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(1)

C

(2)
(3)

SOLUTION

0:

F

2 8cos 0: F 8co 4 ksi 708

PR

Fo on ana of 8 cos 2 A A 220 3cos 2 8 cos 20 A A os 20 sin 20 3 ksi 8 ksi

ROBLEM 7

or the given st n the oblique alysis based o Sec. 7.1A. 20 cos 20 3 0 sin 20 3 0 sin 20 3A 2 3(cos 20

7.1

tate of stress, face of the sh on the equilibr 3 cos 20 sin 2A sin 20 cos 20 cos 20 cos 2 A 2 sin 20 ) 4 determine the haded triangu rium of that el 20 3 sin 20A 2 4sin 20 0 3 sin 20A 4sin 20 cos 20 e normal and ular element s lement, as wa 0 cos 20 4 0 0 sin 20 4A 0 shearing stres shown. Use a as done in the 4 sin 20 sin 2A sin 20 cos 20 A sses exerted method of derivations 20 0

9.46 ksi

0 0

1.013 ksi

(4)

SOOLUTION F

F

60 MPa 90 M 608

PRO

For th the o based 0: A 9 180sin 30 c

0:

A 90 2 90(cos 30 MPa

OBLEM 7.2

he given state blique face of d on the equilib 90 sin 30 cosA cos30 60co 0 sin 30 sin 3A 2 sin 30 )

2

e of stress, det f the shaded t brium of that e s30 90 coA 2 os 30 30 90 cosA 60cos30 sin termine the no triangular elem element, as wa s30 sin 30 30 cos 30 30

ormal and she ment shown. U as done in the 60 cos30 cA 60 cos 30 sinA earing stresses Use a method derivations of cos30 0 3 n 30 0 7 s exerted on d of analysis f Sec. 7.1A. 32.9 M Pa 71.0 M Pa

(5)

SOLUTION 0: F 4co 0: F 4(c 10 758 4 cos1 A A os15 sin15 4 cos1 A A 2 2 cos 15 sin ksi 6 ksi 4 ksi

PROBLEM

For the give exerted on th method of an the derivation 5 sin15 10 2 10cos 15 5 cos15 1 215 ) (10

M 7.3

en state of st he oblique fa nalysis based ns of Sec. 7.1A 0 cos15 cos1A 2 6sin 15 4 0 cos15 sinA 6)cos15 sin1 tress, determi ace of the sha on the equilib A. 15 6 sin1A 4sin15 cos15 15 6 sin1A 5

ine the norma aded triangula brium of that 5 sin15 4A 5 5 cos15 4 al and sheari ar element sho element, as w sin15 cos15 A 1 4 sin15 sin15A 0 ing stresses own. Use a was done in 5 0 10.93 ksi 5 0 0.536 ksi

(6)

SOOLUTION 0 F

0

F

1 80 M 558

P

F o a o Stre 0: A 80A 2 80 cos 55

0:

A 80A 120 cos 55 si MPa 40 MPa

PROBLEM

For the given s on the oblique analysis based of Sec. 7.1A. esses cos55 cos5 A 2 40sin 55 cos 55 sin 5 A n 55

7.4

state of stress e face of the on the equilib Areas 55 40 sin A 55 40 sin A , determine th shaded triang brium of that e Forces 55 sin 55 55 cos 55 he normal and gular element element, as w 0 d shearing stre shown. Use a was done in the

0.5 5 esses exerted a method of e derivations 521 MPa 56.4 MPa

(7)

PROBLEM 7.5

For the given state of stress, determine (a) the principal planes, (b) the principal stresses.

SOLUTION

60 MPa 40 MPa 35 MPa

x y xy (a) tan 2 2 (2)(35) 3.50 60 40 xy p x y 2 p 74.05 p 37.0 , 53.0 (b) 2 2 max, min 2 2 2 2 60 40 60 40 (35) 2 2 50 36.4 MPa x y x y xy max 13.60 MPa min 86.4 MPa 40 MPa 35 MPa 60 MPa

(8)

PROBLEM 7.6

For the given state of stress, determine (a) the principal planes, (b) the principal stresses. SOLUTION 2 ksi x y 10 ksi xy 3 ksi (a) tan 2 2 (2)( 3) 0.750 2 10 xy p x y 2 p 36.87 p 18.4 ,108.4 ◄ (b) 2 2 max,min 2 2 x y x y xy 2 2 2 10 2 10 ( 3) 2 2

6 5 ksi max 11.00 ksi ◄

min 1.000 ksi ◄ 10 ksi

2 ksi 3 ksi

(9)

PROBLEM 7.7

For the given state of stress, determine (a) the principal planes, (b) the principal stresses.

SOLUTION

150 MPa,

x y 30 MPa, xy 80 MPa

(a) tan 2 2 2( 80 MPa) 1.33333 MPa

( 150 MPa 30 MPa) xy p x y 2 p 53.130 and 126.870 26.6 and 63.4 p(b) max,min 2 2 2 x y x y xy 2 2

150 MPa 30 MPa 150 MPa 30 MPa

( 80 MPa) 2 2 90 MPa 100 MPa max 190.0 MPa ◄ min 10.00 MPa ◄ 30 MPa 80 MPa 150 MPa

(10)

PROBLEM 7.8

For the given state of stress, determine (a) the principal planes, (b) the principal stresses. SOLUTION 18 ksi x y 12 ksi xy 8 ksi (a) tan 2 2 (2)(8) 0.5333 18 12 xy p x y 2 p 28.07 p 14.0 ,104.0 ◄ (b) 2 2 max,min 2 2 x y x y xy 2 2 18 12 18 12 (8) 2 2 3 17 ksi max 20.0 ksi ◄ min 14.00 ksi ◄ 12 ksi 8 ksi 18 ksi

(11)

PROBLEM 7.9

For the given state of stress, determine (a) the orientation of the planes of maximum in-plane shearing stress, (b) the maximum in-plane shearing stress, (c) the corresponding normal stress.

SOLUTION

60 MPa 40 MPa 35 MPa

x y xy (a) tan 2 60 40 0.2857 2 (2)(35) x y s xy 2 s 15.95 s 8.0 , 98.0 (b) 2 2 max 2 x y xy 2 2 60 40 (35) 2 max 36.4 MPa (c) ave 60 40 2 2 x y 50.0 MPa 40 MPa 35 MPa 60 MPa

(12)

PROBLEM 7.10

For the given state of stress, determine (a) the orientation of the planes of maximum in-plane shearing stress, (b) the maximum in-plane shearing stress, (c) the corresponding normal stress.

SOLUTION 2 ksi x y 10 ksi xy 3 ksi (a) tan 2 2 10 1.33333 2 (2)( 3) x y s xy 2 s 53.13 26.6 , 63.4 s (b) 2 2 max 2 x y xy 2 2 2 10 ( 3) 2 max 5.00 ksi (c) ave 2 10 2 2 x y

6.00 ksi

10 ksi 2 ksi 3 ksi

(13)

PROBLEM 7.11

For the given state of stress, determine (a) the orientation of the planes of maximum in-plane shearing stress, (b) the maximum in-plane shearing stress, (c) the corresponding normal stress.

SOLUTION 150 MPa, x y 30 MPa, xy 80 MPa (a) tan 2 150 30 0.750 2 2( 80) x y s xy 2 s 36.87 and 216.87 18.4 and 108.4 s (b) 2 2 max 2 x y xy 2 2 150 30 ( 80) 2 max 100.0 MPa (c) ave 2 x y 150 30 2

90.0 MPa

30 MPa 80 MPa 150 MPa

(14)

PROBLEM 7.12

For the given state of stress, determine (a) the orientation of the planes of maximum in-plane shearing stress, (b) the maximum in-plane shearing stress, (c) the corresponding normal stress.

SOLUTION 18 ksi x y 12 ksi xy 8 ksi (a) tan 2 18 12 1.875 2 (2)(8) x y s xy 2 s 61.93 s 31.0 , 59.0 (b) 2 2 max 2 x y xy 2 2 18 12 (8) 2 max 17.00 ksi (c) ave 18 12 2 2 x y 3.00 ksi 12 ksi 8 ksi 18 ksi

(15)

PROBLEM 7.13

For the given state of stress, determine the normal and shearing stresses after the element shown has been rotated through (a) 25 clockwise, (b) 10 counterclockwise.

