solution manual Civil Engineering Materials 1st edition Peter A. Claisse complete chapters

SOLUTION MANUAL

Instructor’s Manual Part 2: Solutions to Tutorial Questions

Introduction This part of the instructor’s manual presents the solutions to the tutorial questions which are at the end of the book. All of the solutions to the questions at the end of the individual chapters are given with the questions.

The majority of these questions were originally set in exams. Typically in a two hour exam five questions are set and the students are required to answer three of them. In a three hour exam they answer five out of eight. An able student can often complete their solutions in far less time than this but some, in particular those whose first language is not English, need all the time.

Solutions to essay questions should typically be about 600 words. Students are advised that there is no need for an introduction (e.g. copying out the question) or conclusions. Marks are simply awarded for each valid point made. Obviously the notes given here do not constitute a full essay solution; they are only intended to give guidance on the directions the discussions should take.

Knowledge of material from other chapters has been assumed in some questions. For example, the concept of the proof stress from chapter 30 is needed in some of the chapter 2 questions.

Chapter 1 1 How many m are there in a m? …………….106 2 How many Pa are there in a MPa? ………….106 3 How many Pa are there in a GPa? ……………….109 4 How many ns are there in one second? …………………109 5 If 1017 is written in non-scientific notation how many zeros are there after the 1? ………………..17 6 If 10-14 is written in non-scientific notation how many zeros are there between the decimal point and the 1? …………………..13 7 How many g are there in 107 kg ………………….1010 8 How many s are there in 10-2 s …………………..104 9. The force required to make a mass m kg accelerate with an acceleration a is ma Newtons. a. What are the units of acceleration ?

b. What are the units of force when expressed in metres, kilogrammes and seconds ?. c. If the force is 3.2  102 kN and the mass is 2  106 kg what is the acceleration ? d. If the force is 4.3  103 MN and the mass is 3 g what is the acceleration ? e. If the force is 2.7  107 N and the mass is 20 mg what is the acceleration ? ………………………………………….. a m/s2 b kg m/s2 c 0.16 m/s2 d 1.43E12 m/s2 e 1.35E12 m/s2 10. The force on a mass m kg due to gravity is mg Newtons where g = 9.81 m/s2 a. If the force is 3.5  103 kN what is the mass ? b. If the mass if 35700 kg what is the force ? c. If the mass is 2  1013 g what is the force ? ………………………….. a 3.57E5 kg b 3.5E5 N c 1.96E11 N 11. The energy needed to apply a force F Newtons over a distance L metres is FL Joules. a. What are the units of energy when expressed in metres, kilogrammes and seconds ? b. If the force is 3.5  108 N and the distance is 5 km what is the energy ?. c. If the force is 2.7  105 kN and the distance is 20 mm what is the energy ? d. If the energy is 2  104 MJ and the distance is 2  107 m what is the force ? ………………………………. a kg m2/s2 b 1.75E12 J c 5.4E6 J d 1000 N 12. The stress on an area A m2 due to a force F Newtons is F/A Pascals. a. What are the units of stress when expressed in metres, kilogrammes and seconds ? b. If the force is 3 GN and the stress is 20 GPa what is the area ? c. If the force is 5  105 N and the stress is 10 kPa what is the area? d. If the force is 2.2  103 N and the area is square with side 100 mm what is the stress ? e. If the force is 20 MN and the stress is 20 mPa on a square area what is the length of the side ? f. If the force is 3.2  107 N and the stress is 2  1015 Pa on a square area what is the length of the side ? a kg/(m s2) b 0.15m2 c 50m2 d 2.2E5 Pa e 3.16E4 m f 1.26E-4 m Chapter 2 1.

a. The following observations are made when a 100mm (3.94 in.) length of 10mm (0.394 in.) diameter steel bar is loaded in tension:

Load kN 0 1.57 4.71 7.85 9.42 15.7 18.84 21.98 23.55 24.5 25.5 25.9 26.5 28.5 31.5

Extension mm 0 0.01 0.03 0.05 0.06 0.1 0.12 0.14 0.15 0.2 0.3 0.365 0.5 1 1.9

extension Load kip in.

0.00 0.0000 0.35 0.0004 1.06 0.0012 1.77 0.0020 2.12 0.0024 3.53 0.0039 4.24 0.0047 4.95 0.0055 5.30 0.0059 5.51 0.0079 5.74 0.0118 5.83 0.0144 5.96 0.0197 6.41 0.0394 7.09 0.0748

Calculate the following: a. the Young's modulus b. the 0.2% proof stress c. the yield stress d. the ultimate stress e. Explain which of these measurements is used for specification of steel and why it is used. …………………………………………………..

MKS solution The student should plot the points cross sectional area of bar = 0.01  0.01  PI/4 = 0.000078 m2 a. The points are linear up to 0.15 mm thus E = 23.55  100 /(0.15  area) = 201 GPa b. A line drawn parallel to the linear portion at 0.2mm greater extension intersects the curve at 25.905 kN. Thus Stress = 25.905/area = 332 MPa c. Yield is at 23.55 Kn thus stress = 23.55/area = 302 MPa d. Ult load = 31.4 kN thus stress = 31.4/area = 402 MPa US unit solution The student should plot the points cross sectional area of bar = 0.394  0.394  PI/4 = 0.122 in2

a. The points are linear up to 0.0059 in thus E = 5.3  3.94 /(0.0059  area) = 29010 ksi 0.2% of 3.94 is 0.0079 in b. A line drawn parallel to the linear portion at 0.0079 in greater extension intersects the curve at 5.84 kip. Thus Stress = 5.84/area = 47.87 ksi c. Yield is at 5.30 kip thus stress = 5.30/area = 43.44 ksi d. Ult load = 7.09 kip thus stress = 7.09/area = 58.11 ksi

2. A cube is made of a material which has no volume change when compressed (rubber comes close to this).

What is its Poisson's ratio ? …………………………………………………….

0.5 (approx.)

If a 1m cube is compressed vertically by 10% volume = .9 * 1.05 * 1.05 = 1 (approx.)

3 a. A concrete cylinder has a length of 300mm (11.81 in.) and a diameter of 100mm (3.94 in.). If the length when supporting a load of 4700 Kg (10363 lb.) is 299.93mm (11.807 in.) what is the Young's modulus? b If the diameter of the cylinder becomes 100.004 mm (3.94016) when the load is applied what is the Poisson's ratio? c If the characteristic strength of the cylinder is 35MPa (5.075 ksi) and a factor of safety of 1.4 is used what is the load, in tonnes (lb), that the cylinder will support? ......................................................................................

MKS solution a area = PI  .05  .05 = 7.8E-3 m2 Load = 47000 N Stress = 47000/7.8E-3 = 6E6 Pa Ext = 7E-2 mm Strain = 7E-2/300 = 2.3E-4 E = 6E6/2.3E-4 = 26GPa b Ext = 4E-3 Strain = 4E-5 Ratio = 4E-5/2.3E-4 = .174 c 3.5E6  7.8E-3 / 1.4 = 1.95E5N = 19.5 T US Unit Solution a area = PI  3.94  3.94/4 = 12.19 in2 Stress = 10363/12.19 = 850 psi Ext = 3E-3 in

Strain = 3E-3/11.81 = 2.5E-4 E = 850/2.5E-4 = 3.4E6 psi b Ext = 1.6E-5 in Strain = 4E-5 Ratio = 4E-5/2.5E-4 = .16 c 5.075E3  12.19 / 1.4 = 44189lb

4 A water tank is supported by four identical timber posts which all carry an equal load. Each post measures 50mm (1.97 in.) by 100mm (3.94 in.) in cross section and is 1.0m (39.4 in.) long. When 1m3 (1.31 yd3) of water is pumped into the tank the posts get 0.08mm (0.003in) shorter. a. What is the Young's modulus of the timber in the direction of loading? b. If the cross sections measure 100.003mm (3.940118 in.) by 50.00025mm (1.97000985) after loading what are the relevant Poisson's ratios? c. If 300 litres (0.393 yd3) of water is now pumped out of the tank what are the new dimensions of the posts ? (assume that the strain remains elastic) .......................................................................

MKS Solution a. Stress = 10000/(4*0.1*0.05) = 5E5Pa Strain = 0.08/1000 = 8E-5 Modulus = 6.2E9 Pa b. Strain on long side = 0.003/100 = 3E-5 Thus ratio = 3E-5/8E-5 = 0.375 Strain on short side = .00025/50 = 5E-6 Ratio = .0625 c. Extension will be 0.7 x previous: 0.08*0.7= 0.056 thus length = 1000-0.056 = 999.944 0.003*0.7 = 0.0021 thus long side = 100.0021 0.00025*0.7 = 0.000175 thus short side = 50.000175 US Unit Solution a. Mass of water = 1.31*1686=2209lb.

Stress = 2209/(4*1.97*3.94) = 71.14psi Strain = 0.003/39.4 = 8E-5 Modulus = 8.9E5 psi b. Strain on long side = 0.000118/3.94 = 3E-5 Thus ratio = 3E-5/8E-5 = 0.375 Strain on short side = .00000985/1.97 = 5E-6 Ratio = .0625 c. Extension will be 0.7 x previous:

0.003*0.7= 2.1E-3 thus length = 39.4-2E-3 = 39.4021 0.000118*0.7 = 8.26E-5 thus long side = 3.94+8.2E-5 = 3.9400826 0.00000985*0.7 = 6.89E-6 thus short side = 1.97 + 6.89E-6 = 1.97000689

5 A 6m (19.68 foot) long section of steel pipe has an outside diameter of 100mm (3.937 in.) and an inside diameter of 90mm (3.543 in.). The steel has a Young's modulus of 200GPa (29000 ksi) and a 0.2% proof stress of 340MPa (49.3 ksi).

Calculate the load in Tonnes (lb.) required to: a. Increase the length to 6.006m (19.6997 feet) b. Increase the length to the point where it has an irreversible plastic extension of 0.2%. c. Reduce the length to 6.015m (19.729 feet) by partially unloading after applying the load in part (b). ..............................................................................

