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9–2. The state of stress at a point in a member is shown on the element. Determine the stress components acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec. 9.1.

8 MPa B

5 MPa 40⬚ A 3 MPa

¢Fx¿ + (8¢A sin 40°) cos 40° - (5¢A sin 40°) cos 50° - (3¢A cos 40°) cos 40° +

a+ ©Fx¿ = 0

(8¢A cos 40°) cos 50° = 0 ¢Fx¿ = - 4.052¢A b+ ©Fy¿ = 0

¢Fy¿ - (8¢A sin 40°) sin 40° - (5¢A sin 40°) sin 50° + (3¢A cos 40°) sin 40° + (8¢A cos 40°) sin 50° = 0 ¢Fy¿ = - 0.4044¢A

sx¿ = lim¢A : 0

tx¿y¿ = lim¢A : 0

¢Fx ¿ = - 4.05 MPa ¢A ¢Fy ¿

Ans.

Ans.

= - 0.404 MPa

¢A

The negative signs indicate that the senses of sx¿ and tx¿y¿ are opposite to that shown on FBD.

Ans: sx¿ = - 4.05 MPa, tx¿y¿ = - 0.404 MPa 837

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9–5. Determine the normal stress and shear stress acting on the inclined plane AB. Solve the problem using the stress transformation equations. Show the results on the sectional element.

15 MPa

B

A

60⬚ 6 MPa

Stress Transformation Equations: u = + 150° (Fig. a)

sx = 0

sy = - 15 MPa

txy = - 6 MPa

We obtain, sx¿ =

=

sx + sy

sx - sy +

2

2

cos 2u + txy sin 2u

0 - ( - 15) 0 + (- 15) + cos 300° + ( -6) sin 300° 2 2 Ans.

= 1.45 MPa tx¿y¿ = -

= -

sx - sy 2

sin 2u + txy cos 2u

0 - (- 15) sin 300° + (- 6) cos 300° 2

= 3.50 MPa

Ans.

The results are indicated on the triangular sectioned element shown in Fig. b.

Ans: sx¿ = 1.45 MPa, tx¿y¿ = 3.50 MPa 840

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9–10. Determine the equivalent state of stress on an element at the same point oriented 60° clockwise with respect to the element shown. Sketch the results on the element.

100 MPa 75 MPa 150 MPa

Stress Transformation Equations: u = - 60° (Fig. a)

sx = 150 MPa

sy = 100 MPa

txy = 75 MPa

We obtain, sx¿ =

=

sx + sy

sx - sy +

2

2

cos 2u + txy sin 2u

150 + 100 150 - 100 + cos ( -120°) + 75 sin (-120°) 2 2

= 47.5 MPa sy¿ =

=

Ans.

sx + sy

sx - sy -

2

2

cos 2u - txy sin 2u

150 - 100 150 + 100 cos ( -120°) - 75 sin (-120°) 2 2 Ans.

= 202 MPa tx¿y¿ = -

= -

sx - sy 2

sin 2u + txy cos 2u

150 - 100 sin ( - 120°) + 75 cos (- 120°) 2

= - 15.8 MPa

Ans.

The negative sign indicates that tx¿y¿ is directed towards the negative sense of the y¿ axis. These results are indicated on the element shown in Fig. b.

Ans: sx¿ = 47.5 MPa, sy¿ = 202 MPa, tx¿y¿ = -15.8 MPa 845

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9–14. The state of stress at a point is shown on the element. Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the element in each case. Show the results on each element.

30 30 ksi MPa 12 12 ksi MPa

sx = - 30 MPa ksi

sy = 0

MPa txy = - 12 ksi 4.21 MPa

a)

15 MPa

sx + sy

s1, 2 =

2

;

C

a

sx - sy

b + txy 2 =

-30 + 0 -30 - 0 2 a ; b + ( -12)2 2 C 2

ksi s1 = 4.21 MPa

Ans.

s2 = - 34.2 MPa ksi

Ans.

2

2

34.2 MPa

34.2 MPa

19.2 MPa 19.2 MPa

4.21 MPa

15 MPa 15 MPa

Orientation of principal stress: txy

tan 2uP =

(sx - sy)>2

uP = 19.33° and

-12 = 0.8 ( -30 - 0)>2

=

- 70.67°

Use Eq. 9-1 to determine the principal plane of s1 and s2. sx + sy

sx¿ =

2

+

sx - sy 2

cos 2u + txy sin 2u

u = 19.33° sx¿ =

-30 - 0 -30 + 0 + cos 2(19.33°) + ( -12)sin 2(19.33°) = - 34.2 MPa ksi 2 2

Therefore uP2 = 19.3°

Ans.

and uP1 = - 70.7°

Ans.

b) tmaxin-plane = savg =

C

a

sx - sy

2

=

- 30 - 0 2 MPa b + ( -12)2 = 19.2 ksi C 2 a

Ans.

- 30 + 0 = - 15 MPa ksi 2

Ans.

2

sx + sy

2

b + txy 2 =

Orientation of max, in - plane shear stress: tan 2uP =

-(sx - sy)>2 txy

uP = - 25.2°

and

=

-( - 30 - 0)>2 = - 1.25 - 12 Ans.

64.3°

By observation, in order to preserve equllibrium along AB, tmax has to act in the direction shown in the figure. Ans. (a) s1 4.21 MPa, s2 34.2 MPa, up2 19.3°, up1 70.7° = 19.2 MPa, ksi savg 15 MPa, (b) tmax in-plane

us = - 25.7°, 64.3° 849

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*9–20. Planes AB and BC at a point are subjected to the stresses shown. Determine the principal stresses acting at this point and find sBC.

A 5 MPa

15 MPa

45⬚

B 6 MPa

C

Stress Transformation Equations: Referring to Fig. a and the established sign convention, u = - 135°

sx = - 15 MPa sy = sAC

txy = 5 MPa

tx¿y¿ = 6 MPa sx¿ = sBC

We have tx¿y¿ = -

6 = -

sx - sy 2

sin 2u + txy cos 2u

-15 - sAC sin (-270°) + 5 cos ( -270°) 2

sAC = - 3 MPa Using this result, s1, 2 =

=

sx + sy 2

;

A

a

sx - sy

2

-15 - (- 3) 2 - 15 + ( - 3) 2 d + 5 ; Ac 2 2

s1 = - 1.19 MPa sBC = sx¿ =

=

2

b + txy2

sx + sy 2

Ans.

s2 = - 16.8 MPa sx - sy +

2

cos 2u - txy sin 2u

-15 + ( - 3) -15 - ( -3) + cos (- 270°) - 5 sin ( -270°) 2 2 Ans.

= - 14.1 MPa

858

sBC

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9–23. The wood beam is subjected to a load of 12 kN. If a grain of wood in the beam at point A makes an angle of 25° with the horizontal as shown, determine the normal and shear stress that act perpendicular and parallel to the grain due to the loading.

I =

12 kN 1m

2m A 25⬚

4m 300 mm

75 mm

200 mm

1 (0.2)(0.3)3 = 0.45(10 - 3) m4 12

QA = yA¿ = 0.1125(0.2)(0.075) = 1.6875(10 - 3) m3 sA =

13.714(103)(0.075) MyA = 2.2857 MPa (T) = I 0.45(10 - 3)

tA =

6.875(103)(1.6875)(10 - 3) VQA = 0.1286 MPa = It 0.45(10 - 3)(0.2)

sx = 2.2857 MPa sx¿ =

sx¿ =

sx + sy

sx - sy +

2

sy = 0

2

txy = - 0.1286 MPa

u = 115°

cos 2u + txy sin 2u

2.2857 + 0 2.2857 - 0 + cos 230° + (- 0.1286)sin 230° 2 2 Ans.

= 0.507 MPa tx¿y¿ = -

sx - sy

= -a

2

sin 2u + txy cos 2u

2.2857 - 0 b sin 230° + ( -0.1286)cos 230° 2

= 0.958 MPa

Ans.

