196634050 Mechanics of Materials

CHAPTER 7

PROBLEM 7.2 For the given state of stress, determine the normal and shearing stresses exerted on the oblique face of the shaded triangular element shown. Use a method of analysis based on the equilibrium of that element, as was done in the derivations of Sec. 7.2.

SOLUTION

Stresses

Areas

Forces

ΣF = 0:

σ A − 80 A cos 55° cos55° + 40 A sin 55° sin 55° = 0 σ = 80 cos 2 55° − 40sin 2 55°

σ = −0.521 MPa 

ΣF = 0: 



τ A − 80 A cos 55° sin 55° − 40 A sin 55° cos 55°  τ = 120 cos 55° sin 55°

τ = 56.4 MPa 



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

SOLUTION

σ x = −60 MPa σ y = −40 MPa τ xy = 35 MPa (a)

tan 2θ p =

2τ xy

σx −σ y

=

(2) (35) = −3.50 −60 + 40

2θ p = −74.05°

(b)

σ max, min =

σx +σy 2

θ p = −37.0°, 53.0°  2

σx −σ y  2 ±   + τ xy 2   2

−60 − 40  −60 + 40  2 ±   + (35) 2 2   = −50 ± 36.4 MPa =

σ max = −13.60 MPa  σ min = −86.4 MPa 

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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

σ x = −8 ksi σ y = 12 ksi τ xy = 5 ksi (a)

tan 2θ s = −

σx − σy −8 − 12 =− = + 2.0 2τ xy 2(5)

2θ s = 63.435°

θ s = 31.7°, 121.7°  2

(b)

σx − σy  2  + τ xy  2  

τ max = 

2

 −8 − 12  2   + (5)  2  



=

(c)

σ ′ = σ ave = =

τ max = 11.18 ksi 

σx + σy 2 −8 + 12 2

σ ′ = 2.00 ksi 



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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

σ x = 0 σ y = −80 MPa τ xy = −50 MPa σx +σy 2

= −40 MPa

σ x′ =

σx +σy 2

τ x′y′ = − σ y′ = (a)

+

σx −σ y 2

σx +σy 2

−

σx −σ y 2 σx −σ y 2

cos 2θ + τ xy sin 2θ

sin 2θ + τ xy cos 2θ

σx −σ y 2

cos 2θ − τ xy sin 2θ

θ = −25° 2θ = −50° σ x′ = −40 + 40 cos ( −50°) − 50 sin (−50°) τ x′y′ = −40 sin ( −50°) − 50 cos ( −50°) σ y′ = −40 − 40 cos (−50°) + 50 sin (−50°)

(b)

= 40 MPa

σ x′ = 24.0 MPa  τ x′y′ = −1.5 MPa  σ y′ = −104.0 MPa 

θ = 10° 2θ = 20° σ x′ = −40 + 40 cos (20°) − 50 sin (20°) τ x′y′ = −40 sin (20°) − 50 cos (20°) σ y′ = −40 − 40 cos (20°) + 50 sin (20°)

σ x′ = −19.5 MPa  τ x′y′ = −60.7 MPa  σ y′ = −60.5 MPa 

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PROBLEM 7.20 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 σ = 800 kPa and τ = 600 kPa, determine the largest centric load P that can be applied.

SOLUTION For plane a-a, θ = 65°.

σ x = 0, τ xy = 0, σ y =

P A

σ = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ = 0 +

P 2 sin 65° + 0 A

Aσ (50 × 10−3 )(80 × 10−3 )(800 × 103 ) = = 3.90 × 103 N sin 2 65° sin 2 65° P τ = −(σ x − σ y ) sin θ cos θ + τ xy (cos2 θ − sin 2 θ ) = sin 65° cos 65° + 0 A Aτ (50 × 10−3 )(80 × 10−3 )(600 × 103 ) P= = = 6.27 × 103 N sin 65° cos 65° sin 65° cos 65° P=

Allowable value of P is the smaller one.