MKS Solution a Area = Pi(502 - 452) = 1491mm2 Strain = .006/6 = 1E-3 Stress = 1E-3 * 200E9 = 200E6 Pa Load = 200E6 * 1491E-6 = 29.8E4 N = 29.8Tonne b Stress = 340E6 Pa load = 340E6 * 1491E-6 = 50.69E4 N = 50.69Tonne c Strain = 0.015/6 = 0.0025 Strain - 0.2% = 0.0005 Stress = 5E-4 * 200E9 = 100E6 Pa Load = 100E6 * 1491E-6 = 14.91E4N = 14.91Tonne US Unit Solution Area = Pi(3.9372 – 3.5432)/4 = 2.315in2 Strain = (19.6997-19.68)/19.68 = 1E-3 Stress = 1E-3 * 29000 = 29 ksi Load = 29E3 * 2.315 = 67135lb b Stress = 49.3 ksi load = 49.3E3 * 2.315 = 114129lb c Strain = (19.729-19.68)/19.68 = 0.0025 Strain - 0.2% = 0.0005 Stress = 5E-4 * 29000E3 = 14500psi Load = 14500 * 2.315 = 33567lb

6

The following observations are made when a 100mm (3.94 in.) length of 10mm (0.394 in) diameter steel bar is loaded in tension: Load kN 0 1.88 5.65 9.42 11.3 18.84 22.61 26.38 28.26 29.4 30.6 31.1 31.8 34.2 37.8

Extension mm 0 0.01 0.03 0.05 0.06 0.1 0.12 0.14 0.15 0.2 0.3 0.365 0.5 1 1.9

extension Load kip in.

0.00 0.0000 0.42 0.0004 1.27 0.0012 2.12 0.0020 2.54 0.0024 4.24 0.0039 5.09 0.0047 5.94 0.0055 6.36 0.0059 6.62 0.0079 6.89 0.0118 7.00 0.0144 7.16 0.0197 7.70 0.0394 8.51 0.0748

Calculate the following: a. the Young's modulus b. the 0.2% proof stress c. the estimated yield stress d. the ultimate stress e. Explain which of these measurements is used for specification of steel and why it is used. ……………………………………………..

MKS solution The student must draw a graph of either load vs extension or stress vs strain Cross section area = pi * 0.01 * 0.01 /4 = 7.85 E -5 a) when load = 18.84 ext = 0.1 Stress = 18.84E3 / 7.85E-5 = 2.4 E8 Pa Strain = 0.1/100 = 1E-3 Thus E = 2.4E8/1E-3 = 240GPa The student may obtain this from a gradient on the graph. b) Load = 31.3 (from graph) Thus stress = 398MPa

c) Load = 30.1 (from graph) Thus Stress = 384 MPa d) Load = 37.8 at failure Thus stress = 482 Mpa e) The proof stress is used because the measurement is clearly defined (repeatable) and close to the yield stress.

US Unit solution The student must draw a graph of either load vs extension or stress vs strain cross sectional area of bar = 0.394  0.394  PI/4 = 0.122 in2 a) when load = 4.24 ext = 0.0039 Stress = 4.24E3 / 0.122 = 34754psi Strain = 0.0039/3.94 = 1E-3 Thus E = 34754/1E-3 = 34754ksi The student may obtain this from a gradient on the graph. b) Load = 7.06 (from graph) Thus stress = 57.87ksi c) Load = 6.74 (from graph) Thus Stress = 55.24 ksi d) Load = 8.51 at failure Thus stress = 69.75 ksi e) The proof stress is used because the measurement is clearly defined (repeatable) and close to the yield stress.

7 (a) A round steel bar with an initial diameter of 25mm and length of 2m is placed in tension supporting a load of 2000kg. If the Young's modulus of the bar is 220GPa what is the length of the bar when supporting the load ? (b) If the Poisson's ratio of the steel is 0.4 what will the diameter of the bar be when supporting the load ? (c) If the 0.2% proof stress of the steel is 150MPa what load would be required to give an irreversible extension of 0.2% in the bar ? (d) Describe what will happen if the load is increased beyond the 0.2% proof stress ………………………………………………………

Area = p1 * 0.025 * 0.025 /4 = 4.9E-4 a) Stress = 2000 * 10 / 4.9E-4 = 4.1E7 Pa Strain = 4.1E7 / 220E9 = 1.8E-4 New length = 2 (1 + 1.8E-4) = 2.00036 m

b) Lateral strain = 1.8E-4 * 0.4 = 7.2E-5 New dia = 25(1 - 7.2E-5) = 24.9982 mm c) Load = 150E6 * 4.9E-4 = 7.3E4 N = 7300kg d) The bar will yield to give a substantial plastic extension before failure. The load will increase slightly due to strain hardening before reducing due to necking.

8 A 2m long tie in a steel frame is made of circular hollow section steel with an outer diameter of 60mm and a wall thickness of 3mm. The properties of the steel are as follows: Young's modulus: 200 GPa Yield stress: 300 MPa 0.2% proof stress 320 MPa Poisson's ratio 0.15 a. What is the extension of the tie when the tensile load in it is 100 kN. b. What is the reduction in wall thickness when this load of 100 kN is applied. c. What load is required to produce an irreversible strain of 0.2%. d. If the load calculated in (c) is applied and then reduced by half what is the extension (while the reduced load is still applied). ……………………………………………………….. a) Area = Pi(30^2 – 27^2) = 537 mm2 stress = 1E5/537E-6 = 186 Mpa Ext = 2 * 186E6/200E9 = 1.86mm b) Reduction = 3 * 0.15 * 186E6/200E9 = 4.18E-4 mm c) Load = 320E6 * 537E-6 = 171 kN d) Stress = 160Mpa, this gives strain = 160E6/200E9 = 8E-4 Thus total strain = 2E-3 + 8E-4 = 2.8E-3 Thus Extension = 2.8E-3 * 2 = 5.6 mm 9 a A flat steel bar measuring 25mm by 210mm by 3m long is placed in tension supporting a load of 4 Tonnes. If the Young's modulus of the steel is 210 GPa what is the length of the bar when supporting the load? b If the Poisson's ratio of the steel is 0.30 what are the dimensions of the steel section when supporting the load? c If the 0.2% proof stress of the steel is 300 MPa what load would be required to give an irreversible extension of 0.2%? (give your answer in tonnes). d Describe what will happen if the stress is increased beyond the proof stress. …………………………………………………. a area = 5.25E-3 m2 Load = 4E4 N Stress = 7.62E6 Pa Strain = 7.62E6/2.1E11 = 3.63E-5 Extension = 3.63E-5  3 = 1.098E-4 m Length = 3.000109 m

b Horiz strain = 3.63E-5  .3 = 1.09E-5 Along (1 – 1.09E-5)  25 = 24.9997 mm Across (1 – 1.09E-5)  210 = 209.9977 mm c 300MPa  5.25E-3 m2 = 157.5E4 N = 157.5T d Plastic deformation. Irreversible. Large increase in strain for little increase in stress to failure.

10 10000

Load N

8000 6000 4000

y = 4971.5x - 7513.4

2000 0 0

2

4

6

8

Displacement mm

Figure Q10 Figure Q10 shows the load-displacement plot when a timber sample measuring 15mm by 16mm by 42mm is tested in compression on end (with the load applied to the 15mm by 16mm faces). The equation on the graph is for the best line fit shown. a. What is the approximate failure load for the sample in N? b. What is the approximate failure stress in MPa? c. What is the Young’s modulus in MPa? d. If a 100mm square, 2m tall column in a building is made from the same timber (with the grain in the same direction) and is subject to a load of 150 kN, what displacement will this cause at the top? (assume it does not buckle) e. If the grain in the timber test sample had be in a different direction what effect would this have had on the failure stress? ...................................................................................................... a. 8200 N b. 8200/(.015 * .016) = 34.16 MPa c. 4971 * 1000 * .042/(.015 * .016) = 869 MPa d. Stress = 150E3/.01 = 15 E6 Pa strain = 15E6/869E6 = 17 E-3 displacement = 2 * 17 E-3 = 34 mm e. If the grain was along the sample to start with, changing it will reduce the failure stress.

11 a) Describe the difference between load and stress. b) A 4 inch diameter concrete cylinder 8 inches long is loaded on the end with 17920 lb. What is the stress in it ? c) The Young's modulus of the cylinder is 3000ksi. What is its height after loading ? d) The Poisson's ratio of the cylinder is 0.17. What is its diameter after loading ? e) A column 6 feet high measuring 12 inches by 20 inches on plan is made with the same concrete as in the cylinder and supports a bridge. In order to prevent damage to the deck the maximum

permitted reduction in length of the column is 0.016 inches. Assuming that the column is not reinforced calculate the reduction in length of the column when it is supporting the full weight of an 89600 lb. lorry and state whether the deck will fail. ………………………………………………… a) stress is load/area and is a material property. b) area = pi * 2 * 2 = 12.6in2 stress = 17920 /12.6 = 1422 psi c) strain = 1422/3000E3 = 4.74E-4 extension = 4.7E-4 * 8 = 3.79E-3 in. new height = 8-3.79E-3 = 7.996 in. d) lateral strain = 4.7E-4 * 0.17 = 8E-5 lateral extension = 4 * 8E-5 = 3.2E-4 new diameter = 4 + 3.2E-4 = 4.00032 e) Stress = 89600/ (12*20) = 373 psi strain = 373/3000E3 = 1.24E-4 Extension = 1.24E-4 * 6 * 12 = .009 in. deck will not fail Chapter 4 1 Two identical water tanks are supported 100mm (3.94 in.) above a flat rigid floor on timbers which are 100mm by 100mm by 750mm (3.94 by 3.94 by 29.5 in.) long. Each tank rests on the whole of one of the 100mm wide edges of two timbers. a. If the tanks are at the same level when empty what is the difference in level when one tank has 2.5m3 (3.275 yd3) more water in it than the other ? b. If 0.7m3 (0.917 yd3) of hot water is added to the tank with less water in it and the temperature of the timbers under it is raised by 25oC (45oF) what does the difference in level become ? c. Describe how the answer to part a would have been affected if the timbers had been (i) wet or (ii) loaded parallel to the grain.

The properties of the timbers are as follows: Young's modulus: 80 N/mm2 ((11600 psi) Coefficient of thermal expansion: 34 s/oC (19 s/oF) .................................................................................................

MKS Solution a) area = 2 * 0.1 * 0.75 = 0.15m2 load = 2500 * 10 = 25000N Stress = 25000/ 0.15 = 170kPa Thus strain = 17E4 / 80E6 = 0.21 E-2 Thus extension = 100 * 0.21E-2 = 0.21mm b) Level change from mass of water = 0.21 * 0.7/2.5 = 0.06mm

Level change from temperature = 25 * 34E-6 * 100 = 0.085mm Thus new level difference = 0.21 - 0.06 + 0.085 = 0.235mm c) (i) wetting timber will reduce modulus so movement will increase (ii) end grain loading gives higher modulus. movement will decrease.

US Unit Solution a) area = 2 * 3.94 * 29.5 = 232.46 in2 mass of water = 3.275 * 1686 = 5522 lb Stress = 5522/ 232.46 = 23.75psi Thus strain = 23.75/ 11600 = 2.05 E-3 Thus extension = 3.94 * 2.05E-3 = 8.077E-3 in b) Level change from mass of water = 8.077E-3 * 0.7/2.5 = 2.26 E-3 Level change from temperature = 45 * 19E-6 * 3.94 = 3.369E-3 in Thus new level difference = 8.077E-3 – 2.26E-3 + 3.369E-3 = 9.186E-3 in c) (i) wetting timber will reduce modulus so movement will increase (ii) end grain loading gives higher modulus. movement will decrease.

