TW21 UNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BEng (HONS) CIVIL ENGINEERING SEMESTER 1 EXAMINATION 2016/2017 MATHEMATICS & STRUCTURAL ANALYSIS MODULE NO: CIE4011 Date: Wednesday 11 th January 2017 Time: 10:00 am 1:00 pm INSTRUCTIONS TO CANDIDATES: There are FIVE questions. Answer ALL FIVE questions. Marks for parts of questions are shown in brackets. This examination paper carries a total of 100 marks. Supplementary formulae sheet is provided on page 8 at the rear of the question paper. Lined Graph Paper is available for use. All working must be shown. A numerical solution to a question obtained by programming an electronic calculator will not be accepted.
Page 2 of 9 SECTION A: STRUCTURAL ANALYSIS Question 1 (a) Figure Q1 (a) shows a simply supported beam with a pin support at A, and a roller support at D. The beam is carrying one vertical point load at C, and a uniform distributed load (UDL) between B and C. (i) Calculate and state the support reactions at A and D. (2 marks) (ii) Daw the Shear Force and Bending Moment Diagrams, showing important values along the beam and state the value of the maximum bending moment and its position along the beam. (8 marks) 40 kn/m 20 kn A 2m B C D 3m 3m 8m Figure Q1 (a) (b) For the pin jointed structure shown in Figure Q1 (b): (i) Calculate and state the support reactions at A and E. (3 marks) (ii) Calculate the axial forces in members AC, CE, and ED and state whether each axial force is in tension or compression. (7 marks) B C 45 0 D 60 kn 4 m A E 4m 4m Figure Q1 (b) Total 20 marks
Page 3 of 9 School of Engineering Please turn the page
Question 2 Page 4 of 9 80mm 50mm 15mm 140mm 25mm 200mm Figure Q2 (i) Figure Q2 (i) shows a cross-section of an asymmetrical steel beam. a) Determine the position of the horizontal neutral axis of the beam. (6 marks) b) What is the value of the second moment of area I about the horizontal neutral axis of the beam section? (9 marks) Question 2 continued over the page
Question 2 continued Page 5 of 9 100mm A 40mm 25mm 250mm B 5 3.5m Figure Q2 (ii) Section through cantilever tee beam Figure Q2 (iii) Elevation on cantilever tee beam Figure Q2 (ii) shows a cross-section of a different asymmetrical cast iron tee beam with a cantilever span of 3.5m; also see elevation in Figure Q2 (iii). The allowable bending stresses in the tee beam are shown in the table below: Maximum stress (N/mm 2 ) Tension 23 Compression 105 The geometrical properties of the tee beam are shown in the table below: Distance of the horizontal neutral axis of the tee beam above 181.6 mm the bottom of the section Second moment of area (I) 8437 cm 4 c) What is the maximum force A that can be applied vertically downward to the cantilever tee beam without exceeding the allowable bending stress in the tee beam (ignore force B)? (9 marks) d) What is the maximum force B that can be applied vertically upward to the cantilever tee beam without exceeding the allowable bending stress in the tee beam (ignore force A)? (6 marks) END OF SECTION A Total 30 marks
Page 6 of 9 Question 3 SECTION B: MATHEMATICS (a) The maximum deflection of a cantilever beam under uniformly varying load is given by: y = wl4 8EI (i) Rearrange the equation by transposition to find an expression for w. (2 marks) (ii) If the deflection of the beam (y) is limited to 13 mm, the elastic modulus (E) is 200 kn/mm 2, the second moment of area (I) is 2356 cm 4 and the length of the beam (L) is 3.0 m, what is the magnitude of the distributed load (w) in (kn/m)? (6 marks) (b) A heating installation for one house consists of 6 radiators and 7 convertor heaters and the cost, including labour, is 3126. In a second house 5 radiators and 6 convertor heaters are used and the cost, including labour, is 2730. In each house the installation costs are 450. Find the cost of a radiator and the cost of a convertor heater. (6 marks) (c) Figure Q3c shows a template whose area is 1378.2 cm 2. Find the value of r. Figure Q3(c) r 60 cm (6 marks) Total 20 marks Please turn the page
f(x) School of Engineering Question 4 Page 7 of 9 A function can be expressed by the following equation: f(x) = 12 + x 2 6x With the same function being graphically represented in Figure Q4. 14 12 10 8 6 4 2 0 0 1 2 3 4 x Figure Q4 (a) Find the approximate area enclosed by the function, the x-axis, and the ordinates at x = 0 and x = 4, using: (i) Trapezoidal rule (4 strips) (4 marks) (ii) Simpson s rule (4 strips) (4 marks) (iii) Simpson s rule (8 strips) (5 marks) (b) Compare your answers and comment on your findings. (2 marks) Total 15 marks Please turn the page
Question 5 Page 8 of 9 A round steel specimen, as shown in Figure Q5, is loaded in tension in compliance with BS EN ISO 377:2013. From the test data, the stress-strain relationship of this elastic material is given in Table Q5. Table Q5 Figure Q5 Strain (x 10-5 ) Stress (N/mm 2 ) 0 0 40.12 92.946 61.64 134.375 81.53 176.606 94.47 205.945 105.38 229.728 123.55 266.015 156.82 337.537 (a) Using the graph paper provided, plot stress versus strain on an appropriately scaled axis. (4 marks) (b) Draw an appropriate trend-line through the points. (4 marks) (c) Use the plot (a) to generate the linear function for the data set. (4 marks) (d) Using the function from (c), give a value for the elastic modulus of steel. (3 marks) Total 15 marks END OF EXAM Please turn the page (for Supplementary Formulae Sheets)
Page 9 of 9 Formula sheet for structural analysis Simply supported and cantilever beams Typical units Mmax (knm) Rsupport (kn) Deflectionmax (mm) Simply supported beam length L Point load P at centre PL 4 P 2 PL 3 48EI Simply supported beam length L UDL w along full length wl 2 8 wl 2 5wL 4 384EI Cantilever beam length L Point load P at tip PL P PL 3 3EI Cantilever beam length L UDL w along full length wl 2 2 wl wl 4 8EI Shape properties Typical units Rectangle with side lengths b and h Area A (mm 2 ) Elastic section modulus Wel (mm 3 ) bh bh 2 6 Plastic section modulus Wpl (mm 3 ) bh 2 4 2nd moment of area I (mm 4 ) bh 3 12 Stresses Bending stress = My I Bending stress = M S Bending stress = M z Radius of gyration = I A Axial stress = P A Average shear stress = V A Complex shapes Centroid Parallel axes theorem x = x ia i A i and y = y ia i A i I xx = (I 0 + Ay 2 )
Page 10 of 9 Formulae sheet