Mechanics of materials 7th edition beer pdf free download

CHAPTER 7

PR ROBLEM 7

4 ksi 3 ksi 8 ksi

Foor the given sttate of stress, determine thee normal and shearing stressses exerted onn the oblique face of the shhaded trianguular element shown. s Use a method of anaalysis based on o the equilibrrium of that ellement, as waas done in the derivations of Sec. A.

SOLUTION

F

0:

A

8 A cos 20 2 cos 20

8cos 2 20

3cos 20 sin 20

3 A cos 20 sin 20 2 3 sin 20 cos 20

3 A sin cos 20 4sin 2 20

4 A sin 20 sin 20 2

0

0

ksi F

0:

A

8 A cos sin 20

8coos 20 sin 20

3(cos2 20

3A A cos 20 cos 20

sin 2 20 )

3 A sin sin 20

4A A sin 20 cos

0

4 20 cos 4sin

ksi

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PRO OBLEM 2

60 MPa

For th he given statee of stress, dettermine the noormal and sheearing stressess exerted on the oblique face off the shaded triangular t elem ment shown. Use U a methodd of analysis based d on the equilibbrium of that element, e as waas done in the derivations off Sec. A.

90 MPa M

SO OLUTION

F

0:

90 9 A sin 30 coss 30

A

sin 30 cos c 30

90 A cos 30 sin 30

60 A cos 30 ccos 30

0

60 coos 2 30 3 M Pa

F

0:

A

90 0 A sin 30 sin 30 3

90(cos 2 30

sin 2 30 )

90 A cos 30 cos 30

60 A cos 30 sinn 30

0

60 cos 30 sin 30 7 M Pa

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PROBLEM M

10 ksi

For the giveen state of sttress, determiine the normaal and sheariing stresses exerted on thhe oblique faace of the shaaded triangulaar element shoown. Use a method of annalysis based on the equilibbrium of that element, as was w done in the derivationns of Sec. A A.

6 ksi



4 ksi

SOLUTION

F

0:

A

4 A cos15 sin15

4 co os15 sin15

10 cos 2 15

A cos15 cos

6sin 2 15

6 A sin15 sin15

4 A sin15 cos

0

4 4sin15 cos 1 ksi

F

0:

A

4(ccos2 15

4 A cos15 cos15

10 A cos15 sin 15

sin 2 15 )

6) cos15 sin15

(10

6 A sin15 cos15

4 A sin15 sin

0

0 ksi

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80 MPa M

PROBLEM P For F the given state s of stress, determine thhe normal andd shearing streesses exerted on o the obliquee face of the shaded trianggular element shown. Use a method of analysis a based on the equilibbrium of that element, e as was w done in thee derivations of o Sec. A.

40 MPa

SO OLUTION

Streesses

F

0 0:

A

80 A cos 55 cos

80 cos 2 55

F

0 0:

A

Areas

Forces

40 A sin 55 sin 55

40sin 2 55

80 A cos 55 sin 55 5

0 MPa

40 A sin 55 cos 55 5 MPa

1 cos 55 sin 55

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PROBLEM

40 MPa 35 MPa 60 MPa

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

SOLUTION 60 MPa

x

(a)

tan 2

2

xy

p x

2

p

y

(2)(35) 60 40

y

40 MPa

xy

35 MPa





,

p 2

(b)

x max, min

y

x

2 60 40 2

y

2 xy

2 60 40 2

2

(35)2

50 MPa max min

MPa MPa

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PROBLEM

10 ksi

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

2 ksi

3 ksi

SOLUTION x

(a)

tan 2

2 ksi

2 p

p

y

3 ksi

xy

(2)( 3) 2 10

xy

x

2

10 ksi

y





p

, ◄

2

(b)

x

max,min

x

y

2 2

10 2

6

y

2 2

10 2

2 xy

2

( 3)2

5 ksi

max

ksi ◄

min

ksi ◄

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PROBLEM

30 MPa

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

MPa

80 MPa

SOLUTION x

(a)

tan 2

MPa,

2 p

2

p

y



80 MPa

xy

2( 80 MPa) ( MPa 30 MPa)

xy

x

30 MPa,

y

MPa

and p

(b)

max,min

x

y

x

2

y

2

MPa

30 MPa 2

90 MPa

and ◄

2 xy

MPa

30 MPa 2

2

( 80 MPa)2

MPa

max min

MPa ◄ MPa ◄

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PROBLEM

12 ksi 8 ksi

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

18 ksi

SOLUTION x

(a)

tan 2

18 ksi

2 p

2

p

(2)(8) 18 12

xy

x

12 ksi

y

y

xy

8 ksi





, ◄

p

2

(b)

max,min

x

y

x

2 18

12 2

y

2 18

12 2

2 xy

2

(8)2

3 17 ksi max min

ksi ◄ ksi ◄

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PROBLEM

40 MPa 35 MPa

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.