P = 3.90 kN 

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PROBLEM 7.26 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 c=

Torsion:

τ =

Bending:

I =

1 1 d = (32) = 16 mm = 16 × 10−3 m 2 2

Tc 2T 2(350 N ⋅ m) = = = 54.399 × 106 Pa = 54.399 MPa 3 3 −3 J πc π (16 × 10 m)

π 4

c4 =

π 4

(16 × 10−3 )4 = 51.472 × 10−9 m 4

M = (0.15 m)(3 × 103 N) = 450 N ⋅ m

σ =−

My (450)(16 × 10−3 ) =− = −139.882 × 106 Pa = −139.882 MPa I 51.472 × 10−9

Top view:

Stresses:

σ x = −139.882 MPa σ ave =

σy = 0

τ xy = −54.399 MPa

1 1 (σ x + σ y ) = (−139.882 + 0) = −69.941 MPa 2 2 2

R=

(a)

σx − σy  2   + τ xy = 2  

(−69.941) 2 + (−54.399) 2 = 88.606 MPa

σ max = σ ave + R = −69.941 + 88.606

σ max = 18.67 MPa 

σ min = σ ave − R = −69.941 − 88.606

σ min = −158.5 MPa 

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PROBLEM 7.26 (Continued)

tan 2θ p =

2τ xy

σx − σy

=

(2)(−54.399) = 0.77778 −139.882

2θ p = 37.88°

θ p = 18.9° and 108.9° 

(b)

τ max = R = 88.6 MPa

τ max = 88.6 MPa 

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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 inplane shearing stress, (c) the corresponding normal stress.

SOLUTION

σ x = 4 ksi, σ y = −12 ksi, τ xy = −15 ksi σ ave =

σx +σy 2

= −4 ksi

Plotted points for Mohr’s circle: X : (σ x , − τ xy ) = (4 ksi, 15 ksi) Y : (σ y , τ xy ) = (−12 ksi, − 15 ksi) C : (σ ave , 0) = ( −4 ksi, 0)

(a)

tan α =

FX 15 = = 1.875 CF 8

α = 61.93° 1 2 β = 180° − α = 118.07° 1 θb = β = 59.04° 2

θ a = − α = −30.96°

θ a = −31.0° 

θb = 59.0° 

R = (CF ) 2 + ( FX ) 2 = (8)2 + (15) 2 = 17 ksi

(b)

(a′)

σ a = σ max = σ ave + R = −4 + 17

σ max = 13.00 ksi 

σ min = σ min = σ ave − R = −4 − 17

σ min = −21.0 ksi 

θ d = θ a + 45° = 14.04°

θ d = 14.0° 

θe = θb + 45° = 104.04°

θe = 104.0° 

τ max = R (b′)

σ ′ = σ ave

τ max = 17.00 ksi  σ ′ = −4.00 ksi 

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PROBLEM 7.41 Solve Prob. 7.19, using Mohr’s circle. PROBLEM 7.19 A steel pipe of 12-in. outer diameter is fabricated from 1 -in.-thick plate by welding along a helix which forms an angle of 22.5° with 4 a plane perpendicular to the axis of the pipe. Knowing that a 40-kip axial force P and an 80-kip ⋅ in. torque T, each directed as shown, are applied to the pipe, determine σ and τ in directions, respectively, normal and tangential to the weld.

SOLUTION 1 d 2 = 6 in., t = 0.25 in. 2 c1 = c2 − t = 5.75 in.

d 2 = 12 in. c2 =

( π J = (c 2

) π − c ) = (6 2

A = π c22 − c12 = π (62 − 5.752 ) = 9.2284 in 2 4 2

4 1

4

− 5.754 ) = 318.67 in 4

Stresses: P 40 =− = −4.3344 ksi A 9.2284 Tc (80)(6) τ =− 2 =− = 1.5063 ksi J 318.67 σ x = 0, σ y = −4.3344 ksi, τ xy = 1.5063 ksi

σ =−

Draw the Mohr’s circle. X : (0, −1.5063 ksi) Y : ( − 4.3344 ksi, 1.5063 ksi) C : (−2.1672 ksi, 0) 1.5063 = 0.69504 α = 34.8° 2.1672 β = (2)(22.5°) − α = 10.8°

tan α =

R = (2.1672) 2 + (1.5063) 2 = 2.6393 ksi

σ w = −2.1672 − 2.6393 cos 10.8° τ w = −2.6393sin10.2°

σ w = −4.76 ksi  τ w = −0.467 ksi 

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PROBLEM 7.44 Solve Prob. 7.22, using Mohr’s circle. PROBLEM 7.22 Two steel plates of uniform cross section 10 × 80 mm are welded together as shown. Knowing that centric 100-kN forces are applied to the welded plates and that the in-plane shearing stress parallel to the weld is 30 MPa, determine (a) the angle β, (b) the corresponding normal stress perpendicular to the weld.

