1 - Geomertry of Areas CE14 Lecture Slides

1. Introduction to Geometry of Areas

In civil engineering, geometry of areas is a fundamental topic that bridges mathematics and engineering applications. Whether we are analyzing the strength of a beam, determining the center of gravity of a structural section, computing hydraulic properties of a river channel, or estimating the cross-sectional area of an embankment, engineers rely on accurate determination of areas and their geometric properties.

The study of geometry of areas focuses not only on measuring areas of plane figures but also on calculating related geometric characteristics such as centroids and moments of inertia. These properties serve as the foundation of structural analysis, material mechanics, hydraulics, geotechnical engineering, and surveying.

For instance:

In structural engineering, the bending stress in a beam depends on the second moment of area of its cross-section.

In hydraulics, the flow capacity of a channel depends on its cross-sectional area and wetted perimeter.

In surveying and geodesy, land areas are calculated using geometric methods.

Thus, Geometry of Areas is both a theoretical and practical tool, connecting mathematics to the real-world problems encountered by civil engineers.

1. CE 14 Solid Mechanics

2. I . SPECIAL TOPICS IN FLEXURE A. Geometry of Areas Definition: Laws of Transformation and Product of inertia Parallel Axis Theorem on Product of Inertia Moments of Inertia About an Inclined Axes Principal Moments of Inertia Mohr’s Circle for Second Moment of Areas B. Flexural Stresses : Unsymmetrical Bending, Inelastic Action, Curved Beams, Non- homogeneous beams by transformed section, Reinforced Concrete beams by transformed sections Unsymmetrical Bending about the Horizontal and Vertical Axes of the Cross Section Unsymmetrical Bending about the Principal Axes

3. I.A Geometry of Areas

4. Product of Inertia for an Area Consider the figure shown below A dA y x y x Product of Inertia of Element dA : Product of Inertia of A wrt x and y axis:

5. Product of Inertia for an Area Consider the figure shown below A dA y x y x Product of Inertia of A wrt x and y axis: NOTE: 1. I xy can be positive, negative or zero. Unit: length 4 – m 4 , mm 4 , ft 4 , in 4 2. The product of inertia of an area wrt any two orthogonal axes is zero when either of the axes is an axis of symmetry.

6. Product of Inertia for an Area Parallel-Axis Theorem dA y’ x’ y x Product of Inertia of Element dA : Product of Inertia of A wrt x and y axis: C x’ y’ d x d y

7. Product of Inertia for an Area Parallel-Axis Theorem dA y’ x’ y x The product of inertia of an area wrt any two perpendicular axes x and y is equal to the product of inertia of the area wrt a pair of centroidal axes parallel to the x and y axes added to the product of the area and the two centroidal distances from the x and y axes. C x’ y’ d x d y

8. Product of Inertia for an Area Example 1 y x b h Determine the following: Product of Inertia, I xy Product of Inertia, I x’y’ , with respect to a pair of centroidal axes x’ and y’ parallel to the given axes x and y

9. Product of Inertia for an Area Example 1 y x b h Solution: Product of Inertia, I xy dA x y dy Consider the strip The area, dA , is equal to Substituting gives

10. Product of Inertia for an Area Example 1 y x b h Solution: Product of Inertia, I xy dA x y dy But x is a function of y, and using similar triangles Substitute x to dI xy gives

11. Product of Inertia for an Area Example 1 y x b h Solution: Product of Inertia, I xy dA x y dy Integrating

12. Product of Inertia for an Area Example 1 y x b/3 2h/3 Solution: b) Product of Inertia, I x’y’ Parallel Axes Theorem C x’ y’

13. Moments of Inertia about Inclined/rotated Axes dA y cos  y’ y x  x’ y’ y x   x sin  x’ x cos  y sin  Transformation Equations: Moments and Product of Inertia of dA wrt x’ and y’ axes:

14. Moments of Inertia about Inclined Axes Expanding and integrating each expression and realizing that These equations may be simplified using the trigonometric identities Gives

15. Moments of Inertia about Inclined/rotated Axes Which then gives A  x’ y’ y x  Moments of Inertia of An Area About an Inclined Axes x’ and y’ in terms of I x , I y , I xy and 

16. Moments of Inertia about Inclined Axes Adding the first and second equations A x’ y’ y x The sum (also called the polar moment of inertia) I x’ + I y’ is a constant. Since sum is constant, I x’ will be maximum and the corresponding I y’ will be minimum for one particular value of . y’’ x’’ o

17. Principal Moments of Inertia First Objective: Determine the orientation of the axes or angle  which the moments I x’ and I y’ are maximum or minimum. A  x’ y x  y’ DEFINITION: Principal Axes of Inertia – the set of axes for which the second moment of areas are maximum or minimum Principal Moments of Inertia – the second moments of the area with respect to the principal axes of the area designates as I max and I min .

18. Principal Moments of Inertia Consider the following derived equation of moments of inertia I x’ Differentiate with respect to , then set to zero gives Where  p represents two values of  that are 90 degrees apart that locate the two principal axes.

