ME 201 Engineering Mechanics: Statics

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1 ME 0 Engineering Mechanics: Statics Unit 9. Moments of nertia Definition of Moments of nertia for Areas Parallel-Axis Theorem for an Area Radius of Gyration of an Area Moments of nertia for Composite Areas

2 Moments of nertia The Moment of nertia,, represents the relative capacity of a section to resist bending or buckling in a direction perpendicular to the neutral axis Based on shape, not on material Also referred to as nd Moment Area x Ay y Ax

3 Moment of nertia of Simple Shapes Rectangle h b x c xc bh Suppose we have a x. Which orientation poses the greater resistance to bending? x c x c xc.67 in xc 0.67 in

4 Moment of nertia of Simple Shapes Triangle h Circle r b x c x c xc xc bh 6 r d 6 Semi-Circle r c y x xc r 8

5 Moment of nertia about Another Axis The Moment of nertia may be computed about another axis using the Parallel Axis Theorem (also called Transfer Formula) x c Ad

6 Parallel Axis Theorem h xc b bh x c x x xc bh bh Ad h bh bh bh

7 Finding of Composite Areas Procedure Divide into numbered, simple areas Find composite centroid Find c of each of each area Use parallel axis theorem to find composite y x

8 Example Problem of Composite Area 0 00 Find: x about centroid 0 0

9 00 0 Find: x about centroid Solution: Find Centroid Example Problem Solution 0 0 A y 0* / 5 A y 0* / 50 X Shape y A ya mm ( by symmetry) ~ ya 6000 Y A 7600 mm

10 00 0 Find: x about centroid Solution: Find Centroid Example Problem Solution 0 0 y y Y 5 x xc Ad mm 0*0 0*00 (600)(5 80.8) (000)( ).E6 mm

11 Radius of Gyration s defined as the distance from its moment of inertia axis at which the entire area could be considered as being concentrated without changing its moment of inertia A compressive member tends to buckle in the direction of its least radius of gyration k A What are the units? in in in

12

13 Centroid Review Find the centroid, y of the shaded area: A.. m B.. m C..57 m D..8 m E. None of the above

14 Moments of nertia The Moment of nertia,, represents the relative capacity of a section to resist bending or buckling in a direction perpendicular to the neutral axis Based on shape, not on material Also referred to as nd Moment Area x Ay y Ax

15 Moment of nertia of Simple Shapes The Moment of nertia for simple geometric shapes about the Centroidal Axis of the shape is shown in the next series of slides Moments of nertia may also be computed about other axes.

16 Moment of nertia of Simple Shapes Rectangle h b x c xc bh Suppose we have a x. Which orientation poses the greater resistance to bending? x c x c xc.67 in xc 0.67 in

17 Moment of nertia of Simple Shapes Triangle h Circle r b x c x c xc xc bh 6 r d 6 Semi-Circle r c y x xc r 8

18 Moment of nertia about Another Axis The Moment of nertia may be computed about another axis using the Parallel Axis Theorem (also called Transfer Formula) x c Ad

19 Parallel Axis Theorem h xc b bh x c x x xc bh bh Ad h bh bh bh

20 Finding of Composite Areas Procedure Divide into numbered, simple areas Find c of each of each area Find composite centroid Use parallel axis theorem to find composite y x

21 Example Problem of Composite Area 0 00 Find: x about centroid 0 0

22 00 0 Find: x about centroid Example 0 0

23 00 0 Find: x about centroid Solution: Find Centroid Example Problem Solution 0 0 A y 0* / 5 A y 0* / 50 X Shape y A ya mm ( by symmetry) ~ ya 6000 Y A 7600 mm

24 00 0 Find: x about centroid Solution: Find Centroid Example Problem Solution 0 0 y y Y 5 x xc Ad mm 0*0 0*00 (600)(5 80.8) (000)( ).E6 mm

25 Radius of Gyration s defined as the distance from its moment of inertia axis at which the entire area could be considered as being concentrated without changing its moment of inertia A compressive member tends to buckle in the direction of its least radius of gyration k A What are the units? in in in

26 n Class Exercise Find: xc & yc (about centroid)

27 Find: xc & yc (about centroid) Solution

28 Find: xc & yc (about centroid) Solution

29 Find: xx, yy k xx, k yy n Class Exercise

30 Find: xx, yy, k xx, k yy Solution

31 Find: xx, yy, k xx, k yy Solution

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