Ans: sx¿ = 0.507 MPa, tx¿y¿ = 0.958 MPa 861

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sy

9–25. The wooden block will fail if the shear stress acting along the grain is 4 MPa. If the normal stress s x = 3 MPa, determine the necessary compressive stress sy that will cause failure.

58⬚

tx¿y¿ = - a 4 = -a

sx - sy

3

2 -3 sy 2

sx ⫽ 3 MPa

b sin 2u + txy cos 2u b sin 296° + 0

sy = - 5.9 MPa

Ans.

Ans: sy = - 5.9 MPa 863

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9–29. The 75-mm diameter shaft is supported by a smooth thrust bearing at A and a smooth journal bearing at B. Determine the principal stresses and maximum in-plane shear stress at a point on the outer surface of the shaft at section a–a.

A

a 7 00 N ⭈ m a

B

7 00 N ⭈ m

1 5 kN

Internal Loadings: Consider the equilibrium of the free-body diagram of the shaft’s right cut segment, Fig. a. ©Fx = 0;

N + 15 kN = 0

N = - 15 kN

©Mx = 0;

T + 700 = 0

T = - 700 N.m

Section Properties: The cross-sectional area and the polar moment of inertia of the shaft’s cross section are A = p(37.5 2) = 4417.86p mm 2 J =

p (37.5 4) = 3106311.1 p mm 4 2

Normal and Shear Stress: The normal stress is contributed by the axial stress only. Thus, s =

N - 15000N = - 3395.31 kPa = 4417.86 A

The shear stress is contributed by the torsional shear stress only. t =

Tc - 700(1000)(37.5) = = 84505.38 kPa J 3106311.1 1

The state of stress at a point on the outer surface of the shaft, Fig. b, can be represented by the element shown in Fig. c. In-Plane Principal Stress: sx = - 3395.31 kPa, sy = 0, and txy = - 84505.38 kPa . We obtain, s1,2 =

=

sx + sy ;

2

Aa

sx - sy 2

2 b + txy 2

2 - 3395.31 + 0 ; a - 3395.31 - 0 b + ( - 84505.38)2 2 A 2

s1 = 84324.78 kPa = 84.324 MPa s 2 = - 84720.1 kPa = - 84.72 kPa

Ans.

Maximum In-Plane Shear Stress: tmax

in-plane

=

A

a

sx - sy 2

b + txy2 = A a 2

- 3395.31 - 0 2 2 b + ( - 84505.38) = 84522.43 kPa 2

= 84.52 MPa

Ans.

Ans: s1 = 84.324 M < Pa, s2 = -84.72 MPa, t max = 84.52 MPa [email protected]

869

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9–31. Determine the principal stress at point A on the cross section of the arm at section a–a. Specify the orientation of this state of stress and indicate the results on an element at the point.

7.5 mm A 50 mm

7.5 mm

Support Reactions: Referring to the free-body diagram of the entire arm shown in Fig. a, ©MB = 0; FCD sin 30°(0.3) - 500(0.65) = 0

FCD = 2166.67 N

+ ©F = 0; : x

Bx - 2166.67 cos 30° = 0

Bx = 1876.39 N

+ c ©Fy = 0;

2166.67 sin 30° - 500 - By = 0

By = 583.33 N

20 mm

Section a – a D

Internal Loadings: Consider the equilibrium of the free-body diagram of the arm’s left segment, Fig. b. + ©F = 0; : x

1876.39 - N = 0

N = 1876.39 N

+ c ©Fy = 0;

V - 583.33 = 0

V = 583.33 N

+ ©MO = 0;

583.33(0.15) - M = 0

M = 87.5N # m

0.15 m

Referring to Fig. c, QA = y¿A¿ = 0.02125(0.0075)(0.02) = 3.1875 A 10 - 6 B m3 Normal and Shear Stress: The normal stress is a combination of axial and bending stress. Thus,

-1876.39

0.5625 A 10

-3

B

MyA N + A I 87.5(0.0175)

+

0.16367 A 10 - 6 B

= 6.020 MPa

The shear stress is caused by transverse shear stress.

tA

583.33 C 3.1875 A 10 - 6 B D VQA = = = 1.515 MPa It 0.16367 A 10 - 6 B (0.0075)

The state of stress at point A can be represented on the element shown in Fig. d. In-Plane Principal Stress: sx = 6.020 MPa, sy = 0, and txy = 1.515 MPa. We have s1,2 =

=

sx + sy 2

;

C

¢

sx - sy 2

2

≤ + txy 2

6.020 - 0 2 6.020 + 0 a ; b + 1.5152 2 C 2

s1 = 6.38 MPa

s2 = - 0.360 MPa

Ans

871

C 0.15 m

0.35 m 500 N

1 1 (0.02) A 0.053 B (0.0125) A 0.0353 B = 0.16367 A 10 - 6 B m4 12 12

=

a

a

A = 0.02(0.05) - 0.0125(0.035) = 0.5625 A 10 - 3 B m2

sA =

60⬚

B

Section Properties: The cross - sectional area and the moment of inertia about the z axis of the arm’s cross section are

I =

7.5 mm

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9–31. Continued Orientation of the Principal Plane: tan 2uP =

txy

A sx - sy B >2

=

1.515 = 0.5032 (6.020 - 0)>2

up = 13.36° and -76.64° Substituting u = 13.36° into sx¿ =

=

sx + sy 2

sx - sy +

2

cos 2u + txy sin 2u

6.020 + 0 6.020 - 0 + cos 26.71° + 1.515 sin 26.71° 2 2

= 6.38 MPa = s1 Thus, A uP B 1 = 13.4 and A uP B 2 = - 76.6°

Ans.

The state of principal stresses is represented by the element shown in Fig. e.

Ans: s1 = 6.38 MPa, s2 = -0.360 MPa, up1 = 13.4° and up2 = - 76.6° 872

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9–35. The square steel plate has a thickness of 10 mm and is subjected to the edge loading shown. Determine the maximum in-plane shear stress and the average normal stress developed in the steel. sx = 5 kPa

sy = - 5 kPa

tmax

sx - sy

in-plane

savg =

a

txy = 0

200 mm

b + t2xy

C

=

5 + 5 2 b + 0 = 5 kPa C 2 a

sx + sy 3

=

50 N/m

2

=

2

50 N/m

Ans. 200 mm

5 - 5 = 0 2

Ans.

Note: tan 2us =

tan 2us =

- (sx - sy)>2 txy -(5 + 5)>2 = q 0

us = 45°

Ans: tmax in-plane

879

= 5 kPa, savg = 0

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P

9–37. The shaft has a diameter d and is subjected to the loadings shown. Determine the principal stress and the maximum in-plane shear stress that is developed at point A. The bearings only support vertical reactions.

F

F A L 2

L 2

Support Reactions: As shown on FBD(a). Internal Forces and Moment: As shown on FBD(b). Section Properties: A =

p 2 d 4

p d 4 p 4 a b = d 4 2 64

I =

QA = 0

Normal Stress: N Mc ; A I

s =

-F p 2 ; 4 d

=

sA =

A d2 B

PL 4 p 64

d4

4 2PL - Fb a d pd2

Shear Stress: Since QA = 0, tA = 0 In-Plane Principal Stress: sx =

4 2PL a - Fb. pd2 d

sy = 0 and txy = 0 for point A. Since no shear stress acts on the element, s1 = sx =

4 2PL a - Fb 2 d pd

Ans.

s2 = sy = 0

Ans.

Maximum In-Plane Shear Stress: Applying Eq. 9-7 for point A, t

max in-plane

=

£

B

a

4 2 pd

sx - sy 2

2

b + t2xy

A 2PL d - FB - 0

=

Q

=

2PL 2 a - Fb d pd2

2

2

≥ + 0

Ans.