SOLUTION

σx =

P 100 × 103 = = 125 × 106 Pa = 125 MPa A (10 × 10−3 )(80 × 10−3 )

σy = 0

τ xy = 0

From Mohr’s circle: 30 = 0.48 62.5

(a)

sin 2β =

(b)

σ = 62.5 + 62.5cos 2β

β = 14.3° 

σ = 117.3 MPa 

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PROBLEM 7.45 Solve Prob. 7.23, using Mohr’s circle. PROBLEM 7.23 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: V = 400 lb

M = (400)(6) = 2400 lb ⋅ in

T = (400)(2) = 800 lb ⋅ in

Shaft cross section

d = 1 in. J =

π 2

c=

1 d = 0.5 in. 2

c 4 = 0.098175 in 4

1 J = 0.049087 in 4 2

Tc (800)(0.5) = = 4.074 × 103 psi = 4.074 ksi 0.098175 J

Torsion:

τ =

Bending:

σ =

Transverse shear:

Stress at point H is zero.

Resultant stresses:

σ x = 24.446 ksi, σ ave =

I =

Mc (2400)(0.5) = = 24.446 × 103 psi = 24.446 ksi 0.049087 I

σ y = 0,

τ xy = 4.074 ksi

1 (σ x + σ y ) = 12.223 ksi 2 2

R=

=



σx − σy  2   + τ xy 2  

(12.223) 2 + (4.074) 2 = 12.884 ksi

σ a = σ ave + R

σ a = 25.107 ksi 

σ b = σ ave − R

σ b = −0.661 ksi 

τ max = R 

τ max = 12.88 ksi 



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PROBLEM 7.47 Solve Prob. 7.25, using Mohr’s circle. PROBLEM 7.25 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 do 102 = = 51 mm 2 2

ro = J = I =

π 2

(r

4 o

ri = ro − t = 45 mm

)

− ri4 = 4.1855 × 106 mm 4 = 4.1855 × 10−6 m 4

1 J = 2.0927 × 10−6 m 4 2

Force-couple system at center of tube in the plane containing points H and K: Fx = 10 × 103 N M y = (10 × 103 )(200 × 10−3 ) = 2000 N ⋅ m M z = −(10 × 103 )(150 × 10−3 ) = −1500 N ⋅ m

Torsion:

T = M y = 2000 N ⋅ m c = ro = 51 × 10−3 m

τ xy =

Tc (2000)(51 × 10−3 ) = = 24.37 MPa J 4.1855 × 10−6

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

Stress due to V = Fx is zero at point K.

Bending:

σy =

Mz c (1500)(51 × 10−3 ) = = 36.56 MPa I 2.0927 × 10−6

Point K lies on compression side of neutral axis.

σ y = −36.56 MPa

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PROBLEM 7.47 (Continued)

Total stresses at point K:

σ x = 0,

σ y = −36.56 MPa,

σ ave =

1 (σ x + σ y ) = −18.28 MPa 2

τ xy = 24.37 MPa

2

R=

σx − σy  2   + τ xy = 30.46 MPa 2  

σ max = σ ave + R = −18.28 + 30.46 σ max = 12.18 MPa  

σ min = σ ave − R = −18.28 − 30.46 σ min = −48.74 MPa  τ max = R

τ max = 30.46 MPa 



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PROBLEM 7.61 For the element shown, determine the range of values of τ xy for which the maximum tensile stress is equal to or less than 60 MPa.

SOLUTION

σ x = −20 MPa σ y = −120 MPa 1 2

σ ave = (σ x + σ y ) = −70 MPa Set

σ max = 60 MPa = σ ave + R R = σ max − σ ave = 130 MPa

But 2

σ −σx  2 + τ xy R=  x  2   σx −σx   2  

2

τ xy = R 2 − 

= 1302 − 502 = 120 MPa Range of τ xy :

−120 MPa ≤ τ xy ≤ 120 MPa 

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PROBLEM 7.66 For the state of plane stress shown, determine the maximum shearing stress when (a) σ x = 6 ksi and σ y = 18 ksi , (b) σ x = 14 ksi and σ y = 2 ksi . (Hint: Consider both in-plane and out-of-plane shearing stresses.)

SOLUTION (a)

σ x = 6 ksi, σ y = 18 ksi, τ xy = 8 ksi σ ave =

1 (σ x + σ y ) = 12 ksi 2 2

σx − σy  2   + τ xy 2  

R=

= 62 + 82 = 10 ksi σ a = σ ave + R = 12 + 10 = 22 ksi (max)

σ b = σ ave − R = 12 − 10 = 2 ksi σc = 0 τ max(in-plane) = R = 10 ksi

1 (σ max − σ min ) 2 σ x = 14 ksi, σ y = 2 ksi, τ xy = 8 ksi

τ max =

(b)

(min)

σa =

1 (σ x + σ y ) = 8 ksi 2

R=

σx − σy  2   + τ xy 2  

=

62 + 82 = 10 ksi

τ max = 11 ksi 

2

σ a = σ ave + R = 18 ksi (max) σ b = σ ave − R = −2 ksi (min) σc = 0 σ max = 18 ksi σ min = −2 ksi τ max =

1 (σ max − σ min ) 2

τ max = 10 ksi 

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PROBLEM 7.72 For the state of stress shown, determine the maximum shearing stress when (a) σ z = 0, (b) σ z = +45 MPa, (c) σ z = − 45 MPa.