2 The bricks in a wall have a specific heat of 900 J/kg/oC, (0.216 BTU/lb/oF) a thermal conductivity of 0.7 W/m/oC (0.414 BTU/hr ft oF) and a density of 1700 kg/m3 (2873 lb/yd3). The wall is solid and is 215mm (9 in.) thick, 3m (10 feet) high and 10m (33 feet) long. a. If the wall is at 25oC (77 oF) on the inside and 10oC (50oF) on the outside what is the rate of heat loss through it in Watts? b. If the inside and outside temperatures are lowered by 5oC (9oF) what is the heat loss from the wall in Joules. c. Describe how the thermal performance of walls is improved relative to this example in modern construction. ...............................................................................................

MKS Solution a) Q = 0.7 * 3 * 10 * (25 - 10)/.215 = 1465W b) Heat = 5 * 900 * 1700 * 0.215 * 3 * 10 = 4.9E7 J c) cavities (filled or empty), low k aerated blocks etc. for inner skin, dry lining.

US Unit Solution a) Q = 0.414 * 10 * 33 * (77 - 50)/0.75 = 4918 BTU/hr b) Volume = 0.75 * 10 * 33 = 247.5 ft3 = 9.17 yd3 Heat = 9 * 0.216 * 2873 * 9.17 = 51215 BTU

c) cavities (filled or empty), low k aerated blocks etc. for inner skin, dry lining.

3 The four materials listed below are being considered for the cladding of a building: Material Proposed Thermal Coefficient of Thickness Conductivity thermal expansion mm W/m/oC s/oC Concrete 100 1.4 11 Steel 5 84 11 Aluminium 8 200 24 Brick 215 0.9 8 For each material: a. Calculate the heat loss in Watts through each m2 of the cladding if the temperature difference between the interior of the building and the outside air is 10oC. b State the assumptions you made in your calculation in section (a) and indicate whether they are realistic. c. Calculate the spacing of the expansion joints in the cladding if the maximum acceptable expansion of a panel between joints is 2mm for a 20oC temperature rise. ........................................................................... a Loss = k  1  10/d = 10k/d Concrete: loss = 101.4/.07 = 140W Steel: 1.68E5W Aluminium: 2.5E5W Brick: 42W b assumes that surfaces of cladding are at temperature of environment, i.e. all of the temperature drop is in the cladding. Generally unrealistic particularly for the steel and aluminium. c Coeff  spacing  20 = 0.002 Thus Spacing = 1E-4/coeff Concrete: spacing = 1E-4/11E-6 = 9m Steel: 9m Aluminium: 4.1m Brick: 12.5m 4 a) A 250mm thick concrete wall is cast using a mix containing 350kg/m 3 of cement. The shuttering on each side of the wall is 20mm ply wood.

If the heat of hydration is being generated at a rate of 7W per kg of cement and is all being lost through the shutters what is the temperature drop across them. (the thermal conductivity of ply wood = 0.15 W/m2/oC) b) Describe two methods for reducing the heat of hydration in a concrete mix. …………………………………………………………………………………….. a) Heat per m2 = .25 * 350 * 7 = 612.5 W 612.5 = 0.15 * 2 * TDIFF /0.02 TDIFF = 41 degrees b) Two of the following:

Reduce cement content by using plasticiser.

Use PFA or GGBS Cool concrete in mixer Use low heat cement Use retarder 5 A building has a 120mm thick concrete floor slab with rooms above and below it. The slab is 5m long and 4m wide. The properties of the concrete are as follows: Thermal conductivity: 1.5W/m/oC Density: 2300Kg/m3 Specific Heat: 800J/kg/oC Coefficient of thermal expansion: 8s/oC a. What is the heat transmission through the floor in W if the temperature difference between the two rooms is 15oC? b. What is the energy in J which will be absorbed by the slab if the average temperature of the rooms rises by 12oC on a hot day? c. What is the increase in length of the slab when the temperature is increased by 10oC? d. Describe two methods for reducing the heat transmission ........................................................................... a Q = 1.5  (54) 15 /0.12 = 3750W b Mass = 5  4  .12  2300 = 5520kg Energy = 5520  800 12 = 5.29E7 J c Expansion = 5 8E-6 10 = 0.4mm d 2 of: Light weight aggregate No fines concrete Insulation on underside Hollow core slabs Chapter 5 1. What is the velocity and wavelength of ultrasound with a frequency of 5kHz in a thin concrete element with density 2400kg/m3, a Young's modulus 50GPa and a Poisson's ratio of 0.15. ………………………………………

50E9 = v2 * 2400 * (1+.15)(1-.3)/(1-.15) v=4690m/s = wavelength * 5000 wavelength = 0.94m 2 a) 4 litres of an ideal gas is at a pressure of 1 atmosphere and a temperature of 0oC. (i) If the volume is reduced to 3 litres and the temperature rises to 20oC what does the pressure become? (ii) If the molecular weight of the gas is 18 what is its mass? b) A pulse of ultrasound takes 1.4  10-4s to travel through a concrete wall 400mm thick. If the Poisson's ratio of the concrete is 0.13 and the density is 2400kg/m3 what is the Young's modulus? ....................................................................................

a (i) P1 V1/T1 = 1  4/273 = P2  3/293 P2 = 1.43 atm (ii) PV = nRT 1E5  4E-3 = n  8.31  273 n = .176 mass = 18  .176 = 3.17 g b v = .4/1.4E-4 = 2857 m/s E = (2857)^2  2400  ((1+.13)  (1-.26) / (1-.13)) = 18.6 GPa 3 A 200mm thick concrete wall is tested using ultrasonics. The measured transit time is 50s. a. Calculate the Young's modulus of the concrete assuming a density of 2300 kg/m3 and a Poisson's ratio of 0.11.

What is the percentage error in the answer in (a) if: b. The actual density is 2380 kg/m3 c. The actual Poisson's ratio is 0.13 d. The actual wall thickness is 220 mm. e. Discuss the consequences of these percentage errors on the precautions that should be taken when carrying out ultrasonic testing. ..............................................................................

a b c d

Thickness time velocity density poisson E 0.2 5.00E-05 4.00E+03 2300 0.11 0.2 5.00E-05 4.00E+03 2380 0.11 0.2 5.00E-05 4.00E+03 2300 0.13 0.22 5.00E-05 4.40E+03 2300 0.11

% difference 3.58E+10 3.70E+10 3.54E+10 4.33E+10

3.5 -1.2 21.0

e It is very important to measure the path length accurately.

4 a. A pulse of ultrasound takes 10-4s to travel through a concrete wall 400mm thick. If the Poisson's ratio of the concrete is 0.13 and the density is 2400kg/m3 what is its Young's Modulus. b. A container of ideal gas at atmospheric pressure is heated from 20oC to 50oC. What is the final pressure ? c. 1 litre of ideal gas at atmospheric pressure is compressed into 0.3 litres at the same temperature.

What is the final pressure? d. A fixed volume of ideal gas at 30oC and 1MPa is heated to 100oC. What is the final pressure? ……………………………………………………………………………. a. v = .4/1E-4 = 4000 E = 4000^2 x 2400 x ((1.13 x .74)/ .87) = 3.7 E10 Pa b. 1.01E5 x 323/293 = 1.113 E 5 Pa

c. 1.01E5 x 1/.3 = 3.366 E5 Pa d. 1 x 373/303 = 1.23 MPa Chapter 6 1 A 4mm diameter copper wire has a resistivity of 1.9  10-8 m and is carrying a current of 100 Amps. a. What is the power loss per m length in the wire ? b. If the wire has a specific heat of 370 J/kgoC and no heat is lost from it what is its temperature rise per minute. (density of copper = 8900 kg/m3). c. If the coefficient of thermal expansion of the wire is 17  10-6/oC what is the percentage expansion after 3 minutes. …………………………………………………………………. a) resistance per m = 1.9E-8 / (Pi * 0.002^2) = 0.0015 power loss = 100 * 100 * 0.0015 = 15W b) T2-T1 = 15*60/(370*8900*Pi * 0.002^2) = 22 degrees c) 17E-6 * 22 * 3 = 0.11% 2 A current of 20 A is flowing first through a copper conductor, then through an aluminium one and finally through a lead one. All of the conductors have a circular cross section with a diameter of 4 mm. The properties of the conductors are as follows: Material Resistivity m Density kg/m3 Specific Heat J/kg/oC

Copper 1.710-8 8800 390

Aluminium 2.810-8 2600 880

Lead 1.810-7 11000 200

For each metre length of each conductor calculate the following: a. The resistance. b. The voltage drop c. The power loss d. The mass e. The temperature rise in 10 minutes (assume no heat loss) ............................................................................ current diameter area

20 0.004 1.25664E-05

Material

Copper

Aluminium

Lead

Resistivity Ωm

1.70E-08

2.80E-08

1.80E-07

Density kg/m3

8800

2600

11000

Specific Heat J/kg/oC

390

880

200

Resistance = resistivity/area V drop = IR

1.35E-03 0.027

2.23E-03 0.045

1.43E-02 0.286

Power = VI Mass = density × area T rise = 600s × Power/(sp ht × mass)

0.54 0.111 8

0.89 0.033 19

5.73 0.138 124

3 a) State what is meant by the following terms in electrical circuits: (i) Current (ii) Resistance (iii) Resistivity b) The maximum acceptable voltage drop on a 32kV transmission line is 1% per km. (i) What cross section area of copper wire is required for a power of 50MW ? (ii) How much would the wire between two pylons 100m apart weigh ? (iii) What would the weight be if aluminium wire was used ?

