Worked Examples of mathematics used in Civil Engineering Worked Example 1: Stage 1 Engineering Surveying (CIV_1010) Tutorial - Transition curves and vertical curves. Worked Example 1 draws from CCEA Advanced Subsidiary (As) and Advanced GCE (A2) Mathematics modules; Module M2- Mechanics 2, topic 5 in relation to motion in a horizontal circle Module C2- As core Mathematics 2, topic 1 in relation to the co-ordinate geometry of a circle including use of circle properties Module C2- As core Mathematics 2, topic 3 in relation to radian measure including use for calculation of arc length A transition curve is a curve of gradually varying radius used in highway design to join a straight section of road to a circular curved section. Transition curves are used to reduce the shock lateral loading imposed on the vehicle by allowing the radial force to build up slowly rather than instantaneously. Source: Whyte, W. and Paul, R. (1997, 4 th Edn.). Basic Surveying, Elsevier, London, pp. 289-291. Question: It is required that two intersecting straights are joined by a circular arc and transition curves. Use the following data to select a suitable road alignment: I=43 0 20 10 The minimum radius (R) which can be used is 420m Rate of change of radial acceleration, r= 0.3m/s 3 Deflection angle between 2 tangents, I= 43 0 20 10 Equation relating length of transition curve to angle; so that total length of curve, Design Speed of Road, V= 80km/hr Solution: Change design speed into m/s (M1.1); Δ θ R Δ L = 22.2m/s, V=22.22 m/s 1
Calculate L, where L is total length of each transition curve (M1.1; M2.3; M2.4; M2.5); (Minimum length to satisfy rate of change of radial acceleration) Calculate Δ, corresponding to the total angle traversed by each transition curve (i.e. angle corresponding to L) (C3.2; M2.4; FP2.5): Use transition curve equation; Rearrange to make the subject Apply boundary condition that at, Reducing equation to; Therefore or Total angle corresponding to 2 transition curves= Calculate length of central circular arc (C2.1; C2.3; FP1.6; FP3.7); Or Calculate total arc length if curve totally transitioned (C2.1; C2.3; FP1.6; FP3.7); If curve totally transitioned 2
Worked Example 2: Stage 1 Fluids 1 (CIV_1008) Tutorial- Buoyancy Worked Example 2 draws from CCEA Advanced Subsidiary (As) and Advanced GCE (A2) Mathematics modules; Module M1- Mechanics 1, topic 4 in relation to the equilibrium of a particle Module M1- Mechanics 1, topic 7 in relation to mass and acceleration Stable equilibrium of a floating body, such as a ship, depends on the relative lines of action and resulting moment of the upthrust force (acting upwards) and weight of the body (acting downwards). The weight of the body acts through its centre of gravity which is fixed. Whereas, the upthrust force acts through the centre of buoyancy of the floating body which can move relative to the body. Question: An oil tanker in a state of stable equilibrium can carry 0.5x10 9 kg of oil of relative density 0.85. The ship can be considered as a rectangular in shape, length 380m and width 55m. The mass of the ship is 190x10 6 kg. Calculate the draught of the fully loaded ship in seawater. (, the draught of a vessel is the depth to which it is immersed in the water). When the ship is unloaded, it is necessary to carry seawater ballast in order to keep the propeller submerged. A minimum draught of 20m must be maintained. What volume of seawater must be added to meet this requirement, and what fraction of the ships capacity will be filled with seawater ballast? Solution: For the ship to float (M1.4; M1.5; M3.1); W Calculate the weight of the ship (M1.2; M1.7); d Let d be the draught of the ship in seawater; U 55m Calculate upthrust; 3
But, therefore; Calculate total weight of the vessel to maintain a draught, d, of 20m; The required weight of seawater ballast is; The required volume of seawater is; The maximum volumetric capacity of the ship is; The fraction of the total capacity occupied by the seawater ballast is; 4