SOLUTION 0 8 ksi 5 ksi 4 ksi 4 ksi 2 2 cos 2 + sin 2 2 2 sin 2 + cos 2 2 cos 2 sin 2 2 2 x y xy x y x y x y x y x xy x y x y xy x y x y y xy (a) 25 2 50 4 4 cos ( 50°) + 5 sin ( 50°) x x 2.40 ksi 4 sin ( 50 ) 5 cos ( 50 ) x y x y 0.1498 ksi 4 4 cos ( 50 ) 5 sin ( 50) y y 10.40 ksi (b) 10 2 20 4 4 cos (20°) + 5 sin (20°) x x 1.951 ksi 4 sin (20°) + 5 cos (20°) x y x y 6.07 ksi 4 4 cos (20°) 5 cos (20°) y y 6.05 ksi 8 ksi 5 ksi

(16)

PROBLEM 7.14

For the given state of stress, determine the normal and shearing stresses after the element shown has been rotated through (a) 25 clockwise, (b) 10 counterclockwise.

SOLUTION

60 MPa 90 MPa 30 MPa

15 MPa 75 MPa 2 2 cos 2 + sin 2 2 2 sin 2 + cos 2 2 cos 2 sin 2 2 2 x y xy x y x y x y x y x xy x y xy xy x y x y y xy (a) 25 2 50 15 75 cos ( 50 ) 30 sin ( 50 ) x x 56.2 MPa 75 sin ( 50 ) 30 cos ( 50 ) x y x y 38.2 MPa 15 75 cos ( 50 ) 30 sin ( 50 ) y y 86.2 MPa (b) 10 2 20 15 75 cos (20°) + 30 sin (20°) x x 45.2 MPa 75 sin (20°) + 30 cos (20°) x y x y 53.8 MPa 15 75 cos (20°) 30 sin (20°) y y 75.2 MPa 90 MPa 30 MPa 60 MPa

(17)

PROBLEM 7.15

For the given state of stress, determine the normal and shearing stresses after the element shown has been rotated through (a) 25 clockwise, (b) 10 counterclockwise.

SOLUTION

8 ksi 12 ksi 6 ksi

2 ksi 10 ksi 2 2 cos 2 + sin 2 2 2 sin 2 + cos 2 2 cos 2 sin 2 2 2 x y xy x y x y x y x y x xy x y x y xy x y x y y xy (a) 25 2 50 2 10 cos ( 50 ) 6 sin ( 50 ) x x 9.02 ksi 10 sin ( 50 ) 6 cos ( 50 ) x y x y 3.80 ksi 2 10 cos ( 50 ) 6 sin ( 50 ) y y 13.02 ksi (b) 10 2 20 2 10 cos (20°) 6 sin (20°) x x 5.34 ksi 10 sin (20°) 6 cos (20°) x y x y 9.06 ksi 2 10 cos (20°) + 6 sin (20°) y y 9.34 ksi 12 ksi 6 ksi 8 ksi

(18)

PROBLEM 7.16

For the given state of stress, determine the normal and shearing stresses after the element shown has been rotated through (a) 25 clockwise, (b) 10 counterclockwise.

SOLUTION 0 80 MPa 50 MPa 40 MPa 40 MPa 2 2 cos 2 sin 2 2 2 sin 2 + cos 2 2 cos 2 sin 2 2 2 x y xy x y x y x y x y x xy x y x y xy x y x y y xy (a) 25 2 50 40 40 cos ( 50 ) 50 sin ( 50°) x x 24.0 MPa 40 sin ( 50°) 50 cos ( 50 ) x y x y 1.498 MPa 40 40 cos ( 50 ) 50 sin ( 50 ) y y 104.0 MPa (b) 10 2 20 40 40 cos (20°) 50 sin (20°) x x 19.51 MPa 40 sin (20°) 50 cos (20°) x y x y 60.7 MPa 40 40 cos (20°) + 50 sin (20°) y y 60.5 MPa 80 MPa 50 MPa

(19)

PROBLEM 7.17

The grain of a wooden member forms an angle of 15° with the vertical. For the state of stress shown, determine (a) the in-plane shearing stress parallel to the grain, (b) the normal stress perpendicular to the grain.

SOLUTION

0

x y xy 250 psi 15

(a) sin 2 cos 2

2 x y x y xy 0 250cos( 30 ) 217 psi x y (b) cos 2 sin 2 2 2 x y x y x xy 0 0 250sin( 30 ) 125.0 psi x 250 psi 158

(20)

PROBLEM 7.18

The grain of a wooden member forms an angle of 15° with the vertical. For the state of stress shown, determine (a) the in-plane shearing stress parallel to the grain, (b) the normal stress perpendicular to the grain.

SOLUTION

3 MPa

x y 1.8 MPa xy 0

15 2 30

(a) sin 2 sin 2

2 x y x y xy 3 1.8 sin( 30 ) 0 2 0.300 MPa x y (b) cos 2 sin 2 2 2 x y x y x xy 3 1.8 3 1.8 cos( 30 ) 0 2 2 2.92 MPa x 1.8 MPa 3 MPa 158

(21)

SO The P' 80 m LUTION e smaller value mm b 120 all a (8 A N 0: y F all a S 0: x F e for P govern b 0 mm

P

Tw se th ar an ce Forces all 0)(120) 9.6 (4 /sinA sin N P all (6 /sinA : S Pcos ns. P

PROBLEM 7

wo wooden m ection are join

hat 22 a re, respectivel nd 600 kPa in entric load P th 3 2 3 10 mm 9 400 10 )(9.6 sin 22 0 sin N P 3 600 10 )(9.6 sin 22 0 co P

7.19

members of 80 ed by the simp and that the m

ly, 400 kPa in n shear (parall

hat can be app

Are 3 2 3 .6 10 m 10 ) 10.2 10.251 n sin 22 N 3 10 ) 15.3 15.376 os cos 2 S 0 120-mm u ple glued scar maximum allow n tension (per el to the splic plied. eas 3 251 10 N 3 10 27.4 1 2 3 376 10 N 3 10 16.58 22 uniform rectan rf splice shown wable stresses rpendicular to ce), determine 3 10 N 3 10 N 1 P ngular cross n. Knowing in the joint the splice) e the largest 6.58 kN

(22)

SO (a) (b) P' 80 OLUTION x F y F mm 12 0: S Pcos (9 /sin N A 0: N Psin (4 /sin N A b 20 mm

P

Tw se th ap sh pe Forces (80)(12 A 0 S P 3 3 .063 10 )sin 9.6 10 n 0 N 3 3 .226 10 )sin 9.6 10 P

ROBLEM 7

wo wooden m ction are joine at 25 a pplied to the hearing stress erpendicular to s 3 20) 9.6 10 m cos (10 P n 25 399 10 sin (10 P 25 186.0

7.20

members of 80 ed by the simp and that centri

members as s parallel to o the splice. A 2 mm 9.6 10 3 10 )cos 25 3 0 Pa 3 10 )sin 25 3 10 Pa 0 120-mm u mple glued scar

ic loads of m s shown, det the splice, Areas 3 2 0 m 3 9.063 10 N 3 4.226 10 N uniform rectan rf splice show magnitude P termine (a) t (b) the no N N 1 ngular cross wn. Knowing 10 kN are the in-plane ormal stress 399 kPa 186.0 kPa

(23)

SO (a) (b) a ! LUTION From the M 2( 1 co P A P a

PROBL

The centr plane a-a with the h 0 0 x xy y P Mohr’s circle, ) (2)(1 os 2 1 cos

LEM 7.21

ric force P is are 15 horizontal, (b) / P A tan 15) s 2 applied to a ksi and 5 the maximum 5 0.33 15 2 2 P P A A short post as 5 ksi, determin m compressive 333 cos 2 shown. Know ne (a) the ang e stress in the p

wing that the gle that plan

post. 1 P A stresses on ne a-a forms 18.4 16.67 ksi

(24)

SO For All 50 m OLUTION r plane a-a, lowable value P mm a

PR

Two a-a stres cent 65 . 0 s s x P P of P is the sm a 25"

OBLEM 7.2

o members of that forms an sses for the glu tric load P that

2 2 0, 0, cos (50 sin 65 ( )sin sin 65 cos 65 xy x y x y A A maller one.