19. Second Objective: Determine the equations for the principal moments of inertia I max and I min . Principal Moments of Inertia Substitute the above sine and cosine equations in I x’

20. Principal Moments of Inertia Gives the equations for the principal moments of inertia of an area Principal Moments of Inertia in terms of I x , I y and I xy

21. Principal Moments of Inertia Now, consider the following derived equation of moments of inertia I x’y’ Substitute in the above the derived equations for sine and cosine gives, The product of inertia with respect to the principal axes is zero. Since the product of inertia is zero with respect to any axis of symmetry, it follows that any axis of symmetry must be a principal axis for any point on the axis of symmetry.

22. Mohr’s Circle for Moments of Inertia Involves the construction of a circle in a rectangular coordinate system such that the abscissa represents the moment of inertia I , and the ordinate represents the product of inertia I xy . (abscissa) (ordinate) C R Each point on the circle represents (I x or I y , I xy )

23. Mohr’s Circle for Moments of Inertia Consider again the derived equations for the moments of inertia Squaring the first and third equations and adding, it is found that Equation of a Circle Center C at Radius =

24. Mohr’s Circle for Moments of Inertia PROCEDURE FOR ANALYSIS 1. Compute I x , I y and I xy with respect to the given x and y axes of the area. 2 . Construct the Circle. 3 . Determine the principal moments of inertia, I max and I min . C R A . Establish the rectangular coordinate system. B. Determine the center of circle C. Plot the reference point A having coordinates (I x , I xy ) A (I x , I xy ) D. Connect A and C and determine the distance by trigonometry. This distance represents the radius. E . Draw the circle. I max I min

25. Mohr’s Circle for Moments of Inertia Example 2 60 mm C x y 80 mm 26.54 mm 16.54 mm 10 mm 10 mm Determine the Principal Moments of Inertia using Mohr’s Circle.

26. Mohr’s Circle for Moments of Inertia Example 2 60 mm C x y 80 mm 26.54 mm 16.54 mm 10 mm 10 mm Solution: STEP 1 : Compute for I x , I y , I xy

27. Mohr’s Circle for Moments of Inertia Example 2 60 mm C x y 80 mm 26.54 mm 16.54 mm 10 mm 10 mm Solution: STEP 1 : Compute for I x , I y , I xy But Because the centroidal axis of a rectangle is an axis of symmetry

28. Mohr’s Circle for Moments of Inertia Example 2 Solution: STEP 2 : Construct Mohr’s Circle A B -32.3 80.8 32.3 38.8

29. Mohr’s Circle for Moments of Inertia Example 2 Solution: STEP 2 : Construct Mohr’s Circle C A B 2  R Connect A and B to establish the location of C -32.3 80.8 32.3 38.8

30. I max I min Mohr’s Circle for Moments of Inertia Example 2 Solution: STEP 3 : Compute I max and I min C A B 2  R -32.3 80.8 32.3 38.8

31. I max I min Mohr’s Circle for Moments of Inertia Example 2 Solution: STEP 3 : Compute I max and I min C A B 2  R -32.3 80.8 32.3 38.8

32. Mohr’s Circle for Moments of Inertia Example 2 Solution: STEP 3 : Compute I max and I min Counterclockwise from x-axis 60 mm C x y 80 mm 26.54 mm 16.54 mm 10 mm 10 mm max min 

33. B. FLEXURAL STRESSES B.1 Unsymmetrical Bending

34. Unsymmetrical Bending RECALL: The discussion in ES 13 were limited to beams with at least one longitudinal plane of symmetry and with the load applied in the plane of symmetry . Plane x-y is the plane of symmetry in the figures shown.

35. Unsymmetrical Bending NEW TOPIC: In CE 14 , pure bending (bent with couples only; no transverse forces) of 1) Beams with a plane of symmetry but with the load (couple) applied not in or parallel to the plane of symmetry. 2) Beams with no plane of symmetry.

36. Unsymmetrical Bending COMPARISON ES 13 CE 14 Neutral Axis is the Centroidal Axis. Where is the location of neutral axis?

37. Unsymmetrical Bending Consider a beam with unsymmetrical cross section loaded with a couple M in a plane making an angle  with the xy plane, x y z Plane of loading M n.a . n.a . Neutral Surface y z Assumptions 1) Beam is straight and of uniform cross section. 2) A plane cross section remains a plane after bending. 3) Material and behavior /Action is linearly elastic. C

38. Unsymmetrical Bending Note: Since the orientation of the neutral axis is not known, the flexural stress distribution function cannot be expressed in terms of one variable as before in ES 13 . x y z Plane of loading M n.a . n.a . Neutral Surface y z C Not Applicable! However, since the plane section remains plane, the stress variation can be written as (equation 1)

39. Unsymmetrical Bending x y z Plane of loading M n.a . n.a . Neutral Surface y z C For Equilibrium a must be equal to zero

40. Unsymmetrical Bending y z M n.a . n.a . C

41. Unsymmetrical Bending y z M n.a . n.a . C

42. Unsymmetrical Bending y z M n.a . n.a . C where I y and I z are the moments of inertia of the cross sectional area wrt the y and z axes, and I yz is the product of inertia. T he components of the applied moment are equal to: M z =+ Mcos (α) M y =- Msin (α)

43. Unsymmetrical Bending O r rearranging T he flexural loadings stresses for unsymmetrical bending of beams Sign Convention +  = TENSILE - = COMPRESSIVE

44. Unsymmetrical Bending To determine the orientation of the neutral axis, equate the flexural stress formula to zero, which then gives SIGN CONVENTION 1) M is positive when M cos  = M z produces tension in the bottom of the beam. 2) Since y is taken as positive downwards,  and  (measured from z) are positive clockwise.

