CHAPTER 4 Stress Transformation


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1 CHAPTER 4 Stress Transformation
2 ANALYSIS OF STRESS For this topic, the stresses to be considered are not on the perpendicular and parallel planes only but also on other inclined planes. A P a a b b P z y x (A body subjected to load P)
3 ANALYSIS OF STRESS On plane aa, normal force, N produces normal stress. σ = N A = P A On plane bb, normal force, N produces normal stress and shear force, V produces shearing stress. P Area = A/cos N = P cos P V = P sin
4 ANALYSIS OF STRESS σ n = P cosθ A cosθ = P A cos θ = σ x cos θ τ = P sinθ A cosθ = P A sinθ cosθ = σ x sinθ cosθ
5 ANALYSIS OF STRESS On an element, there are 6 components of stress: x, y, z, xy, yz, and zx y z x If the z axis is not considered or the stresses are independent with the z axis, then there exist only stresses in the x and y directions.
6 ANALYSIS OF STRESS UNDER D This state of dimensional stress is known as plane stress. z = xz = yz = 0 y y y x x x a xy x a All directions shown (axes and on element) are taken as positive. y
7 STRESS TRANSFORMATION Stresses on an element can be transformed using methods: (i) Equations Method (ii) Mohr Circle
8 STRESS TRANSFORMATION i) Equations Method Consider an element rotated an amount of about the zaxis. Stresses on an inclined plane will be yielded and can be expressed in terms of x, y, xy and. x A cos y xy x y y x x x A A sin xy (A sin) cos
9 Sign Convention: STRESS TRANSFORMATION Positive if counterclockwise and usually taken from the vertical surface (xplane) to the intended plane Positive if tension or in the direction of positive axis Positive if in the direction of positive shear (counterclockwise) xy = yx + y + xy + x
10 Normal Stress: STRESS TRANSFORMATION σ x = σ x + σ y + σ x σ y cosθ + τ xy sinθ σ y = σ x + σ y σ x σ y cosθ τ xy sinθ Shear Stress: τ x y = σ x σ y sinθ + τ xy cosθ Note: If is clockwise, then put a negative sign in the equation(s)
11 STRESS TRANSFORMATION y + 90 y x y x x x
12 STRESS TRANSFORMATION InPlane Principle Stress: Taking dσ x dθ = 0; tanθ p = τ xy σ x σ y σ max & σ min = σ x+σ y ± σ x σ y + τxy
13 STRESS TRANSFORMATION Maximum InPlane Shear Stress: Taking dτ x y dθ = 0; tanθ s = σ x σ y τ xy τ max & τ min = ± σ x σ y + τxy
14 STRESS TRANSFORMATION Average Normal Stress: When substituting the values for s into the equation for normal stress (σ x ), there is also a normal stress on the plane of maximum inplane shear stress, which can be determined by: σ avg = σ x + σ y
15 EXAMPLE 1 The state of plane stress at a point on a body is shown on the element in the Figure. Represent this stress state in terms of the principal stresses. 90 MPa 60 MPa x 0 MPa xy y
16 EXAMPLE 1 Solution 90 MPa 60 MPa x 0 MPa xy x = 0 MPa (Compression) y y = 90 MPa (Tension) xy = 60 MPa (Clockwise)
17 Principal Stresses EXAMPLE 1 Solution σ max & σ min = σ x + σ y ± σ x σ y + τ xy σ max & σ min = ± σ max & σ min = 35 ± 81.4 max = = 116 MPa min = = 46.4 MPa
18 Orientation of Element EXAMPLE 1 Solution tanθ p = τ xy σ x σ y tanθ p = p = p = 3.7 p1 = = p1 = 66.3
19 EXAMPLE 1 Solution The principal plane on which each normal stress acts can be determined by applying: σ x = σ x + σ y + σ x σ y cosθ + τ xy sinθ σ x = + cos sin x = 46.4 MPa Hence, σ min = 46.4 MPa acts on the plane defined by θ p = 3.7, whereas σ max = 116 MPa acts on the plane defined by θ p1 = 66.3.
20 EXAMPLE 1 Solution By replacing the θ p1 and θ p into the equation for shear stress: τ x y = σ x σ y sinθ + τ xy cosθ 0 90 τ x y = sin τ xy cos x y = 0 MPa No shear stress acts on this element.