22

uniform cross n angle of 25 ued joint are

t can be applie 2 3 2 3 sin 2 10 )(80 1 sin 6 cos ( (50 10 )( y y xy xy P A sin s section 50 5 with the h 800 kPa a ed. 3 3 2 2 sin cos 0 )(800 10 65 (cos sin (80 10 )(603 n 65 cos 65 80 mm are g horizontal. Kn and 600 k 2 3 3 0 sin 65 ) 3.90 10 ) sin 65 P A P A 3 0 10 ) 6.27 glued together nowing that th kPa, determin 5 0 N cos 65 0 3 7 10 N P r along plane he allowable ne the largest 3.90 kN

(25)

PROBLEM 7.23

The axle of an automobile is acted upon by the forces and couple shown. Knowing that the diameter of the solid axle is 32 mm, determine (a) the principal planes and principal stresses at point H located on top of the axle, (b) the maximum shearing stress at the same point. SOLUTION 3 1 1(32) 16 mm 16 10 m 2 2 c d Torsion: 6 3 3 3 2 2(350 N m) 54.399 10 Pa 54.399 MPa (16 10 m) Tc T J c Bending: 4 3 4 9 4 3 3 6 9 (16 10 ) 51.472 10 m 4 4 (0.15m)(3 10 N) 450 N m (450)(16 10 ) 139.882 10 Pa 139.882 MPa 51.472 10 I c M My I

Top view: Stresses:

ave 2 2 2 2 139.882 MPa 0 54.399 MPa 1 1 ( ) ( 139.882 0) 69.941 MPa 2 2 ( 69.941) ( 54.399) 88.606 MPa 2 x y xy x y x y xy R

(a) max ave R 69.941 88.606 max 18.67 MPa

min ave R 69.941 88.606 min 158.5 MPa

3 kN 3 kN 350 N · m 0.15 m H 0.2 m

(26)

PROBLEM 7.23 (Continued) 2 (2)( 54.399) tan 2 0.77778 2 37.88 139.882 xy p p x y 18.9 and 108.9° p

(27)

PROBLEM 7.24

A 400-lb vertical force is applied at D to a gear attached to the solid l-in. diameter shaft AB. Determine the principal stresses and the maximum shearing stress at point H located as shown on top of the shaft.

SOLUTION

Equivalent force-couple system at center of shaft in section at point H:

400 lb (400)(6) 2400 lb in.

(400)(2) 800 lb in.

V M

T Shaft cross section:

4 4 4 1 1 in. 0.5 in. 2 1 0.098175 in 0.049087 in 2 2 d c d J c I J

Torsion: (800)(0.5) 4.074 10 psi 4.074 ksi3

0.098175 Tc

J

Bending: (2400)(0.5) 24.446 10 psi3 24.446 ksi

0.049087 Mc

I

Transverse shear: Stress at point H is zero.

ave 2 2 2 2 24.446 ksi, 0, 4.074 ksi 1 ( ) 12.223 ksi 2 (12.223) (4.074) 2 12.884 ksi x y xy x y x y xy R ave a R a 25.1 ksi ave b R b 0.661 ksi

max R max 12.88 ksi

6 in. 2 in. D A B H C 400 lb

(28)

PROBLEM 7.25

A mechanic uses a crowfoot wrench to loosen a bolt at E. Knowing that the mechanic applies a vertical 24-lb force at A, determine the principal stresses and the maximum shearing stress at point H located as shown as on top of the 3

4-in. diameter shaft.

SOLUTION

Equivalent force-couple system at center of shaft in section at point H:

24 lb (24)(6) 144 lb in.

(24)(10) 240 lb in.

V M

T Shaft cross section:

4 4 4 1 0.75 in., 0.375 in. 2 1 0.031063 in 0.015532 in 2 2 d c d J c I J

Torsion: (240)(0.375) 2.897 10 psi3 2.897 ksi

0.031063 Tc

J

Bending: (144)(0.375) 3.477 10 psi 3.477 ksi3

0.015532 Mc

I

Transverse shear: At point H, the stress due to transverse shear is zero. Resultant stresses: ave 2 2 2 2 3.477 ksi, 0, 2.897 ksi 1 ( ) 1.738 ksi 2 1.738 2.897 3.378 ksi 2 x y xy x y x y xy R ave a R a 5.12 ksi ave b R b 1.640 ksi

max R max 3.38 ksi

24 lb 10 in. 6 in. E B A H

(29)

SO Forc Tor 10 kN LUTION ce-couple syst sion: At po 200 mm 6 m z N C tem at center o oint K, place l mm 1 T A y K H A B

P

Th th de at 4 1 2 2 4.1855 1 2 o o o d r J r I J of tube in the p 10 kN 10 10 (10 1 2000 N (10 1500 x y z F M M local x-axis in 51 ( 24.37 24.37 y o xy T M c r Tc J 150 mm 51 mm x D

PROBLEM 7

he steel pipe A hickness. Kno etermine the p t point K. 4 6 4 6 102 51 mm 2 4.1855 5 10 m 2.0927 10 m i r plane containi 3 3 3 3 0 N 0 )(200 10 N m 10 )(150 10 0 N m negative glob 3 6 6 2000 N m 1 10 m (2000)(51 10 4.1855 10 10 Pa MPa

7.26

AB has a 102 wing that arm principal stres 6 4 4 5 10 mm m i o r r t ing points H a 3 3 ) ) bal z-direction 3 6 0 ) 2-mm outer di m CD is rigid sses and the m

45 mm and K: n. ameter and a dly attached t maximum she 6-mm wall to the pipe, aring stress

(30)

PROBLEM 7.26 (Continued)

Transverse shear: Stress due to transverse shear V F is zero at point K. x Bending: 3 6 6 | | (1500)(51 10 ) | | 36.56 10 Pa 36.56 MPa 2.0927 10 z y M c I

Point K lies on compression side of neutral axis.

36.56 MPa y

Total stresses at point K:

ave 2 2 0, 36.56 MPa, 24.37 MPa 1 ( ) 18.28 MPa 2 30.46 MPa 2 x y xy x y x y xy R

max ave R 18.28 30.46 max 12.18 MPa

min ave R 18.28 30.46 min 48.7 MPa

(31)

PROBLEM 7.27

For the state of plane stress shown, determine the largest value of yfor which the maximum in-plane shearing stress is equal to or less than 75 MPa.

SOLUTION 60 MPa, ?, 20 MPa x y xy Let . 2 x y u Then 2 2 2 2 2 2 2 75 MPa 75 20 72.284 MPa 2 60 (2)(72.284) 84.6 MPa or 205 MPa y x xy xy y x u R u u R u

Largest value of y is required. y 205 MPa

60 MPa 20 MPa #y

(32)

PROBLEM 7.28

For the state of plane stress shown, determine (a) the largest value of xy for which the maximum in-plane shearing stress is equal to or less than 12 ksi, (b) the corresponding principal stresses.