Use the following: Copper Aluminium Resistivity m 1.7  10-8 2.8  10-8 3 Density kg/m 8900 7800 …………………………………………………………………………………….. a i) Current is the movment of charge in Coulombs/sec ii) Resistance is voltage/current iii) Resistivity is resistance per unit area per unit length. b i) Voltage drop over 1km is 32000/100 = 320V Current = 50E6/32E3 = 1.56E3 Resistance = 320/1.56E3 = 0.205 Area = 1.7E-8 * 1000/0.205 = 8.3E-5 m2 = 83mm2 ii) Mass = 8.3E-5 * 100 * 8900 = 74kg iii) Area = 2.8E-8 * 1000/0.205 = 13.6E-5m2 Mass = 13.6E-5 * 100 * 7800 = 106kg 4 A 4mm diameter copper wire is carrying a current of 20 A and has a resistivity 1.8  10-8 m. a What is the power loss per m length in the wire? b What is the voltage drop per m length along the wire? c If the wire is 20m long and the supply voltage is 240V what is the percentage power loss? ......................................................................................... a. area = 1.26E-5 m2 R = 1.8E-8 /1.26 E-5 = 1.43 E-3 Ohm/m Power = 20  20  1.43E-3 = 0.572 W/m b. V = 20  1.43E-3 = 2.86 E-2 V/m

c. Power in = 240  20 = 4800 W Loss = .57  20 = 11.4 W 11.4/4800 = .24% 5 a) Describe the difference between resistance and resistivity. b) A 3mm diameter copper wire has a resistivity of 1.9  10-8 m what is its resistance per m length. c) If the wire is carrying a current of 100 Amps what is the power loss per m length in the wire ? d). If the wire has a specific heat of 370 J/kgoC and no heat is lost from it what is its temperature rise per minute. (density of copper = 8900 kg/m3). e). If the coefficient of thermal expansion of the wire is 17  10-6/oC what is the percentage expansion after 3 minutes. …………………………………………………………………. a) Resistance is resistivity  length/area. Resistivity is a material property b) resistance per m = 1.9E-8 / (Pi * 0.0015^2) = 0.0027  c) power loss = 100 * 100 * 0.0027 = 27W d) T2-T1 = 27*60/(370*8900*Pi * 0.0015^2) = 70 degrees e) 17E-6 * 70 * 3 = 0.36%

Chapter 7 1 Give the meaning of the following terms. Where possible give examples.: a. The atomic number of an element b. A molecule c. An exothermic reaction d. An acid e. pH f. Hard water ........................................................................... a. Number of protons in the nucleus = the number of electrons in stable state. e.g. Oxygen = 16 b. A number of atoms which combine to form a stable particle which can exist in isolation, e.g. H2O. c. A reaction that gives off heat. d. A chemical with a surplus of hydrogen ions. e. A measure of acidity = -log(1/H) where H is the number of hydrogen ions per litre. f. Alkaline water 2 a. Calculate the volume of 3 mols of a gas at 20oC and a pressure of 3 Bar.

b. Explain what a catalyst is and give and example of one. c. List three factors which will increase the rate of a chemical reaction. d. Give an example of how one of the factors from part (c) has a significant effect in construction. .............................................................................. a.

PV = nRT 1Bar = 100kPa. 300E3 * vol = 3 * 8.31 * 293 thus vol = .024m3 b.

A catalyst accelerates a chemical reaction without being used up in it. example: lime in hydration of GGBS, other examples could be platinum in car exhaust etc. c. Any 3 of the following:     

The temperature. A rough rule is that an increase of 20oC doubles the rate of reaction.

The pressure, high pressures will increase reaction rates.

The energy released by the reaction.

The physical consistency of the materials. A fine powder will react faster than a coarse one (e.g. rapid hardening cement is finely ground).

The presence of a catalyst. This is a material which is not used up by the reaction but makes is proceed faster.

d. e.g. heat/fineness affects cement hydration heat affects setting of epoxy Catalyst affects polyester resin and GGBS in concrete.

3 a) State typical values for the chemical pH of the following: (i) A strong acid (ii) Concrete (iii) “Hard” water (iv) “Soft” water b) (i) Describe the process in which lime is produced from limestone. (ii) Describe the process in which lime is carbonated in concrete. ........................................................................................................................................ a) (i) 1-4 (ii) 11 - 12.5 (iii) 7 - 10 (iv) 4-7 marks will be given for getting these in the correct order. b) (i). calcination: This is carried out in kilns.

Calcium carbonate (limestone)  Calcium oxide (quicklime) + Carbon dioxide CaCO3  CaO + CO2 requires substantial heat input. (ii) the carbonation reaction: Hydrated lime + Carbon dioxide  Calcium carbonate + water Ca(OH) + CO2  CACO3 + H2O proceeds slowly in the presence of atmospheric carbon dioxide.

4 a) Describe three factors which affect the rate of a chemical reaction. Give examples of reactions in which each of the factors is significant. b) Describe the “lime cycle” and its relevance to construction processes. ………………………………………………………………………………………………. a) Reaction rates are affected by the following:    

The temperature. A rough rule is that an increase of 20oC doubles the rate of reaction (e.g. hydration of cement).

The energy released by the reaction (e.g. combustion) The physical consistency of the materials. A fine powder will react faster than a coarse one (e.g. rapid hardening cement is finely ground).

The presence of a catalyst. This is a material which is not used up by the reaction but makes is proceed faster (e.g. lime as a catalyst in hydration of GGBS)

b) The lime cycle The first process is calcination. This is carried out in kilns.

Calcium carbonate (limestone)  Calcium oxide (quicklime) + Carbon dioxide The quicklime is highly reactive. Some of it is used directly (e.g. in concrete block production) but most of it is slaked in the hydration reaction which is quite violent and gives off a lot of heat.

Quicklime + water  Hydrated (slaked) lime Hydrated lime is sold at builders merchants for use in bricklaying mortar. Traditional mortars set by the carbonation reaction: Hydrated lime + Carbon dioxide  Calcium carbonate + water This reaction takes place in concrete when it carbonates.

Chapter 8

1 a (i) A brick is fully saturated and has a density of 1800 kg/m3. When it is dried it has a density of 1500 kg/m3 What is its porosity? (ii) If the brick is cut in half what proportion of the cut surface will be made up of pores? b Grout is being pumped along a pipe. All of the grout which is more than 10mm from the wall of the pipe is moving at 0.3 m/s and the material in contact with the pipe wall is not moving. Assuming that the grout has a viscosity of 0.15 PaS calculate the drag on each m2 of the surface of the pipe. ............................................................................................ a (i) 1800 kg/m3 = 1500 kg brick + 300 kg Water = 300 l Water Porosity = 30% (ii) 30% b 0.15 = shear  0.01/0.3 Shear = 4.5 N/m2 2 a) A sample of wood is fully saturated and has a density of 800 kg/m3. It is dried completely and the density is reduced to 600 kg/m3. What is the porosity ? b) On a cold morning the air in a building is at 100% RH and the partial pressure of the water vapour in it is 0.01 bar. As the building warms up the saturation pressure for water in the air in it is increased to 0.07 bar. Assuming that no water evaporates from surfaces in the building and the partial pressure of the water vapour remains constant what will the humidity be ? c) Discuss how the conditions in the building will change as the temperature rises. ………………………………………………………………………… a) mass of water in 1 m3 is 200kg vol of water is 200 l porosity = 0.2 b) at 100%RH partial pressure = saturation pressure = 0.01bar when warm partial/saturation = 0.01/0.07 = 14% c) no water is lost from the building – some will be gained from evaporation but it will feel less damp due to reduction in humidity.

3 a) What would the change in temperature of a 2 m3 block of concrete with a density of 2300 kg/m3 and a specific heat of 850 J/kg/oC be if it was heated by a 4 kW heater for 30 minutes. (assume no heat loss) b) If the floors of a building are constructed with a material with a high specific heat what effect is this likely to have on the temperature of the rooms. c) (i) A brick is fully saturated and has a density of 1850 kg/m3. When it is dried it has a density of 1550 kg/m3 What is its porosity? (ii) If the brick is cut in half what proportion of the cut surface will be made up of pores? ……………………………………………………….. a

4000 x 30 x 60 / (850 x 2300 x 2) = 1.84 oC b The temperature will be more stable – cool in the day and warm at night. c (i) 1850 kg/m3 = 1550 kg brick + 300 kg Water = 300 l Water Porosity = 30% (ii) 30%

Chapter 9 1 A brick is 215mm long by 65mm high by 102mm wide and has no frog or perforations. The brick is fully saturated with water and weighs 2.4 kg. The brick is then dried and weighs 2.17 kg. What is: a. The bulk dry density of the brick ? b. The skeletal density of the solid material in it ? c. The porosity of the brick ? d. The brick is now placed in a high humidity environment so 70% of its porosity is full of water.

What will its mass be ? e. A brick wall is built without a damp proof course and the bottom of it is exposed to water. If the typical pore radius in the bricks is 0.13m what is the theoretical height to which the water will rise ? f. Why will it not rise to this height in practice ? (surface tension of water: s = 0.073 N/m) ........................................................................ vol of brick = .215 * .065 * .102 = 1.42E-3 m3 a) dry density = 2.17 / 1.42E-3 = 1528 kg/m3 b) vol of water = mass of water = 2.4-2.17 = 0.23 litres net vol = 1.42 - 0.23 = 1.19 E-3 m3 net density = 2.17 / 1.19E-3 = 1823 kg/m3 c) porosity = 0.23/1.42 = 16.2% d) mass = 2.17 + (0.23 * 0.7) = 2.33 kg e) H = 2 * 0.073 /(1.3E-7 * 1000 * 9.8) = 114m f) because the pores are not uniform (“ink bottle“ pores etc.)

2 a.) The flow through a concrete plug in a pipe is 2ml per day. What is the flow in m3/s? b) A 300mm diameter pipe is blocked with a concrete plug 100mm thick. A 20m head of water is applied to one side of the plug and the other side is open to the atmosphere. The flow rate through the plug is 2ml per day, what is the coefficient of permeability of the concrete? c) The viscosity of the water is 10-3 PaS. What is the intrinsic permeability of the concrete? d) Describe what happens when ordinary concrete is exposed to sulphates. e) Describe two methods of preventing the effect of sulphate on concrete

............................................................................ a Flow = 2ml/day = 2.31E-11 m3/s b Area = 0.07 m2 V = 2.31E-11/0.07 = 3.3 E-10 m/s = k(20-0)/.1 k = 1.64 E-12 m/s c 20 m head = 20  10  1000 = 2E5 Pa V = 3.3 E-10 = K/1E-3  2E5/.1 K = 1.64 E-19 m2 d Sulphate reacts with the aluminate phases in cement and causes expansion and then crumbling. e.

Can be prevented with sulphate resisting cement or GGBS. 3 a. A river has been diverted to flow through a rectangular concrete culvert with external dimensions 4m high and 2m wide and with a wall thickness of 120mm contains water at an average pressure head of 20m. If there is no water pressure on the outside of the culvert and the coefficient of permeability of the concrete is 1.3  10-12 m/s what is the flow in ml/day through the walls of each m length of culvert ?. b. Describe the effect on the concrete if the water contains salt. ……………………………………………………………………………………….. a) V = 1.3E-12 * 20/.12 = 2.166E-10 m/s Area per m length = 12 m2 Q = 2.166E-10 * 12 = 2.6E-9 m3/s = 2.6E-9 * 1E6 * 60 * 60 * 24 = 224 ml/day b) salt will not affect concrete unless there is rapid evaporation from outer surface in which case crystallisation may cause spalling. salt will rapidly corrode embedded steel, leading to loss of strength and spalling.