SOLUTION

σ x = 100 MPa,

σ y = 20 MPa,

τ xy = 75 MPa

1 2

σ ave = (σ x + σ y ) = 60 MPa 2

σx −σy  2 R=   + τ xy 2   = 402 + 752 = 85 MPa σ a = σ ave + R = 145 MPa

σ b = σ ave − R = −25 MPa (a)

σ z = 0, σ a = 145 MPa, σ b = −25 MPa 1 2

σ max = 145 MPa, σ min = −25 MPa, τ max = (σ max − σ min ) (b)

σ z = +45 MPa, σ a = 145 MPa, σ b = −25 MPa 1 2

σ max = 145 MPa, σ min = −25 MPa, τ max = (σ max − σ min ) (c)

τ max = 85 MPa 

τ max = 85 MPa 

σ z = −45 MPa, σ a = 145 MPa, σ b = −25 MPa σ max = 145 MPa,

1 2

σ min = −45 MPa, τ max = (σ max − σ min )

τ max = 95 MPa 

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PROBLEM 7.83 The state of plane stress shown occurs in a machine component made of a steel with σ Y = 45 ksi. Using the maximum-distortion-energy criterion, determine whether yield will occur when (a) τ xy = 9 ksi, (b) τ xy = 18 ksi, (c) τ xy = 20 ksi. If yield does not occur, determine the corresponding factor of safety.

SOLUTION

σ x = 36 ksi, σ y = 21 ksi, σ z = 0 For stresses in xy-plane, σ ave =

σx − σy 2

1 (σ x + σ y ) = 28.5 ksi 2

= 7.5 ksi

2

(a)

τ xy = 9 ksi

σx − σy  2   + τ xy = 2  

R=

(7.5) 2 + (9)2 = 11.715 ksi

σ a = σ ave + R = 40.215 ksi, σ b = σ ave − R = 16.875 ksi σ a2 + σ b2 − σ a σ b = 34.977 ksi < 45 ksi F .S . =

(No yielding)

45 39.977

F .S . = 1.287  2

(b)

σx − σy  2  + τ xy = 2  

τ xy = 18 ksi R = 

(7.5) 2 + (18)2 = 19.5 ksi

σ a = σ ave + R = 48 ksi, σ b = σ ave − R = 9 ksi σ a2 + σ b2 − σ a σ b = 44.193 ksi < 45 ksi F .S . =

(No yielding)

45 44.193

F .S . = 1.018  2

(c)

τ xy = 20 ksi

R=

σx − σy  2   + τ xy = 2  

(7.5)2 + (20) 2 = 21.36 ksi

σ a = σ ave + R = 49.86 ksi, σ b = σ ave − R = 7.14 ksi σ a2 + σ b2 − σ a σ b = 46.732 ksi > 45 ksi

(Yielding occurs) 

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PROBLEM 7.84 Solve Prob. 7.83, using the maximum-shearing-stress criterion. PROBLEM 7.83 The state of plane stress shown occurs in a machine component made of a steel with σ Y = 45 ksi. Using the maximum-distortion-energy criterion, determine whether yield will occur when (a) τ xy = 9 ksi, (b) τ xy = 18 ksi, (c) τ xy = 20 ksi. If yield does not occur, determine the corresponding factor of safety.

SOLUTION

σ x = 36 ksi, σ y = 21 ksi, σ z = 0 σ ave =

For stress in xy-plane,

1 (σ x + σ y ) = 28.5 ksi 2

σx − σy 2

= 7.5 ksi

2

(a)

τ xy = 9 ksi

σx − σy  2   + τ xy = 11.715 ksi 2  

R=

σ a = σ ave + R = 40.215 ksi, σ b = σ ave − R = 16.875 ksi τ max = 34.977 ksi, σ min = 0

2τ max = σ max − σ min = 40.215 ksi < 45 ksi F .S . =

45 40.215

(No yielding) F .S . = 1.119 

2

(b)

σx − σy  2  + τ xy = 19.5 ksi 2  

τ xy = 18 ksi R = 

σ a = σ ave + R = 48 ksi, σ b = σ ave − R = 9 ksi σ max = 48 ksi σ min = 0 2τ max = σ max − σ min = 48 ksi > 45 ksi