60 MPa

SOLUTION x

(a)

tan 2 2

x s

s

60 40 (2)(35)

y

2

xy

60 MPa

y

40 MPa

xy

35 MPa





s

,

2

(b)

(c)

x max

y

2 xy

2 60 40 2

2

x

y

ave

2

(35) 2

max

60 40 2

MPa

MPa

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PROBLEM

10 ksi

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.

2 ksi

3 ksi

SOLUTION 2 ksi

x

(a)

tan 2 2

y

x

s

2 10 (2)( 3)

y

2

10 ksi

xy

xy

3 ksi





s

s

,

2

(b)

x

max

y

2 xy

2 2

10

2

( 3)2

2

max

(c)

ave

x

y

2

2

ksi

10 2

ksi

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PROBLEM

30 MPa

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.

MPa

80 MPa

SOLUTION MPa,

x

(a)

tan 2 2

x

s

30 MPa,

30 2( 80)

y

2

y

xy

xy

80 MPa



and

s

s

and

2

(b)

x

max

y

2 xy

2

30

2

( 80)2

2

max

(c)

ave

x

MPa

y

2 30 2

MPa

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PROBLEM

12 ksi 8 ksi

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.

18 ksi

SOLUTION 18 ksi

x

(a)

tan 2 2

x

s

18 12 (2)(8)

y

2

12 ksi

y

xy

xy

8 ksi





s

,

s

2

(b)

x

max

y

2 xy

2 18

12

2

(8)2

2

max

(c)

ave

x

y

2

ksi

18 12 2

ksi

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PROBLEM

8 ksi 5 ksi

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 x x

y

2

0

8 ksi

y

x

4 ksi y

25

2

x

xy y

y

2

x

y

2

4 sin ( 50 ) 5 cos ( 50 )

xy

xy

sin 2

cos 2

cos 2

xy

sin 2

ksi

x

ksi

xy

4 4 cos ( 50 ) 5 sin ( 50)

y

2

cos 2 +

sin 2 +

4 4 cos ( 50°) + 5 sin ( 50°)

xy

10

y

50

x

(b)

y

2 x

4 ksi

2

x

(a)

x

2

xy

y

y

2

x x

5 ksi

xy

ksi

y

20 4 4 cos (20°) + 5 sin (20°)

4 sin (20°) + 5 cos (20°) 4 4 cos (20°)

5 cos (20°)

x

ksi

xy

ksi

y

ksi

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PROBLEM

90 MPa 30 MPa

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.

60 MPa

SOLUTION x x

y

2

60 MPa

x

y

(a)

25

2

x

2

sin 2 +

y

2

x

y

2

xy

xy

sin 2

cos 2

cos 2

xy

sin 2

x

MPa

xy

MPa

15 75 cos ( 50 ) 30 sin ( 50 )

y

MPa

20 15 75 cos (20°) + 30 sin (20°)

75 sin (20°) + 30 cos (20°)

xy y

y

75 sin ( 50 ) 30 cos ( 50 )

y

2

cos 2 +

15 75 cos ( 50 ) 30 sin ( 50 )

xy

10

y

50

x

(b)

x

30 MPa

xy

75 MPa

2

x

x

y

2

2

xy

y

x

15 MPa

x

90 MPa

y

15 75 cos (20°) 30 sin (20°)

x

MPa

xy

MPa

y

MPa

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PROBLEM

12 ksi

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.

8 ksi

6 ksi

SOLUTION x x

y

2

8 ksi

x

2 ksi x

x

y

2

x

2

y

2

xy

xy

sin 2

cos 2

cos 2

xy

sin 2

2 10 cos ( 50 ) 6 sin ( 50 )

ksi

x

10 sin ( 50 ) 6 cos ( 50 )

y

2

y

sin 2 +

cos 2 +

2 10 cos ( 50 ) 6 sin ( 50 )

xy

10

y

6 ksi

50

x

(b)

y

2 2

x

25

x

xy

10 ksi

2

x

(a)

y

2

xy

y

12 ksi

y

xy

ksi

ksi

y

20

x xy

y

2 10 cos (20°) 6 sin (20°)

x

10 sin (20°) 6 cos (20°)

2 10 cos (20°) + 6 sin (20°)

ksi

xy

ksi

y

ksi

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PROBLEM

80 MPa

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. 50 MPa

SOLUTION x x

y

2

0

80 MPa

y

x x

y

(a)

25

2

2

sin 2 +

y

2

x

y

2

xy

xy

sin 2

xy

sin 2

cos 2

cos 2

40 sin ( 50°) 50 cos ( 50 ) 40 40 cos ( 50 ) 50 sin ( 50 )

y

2

y

cos 2

40 40 cos ( 50 ) 50 sin ( 50°)

xy

10

y

40 MPa

50 x

(b)

x

2

x

x

y

2

2

xy

y

x

40 MPa

50 MPa

xy

x

MPa MPa

xy

y

MPa

x

MPa

xy

MPa

y

MPa

20 x

xy y

40 40 cos (20°) 50 sin (20°)

40 sin (20°) 50 cos (20°) 40 40 cos (20°) + 50 sin (20°)

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PROBLEM

psi

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 x

(a)

0

y

x

xy

0

psi

xy

y

sin 2 2 cos( 30 )

15

xy cos 2

xy

(b)

x

x

0

y

2 0

x

y

cos 2 2 sin( 30 )

psi

xy sin 2

x

psi

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PROBLEM

MPa

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.

