Mechanics of Materials - Problems - Solution Manual Part 11 doc

Determine the principal stresses and the maximum shearing stress at point 1 located as shown on top of the shaft.. Knowing that the mechanic applies a vertical 24-lb force at 4, determin

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

Stvess at peist H is zero

Gee 24.AUE ksi, Gy O, Tự 7 HO? kei

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

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)» 10" ps BAT7 ks:

Transverse Shear: At pomt H_ stress dve fo tuansverse sheav is zero

Resudtant stresses: 6 = 3.477 kei, 6) = O , Dy = 2297 ksi

PROBLEM 7.42 7.49 Solve Prob 7.27, using Mohr’s circle

9.27 For the state of plane stress shown, determine the largest value of a, for which

stress is equal to or less than 15 ksi

*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

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

(a) The fangest atbouahte valve

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 chìa 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

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

Thus ISMPa < 6, € 135 MPa

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 ¢, for which the

in-plane shearing stress parallel to the weld is zero, (6) the corresponding principal

so that sỹ z 2 MPa The

coovelinafes oF C cue

PROBLEM 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 7.30 For the state of plane stress.shown, determine @) the value of ¢, for which the in-plane shearing stress parallel to the weld is zero, (6) the corresponding principal

7.54 throngh 7.57 Determine the principal planes and the principal stresses for the

PROBLEM 7.56 state of plane stress resulting from the superposition of the two states of stress shown

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

8, = fo MPa , G= Oo

Try * ~ GO MPa

Cave * 4(Ge4 Gy) = 4S MPa

oe Gy € SO MPa for states of stress

£ Mohr's civele From the civede 4s aye 1s —

UAV & Mohv's circte The

ang te @ is cadeu Pat_ct From

PROBLEM 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

Range of Try ~120 MPa € Ty € 120 MPa mm

PROBLEM 7.62 mee rer ke m hung shown, ¢ seen ne m of ales of fy for which the

|#zÌ* ý K* (SY = 7150*- So" = 141.4 MPa

Range of Try - LY MPa € Ty S 1 MPA =

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, ø, = 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 kí, - Sue = 2H VF ILS ksi

Grin = Cave -R R= Gave ~ Sa

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, - 2Ø) 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.]

1 OE + €X” ( co 2Ðpces2Ð + sìa 28; sia 4©)

= OC + CX’ cos 2Ð cas2e + CX! sin 29; sìa 2©

= Sat 6Š cà 20 4 Cy sin 20 ~ sin (2Op- 20) = TX’ (sin 20, cos 20 ~ ccs 2Êp sia 4©

sin 20, cos 28 - TX'cos 2Op sin 20

PROBLEM 7.65 7.65 (đ) Prove that the expression Ø 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,

to the civede ot KL Twangle OCK

Coan = ¥ (One - Gorn) = 100 MPa —_

S& = Bare -f 60 MPa

Toman, = (Coan Gan) = tb ksi —

(be) GQ = 14 ker & = 2 ksi Ty > 8 ks #3 0x)

¬

PROBLEM 7.69 when (a) g, = 0 and a, = 12 ksi, (6) @, = 21 ksi and g, = 9 ksi (Hint: Consider both 7.69 For the state of plane stress shown, determine the maximum shearing stress

in-plane and out-of-plane shearing stresses.) SOLUTION

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

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 > £ (Bray - Son) = 85 MPa

(c) G6,=-4S MPa, 6, WS MPa, GL= -25 MPa

Omron = YS MPa, Gin = = 4SMPA ae Soni) F

7.70 and 7.71 For the state of stress shown, determine the maximum shearing stress

PROBLEM 7.71 when (a) ø,= 0, (b) 0,= +45 MPa, (c) o,=-45 MPa

ta) 6° oO 2 6.2 19S MPa, OL = Z5 MPa

6G „+ 134$ HỮa , „ve O, - Tran? Ạ,„À-6„) > 97.5 MPa -

Gone? 19S MPa, Gai = 4S MP4, 7x" 1 „-6 )= 39 MP4 z

Gunn? 195 MPA, Cain = ~4SMPQ , Tom? KBoug"See)® 120 MPa =e

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 ® $ (Grau Sain) = 6S Koi ~

(o) 6 =-dksỉ, Ø.z HH kí, 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 _

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

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\ì 6z O, Gur HM kai , 6,7 | ksi