(Yielding occurs) 

2

(c)

τ xy = 20 ksi R =

σx − σy  2   + τ xy = 21.36 ksi 2  

σ a = σ ave + R = 49.86 ksi σ b = σ ave − R = 7.14 ksi τ max = 49.86 ksi σ min = 0

2τ max = σ max − σ min = 49.86 ksi > 45 ksi

(Yielding occurs) 

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PROBLEM 7.85 The 36-mm-diameter shaft is made of a grade of steel with a 250-MPa tensile yield stress. Using the maximum-shearing-stress criterion, determine the magnitude of the torque T for which yield occurs when P = 200 kN.

SOLUTION P = 200 kN = 200 × 103 N

c=

1 d = 18 mm = 18 × 10−3 m 2

A = π c 2 = π (18 × 10−3 ) 2 = 1.01788 × 10−3 m 2 P 200 × 103 =− = 196.488 × 106 Pa A 1.01788 × 10−3 = 196.488 MPa

σy = −

σx = 0

σ ave =

1 1 (σ x + σ y ) = σ y = 98.244 MPa 2 2 2

R=

σx − σy  2   + τ xy = 2  

2 (98.244)2 + τ xy

σ a = σ ave + R (positive) σ b = σ ave − R (negative) σ a − σ b = 2R

σa − σb > σa

σa − σb > σb

Maximum shear stress criterion under the above conditions:

σ a − σ b = 2R = σ Y = 250 MPa

R = 125 MPa

Equating expressions for R, 125 =

2 (98.244)2 + τ xy

τ xy = (125) 2 − (98.244) 2 = 77.286 MPa = 77.286 × 106 Pa Torsion:

J =

π

c4 =

π

(18 × 10−3 ) 4 = 164.896 × 10−9 m 4

2 2 Tc τ xy = J Jτ xy (164.846 × 10−9 )(77.286 × 106 ) = T = c 18 × 10−3 = 708 N ⋅ m

T = 708 N ⋅ m 

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PROBLEM 7.86 Solve Prob. 7.85, using the maximum-distortion-energy criterion. PROBLEM 7.85 The 36-mm-diameter shaft is made of a grade of steel with a 250-MPa tensile yield stress. Using the maximum-shearing-stress criterion, determine the magnitude of the torque T for which yield occurs when P = 200 kN.

SOLUTION 1 d = 18 mm = 18 × 10−3 m 2 A = π c 2 = π (18 × 10−3 ) 2 = 1.01788 × 10−3 m 2

P = 200 kN = 200 × 103 N

c=

P 200 × 103 =− = 196.488 × 106 Pa −3 A 1.01788 × 10 = 196.448 MPa

σy = −

1 2

σx = 0

1 2

σ ave = (σ x + σ y ) = σ y = 98.244 MPa

σx −σ y R=  2  σ a = σ ave + R

2

 2 2 2  + τ xy = (98.244) + τ xy  σ b = σ ave − R

Distortion energy criterion:

σ a2 + σ b2 − σ aσ b = σ Y2 (σ ave + R )2 + (σ ave − R )2 − (σ ave + R )(σ ave − R ) = σ Y2 2 σ ave + 3R 2 = σ Y−2 2 (98.244) 2 + (3)[(98.244)2 + τ xy ] = (250) 2

τ xy = ±89.242 MPa Torsion:

J=

τ xy

π

c4 =

2 Tc = J

Jτ xy

π 2

(18 × 10−3 ) 4 = 164.846 × 10−9 m 4

(164.846 × 10−9 )(89.242 × 106 ) c 18 × 10−3 = 818 N ⋅ m

T=

=

T = 818 N ⋅ m 

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PROBLEM 7.98 A spherical gas container made of steel has a 5-m outer diameter and a wall thickness of 6 mm. Knowing that the internal pressure is 350 kPa, determine the maximum normal stress and the maximum shearing stress in the container.

SOLUTION d = 5 m t = 6 mm = 0.006 m, r =

σ =

d − t = 2.494 m 2

pr (350 × 103 Pa)(2.494 m) = = 72.742 × 106 Pa 2t 2(0.006 m)

σ = 72.7 MPa  σ max = 72.742 MPa σ min ≈ 0 (Neglecting small radialstress) τ max =

1 (σ max − σ min ) 2

τ max = 36.4 MPa 

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PROBLEM 7.104 A steel penstock has a 750-mm outer diameter, a 12-mm wall thickness, and connects a reservoir at A with a generating station at B. Knowing that the density of water is 1000 kg/m3 , determine the maximum normal stress and the maximum shearing stress in the penstock under static conditions.

