P4 Stress and Strain Dr. A.B. Zavatsky HT08 Lecture 5 Plane Stress Transformation Equations

Stress elements and plane stress. Stresses on inclined sections. Transformation equations. Principal stresses, angles, and planes. Maximum shear stress.

1 Normal and shear stresses on inclined sections

To obtain a complete picture of the stresses in a bar, we must consider the stresses acting on an “inclined” (as opposed to a “normal”) section through the bar. Inclined section Normal section

P P

Because the stresses are the same throughout the entire bar, the stresses on the sections are uniformly distributed.

Inclined Normal P section P section

2 y N θ

P x P V The force P can be resolved into components: Normal force N perpendicular to the inclined plane, N = P cos θ Shear force V tangential to the inclined plane V = P sin θ

If we know the areas on which the forces act, we can calculate the associated stresses.

y y

σθ τθ area x area x A A

area (A / cos θ) area (A / cos θ)

3 y σθ τθ θ x

σ σForce N P cos θ P 2 θσ = = σ = = cos Area θArea A/cosθ A θ = cos2 = x ()1+ cos 2θ θ x 2

σmax = σx occurs when θ = 0° θ σ Force −V − P sin θ P θ θ τ = =θ = = − sin cos θ Area Areaθ Aσ/cos A τ = − sin cos = − x ()sin 2θ θ x 2

τmax = ± σx/2 occurs when θ = -/+ 45°

4 Introduction to stress elements Stress elements are a useful way to represent stresses acting at some point on a body. Isolate a small element and show stresses acting on all faces. Dimensions are “infinitesimal”, but are drawn to a large scale. y

P x σx = P / A

Area A y y

σx σx = PA/ σx σx = PA/

x x

z

5 Maximum stresses on a bar in tension

P P a

σ x σx = σmax = P / A

a Maximum normal stress, Zero shear stress

6 Maximum stresses on a bar in tension

P P ab

σx/2 θ = 45°

τmax = σx/2

σx/2 b Maximum shear stress, Non-zero normal stress 7 Stress Elements and Plane Stress

When working with stress elements, keep in mind that only one intrinsic state of stress exists at a point in a stressed body, regardless of the orientation of the element used to portray the state of stress.

We are really just rotating axes to represent stresses in a new coordinate system.

y y1

x1

θ σx σx x

8 y σy τ Normal stresses σx, σy, σz yx (tension is positive) τyz τxy τzy σx σx x τzx τxz

σz Shear stresses τxy = τyx, z τxz = τzx, τyz = τzy

σy

Sign convention for τab Subscript a indicates the “face” on which the stress acts (positive x “face” is perpendicular to the positive x direction) Subscript b indicates the direction in which the stress acts

Strictly σx = σxx, σy = σyy, σz = σzz 9 When an element is in plane stress in the xy plane, only the x and y faces are subjected to stresses (σz = 0 and τzx = τxz = τzy = τyz = 0).

Such an element could be located on the free surface of a body (no stresses acting on the free surface).

Plane stress element in 2D

σy

σx, σy τyx τ = τ τ xy yx y xy σ = 0 z x σx σx

τxy

τyx

σy 10 Stresses on Inclined Sections y y1 σy τ τyx σy1 y11x τx11y x1 τxy y σx1 x θ σx σx x

τ σx1 xy τy11x

τyx τx11y σy1

The stress system is known in terms of coordinate system xy. We want to find the stresses in terms of the rotated coordinate system x1y1.

Why? A material may yield or fail at the maximum value of σ or τ. This value may occur at some angle other than θ = 0. (Remember that for uni- axial tension the maximum shear stress occurred when θ = 45 degrees. ) 11 Transformation Equations

Stresses Forces y y y 1 y1

x1 x τ τ A sec θ 1 θ x11y θ xy11 σx σ A σx1 θ x θ σ A sec θ x x1 x τxy τxy A A tan τyx τθyx σ y σy A tan θ

Forces can be found from stresses if Left face has area A. the area on which the stresses act is known. Force components can then Bottom face has area A tan θ. be summed. Inclined face has area A sec θ.