4 (a) Why is concrete not waterproof? (b) A concrete wall 150mm thick forms the side of a water tank with the outside of the wall open to the atmosphere. If the coefficient of permeability of the concrete is 6.5  10-12 m/s calculate the velocity of the flow of water through the wall in m/s at a depth of 5m below the water surface. (c) Calculate the rate of flow of water in ml/m2/s through the wall. (d) If the tank contains sea water with chlorides in it what effect will they have on the concrete and the reinforcing steel in it ? (e) If the water contains sulphates what effect will they have? .....................................................................................................

a) Because it contains pores which are formed by mix water which has not been used in the hydration process. b) v = 6.5E-12 * 5/0.15 = 2.166E-10 m/s c) thus flow = 21.66E-11 m3/m2/s = 2.166E-4 ml/m2/s d) Chlorides will not affect concrete but will cause corrosion of steel. e) Sulphates will also cause some steel corrosion and will also cause expansion/crumbling of concrete

Chapter 10 1 a) A 200mm thick concrete wall is fully saturated with water with no pressure differential. On one side of it the water contains salt at a concentration of 10% by mass and on the other side the water is kept at zero concentration. The capacity factor for the salt is 0.5, the porosity of the concrete is 7% and the apparent diffusion coefficient is 10-12 m2/s. Describe the process which transports the salt through the concrete and draw three graphs showing the changes in salt concentration across the thickness of the wall: (i). Early in the transient phase (ii). Late in the transient phase (iii). In the steady state On each graph show the concentration in solution and the concentration adsorbed onto the matrix. b) Calculate the total mass of salt in each m2 of wall in the steady state. c) Calculate the mass of salt leaving each m2 of the wall per second in the steady state. ...............................................................................

Cl on the high concentration side = 100kg/m3 Cs = Cl = 50 kg/m3 Total concentration /m3 = 100  7% = 7kg in solution + 43 kg adsorbed. note that the 7kg is not Cl because it is the concentration per m3 of solid, not per m3 of liquid In the figures the ratio adsorbed/in solution is always constant.

Concentration kg/m3

Early in transient 45 40 35 30 25 20 15 10 5 0 -5 0

50

100

150

200

250

200

250

200

250

Distance mm in solution

adsorbed

Concentration kg/m3

Late in transient 50 45 40 35 30 25 20 15 10 5 0 0

50

100

150

Distance mm in solution

adsorbed

Concentration kg/m3

Steady State 50 45 40 35 30 25 20 15 10 5 0 0

50

100

150

Distance mm in solution

adsorbed

b.

Total on high side = 50 kg/m3 average = 25 kg/m3 thus total in wall = 25  0.2 = 5 kg c.

F = Da dCs/dx = 10-12  50/0.2 = 2.5  10-10 kg/m2/s 2 a. What is the theoretical height to which water will rise in concrete due to capillary suction ?. b. A 150mm concrete cube is placed in water with the water level just above the base of the cube and the top surface is kept dry by evaporation. The sides of the cube are sealed. What is the theoretical rate of flow of water up the cube in ml/s ? c. If the water contains 8% salt what will be the flux of salt up the cube in the steady state. d. If a second cube is tested in similar conditions but with a constant flow of clean water across the top what will the steady state flux now be? e. For each experiment describe the following: the effect of changing the capacity factor of the concrete. the salt concentration in the cube in the steady state. the ultimate effect on the cube if the experiment is continued for a long time For this question use the following data: typical diameter of pores in concrete = 0.015m surface tension of water = 0.073 N/m coefficient of permeability of concrete = 5  10-12m/s porosity of concrete = 8% Intrinsic diffusion coefficient of salt in water= 10-12 m2/s ............................................................................ a.

H = 2  0.073 /(1.5/2  10-8  1000  9.81) = 1984 m note that this is highly unrealistic. b V = 5 10-12  1984 / 0.15 = 6.61  10-8 m/s Flow = 6.61  10-8  0.15  0.15 = 1.488  10-9 m3/s = 1.488  10-3 ml/s = 1.488  10-6 l/s c 1.488  10-6 kg/s  8% = 1.19  10-7 kg/s d Diffusion: Cl=80 kg/m3 F x area = .08  10-12  80/.15  .15  .15 = 9.6  10-13 kg/s e (i) increasing the capacity factor will increase the length of the transient in both cubes but will not affect the steady state. (ii) the first cube will be fully saturated to the concentration of the solution but the second will have a linear concentration drop to the top.

(iii) In the first cube the salt will build up near the top of the cube and, as the pressure from it increases, the corners will fall off. Nothing will happen to the second 3 (a) State why the process of diffusion is important to the durability of concrete.

The mix design of concrete elements which are exposed to chlorides is changed to include PFA. This increases the capacity factor from 1 to 5 and reduces the porosity from 10% to 8%. The density remains unchanged at 2300 kg/m3 and the intrinsic diffusion coefficient is 5  10-12 and does not change. (b) Calculate apparent diffusion coefficient before and after the change is made. (c) Describe the changes to the flux of chloride ions both in the transient phase (shortly after exposure) and in the long term steady state. ……………………………………………………………………………………………………. (a) Diffusion is the movement of ions in solution with a concentration gradient.

Key process for chloride ingress (corrosion), sulphate attack, carbonation.

Occurs whenever concrete is exposed to chloride in the environment (b) Di is 5  10-12 Da = Di × / thus Da is 5E-13 before and 8E-14 after (c) During the transient phase the concentration of chlorides in the pore solution is increasing. Thus, if the distribution ratio is high, substantial adsorption of chlorides onto the cement matrix will occur. This will, in turn, control the concentrations in solution and restrict the rate of penetration.

Thus in the transient (which will last for working life of structure) major reduction in flow rates because most of the chlorides entering the element are absorbed onto the matrix.

In the steady state there is no change in the concentration in the pore solution so the distribution ratio is not relevant. The ions will diffuse through the concrete Thus the flux depends on Di, i.e. reduced by 20%.

Chapter 11 1 a. Describe the different situations in which ionising radiation might be encountered in construction. b. A concrete wall is being irradiated with 50W of radiation. If the power is reduced to 25W at a depth of 200mm what is the power at a depth of 300mm ? ........................................................................... a.

1. Natural radiation (e.g. radon in houses) 2. Radioactive sources (e.g. weld testing) 3. Radioactive waste from demolition or disposal sites 4. Accelerators (e.g. Xray machines) 5. Reactors 6. Secondary radiation e.g. from irradiated shielding.

(descriptions in section 11.3) b.

25 = 50 * exp(-k * 200) k = ln(2)/200 = 3.46E-3 power = 50 * exp(-k * 300)= 18W Alternatively it may be seen that: Power =50  e-3/e-2 = 50/e = 18 W 2 A radioactive source has an initial power of 10W and a power of 3W after 1 year. What is its half life and what will the power be after 2 years ? …………………………………………

3 = 10 * exp(-k*1) k = ln(10/3) = 1.204 T0.5 = ln(2)/1.204 = 0.575 power = 10*exp(-1.204*2) = 0.9W 3 A 5mm steel plate reduces the power of incident radiation from 10W to 0.1W. What thickness of plate must be added to this to reduce the power to 0.001 W? ………………………………………….

0.1=10*exp(-k*5) k = ln(10/0.1)/5 = 0.921 0.001 = 10*exp(-0.921*x) x = ln(10/0.001)/.921 = 10mm total thickness 5 mm must be added.

This may be seen by inspection: 5 mm of steel is needed to reduce the power by a factor of 100.

4 a) Describe three different types of ionising radiation b) A concrete wall is being irradiated with 40W of radiation. If the power is reduced to 15W at a depth of 150mm what is the power at a depth of 200mm. ……………………………………………………………… a see section 11.2 b X=A  exp(-kd) A = 40 k = ln(40/15)/150 = .0065 X = 10.9 W Chapter 12 1 a) Explain why a single low result from testing a construction material should not normally result in the rejection of a batch of it. b) 5 bricks are tested for strength. If the average is 20MPa and the standard deviation is 3MPa what is the strength below which 5% would be expected to lie ? c) What is the probability of one strength in the set of five being below the 5% level. ?

d) If three further sets of five are tested what is the probability of all three sets having one each below the 5% level ? …………………………………………………………………………………. a) Test results will normally form a normal distribution in which some low values are expected. b) 20 – 1.64 * 3 = 15.08 c) 5 * .95^4 * .05 = 0.203 d) .203^3 = 0.008 2 a) Why is the failure of a single concrete cube not necessarily a cause for concern about the quality of the structure in which the concrete was used? b) Concrete cubes are being tested and the mean strength is 40MPa and the standard deviation is 5MPa. What is the strength below which 5% of the strengths will lie? c) Four concrete cubes are tested with probability of failure of 5%.

What is the probability of: (i) 1 failure ? (ii) no failures ?. ……………………………………………………… a.

Cube results form a statistical distribution. 5% failures is normally expected. b 40 – 5 x 1.64 = 31.8 Mpa c i) .95 x .95 x .95 x .05 x 4 = 0.17 ii) .95^4 = 0.81 3 Concrete is delivered to site with a characteristic strength of 30 MPa and a percentage defect rate of 5%. If 4 cubes are made what is the probability of: a. No failures b. 1 failure c. If 4 cubes are made each day what is the probability of getting 1 failure per day on each of 3 consecutive days. d. If the observed standard deviation of the test results is 7 MPa what is the mean strength. e. State three causes for variations in test results. .................................................................... a. 0.954 = 0.814 b. 0.953  0.05  4 = 0.171 c. .1713 = 0.005 d. 30 + (7  1.64) = 41.5 MPa e. Any 3 from the following: Variations in materials Inaccuracies in batching Temperature changes Changes in haul times Sampling variations in load

Testing variations (see section 13.2) 4 Concrete is delivered to site with a characteristic strength of 30 MPa and a percentage defect rate of 5%. If 5 cubes are made what is the probability of: a. No failures b. 1 failure c. If 5 cubes are made each day what is the probability of getting 1 failure per day on each of 3 consecutive days. d. If the observed standard deviation of the test results is 6 MPa what is the mean strength. .................................................................... a. 0.955 = 0.773 b. 0.954  0.05  5 = 0.204 c. .2043 = 0.0085 d. 30 + (6  1.64) = 39.8 MPa Chapter 13 1..

After one year of production with satisfactory results an increasing number of cube failures are reported from a concrete batching plant. a. Describe how you would determine whether the trend is significant. b. Describe the various possible causes for the change and how you would identify them. ……………………………………………………………………………….. a.

CUSUM (section 13.3) should be described Multivariate analysis (section 13.5) should be discussed b. Changes in materials, e.g. cement, aggregate identified by testing them or trying different ones.