3 MPa



SOLUTION x

3 MPa

15 (a)

2

x

xy

MPa

y

y

2 3

2

xy

0

30

sin 2

xy sin 2

sin( 30 )

0

MPa

xy

(b)

x

x

y

x

2 3

y

2

2

3

2

cos 2 cos( 30 )

xy sin 2

0 x

MPa

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P'

80 mm m

P PROBLEM 7

mm

Tw wo wooden members m of mm uniform u rectanngular cross seection are joined by the simpple glued scarrf splice shownn. Knowing 22 and thhat a that the maximum m allow wable stresses in the joint arre, respectivelly, kPa inn tension (perrpendicular to the splice) annd kPa inn shear (parallel to the splicce), determinee the largest ceentric load P thhat can be appplied.

b P

SOLUTION

Forces

Areeas

A (80) () mm 2 N all

a all

Fy

0: N

Sall

Fx

aall

A/sin

P sin

A/sin

S

P cos

10 3 m 2

( )( 10 3 ) N sin 22 0

P

N sinn

sin

N

( )( 10 3 ) N sin 22

0

P

S coos

Thee smaller valuee for P governns.

cos 22 2

N P kN

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P'

PROBLEM 7

mm

80 mm

Tw wo wooden members m of mm uniform u rectanngular cross section are joineed by the simp mple glued scarrf splice show wn. Knowing 25 and a that centriic loads of magnitude m P 10 kN are that t in-plane appplied to the members ass shown, dettermine (a) the shhearing stresss parallel to the splice, (b) the noormal stress peerpendicular too the splice.

b P

SO OLUTION

Forcess

A Areas

A (80)() mm m 2 (a)

Fx

0: S N A/sin

(b)

Fy

0: N N A/sin

P cos

0

S

P cos

( )sinn 25 10 3

P sin n

0

N

P sin

( )sin 25

3 m 2

(10 ) cos 25

N

Pa

(10 )sin 25 Pa

kPa

N 1 kPa

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P

PROBL LEM The centrric force P is applied to a short post as shown. Know wing that the stresses on 5 ksi, determinne (a) the anggle that planne a-a forms 15 ksi and plane a-a are with the horizontal, h (b) the maximum m compressivee stress in the post. p

a

!

a

SOLUTION

x

0

xy

0

y

(a)

P/ A

From the Mohr’s M circle,

tan

5 15

P 2A (b)

P A

2( ) 1 co os 2





P cos 2 2A

(2)() 1 coss 2

P 1 ksi A

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PROBLEM 22 a a

25"

50 mm m

Two o members of uniform crosss section 50 80 mm are glued g togetherr along plane a-a that forms ann angle of with the horizontal. h Knnowing that thhe allowable kPa and a kPa, k determinne the largest stressses for the gluued joint are centtric load P thatt can be applieed.

P

SO OLUTION Forr plane a-a,

65 .

x

0, 0 x

P

0,

cos 2

x

y

y

P A

sin 2

2

xy

sin

cos

0

(50 10 3 )(80 10 3 )( ) sin 2 65 6

A sin s 2 65 (

P

xy

y )sin

A sin s 65 cos 65

P 2 sin A

0

N

P sin 65 cos 65 0 A (50 10 3 )((80 10 3 )( ) N sinn 65 cos 65

cos

( xy (cos

Alllowable value of P is the sm maller one.

2

sin 2 )

P

kN

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PROBLEM

m m 3 kN

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.

H

N · m 3 kN

SOLUTION 1 d 2

c Tc J

Torsion:

I

Bending:

4

2T c3

c4

4

1 (32) 2

2( N m) (16 10 3 m)3

(16 10 3 )4

(m)(3 N)

M

16 mm 16 10 3 m MPa

10 9 m 4 N m

()(16 10 3 ) 10 9

My I

Pa

Top view:

Pa

MPa

Stresses:

x

MPa 1 ( 2

x

y)

R

x

y

ave

y

0

1 ( 2

xy

0)

MPa MPa

2

(a)

2

max

ave

R





min

ave

R





2 xy

( )2

( ) 2

MPa

max min

MPa MPa

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PROBLEM (Continued)

tan 2

2 p x

xy y

(2)( )



2

p



p

(b)

max

R

MPa



max

and °

MPa

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6 in.

PROBLEM

C H

A 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.

B

A D 2 in. lb

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

Shaft cross section:

V

lb

T

()(2)

d

1 in. c

J

c4

2

1 d 2

()(6)

lb in.

lb in.

in.

in 4

Torsion:

Tc J

()()

Bending:

Mc I

()()

Transverse shear:

M

1 J 2

I

in 4

psi

ksi

psi

ksi

Stress at point H is zero. x ave

ksi, 1 ( 2

x

y)

x

y

y

0,

xy

ksi

ksi 2

R

2

2 xy

() 2

() 2

ksi a

ave

R

b

ave

R

max

R

a

ksi

b

ksi

max

ksi

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PROBLEM

H

E

A mechanic uses a crowfoot wrench to loosen a bolt at E. Knowing that the mechanic applies a vertical lb force at A, determine the principal stresses and the maximum shearing stress at point H located as shown as on top of the 34 -in. diameter shaft.