12 y

y1 x τ A sec θ 1 θ xy11 θ σ A σ x θ σ A sec θ θ x1 x τ τ τxy A θ θ τθA tan σ σyx θ σ A tan θ θ τ y θ Sum forces in the x direction : θ τ 1 σ θ () ( ) ( ) σσ x1 Asec − ()x A cos − xy A sin − y Atan θsin − yx Atan cosθ = 0 θ σ τ Sum forces inθ the y1 direction : τ σ θ x1y1 Asecσ + ()x A sin θ− ()xy A cos − y Atan (cos − yx Atan )sinθ = 0 ( ) σ τ Using τ xy =τ yx and simplifyinθ g givesθ : θ θ = cos2 + sin 2 + 2 τ sin cos x1 x y xy θ 2 2 x1y1 = −()x − y sin cos + xy ()cos − sin θ

13 θ θ

θ θ

Using the following trigonometric identities θ θ σ 2σ 1+ cos 2 2 1− cos 2 sin 2θ cos = σ sin = sin cos = 2 σ 2 2 σ τ gives the σtransformation equations for plane stress : + σ − θ x y x y τ x1 = + cos 2 + xy sin 2 2 2θ θ σ σ ()− τ HLT, page 108 = − σx y sin 2 + cos 2θ x1y1 σ xy 2 σ θ For stresses on the y face, substitute + 90° forθ : 1 θ + − τ = x y − x y cos 2 − sin 2θ y1 2 2 xy σ Summingσ the expressions for x and y gives : σ 1 1 x1 + y1 = x +σ y Can be used to find σy1, instead of eqn above.

14 Example: The state of plane stress at a point is represented by the stress element below. Determine the stresses acting on an element oriented 30° clockwise with respect to the original element. 50 MPa Define the stresses in terms of the established sign convention: y σx = -80 MPa σy = 50 MPa 80 MPa x 80 MPa τxy = -25 MPa We need to find σ , σ , and 25 MPa x1 y1 τ when θ = -30°. 50 MPa x1y1

Substitute numerical values into the transformation equations: + − σ =σ x y + x y cos 2 + sin 2 x1 σ2 2 xy σ − 80 + 50 −σ80 − 50 σ = + cos 2()()()− 30° + − 25 sin 2 − 30° = −25.9 MPa x1 2 2 θ τ 15 θ σ +σ σ −σ σ = x y − x y cos 2 − sin 2 y1 2 2 xy −80 + 50 −80 − 50 σ = − cos 2()()()− 30° − − 25 sin 2 − 30° = −4.15 MPa y1 2 2 (σ −σ ) θ τ = − x y sin 2 + cosτ 2 x1y1 2 xy θ ()− 80 − 50 τ x1y1 = − sinθ 2()()()− 30° + − 25 cos 2 − 30° = −68.8 MPa 2 τ y y 1 θ 4.15 MPa 25.8 MPa Note that σy1 could also be obtained (a) by substituting +60° into the equation for σ or (b) by using the +60° x1 x equation σ + σ = σ + σ -30o x y x1 y1 25.8 MPa x 4.15 MPa 1 68.8 MPa (from Hibbeler, Ex. 15.2) 16 Principal Stresses

The maximum and minimum normal stresses (σ1 and σ2) are known as the principal stresses. To find the principal stresses, we must differentiate the transformation equations. σ +σ σ −σ σ σ = x y + x y cos 2 + sin 2 xσ1 2 2 xy θ σ θ d − τ σ x1 = x y ()()− 2sin 2 + 2cos 2 = 0 d 2 θ xy θ θ σ τ d x1 σ = − ()x − y sin 2 + 2 xy cosθ2 = 0 d θ θ τ2 τ xy θ are principal angles associated with tan 2 p = σ p θ x −σ y the principal stresses (HLT, page 108)

There are two values of 2θp in the range 0-360°, with values differing by 180°. There are two values of θp in the range 0-180°, with values differing by 90°. So, the planes on which the principal stresses act are mutually perpendicular. 17 We canσ now solve for the principal stresses by substituting for θp in the stressσ transformation equation for σ . This tells us which σ σ x1 principal stressσ is associated with which principal angle. σ 2τ σ tan 2θ = xy p − x y θ τ x + y x − y x1 = + cos 2 + xy sin 2θ 2 2 σ 2 ⎛σ − ⎞ 2 ⎜ x y ⎟ 2 R = ⎜ ⎟ + xy R ⎝ 2 τ⎠ σ τxy θ − σ σ cos 2 = xσ y σ p 2R 2θ p σ θ τ σ sin 2 = xy (σ – σ ) / 2 σ p x y σ R + − ⎛ − ⎞ ⎛τ ⎞ x y x y ⎜ x y ⎟ τ ⎜ xy ⎟ 1 = + ⎜ ⎟ + xy ⎜ ⎟ 2 2 ⎝ 2R ⎠ ⎝ R ⎠ 18 σ σ Substitutingσ for R and re-arranging gives the larger of the two principal stresses: σ σ 2 σ + ⎛ − ⎞ = σ x y + ⎜ x y ⎟ +τ 2 1 2σ ⎜ 2 ⎟ xy σ ⎝σ ⎠ σ To find the smaller principal stress, use σ + σ = σ + σ . σ 1 2 x y σ 2 + ⎛ − ⎞ x y ⎜ x y ⎟ 2 2 = x + y − 1 = − ⎜ ⎟ +τ xy σ σ 2 ⎝ 2 ⎠ σ These equations can be combined to give: σ σ 2 + ⎛ − ⎞ Principal stresses x y ⎜ x y ⎟ 2 1,2 = ± ⎜ ⎟ +τ xy 2 ⎝ 2 ⎠ (HLT page 108)