SOLUTION 1 1 d − t = (750) − 12 = 363 mm = 363 × 10−3 m 2 2 t = 12 mm = 12 × 10−3 m

r=

p = ρ gh = (1000 kg/m3 )(9.81 m/s 2 )(300 m) = 2.943 × 106 Pa

σ1 =

pr (2.943 × 106 )(363 × 10−3 ) = = 89.0 × 106 Pa t 12 × 10−3

σ max = σ 1

σ max = 89.0 MPa 

σ min = − p ≈ 0 1 2

τ max = (σ max − σ min )

τ max = 44.5 MPa 

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PROBLEM 7.114 For the tank of Prob. 7.112, determine the range of values of β that can be used if the shearing stress parallel to the weld is not to exceed 12 MPa when the gage pressure is 600 kPa. PROBLEM 7.112 The pressure tank shown has a 8-mm wall thickness and butt-welded seams forming an angle β = 20° with a transverse plane. For a gage pressure of 600 kPa, determine (a) the normal stress perpendicular to the weld, (b) the shearing stress parallel to the weld.

SOLUTION d = 1.6 m t = 8 × 10−3 mm r =

1 d − t = 0.792 m 2

pr (600 × 103 )(0.792) = t 8 × 10−3 = 59.4 × 106 Pa = 59.4 MPa pr σ2 = = 29.7 MPa 2t σ −σ2 R= 1 = 14.85 MPa 2 τ w = R |sin 2 β |

σ1 =

|sin 2β a | =

τN R

=

12 = 0.80808 14.85

2 β a = −53.91°

β a = +27.0°

2 βb = + 53.91°

βb = 27.0°

2 β c = 180° − 53.91° = +126.09°

β c = 63.0°

2 β d = 180° + 53.91° = + 233.91°

β d = 117.0°

Let the total range of values for β be Safe ranges for β: 

−180° < β ≤ 180° −22.0°≤ β ≤ 27.0° 

and 63.0°≤ β ≤ 117.0° 

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PROBLEM 7.116 The pressurized tank shown was fabricated by welding strips of plate along a helix forming an angle β with a transverse plane. Determine the largest value of β that can be used if the normal stress perpendicular to the weld is not to be larger than 85 percent of the maximum stress in the tank.

SOLUTION

σ1 =

pr t

σ2 =

pr 2t

1 2

3 pr 4 t σ1 − σ 2 1 pr = R= 2 4 t σ w = σ ave − R cos 2β

σ ave = (σ1 + σ 2 ) =

0.85

pr  3 1  pr =  − cos 2β  t 4 4  t

3  cos 2 β = −4  0.85 −  = −0.4 4  2 β = 113.6°

β = 56.8° 

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PROBLEM 7.124 A pressure vessel of 10-in. inner diameter and 0.25-in. wall thickness is fabricated from a 4-ft section of spirally-welded pipe AB and is equipped with two rigid end plates. The gage pressure inside the vessel is 300 psi and 10-kip centric axial forces P and P′ are applied to the end plates. Determine (a) the normal stress perpendicular to the weld, (b) the shearing stress parallel to the weld.

SOLUTION 1 1 d = (10) = 5 in. t = 0.25 in. 2 2 pr (300) (5) = = 6000 psi = 6 ksi σ1 = t 0.25 pr (300)(5) = = 3000 psi = 3 ksi σ2 = 2t (2)(0.25) r0 = r + t = 5 + 0.25 = 5.25 in. r=

(

)

A = π r02 − r 2 = π (5.252 − 5.002 ) = 8.0503 in 2

σ =− Total stresses.

Longitudinal:

P 10 × 103 =− = −1242 psi = −1.242 ksi A 8.0803

σ x = 3 − 1.242 = 1.758 ksi

Circumferential: σ y = 6 ksi Shear:

τ xy = 0

Plotted points for Mohr’s circle: X : (1.758, 0) Y : (6, 0) C : (3.879) 1 2

σ ave = (σ x + σ y ) = 3.879 ksi 2

σx −σ y  2 R=   + τ xy 2   2

 (1.758 − 6)  =   + 0 = 2.121 ksi 2  

(a)

σ x′ = σ ave + R cos 70° = 3.879 − 2.121 cos 70°

(b)

|τ xy | = R sin 70° = 2.121 sin 70°

σ x′ = 3.15 ksi  |τ x′y ′ | = 1.993 ksi 

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PROBLEM 7.128 For the given state of plane strain, use the method of Sec. 7.10 to determine the state of plane strain associated with axes x′ and y ′ rotated through the given angle θ .