6 in.

B 24 lb A

10 in.

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

Shaft cross section:

d

V

24 lb M

T

(24)(10)

1 d 2

in., c

J

2

c4

Tc J

()()

Bending:

Mc I

()()

Transverse shear: Resultant stresses:

lb in.

lb in.

in. 1 J 2

in 4 I

Torsion:

(24)(6)

in 4

psi

ksi

psi ksi

At point H, the stress due to transverse shear is zero. x ave

ksi, 1 ( 2

y

x

y)

x

y

0,

ksi

xy

ksi 2

R

2

a

ave

R

b

ave

R

max

2 xy



2

ksi

a

ksi

b

R

ksi

max

ksi

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P PROBLEM 7

y

m 6 mm A mm

Thhe steel pipe AB A has a mm outer diameter and a 6-mm wall thhickness. Knowing that arm m CD is rigiddly attached to t the pipe, deetermine the principal p stressses and the maximum m shearing stress att point K.

51 mm

A T

D

10 kN N C

1 mm H

K

B x

z

SOLUTION ro J

I

do 2

2

51 mm

ri

ro

t

45 mm

ro4 ri4 mm 4 2 10 6 m 4 1 J 2

10 6 m 4

Forcce-couple systtem at center of o tube in the plane p containiing points H and a K:

Fx

10 kN 10 N

My

(10 )( 10 3 ) N m

Mz

(10 )( 10 3 ) N m

Torsion:

At po oint K, place local l x-axis in negative globbal z-directionn. T

My

c

ro

xy

N m 10 3 m

(()(51 3 ) Pa MPa Tc J

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PROBLEM (Continued)

Transverse shear:

Stress due to transverse shear V

Fx is zero at point K.

Bending: |

y|

()(51 10 3 ) 10 6

|M z |c I

Pa

MPa

Point K lies on compression side of neutral axis.

MPa

y

Total stresses at point K: x ave

0,

MPa,

y

1 ( 2

x

y)

x

y

xy

MPa

MPa 2

R

2

2 xy

MPa

max

ave

R



max

MPa

min

ave

R



min

MPa

max

R

max

MPa

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PROBLEM

#y 20 MPa 60 MPa

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 x

60 MPa,

y

?,

xy

20 MPa

Let

u

x

y

y

x

R

u2

2

.

Then

Largest value of

y

2 xy

R2

u y

2u

x

2u

75 MPa 2 xy





60 (2)()

is required.

MPa MPa or MPa y

MPa

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PROBLEM

8 ksi

$xy 10 ksi

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 x

10 ksi,

8 ksi,

y

xy

2 max

x

R

z 92



(a)

xy

(b)

ave

1 ( 2

y

2 xy

2 xy

? 10 ( 8) z

2 2 xy

12 ksi

92

xy y)

x

ksi

1 ksi

a

ave

R 1 12 13 ksi

b

ave

R 1 12

a

11 ksi

b

ksi ksi

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P PROBLEM

2 MPaa

$xy

12 MPa

75"

For the state of plane stress shown, determ F mine (a) the vaalue of xy foor which the inn-plane shearring stress paarallel to the weld is zeroo, (b) the corrresponding p principal stressses.

SOLUTION x

Sincce

xy

12 MPa,

y

tan 2

?

15 2

1 ( 2

x

y)

tan 2

x

y

2 xyy

y)

7 MPa M

p

xy

p x

xy

xy

0, x -direction is a principal direection. p

(a)

2 MPaa,

y

1 (12 2)) tan( 30 ) 2

xy

MPa

2

R

ave

(b)

2 1 ( 2

x

52



MPa M

a

ave

R

7

a

MPa

b

ave

R

7

b

MPa

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PROBLEM

15 ksi 8 ksi

Determine the range of values of is equal to or less than 10 ksi.

x

for which the maximum in-plane shearing stress

#x

SOLUTION x

Let u

R

x

x

2 u2

2 xy

R2

u x

y

y

max 2 xy

y

?,

y

15 ksi,

xy

8 ksi

2u

10 ksi

2u 15 (2)(6)

82 z

6 ksi

27 ksi or 3 ksi 3 ksi

Allowable range:

x

27 ksi

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PR ROBLEM 7 40 MPaa

Soolve Probs. and , usinng Mohr’s circcle.

MPa

PR ROBLEM through For the givven state of stress, s determ mine (a) the priincipal planes, (b) the principal stresses.

60 MPa

PR ROBLEM through For the giiven state of stress, determ mine (a) the oriientation of thhe planes of maximum m in-pplane shearing stress, (b) thee maximum in--plane shearinng stress, (c) thhe correspondiing normal strress.