SOLUTION 2 2 2 2 max 2 2 10 ksi, 8 ksi, ? 10 ( 8) 9 12 ksi x y xy x y xy xy xy R z z (a) 122 92 xy xy 7.94 ksi (b) ave 1( ) 1 ksi 2 x y ave 1 12 13 ksi a R a 13.00 ksi ave 1 12 11 ksi b R b 11.00 ksi 8 ksi 10 ksi $xy

(33)

SO Sinc (a) (b) LUTION ce x y 0, x 1 ( 2 xy x ave 1 ( 2 x R ave a ave b 2 MPa 75"

P

F in p -direction is a ) tan 2 x y 2 2 2 ) 7 M x y xy x y 7 5.77 R 7 5.77 R $xy 12 MPa a

PROBLEM

For the state o n-plane shear principal stress 12 MPa, x a principal dire tan 2 p p 1 (12 2) 2 p 2 5 2.89 MPa y 735 735

7.29

f plane stress ring stress pa ses. 2 MPa y ection. 15 2 xy x y ) tan( 30 ) 2 9 5.7735 M shown, determ arallel to the a, xy ? MPa

mine (a) the va weld is zero alue of xy fo o, (b) the cor 2. xy 12. a 1.2 b or which the rresponding 89 MPa .77 MPa 226 MPa

(34)

PROBLEM 7.30

Determine the range of values of x for which the maximum in-plane shearing stress is equal to or less than 10 ksi.

SOLUTION ?, 15 ksi, 8 ksi x y xy Let 2 2 max 2 2 2 2 2 2 10 ksi 10 8 6 ksi 2 15 (2)(6) 27 ksi or 3 ksi x y x y xy z xy x y u u R u u R u

Allowable range: 3 ksi x 27 ksi

15 ksi 8 ksi

(35)

SO Plot (a) (b) (a ) (b ) (c ) LUTION

tted points for X Y C tan GX CG 74 1 2 b 18 1 2 a R C min a max a d B e A max R a 40 MPa 35

PR

So PR pri PR ori in-ave 6 4 35 x y xy r Mohr’s circle ave : ( , ) : ( , ) ( : ( , 0) ( x xy y xy X Y C 35 3.50 10 X G 4.05 1 37.03 2 0 105.9 52.97 2 2 CG GX ave R 50 ave R 50 45 7.97 B 45 97.97 A 36.4 MPa ave 50 MPa a 5 MPa 60 MPa

ROBLEM 7

olve Probs. 7.5 ROBLEM 7.5 incipal planes ROBLEM 7.9 ientation of th -plane shearin 60 MPa, 40 MPa, 5 MPa 50 2 x y e: ( 60 MPa, ( 40 MPa, 35 ( 50 MPa, 0) 00 95 2 2 10 35 36.4 36.4 7 a

7.31

5 and 7.9, usin 5 through 7. , (b) the princ 9 through 7. he planes of m ng stress, (c) th MPa 35 MPa) MPa) 36.4 MPa ng Mohr’s circ .8 For the giv

ipal stresses. 12 For the gi maximum in-p he correspondi cle. ven state of s iven state of plane shearing ing normal str stress, determ stress, determ stress, (b) the ress. b a min 86 max 13 e max 36 50

mine (a) the mine (a) the e maximum 37.0 53.0 6.4 MPa 3.60 MPa 8.0 d 98.0 e 6.4 MPa 0.0 MPa

(36)

PROBLEM 7.32

Solve Probs. 7.7 and 7.11, using Mohr’s circle.

PROBLEM 7.5 through 7.8 For the given state of stress, determine (a) the principal planes, (b) the principal stresses.

PROBLEM 7.9 through 7.12 For the given state of stress, determine (a) the orientation of the planes of maximum in-plane shearing stress, (b) the maximum in-plane shearing stress, (c) the corresponding normal stress.

SOLUTION 150 MPa x 30 MPa y 80 MPa xy ave 90 MPa 2 x y

Plotted points for Mohr’s circle:

: ( ,x xy) (150 MPa, 80 MPa) X : ( ,y xy) (30 MPa, 80 MPa) Y ave : ( , 0) (90 MPa, 0) C (150 30) 60 2 2 x y 2 2 (60) (80) 100 R (a) tan 2 80 60 p 2 p 53.130 p 26.6 and 63.4

(b) max ave R 90 100 max 190.0 MPa

min ave R 90 100 min 10.00 MPa

(a′) s p 45 s 18.4 and 108.4

(b′) max R max 100.0 MPa

(c ) ave 90.0 MPa

30 MPa

80 MPa 150 MPa

(37)

PROBLEM 7.33

Solve Prob. 7.10, using Mohr’s circle.

PROBLEM 7.9 through 7.12 For the given state of stress, determine (a) the orientation of the planes of maximum in-plane shearing stress, (b) the maximum in-plane shearing stress, (c) the corresponding normal stress.

SOLUTION 2 ksi x y 10 ksi xy 3 ksi ave 2 10 6 ksi 2 2 y x

Plotted points for Mohr’s circle:

: ( ,x xy) (2 ksi, 3 ksi) X : ( ,y xy) (10 ksi, 3 ksi) Y ave : ( , 0) (6 ksi, 0) C 3 tan 0.75 4 FX FC 36.87 1 18.43 2 B (a) D B 45 26.6 D 26.6 45 63.4 E B E 63.4 2 2 2 2 4 3 5 ksi R CF FX

(b) max R 5.00 ksi max 5.00 ksi

(c) ave 6.00 ksi 6.00 ksi

10 ksi

2 ksi 3 ksi

(38)

PROBLEM 7.34

Solve Prob. 7.12, using Mohr’s circle.

PROBLEM 7.9 through 7.12 For the given state of stress, determine (a) the orientation of the planes of maximum in-plane shearing stress, (b) the maximum in-plane shearing stress, (c) the corresponding normal stress.

SOLUTION

18 ksi

x y 12 ksi xy 8 ksi

ave x 2 y 3 ksi Plotted points for Mohr’s circle:

: ( ,x xy) (18 ksi, 8 ksi) X : ( ,y xy) ( 12 ksi, 8 ksi) Y ave : ( , 0) (3 ksi, 0) C 8 tan 0.5333 15 FX CF 28.07 1 14.04 2 A (a) D A 45 59.0 D 59.0 45 30.1 E A E 30.1 2 2 2 2 15 8 17 ksi R CF FX

(b) max R 17.00 ksi max 17.00 ksi

(c) ave 3.00 ksi 3.00 ksi

12 ksi 8 ksi

(39)

SO LUTION 8 ksi 5 ksi

PROB

Solve Pr PROBL shearing (b) 10 c

LEM 7.35

rob. 7.13, usin LEM 7.13 thr g stresses after counterclockw Plot (a) (b) ng Mohr’s circ ough 7.16 Fo r the element wise.

tted points for

tan 2 2 p p R 25 10 cle.

r the given sta shown has be ave 0 8 5 x y xy r Mohr’s circle : (0, : (8 k : (4 k X Y C 2 5 1 4 51.34 FX FC FC FX . 2 5 5 x x y R y . 2 2 5 x x y R y ate of stress, d een rotated thr 0, 8 ksi, 5 ksi 4 ks 2 x y e: 5 ksi) ksi, 5 ksi) ksi, 0) 2 2 2 1.25 4 5 X 0 1.34 50 ave Rcos sin R ave Rcos 0 1.34 20 ave Rcos sin R ave Rcos determine the rough (a) 25 si 6.4031 ksi 1.34 x 0. x y 1 y 71.34 1 x x y 6 y normal and clockwise, 2.40 ksi 1497 ksi 0.40 ksi 1.951 ksi 6.07 ksi 6.05 ksi

(40)

SO Plo (a) OLUTION otted points fo 25 90 MP 3 r Mohr’s circl : : : X Y C tan 2 p 2 p R . 2 5 2 x x y y Pa 30 MPa 60 MPa

PROBLEM

Solve Prob. PROBLEM normal and s (a) 25 clock ave 60 MP 90 MPa, 30 MPa 2 x y xy x y le: : ( 60 MPa, : (90 MPa, 30 : (15 MPa, 0) 30 0 75 FX FC 21.80 P 2 FC FX 50 2 2 P 50 ave Rcos sin R ave Rcos

M 7.36

7.14, using M M 7.13 throug shearing stres kwise, (b) 10 Pa, , 15 MPa 30 MPa) 0 MPa) 0.4 10.90 2 2 75 30 X 21.80 28 Mohr’s circle. gh 7.16 For th

ses after the e counterclock 2 0 80.78 MP 8.20 he given state element shown kwise. Pa e of stress, de n has been rota

5 x 3 x y 8 y etermine the ated through 56.2 MPa 38.2 MPa 86.2 MPa

(41)

