1. CIVE1620- Engineering Mathematics 1.1 Lecturer: Dr Duncan Borman Integration - approaches Integration by parts Applications of integration Lecture 8

2. Essential concept in Civil Engineering; Integration Calculating Pressure on an object Calculating work done Centre of Mass (complex 3D shapes) Work Done = F x D ©Cesar Harada 2009, sourced from http://www.flickr.com/photos/worldworldworld/4095737132/ Available under Creative commons license

3. Essential concept in Engineering; Integration ©Nuala 2005, sourced from http://www.flickr.com/photos/nualabugeye/753268371/ Available under creative commons license Fundamental in: CFD and Finite Element Modelling Calculating Energy and work done Centre of Mass (complex 3D shapes) Work Done = F x D ©Curt Smith 2008, sourced from http://www.flickr.com/photos/curtsm/3005592094/ Available under creative commons license

4. Quick Recap Integration is area under the curve Basic integration- just reverse of differentiation (antiderivative) x 3 + k More complex integration (still reverse differentiation) We need variety of techniques to help us (more of an art to doing these) - not fixed rules for each case as in differentiation - needs some thinking/problem solving ability - this is why integration can seem tough Techniques -Substitution -Trig identities -Integration by Parts -Integration by Partial fractions

5. Integration by Substitution (change of variable) Substitution ? Simplifies a function into a form we can integrate

6. Integrating trig functions Trig Identities 1 - cos2x sin(2x) ==== 2sin 2 x 2sin(x)cos(x) 1 + cos2x=2cos 2 x cosec 2 x=cot 2 x + 1 sec 2 x=tan 2 x + 1 1=sin 2 x + cos 2 x Last lecture we developed this to integrate more complex functions

7. Techniques -Substitution -Trig identities -Integration by Parts -Integration by Partial fractions

8. What about There is no substitution that will simplify this into a form we can integrate Integration by parts Integrate both sides Try this one Let u=x 2 and dv=cos(x)

9. What about There is no substitution that will simplify this into a form we can integrate Integration by parts Integrate both sides Try this one Let u=x 2 and dv=cos(x) Easier version to remember (not as mathematically correct)

10. Integration by parts is particularly useful when our integral is the product of: something that is easy to integrate and a power of x. e.g. (will need to use integration by parts n times) The basic rules for choosing u and are: a) Choose u so that is of a simpler form than u. b) Choose so that it is easily integrated.

11. Multiple choice Choose A,B,C or D for each of these: Find the indefinte integral 1) A B C D

12. Multiple choice Choose A,B,C or D for each of these: Which of these substitutions would be suitable to help integrate 2) A B C D

13. Multiple choice Choose A,B,C or D for each of these: Which trig identity would I be best using to integrate: 3) A B C D

14. Multiple choice Choose A,B,C or D for each of these: Which trig identity would I start with to integrate: 4) A B C D

15. Integration when one “part” is an exponential e.g. Has this helped?? “by parts again”

16. Integration when one “part” is an exponential Has this helped?? Let

17. Integration when one “part” is an exponential Has this helped?? Let

18. APPLICATIONS OF INTEGRATION Physical applications Integration arises naturally in the study of many physical phenomena such as the notions of ‘work’ and ‘force’. These typically involve direct application of the techniques we have covered. Example The total force (in Newtons) acting on the end of a water tank can be shown to be r=2.5m F Where h is the height of water in the tank and r is radius of the tank If the radius of the tank is 2.5m, what is the force due to the water when the height of the water is 4m? r=2.5m y

19. F y 4m m=volume x density m=volume x 1000kg/m 3 y Area of disc 5m We can find volume using integration h volume “Sum up all discs” F=volume x 9800 N F=mg =9.8m N a=9.8m/s 2

20. r=2.5m F 4m y 5m h (m)F (MN) 00 10.067 20.23 30.41 40.57 50.64

21. Question For another geometry water tower the force (in Newtons) acting on the base can be shown to be: What is the force due to the water when the height of the water is 5m? F where h is the height of water in the tank y Use the substitution: Sub into the given equation ©Benton Greene 2008, sourced from http://www.flickr.com/photos/beezum88/3027351751/ Available under creative commons license

22. You can use a similar approach if you want to find volume wine in a wine glass! A cylinder in the middle, + A sphere at either side Propane tank Or a bit more complex - volume of liquid propane in tank © Philip Serracino Inglott 2008, sourced from http://commons.wikimedia.org/wiki/File:Wine_Glass_Splash.JPG Available under creative commons license

23. 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 (Note: larger I means more resistant to bending) ©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

24. B d NA y dy To find the second moment of inertia for the flange section, the following substitution would be used... NA d b The same principles can be applied towards the web of the section dA Now that they have been derived, 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

25. These principles can apply for all different shapes and sizes of beams or columns Circular Hollow sectionRectangular section Triangular section

26. 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

27. 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

28. 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

29. 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)

30. 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:

31. 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:

32. Essential concept in Civil Engineering; Integration Calculating Pressure on an object Calculating work done Centre of Mass (complex 3D shapes) Work Done = F x D

33. The Pelamis Wave Power Generator The Pelamis Wave power generator is a revolutionary piece of technology, capable of taking the kinetic energy generated in waves of water and converting it into electricity. They work by having different sections moving up and down, in a snake like motion, in order to accommodate for the forces from the wave, the hinges then generate the electricity due to the movement. ©Jumanji Solar 2010, sourced from http://www.flickr.com/photos/jumanjisolar/4377449313/ Available under creative commons license ©Jumanji Solar 2004, sourced from http://www.flickr.com/photos/jumanjisolar/4377442735/ Available under creative commons license One ‘wave farm’ of 40 generators is capable of supplying electricity for 20,000 homes. Therefore the amount of energy that is produced is dependant on how much work is done by the force of the wave.

34. Work Done = F x x F=cos( x ).e sin( x ) Distance (m) x F The graph shows the resultant Force applied by a wave on a piston as it moves a distance of one metre (from x =0 to x =1). In this region, the relationship between Force, F and displacement, x is given by the following equation. Force (kN) Calculate the work done by the wave assuming all the energy is used in moving the piston. substitution

35. Integration by parts Integration requires practice – you need experience at types of problem to know which approach to use. Techniques -Substitution -Trig identities -Integration by Parts -Integration by Partial fractions Remember the Mathlab task