Worked Example 3: Adapted from Stage 1 Solids & Structures 1 (CIV 1001 ) 2004 Exam Paper Question 4 Worked Example 3 draws from CCEA Advanced Subsidiary (As) and Advanced GCE (A2) Mathematics modules; Module C2- As Core Mathematics 2, topic 5 in relation to the integration of x n and related sums and differences A cantilever is a beam rigidly secured at only one end. The applied load is carried to the fixed support where it is resisted by bending moment and shear stress. Question: A diver of mass 75 kg stands on the end of a fibre glass (Youngs Modulus, E=8Gpa) diving board, 3 m in length. By modelling the diving board as a simple cantilever calculate the deflection at the free end of the diving board. Repeat the calculation for two other divers with masses of 50 kg and 100 kg. Solution: d=40mm b=300mm P Beam Section x L x -P Shear Force Diagram; where shear force is plotted against length, x, from free end x -PL Bending Moment Diagram; where bending moment is plotted against length, x, from free end Use equation; Where E and I correspond to the youngs modulus and second moment of area, specific properties of the diving board 5
Integrate to get deflected slope of board (C2.5; C4.5); Where c 1 is an arbitrary constant of integration Apply boundary condition when x=l, slope is fixed =0 (FP2.7); Integrate again to get an equation for displacement, v in terms of distance, x from free end of board (C2.5; C4.5); Apply boundary condition when x=l, v =0 (FP2.7); =0 Therefore displacement of diving board at tip, when x=0 is; Substitute diving board properties into deflection equation to determine tip deflection; P=75g N L= 3 m b=300x10-3 m d=40x10-3 m E=8x10 9 kn/m 2 6
Worked Example 4: Stage 1 Solids & Structures (CIV 1001) Tutorial- Pin jointed frames Worked Example 4 draws from CCEA Advanced Subsidiary (As) and Advanced GCE (A2) Mathematics modules; Module M4- Mechanics 4, topic 3 in relation to the analysis of light pin-jointed frameworks Module M1- Mechanics 1, topic 2 in relation to the resolution of component forces Module M1- Mechanics 1, topic 5 in relation to the calculation of the sum of moments about a point 6.0m 4.0m 1.0m 2.0m A B Question: The car has mass 1750 kg and the bridge can be taken to have a self mass of 250 kg per unit length of section for the members making up the deck of the bridge and 100 kg per unit length for the other members. The structure is to be analysed as a pin jointed truss and consequently the loads have to be applied at the joints of the structure. Apply the loads in the usual manner and calculate the resultant horizontal and vertical forces at the restraints. Solution (M4.3): Length/m Weight/kN Deck members 6 15 Other horizontal members 5 6 Diagonal members 5 5 Calculate distribution of loads for the car; 1.0m 2.0m 1.0m 7
R C 8.75kN 8.75kN Resolve (M1.2; M1.5; M1.6; M4.2); R D Take moments about C; Therefore; 5+3 5+3 5+3 5+3 2.5+7.5 7.5+5 7.5+5 7.5+5 2.5+7.5 10.21 7.29 R A R B 8 8 8 8 10 12.5 22.71 19.79 10 R A R B Resolve (M1.2; M1.5; M1.6; M4.2); Take moments about A; Therefore; 8
Worked Example 5: Stage 1 Mathematics (CIV 1015) Mathematics 2C Exam May 2004, Question 1 Worked Example 5 draws from CCEA Advanced Subsidiary (As) and Advanced GCE (A2) Mathematics modules; Module M1- Mechanics 1, topic 1 in relation to the application of differentiation to kinematic problems Module M1- Mechanics 1, topic 7 in relation to the application of Newton s second law of motion Module M3- Mechanics 3, topic 4 in relation to analytically modelling the motion of elastic springs Module FP1- Further Pure Maths 1, topic 1 in relation to the addition and multiplication of matrices Question: For one dimensional simple harmonic motion, the motion for the system can be represented by a second order differential equation (presented below). The equations can be obtained using Newton s second law of motion (F=ma, where; F- Force acting, m-mass, a-acceleration) and Hooke s Law (F=-ku, where; F-Force acting, k-rate of spring constant, u-displacement of spring) (C1.6; C4.4; M1.7; M3.4); Where; m= mass, u= displacement, k= Rate of spring constant and subscripts 1 and 2 denote particles 1 and 2 respectively Given that k=1 and m=1, show that the natural frequencies of vibration are given by; 9
Solution: Represent the two systems of differential equations in matrix form; Where denotes the second order differential equation of displacement, u with respect to time, t When m=1 and k=1; Try a solution of the form; Where denotes the angular frequency Differentiate twice (C3.6; M1.1; FP2.7) Substitute into system of equations (FP1.4; This system will have a non-trivial solution when; Expanding to obtain the characteristic equation; 10