(b) 10 . 2 20 2 x x y R y PROB 0 2 21.8 p ave Rcos sin R ave Rcos BLEM 7.36 ( 80 20 41 (Continued) .80 d) 45 x 53 x y 75 y 5.2 MPa 3.8 MPa 5.2 MPa

(42)

SOOLUTION 12 ksi 6

PR

Solv PRO and cloc 6 ksi 8 ksi

ROBLEM 7.

ve Prob. 7.15, OBLEM 7.13 shearing stre ckwise, (b) 10 Plo (a) (b)

.37

using Mohr’s 3 through 7.1 esses after the

counterclock otted points fo tan 2 2 25 10 s circle. 6 For the give e element sho kwise. ave 8 x y xy r Mohr’s circl : (8 : ( : ( 2 X Y C 2 6 10 30.96 p p FX CF R CF . 2 5 5 x x y R y . 2 2 3 x x y y en state of str own has been

ksi, 12 ksi, 6 ksi 2 k 2 x y le: ksi, 6 ksi) 12 ksi, 6 ksi) 2 ksi, 0) 2 2 6 0.6 0 10 FX 50 50 30.96 ave Rcos sin R ave Rcos 20 30.96 20 ave Rcos sin R ave Rcos ess, determine n rotated thro ksi ) 2 6 11.66 k 19.04 x x y y 50.96 x x y y e the normal ugh (a) 25 ksi 9.02 ksi 3.80 ksi 13.02 ksi 5.34 ksi 9.06 ksi 9.34 ksi

(43)

SO LUTION 80 MPa 50 MP

PROB

Solve P PROBL shearin (b) 10 a

BLEM 7.38

Prob. 7.16, usi LEM 7.13 thr ng stresses afte counterclock Plotted (a) (b)

ing Mohr’s cir rough 7.16 Fo er the element kwise. points for Mo tan 25 . 10 . rcle. or the given st t shown has b ave 0, 80 M 50 M 2 x y xy x ohr’s circle: : (0, 50 M : ( 80 M : ( 40 M X Y C 2 n 2 2 51.34 64.03 p p FX CF R CF 2 50 51.34 ave x sin x y R ave y 2 20 51.34 ave x sin x y R ave y tate of stress, d een rotated th MPa, MPa 40 MP y MPa) MPa, 50 MPa) MPa, 0) 2 2 50 1.25 40 4 1 MPa FX 50 1.34 cos R n cos R 20 71.3 cos R n cos R determine the hrough (a) 25 a ) 2 2 0 50 24 x 1.4 x y 104 y 4 19 x 60 x y 60 y normal and clockwise, 4.0 MPa 497 MPa 4.0 MPa 9.51 MPa 0.7 MPa 0.5 MPa

(44)

PROBLEM 7.39

Solve Prob. 7.17, using Mohr’s circle.

PROBLEM 7.17 The grain of a wooden member forms an angle of 15° with the vertical. For the state of stress shown, determine (a) the in-plane shearing stress parallel to the grain, (b) the normal stress perpendicular to the grain.

SOLUTION 0 y x 250 psi xy Plotted points for Mohr’s circle:

(0, 250 psi) X (0, 250 psi) Y (0, 0) C (a) cos 2 (250 psi)cos30 217 psi x y R 217 psi x y (b) sin 2 (250 psi)sin 30 125.0 psi x R 125.0 psi x 250 psi 158

(45)

PROBLEM 7.40

Solve Prob. 7.18, using Mohr’s circle.

PROBLEM 7.18 The grain of a wooden member forms an angle of 15° with the vertical. For the state of stress shown, determine (a) the in-plane shearing stress parallel to the grain, (b) the normal stress perpendicular to the grain.

SOLUTION 3 MPa x y 1.8 MPa xy 0 ave 2.4 MPa 2 x y Points. : ( ,x xy) ( 3 MPa, 0) X : ( ,y xy) ( 1.8 MPa, 0) Y ave : ( , 0) ( 2.4 MPa, 0) C 15 2 30 0.6 MPa CX R 0.6 MPa

(a) x y CX sin 30 Rsin 30 0.6sin 30 0.300 MPa x y 0.300 MPa

(b) x ave CX cos30 2.4 0.6cos30 2.92 MPa x 2.92 MPa

1.8 MPa

3 MPa

(46)

PROBLEM 7.41

Solve Prob. 7.19, using Mohr’s circle.

PROBLEM 7.19 Two wooden members of 80 120-mm uniform rectangular cross section are joined by the simple glued scarf splice shown. Knowing that 22 and that the maximum allowable stresses in the joint are, respectively, 400 kPa in tension (perpendicular to the splice) and 600 kPa in shear (parallel to the splice), determine the largest centric load P that can be applied.

SOLUTION

, x

P

A y 0 xy 0

Plotted points for Mohr’s circle:

: P, 0 , X A Y: (0, 0) : , 0 2 P C 2 P R CX A Coordinates of point Y : (1 cos 2 ) 2 P A sin 2 2 P A Data: A (80)(120) 9.6 10 mm3 2 9.6 10 m3 2 If 400 kPa 400 10 Pa, 3 3 3 2 (2)(9.6 10 )(400 10 ) 1 cos 2 (1 cos 44 ) A P 3 27.4 10 N 27.4 kN If 600 kPa 600 10 Pa, 3 3 3 2 (2)(9.6 10 )(600 10 ) sin 2 (sin 44 ) A P 3 16.58 10 N 16.58 kN

The smaller value of P governs. P 16.58 kN

P'

P

80 mm b

(47)

PROBLEM 7.42

Solve Prob. 7.20, using Mohr’s circle.

PROBLEM 7.20 Two wooden members of 80 120-mm uniform rectangular cross section are joined by the simple glued scarf splice shown. Knowing that 25 and that centric loads of magnitude P 10 kN are applied to the members as shown, determine (a) the in-plane shearing stress parallel to the splice, (b) the normal stress perpendicular to the splice.

SOLUTION

x

P

A y 0 xy 0

Plotted points for Mohr’s circle:

: P, 0 X A

Y

: (0, 0)

: , 0 2 P C A 2 P R CX A Coordinates of point Y: (1 cos 2 ) 2 P A sin 2 2 P A Data: A (80)(120) 9.6 10 mm3 2 9.6 10 m3 2 (a) 3 3 3 (10 10 )sin 50 399 10 Pa 399 kPa (2)(9.6 10 ) (b) 3 3 3 (10 10 )(1 cos50 ) 186.0 10 Pa 186.0 kPa (2)(9.6 10 ) P' P 80 mm b 120 mm

(48)

PROBLEM 7.43

Solve Prob. 7.21, using Mohr’s circle.

PROBLEM 7.21 The centric force P is applied to a short post as shown. Knowing that the stresses on plane a-a are 15 ksi and 5 ksi, determine (a) the angle that plane a-a forms with the horizontal, (b) the maximum compressive stress in the post.

SOLUTION 0 0 x xy y P A

(a) From the Mohr’s circle, 5 tan 0.3333 15 18.4 (b) cos 2 2 2 P P A A 2( ) (2)(15) 1 cos 2 1 cos 2 P A 16.67 ksi 16.67 ksi P a ! a

(49)

PROBLEM 7.44

Solve Prob. 7.22, using Mohr’s circle.

PROBLEM 7.22 Two members of uniform cross section 50 80 mm are glued together along plane a-a that forms an angle of 25 with the horizontal. Knowing that the allowable stresses for the glued joint are 800 kPa and 600 kPa, determine the largest centric load P that can be applied.

SOLUTION 3 3 3 2 0 0 / (50 10 )(80 10 ) 4 10 m (1 cos50 ) 2 2 1 cos50 x xy y P A A P A A P 3 3 3 (2)(4 10 )(800 10 ) 1 cos50 3.90 10 N P P 3 3 3 2 (2)(4 10 )(600 10 ) sin 50 6.27 10 N 2 sin 50 sin 50 P A P A

Choosing the smaller value, P 3.90 kN

P

a 25" 50 mm

(50)

PROBLEM 7.45

Solve Prob. 7.23, using Mohr’s circle.