Variations in batching procedure may be identified by increasing range of results (some high and some low). Sampling and testing variations may be identified by plotting CUSUM for range of groups of samples from the same batch (see section 13.4). May also be identified by testing concrete in the structures (will not have been affected). Appropriate tests would be pundit and rebound hammer.

2 a. Explain why cube failures from the concrete delivered to a site may not be significant. b. Describe the normal method used to determine whether the failures are significant. c. List the different causes which might be responsible for cube failures and describe how the relevant one may be identified when significant failures occur on site. Include a brief description of any test methods you suggest should be used. …………………………………………………………… a All concrete strengths are specified on a statistical basis and 5% failure rates are normally acceptable. The difference between cube results and the insitu strength of concrete should be emphasised. b Description of CUSUM method should include how graph is drawn and use of mask. c

The variations come from a large number of different sources described in section 13.2 Initial investigation should include analysis of range of cube results in sets (using CUSUM) and insitu testing of concrete (student should describe insitu test methods).

If range of test results has increased cause is related to test methods and could be identified by monitoring of method, testing repeat sets or using another lab.

If insitu test results are low then the cause is in concrete supply. The normal cause would be batching error but it could be materials or temperature.

If both insitu results and range are ok it may be a systematic error in testing (e.g. calibration of machine or curing temperature) or it may be bad concrete supply, but it requires more insitu testing to find it.

Chapter 14 1. a. What is the difference between a material which carries a British Standard "kitemark" and one which is supplied as complying with the relevant British Standard but does not carry a kitemark. b. What are the main implications of the introduction of European Standards for specifiers and purchasers of materials in the UK. ……………………………………………………………………………………………………. a) A kitemark indicates that the material has been produced by a BS approved supplier who has been and if found to be satisfactory. These organisations become BSI registered firms. The firms are required to have an approved QA system and are subject to repeat inspections at regular intervals.

Any other supplier may state that their product complies with any standard and is subject to normal legal sanctions if the claim is false. c) see discussion in 14.3.3.

2. a. Describe what is meant by the terms "Quality Assurance" and "Quality Control" and state the difference between them. b. Discuss the differences between an agrément certificate and a National Standard. …………………………………………………………………. a. see definition in 14.6.1 b. see discussion in 14.3.2

3.

Explain the meaning of the following terms: a British Standard b ASTM Standard c ISO Standard d Euro Norm e Agrément Certificate f Quality assurance ……………………………………………………………………. see definitions in 14.3.2 and 14.6.1 Chapter 16 1.

Discuss the merits and weaknesses of current research which is being carried out into new materials and methods for concrete. Have the major failures such as widespread corrosion from carbonation, High Alumina Cement and Calcium Chloride been caused by poor research? – discuss ………………………………………………………………..

Students should discuss the failure of site practice to change in response to research findings. Is this caused by poor research or poor communication with practitioners? Has the internet helped or made matters worse with the volume of material now available and the general increase in the numbers of papers published? It can be argued that over-reliance on electrical chloride tests (see section 22.8.3) will lead to further widespread failures, particularly in the Middle East.

2 With the decline in coal burning power generation in the UK there is likely to be a shortage of PFA. Describe a 3 year research programme to evaluate imported PFA from Eastern Europe to assess its suitability for use in concrete in highway structures. You may assume that you have a good supply of the material and that you may obtain access to structures built with it in the country where it is produced. ……………………………………………………………………………………..

Defining the General Objectives.

It is quite rare to carry out an experiment simply to see if one material is adequate for an application.

Thus the aim is to see if the new method works better than the old one. The objective might be: "Will concrete mixes made with the imported PFA have the same durability as those made with PFA from UK sources?"

Often more complex objectives may be needed such as: "will this change to the concrete mix make it's susceptibility to sulphate attack more sensitive to poor curing" Reviewing previous work This may be done from literature searches, national standards etc. Methods and results are both of interest. Some references are more reliable than others. The best are from refereed journals. The essential point is never to rely on references from a single source, always look for independent confirmation of results.(see section 16.3) Resources and time limits Note that concrete takes several weeks to achieve its design properties. This time must be allowed for.

In particular for durability measurements exposure times of one or two years may be needed.

Choosing a method The three objectives are: 1. Make it realistic 2. Make it fast 3. Make it cheap These three are conflicting.

The samples.

In a real construction environment materials are never in optimum condition. Materials prepared on site (e.g. concrete) suffer most.

The difficulties with simulating site conditions are that all sites are different and that all scientific experiments must be designed to be repeatable.

One solution is to try to simulate both best and worst conditions. It is important to ensure that preparation methods do not bias the experiment towards one type of sample.

The geometry of the sample may affect durability - for example a high surface area/volume ratio will promote sulphate attack on concrete samples. The effect of this may be easily calculated so it may be used to accelerate the experiment.

The Environment.

In general real exposure experiments are not very useful because they are too slow. Deterioration may be accelerated with, for example, heat, pressure, applied voltages, or pre-contaminating the samples.

Each of these methods must be used with care. Heat slows down sulphate attack. Mixing chlorides into wet concrete makes it less permeable. etc. In this programme the existing structures should be tested but the data from them must be treated with caution. On any site (and more so overseas) it is difficult to get accurate data on the exact materials and methods used on a structure. The exposure conditions may also be difficult to estimate.

The null hypothesis The null hypothesis for this programme is that changing the source of pfa does not harm the concrete. If this is shown to be false with a degree of certainty of, say, 5% the new pfa cannot be used with current methods.

The detailed Programme A possible selection of tests would be: Workability Compressive Strength Permeability Chloride diffusion Freeze-thaw Sulphate attack The sensitivity of these to w/c, curing, admixtures, aggregate type etc. should also be checked. A full programme of control tests with UK pfa must be included.

3 A slag arising from the production of a non-ferrous metal has been found to exhibit cementitious properties and is being proposed for use in concrete. You have been awarded a 6 month contract to evaluate the material for use in mass concrete foundations. Describe the major parts of your programme. …………………………………………………………………………………………………………

The major parts are: 1. Define the general objectives. e.g. which properties are important (relate to use as mass foundation, e.g. no reinforcement to protect) 2. Review previous work.

Should describe types and sources of literature and precautions to be taken (e.g. try to find two sources for important items) 3. Identify available resources and time limits.

4. Choose a general method Student should indicate tests to be carried out, e.g. cube strength, sulphate resistance etc.

5. Establish a null hypothesis

“This material is as good as OPC” 6. Write the detailed method to prove the null hypothesis Student should outline purpose of statistical analysis but need not specify statistical methods.

4 When designing a large concrete structure you find a substantial source of secondary (recycled) aggregate close to the site and are considering it in the concrete in order to reduce costs. a. Describe how you would carry out a literature search to find information on the performance of concrete made with the aggregate. Describe the type of literature you would expect to find indicating the relative merits of each type. b. Describe two experiments that you would propose to determine the durability of the concrete and give reasons why the two you have chosen are the most suitable tests for this purpose. These may be either laboratory or insitu tests. ............................................................................. a see section 16.3 b Possible tests (the student should describe two tests): Tests on the aggregate (e.g. percentage fines or water absorption) Tests on concrete made with the aggregate (old samples if possible): Strength, workability etc.

Permeability to liquid/gas: lab test - good indicator of durability Surface absorption (ISAT): probably used insitu - concrete must be dry Phenolphthalein test for carbonation - good simple insitu test 5 When designing a large concrete structure you find a substantial source of mineral waste close to the site and are considering it as aggregate in the concrete. Describe how you would assess the material to see if it was suitable. Justify your choices. You may assume that you have about one year, full laboratory facilities, and access to some old concrete structures built with the aggregate. ............................................................................. literature survey: look for evidence of problems. experimental design: must compare new aggregate with realistic alternative. Aim is to show statistically that it is not any worse. All experiments to have control samples. start with standard tests, e.g. workability and strength. check to see that it can be crushed to a consistent grading.

Cast mortar bars to test for expansion.

Possible tests for durability: Permeability to liquid/gas: lab test - good indicator of durability

Surface absorption (ISAT): probably used insitu, i.e. on the old structure - concrete must be dry Gas transport (Figg) - good insitu test Chapter 18 1 a. Describe FOUR different types of cement and give typical applications for them. b. List TWO different types of cement replacement material and describe how they are produced and used and what properties they give to concrete. ........................................................................... a Four of: Ordinary Portland Cement, OPC .. Now called PC Sulphate Resisting Cement, SRPC Rapid Hardening Portland Cement, RHPC Low Heat Portland Cement, LHPC White Cement Hydrophobic Portland Cement Masonry Cement See section 18.2.5 for details b Two of: GGBS (Ground Granulated Blastfurnace Slag) PFA (Pulverised Fuel Ash) CSF (Condensed Silica Fume) Natural Pozzolans See section 18.3 for details 2 a. Describe the different types of cement which are used to make concrete and how they may be identified from an analysis of dust samples drilled from a concrete structure. b. A 50mm thick concrete barrier has pure water on one side and 80kg/m3 salt solution on the other side. Di is 2 10-12 and the porosity is 15%. What is the flow in kg/m2/s in the steady state? What is the velocity of the ions in solution 20mm from the pure water side? ...........................................................................

See section 18.2.5 Identification.

This is done with an oxide analysis. Can be done with wet chemistry but normally XRD. The cement compounds are calculated from the oxides with the Bogue equations.

Sulphate Resisting Cement ... low C3A Rapid Hardening Cement ... High C3S White cement ... Low C4AF Marks will also be given for a description of portland-pozzolanic cements together with comments on their oxide percentages. b.

Flux F= Di dCl/dx = .15  2E-12  80/.05 = 4.8E-10 kg/m2/s

Flux = concentration × seepage velocity × porosity Cl will have linear gradient because F is constant (steady state) thus Cl = 80 × 20/50 = 32 kg/m2 thus v = 4.8E-10/(32 × .15) = 1 E-10 m/s 3 a. Identify the types of cement which have: (i) a high C3S content. (ii) a low C3A content. (iii) a low C4AF content. for each one explain why they are made in this way. b. What are pozzolanic materials and what are the effects of the pozzolanic reaction in concrete? c. Identify two pozzolanic materials that are used in concrete production and describe the properties of concrete made with them and applications where these properties are used. ....................................................................... a i Rapid hardening cement has a high C3S content because C3S is responsible for the early strength. ii Sulphate resisting cement, because it is the C3A that reacts with the sulphate. iii White cement, because the C4AF gives cement its grey colour. b A pozzolanic material reacts with lime (usually produced by the cement hydration reaction) to form additional cementing compounds.

See section 18.3.3 for a list of effects. c PFA Properties: Good workability, low heat, low permeability, good durability.

Application: Mass pours, marine works.

CSF Properties: Very high strength, high cohesion Application: Special high strength columns.

Natural Pozzolans Properties: Slow strength gain, good workability, good durability.