Ans: s1 =

4 2PL - Fb , s2 = 0, a d pd2

tmax in-plane

881

=

2PL 2 a - Fb d pd2

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9–38. A paper tube is formed by rolling a paper strip in a spiral and then gluing the edges together as shown. Determine the shear stress acting along the seam, which is at 30° from the vertical, when the tube is subjected to an axial force of 10 N. The paper is 1 mm thick and the tube has an outer diameter of 30 mm. s =

P = A

p 4

= -

10 N

10 N 30 mm

10 = 109.76 kPa (0.03 - 0.0282) 2

sx = 109.76 kPa tx¿y¿ = -

30⬚

sx - sy 2

sy = 0

txy = 0

u = 30°

sin 2u + txy cos 2u

109.76 - 0 sin 60° + 0 = - 47.5 kPa 2

Ans.

Ans: tx¿y¿ = -47.5 kPa 882

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1500 7.5 kNlb

pipe has hasan anouter outerdiameter diameterofof3 in., 75 mm, 9–42. The drill pipe a walla wall thickness of 6 in., mm,and andaa weight weight of 0.8 kN/m.IfIf itit is 50 lb>ft. thickness of 0.25 subjected to a torque and axial load as shown, determine (a) the principal stress and (b) the maximum in-plane shear stress at a point on its surface at section a.

800kNm lbft 1.2 620mft

aa 620mft

Internal Forces and Torque: As shown on FBD(a). Section Properties: A =

p 22 2.522)B==414p 0.6875p (75 mm2in2 A 3 -– 63 4

J =

p 6 4 1.254)B ==1.560(10 4.1172 in (37.5 ) mm4 A 1.54 4-– 31.5 2

7.5 kN

Normal Stress:

1.2 kN · m 0.8(6) = 4.8 kN

–12.3(10 N -2500 3) = = - 1157.5 –9.457 psi MPa A 0.6875p 414p

s =

Shear Stress: Applying the torsion formula. t =

6 800(12)(1.5) )(37.5) T c 1.2(10 = =3497.5 28.846psi MPa J 4.1172 6) 1.560(10

N = 12.3 kN T = 1.2 kN · m

- 1157.5MPa psi and = 28.846 3497.5MPa psi for a) In - Plane Principal Stresses: sx = 0, sy = –9.457 and ttxy for xy = any point on the shaft’s surface. Applying Eq. 9-5. s1,2 = =

sx + sy 2

;

C

a

sx - sy 2

2

b + t2xy

0 - ((–9.457) - 1157.5) 2 0 + (1157.5) (–9.457) (28.846)2 ; a b + (3497.5) 2 C 2

28.846 MPa 9.457 MPa

= 578.75 ±;29.2310 3545.08 –4.7285 MPa 2966 psi = 2.97 ksi s1 = 24.50

Ans.

–33.96 MPa= - 4.12 ksi s2 = 4124 psi

Ans.

b) Maximum In - Plane Shear Stress: Applying Eq. 9-7 t

max in-plane

= =

C

a

sx - sy 2

2

b + t2xy

0 - ((–9.457) - 1157.5) 2 (3497.5)2 ≤ + (28.846) C 2

¢

= 29.23 3545 psi = 3.55 ksi MPa

Ans.

Ans: s1 24.51 MPa, s2 33.96 MPa, tmax = 29.24 3.55 ksi MPa in-plane

886

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9–56. Determine the principal stress, the maximum in-plane shear stress, and average normal stress. Specify the orientation of the element in each case. 105 MPa 15 ksi 35 MPa 5 ksi

Construction of the Circle: In accordance with the sign convention, sx = 15 , 105ksi MPa, . Hence, sy = 0 and txy = –35 - 5 ksi MPa. Hence, sx + sy

savg =

2

=

105 15 ++ 00 = 52.5 7.50 MPa ksi 2

105 63.097

Ans.

35 (MPa)

The coordinates for reference point A and C are A(15, 5) A(105,-–35)

C(7.50, C(52.5,0)0) (MPa)

10.6 MPa

The radius of the circle is R = 2(15 7.50)22 ++3552 ==63.097 9.014 ksi (105 -– 52.5) = 63.10 MPa 115.6 MPa

a) 52.5 MPa

In - Plane Principal Stress: The coordinates of points B and D represent s1 and s2, respectively. MPa 7.50 ++ 63.097 9.014 == 115.60 16.5 ksi s1 = 52.5

63.1 MPa

Ans.

52.5 –-63.097 s2 = 7.50 9.014 == –10.60 - 1.51 MPa ksi

Ans.

Orientation of Principal Plane: From the circle tan 2uP1 =

35 5 = 0.6667 15 -– 52.5 7.50 105

uP1 = 16.8° (Clockwise)

Ans.

b) Maximum In - Plane Shear Stress: Represented by the coordinates of point E on the circle. tmax

in-plane

–63.1 = -R = 9.01 MPa ksi

Ans.

Orientation of the Plane for Maximum In - Plane Shear Stress: From the circle tan 2us =

52.5 MPa

15 -– 52.5 7.50 105 = 1.500 5 35

us = 28.2° (Counterclockwise)

Ans.

900

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9–62. The grains of wood in the board make an angle of 20° with the horizontal as shown. Using Mohr’s circle, determine the normal and shear stresses that act perpendicular and parallel to the grains if the board is subjected to an axial load of 250 N.

sx =

300 mm

60 mm 250 N

250 N 20⬚

25 mm

250 P = = 166.67 kPa A (0.06)(0.025)

R = 83.33 Coordinates of point B: sx¿ = 83.33 - 83.33 cos 40° sx¿ = 19.5 kPa

Ans.

tx¿y¿ = - 83.33 sin 40° = - 53.6 kPa

Ans.

Ans: sx¿ = 19.5 kPa, tx¿y¿ = - 53.6 kPa 906

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9–66. Determine the principal stress and maximum in-plane shear stress at point A on the cross section of the pipe at section a–a.

a 300 mm a 300 mm b

30 mm

A B

200 mm

300 N

Internal Loadings: Considering the equilibrium of the free-body diagram of the assembly’s segment, Fig. a, ©Fx = 0; N - 450 = 0

N = 450 N

©Fy = 0; Vy = 0 ©Fz = 0; Vz + 300 = 0

Vz = - 300 N

©Mx = 0; T + 300(0.2) = 0

T = - 60 N # m

©My = 0; My - 450(0.3) + 300(0.3) = 0

My = 45 N # m

©Mz = 0; Mz + 450(0.2) = 0

Mz = - 90 N # m

Section Properties: The cross-sectional area, the moment of inertia about the y and z axes, and the polar moment of inertia of the pipe’s cross section are A = p(0.032 - 0.022) = 0.5p(10-3) m2 Iy = Iz = J =

p (0.034 - 0.024) = 0.1625p(10-6) m4 4

p (0.034 - 0.024) = 0.325p(10-6) m4 2

Referring to Fig. b, (Qy)A = 0 (Qz)A =

4(0.03) p 4(0.02) p c (0.03)2 d c (0.022) d = 12.667(10-6) m3 3p 2 3p 2

Normal and Shear Stress: The normal stress is a combination of axial and bending stress. sA =

MzyA ( - 90)(- 0.03) N 450 = -3 A Iz 0.5p(10 ) 0.1625p(10-6)

= - 5.002 MPa Since Vy = 0, [(txy)V]A = 0. However, the shear stress is the combination of torsional and transverse shear stress. Thus, (txz)A = [(txz)T]A - [(txz)V]A

=

Vz(Qz)A 60(0.03) 300[12.667(10-6)] Tc = 1.391 MPa = -6 J Iy t 0.325p(10 ) 0.1625p(10-6)(0.02)

The state of stress at point A is represented by the element shown in Fig. c.

911

450 N

20 mm Section a – a

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9–66. Continued Construction of the Circle: sx = - 5.002 MPa, sz = 0, and txz = 1.391 MPa. Thus, savg =

sx + sz 2

=

- 5.002 + 0 = - 2.501 MPa 2

The coordinates of reference point A and the center C of the circle are A(- 5.002, 1.391)

C(- 2.501, 0)

Thus, the radius of the circle is R = CA = 2[- 5.002 - ( -2.501)]2 + 1.3912 = 2.862 MPa Using these results, the circle is shown in Fig. d. In-Plane Principal Stresses: The coordinates of reference points B and D represent s1 and s2, respectively. s1 = - 2.501 + 2.862 = 0.361 MPa

Ans.

s2 = - 2.501 - 2.862 = - 5.36 MPa

Ans.