PROBLEM 7.23 The axle of an automobile is acted upon by the forces and couple shown. Knowing that the diameter of the solid axle is 32 mm, determine (a) the principal planes and principal stresses at point H located on top of the axle, (b) the maximum shearing stress at the same point.

SOLUTION 3 1 1 (32) 16 mm 16 10 m 2 2 c d Torsion: Tc 2T3 J c 6 3 3 2(350 N m) 54.399 10 Pa 54.399 MPa (16 10 m) Bending: 4 (16 10 )3 4 51.472 10 m9 4 4 4 I c 3 (0.15m)(3 10 N) 450 N m M 3 6 9 (450)(16 10 ) 139.882 10 Pa 139.882 MPa 51.472 10 My I

Top view Stresses

139.882 MPa, 0, 54.399 MPa x y xy Plotted points: X: ( 139.882, 54.399); Y: (0, 54.399); C: ( 69.941, 0) ave 2 2 2 2 1 ( ) 69.941 MPa 2 2 139.882 (54.399) 88.606 MPa 2 x y x y xy R 3 kN 3 kN 350 N · m 0.15 m H 0.2 m

(51)

PROBLEM 7.45 (Continued) 2 (2)( 54.399) tan 2 139.882 0.77778 xy p x y (a) a 18.9 , b 108.9 ave 69.941 88.606 a R a 158.5 MPa ave 69.941 88.606 b R b 18.67 MPa

(52)

PROBLEM 7.46

Solve Prob. 7.24, using Mohr’s circle.

PROBLEM 7.24 A 400-lb vertical force is applied at D to a gear attached to the solid 1-in.-diameter shaft AB. Determine the principal stresses and the maximum shearing stress at point H located as shown on top of the shaft.

SOLUTION

Equivalent force-couple system at center of shaft in section at point H:

400 lb (400)(6) 2400 lb in.

V M

(400)(2)

800 lb in.

T

Shaft cross section: 1 in. 1 0.5 in.

2

d c d

4 0.098175 in4 1 0.049087 in4

2 2

J c I J

Torsion: (800)(0.5) 4.074 10 psi3 4.074 ksi

0.098175 Tc

J

Bending: (2400)(0.5) 24.446 10 psi3 24.446 ksi

0.049087 Mc

I

Transverse shear: Stress at point H is zero.

Resultant stresses: ave 2 2 2 2 24.446 ksi, 0, 4.074 ksi 1 ( ) 12.223 ksi 2 2 (12.223) (4.074) 12.884 ksi x y xy x y x y xy R ave a R a 25.1 ksi ave b R b 0.661 ksi

max R max 12.88 ksi

6 in. 2 in. D A B H C 400 lb

(53)

PROBLEM 7.47

Solve Prob. 7.25, using Mohr’s circle.

PROBLEM 7.25 A mechanic uses a crowfoot wrench to loosen a bolt at E. Knowing that the mechanic applies a vertical 24-lb force at A, determine the principal stresses and the maximum shearing stress at point H located as shown as on top of the 3

4-in.-diameter shaft.

SOLUTION

Equivalent force-couple system at center of shaft in section at point H:

24 lb

(24)(6) 144 lb in.

V

M

(24)(10) 240 lb in.

T

Shaft cross section: 0.75 in. 1 0.375 in.

2

d c d

4 0.031063 in4 1 0.015532 in4

2 2

J c I J

Torsion: (240)(0.375) 2.897 10 psi3 2.897 ksi

0.031063 Tc

J

Bending: (144)(0.375) 3.477 10 psi3 3.477 ksi

0.015532 Mc

I

Transverse shear: At point H, stress due to transverse shear is zero. Resultant stresses: ave 2 2 2 2 3.477 ksi, 0, 2.897 ksi 1 ( ) 1.738 ksi 2 2 1.738 2.897 3.378 ksi x y xy x y x y xy R ave a R a 5.12 ksi ave b R b 1.640 ksi

max R max 3.38 ksi

24 lb 10 in. 6 in. E B A H

(54)

PROBLEM 7.48

Solve Prob. 7.26, using Mohr’s circle.

PROBLEM 7.26 The steel pipe AB has a 102-mm outer diameter and a 6-mm wall thickness. Knowing that arm CD is rigidly attached to the pipe, determine the principal stresses and the maximum shearing stress at point K.

SOLUTION 102 51 mm 45 mm 2 2 o o i o d r r r t 4 4 4.1855 10 mm6 4 4.1855 10 m6 4 2 o i J r r 6 4 1 2.0927 10 m 2 I J

Force-couple system at center of tube in the plane containing points H and K:

3 3 3 3 3 10 10 N (10 10 )(200 10 ) 2000 N m (10 10 )(150 10 ) 1500 N m x y z F M M Torsion: 3 3 6 2000 N m 51 10 m (2000)(51 10 ) 24.37 MPa 4.1855 10 y o xy T M c r Tc J

Note that the local x-axis is taken along a negative global z direction.

Transverse shear: Stress due to V F is zero at point K. x

Bending: (1500)(51 10 )63 36.56 MPa 2.0927 10 z y M c I

Point K lies on compression side of neutral axis. y 36.56 MPa

200 mm 6 mm 150 mm 51 mm z x T A y D K H 10 kN A B C

(55)

PROBLEM 7.48 (Continued)

Total stresses at point K: x 0, y 36.56 MPa, xy 24.37 MPa

ave 1 ( ) 18.28 MPa 2 x y 2 2 30.46 MPa 2 x y xy R max ave R 18.28 30.46 max 12.18 MPa min ave R 18.28 30.46 min 48.7 MPa

(56)

PROBLEM 7.49

Solve Prob. 7.27, using Mohr’s circle.

PROBLEM 7.27 For the state of plane stress shown, determine the largest value of y for which the maximum in-plane shearing stress is equal to or less than 75 MPa.

SOLUTION 60 MPa, ?, 20 MPa x y xy Given: max 2 2 2 2 75 MPa 2 150 MPa (2)( ) 40 MPa 150 40 144.6 MPa xy R XY R DY XD XY DY 60 144.6 y x XD y 205 MPa 60 MPa 20 MPa #y

(57)

PROBLEM 7.50

Solve Prob. 7.28, using Mohr’s circle.

PROBLEM 7.28 For the state of plane stress shown, determine (a) the largest value of xy for which the maximum in-plane shearing stress is equal to or less than 12 ksi, (b) the corresponding principal stresses.

SOLUTION

The center of the Mohr’s circle lies at point C with coordinates 10 8

, 0 , 0 (1 ksi, 0).

2 2

x y

The radius of the circle is max(in-plane) 12 ksi.

The stress point ( ,x xy) lies along the line X X of the Mohr circle diagram. The extreme points with 1 2 12 ksi

R are X and 1 X 2.

(a) The largest allowable value of xy is obtained from triangle CDX.

2 2 2 2 1 2 1 DX DX CX CD 2 2 12 9 xy xy 7.94 ksi

(b) The principal stresses are a 1 12 a 13.00 ksi

1 12

b b 11.00 ksi

8 ksi

10 ksi $xy

(58)

PROBLEM 7.51

Solve Prob. 7.29, using Mohr’s circle.

PROBLEM 7.29 For the state of plane stress shown, determine (a) the value of xy for which the in-plane shearing stress parallel to the weld is zero, (b) the corresponding principal stresses.

SOLUTION

Point X of Mohr’s circle must lie on X X so that x 12 MPa. Likewise, point Y lies on line Y Y so that y 2 MPa. The coordinates of C are

2 12 , 0 (7 MPa, 0). 2

Counterclockwise rotation through 150° brings line CX to CB, where 0.

12 2

sec 30 sec 30 5.7735 MPa

2 2 x y R (a) tan 30 2 x y xy 12 2 tan 30 2 xy 2.89 MPa (b) a ave R 7 5.7735 a 12.77 MPa ave 7 5.7735 b R b 1.226 MPa $xy 12 MPa 2 MPa 75"

(59)

PROBLEM 7.52

Solve Prob. 7.30, using Mohr’s circle.