SOLUTION x

6 MPa, 60

y

4 MPa, 40 MPa

xy

x ave

y

2

50 MPa

Plottted points forr Mohr’s circlee:

(a)

a

R

(a )

(b ) (c )

x,

xy )

Y:(

y,

xy )

( 40 MPa, 35 MPa)

C:(

ave ,

0)

( 50 MPa, 0)

X 35 GX CG G 10 74

tan

b

(b)

X :(

1 2 1 2

( 60 MPa, 35 MPa)





CG C

2

GX

2

10 2

ave a

R

50

max

a ave

R

50

d

B

45



e

A

45





MPa min max

86 MPa 13 MPa d e

R MPa ave a



a

min

max



b

max

50 MPaa



36 MPa 50 MPa

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PROBLEM 30 MPa

Solve Probs. and , using Mohr’s circle.

MPa

80 MPa

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

MPa

y

30 MPa 80 MPa

xy

x

ave

y

2

90 MPa

Plotted points for Mohr’s circle:

x

tan 2

p

2

p

y , xy )

C:(

ave ,

30)

(60)2

xy )

0)

(80)2



p

90



min

ave

R

90



max

(90 MPa, 0)

R

(b′)

(30 MPa, 80 MPa)

80 60

ave

s

( MPa, 80 MPa)

60

max

(a′)

(c )

Y:(

2

R

(b)

x,

(

y

2

(a)

X :(

p

max min

45

s

R

max

and MPa MPa and MPa

MPa

ave

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PROBLEM

10 ksi

Solve Prob. , using Mohr’s circle.

2 ksi

PROBLEM through 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.

3 ksi

SOLUTION x

2 ksi y

x

ave

10 ksi

y

2

2

10 2

xy

3 ksi

6 ksi

Plotted points for Mohr’s circle:

FX FC

tan

X: (

x,

Y: (

y , xy )

(10 ksi, 3 ksi)

C: (

ave ,

(6 ksi, 0)

3 4

xy )

0)

(2 ksi, 3 ksi)



B

(a)

(b) (c)

1 2



D

B

45

E

B

45

R

CF

max

R

ksi

ave

ksi

2



FX

2



D E

42

32



5 ksi max

ksi

ksi

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PROBLEM

12 ksi 8 ksi

Solve Prob. , using Mohr’s circle. 18 ksi

PROBLEM through 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

x

ave

12 ksi

y

y

xy

8 ksi

3 ksi

2

Plotted points for Mohr’s circle:

FX CF

tan

X: (

x,

Y: (

y,

xy )

C: (

ave ,

0)

8 15

xy )

(18 ksi, 8 ksi) ( 12 ksi, 8 ksi) (3 ksi, 0)



A

(a)

(b) (c)

1 2



D

A

45

E

A

45

R

CF

max

R

ksi

ave

ksi

2



D



FX

2

E



82



17 ksi max

ksi

ksi

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PROBLEM

8 ksi 5 ksi

Solve Prrob. , usinng Mohr’s circcle.

PROBL LEM through For the given staate of stress, determine d the normal and shearingg stresses afterr the element shown has beeen rotated thrrough (a) 25 clockwise, (b) 10 counterclockw c wise.

SOLUTION x

0 0,

y

8 ksi,

xy

5 ksi x

ave

y

4 kssi

2

Plottted points forr Mohr’s circlee:

X : (0, 5 ksi) k 5 ksi) Y : (8 ksi, k 0) C : (4 ksi, tan 2

p

2

p

FX 5 1 FC 4

R (a)

25

FC .

2

X FX

2

xy

y

10

.

42

52

ksi

50

x

(b)

2

2



50

ave

R cos



R sin ave

ksi

x

ksi

xy

R cos

y

ksi

20

x xy y

ave

20 R cos

R sin ave

R cos

x

1 ksi

xy

ksi

y

6 ksi

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PROBLEM M

90 MP Pa 3 MPa 30

Solve Prob. , using Mohr’s M circle.

60 MPa

PROBLEM M througgh For thhe given statee of stress, deetermine the normal and shearing s stresses after the element e shownn has been rotaated through (a) 25 clockkwise, (b) 10 counterclockkwise.

SO OLUTION 60 MP Pa,

x y

90 MPa,,

xy

30 MPa x

ave

y

15 MPa

2

Plootted points for Mohr’s circlle:

X : ( 60 MPa, 30 MPa) Y : (90 MPa, MPa) C : (15 MPa, 0) tan 2

p

2

p

FX FC



R

(a)

25

.

30 75

FC

2

0



P

FX X

2



2

MP Pa

50 5

2

2 x

xy y

2 ave

P

50





R cos

R sin ave

R cos

x

5 MPa

xy

3 MPa y

8 MPa

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PROB BLEM (Continued) ( d)

(b)

10

.

2



2 x

p ave

xy

R sin

y

ave

2



20



R cos

R cos

x

45 MPa

xy

53 MPa

y

MPa

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PR ROBLEM

12 ksi

8 ksi

6 ksi

Solv ve Prob. , using Mohr’ss circle.

PRO OBLEM through For the giveen state of stress, determinee the normal and shearing streesses after thee element shoown has beenn rotated through (a) 25 clocckwise, (b) 10 counterclockkwise.