In-Plane Maximum Shear Stress: The coordinates of point E represent the state of maximum shear stress. Thus, tmax

= |R| = 2.86 MPa

Ans.

in-plane

Ans: s1 = 0.361 MPa , s2 = - 5.36 MPa, tmax = 2.86 MPa in-plane

912

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*9–68. The rotor shaft of the helicopter is subjected to the tensile force and torque shown when the rotor blades provide the lifting force to suspend the helicopter at midair. If the shaft has a diameter of 150 mm, determine the principal stress and maximum in-plane shear stress at a point located on the surface of the shaft.

225 kN 15 kN⭈m

Internal Loadings: Considering the equilibrium of the free-body diagram of the rotor shaft’s upper segment, Fig. a, ©Fy = 0;

N - 225000 = 0

N = 225000 N

©My = 0;

T - 15 = 0

T = 15 kN.m

Section Properties: The cross-sectional area and the polar moment of inertia of the rotor shaft’s cross section are A = p(752) = 17671.45 mm 2 J =

p (75 4 ) = 49700977.53 mm4 2

Normal and Shear Stress: The normal stress is contributed by axial stress only. sA =

N 225000 = = 12.73 MPa A 17671.45 .45

The shear stress is contributed by the torsional shear stress only. tA =

15(1000 × 1000)(75) Tc = = 22.63 MPa J 49700977.53

The state of stress at point A is represented by the element shown in Fig. b.

915

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9–68. Continued Construction of the Circle: sx = 0, sy = 12.73 MPa, and txy = 22.63 MPa. Thus, savg =

sx + sy 2

=

0 + 12.73 = 6.365 MPa 2

The coordinates of reference point A and the center C of the circle are A(0, 22.63)

C(6.365, 0)

Thus, the radius of the circle is R = CA = 2(0 - 6.365)2 + 22.632 = 23.51 MPa Using these results, the circle is shown in Fig. c. In-Plane Principal Stress: The coordinates of reference points B and D represent s1 and s2, respectively. s1 = 6.365 + 23.51 = 29.87 MPa

Ans.

s2 = 6.365 - 23.51 = - 17.16 MPa

Ans.

Maximum In-Plane Shear Stress: The state of maximum shear stress is represented by the coordinates of point E, Fig. a. tmax

= R = 23.51 MPa

Ans.

in-plane

916

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*9–69. The pedal crank for a bicycle has the cross section shown. If it is fixed to the gear at B and does not rotate while subjected to aa force force of of 400 75 lb, N, determine the principal stress in the material on the cross section at point C.

400 N 75 75 lb lb B B

A A 100 mm 4 in.

753mm in.

C C

0.2 in. 5 mm 0.3mm in. 7.5

Internal Forces and Moment: As shown on FBD Section Properties:

0.4mm in. 10 0.4mm in. 10

400 N 100 mm

1 4 3 I = (0.3) A 0.833) B==5000 0.0128 (7.5)(20 mmin 12

M = 40000 N · mm

7.5(5)(7.5) = 281.25 mm QC = y¿A¿ = 0.3(0.2)(0.3) = 0.0180 in33

V = 400 N 7.5 mm

Normal Stress: Applying the flexure formula.

5 mm

7.5 mm

5 mm

My - 300(0.2) –40000(5) 40 MPa psi = 4.6875 ksi sC = = = 4687.5 5000 I 0.0128

10 mm

Shear Stress: Applying the shear formula.

3 MPa

VQC 75.0(0.0180) 400(281.25) tC = = =3351.6 MPa psi = 0.3516 ksi It 0.0128(0.3) 5000(7.5)

40 MPa

Construction of the Circle: In accordance with the sign convention, sx = 40 4.6875 ksi, MPa, . Hence, sy = 0, and txy = 0.3516 3 MPa.ksi Hence, savg =

sx + sy 2

=

20

40 +0 + 0 4.6875 = 2.34375 ksi 20 MPa 2 2

3 (MPa)

The coordinates for reference points A and C are A(40, 3) 0.3516) C(20, 0) C(2.34375, 0) A(4.6875, The radius of the circle is

40

2 (40 – 20)-2 +2.34375) 32 = 20.224 MPa 2 = 2.3670 ksi R = 2(4.6875 + 0.3516

(MPa)

In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2, respectively. 2.34375 + 2.3670 = MPa 4.71 ksi s1 = 20 + 20.224 = 40.223

Ans.

– 20.224 = –0.224 s2 = 20 2.34375 - 2.3670 = MPa - 0.0262 ksi

Ans.

Ans: s1 = 40.223 MPa, s2 = - 0.224 MPa 917

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•9–70.

A spherical pressure vessel has an inner radius of 51.5 ftm and in. Draw anda awall wallthickness thicknessof of0.5 12 mm. DrawMohr’s Mohr’scircle circlefor for the state of stress at a point on the vessel and explain the significance of the result. The vessel is subjected to an 0.56 MPa. internal pressure of 80 psi.

Normal Stress: s1 = s2 =

80(5)(12) pr 0.56(1.5)(1000) = = 4.80ksi 35 MPa 2t 2(0.5) 2(12)

Mohr’s circle: A(35, 0)0) A(4.80,

B(35, 0) 0) C(35, 0) B(4.80, C(4.80, 0)

Regardless of the orientation of the element, the shear stress is zero and the state of stress is represented by the same two normal stress components.

35 MPa

35 MPa

(35, 0)

Ans: s1 s2 35 MPa

918

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z

9–83. The state of stress at a point is shown on the element. Determine the principal stress and the absolute maximum shear stress. x

For y - z plane: A(5, - 4)

B( - 2.5, 4)

y

2.5 kPa ksi

C(1.25, 0) 4 kPa ksi

R = 23.752 + 42 = 5.483 ksi s1 = 1.25 + 5.483 = 6.733 kPa

5 kPa ksi

s2 = 1.25 - 5.483 = - 4.233 kPa ksi Thus,

savg = tabs

max

=

ksi s1 = 6.73 kPa

Ans.

s2 = 0

Ans.

s3 = - 4.23 ksi kPa

Ans.

6.73 + (- 4.23) = 1.25 kPa ksi 2 6.73 - (- 4.23) smax - smin kPa = = 5.48 ksi 2 2

Ans. 25 kPa 4 kPa 5 kPa

423 kPa

673 kPa

Ans: s1 = 6.73 kPa, s2 = 0, s3 = - 4.23 kPa, tabs = 5.48 kPa, max

935

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9–86. The plate is subjected to a tensile force P = 5 kN. If it has the dimensions shown, determine the principal stresses and the absolute maximum shear stress. If the material is ductile it will fail in shear. Make a sketch of the plate showing how this failure would appear. If the material is brittle the plate will fail due to the principal stresses. Show how this failure occurs.

s =

P 25 kN

50 mm P 25 kN 50 mm 300 mm

12 mm

P 25000N = = 20833.33 kPa = 20.833 MPa A (100)(12) s1 = 20.833 MPa

Ans.

s2 = s3 = 0

Ans.

tabs = max

s1 = 10.41 MPa 2

Ans.

Failure by shear:

Failure by principal stress:

Ans: s1 = 20.833 MPa, s2 = s3 = 0, tabs = 10.41 MPa max

937

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300 mm

9–87. Determine the principal stresses and absolute maximum shear stress developed at point A on the cross section of the bracket at section a–a. 150 mm a

5

3 4

a 12 mm B

2.5 kN

6 mm A

6 mm

6 mm 36 mm 36 mm

Section a – a

Ans: smax 5.06 MPa, sint 0, smin 8.04 MPa, tabs = 6.55 755 psi MPa m ax

938

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9–2. The state of stress at a point in a member is shown on the element. Determine the stress components acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec. 9.1.