PROBLEM 7.30 Determine the range of values of x for which the maximum in-plane shearing stress is equal to or less than 10 ksi.

SOLUTION

For the Mohr’s circle, point Y lies at (15 ksi, 8 ksi). The radius of limiting circles is R 10 ksi. Let C1 be the location of the leftmost limiting circle and C2 be that of the rightmost one.

1 2 10 ksi 10 ksi C Y C Y

Noting right triangles C DY and 1 C DY 2 ,

2 2 2 2 2 2

1 1 1 8 10 1 6 ksi

C D DY C Y C D C D

Coordinates of point C1 are (0, 15 6) (0, 9 ksi).

Likewise, coordinates of point C2 are (0, 15 6) (0, 21 ksi).

Coordinates of point X1: (9 6, 8) (3 ksi, 8 ksi)

Coordinates of point X2: (21 6, 8) (27 ksi, 8 ksi)

The point ( ,x xy) must lie on the line X1 X2.

Thus, 3 ksi x 27 ksi

15 ksi 8 ksi

(60)

PROBLEM 7.53

Solve Problem 7.29, using Mohr’s circle and assuming that the weld forms an angle of 60 with the horizontal.

PROBLEM 7.29 For the state of plane stress shown, determine (a) the value of xy for which the in-plane shearing stress parallel to the weld is zero, (b) the corresponding principal stresses.

SOLUTION

Locate point C at 12 2 7 MPa

2 with 0. Angle XCB 120 12 2 2 2 5 MPa x y 5sec 60 10 MPa R 5tan 60 xy 8.66 MPa xy ave a R 7 10 a 17.00 MPa ave b R 7 10 b 3.00 MPa $xy 12 MPa 2 MPa 75"

(61)

PROBLEM 7.54

Determine the principal planes and the principal stresses for the state of plane stress resulting from the superposition of the two states of stress shown.

SOLUTION

Consider state of stress on the right. We shall express it in terms of horizontal and vertical components.

We now can add the two stress elements by superposition.

Principal planes and principal stresses:

3 ksi 5 ksi 6 ksi 2 ksi 4 ksi

+

458

(62)

PROBLEM 7.54 (Continued) ave 1 (6 2) 2 2 1 (6 2) 4 2 2 x y 2 2 (4) (3) 5 R 3 tan 2 4 p 2 p 36.87 18.4 , 108.4 p ave max R 2 5 max 7.00 ksi min ave R 2 5 min 3.00 ksi

(63)

PROBLEM 7.55

Determine the principal planes and the principal stresses for the state of plane stress resulting from the superposition of the two states of stress shown.

SOLUTION

Consider the state of stress on the left. We shall express it in terms of horizontal and vertical components.

Principal axes and principal stress:

ave 1 (118.3 56.7) 87.5 2 1 (118.3 56.7) 30.8 2 2 y x 2 2 (30.8) (75) 81.08 R 75 tan 2 2 67.67 30.8 p p p 33.8 , and 123.8 ave

max R 87.5 81.08 max 168.6 MPa

in ave m R 87.5 81.08 min 6.42 MPa 100 MPa 50 MPa 50 MPa 75 MPa

+

308 50cos30 43.30 43.30 50sin 30 25.0 x y xy

(64)

SO

Exp

Ad

OLUTION

press each sta

dding the two s

#0 te of stress in states of stress # #0 30" terms of horiz s, #0

zontal and ver

#0 30"

PROBLEM

Determine th stresses for t the superposi rtical compone

M 7.56

he principal p the state of pl ition of the two

ents.

planes and th lane stress res o states of stre 0 a p m he principal sulting from ess shown. and 90° max 0 min 0

(65)

PROBLEM 7.57

Determine the principal planes and the principal stresses for the state of plane stress resulting from the superposition of the two states of stress shown.

SOLUTION

Mohr’s circle for 2nd state of stress:

0 0 0 x y x y 0 0 0 0 0 0 3 3 sin 60 sin 60 2 2 1 cos 60 2 x y xy Resultant stresses: 0 0 0 0 0 0 0 3 3 3 3 0 0 2 2 2 2 1 3 2 2 x y xy ave 2 2 2 2 0 0 0 1 ( ) 0 2 3 3 3 2 2 2 x y x y xy R 3 (2) 2 2 tan 2 3 3 xy p x y 2 p 60 b 30 a 60 ave a R a 3 0 ave b R b 3 0 $0 $0 30"

+

(66)

PROBLEM 7.58

For the element shown, determine the range of values of xy for which the maximum tensile stress is equal to or less than 60 MPa.

SOLUTION ave 20 MPa 120 MPa 1 ( ) 70 MPa 2 x y x y

Set max ave

max ave 60 MPa 130 MPa R R But 2 2 2 2 2 2 2 2 130 50 120.0 MPa x x xy x x xy R R

Range of xy: 120.0 MPa xy 120.0 MPa

$xy

120 MPa

(67)

PROBLEM 7.59

For the element shown, determine the range of values of xy for which the maximum in-plane shearing stress is equal to or less than 150 MPa.

SOLUTION 20MPa 120 MPa 1 ( ) 50 MPa 2 x y x y

Set max (in-plane) R 150 MPa

But 2 2 2 2 2 2 2 2 150 50 141.4 MPa x y xy x y xy R R

Range of xy: 141.4 MPa xy 141.4 MPa

$xy

120 MPa

(68)

PROBLEM 7.60

For the state of stress shown, determine the range of values of for which the magnitude of the shearing stress x y is equal to or less than 8 ksi. SOLUTION ave 2 2 2 2 16 ksi, 0 6 ksi 1 ( ) 8 ksi 2 2 ( 8) (6) 10 ksi 2 (2)(6) tan 2 0.75 16 2 36.870 18.435 x y xy x y x y xy xy p x y p b R 8 ksi

x y for states of stress corresponding to arcs HBK and UAV of Mohr’s circle. The angle is calculated from 8 sin 2 8 sin 2 0.8 10 R 2 53.130 26.565 18.435 26.565 45 18.435 26.565 8.13 90 45 90 98.13 k b k b u h v k Permissible range of : h k 45 8.13 u v 45 98.13 Also, 135 188.13 and 225 278.13 !y' !x' "x'y' 16 ksi 6 ksi #

(69)

PROBLEM 7.61

For the state of stress shown, determine the range of values of for which the normal stress x is equal to or less than 50 MPa.

SOLUTION ave 2 2 2 2 90 MPa, 0 60 MPa 1 ( ) 45 MPa 2 2 45 60 75 MPa 2 (2)( 60) 4 tan 2 90 3 2 53.13 26.565 x y xy x y x y xy xy p x y p a R 50 MPa

x for states of stress corresponding to the arc HBK of Mohr’s circle. From the circle,

cos 2 50 45 5 MPa R 5 cos 2 0.066667 75 2 86.177 43.089 26.565 43.089 16.524 2 2 360 4 32.524 360 172.355 220.169 110.085 h a k h k Permissible range of : h k 16.5 110.1 Also, 196.5 290.1 90 MPa 60 MPa #y' #x' $x'y' %

(70)

PROBLEM 7.62

For the state of stress shown, determine the range of values of for which the normal stress x is equal to or less than 100 MPa.

SOLUTION ave 2 2 2 2 90 MPa, 0 60 MPa 1 ( ) 45 MPa 2 2 45 60 75 MPa 2 (2)( 60) 4 tan 2 90 3 2 53.13 26.565 x y xy x y x y xy xy p x y p a R 100 MPa

x for states of stress corresponding to arc HBK of Mohr’s circle. From the circle,

cos 2 100 45 55 MPa R 55 cos 2 0.73333 75 2 42.833 21.417 26.565 21.417 5.15 2 2 360 4 10.297 360 85.666 264.037 132.02 h a k h k Permissible range of is h k 5.1 132.0 Also, 174.8 312.0 90 MPa 60 MPa #y' #x' $x'y' %

(71)

PROBLEM 7.63

For the state of stress shown, it is known that the normal and shearing stresses are directed as shown and that x 14 ksi, y 9 ksi, and min 5 ksi. Determine (a) the orientation of the principal planes, (b) the principal stress max, (c) the

maximum in-plane shearing stress.

