1. CIVE1620 - Engineering Mathematics 1.1 Length of a curve Double Integrals - for calculating areas - for calculating volumes (Estimating errors using the differential) Lecture 11

2. Length of a curve A B S Q: Find length of the curve between x=-1 and x=2 for: Example: Find length of the curve between x=0 and x=4 for: Identity you may want to use:

3. Length of a curve A B S Q: Find length of the curve between x=-1 and x=2 for: Example: Find length of the curve between x=0 and x=4 for: Identity you may want to use:

4. Q: Find length of the curve between x=-1 and x=2 for: Identity you may want to use:

5. Double Integrals

6. Quick reminder

7. Double Integrals – Area between lines

9. Find area under the curve, and the

10. Double Integrals – Area between lines 0 5 Find area bounded by, the x-axis and the line x=5.

11. Double Integrals – Area between lines Find area under the curve and

12. Multiple choice Choose A,B,C or D for each of these: Differentiating wrt x 1) A B C D

13. Multiple choice Choose A,B,C or D for each of these: Differentiating wrt x 2) A B C D

14. Multiple choice Choose A,B,C or D for each of these: Integrate the following 3) A B C D

15. a b y x y = f(x) r h dx f(x) The volume of revolution Double Integrals- Volume of more complex shapes

16. Functions that depend on one dimension (1-D) (eg. x ) Area under the curve What does Integration mean in 1D?

17. Functions that depend on one dimension (1-D) (eg. x ) Functions that depend on two or more dimensions (2-D) or (3-D) (eg. x,y,z ) In 1-D y= f(x) is a curve In 2-D z=f(x,y) is a surface

18. What does Integration mean in 2-D? 1D =Area under the curve 2D =Volume under the curve We need to use double or triple integrals for this

19. Integration (1D) – AREA UNDER A CURVE y=f(x)

20. = volume below graph z = f(x,y) over plane region R. Double Integrals

21. Double Integrals over a Rectangular Region

22. Find the volume under the surface for f(x,y) for x=0 to 2y and y=0 to 1 where:

23. Find the volume under the surface for f(x,y) for x=0 to 1 and y=0 to x where:

25. Second Moment of Inertia Calculating Beam deflection is an important part of engineering, it can be used to find deflections amongst others in key member analysis. A key factor in these calculations is the second moment of inertia (I), this is dependant upon the shape of the section. For example... ( i.e. I = a property of a shape that can be used to predict deflections) H NA The General formula to find the second moment of area is... y Therefore Integration can be used to find the formulae for the second moment of inertia for each type of section ©Elliot Brown 2010, sourced from http://www.flickr.com/photos/ell-r-brown/4586752387/ Available under creative commons licence ©Daily Sunny 2006, sourced from http://www.flickr.com/photos/53558245@N02/4978362207/ Available under creative commons licence dy dA

26. B d NA y To find the second moment of inertia for the flange section, the following substitution would be used... NA h b The same principles can be applied towards the web of the section dA Now that they have been derived using integration, these formulae can be combined to find the overall second moment of inertia of the I or H section, an important aspect in beam analysis Using integration to derive Second moment of inertia equation

27. These principles can apply for all different shapes and sizes of beams or columns (in the end we normally look this up in Eurocodes – but should know where they come from) Circular Hollow sectionRectangular section Triangular section

28. Second moment of inertia can be used in various important engineering calculations. For example, the deflection of a beam can be found by integrating the following formula: Where: R = The radius of the curvature beam at a distance x from the origin ©Les Chatfield 2006, sourced from http://www.flickr.com/photos/elsie/172754915/ Available under creative commons license ©John Keogh 2004, sourced from http://www.flickr.com/photos/jvk/2584122/ Available under creative commons license E = The Elastic or Young’s modulus of the beam. Can be assumed to be 210 kN/mm2. I = The Second Moment of Area of the beam’s cross-section. This value can be found using the formulae found using principles covered in past slides M = The Bending Moment at the section, distance x from the origin v = The vertical deflection at the section distance x from the origin Remember: Curvature of a ‘curve’ y=f(x) is equivalent to f’’(x)

29. Example: For the beam shown below, find in terms of x and EI, the deflection of the beam: Bending moment (M) can be found: Integrating once gives gradient: And again gives deflection, v: from Dr Sheng’s lectures Load = 20kN/m: Total load =100kN Reaction =50kN We can find the deflection of a beam integrating the following formula:

30. After combining the equations, it can be shown that: We assume no deflection (v=0) when at x=0 and x=5: Hence, after integration, the formula for the deflection, v, in the beam is: (with loading of 10kN/m from weight) Using integration to find the formula for deflection in a beam Finding constants of integration A&B When v=0, x=0: When v=0, x=5:

31. Ideas can be developed to triple Integrals http://www.youtube.com/watch?v=OuKYSZ8g_mM http://www.youtube.com/watch?v=vxQvL_WhBGU

32. Length of a curve Double integrals - alternative method to find area under a curve (1D) - Volume under a surface