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####### Russell C. HibbelerChapter 9: Stress TransformationChapter 9: Stress TransformationChapter 9: Stress TransformationMechanics of Material 7Mechanics of Material 7ththEditionEditionPlane-Stress Transformationy General state of stress at a point is characterized by6 independent normal and shear...

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RussellC.Hibbeler Chapter9:StressTransformation

Plane-Stress Transformation y General state of stress at a point is characterized by 6 independent normal and shear stress components. y It can be analyzed in a single plane of a body, the material can said to be subjected to plane stress.

Chapter 9: Stress Transformation th Edition Mechanics of Material 7th © 2008 Pearson Education South Asia Pte Ltd

Plane-Stress Transformation y Stress components from one orientation of an element can transform to an element having a different orientation.

=

Chapter 9: Stress Transformation th Edition Mechanics of Material 7th © 2008 Pearson Education South Asia Pte Ltd

Example 9.1 The state of plane stress at a point on the surface of the airplane fuselage is represented on the element oriented as shown. Represent the state of stress at the point on an element that is oriented 30°clockwise from the position shown.

Solution: The element is sectioned by the line a–a.

The free-body diagram of the segment is as shown.

Chapter 9: Stress Transformation th Edition Mechanics of Material 7th © 2008 Pearson Education South Asia Pte Ltd

Solution: Applying the equations of force equilibrium in the x’ and y’ direction,

+ ∑ Fx ' = 0;

σ x' ΔA − (50ΔA cos 30°) cos 30° + (25ΔA cos 30° )sin 30° + (80ΔA sin 30° )sin 30° + (25ΔA sin 30° )cos 30° = 0

σ x ' = − 4.15 MPa (Ans)

+ ∑ Fy' = 0;

τ x'y'ΔA − (50ΔA cos 30° ) sin 30° − (25ΔA cos 30°) cos 30° − (80ΔA sin 30°) cos 30° + ( 25ΔA cos 30°) sin 30° = 0 τ x'y' = 68.8 MPa (Ans)

Chapter 9: Stress Transformation th Edition Mechanics of Material 7th © 2008 Pearson Education South Asia Pte Ltd

Solution: Repeat the procedure to obtain the stress on the perpendicular plane b–b.

+ ∑ Fx ' = 0;

σ x ' ΔA − ( 25ΔA cos 30°) sin 30° + (80ΔA cos 30°) cos 30°

+ ∑ Fy' = 0;

-τ x'y' ΔA + (25ΔA cos 30°) cos 30° + (80ΔA cos 30°) sin 30°

− ( 25ΔA cos 30°) cos 30° − (50ΔA sin 30°) sin 30° = 0 σ x' = − 25.8 MPa (Ans)

− (25ΔA sin 30° ) sin 30° + (50ΔA sin 30°) cos 30° = 0 τ x'y' = 68.8 MPa (Ans)

The state of stress at the point can be represented by choosing an element oriented.

Chapter 9: Stress Transformation th Edition Mechanics of Material 7th © 2008 Pearson Education South Asia Pte Ltd

General Equations of Plane-Stress Transformation y Positive normal stress acts outward from all faces and positive shear stress acts upward on the right-hand face of the element.

σ x' =

σ x +σ y

τ x'y ' = −

2

+

σ x −σ y 2

σ x −σ y 2

cos 2θ + τ xy sin 2θ

sin 2θ + τ xy cos 2θ

Chapter 9: Stress Transformation th Edition Mechanics of Material 7th © 2008 Pearson Education South Asia Pte Ltd

Example 9.2 The state of plane stress at a point is represented by the element. Determine the state of stress at the point on another element oriented 30° clockwise from the position shown.

