Principal stresses, maximum shear stresses & Mohrs circle

Bending and shear stresses Tension or compression

Combination of stresses - RevisionConsider a shaft (as in most machine parts) subjected to a combination of bending moment M, vertical force F, axial force P, and torque T.

Failure of materialFailure of brittle material in torsionFor brittle material: tension

xxyxyxyyStresses on oblique sectionsConsider an element under plane (normal) stress and shear stresses as shown

Principal stresses deriving equationsConsider normal stress along the normal direction:In order to determine equilibrium of the oblique section along n-direction, all stresses have to be converted to forces first :Some examples: Forces from n-axis: n dAn Forces from x-axis: x dAx x dAn cos Forces from y-axis: y dAy y dAn sin

Principal stresses deriving equations (2)Complete equilibrium of forces in the n-direction:n dAn = y dAy sin + x dAx cos - xy dAx sin - xy dAy cos

Substituting for dAx and dAy :n dAn = y dAn sin2 + x dAn cos2 - 2xy dAn sin cos and removing dAn :n = y sin2 + x cos2 - 2xy sin cos Modifying gives :n = y [(1 - cos2)] + x [(1 + cos2)] - xy sin 2

Further simplification :n = [x + y] + [x - y] cos2 + xy sin 2

Principal stresses deriving equations (3)To obtain equation for direction of maximum normal stresses :From the previously derived equation :n = [x + y] + [x - y] cos2 + xy sin 2

where is the angle in which (n )max or (n )min (or Principal stresses) is acting on an element under stress

Principal stresses deriving equations (4)Features of equation for Tan (2):(n)max and (n)min are called the Principal Stresses

Principal stresses deriving equations (5)Plotting Tan(2) vs :

Maximum shear stresses deriving equationsConsider shear stress along the plane A-B:xxynyn-directionABnxzyIn order to determine equilibrium of the oblique section along A-B plane, all stresses have to be converted to forces :Examples: Forces from n-axis: n dAn Forces from x-axis: x dAx x dAn cos Forces from y-axis: y dAy y dAn sin

Maximum shear stresses deriving equations (2)Complete equilibrium of forces in the A-B plane:n dAn = -x dAx sin + y dAy cos + xy dAx cos - xy dAy sin

Substituting for dAx and dAy :n dAn = -x dAn sin cos + y dAn sin cos + xy dAn cos2 - xy dAn sin2 and removing dAn :n = -x sin cos + y sin cos + xy cos2 - xy sin2

ABdAndAydAxnxxyynModifying gives :n = - (x - y ) sin cos + xy (cos2 - sin2 )

Further simplification :n = - [x - y] sin 2 + xy cos 2

Maximum shear stresses deriving equations (3)To obtain equation for direction of maximum shear stresses :From the previously derived equation : n = [x - y] sin 2 + xy cos 2

where ( - ) is the angle in which max or min is acting on an element under stress

Maximum shear stresses deriving equations (4)Features of equation for Cot (2):max and minare the Maximum and Minimum ShearStresses

Maximum shear stresses deriving equations (5)Plotting Cot(2) vs :

Equation of circleGeneral equation of circle: (x xo)2 + (y yo)2 = r2(xo, yo) is the centre of circler is the radius of circle

Apply equation of circle to a plot of vs (o, o) is the centre of circler is the radius of circle

Equation of circle (2)Referring to the sets of equations for principal stresses and maximum shear stresses:

Graphical representationThe stresses at any angle may be represented graphically by Mohrs stress circle as shown:

Graphical representation (2)Orientation to principal stress plane or directionNote: A rotation of angle in the element invokes a corresponding rotation of angle 2 in the Mohrs circle

Graphical representation (3)Orientation to maximum shear stress plane or directionNote: A rotation of angle B in the element invokes a corresponding rotation of angle 2B in the Mohrs circle

Mohrs stress circleWorked exampleDraw the Mohrs stress circle for the following cases and evaluate:(a) Principal stresses(b) Maximum shear stresses

Case 1: x = 10 MPa; y = 5 MPa; xy = 5 MPaCase 2: x = -5 MPa; y = 15 MPa; xy = -5 MPa

Mohrs stress circle (2)Solution (for Case 1):Construction of Mohrs circle:Case 1: x = 10 MPa; y = 5 MPa; xy = 5 MPa

Mohrs stress circle (3)Solution (for Case 1)Principal stresses

Mohrs stress circle (4)Solution (for Case 1)Maximum shear stresses

Mohrs stress circle (5)Solution (for Case 2):Construction of Mohrs circle:Case 2: x = - 5 MPa; y = 15 MPa; xy = - 5 MPa

Mohrs stress circle (6)Solution (for Case 2)Principal stresses

Mohrs stress circle (7)Solution (for Case 2)Maximum shear stresses

Summary - 1an angle p where the shear stress x'y' becomes zero, i.e. principal stresses acts:

the principal directions where the only stresses are normal stresses, given by:

Summary - 2The transformation to the principal directions can be illustrated as:

Summary - 3Angle, s, is where the maximum shear stress max occurs, given by:

The maximum shear stress is equal to one-half the difference between the two principal stresses

Summary - 4The transformation to the maximum shear stress direction can be illustrated as:

Summary - 5Mohr's circles representing different stress regimes are shown below:

3D Mohrs circle - 3 Dimensional Systems solids are actually three dimensional, and there are always three principal stresses - (though one of them may be zero)

3D Mohrs circleExample:

3D Mohrs CirclePure shearUniaxial tensionUniaxial compressionPlane strainPlane stressTriaxial tensionHydrostatic compression

End of this lecture

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