45. Unsymmetrical Bending If we select y and z axes to be principal axes Y and Z , I YZ = 0 , the formulas reduces to NOTE: The last equation indicates that the neutral axis N.A. is perpendicular to the plane of loading when, 1) the angle  is zero, in which case the plane of loading is (or is parallel) a principal plane. 2) I Z = I Y which is a special case that is true for circles, squares, etc.

46. Unsymmetrical Bending Example 3 240 mm z 90 mm 60 mm 150 mm y 30 mm a a M R A beam with the T-cross-section shown is subjeted to a moment M w/c has a magnitude of 20 kN-m. The resisting moment makes an angle alpha=10 o w rt the z-axis, as shown. T he orientation of the neutral axis T he max. tensile and compressive flexural stresses in the beam

47. Unsymmetrical Bending Example 4 6” z 4” 1” 1” y 1” a a M 3 4 Given: M = +10,000 in-lb applied in a-a Determine: a) The magnitude of the maximum flexural stress b) The orientation of the neutral axis

48. Unsymmetrical Bending Example 4 6” z 4” 1” 1” y 1” a a M 3 4 Solution: IMPORTANT! Note that y and z are axes of symmetry, therefore they are PRINCIPAL AXES. We can use Y Z

49. Unsymmetrical Bending Example 4 6” z 4” 1” 1” y 1” Solution: Y Z Compute I Y and I Z

50. Unsymmetrical Bending Example 4 6” z 4” 1” 1” y 1” Solution: Y Z Compute M Y and M Z a a M 3 4   M Z M Y

51. Unsymmetrical Bending Example 4 6” z 4” 1” 1” y 1” Solution: Y Z Compute Maximum Flexural Stress a a M 3 4   This will occur at the farthest point from NA, say point A A NA (assumed orientation)

52. Unsymmetrical Bending Example 4 6” z 4” 1” 1” y 1” Solution: Y Z Compute Maximum Flexural Stress at A a a M 3 4   A NA (assumed orientation)

53. Unsymmetrical Bending Example 4 6” z 4” 1” 1” y 1” Solution: Y Z Orientation of NA a a M 3 4   A  Counterclockwise from Z (or z): NA

54. Unsymmetrical Bending Example 4 NOTE: In this example, it is better to solve first for the orientation, , of the NA rather than assuming its orientation. Afterwards, determine the farthest point from NA, then solve for the fiber stress, , at this point. 6” z 4” 1” 1” y 1” Y Z a a M 3 4   A  NA

55. Unsymmetrical Bending Example 4 1519.5 psi (T) NA 1519.5 psi (C) z A Fiber Stress Distribution,  : y

56. Unsymmetrical Bending Example 5 Given: L8x8x1 in M = +7,500 ft-lb applied in xy - plane I y = I z = 89 in 4 and I min = 36.5 in 4 Determine: A ) The flexural stress at A B ) The maximum flexural stress and its location on the cross section C ) The orientation of the neutral axis z y 8” 2.37” A M

57. Unsymmetrical Bending Example 5 z y 8” 2.37” A Solution: Note that y and/or z are NOT axes of symmetry, so we will use

58. Unsymmetrical Bending Example 5 Solution: Compute I yz (Since I y and I z are already given) I min C (89, I yz ) 89 36.5 (89, - I yz ) A B Mohr’s Circle Given Points Line Segment AB is a diameter of Mohr’s Circle. The intersection with I-axis locates the Center C. Therefore,

59. Unsymmetrical Bending Example 5 Solution: Compute I yz (Since I y and I z are already given) I min C (89, I yz ) 89 36.5 (89, - I yz ) A B Since R I yz From the figure,

60. Unsymmetrical Bending Example 5 Solution: Compute M y and M z Since M is applied at xy -plane z y 8” 2.37” A This simplifies the fiber stress equation to

61. Unsymmetrical Bending Example 5 Solution: A) Flexural Stress at A Coordinates of A are z y 8” 2.37” A Solving,

62. Unsymmetrical Bending Example 5 Solution: B ) Maximum Flexural Stress and its Location Solve first for the orientation of NA (to be able to locate farthest point) z y 8” 2.37” A Since M y is zero, the equation becomes  NA

63. Unsymmetrical Bending Example 5 Solution: B ) Maximum Flexural Stress and its Location Farthest could be A or B from NA. Solving for the fiber stress at B z y 8” 2.37” A  NA B

64. Unsymmetrical Bending Example 5 Solution: C ) The Orientation of the NA and Show its Location z y 8” 2.37” A From B)  NA Clockwise from z-axis

65. The End