ME 243. Lecture 10: Combined stresses

Transcription

1 ME 243 Mechanics of Solids Lecture 10: Combined stresses Ahmad Shahedi Shakil Lecturer, Dept. of Mechanical Engg, BUET Website: teacher.buet.ac.bd/sshakil

2 Stress at a point The most general state of stress at a point may be represented by 6 components,,, x y z normal stresses xy, yz, zx shearing stresses (Note : xy yx, yz zy, zx xz ) Same state of stress is represented by a different set of components if axes are rotated. We are concerned with how the components of stress are transformed under a rotation of the coordinate axes.

3 Plane stress Plane Stress - state of stress in which two faces of the cubic element are free of stress. For the illustrated example, the state of stress is defined by,, x y xy and z zx zy 0. State of plane stress occurs in a thin plate subjected to forces acting in the midplane of the plate. State of plane stress also occurs on the free surface of a structural element or machine component.

4 Plane stress The general state of plane stress at a point is therefore represented by a combination of two normal-stress components σ x and σ y, and one shear stress component τ xy, which act on four faces of the element (fig a). If this state of stress is defined on an element having a different orientation as in Fig. b, then it will be subjected to three different stress components defined as σ x, σ y and τ xy.

5 Plane stress transformation The element in Fig. a is sectioned along the inclined plane and the segment shown in Fig. b is isolated. Assuming the sectioned area is A, then the horizontal and vertical faces of the segment have an area of A sin θ and A cos θ respectively.

6 The resulting free-body diagram of the segment is shown in Fig. c Applying the equations of equilibrium to determine the unknown normal and shear stress components σ x and τ xy we have, Or, Or, Or,

7 (a) (b) If the normal stress acting in the y direction is needed, it can be obtained by simply substituting θ=θ+90 o for θ into Eq. (a), which gives (C)

8

9 Principal stresses In-Plane Principal Stresses: To determine the maximum and minimum normal stress, we must differentiate Eq. (a) with respect to θ and set the result equal to zero. This gives, Solving this equation we obtain the orientation θ = θ p of the planes of maximum and minimum normal stress.

10 Principal stresses The solution has two roots θ p1 and θ p2. Specifically, the values of 2θ p1 and 2θ p2 are 180 apart, so θ p1 and θ p2 will be 90 apart. Now, 2θ p

11 Principal stresses Substituting these trigonometric values into Eq. (a) and simplifying, we obtain, This result gives the maximum or minimum in-plane normal stress acting at a point, where σ 1 σ 2 This particular set of values are called the in-plane principal stresses, and the corresponding planes on which they act are called the principal planes of stress, From eqn. (b), for θ = θ p, τ xy = 0. So, no shear stress acts on the principal planes.

12 Principal stresses

13 Principal stresses Maximum In-Plane Shear Stress: The orientation of an element that is subjected to maximum shear stress on its sides can be determined by taking the derivative of Eq. (b) with respect to θ and setting the result equal to zero. This gives, The solution has two roots θ s1 and θ s2 tan 2θ s is the negative reciprocal of tan 2θ p and so each root 2θ s is 90 from 2θ p, and the roots θ p and θ s are 45 apart. Therefore, an element subjected to maximum shear stress will be 45 from the position of an element that is subjected to the principal stress.

14 Principal stresses Substituting the trigonometric values of sin 2θ s and cos 2θ s into Eq. (b) and simplifying, we obtain, Here, τ max is the maximum in-plane shear stress. Substituting the trigonometric values of sin 2θ s and cos 2θ s into Eq. (b) and simplifying, we obtain, Here, σ avg is the average normal stress on the planes of maximum normal stress.

15 Principal stresses σ avg = σ =

16 Problem# 7.01 (Beer-Johnston) For the state of plane stress shown, determine (a) the principal panes, (b) the principal stresses, (c) the maximum shearing stress and the corresponding normal stress.