Solution: From the sign convention we have,

σ x = − 80 MPa

σ y = 50 MPa τ xy = −25 MPa θ = −30°

To obtain the stress components on plane CD, σ +σ y σ x − σ y cos 2θ +τ xy sin 2θ = −25.8 MPa (Ans) σ x' = x + 2 2 σ −σy τ x'y ' = − x sin 2θ + τ xy cos 2θ = −68.8 MPa (Ans) 2

Chapter 9: Stress Transformation th Edition Mechanics of Material 7th © 2008 Pearson Education South Asia Pte Ltd

Solution: To obtain the stress components on plane BC,

σ x = − 80 MPa σ x' =

σ y = 50 MPa τ xy = −25 MPa θ = −60°

σ x +σ y σ x − σ y

τ x'y ' = −

+

2

σx −σy 2

2

cos 2θ +τ xy sin 2θ = −4.15 MPa (Ans)

sin 2θ +τ xy cos 2θ = 68.8 MPa (Ans)

The results are shown on the element as shown.

Chapter 9: Stress Transformation th Edition Mechanics of Material 7th © 2008 Pearson Education South Asia Pte Ltd

Principal Stresses and Maximum In-Plane Shear Stress In-Plane Principal Stresses y Orientation of the planes will determine the maximum and minimum normal stress. tan 2θ p =

τ xy (σ x − σ y )/ 2

Chapter 9: Stress Transformation th Edition Mechanics of Material 7th © 2008 Pearson Education South Asia Pte Ltd

. y The solution has two roots, thus we obtain the following principle stress.

σ 1, 2 =

σ x +σ y 2

⎛ σ x −σ y ± ⎜⎜ ⎝ 2

Chapter 9: Stress Transformation th Edition Mechanics of Material 7th © 2008 Pearson Education South Asia Pte Ltd

2

⎞ ⎟⎟ + τ xy 2 where σ 1 > σ 2 ⎠

Principal Stresses and Maximum In-Plane Shear Stress Maximum In-Plane Shear Stress y Orientation of an element will determine the maximum and minimum shear stress. tan 2θs =

− (σ x − σ y )/ 2

τ xy

y The solution has two roots, thus we obtain the maximum in-plane shear stress and averaged normal stress. 2

⎛ σ −σ y ⎞ ⎟⎟ + τ xy2 τ max in -plane = ⎜⎜ x 2 ⎠ ⎝ Chapter 9: Stress Transformation th Edition Mechanics of Material 7th © 2008 Pearson Education South Asia Pte Ltd

σ avg =

σ x −σ y 2

Example 9.3 When the torsional loading T is applied to the bar it produces a state of pure shear stress in the material. Determine (a) the maximum in-plane shear stress and the associated average normal stress, and (b) the principal stress.

Solution: From the sign convention we have, σ x = 0

σy = 0

τ xy = −τ

a) Maximum in-plane shear stress is 2

σ +σ y ⎛σ −σ y ⎞ τ max in -plane = ⎜⎜ x σ avg = x ⎟⎟ + τ xy 2 = ±τ = 0 (Ans) 2 2 ⎝ ⎠ b) For principal stress, τxy σ σ tan 2θ p = (σ x − σ y ) / 2 ⇒ p2 = 45°, p1 = 135° σ 1, 2 =

σ x +σ y 2

2

⎛σ x−σ y ⎞ ⎟⎟ + τ xy2 = ±τ ± ⎜⎜ ⎝ 2 ⎠

Chapter 9: Stress Transformation th Edition Mechanics of Material 7th © 2008 Pearson Education South Asia Pte Ltd

(Ans)

Example 9.6 The state of plane stress at a point on a body is represented on the element. Represent this stress state in terms of the maximum in-plane shear stress and associated average normal stress.

Solution: Since σ x = −20, σ y = 90, τ xy = 60 we have tan 2θ s =

− (σ x − σ y ) / 2

⇒ σ s 2 = 21.3°,σ s1 = 111.3° τ xy The maximum in-plane shear stress and average normal stress is 2

⎛σ −σy ⎞ ⎟⎟ + τ xy 2 = 81.4 MPa (Ans) τ max in -plane = ⎜⎜ x ⎝ 2 ⎠ σ +σ y σavg = x = 35 MPa (Ans) 2 Chapter 9: Stress Transformation th Edition Mechanics of Material 7th © 2008 Pearson Education South Asia Pte Ltd

Mohr’s Circle—Plane Stress y Plane stress transformation is able to have a graphical solution that is easy to remember.