8 MPa B

5 MPa 40⬚ A 3 MPa

¢Fx¿ + (8¢A sin 40°) cos 40° - (5¢A sin 40°) cos 50° - (3¢A cos 40°) cos 40° +

a+ ©Fx¿ = 0

(8¢A cos 40°) cos 50° = 0 ¢Fx¿ = - 4.052¢A b+ ©Fy¿ = 0

¢Fy¿ - (8¢A sin 40°) sin 40° - (5¢A sin 40°) sin 50° + (3¢A cos 40°) sin 40° + (8¢A cos 40°) sin 50° = 0 ¢Fy¿ = - 0.4044¢A

sx¿ = lim¢A : 0

tx¿y¿ = lim¢A : 0

¢Fx ¿ = - 4.05 MPa ¢A ¢Fy ¿

Ans.

Ans.

= - 0.404 MPa

¢A

The negative signs indicate that the senses of sx¿ and tx¿y¿ are opposite to that shown on FBD.

Ans: sx¿ = - 4.05 MPa, tx¿y¿ = - 0.404 MPa 837

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9–5. Determine the normal stress and shear stress acting on the inclined plane AB. Solve the problem using the stress transformation equations. Show the results on the sectional element.

15 MPa

B

A

60⬚ 6 MPa

Stress Transformation Equations: u = + 150° (Fig. a)

sx = 0

sy = - 15 MPa

txy = - 6 MPa

We obtain, sx¿ =

=

sx + sy

sx - sy +

2

2

cos 2u + txy sin 2u

0 - ( - 15) 0 + (- 15) + cos 300° + ( -6) sin 300° 2 2 Ans.

= 1.45 MPa tx¿y¿ = -

= -

sx - sy 2

sin 2u + txy cos 2u

0 - (- 15) sin 300° + (- 6) cos 300° 2

= 3.50 MPa

Ans.

The results are indicated on the triangular sectioned element shown in Fig. b.

Ans: sx¿ = 1.45 MPa, tx¿y¿ = 3.50 MPa 840

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9–10. Determine the equivalent state of stress on an element at the same point oriented 60° clockwise with respect to the element shown. Sketch the results on the element.

100 MPa 75 MPa 150 MPa

Stress Transformation Equations: u = - 60° (Fig. a)

sx = 150 MPa

sy = 100 MPa

txy = 75 MPa

We obtain, sx¿ =

=

sx + sy

sx - sy +

2

2

cos 2u + txy sin 2u

150 + 100 150 - 100 + cos ( -120°) + 75 sin (-120°) 2 2

= 47.5 MPa sy¿ =

=

Ans.

sx + sy

sx - sy -

2

2

cos 2u - txy sin 2u

150 - 100 150 + 100 cos ( -120°) - 75 sin (-120°) 2 2 Ans.

= 202 MPa tx¿y¿ = -

= -

sx - sy 2

sin 2u + txy cos 2u

150 - 100 sin ( - 120°) + 75 cos (- 120°) 2

= - 15.8 MPa

Ans.

The negative sign indicates that tx¿y¿ is directed towards the negative sense of the y¿ axis. These results are indicated on the element shown in Fig. b.

Ans: sx¿ = 47.5 MPa, sy¿ = 202 MPa, tx¿y¿ = -15.8 MPa 845

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9–14. The state of stress at a point is shown on the element. Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the element in each case. Show the results on each element.

30 30 ksi MPa 12 12 ksi MPa

sx = - 30 MPa ksi

sy = 0

MPa txy = - 12 ksi 4.21 MPa

a)

15 MPa

sx + sy

s1, 2 =

2

;

C

a

sx - sy

b + txy 2 =

-30 + 0 -30 - 0 2 a ; b + ( -12)2 2 C 2

ksi s1 = 4.21 MPa

Ans.

s2 = - 34.2 MPa ksi

Ans.

2

2

34.2 MPa

34.2 MPa

19.2 MPa 19.2 MPa

4.21 MPa

15 MPa 15 MPa

Orientation of principal stress: txy

tan 2uP =

(sx - sy)>2

uP = 19.33° and

-12 = 0.8 ( -30 - 0)>2

=

- 70.67°

Use Eq. 9-1 to determine the principal plane of s1 and s2. sx + sy

sx¿ =

2

+

sx - sy 2

cos 2u + txy sin 2u

u = 19.33° sx¿ =

-30 - 0 -30 + 0 + cos 2(19.33°) + ( -12)sin 2(19.33°) = - 34.2 MPa ksi 2 2

Therefore uP2 = 19.3°

Ans.

and uP1 = - 70.7°

Ans.

b) tmaxin-plane = savg =

C

a

sx - sy

2

=

- 30 - 0 2 MPa b + ( -12)2 = 19.2 ksi C 2 a

Ans.

- 30 + 0 = - 15 MPa ksi 2

Ans.

2

sx + sy

2

b + txy 2 =

Orientation of max, in - plane shear stress: tan 2uP =

-(sx - sy)>2 txy

uP = - 25.2°

and

=

-( - 30 - 0)>2 = - 1.25 - 12 Ans.

64.3°

By observation, in order to preserve equllibrium along AB, tmax has to act in the direction shown in the figure. Ans. (a) s1 4.21 MPa, s2 34.2 MPa, up2 19.3°, up1 70.7° = 19.2 MPa, ksi savg 15 MPa, (b) tmax in-plane

us = - 25.7°, 64.3° 849

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*9–20. Planes AB and BC at a point are subjected to the stresses shown. Determine the principal stresses acting at this point and find sBC.

A 5 MPa

15 MPa

45⬚

B 6 MPa

C

Stress Transformation Equations: Referring to Fig. a and the established sign convention, u = - 135°

sx = - 15 MPa sy = sAC

txy = 5 MPa

tx¿y¿ = 6 MPa sx¿ = sBC

We have tx¿y¿ = -

6 = -

sx - sy 2

sin 2u + txy cos 2u

-15 - sAC sin (-270°) + 5 cos ( -270°) 2

sAC = - 3 MPa Using this result, s1, 2 =

=

sx + sy 2

;

A

a

sx - sy

2

-15 - (- 3) 2 - 15 + ( - 3) 2 d + 5 ; Ac 2 2

s1 = - 1.19 MPa sBC = sx¿ =

=

2

b + txy2

sx + sy 2

Ans.

s2 = - 16.8 MPa sx - sy +

2

cos 2u - txy sin 2u

-15 + ( - 3) -15 - ( -3) + cos (- 270°) - 5 sin ( -270°) 2 2 Ans.

= - 14.1 MPa

858

sBC

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9–23. The wood beam is subjected to a load of 12 kN. If a grain of wood in the beam at point A makes an angle of 25° with the horizontal as shown, determine the normal and shear stress that act perpendicular and parallel to the grain due to the loading.

I =

12 kN 1m

2m A 25⬚

4m 300 mm

75 mm

200 mm

1 (0.2)(0.3)3 = 0.45(10 - 3) m4 12

QA = yA¿ = 0.1125(0.2)(0.075) = 1.6875(10 - 3) m3 sA =

13.714(103)(0.075) MyA = 2.2857 MPa (T) = I 0.45(10 - 3)

tA =

6.875(103)(1.6875)(10 - 3) VQA = 0.1286 MPa = It 0.45(10 - 3)(0.2)

sx = 2.2857 MPa sx¿ =

sx¿ =

sx + sy

sx - sy +

2

sy = 0

2

txy = - 0.1286 MPa

u = 115°

cos 2u + txy sin 2u

2.2857 + 0 2.2857 - 0 + cos 230° + (- 0.1286)sin 230° 2 2 Ans.

= 0.507 MPa tx¿y¿ = -

sx - sy

= -a

2

sin 2u + txy cos 2u

2.2857 - 0 b sin 230° + ( -0.1286)cos 230° 2

= 0.958 MPa

Ans.