ε x = −500μ , ε y = +250 μ , γ xy = 0, θ = 15°

SOLUTION

θ = +15° εx + ε y 2

εx − ε y

= −125μ

ε x′ =

εx + ε y

2 +

εx − ε y

cos 2θ +

εx + ε y

−

εx − ε y

cos 2θ −

2

γ xy

2 2 2 = −125μ + (−375μ ) cos 30° + 0

ε y′ =

γ xy

= −375μ

γ xy

2 2 2 = −125μ − (−375μ ) cos 30° − 0

=0

sin 2θ

ε x′ = −450μ  sin 2θ

ε y′ = 200μ 

γ x′y′ = −(ε x − ε y ) sin 2θ + γ xy cos 2θ = −(−500 μ − 250μ ) sin 30° + 0

γ x′y′ = 375μ 

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PROBLEM 7.136 The following state of strain has been measured on the surface of a thin plate. Knowing that the surface of the plate is unstressed, determine (a) the direction and magnitude of the principal strains, (b) the maximum inplane shearing strain, (c) the maximum shearing strain. (Use v = 13 . )

ε x = −260μ , ε y = −60 μ , γ xy = +480μ

SOLUTION For Mohr’s circle of strain, plot points: X : (−260μ , − 240μ ) Y : (−60μ , 240 μ ) C : (−160μ , 0) tan 2θ p =

γ xy εx − ε y

=

480 = −2.4 −260 + 60

2θ p = −67.38°

θb = −33.67°  θ a = 56.31° 

R = (100μ )2 + (240μ )2 R = 260μ

(a)

(b)

ε a = ε ave + R = −160 μ + 260μ

ε a = 100 μ 

ε b = ε ave − R = −160 μ − 260μ

ε b = −420μ 

1 γ max (in-plane) = R γ max (in-plane) = 2 R 2 v v 1/3 εc = − (ε a + ε b ) = − (ε x + ε y ) = − (−260 − 60) 1− v 1− v 2/3 = 160μ

γ max (in-plane) = 520μ 

ε max = 160μ ε min = −420μ (c)

γ max = ε max − ε min = 160μ + 420μ

γ max = 580 μ 

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PROBLEM 7.140 For the given state of plane strain, use Mohr’s circle to determine (a) the orientation and magnitude of the principal strains, (b) the maximum in-plane shearing strain, (c) the maximum shearing strain.

ε x = +60μ , ε y = +240 μ , γ xy = −50μ

SOLUTION Plotted points: X : (60μ , 25μ ) Y : (240μ , −25μ ) C : (150μ , 0) tan 2θ p =

γ xy εx − ε y

=

−50 = 0.277778 60 − 240

2θ p = 15.52°

θ a = 97.76°  θb = 7.76° 

R = (90 μ ) 2 + (25μ ) 2 = 93.4μ

(a)

ε a = ε ave + R = 150 μ + 93.4μ

ε a = 243.4μ 

ε b = ε ave − R = 150μ − 93.4 μ

ε b = 56.6μ 

(b)

γ max (in-plane) = 2R

(c)

ε c = 0, ε max = 243.4μ , ε min = 0 γ max = ε max − ε min

γ max (in-plane) = 186.8μ 

γ max = 243.4 

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PROBLEM 7.147 The strains determined by the use of the rosette attached as shown during the test of a machine element are

ε1 = −93.1 × 10−6 in./in. ε 2 = +385 × 10−6 in./in. ε 3 = +210 × 10−6 in./in. Determine (a) the orientation and magnitude of the principal strains in the plane of the rosette, (b) the maximum in-plane shearing strain.

SOLUTION Use ε x′ =

γ xy 1 1 (ε x + ε y ) + (ε x − ε y ) cos 2θ + sin 2θ 2 2 2

where and

θ = −75°

for gage 1,

θ =0

for gage 2,

θ = +75°

for gage 3.