To find out which principal stress goes with which principal angle, we could use the equations for sin θp and cos θp or for σx1. 19 The planes on which the principal stresses act are called the τ principalσ planes. What shear stresses act on the principal planes? σ

θ Compareσ the equations for τ x1y1 = 0 and dσ x1 dθ = 0 σ σ τ ()− θ = − x y sinθ2 + cos 2 = 0 x1yθ1 σ τ xy 2 σ θ − ()x − y sin 2θ + 2 xy cos 2 = 0 d τ x1 = − ()− sin 2 + 2 cos 2θ = 0 d x y xy

Solving either equation gives the same expression for tan 2θp Hence, the shear stresses are zero on the principal planes.

20 y σy

τyx σ2

τxy y σ1 θp2

x θp1 σx σx x

σ1 τxy τ σ yxσ σ 2 σy σ Principal Stresses σ σ 2 + ⎛ − ⎞ x y ⎜ x y ⎟ 2 1,2 = ± ⎜ ⎟ +τ xy 2 ⎝ 2 ⎠

Principal Angles defining the Principal Planes θ τ 2 xy tan 2 p = σ x −σ y

21 Example: The state of plane stress at a point is represented by the stress element below. Determine the principal stresses and draw the corresponding stress element. 50 MPa

y Define the stresses in terms of the 80 MPa x 80 MPa established sign convention: σx = -80 MPa σy = 50 MPa τxy = -25 MPa 25 MPa σ σ σ 50 MPa 2 + σ ⎛ − ⎞ = x y ± ⎜σ x y ⎟ + 2 σ 1,2 2 ⎜ 2 ⎟ xy ⎝ τ⎠ 2 σ − 80 + 50 ⎛ − 80 − 50 ⎞ 2 1,2 = ± ⎜ ⎟ + ()− 25 = −15 ± 69.6 2 ⎝ 2 ⎠

1 = 54.6 MPa σ 2 = −84.6 MPa 22 y 2τ xy tan 2θ p = − 54.6 MPa σ x y σ 2()− 25 tanθ 2 = = 0.3846 100.5o 84.6 MPa p − 80 − 50 84.6 MPa x 10.5o

θ2 p = 21.0° and 21.0 +180° 54.6 MPa θ p = 10.5°,100.5°

But we must check which angle goes with which principal stress.

+ − σ =σ x y + x y cos 2 + sin 2 x1 σ2 2 xy σ − 80 + 50 −σ80 − 50 σ = + cos 2()()()10.5° + − 25 sin 2 10.5° = −84.6 MPa x1 2 2 θ τ σ1 = 54.6 MPa with θp1 = 100.5° θ σ2 = -84.6 MPa with θp2 = 10.5° 23 The two principal stresses determined so far are the principal stresses in the xy plane. But … remember that the stress element is 3D, so there are always three principal stresses.

yp y

σ2 σy

xp τ σ1 τ σ σ x x x

σ1 σ = 0 τ 3 τ σ2 zp σy

σx, σy, τxy = τyx = τ σ1, σ2, σ3 =0

Usually we take σ1 > σ2 > σ3. Since principal stresses can be com- pressive as well as tensile, σ3 could be a negative (compressive) stress, rather than the zero stress. 24 Maximum Shear Stress

To find the maximum shear stress, we must differentiate the trans- formation equation for shear. (σ −σ ) τ τ = − x y sin 2 + cos 2 x1y1 2 θ xy θ τ d σ θ x1y1 = −()σ− cos 2 − 2 sin 2 = 0 d x y θ xy θ σ τ ⎛ −σ ⎞ tan 2 = −⎜ x y ⎟ θ s ⎜ ⎟ ⎝ 2τ xy ⎠

There are two values of 2θs in the range 0-360°, with values differing by 180°. There are two values of θs in the range 0-180°, with values differing by 90°. So, the planes on which the maximum shear stresses act are mutually perpendicular. Because shear stresses on perpendicular planes have equal magnitudes, the maximum positive and negative shear stresses differ only in sign. 25 θ σ We can now solveσ for the maximum shear stress by substituting for θ in the stressτ transformation equation for τ . τs x1y1 σ ⎛ σ− ⎞ tan 2 = −⎜ x y ⎟ s ⎜ 2 ⎟ ⎝ xy ⎠ θ τ ()x − y σ x1y1 = − sin 2 + xy cos 2θ σ 2 2 ⎛ − ⎞ R2 = ⎜ x y τ⎟ + 2 ⎜ 2 ⎟ xy θ ⎝ τ ⎠

R ) / 2 y xy σ cos 2 s = – R

τ x σ θ σ

2θ (σ x −σ y s σ sin 2 = − s 2R τxy τ

2 τ ⎛ − ⎞ ⎜ x y ⎟ 2 max = ⎜ ⎟ + xy min = −τ max ⎝ 2 ⎠

26 Use equations for sin θs and cos θs or τx1y1 to find out which face has the positive shear stress and which the negative.