Ans: sx¿ = 0.507 MPa, tx¿y¿ = 0.958 MPa 861

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sy

9–25. The wooden block will fail if the shear stress acting along the grain is 4 MPa. If the normal stress s x = 3 MPa, determine the necessary compressive stress sy that will cause failure.

58⬚

tx¿y¿ = - a 4 = -a

sx - sy

3

2 -3 sy 2

sx ⫽ 3 MPa

b sin 2u + txy cos 2u b sin 296° + 0

sy = - 5.9 MPa

Ans.

Ans: sy = - 5.9 MPa 863

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9–29. The 75-mm diameter shaft is supported by a smooth thrust bearing at A and a smooth journal bearing at B. Determine the principal stresses and maximum in-plane shear stress at a point on the outer surface of the shaft at section a–a.

A

a 7 00 N ⭈ m a

B

7 00 N ⭈ m

1 5 kN

Internal Loadings: Consider the equilibrium of the free-body diagram of the shaft’s right cut segment, Fig. a. ©Fx = 0;

N + 15 kN = 0

N = - 15 kN

©Mx = 0;

T + 700 = 0

T = - 700 N.m

Section Properties: The cross-sectional area and the polar moment of inertia of the shaft’s cross section are A = p(37.5 2) = 4417.86p mm 2 J =

p (37.5 4) = 3106311.1 p mm 4 2

Normal and Shear Stress: The normal stress is contributed by the axial stress only. Thus, s =

N - 15000N = - 3395.31 kPa = 4417.86 A

The shear stress is contributed by the torsional shear stress only. t =

Tc - 700(1000)(37.5) = = 84505.38 kPa J 3106311.1 1

The state of stress at a point on the outer surface of the shaft, Fig. b, can be represented by the element shown in Fig. c. In-Plane Principal Stress: sx = - 3395.31 kPa, sy = 0, and txy = - 84505.38 kPa . We obtain, s1,2 =

=

sx + sy ;

2

Aa

sx - sy 2

2 b + txy 2

2 - 3395.31 + 0 ; a - 3395.31 - 0 b + ( - 84505.38)2 2 A 2

s1 = 84324.78 kPa = 84.324 MPa s 2 = - 84720.1 kPa = - 84.72 kPa

Ans.

Maximum In-Plane Shear Stress: tmax

in-plane

=

A

a

sx - sy 2

b + txy2 = A a 2

- 3395.31 - 0 2 2 b + ( - 84505.38) = 84522.43 kPa 2

= 84.52 MPa

Ans.

Ans: s1 = 84.324 M < Pa, s2 = -84.72 MPa, t max = 84.52 MPa [email protected]

869

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9–31. Determine the principal stress at point A on the cross section of the arm at section a–a. Specify the orientation of this state of stress and indicate the results on an element at the point.

7.5 mm A 50 mm

7.5 mm

Support Reactions: Referring to the free-body diagram of the entire arm shown in Fig. a, ©MB = 0; FCD sin 30°(0.3) - 500(0.65) = 0

FCD = 2166.67 N

+ ©F = 0; : x

Bx - 2166.67 cos 30° = 0

Bx = 1876.39 N

+ c ©Fy = 0;

2166.67 sin 30° - 500 - By = 0

By = 583.33 N

20 mm

Section a – a D

Internal Loadings: Consider the equilibrium of the free-body diagram of the arm’s left segment, Fig. b. + ©F = 0; : x

1876.39 - N = 0

N = 1876.39 N

+ c ©Fy = 0;

V - 583.33 = 0

V = 583.33 N

+ ©MO = 0;

583.33(0.15) - M = 0

M = 87.5N # m

0.15 m

Referring to Fig. c, QA = y¿A¿ = 0.02125(0.0075)(0.02) = 3.1875 A 10 - 6 B m3 Normal and Shear Stress: The normal stress is a combination of axial and bending stress. Thus,

-1876.39

0.5625 A 10

-3

B

MyA N + A I 87.5(0.0175)

+

0.16367 A 10 - 6 B

= 6.020 MPa

The shear stress is caused by transverse shear stress.

tA

583.33 C 3.1875 A 10 - 6 B D VQA = = = 1.515 MPa It 0.16367 A 10 - 6 B (0.0075)

The state of stress at point A can be represented on the element shown in Fig. d. In-Plane Principal Stress: sx = 6.020 MPa, sy = 0, and txy = 1.515 MPa. We have s1,2 =

=

sx + sy 2

;

C

¢

sx - sy 2

2

≤ + txy 2

6.020 - 0 2 6.020 + 0 a ; b + 1.5152 2 C 2

s1 = 6.38 MPa

s2 = - 0.360 MPa

Ans

871

C 0.15 m

0.35 m 500 N

1 1 (0.02) A 0.053 B (0.0125) A 0.0353 B = 0.16367 A 10 - 6 B m4 12 12

=

a

a

A = 0.02(0.05) - 0.0125(0.035) = 0.5625 A 10 - 3 B m2

sA =

60⬚

B

Section Properties: The cross - sectional area and the moment of inertia about the z axis of the arm’s cross section are

I =

7.5 mm

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9–31. Continued Orientation of the Principal Plane: tan 2uP =

txy

A sx - sy B >2

=

1.515 = 0.5032 (6.020 - 0)>2

up = 13.36° and -76.64° Substituting u = 13.36° into sx¿ =

=

sx + sy 2

sx - sy +

2

cos 2u + txy sin 2u

6.020 + 0 6.020 - 0 + cos 26.71° + 1.515 sin 26.71° 2 2

= 6.38 MPa = s1 Thus, A uP B 1 = 13.4 and A uP B 2 = - 76.6°

Ans.

The state of principal stresses is represented by the element shown in Fig. e.

Ans: s1 = 6.38 MPa, s2 = -0.360 MPa, up1 = 13.4° and up2 = - 76.6° 872

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9–35. The square steel plate has a thickness of 10 mm and is subjected to the edge loading shown. Determine the maximum in-plane shear stress and the average normal stress developed in the steel. sx = 5 kPa

sy = - 5 kPa

tmax

sx - sy

in-plane

savg =

a

txy = 0

200 mm

b + t2xy

C

=

5 + 5 2 b + 0 = 5 kPa C 2 a

sx + sy 3

=

50 N/m

2

=

2

50 N/m

Ans. 200 mm

5 - 5 = 0 2

Ans.

Note: tan 2us =

tan 2us =

- (sx - sy)>2 txy -(5 + 5)>2 = q 0

us = 45°

Ans: tmax in-plane

879

= 5 kPa, savg = 0

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P

9–37. The shaft has a diameter d and is subjected to the loadings shown. Determine the principal stress and the maximum in-plane shear stress that is developed at point A. The bearings only support vertical reactions.

F

F A L 2

L 2

Support Reactions: As shown on FBD(a). Internal Forces and Moment: As shown on FBD(b). Section Properties: A =

p 2 d 4

p d 4 p 4 a b = d 4 2 64

I =

QA = 0

Normal Stress: N Mc ; A I

s =

-F p 2 ; 4 d

=

sA =

A d2 B

PL 4 p 64

d4

4 2PL - Fb a d pd2

Shear Stress: Since QA = 0, tA = 0 In-Plane Principal Stress: sx =

4 2PL a - Fb. pd2 d

sy = 0 and txy = 0 for point A. Since no shear stress acts on the element, s1 = sx =

4 2PL a - Fb 2 d pd

Ans.

s2 = sy = 0

Ans.

Maximum In-Plane Shear Stress: Applying Eq. 9-7 for point A, t

max in-plane

=

£

B

a

4 2 pd

sx - sy 2

2

b + t2xy

A 2PL d - FB - 0

=

Q

=

2PL 2 a - Fb d pd2

2

2

≥ + 0

Ans.