SO OLUTION 8 ksi,

x

12 ksi,

y

6 ksi

xy

x

y

ave

2 ksi k

2

Plootted points for Mohr’s circlle:

X : (8 ksi, 6 ksi) Y : ( 12 ksi, 6 ksi)) C : ( 2 ksi, 0)

(a)

tan 2

p

2

p

25

FX 6 CF

R

CF

.

2

2



FX

2

x

xy y

10

.

62

ksi k

5 50

5 50

(b)



2

ave





R cos

R sin ave

R cos

x

ksi

xy

ksi ksi

y

2 20

3 x

xy y

20 R cos

ave

R sin ave



R cos

x

ksi

xy

ksi

y

ksi

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80 MPa

PROB BLEM Solve Prob. P , usiing Mohr’s cirrcle.

50 MPa

PROBL LEM thrrough Foor the given sttate of stress, determine d the normal and shearinng stresses afteer the elementt shown has been rotated thhrough (a) 25 clockwise, (b) 10 counterclockkwise.

SOLUTION x

0,

y

M 80 MPa,

xy

50 MPa M x

ave

y

40 MPa

2

Plotted points for Moohr’s circle:

X : (0, 50 MPa) M MPa, 50 MPa)) Y : ( 80 M M 0) C : ( 40 MPa,

tann 2

p

2

p

R (a)

25

.

2

FX 50 CF 40 2

FX CF MPa

x

xy y

10

.





50



(b)

2

2

x

xy y

50



R cos

ave

x

R sinn

MPa

xy

R cos

ave

MPa

y

MPa

x

MPa

xy

MPa

y

MPa 60

20



20

ave

R cos

R sinn ave

R cos



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psi

PROBLEM Solve Prob. , using Mohr’s circle.



PROBLEM 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 x xy

y

0

psi

Plotted points for Mohr’s circle:

(a)

xy

X

(0, psi)

Y

(0, psi)

C

(0, 0)

R cos 2 ( psi)cos30 psi xy

(b)

x

psi

R sin 2 ( psi) sin 30 psi x

psi

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PROBLEM

MPa

Solve Prob. , using Mohr’s circle. 3 MPa

PROBLEM 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 x

ave

3 MPa x

MPa

y

y

xy

0

MPa

2

Points. X: (

x,

xy )

( 3 MPa, 0)

Y: (

y,

xy )

( MPa, 0)

C: (

ave ,

0)

( MPa, 0)

15 CX

(a)

xy

CX sin 30

(b)

x

ave

CX cos 30

2

MPa

R sin 30

30

R

MPa

sin 30 cos 30

MPa MPa

MPa

xy x

MPa

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PROBLEM P'

80 mm

Solve Prob. , using Mohr’s circle.

mm

b P

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

SOLUTION P , A

x

0

y

xy

0

Plotted points for Mohr’s circle:

X:

P ,0 , A

C:

P ,0 2

R

CX

Y : (0, 0)

P 2A

Coordinates of point Y : P (1 cos 2 ) 2A P sin 2 2A

Data:

A

If

(80)() kPa

P

mm 2 Pa,

(2)( 10 3 )( ) (1 cos 44 )

2A 1 cos 2 N

If

kPa

P

2A sin 2

10 3 m 2

kN

Pa,

(2)( 10 3 )( ) (sin 44 )

N

kN

The smaller value of P governs.

P

kN

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PROBLEM P'

80 mm

Solve Prob. , using Mohr’s circle.

mm

b P

PROBLEM Two wooden members of 80 mm uniform rectangular cross section are joined by the simple glued scarf 25 and that centric loads of splice shown. Knowing that 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

0

y

xy

0

Plotted points for Mohr’s circle:

X:

P ,0 A

Y : (0, 0)

C:

P ,0 2A

R

CX

P 2A

Coordinates of point Y: P (1 cos 2 ) 2A P sin 2 2A

Data:

A

(80)()

mm 2

(a)

(10 )sin 50 (2)( 10 3 )

(b)

(10 )(1 cos 50 ) (2)( 10 3 )

10 3 m 2

Pa

kPa

Pa

kPa

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PROBLEM

P

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

a

!

a

SOLUTION

x

0

xy

0

y

P A

(a)

From the Mohr’s circle, tan

(b)

P A

5 15 P P cos 2 2A 2A

1

2( ) cos 2

1



(2)(15) cos 2

ksi ksi

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PROBLEM Solve Prob. , using Mohr’s circle.

a a

25"

50 mm

PROBLEM 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 kPa and kPa, determine the largest centric load P that can be applied.

P

SOLUTION

x

0

xy

0

y

P/A

A (50 10 3 )(80 10 3 ) 4 10 3 m 2

P

P (1 cos50 ) 2A 2A 1 cos 50

(2)(4 10 3 )( ) 1 cos 50 P N P

P sin 50 2A

P

2A sin 50

Choosing the smaller value,

(2)(4 10 3 )( ) sin 50

N P

kN

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PROBLEM

m m 3 kN

Solve Prob. , using Mohr’s circle.