Application: Footings, building mortars.

4 (a) How are the proportions of the cement compounds adjusted to give the following properties: 1. Low heat 2. Rapid hardening 3. Sulphate resistance 4. White cement (b) Describe TWO cement replacement materials and describe how they may be used and what properties they give to concrete ……………………………………………………………….. (i) Low C3S

(ii) High C3S (but mainly fine grinding) (iii) Low C3A (iv) Low C4AF b) The student will probably describe PFA and GGBS but CSF, natural pozzolans etc. would also be acceptable.

See section 18.3 for descriptions 5 State the full names of the following materials and describe their production and their uses in concrete construction: a. PFA b. GGBS c. CSF d. CaCO3 ………………………………………………………………………….. a) Pulverised Fuel Ash Produced in power stations from burning pulverised coal. Fly ash is precipitated and classified to select particle sizes.

Used as cement replacement at levels up to 40%. Gives slower early strength gain but good durability and generally lower cost. b) Ground Granulated Blastfurnace Slag Produced by granulating (by rapid cooling) and then grinding slag from steel production is blastfurnaces.

Used as cement replacement. Can be at higher percentages than PFA but strength gain becomes very slow. Gives good durability and white colour. c) Condensed Silica Fume Produced from silicon steel manufacture. Collected in bag filters.

Used as a cement replacement at low levels (below 10%). Gives high strength and good durability if used very carefully. Can also be used to make underwater concrete. Needs a superplasticiser for effective use. Should not be used for ordinary concrete (special applications only). d) Calcium carbonate (limestone) Produced from quarries.

Used as an aggregate and also as limestone fines in cement.

6 a Calculate the compound compositions for Portland cements with the following oxide analyses. (8 marks) Sampl e C S A F S

A

B

C

63 20 6 3 2.5

64.15 21.87 5.35 3.62 2.53

62 22.5 3 4.2 4

b Calculate the total heat generated by 1m3 of concrete made with each of the cements with a cement content of 300 kg/m3 c Outline the main consequences of the heat of hydration in concrete You may use the data in the following table: Compound Heat of Hydration J/g C 3S 502 C 2S 251 C 3A 837 C4AF 419 ……………………………………………….

C3S C2S C3A C4AF

52.7 17.7 10.8 9.1

b Heat J A 1.31E+08

B C 1.25E+08 1.08E+08

46.5 27.7 8.1 11.0

43.8 31.6 0.9 12.8

c Thermal cracking caused by differential or restrained thermal movement Excessive heat (over 80) caused by thermal runaway will damage hydrates.

Some heat is helpful to protect from frost.

Chapter 19 1 The following results are obtained after sieving three different samples of aggregate: Mass retained in sieve g Sieve Sample Sample Sample C A B 20 mm 0 0 0 9.5 mm 150 0 1100 4.75 mm 200 0 0 2.36 mm 170 100 0 1.18 mm 200 200 0 600 m 150 200 150 300 m 200 100 200 150 m 50 100 100 Base 20 35 20 a. Plot the grading curves b. Identify which is (i) sand, (ii) gap-graded aggregate and (iii) all in aggregate. c. Discuss the problems that occur with aggregate shrinkage and how they may be avoided by using standard tests. ...................................................................... a

120.00 100.00 80.00

A

60.00

B

40.00

C

20.00 0.00

b A is all in B is sand C is gap graded c see section 19.6.3 2 a The following results are obtained after sieving three different samples of aggregate.

Sieve 20 mm 9.5 mm 4.75 mm 2.36 mm 1.18 mm 600 m 300 m 150 m Base

Mass retained in sieve g Sample Sample Sample A B C 0 0 0 1200 0 300 0

0

100

0

80

200

0 150 200 100 54

100 200 200 50 35

300 200 100 20 16

(i) Plot the grading curves. (ii). Identify which samples are (i) gap graded aggregate and (ii) all in aggregate (iii) sand. b Describe two different major sources of aggregate for concrete. ……………………………………………………………

120.00

100.00 80.00

A

60.00

B

40.00

C

20

9.5

4.75

2.36

1.18

0.6

0.3

0.15

0.00

Base

20.00

(ii) b is sand, a is gap graded, c is all in b. see sections 19.4 and 19.5 Chapter 20 (For these questions assume the specific gravity of the unhydrated cement is 3.15 kg/l and the for the hydrated cement it is 2.15 kg/l) 1 Define the following for cementitious materials: a. Porosity b. Specific gravity. c. A concrete sample is made with the following proportions: water/cement ratio: 0.45 Cement content: 350 kg/m3 Aggregate content: 1800 kg/m3 The sample is cured and dried and the density after drying is 2200 kg/m3. What is the porosity ? ............................................................................ a Porosity = volumetric proportion of voids b SG = skeletal density c

Cement Water Hyd Cem Pores Agg Total

Before Hydration Mass kg Vol l 350 111 158 158 0 0 0 0 1800 731 2308 1000

Water lost

108

After Hydration Mass kg vol l 150 48 250 1800 2200

116 105 731 1000

Water in HC Cem in HC Total HC

50 200 250

Porosity %

10.5

2 (a) A concrete sample is made with a w/c ratio of 0.53 and a cement content of 370 kg/m 3 and an aggregate content of 2000 kg/m3. A cube of the concrete is dried and the weight loss on drying is 4.5%. . What percentage of the cement has hydrated and what is the porosity. (b). Describe the effects of the heat of hydration of cement in concrete. (c). Describe the different methods that may be used to reduce the heat of hydration and the effect it has on the concrete. …………………………………………………………………………….. a) Before Hydration After Hydration Mass kg Vol l Mass kg vol l Cement 370 117 47 15 Water 196 196 Hyd Cem 0 0 403 188 Pores 0 0 111 Agg 2000 686 2000 686 Total 2566 1000 2451 1000 Water lost Water in HC Cem in HC Total HC

115 81 323 403

Porosity % % hydrated

11.1 87

b) Heat only causes chemical effects over 80C, the main effect is thermal movement which will cause build up of stresses and possible cracking. Heat can help with frost protection. c) Heat is reduced by: reducing cement content using cement replacements (reduces rate of heating, not total heat) Low heat cement (if available) pre-cooling to slow reaction rate using retarder effect of heat is reduced by: cooling (e.g. ice in mix) reducing thermal gradient (e.g. insulating shutters) expansion joints 3 a. Explain why a cement with a high w/c ratio will have poor durability. b. 5 kg of cement is mixed with 2.5 kg of water. The resulting paste sample is cured and dried and its final mass is 6.2 kg. Calculate the porosity. ………………………………………………………………………………………… a)

High w/c will leave free water which will not be used in the hydration reaction (normally added for workability). The volume occupied by this free water will become capillary pores which will normally be connected at above 10% paste porosity. This will provide pathways for aggressive species (and water for freeze/thaw) to move in and cause deterioration. b) Before Hydration

After Hydration

Cement

Mass kg 5.000

Vol l 1.587

Mass kg 0.200

vol l 0.063

Water Hyd Cem

2.500 0.000

2.500 0.000

6.000

2.791

Pores

0.000

0.000

Total

7.500

4.087

Water lost Water in HC

1.300 1.200

Cem in HC Total HC

4.800 6.000

1.233 6.200

4.087

Porosity %

30.169

4 10 kg of cement is mixed with 3 kg of water. The resulting paste sample is cured and dried and its final mass is 12 kg. Calculate the porosity ………………………………………………………………………

Before Hydration After Hydration Mass kg Vol l Mass kg vol l Cement 10.000 3.175 2.000 0.635 Water 3.000 3.000 Hyd Cem 0.000 0.000 10.000 4.651 Pores 0.000 0.000 0.889 Total 13.000 6.175 12.000 6.175 Water lost Water in HC Cem in HC Total HC

1.000 2.000 8.000 10.000

Porosity %

14.390

Chapter 21 To use US method for these questions assume: Fineness modulus of fine aggregate

2.8

Dry-rodded density of coarse aggregate

1600kg/m3 (2700 lb/yd3)

Absolute density of aggregate

2600kg/m3 (4400 lb/yd3) (for question 1)

The cylinder strength should be taken as the characteristic strength given.

1 Calculate the mix quantities for a 0.2 m3 trial concrete mix to the following specification: Characteristic strength

30 MPa (4350 psi) at 28 days

No. of test samples

5

Cement

OPC

Percent defectives

5%

Target slump

100 mm (4 in.)

Coarse aggregate

10 mm (3/8 in.) crushed

Fine aggregate

crushed, 40% passing a 600m sieve

....................................................................................

UK Method TMS = 30 + 1.64 * 8 = 43.12 Table 2 strength = 49 w/c = 0.54 w=250 c=462 % fine agg = 57 Cem W per m3 462 250 per .2m3 92 50

FA 927 185

CA 701 140

US Method Step 1. The slump is 75 to 100mm Step 2. The maximum aggregate size is approximately 9.5mm Step 3. The estimated water content is 228kg/m3 and the entrapped air is 3% (table 21.4) Step 4. The w/c = 0.54 (table 21.5) Step 5. The cement content = 228/.54 = 422kg/m3 Step 6. Dry-rodded volume of coarse aggregate per unit volume of concrete = 0.46 (table 21.6) Thus mass of coarse aggregate = 0.46 × 1600 = 736kg/m3 Step 7. (using a mass basis) Estimated density = 2280 kg/m3 (table 21.7) Thus fine aggregate = 2280 – 228 – 422 – 736 = 894kg/m3 Step 7 (using an absolute volume basis) Volume of water = 228/1000 = 0.228m3 Volume of cement = 422/3150 = 0.134m3 Volume of coarse aggregate = 736/2600 = 0.283m3 Volume of entrapped air = 0.03 × 1 = 0.03m3 Thus volume of fine aggregate = 1 – 0.228 – 0.134 – 0.283 – 0.03 = 0.325m3 Mass of fine aggregate = 0.325 × 2600 = 845kg/m3 Thus the final quantities in kg/m3 are: water cement

fine aggregate

coarse aggregate

mass basis volume basis

228 228

422 422

894 845

736 736

US method with US units Step 1. The slump is 3-4 in.

Step 2. The maximum aggregate size is approximately 3/8 in.