21 EXAMPLE The state of plane stress at a point on a body is represented on the element shown in the Figure. Represent this stress state in terms of the maximum inplane shear stress and associated average normal stress. 90 MPa 60 MPa x 0 MPa xy y
22 EXAMPLE Solution Maximum InPlane Shear Stresses max x y xy max max MPa 60
23 Orientation of Element tan 0 90 tan s s s1 s s1 s x EXAMPLE Solution xy y
24 The proper direction of max on the element can be determined by applying the equation: x' y' x' y' x' y' x EXAMPLE Solution y sin xy cos 0 90 sin cos MPa
25 EXAMPLE Solution Average Normal Stress Besides the maximum shear stress, as calculated above, the element is also subjected to an average normal stress determined from the equation: avg avg avg x y MPa
26 ii) Mohr Circle STRESS TRANSFORMATION In this method, stresses on a plane are drawn as one point on a Mohr circle. From the equations, it can be shown that: σ x σ x + σ y + τ x y = σ x σ y + τ xy Centre of Circle R = max
27 STRESS TRANSFORMATION Steps for constructing Mohr Circle: 1. Determine centre of circle, C (x, y) x = avg, y = 0. Determine point A ( x, xy ) : coordinate when = 0 x = +ve (tension) or ve (comp) xy = +ve (counter clockwise) or ve (clockwise) 3. Determine point A ( y, yx ) : coordinate when = Draw circle connecting A and A with C as the centre 5. The datum (reference line) in Mohr circle is line AC 6. All angles must be determined from line AC ( = 0)
28 STRESS TRANSFORMATION y Y(+ y,  yx ) yx max Y(+ y,  yx ) 0 min ( 1 ) ( x + y )/ p (min) p (max) max ( ) (+ve) yx y s (max) X(+ x, + xy ) X(+ x, + xy ) xy x x (+ve) xy
29 EXAMPLE 3 The state of plane stress at a point on a body is represented on the element shown in the Figure. Determine the principal stresses and the orientation acting at this point. 90 MPa 60 MPa x 0 MPa xy y
30 Construction of the Circle x = 0 MPa (Compression) y = 90 MPa (Tension) τ xy = 60 MPa (Clockwise) EXAMPLE 3 Solution The Centre of Circle is at: σ avg = σ x + σ y = = 35 MPa The Radius of the Circle is: R = σ x σ y + τ xy = = 81.4 MPa
31 EXAMPLE 3 Solution A (90,  60) σ min = 46.4 MPa θ p C = (35, 0) θ p, max = 13.6 σ max = MPa 60 MPa 35 MPa A (0, 60) 0 MPa tan θ p = 60 / 55 θ p = tan θ p = θ p = 3.74
32 EXAMPLE 4 The state of plane stress at a point on a body is represented on the element shown in the Figure. Determine the maximum inplane shear stresses and the orientation of the element upon which they act. 90 MPa 60 MPa x 0 MPa xy y
33 EXAMPLE 4 Solution tan θ s1 = 55 / 60 θ s1 = tan θ s1 = 4.5 θ s1 = 1.3 τ min = 81.4MPa A (90, 60) θ s1 = MPa A (0, 60) 0 MPa τ max = 81.4MPa
34 EXAMPLE 4 Solution y 81.4 MPa x 35 MPa 1.3 x
35 EXAMPLE 5 The state of plane stress at a point on a body is represented on the element shown in the Figure. Represent this state of stress on an element oriented 30 counterclockwise from the position shown. 90 MPa 60 MPa x 0 MPa xy y
36 EXAMPLE 5 Solution tan θ = 55 / 60 θ = tan θ = 4.5 θ = 1.3 B = = 17.5 At Point B: x = sin17.5 C x = MPa x y = 81.4 cos = 17.5 x y x y = MPa A (0, 60) x B 0 MPa 35 MPa
37 EXAMPLE 5 Solution At Point B : y = 81.4 sin B x y = 10.5 MPa x y = 81.4 cos 17.5 x y = MPa x y = 17.5 C 4.5 = 17.5 x y A (0, 60) x 0 MPa 35 MPa B
38 EXAMPLE 5 Solution y x 30 x
39 CLASS EXERCISE An element is subjected to the plane stresses shown in the figure. (a) Determine normal stress and shear stress acting on the plane that is inclined at 0 o as shown in the figure. (b) Determine the maximum normal stress and its orientation. (c) Sketch the plane of the maximum normal stress by showing its values and orientation. (d) Determine the maximum shear stress. 50 N/mm 30 N/mm 30 N/mm 0 o 50 N/mm
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