SOLUTION

ave min ave ave min

2 2

2

2 2 2

1

14 ksi, 9 ksi, ( ) 11.5 ksi

2 11.5 5 6.5 ksi 2 6.5 2.5 6 ksi 2 x y x y x y xy x y xy R R R R

But it is given that xy is positive, thus xy 6 ksi.

(a) tan 2 2 (2)(6) 2.4 5 2 67.38 xy p x y p 33.7 a 123.7 b (b) max ave R max 18.00 ksi (c) max(in-plane) R max(in-plane) 6.50 ksi $xy #y #x

(72)

PROBLEM 7.64

The Mohr’s circle shown corresponds to the state of stress given in Fig. 7.5a and b. Noting that x OC (CX )cos(2 p 2 ) and that x y (CX )sin (2 p 2 ), derive the expressions for x and x y given in Eqs. (7.5) and (7.6), respectively. [Hint: Use sin(A B) sin cosA B cos sinA B and cos (A B) cos cosA B sinAsin .]B SOLUTION 1 ( ) 2 cos 2 cos 2 2 sin 2 sin 2 x y x y p p p p xy OC CX CX CX CX CX CX cos (2 2 )

(cos 2 cos 2 sin 2 sin 2 )

cos 2 cos 2 sin 2 sin 2

cos 2 sin 2 2 2 x p p p p p x y x y xy OC CX OC CX OC CX CX

sin (2 2 ) (sin 2 cos 2 cos 2 sin 2 )

sin 2 cos 2 cos 2 sin 2

cos 2 sin 2 2 x y p p p p p x y xy CX CX CX CX !x'y' !xy "y "y' "x' "x " X' Y Y' O C X ! 2#p 2#

(73)

PROBLEM 7.65

(a) Prove that the expression x y x y2 , where x, y, and x y are components of the 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. (b) Using the invariance property established in part a, express the shearing stress xy in terms of x, y, and the principal stresses max and min.

SOLUTION

(a) From Mohr’s circle,

ave ave sin 2 cos 2 cos 2 x y p x p y p R R R 2 x y x y 2 2 2 2 2

ave R cos 2 p R sin 2 p

2 2

ave R ; independent of p.

Draw line OK from origin tangent to the circle at K. Triangle OCK is a right triangle.

2 2 2 2 2 2 2 2 ave 2 x y x y OC OK CK OK OC CK R

(b) Applying above to x, y, and xy, and to a, b,

2 2 2 2

ave

x y xy a b ab R

But ab 0, a max, b min

2 max min x y xy 2 max min max min xy x y xy x y

(74)

PROBLEM 7.66

For the state of plane stress shown, determine the maximum shearing stress when (a) x 14 ksi and y 4 ksi, (b) x 21 ksi and y 14 ksi. (Hint: Consider both in-plane and out-of-plane shearing stresses.)

SOLUTION (a) ave 1 (14 4) 9 2 1( ) 1(14 4) 5 2 x y 2 2 2 (5) (12) 13 R ave max R 9 13 22 in ave m R 9 13 4

Since max and min have opposite signs, the maximum shearing stress is equal to the maximum in-plane shearing stress.

max R 13.00 ksi max 13.00 ksi

12 ksi y z x σy σx

(75)

PROBLEM 7.66 (Continued) (b) ave 1 (21 14) 17.5 2 1( ) 1(21 14) 3.5 2 x y 2 2 2 (3.5) (12) 12.5 R ave max R 17.5 12.5 30 in ave m R 17.5 12.5 5

Since max and min have the same sign, max is out of the plane of stress. Using Mohr’s circle through O and A, we have

max max

1 1

(30 ksi)

(76)

PROBLEM 7.67

For the state of plane stress shown, determine the maximum shearing stress when (a) x 20 ksi and y 10 ksi, (b) x 12 ksi and y 5 ksi. (Hint: Consider both in-plane and out-of-plane shearing stresses.)

SOLUTION (a) ave 1(20 10) 15 2 1 1 ( ) (20 10) 5 2 x y 2 2 2 (5) (12) 13 R ave max R 15 13 28 in ave m R 15 13 2

Since max and min have the same sign, max is out of the plane of stress. Using Mohr’s circle through O and A, we have max max 1 1 (28 ksi) 2 2 max 14.00 ksi 12 ksi y z x σy σx

(77)

PROBLEM 7.67 (Continued) (b) ave 1 (12 5) 8.5 2 1 1 ( ) (12 5) 3.5 2 x y 2 2 2 (3.5) (12) 12.5 R ave max R 8.5 12.5 21 in ave m R 8.5 12.5 4

Since max and min have opposite signs, the maximum shearing stress is equal to the maximum in-plane shearing stress.

(78)

PROBLEM 7.68

For the state of stress shown, determine the maximum shearing stress when (a) y 40 MPa, (b) y 120 MPa. (Hint: Consider both in-plane and out-of-plane shearing stresses.)

SOLUTION

(a)

ave

140 MPa, 40 MPa, 80 MPa

1 ( ) 90 MPa 2 x y xy x y 2 2 502 802 94.34 MPa 2 x y xy R ave ave 184.34 MPa (max) 4.34 MPa (min) 0 a b c R R max(in-plane) 1 ( ) 94.34 MPa 2 a b R

max max min

1 1

( ) ( ) 94.3 MPa

2 2 a b max 94.3 MPa

(b)

ave

140 MPa, 120 MPa, 80 MPa

1 ( ) 130 MPa 2 x y xy x y 2 2 102 802 80.62 MPa 2 x y xy R ave ave 210.62 MPa (max) 49.38 MPa 0 (min) a b c R R

max a 210.62 MPa min c 0

max(in-plane) R 86.62 MPa max max min

1 ( ) 105.3 MPa 2 max 105.3 MPa 80 MPa y z x 140 MPa σy

(79)

PROBLEM 7.69

For the state of stress shown, determine the maximum shearing stress when (a) y 20 MPa, (b) y 140 MPa. (Hint: Consider both in-plane and out-of-plane shearing stresses.)

SOLUTION

(a)

ave

140 MPa, 20 MPa, 80 MPa

1 ( ) 80 MPa 2 x y xy x y 2 2 602 802 100 MPa 2 x y xy R ave ave 80 100 180 MPa (max) 80 100 20 MPa (min) 0 a b c R R max(in-plane) 1 ( ) 100 MPa 2 a b

max max min

1

( ) 100 MPa

2 max 100.0 MPa

(b)

ave

140 MPa, 140 MPa, 80 MPa

1 ( ) 140 MPa 2 x y xy x y 2 2 0 802 80 MPa 2 x y xy R ave ave 220 MPa (max) 60 MPa 0 (min) a b c R R max (in-plane) 1 ( ) 80 MPa 2 a b

max max min

1 ( ) 110 MPa 2 max 110.0 MPa 80 MPa y z x 140 MPa σy

(80)

PROBLEM 7.70

For the state of stress shown, determine the maximum shearing stress when (a) z 0, (b) z 60 MPa, (c) z 60 MPa.

SOLUTION

The z axis is a principal axis. We determine the other two principal axes by drawing Mohr’s circle for a rotation in the xy plane.

ave 1(30 100) 65 2 1 1 ( ) (30 100) 35 2 x y 2 2 2 (35) (84) 91 R ave 65 91 156 MPa A R ave 65 91 26 MPa B R

(a) z 0. Point Z corresponding to the z axis is located at O between A and B. Therefore, the largest of the 3 Mohr’s circles is the circle we drew through A and B. We have

max R 91.0 MPa

(b) z 60 MPa. Point Z is located between A and B. The largest of the 3 circles is still the circle through A and B, and we still have

max R 91.0 MPa

(c) z 60 MPa. Point Z is now outside the circle through A and B. The largest circle is the circle through Z and A. max 1 1 ( ) (60 156) 2 ZH 2 max 108.0 MPa z x 84 MPa y 30 MPa 100 MPa σz

References

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