What normal stresses act on the planes with maximum shear stress? σ Substitute for θ in the equationsσ σ for σ and σ to get s σx1 y1 + = = x y = σ x1 y1 2 s y

σy σ τyx s τmax

τxy τmax y σs x θs σx σx x σ τxy s τmax τyx τmax σs σy

27 Example: The state of plane stress at a point is represented by the stress element below. Determine the maximum shear stresses and draw the corresponding stress element. 50 MPa

y Define the stresses in terms of the 80 MPa x 80 MPa established sign convention: σx = -80 MPa σy = 50 MPa τxy = -25 MPa 25 MPa 50 MPa

2 σ x +σ y τ ⎛ x − y ⎞ 2 = ⎜ ⎟ + σ s = max σ ⎜ ⎟ xy 2 ⎝ σ 2 ⎠ − 80 + 50 2 τ σ s = = −15 MPa ⎛ − 80 − 50 ⎞ 2 2 τ max = ⎜ ⎟ + ()− 25 = 69.6 MPa ⎝ 2 ⎠

28 ⎛ − ⎞ ⎛ −80 − 50 ⎞ tan 2 = −⎜ x y ⎟ = −⎜ ⎟ = −2.6 θ s σ ⎜ ⎟ ⎜ ⎟ σ2 xy ⎝ 2()− 25 ⎠ ⎝ ⎠ y θ2 s = −69.0°τand − 69.0 +180° 15 MPa 15 MPa θs = −34.5°, 55.5°

55.5o But we must check which angle goes o x -34.5 with which shear stress. 15 MPa (σ −σ ) 15 MPa τ = − x y sin 2 + cos 2 x1y1 2 xy 69.6 MPa ()− 80 − 50 τ x1y1 = − sinθ 2()()()− 34.5 + − 25 cos 2 − 34.5 = −69.6 MPa 2 τ θ τmax = 69.6 MPa with θsmax = 55.5° τmin = -69.6 MPa with θsmin = -34.5°

29 Finally, we can ask how the principal stresses and maximum shear stresses are related and how the principal angles and maximum shear anglesσ areσ related. σ 2 + ⎛ − ⎞ = x y σ± ⎜ x y ⎟ + 2 1,2 σ⎜ ⎟ xy 2 ⎝ 2 ⎠ σ σ 2 τ σ ⎛ − ⎞ − = 2 ⎜ σx y ⎟ + 2 σ 1 2 ⎜ 2 ⎟ xy σ ⎝ ⎠ τ τ 1 − 2 = 2 max τ σ −σ = 1 2 θ max σ 2 σ ⎛ τ − ⎞ 2 ⎜ x y ⎟ xy tan 2θs = − tan 2 p = ⎜ ⎟ θ ⎝ 2 xy ⎠ τ x − y θ σ −1 σ tan 2 s = = −cot 2θ p tan 2 p 30 tan 2θ + cot 2θ = 0 θ s p sin 2 cos 2 θ s θ+ p = 0 cos 2 sin 2 θ s θ p sin 2 θs sin 2 p + cos 2 s cos 2 p = 0 θ θ cos ()2 sθ− 2 p = 0 θ θ 2 −θ 2 = ± 90° θ s p θ s − p = ± 45° θ s = θ p ± 45°

So, the planes of maximum shear stress (θs) occur at 45° to the principal planes (θp).

31 Original Problem Principal Stresses y 50 MPa 54.6 MPa y o 84.6 MPa 80 MPa x 80 MPa 100.5 84.6 MPa x 10.5 o

25 MPa 50 MPa 54.6 MPa

σx = -80, σy = 50, τxy = 25 σ1 = 54.6, σ2 = 0, σ3 = -84.6 Maximum Shear y 15 MPa 15 MPa σ1 −σ 2 τ = θs = θ p ± 45° max 2 o θ =10.5° ± 45° 55.5 54.6 − ()−84.6 s o x τ -34.5 max = 2 θs = −34.5°, 55.5° 15 MPa τ = 69.6 MPa 15 MPa max 69.6 MPa τ = 69.6, σ = -15 max s 32