1 2

1 2

1 2

1 2

1 2

1 2

ε1 = (ε x + ε y ) + (ε x − ε y ) cos (−150°) + ε 2 = (ε x + ε y ) + (ε x − ε y ) cos 0 +

γ xy 2

γ xy 2

sin (−150°)

sin 0

ε 3 = (ε x + ε y ) + (ε x − ε y ) cos (150°) +

γ xy 2

sin (150°)

(1) (2) (3)

ε x = ε z = 385 × 10−6 in/in

From Eq. (2), Adding Eqs. (1) and (3),

ε1 + ε 3 = (ε x + ε y ) + (ε x − ε y ) cos 150° = ε x (1 + cos 150°) + ε y (1 − cos 150°) ε + ε − ε (1 + cos 150°) εy = 1 3 x (1 − cos 150°)

=

−93.1 × 10−6 + 210 × 10−6 − 385 × 10−6 (1 + cos 150°) 1 − cos 150°

= 35.0 × 10−6 in/in

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PROBLEM 7.147 (Continued)

Subtracting Eq. (1) from Eq. (3),

ε 3 − ε1 = γ xy sin 150° γ xy =

ε 3 − ε1 sin 150°

=

210 × 10−6 − ( −93.1 × 10−6 ) sin 150°

= 606.2 × 10−6 in/in tan 2θ p =

γ xy εx − εy

=

606.2 × 10−6 = 1.732 385 × 10−6 − 35.0 × 10−6

(a) θ a = 30.0°, θb = 120.0° 

1 1 2 2 −6 = 210 × 10 in/in

ε ave = (ε x + ε y ) = (385 × 10−6 + 35.0 × 10−6 )

2

 ε x − ε y   γ xy  R=   +   2   2 

2

 385 × 10−6 − 35.0 × 10−6 =  2 

(b)

γ max (in-plane) 2

2

  606.2 2 −6  +   = 350.0 × 10 2   

ε a = ε ave + R = 210 × 10−6 + 350.0 × 10−6

ε a = 560 × 10−6 in/in 

ε b = ε ave − R = 210 × 10−6 − 350.0 × 10−6

ε b = −140.0 × 10−6 in/in 

= R = 350.0 × 10−6 in/in

γ max (in-plane) = 700 × 10−6 in/in 

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PROBLEM 7.150 A single strain gage is cemented to a solid 4-in.-diameter steel shaft at an angle β = 25° with a line parallel to the axis of the shaft. Knowing that G = 11.5 × 106 psi, determine the torque T indicated by a gage reading of 300 × 10−6 in./in.

SOLUTION For torsion,

σ x = σ y = 0, τ = τ 0 1 (σ x − vσ y ) = 0 E 1 ε y = (σ y − vσ x ) = 0 E τ0 1 τ γ xy = γ xy = 0 G 2 2G

εx =

Draw the Mohr’s circle for strain. R=

τ0 2G

ε x′ = R sin 2 β = But

τ0 = T= =

τ0 2G

sin 2β

2G ε x′ Tc 2T = 3 = J πc sin 2 β

π c3Gε x′ sin 2 β π (2)3 (11.5 × 106 )(300 × 10−6 ) sin 50°

= 113.2 × 103 lb ⋅ in

T = 113.2 kip ⋅ in 

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PROBLEM 7.157 Solve Prob. 7.156, assuming that the rosette at point A indicates the following strains:

ε1 = −30 × 10−6 in./in. ε 2 = +250 × 10−6 in./in. ε 3 = +100 × 10−6 in./in. PROBLEM 7.156 A centric axial force P and a horizontal force Qx are both applied at point C of the rectangular bar shown. A 45° strain rosette on the surface of the bar at point A indicates the following strains:

ε1 = −60 × 10−6 in./in. ε 2 = +240 × 10−6 in./in. ε 3 = +200 × 10−6 in./in. Knowing that E = 29 × 106 psi and v = 0.30, determine the magnitudes of P and Qx.

SOLUTION

ε x = ε1 = −30 × 10−6 ε y = ε 3 = +100 × 10−6 γ xy = 2ε 2 − ε1 − ε 3 = 430 × 10−6 E 29 (ε x + vε y ) = [ −30 + (0.3)(100)] 2 1− v 1 − (0.3) 2 =0 E 29 σy = (ε y + vε x ) = [100 + (0.3)(−30)] 2 1− v 1 − (0.3) 2

σx =

= 2.9 × 103 psi P =σy A

P = Aσ y = (2)(6)(2.9 × 103 ) = 34.8 × 103 lb

P = 34.8 kips 

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PROBLEM 7.157 (Continued)

G=

E 29 × 106 = = 11.1538 × 106 psi 2(1 + v) (2)(1.30)

τ xy = Gγ xy = (11.1538)(430) = 4.7962 × 103 psi 1 3 1 bh = (2)(6)3 = 36 in 4 12 12 ˆ Q = A y = (2)(3)(1.5) = 9 in 3 t = 2 in. VQˆ τ xy = It Itτ xy (36)(2)(4.7962 × 103 ) V= = = 38.37 × 103 lb ˆ 9 Q I=

Q =V

Q = 38.4 kips 

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