H

PROBLEM 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.

N · m 3 kN

SOLUTION c

Torsion:

1 d 2

1 (32) 2

Tc J

2T c3

16 mm 16 10 3 m

2( N m) (16 10 3 m)3 Bending:

I

M

4

c4

4

Pa

(16 10 3 )4

(m)(3 N) My I

MPa

10 9 m 4

N m

()(16 10 3 ) 10 9

Pa

Top view

Stresses

x

Plotted points:

MPa

MPa,

y

X : ( , ); ave

1 ( 2

0,

xy

MPa

Y: (0, ); C: ( , 0)

x

y)

x

y

MPa 2

R

2 xy

2 2

2

()2

MPa

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PROBLEM (Continued)

tan 2

2 p

xy

x

y

(2)( )

(a)

(b)

a a

ave

R





b

ave

R





max

R



,

b

MPa

a b

max

MPa MPa

PROPRIETARY MATERIAL. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

6 in.

PROBLEM

C H

Solve Prob. , using Mohr’s circle.

B

PROBLEM A 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.

A D 2 in. lb

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

V

lb

M

T

()(2)

lb in.

Shaft cross section:

()(6)

lb in.

d

1 in.

J

c4

Resultant stresses:

()()

Mc I

Bending:

in.

in 4

Tc J

Torsion:

Transverse shear:

2

1 d 2

c

I

1 J 2

in 4

psi

()()

ksi

psi

ksi

Stress at point H is zero. x ave

ksi, 1 ( 2

x

y)

x

y

y

0,

xy

ksi

ksi 2

R

2 () 2

a

ave

R

b

ave

R

max

R

2 xy

() 2

ksi a

ksi

b

ksi

max

ksi

PROPRIETARY MATERIAL. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

PROBLEM

H

Solve Prob. , using Mohr’s circle.

E

PROBLEM A mechanic uses a crowfoot wrench to loosen a bolt at E. Knowing that the mechanic applies a vertical lb force at A, determine the principal stresses and the maximum shearing stress at point H located as shown as on top of the 34 -in.-diameter shaft.

6 in.

B 24 lb 10 in.

A

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

V

24 lb

M

T

(24)(10)

lb in.

Shaft cross section:

(24)(6)

lb in.

d

in.

J

c4

Resultant stresses:

1 J 2

I

()()

Mc I

Bending:

in.

in 4

Tc J

Torsion:

Transverse shear:

2

1 d 2

c

in 4

psi

()()

psi

ksi

ksi

At point H, stress due to transverse shear is zero. x ave

ksi, 1 ( 2

0,

y

x

y)

x

y

xy

ksi

ksi 2

R

2

a

ave

R

b

ave

R

max

R



xy

2

ksi a

ksi

ksi

b max

ksi

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PROBLEM

y

6 mm A mm

Solve Prob. , using Mohr’s circle.

51 mm

A T

PROBLEM The steel pipe AB has a 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.

D

10 kN C

mm H

K

B z

x

SOLUTION ro

J I

do 2

2

ro4

2

51 mm

ri4

1 J 2

ri

ro

t

mm 4

45 mm

10

6

m4

10 6 m 4

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

10 N (10 )( 10 3 ) (10 )( 10 3 )

T

My

c

ro

xy

N m N m

N m 51 10 3 m ()(51 10 3 ) 10 6

Tc J

MPa

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

Stress due to V y

Fx is zero at point K. ()(51 10 3 ) 10 6

Mz c I

Point K lies on compression side of neutral axis.

y

MPa

MPa

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PROBLEM (Continued)

Total stresses at point K:

x

ave

0, 1 ( 2

MPa,

y

x

y)

x

y

xy

MPa

MPa 2

R max

min

max

2 ave

ave

R

R

R

2 xy





MPa

max

MPa

min

MPa



max

MPa

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PROBLEM

#y 20 MPa 60 MPa

Solve Prob. , using Mohr’s circle. PROBLEM 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 x

60 MPa,

y

?,

xy

20 MPa

Given: max

R

XY

2 R MPa

DY

(2)(

XD y

75 MPa

XY x

xy ) 2

40 MPa DY

2



XD 60



MPa y

MPa

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PROBLEM

8 ksi

$xy 10 ksi

Solve Prob. , using Mohr’s circle. PROBLEM 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 x

y

2 The radius of the circle is

max (in-plane)

10 8 , 0 2

,0

(1 ksi, 0).

12 ksi.

The stress point ( x , xy ) lies along the line X1 X 2 of the Mohr circle diagram. The extreme points with R 12 ksi are X 1 and X 2 . (a)

The largest allowable value of

xy 2

(b)

The principal stresses are

is obtained from triangle CDX. 2

DX 1

DX 2

xy



a

1 12

b

1 12

2

CX 1

CD

2

92

xy a b

ksi ksi ksi

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2 MPa

PROBLEM $xy

Solve Prob. , using Mohr’s circle. 12 MPa

75"

PROBLEM 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 that y 2 MPa. The coordinates of C are

so that

2 12 , 0 2

x

12 MPa. Likewise, point Y lies on line Y Y

(7 MPa, 0).

