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# 6161103 10.8 mohr’s circle for moments of inertia

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### 6161103 10.8 mohr’s circle for moments of inertia

1. 1. 10.8 Mohr’s Circle for Moments of InertiaIt can be found that 2 2  Ix + I y   Ix − Iy   Iu −   + I uv =  2  + I xy 2  2    2   In a given problem, Iu and Iv are variables and Ix,Iy and Ixy are known constants (I u − a )2 + I uv = R 2 2When this equation is plotted on a set of axesthat represent the respective moment of inertiaand the product of inertia, the resulting graphrepresents a circle
2. 2. 10.8 Mohr’s Circle for Moments of InertiaThe circle constructed is known as a Mohr’scircle with radius 2  Ix + Iy R=   2   + I xy 2  and center at (a, 0) wherea = (I x + I y ) / 2
3. 3. 10.8 Mohr’s Circle for Moments of InertiaProcedure for AnalysisDetermine Ix, Iy and Ixy Establish the x, y axes for the area, with the origin located at point P of interest and determine Ix, Iy and IxyConstruct the Circle Construct a rectangular coordinate system such that the abscissa represents the moment of inertia I and the ordinate represent the product of inertia Ixy
4. 4. 10.8 Mohr’s Circle for Moments of InertiaProcedure for AnalysisConstruct the Circle Determine center of the circle O, which is located at a distance (Ix + Iy)/2 from the origin, and plot the reference point a having coordinates (Ix, Ixy) By definition, Ix is always positive, whereas Ixy will either be positive or negative Connect the reference point A with the center of the circle and determine distance OA (radius of the circle) by trigonometry Draw the circle
5. 5. 10.8 Mohr’s Circle for Moments of InertiaProcedure for AnalysisPrincipal of Moments of Inertia Points where the circle intersects the abscissa give the values of the principle moments of inertia Imin and Imax Product of inertia will be zero at these pointsPrinciple Axes To find direction of major principal axis, determine by trigonometry, angle 2θp1, measured from the radius OA to the positive I axis
6. 6. 10.8 Mohr’s Circle for Moments of InertiaProcedure for AnalysisPrinciple Axes This angle represent twice the angle from the x axis to the area in question to the axis of maximum moment of inertia Imax Both the angle on the circle, 2θp1, and the angle to the axis on the area, θp1must be measured in the same sense The axis for the minimum moment of inertia Imin is perpendicular to the axis for Imax
7. 7. 10.8 Mohr’s Circle for Moments of InertiaExample 10.10Using Mohr’s circle, determine the principlemoments of the beam’s cross-sectional areawith respect to an axispassing through thecentroid.
8. 8. 10.8 Mohr’s Circle for Moments of InertiaSolutionDetermine Ix, Iy and Ixy Moments of inertia and the product of inertia have been determined in previous examples ( ) I x = 2.90 109 mm4 I y = 5.60(109 )mm4 I xy = −3.00(109 )mm4Construct the Circle Center of circle, O, lies from the origin, at a distance (I x + I y )/ 2 = (2.90 + 5.60) / 2 = 4.25
9. 9. 10.8 Mohr’s Circle for Moments of InertiaSolution With reference point A (2.90, -3.00) connected to point O, radius OA is determined using Pythagorean theorem OA = (1.35)2 + (− 3.00)2 = 3.29Principal Moments of Inertia Circle intersects I axis at points (7.54, 0) and (0.960, 0)
10. 10. 10.8 Mohr’s Circle for Moments of InertiaSolution ( ) I max = 7.54 109 mm4 ( ) I min = 0.960 109 mm 4Principal Axes Angle 2θp1 is determined from the circle by measuring CCW from OA to the direction of the positive I axis −1 BA  2θ p1 = 180 − sin  o  OA     o −1 3.00  o = 180 − sin   = 114.2  3.29 
11. 11. 10.8 Mohr’s Circle for Moments of InertiaSolution The principal axis for Imax = 7.54(109) mm4 is therefore orientated at an angle θp1 = 57.1°, measured CCW from the positive x axis to the positive u axis v axis is perpendicular to this axis