Sra 2 1 Ms, Shin = 0, Conan ® ECCmast Crain) = 7 kes =

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)* -¬ 6.56 MP2

64 ~ G„+f = 68.44 MPa, GL = we RF = B56 MPa

6 =0 Ginn © 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 = -208.4¢.MPa, 6.20 , Gum? O

6+ -208.44 MĐA, 2> +ÁG-„- 6a) > 104.22 MPa #75 MPa

Case (2) Assume ve ©2424: 1-6 vì 75 HPA

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,

1.15 For the state ofstress shown, determine two values of ø, for which the maximum

PROBLEM 7.75 shearing stress is 7.5 ksi

6, = 10 ksi, Ty 7 kes , 2 1ẨS ksi

let us Sx Oe G7 +6,

(a) Bt +45 ksi Gy? Ro + 6y = 19 ksi ve¡e<Ì

Gn = ‡1(6x+6š }= I46 Mái, 6à* Cant Re 22ksi, Ge Sue ~ R= 7 kev

Sve 22K, Sain O, Tome? ECan Gece) = MH ko Z5 hese

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"

(6„-Š, -03* + Ure Ty”

(6.-G„)* - 2(-6x)u +ưД + se + Ty”

(6.-Gat= Ty* _ (s-I!9)”- 6Ì 2 Lg aks:

4u 7 6.- Oy is - 10

voto bd ksi G7 20+ Gye 78 ksi =

Sue = £(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

7.76 For the state of stress shown, determine the value of «, for which the maximum

PROBLEM 7.76 shearing stress is 80 MPa

Assume Guta =O Gram = ZT = 160 MPa

R= Gu4- Gi = 160-95 = 6S MPa

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

Center of Mobrs ccvede

Mes at pect Ce Lines

markect (a) show the

Reacts on Tae Limit

Center os Mohr's crete Des at point CQ R= ig ks:

Try = 4fR* -ot = £1.24 ksi _

= |2O - (2X8oÌ* - do HA =8

7.79 For the state of stress shown, determine the range of values of ¢,, for which the

PROBLEM 7.79 maximum shearing stress is equal to or less than 60 MPa

Assume ve 6 = foo MPa

Sa * Oo, = Siren - 12% + CHPa^

=> + 6o* - $o* = +oHPa

~4o MPa € Te < 4o MPa — 320 l—————— [00

*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

Then Ene (inptened is minimum iF yo zo

ốy: G~20 = Oe = HHO MPa, Bae ® 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 = Ờ, - 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

đu! TH ẹ

Optimum vadvye Por ve occurs when TumeCovteot plane) © Coen cineploue)

“ +(6.+ R)=R ov (Sat R o 6,-U 2 2Jo* + 2%

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

i GS, + BE" ks Gy = al kes 6,=0

For stresses fy xy- plane Case 7 + Out Sy VF 28.5 ks¿

6xx ¬

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 =

(bì Gyr 18 ksi R+ J(Šš5}'4 2` * (Z5Y+t0§SŸ = 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

Pes 4 6-66, = 46.732 ks > 4S ksi (Yielding occurs )

7.81 The state of plane stress shown occurs in a machine component made of a steel

PROBLEM 7.82 with ø = 45 ksi Using the maximum-distortion-energy criterion, determine whether

yield occurs when (a) ¢, = 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

= 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? Ố»v- San = 4O 21S koi < 4S ksi (Ne yiedeling )

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 * Ó

2-9? „6x * d3 Kes? > 45 ks; CYielzbna vecuvs )

Cì 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 )

7.83 The state of plane stress shown occurs in a machine component made of a steel

PROBLEM 7.83 with ¢,=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 )

(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

©) 6 = 280 MPa, G„ = -28o MPa

Gat Gaet Rt -180MPa , Gy > Sae - R= - 380 MPa

Sine = OC, Gunn = - 38O MPa

20 = G.„- Ô~ = 380 MPa > B25 MPa = (Yielehing ocews )

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 )

(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

(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 )

6, > + : “ng FRING RIOPA = 211.6 MPa

6 = 0 Cae = 4 (Ge Gy) = AG

PROBLEM 7.87 7.87 The 1.5-in-diameter shaft 4B is made of a grade of steel for which the yield

+ strength is Øy = 42 ksi Using the maximum-shearing-stress criterion, detcrmine the

magnitude of the torque T for which yield occurs when P = 60 kips,