Ans: s1 =

4 2PL - Fb , s2 = 0, a d pd2

tmax in-plane

881

=

2PL 2 a - Fb d pd2

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9–38. A paper tube is formed by rolling a paper strip in a spiral and then gluing the edges together as shown. Determine the shear stress acting along the seam, which is at 30° from the vertical, when the tube is subjected to an axial force of 10 N. The paper is 1 mm thick and the tube has an outer diameter of 30 mm. s =

P = A

p 4

= -

10 N

10 N 30 mm

10 = 109.76 kPa (0.03 - 0.0282) 2

sx = 109.76 kPa tx¿y¿ = -

30⬚

sx - sy 2

sy = 0

txy = 0

u = 30°

sin 2u + txy cos 2u

109.76 - 0 sin 60° + 0 = - 47.5 kPa 2

Ans.

Ans: tx¿y¿ = -47.5 kPa 882

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1500 7.5 kNlb

pipe has hasan anouter outerdiameter diameterofof3 in., 75 mm, 9–42. The drill pipe a walla wall thickness of 6 in., mm,and andaa weight weight of 0.8 kN/m.IfIf itit is 50 lb>ft. thickness of 0.25 subjected to a torque and axial load as shown, determine (a) the principal stress and (b) the maximum in-plane shear stress at a point on its surface at section a.

800kNm lbft 1.2 620mft

aa 620mft

Internal Forces and Torque: As shown on FBD(a). Section Properties: A =

p 22 2.522)B==414p 0.6875p (75 mm2in2 A 3 -– 63 4

J =

p 6 4 1.254)B ==1.560(10 4.1172 in (37.5 ) mm4 A 1.54 4-– 31.5 2

7.5 kN

Normal Stress:

1.2 kN · m 0.8(6) = 4.8 kN

–12.3(10 N -2500 3) = = - 1157.5 –9.457 psi MPa A 0.6875p 414p

s =

Shear Stress: Applying the torsion formula. t =

6 800(12)(1.5) )(37.5) T c 1.2(10 = =3497.5 28.846psi MPa J 4.1172 6) 1.560(10

N = 12.3 kN T = 1.2 kN · m

- 1157.5MPa psi and = 28.846 3497.5MPa psi for a) In - Plane Principal Stresses: sx = 0, sy = –9.457 and ttxy for xy = any point on the shaft’s surface. Applying Eq. 9-5. s1,2 = =

sx + sy 2

;

C

a

sx - sy 2

2

b + t2xy

0 - ((–9.457) - 1157.5) 2 0 + (1157.5) (–9.457) (28.846)2 ; a b + (3497.5) 2 C 2

28.846 MPa 9.457 MPa

= 578.75 ±;29.2310 3545.08 –4.7285 MPa 2966 psi = 2.97 ksi s1 = 24.50

Ans.

–33.96 MPa= - 4.12 ksi s2 = 4124 psi

Ans.

b) Maximum In - Plane Shear Stress: Applying Eq. 9-7 t

max in-plane

= =

C

a

sx - sy 2

2

b + t2xy

0 - ((–9.457) - 1157.5) 2 (3497.5)2 ≤ + (28.846) C 2

¢

= 29.23 3545 psi = 3.55 ksi MPa

Ans.

Ans: s1 24.51 MPa, s2 33.96 MPa, tmax = 29.24 3.55 ksi MPa in-plane

886

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9–56. Determine the principal stress, the maximum in-plane shear stress, and average normal stress. Specify the orientation of the element in each case. 105 MPa 15 ksi 35 MPa 5 ksi

Construction of the Circle: In accordance with the sign convention, sx = 15 , 105ksi MPa, . Hence, sy = 0 and txy = –35 - 5 ksi MPa. Hence, sx + sy

savg =

2

=

105 15 ++ 00 = 52.5 7.50 MPa ksi 2

105 63.097

Ans.

35 (MPa)

The coordinates for reference point A and C are A(15, 5) A(105,-–35)

C(7.50, C(52.5,0)0) (MPa)

10.6 MPa

The radius of the circle is R = 2(15 7.50)22 ++3552 ==63.097 9.014 ksi (105 -– 52.5) = 63.10 MPa 115.6 MPa

a) 52.5 MPa

In - Plane Principal Stress: The coordinates of points B and D represent s1 and s2, respectively. MPa 7.50 ++ 63.097 9.014 == 115.60 16.5 ksi s1 = 52.5

63.1 MPa

Ans.

52.5 –-63.097 s2 = 7.50 9.014 == –10.60 - 1.51 MPa ksi

Ans.

Orientation of Principal Plane: From the circle tan 2uP1 =

35 5 = 0.6667 15 -– 52.5 7.50 105

uP1 = 16.8° (Clockwise)

Ans.

b) Maximum In - Plane Shear Stress: Represented by the coordinates of point E on the circle. tmax

in-plane

–63.1 = -R = 9.01 MPa ksi

Ans.

Orientation of the Plane for Maximum In - Plane Shear Stress: From the circle tan 2us =

52.5 MPa

15 -– 52.5 7.50 105 = 1.500 5 35

us = 28.2° (Counterclockwise)

Ans.

900

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9–62. The grains of wood in the board make an angle of 20° with the horizontal as shown. Using Mohr’s circle, determine the normal and shear stresses that act perpendicular and parallel to the grains if the board is subjected to an axial load of 250 N.

sx =

300 mm

60 mm 250 N

250 N 20⬚

25 mm

250 P = = 166.67 kPa A (0.06)(0.025)

R = 83.33 Coordinates of point B: sx¿ = 83.33 - 83.33 cos 40° sx¿ = 19.5 kPa

Ans.

tx¿y¿ = - 83.33 sin 40° = - 53.6 kPa

Ans.

Ans: sx¿ = 19.5 kPa, tx¿y¿ = - 53.6 kPa 906

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9–66. Determine the principal stress and maximum in-plane shear stress at point A on the cross section of the pipe at section a–a.

a 300 mm a 300 mm b

30 mm

A B

200 mm

300 N

Internal Loadings: Considering the equilibrium of the free-body diagram of the assembly’s segment, Fig. a, ©Fx = 0; N - 450 = 0

N = 450 N

©Fy = 0; Vy = 0 ©Fz = 0; Vz + 300 = 0

Vz = - 300 N

©Mx = 0; T + 300(0.2) = 0

T = - 60 N # m

©My = 0; My - 450(0.3) + 300(0.3) = 0

My = 45 N # m

©Mz = 0; Mz + 450(0.2) = 0

Mz = - 90 N # m

Section Properties: The cross-sectional area, the moment of inertia about the y and z axes, and the polar moment of inertia of the pipe’s cross section are A = p(0.032 - 0.022) = 0.5p(10-3) m2 Iy = Iz = J =

p (0.034 - 0.024) = 0.1625p(10-6) m4 4

p (0.034 - 0.024) = 0.325p(10-6) m4 2

Referring to Fig. b, (Qy)A = 0 (Qz)A =

4(0.03) p 4(0.02) p c (0.03)2 d c (0.022) d = 12.667(10-6) m3 3p 2 3p 2

Normal and Shear Stress: The normal stress is a combination of axial and bending stress. sA =

MzyA ( - 90)(- 0.03) N 450 = -3 A Iz 0.5p(10 ) 0.1625p(10-6)

= - 5.002 MPa Since Vy = 0, [(txy)V]A = 0. However, the shear stress is the combination of torsional and transverse shear stress. Thus, (txz)A = [(txz)T]A - [(txz)V]A

=

Vz(Qz)A 60(0.03) 300[12.667(10-6)] Tc = 1.391 MPa = -6 J Iy t 0.325p(10 ) 0.1625p(10-6)(0.02)

The state of stress at point A is represented by the element shown in Fig. c.

911

450 N

20 mm Section a – a

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9–66. Continued Construction of the Circle: sx = - 5.002 MPa, sz = 0, and txz = 1.391 MPa. Thus, savg =

sx + sz 2

=

- 5.002 + 0 = - 2.501 MPa 2

The coordinates of reference point A and the center C of the circle are A(- 5.002, 1.391)

C(- 2.501, 0)

Thus, the radius of the circle is R = CA = 2[- 5.002 - ( -2.501)]2 + 1.3912 = 2.862 MPa Using these results, the circle is shown in Fig. d. In-Plane Principal Stresses: The coordinates of reference points B and D represent s1 and s2, respectively. s1 = - 2.501 + 2.862 = 0.361 MPa

Ans.

s2 = - 2.501 - 2.862 = - 5.36 MPa

Ans.

