = WWO-80 = 60 MPa
Gime, = 220 MPa , Simin = ễỉ, - 7> #(G.~ 6S )= HO HP^ Assumption is in `, Assume GuE 6a = Gae +R = Ốy¿-U +3 0z 22* Surg % O Toray 2 4 Cray - Gem) * 46 d6 = : +O nọ mincmdwa } đu! TH ẹ
Trang 34
7.81 The state of plane stress shown occurs in a machine component made of.a steel
PROBLEM 7.81 wit a= 45 isi Using the rumination serpy ito determine whether
21 ksi occur, determine the corresponding factor of safety
{ SOLUTION
i GS, + BE" ks Gy = al kes 6,=0
For stresses fy xy- plane Case 7 + Out Sy VF 28.5 ks¿
6xx ơ
AY Ty 4 sĩ Re IEEY Ty = Y(7S) 4 YF 11.715 kai
Gar Git R= 40.215 ksi, Sy = Gaee- R= 16.875 ksi
Vl 65+ 6-66, = 34.977 ksi < 4S kee (Mo yielding ) F.S * gpagp 5 1287 =
(bỡ Gyr 18 ksi R+ J(Šš5}'4 2` * (Z5Y+t0ĐSŸ = 14.6 kei Gar GutR = 4B ksi, 6 Gn - Re ksi
4 6° +6, - 6.6, + 44.193 bei << 4S si (No yieleling ) | (C) By = 20 kes Re f( Meh) tr = 7S) + Go) = 21.36 ksi
B= Sant R= 49.86, Gur Sue - Ro 7.H ksị
Trang 35
7.81 The state of plane stress shown occurs in a machine component made of a steel
PROBLEM 7.82 with ứ = 45 ksi Using the maximum-distortion-energy criterion, determine whether yield occurs when (a) Â, = 9 ksi, (4) ng 18 ksi, (c) „ = 20 ksi If yield does not
{ 21 ksi occur, determine the corresponding factor of safety —>
7.82 Solve Prob 7.81, using the maximum-shearing-stress criterion
_Ể_— SOLUTION
= 6, = 36 ksi GF 2) ks 620
| Fow stresses in Xy- plane G2 = 4 (Ố,ty}= 28.5 ks: &x -Šy Fo ZS ksi
(Q) Ty = F ksi KzJ/(SSẼ)”+ tự r 1.716 ksi
Git Gan + RF 09,Ä1Eksi, Sy = 64 TT c 16.875 kes!
Gino = 34.977 ksi , Guin = O
A421? Ốằv- San = 4O 21S koi < 4S ksi (Ne yiedeling )
~ 45 _ —_
FS tage = 1.114
(b) Tey = l8 ke: + J(SzS5)`+ 2z r 148 kei
Gaz Saet RF 4B ksi , Gur Gae-R = 26
Sime = 42 ksi va * ể
2-9? „6x * d3 Kes? > 45 ks; CYielzbna vecuvs )
Cỡ Ty = 20 ksi R= f[(SESY + TZ = 21.4 ksi
6% Gare t R = 49.86 ksi G& = Sue -R= 714 ksi
Sine = 49.86 ksi Cun = O
2124 * Corina ~ Sin = 4986 Ks > 4S lest CYieleing occurs )
Trang 36
C1 (TT E`] C3 = “T1 LEL ƒ L- on t—~
7.83 The state of plane stress shown occurs in a machine component made of a steel
PROBLEM 7.83 with Â,=325 MPa Using the maximum-shearing-stress criterion, determine whether
yield occurs when (a) % = 200 MPa, (6) % = 240 MPa, (c) a = 280 MPa If yield I" does not occur, determine the corresponding factor of safety
—— e100 MPa SOLUTION
Gave = - Oe Re (Sz & Jr + Tyt = 100 MPa
(ay G2 200 MPa, Caz = - 200 MPa
6 = Sue tR = -100 MPa 6,7 Gue- R* - 300MPa
Grmug = O Guin 2 - 300 MPa
xe = Sonn Gain = ZOOMPa = 325 MPa (No yiedehing )
F.S = 345 =e joo 1,083
(b) Qr 240 MPa ) Gas = - 240 MPa
62° Ga t R= - 140 MPa ,
G = O 3 Guin = - 340 MPa
RC ow * Coan ~ Sain = 340 MPa > BAS MPa (vielliaa aecovs
Gu: Gin -R = -340 MPa
â) 6 = 280 MPa, G„ = -28o MPa
Gat Gaet Rt -180MPa , Gy > Sae - R= - 380 MPa
Sine = OC, Gunn = - 38O MPa
20 = G.„- ễ~ = 380 MPa > B25 MPa = (Yielehing ocews )
Trang 37
7.83 The state of plane stress shown occurs in a machine component made ofa steel
PROBLEM 7.84 with g,=325MPa, Using the maximum-shearing-stress criterion, determine whether
yield occurs when (2) ơ = 200 MPa, (5) a = 240 MPa, (c) 4 =280 MPa If yield does not occur, determine the corresponding factor of safety
7.84 Solve Prob 7.83, using the maximum-distortion-energy criterion
SOLUTION
Gave = - & R= (Sep Se)" 4 Dy" = 100 MPa
(ad 6 = 200 MPa Gan = - 200 MPa
6 = Sue +t Ro -100 MPa , 6, = Gae- Ro 7300 MPa
G` + 6G *°-Gđể 7 264.56 MPa < 325 MPa (No yiedohing )
F.S.= sốc 7 1.443 ~
(b) 6, = 240 MPa Gre = ~24O MPa
Git 6 +R 7-10 MPa , 6, = Gue-R = - 340 MPa
+ 6&* + @`- GỐy => 295.97 MPa < 225 MP (No yielding )
_ 32G
F.S * 2422 1,098 ~
(cÀ 6,7 280 MPa Gave = ~280 MPa
6 = Gan tR = -180 MPa, 6, 2 Gaue — R= - 380 MPa
4 &*+ G(*°- 6A6, r 3244.21 MP4 > 325 MP2 Criedding occurs )
Trang 38
| pm | No h | ie L = - | I | _ } DS ) { | io Ạ
7.85 The 38-mm-diameter shaft AB is made of a grade of steel for which the yield PROBLEM 7.85 strength is o,=250 MPa Using the maximum-shearing-stress criterion, determine the
magnitude of the torque T for which yield occurs when P = 240 KN SOLUTION P A 240x107 N “sa Edt = (says 1.134) x potme = 1.134/*/0° m s
6, > + : “ng FRING RIOPA = 211.6 MPa
Trang 40Cd Cy Co co = im: =, Mr (] oO Ít # L7 L1 ÊE] 6
PROBLEM 7.87 7.87 The 1.5-in-diameter shaft 4B is made of a grade of steel for which the yield