Counterclockwise rotation through ° brings line CX to CB, where

R (a)

x xy

y

2

x

y

2

sec 30

12 2 sec 30 2

0.

MPa

tan 30

12 2 tan 30 2

(b)

so

xy

MPa

a

ave

R

7

a

MPa

b

ave

R

7

b

MPa

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PROBLEM

15 ksi 8 ksi

Solve Prob. , using Mohr’s circle. #x

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

x

for which the maximum

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.

C1Y

10 ksi

C2Y

10 ksi

Noting right triangles C1 DY and C2 DY ,

C1D

2

DY

2

C1Y

2

C1D

2

82



C1D

6 ksi

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. 3 ksi

Thus,

x

27 ksi

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2 MPa

PROBLEM $xy

75"

Solve Problem , using Mohr’s circle and assuming that the weld forms an angle of 60 with the horizontal. 12 MPa

PROBLEM 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 12

Locate point C at Angle XCB x

y

2

2

7 MPa with

2

0.



12

2

2 5 MPa

R

5sec 60 10 MPa

5 tan 60

xy

MPa

xy ave

a

7

10 ave

b

7

R

10

a

MPa

b

MPa

R

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3 ksi

6 ksi

5 ksi

+



PROBLEM 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.

2 ksi 4 ksi

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:

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PROBLEM (Continued)

ave

x

y

2

1 (6 2

2)

1 (6 2

2)

(4)2

R tan 2

p

2

p

2

4

(3)2

5

3 4

p

max

ave

2

R

,

5 max

min

ave

R

2

ksi

5 min

ksi

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PROBLEM

MPa 50 MPa 50 MPa

+



75 MPa

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.

x

50 cos 30

y



xy

50sin 30

Principal axes and principal stress:

ave

y

x

2 R tan 2

p

1 ( 2

)

1 ( 2

)

()2

(75)2

75

2

p







max

ave

R





min

ave

R





p



, and max min

MPa MPa

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#0 #0

PROBLEM M

#0 #0

Determine thhe principal planes p and thhe principal stresses for the t state of pllane stress ressulting from the superposiition of the twoo states of streess shown.

30" 30"

SO OLUTION Exppress each state of stress in terms of horizzontal and verrtical componeents.

s of stresss, Addding the two states

p

0 and a 90° max m min

0 0

PRO OPRIETARY MATERIAL. Copyriight © McG Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use. Not authorized for salle or distribution in i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted on a website, in wholee or part.

PROBLEM

$0

$0

30"

+

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: x

0

y

0 0

xy

xy

3 2

0 sin 60

x

1 2

0 cos 60

0

y

0

y

3 2

0 sin 60

0

0

Resultant stresses: x

3 2 1 2

0

xy

0

ave

1 ( 2

3 2

0

3 2

0

x

y)

x

y

tan 2

2 x

2

2 xy

(2)

xy y

p

60

a

ave

R

b

ave

R

0

0

2

p

3 2

0

0

2

R

3 2

0

3 2 3

3 2

2 0

3 2

2 0

3

0

3 b

30

a

a b

60

3

0

3

0

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PROBLEM

MPa

$xy

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

xy

for which the

20 MPa

SOLUTION x ave

Set

max

R

20 MPa 1 ( 2

y)

x

60 MPa max

MPa

y

70 MPa

R

ave

MPa

ave

But 2 x

R

2 xy

x

2

2 xy

R

2

x

x

2

MPa Range of

xy :

MPa

xy

MPa

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PROBLEM

MPa

$xy

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 MPa. 20 MPa

SOLUTION 20 MPa

x

1( 2

y)

x

Set

max (in-plane)

But

R

MPa

y

50 MPa

R MPa 2 x

y

2 xy

2

2 xy

R

2



x

y

2

MPa

Range of

xy :

MPa

xy

MPa

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!y' 6 ksi

PROBLEM

"x'y'

!x' #

16 ksi

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 x xy ave

16 ksi,

0

y

6 ksi 1 ( 2

x

y)

x

y

8 ksi 2

R

( 8)2 tan 2

2 p x

2

2 xy

2 (6) 2

xy y

p



b



10 ksi

(2)(6) 16



8 ksi for states of stress corresponding to arcs HBK and UAV of Mohr’s circle. The angle

xy

is

calculated from R sin 2

2

8

8 10

sin 2







k

b





45

k

b







u

h

90

45

v

k

90



Permissible range of : Also,

h

k

u

v



45 45



and





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PROBLEM

#y' #x'

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.

%

90 MPa

$x'y' 60 MPa

SOLUTION x

90 MPa,

0

60 MPa

xy ave

y

1 ( 2

x

y)

x

y

45 MPa 2

R

2

tan 2

p



xy

x

2

x

2 xy

2

y

p



a



75 MPa (2)( 60) 90

4 3

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

R cos 2

50

cos 2

5 75

2

5 MPa



h

2

45



a

k

2

k



h



Permissible range of

:

4









h





k



Also,