In-Plane Maximum Shear Stress: The coordinates of point E represent the state of maximum shear stress. Thus, tmax

= |R| = 2.86 MPa

Ans.

in-plane

Ans: s1 = 0.361 MPa , s2 = - 5.36 MPa, tmax = 2.86 MPa in-plane

912

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*9–68. The rotor shaft of the helicopter is subjected to the tensile force and torque shown when the rotor blades provide the lifting force to suspend the helicopter at midair. If the shaft has a diameter of 150 mm, determine the principal stress and maximum in-plane shear stress at a point located on the surface of the shaft.

225 kN 15 kN⭈m

Internal Loadings: Considering the equilibrium of the free-body diagram of the rotor shaft’s upper segment, Fig. a, ©Fy = 0;

N - 225000 = 0

N = 225000 N

©My = 0;

T - 15 = 0

T = 15 kN.m

Section Properties: The cross-sectional area and the polar moment of inertia of the rotor shaft’s cross section are A = p(752) = 17671.45 mm 2 J =

p (75 4 ) = 49700977.53 mm4 2

Normal and Shear Stress: The normal stress is contributed by axial stress only. sA =

N 225000 = = 12.73 MPa A 17671.45 .45

The shear stress is contributed by the torsional shear stress only. tA =

15(1000 × 1000)(75) Tc = = 22.63 MPa J 49700977.53

The state of stress at point A is represented by the element shown in Fig. b.

915

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9–68. Continued Construction of the Circle: sx = 0, sy = 12.73 MPa, and txy = 22.63 MPa. Thus, savg =

sx + sy 2

=

0 + 12.73 = 6.365 MPa 2

The coordinates of reference point A and the center C of the circle are A(0, 22.63)

C(6.365, 0)

Thus, the radius of the circle is R = CA = 2(0 - 6.365)2 + 22.632 = 23.51 MPa Using these results, the circle is shown in Fig. c. In-Plane Principal Stress: The coordinates of reference points B and D represent s1 and s2, respectively. s1 = 6.365 + 23.51 = 29.87 MPa

Ans.

s2 = 6.365 - 23.51 = - 17.16 MPa

Ans.

Maximum In-Plane Shear Stress: The state of maximum shear stress is represented by the coordinates of point E, Fig. a. tmax

= R = 23.51 MPa

Ans.

in-plane

916

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*9–69. The pedal crank for a bicycle has the cross section shown. If it is fixed to the gear at B and does not rotate while subjected to aa force force of of 400 75 lb, N, determine the principal stress in the material on the cross section at point C.

400 N 75 75 lb lb B B

A A 100 mm 4 in.

753mm in.

C C

0.2 in. 5 mm 0.3mm in. 7.5

Internal Forces and Moment: As shown on FBD Section Properties:

0.4mm in. 10 0.4mm in. 10

400 N 100 mm

1 4 3 I = (0.3) A 0.833) B==5000 0.0128 (7.5)(20 mmin 12

M = 40000 N · mm

7.5(5)(7.5) = 281.25 mm QC = y¿A¿ = 0.3(0.2)(0.3) = 0.0180 in33

V = 400 N 7.5 mm

Normal Stress: Applying the flexure formula.

5 mm

7.5 mm

5 mm

My - 300(0.2) –40000(5) 40 MPa psi = 4.6875 ksi sC = = = 4687.5 5000 I 0.0128

10 mm

Shear Stress: Applying the shear formula.

3 MPa

VQC 75.0(0.0180) 400(281.25) tC = = =3351.6 MPa psi = 0.3516 ksi It 0.0128(0.3) 5000(7.5)

40 MPa

Construction of the Circle: In accordance with the sign convention, sx = 40 4.6875 ksi, MPa, . Hence, sy = 0, and txy = 0.3516 3 MPa.ksi Hence, savg =

sx + sy 2

=

20

40 +0 + 0 4.6875 = 2.34375 ksi 20 MPa 2 2

3 (MPa)

The coordinates for reference points A and C are A(40, 3) 0.3516) C(20, 0) C(2.34375, 0) A(4.6875, The radius of the circle is

40

2 (40 – 20)-2 +2.34375) 32 = 20.224 MPa 2 = 2.3670 ksi R = 2(4.6875 + 0.3516

(MPa)

In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2, respectively. 2.34375 + 2.3670 = MPa 4.71 ksi s1 = 20 + 20.224 = 40.223

Ans.

– 20.224 = –0.224 s2 = 20 2.34375 - 2.3670 = MPa - 0.0262 ksi

Ans.

Ans: s1 = 40.223 MPa, s2 = - 0.224 MPa 917

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•9–70.

A spherical pressure vessel has an inner radius of 51.5 ftm and in. Draw anda awall wallthickness thicknessof of0.5 12 mm. DrawMohr’s Mohr’scircle circlefor for the state of stress at a point on the vessel and explain the significance of the result. The vessel is subjected to an 0.56 MPa. internal pressure of 80 psi.

Normal Stress: s1 = s2 =

80(5)(12) pr 0.56(1.5)(1000) = = 4.80ksi 35 MPa 2t 2(0.5) 2(12)

Mohr’s circle: A(35, 0)0) A(4.80,

B(35, 0) 0) C(35, 0) B(4.80, C(4.80, 0)

Regardless of the orientation of the element, the shear stress is zero and the state of stress is represented by the same two normal stress components.

35 MPa

35 MPa

(35, 0)

Ans: s1 s2 35 MPa

918

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z

9–83. The state of stress at a point is shown on the element. Determine the principal stress and the absolute maximum shear stress. x

For y - z plane: A(5, - 4)

B( - 2.5, 4)

y

2.5 kPa ksi

C(1.25, 0) 4 kPa ksi

R = 23.752 + 42 = 5.483 ksi s1 = 1.25 + 5.483 = 6.733 kPa

5 kPa ksi

s2 = 1.25 - 5.483 = - 4.233 kPa ksi Thus,

savg = tabs

max

=

ksi s1 = 6.73 kPa

Ans.

s2 = 0

Ans.

s3 = - 4.23 ksi kPa

Ans.

6.73 + (- 4.23) = 1.25 kPa ksi 2 6.73 - (- 4.23) smax - smin kPa = = 5.48 ksi 2 2

Ans. 25 kPa 4 kPa 5 kPa

423 kPa

673 kPa

Ans: s1 = 6.73 kPa, s2 = 0, s3 = - 4.23 kPa, tabs = 5.48 kPa, max

935

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9–86. The plate is subjected to a tensile force P = 5 kN. If it has the dimensions shown, determine the principal stresses and the absolute maximum shear stress. If the material is ductile it will fail in shear. Make a sketch of the plate showing how this failure would appear. If the material is brittle the plate will fail due to the principal stresses. Show how this failure occurs.

s =

P 25 kN

50 mm P 25 kN 50 mm 300 mm

12 mm

P 25000N = = 20833.33 kPa = 20.833 MPa A (100)(12) s1 = 20.833 MPa

Ans.

s2 = s3 = 0

Ans.

tabs = max

s1 = 10.41 MPa 2

Ans.

Failure by shear:

Failure by principal stress:

Ans: s1 = 20.833 MPa, s2 = s3 = 0, tabs = 10.41 MPa max

937

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300 mm

9–87. Determine the principal stresses and absolute maximum shear stress developed at point A on the cross section of the bracket at section a–a. 150 mm a

5

3 4

a 12 mm B

2.5 kN

6 mm A

6 mm

6 mm 36 mm 36 mm

Section a – a

Ans: smax 5.06 MPa, sint 0, smin 8.04 MPa, tabs = 6.55 755 psi MPa m ax

938