17 solution Find the element orientation for the principal stresses from tan 2 2 p p 2 x xy 53.1, y x y 50 MPa 10 MPa xy 40 MPa p 26.6, Determine the principal stresses from 2 1 1,2 x 2 y x 2 y 2 2 xy MPa 30 MPa

18 solution Calculate the maximum shearing stress with max x 2 y 2 2 xy x y 50 MPa 10 MPa xy 40 MPa max 50 MPa s p 45 s 18.4, The corresponding normal stress is ave x 2 y θ s 20 MPa 7-18

19 Mohr s circle Equations (a) and (b) are the parametric equations of a circle. Now, rearranging and doing (a) 2 +(b) 2, The above eqn. is in the form, where,

20 Mohr s circle Take tensile normal stress as positive. Take clockwise shear as positive. For a known state of plane stress x, y, plot the points X and Y and construct the circle centered at C. x y x y ave R xy xy

21 Mohr s circle The principal stresses are obtained at A and B. 1,2 tan 2 p ave R 2 x xy y Here, σ max = σ 1, σ min = σ 2 AB is the principal plane. The direction of rotation of Ox to Oa is the same as CX to CA.

22 Mohr s circle Maximum in-plane shear stress is found at points D and E. DE is the maximum shear plane. Normal stress at the shear plane is average stress, σ. From Mohr s circle, we see that the angle between principal plane and maximum shear plane is 45 o

23 Mohr s circle With Mohr s circle uniquely defined, the state of stress at other axes orientations may be depicted. For the state of stress at an angle with respect to the xy axes, construct a new diameter X Y at an angle 2 with respect to XY. Normal and shear stresses are obtained from the coordinates X Y.

24 Problem# 7.02 (BEER) For the state of plane stress shown, (a) construct Mohr s circle, determine (b) the principal planes, (c) the principal stresses, (d) the maximum shearing stress and the corresponding normal stress. SOLUTION: Construction of Mohr s circle x y ave 2 2 CF MPa R CX FX 40 MPa MPa 20 MPa

25 Principal planes and stresses OA OC CA MPa 2 OB OC BC MPa 50 tan 2 2 p p p FX CP Here, σ max = σ 1, σ min = σ 2

26 Maximum shear stress s p 45 max R ave s max 50 MPa 20 MPa

27 Problem# 7.2 (Beer) For the state of stress shown, determine (a) the principal planes and the principal stresses, (b) the stress components exerted on the element obtained by rotating the given element counterclockwise through 30 degrees. SOLUTION: Construct Mohr s circle x y 100 ave R MPa CF FX MPa

28 Principal planes and stresses tan 2 2 p p p XF CF clockwise max OA OC CA max 132 MPa max OA OC min 28 MPa BC

29 Stress components after rotation by 30 o Points X and Y on Mohr s circle that correspond to stress components on the rotated element are obtained by rotating XY counterclockwise through x y xy OK OC KC cos52.6 OL OC CL cos52.6 KX 52sin 52.6 x y xy 48.4 MPa MPa 41.3MPa

30 Strain rosette Strain gauges indicate normal strain through changes in resistance. For a general loading on a body, however, the strains at a point on its free surface are determined using a cluster of three electrical-resistance strain gauges, arranged in a specified pattern. This pattern is referred to as a strain rosette Once the normal strains on the three gauges are measured, the data can then be transformed to specify the state of strain at the point. Strain rosette Strain gauge

31 Strain rosette The axes of the three gauges are arranged at the angles θ a, θ b, θ c as shown in fig. If the readings ε a, ε b, ε c are taken, we can determine the strain components ε x, ε y and γ xy at the point by applying the strain-transformation equation,

32 Strain rosette For rectangular of 45o strain rosette, θ a = 0, θ b = 45, θ c = 90 For 60 strain rosette, θ a = 0, θ b = 60, θ c = 120

33 Strain rosette

34 Strain rosette Principal strains can be found by following equations, Maximum in-plane shear strain can be found by following equations,

35 Strain rosette Stresses on x-y plane can be found by following equations, Once the stresses are known, principal stresses and maximum in-plane shear stresses can be found.

36 Problem#