Chapter 9: Mechanics © 2008 Pear

Example 9.8 The torsional loading T produces the state of stress in the shaft as shown. Draw Mohr’s circle for this case.

Solution: We first construct of the circle, σ x = 0,σ y = 0 and τ xy = −τ The center of the circle C is on the axis at σ avg = Point A is the average normal stress and maximum in-plane shear stress, τ max in -plane = −τ , σ avg = 0 Principal stresses are identified as points B and D on the circle.

σ1 = τ , σ 2 = −τ Chapter 9: Stress Transformation th Edition Mechanics of Material 7th © 2008 Pearson Education South Asia Pte Ltd

σ x +σ y 2

=0

Example 9.10 The state of plane stress at a point is shown on the element. Determine the maximum in-plane shear stresses and the orientation of the element upon which they act.

Solution: We first construct of the circle, σ x = − 20, σ y = 90 and τ xy = 60 The center of the circle C is on the axis σavg =

− 20 + 90 = 35 MPa 2

2 2 From point C and the A(-20, 60) are plotted, we have R = 60 + 55 = 81.4 MPa

Max in-plane shear stress and average normal stress are

τ max in -plane = 81.4 MPa , σ avg = 35 MPa (Ans) The counter-clockwise angle is ⎛ 20 + 35 ⎞ 2θs 1 = tan −1 ⎜ ⎟ ⇒ 21.3° (Ans) ⎝ 60 ⎠ Chapter 9: Stress Transformation th Edition Mechanics of Material 7th © 2008 Pearson Education South Asia Pte Ltd

Example 9.12 An axial force of 900 N and a torque 2.5 Nm of are applied to the shaft. The shaft diameter is 40 mm, find the principal stresses at a point P on its surface.

Solution: The stresses produced at point P is

τ=

900 Tc 2.5( 0.02) P 198 . 9 kPa, σ = π = = = = 716.2 kPa 4 2 J A ( ) ( ) 0 . 02 0 . 02 π 2

The principal stresses can be found using Mohr’s circle,

σ avg =

0 + 716.2 = 358.1 kPa 2

Principal stresses are represented by points B and D,

σ 1 = 358.1 + 409.7 = 767.7 kPa (Ans) σ 2 = 358.1− 409.7 = −51.5 kPa (Ans) Chapter 9: Stress Transformation th Edition Mechanics of Material 7th © 2008 Pearson Education South Asia Pte Ltd

Example 9.13 The beam is subjected to the distributed loading of w = 120 kN/m. Determine the principal stresses in the beam at point P, which lies at the top of the web. Neglect the size of the fillets and stress concentrations at this point. I = 67.4(10-6) m4.

Solution: The equilibrium of the sectioned beam is as shown where V = 84 kN At point P,

( ) ( )

− My 30.6 103 0.1 σ= = = −45.4 (Ans) I 67.4 10− 6 VQ 84[( 0.1075)(0.175)(0.015)] τ= = = 35.2 MPa (Ans) 67.4 10 −6 (0.01) It

(

)

Chapter 9: Stress Transformation th Edition Mechanics of Material 7th © 2008 Pearson Education South Asia Pte Ltd

M = 30.6 kNm

Solution: Thus the results are as follows,

45.4 + 0 The centre of the circle is − = −22.7 and point A is (-45.4, -32.5). 2 Thus the radius is calculated as 41.9, thus the principle stresses are

σ 1 = ( 41.9 − 22.7) = 19.2 MPa σ 2 = − ( 22.7 + 41.9) = − 64.6 MPa The counter-clockwise angle is

2θ p 2 = 57.2° ⇒ θ p 2 = 28.6° Chapter 9: Stress Transformation th Edition Mechanics of Material 7th © 2008 Pearson Education South Asia Pte Ltd...