Step 3. The estimated water content is 385lb/yd3 and the entrapped air is 3% (table 21.4) Step 4. The w/c = 0.54 (table 21.5) Step 5. The cement content = 385/.54 = 713lb/yd3 Step 6. Dry-rodded volume of coarse aggregate per unit volume of concrete = 0.46 (table 21.6) Thus mass of coarse aggregate = 0.46 × 2700 = 1242lb/yd3 Step 7. (using a mass basis) Estimated density = 3840 lb/yd3 (table 21.7) Thus fine aggregate = 3840 – 385 – 713 – 1242 = 1500 lb/yd3 Step 7 (using an absolute volume basis) Volume of water = 385/1686 = 0.228yd3 Volume of cement = 713/5311 = 0.134yd3 Volume of coarse aggregate = 1242/4400 = 0.283yd3 Volume of entrapped air = 0.03 × 1 = 0.03m3 Thus volume of fine aggregate = 1 – 0.228 – 0.134 – 0.283 – 0.03 = 0.325yd3 Mass of fine aggregate = 0.325 × 4400 = 1430kg/m3 Thus the final quantities in lb/yd3 are:

mass basis volume basis

water cement 385 713 385 713

fine aggregate 1500 1430

coarse aggregate 1242 1242

2 Calculate the mix quantities for a 0.25m3 trial mix to the following specification: Characteristic Strength

40 MPa (5800 psi) at 28 days

No of test samples

3

Cement

OPC

Proportion defectives

5%

Slump

100mm (4 in.)

Coarse Aggregate

20mm (3/4 in.) crushed

Fine Aggregate

crushed, 60% passing a 600m sieve

Aggregate SSD density

2500 kg/m3 (4225 lb/yd3)

.......................................................................................

UK Method TMS = 35 + 1.64  8 = 53.12 strength from table 2 = 49

w/c = .46 water = 225 c = 489 density = 2260 Agg = 2260 - 489 - 225 = 1546 % fine = 36

Per m 3 Per .25 m3

Cem 489 122

Water FA 225 556 56 139

CA 990 247

US Method Step 1. The slump is 75 to 100mm Step 2. The maximum aggregate size is approximately 19mm Step 3. The estimated water content is 205kg/m3 and the entrapped air is 2% (table 21.4) Step 4. The w/c = 0.42 (table 21.5) Step 5. The cement content = 205/.42 = 488kg/m3 Step 6. Dry-rodded volume of coarse aggregate per unit volume of concrete = 0.62 (table 21.6) Thus mass of coarse aggregate = 0.62 × 1600 = 992kg/m3 Step 7. (using a mass basis) Estimated density = 2345 kg/m3 (table 21.7) Thus fine aggregate = 2345 – 205 – 488 – 992 = 660kg/m3 Step 7 (using an absolute volume basis) Volume of water = 205/1000 = 0.205m3 Volume of cement = 488/3150 = 0.155m3 Volume of coarse aggregate = 992/2500 = 0.397m3 (density of aggregate given as 2500kg/m3) Volume of entrapped air = 0.02 × 1 = 0.02m3 Thus volume of fine aggregate = 1 – 0.205 – 0.155 – 0.397 – 0.02 = 0.223m3 Mass of fine aggregate = 0.223 × 2500 = 557kg/m3 Thus the final quantities in kg/m3 are:

mass basis volume basis

water cement 205 488 205 488

fine aggregate 660 557

coarse aggregate 992 992

US Method with US units Step 1. The slump is 3-4 in.

Step 2. The maximum aggregate size is approximately ¾ in Step 3. The estimated water content is 340 lb/yd3 and the entrapped air is 2% (table 21.4) Step 4. The w/c = 0.42 (table 21.5) Step 5. The cement content = 340/.42 = 810kg/m3 Step 6. Dry-rodded volume of coarse aggregate per unit volume of concrete = 0.62 (table 21.6) Thus mass of coarse aggregate = 0.62 × 2700 = 1674 lb/yd3 Step 7. (using a mass basis) Estimated density = 3960 lb/yd3 (table 21.7) Thus fine aggregate = 3960 – 340 – 810 - 1674 = 1136 lb/yd3 Step 7 (using an absolute volume basis) Volume of water = 340/1686 = 0.202yd3

Volume of cement = 810/5311 = 0.153m3 Volume of coarse aggregate = 1674/4225 = 0.396yd3 (density of aggregate given as 4225 lb/yd3) Volume of entrapped air = 0.02 × 1 = 0.02yd3 Thus volume of fine aggregate = 1 – 0.202 – 0.153 – 0.396 – 0.02 = 0.229m3 Mass of fine aggregate = 0.229 × 4225 = 968 lb/yd3 Thus the final quantities in lb/yd3 are:

mass basis volume basis

water cement 340 810 340 810

fine aggregate 1136 968

coarse aggregate 1674 1674

Chapter 22 1 a. Describe the precautions which should be taken when carrying out a slump test on site. b. Rheological tests are carried out on three different concrete mixes and the results are as follows:

Shear rate (relative units) 0 0.2 0.4 0.6 0.8 1.0

Shear Stress (relative units) Control Mix A Mix B mix

0.5 0.62 0.73 0.83 0.91 1.0

0.3 0.41 0.52 0.61 0.7 0.8

0.4 0.5 0.6 0.62 0.64 0.66

Discuss the conclusions that may be drawn about the performance of these mixes in the following: i A slump test ii A concrete pump iii Placement from a skip ………………………………………………………………………………………………………………. a) see bullet list in section 22.2.3 b) i .Slump works at low shear so mix A will have high slump ii Concrete pump will pump fastest with mix B but A will be best if line gets blocked iii Placement from skip will be best with A because it will not get stuck (low shear).

2 a. What is the difference between air entrained concrete and foamed concrete? b. When the air content of a sample of concrete is measured using a pressure type air meter the following observations are made: Volume of concrete: 15 litres Initial pressure: open to atmosphere Pressure change: 1.3 Atmospheres

Volume change: 160 ml What is the percentage of entrained air? c. Describe the applications and correct methods of use of air entrained concrete. …………………………………………………………………. a air entrained has about 4% air for frost resistance etc. foamed has about 30-60% air and no coarse aggregate and is for trench backfill etc. b v2/v1 = 1/2.3 = .435 v1 - v2 = 60 thus v1 = 283 thus air = 283/15000 = 1.88% c Applications Frost resistance: roads etc.

Cohesion enhancement to reduce bleed/plastic settlement etc.

Methods. must be added at plant for quality assurance minimise haul times check air content on site watch out for PFA Chapter 23 1 Describe the different types of non-structural cracks that might occur in a concrete road bridge within the first year after casting. Describe the causes of the cracking and the methods that may be used to prevent it.

Indicate which parts of the bridge might be affected by each type of crack …………………………………………………………………………………

The student should discuss several of the types of cracking in sections 23.3.4 to 23.3.10.

For example: Drying shrinkage in an in-situ deck Early thermal cracking in abutment walls.

Plastic settlement in the deck (particularly with downstand beams) Plastic shrinkage in the deck Crazing in the columns if steel shutters used.

2 Define the following terms in concrete construction: (a) Segregation (b) Honeycombing (c) Bleeding (d) Plastic settlement (e) Plastic cracking (f) Describe two different methods which could be used to rectify plastic cracking. (g) Cracks have been observed in a suspended concrete floor slab in a new residential development.

Outline a programme of investigation to determine the cause of the cracks. Justify your choices.

……………………………………………………………. (a) Segregation is the separation of the constituents of concrete after mixing, e.g. bleeding. (b) Honeycombing is the visible effect of segregation, i.e. coarse aggregate with no fines visible when shutters removed. (c) Bleeding is the loss of water from the top of the pour. (d) Plastic settlement is the settlement due to loss of volume in bleeding. (e) Plastic cracking is the cracking caused by plastic settlement when restrained, e.g. by rebars. (f) (two of:) re-vibration after 2-4 hours air entrainment reduction of water content Improve curing (g) The key observations are: type of cracks (e.g. crazing is fine map pattern) location of cracks, e.g. over downstand beams time of occurrence (and growth) construction details (e.g. mix design) The student should discuss the causes of cracking in section 23.3. e.g. if the cracks occur in the first few hours the cause will be plastic settlement or plastic shrinkage. The concrete should be checked for bleeding or rapid early drying. The cracks should then be checked to see if they are over the reinforcement or any major change in section, waffle sections or columns.

Structural causes (e.g. early loading and shutter movement) should also be described.

Chapter 24 1 a. Describe the consequences of using concrete with insufficient workability in construction. b) Describe two ways in which the workability of a concrete mix may be increased and state the advantages and disadvantages of each. c) Describe the precautions which should be taken when carrying out slump tests on site to ensure that an accurate result is obtained. …………………………………………………………………….. a) Insufficient workability will lead to: Excessive work required for placing.

Voidage: blowholes and honeycombing.

Poor surface finish. b)

i add water. cheap but lowers strength and durability. ii add water and cement. Expensive but strength and durability maintained. May lead to excessive heat of hydration. iii add plasticiser/superplasticiser. Some additional cost. c) see section 22.2.3 2 a) Describe what calcium chloride is used for in concrete and give examples of where it should and should not be used. b) Discuss the difficulties which occur when frost resistance is required in readymixed concrete which must be batched at a plant which is a substantial distance from the site. Your discussion should include a description of methods which may be used to overcome the difficulties. c) Discuss the different methods which may be used to achieve a high early strength of concrete.

Describe the circumstances under which each method should be used. ………………………………………………………………………………. a) accelerator (admixture). Corrodes reinforcement so should only be used in unreinforced concrete except special circumstances, e.g. temporary works or permanently frozen concrete. Also causes efflorescence. b) Frost resistance requires air entrainer and air content may be lost in long haulage. Some research suggests that GGBS may give frost resistance. Air entrainer should be added at site – this requires well organised quality system which ensures proper supervision at all times and should also include dispensing system which places the admixture well back in the drum. c) Superplasticiser with reduced water – generally suitable, possibly expensive. accelerator – ok but no chlorides with steel reinforcement silica fume – not generally a good idea warm curing – ok for small sections – never go above 70 C rapid hardening cement - ok 3 How may admixtures be used to achieve the following in concrete? a High slump b Frost resistance c Low heat of hydration d High ultimate strength e High early strength f Reduced cost g Reduced plastic cracking h Reduced bleeding i Improved durability j Improved pumping …………………………………………………………………… a High slump Add superplasticiser (SP) b Frost resistance

Air entrainer c Low heat of hydration Add SP and remove cement and water or possibly use retarder (delays heat and reduces max temp) d High ultimate strength Add SP and remove water e High early strength Add SP and remove water or use accelerator f Reduced cost Add plasticiser and remove cement and water g Reduced plastic cracking Air entrainer h Reduced bleeding Air entrainer or reduce water with SP i Improved durability Add SP and reduce water j Improved pumping Add SP (one sold as a pumping aid which works at high shear) 4 State how a typical concrete mix may be modified by the use of admixtures to achieve the following and in each case describe the effect on the durability of the mix: (a) High workability. (b) High strength (c) Low cost (d) Self Compacting Concrete ........................................................................................................... a) add plasticiser (p) or superplasticiser (sp) but leave mix design unchanged. May help durability by improving compaction but could cause segregation. b) add sp and remove some water. Will greatly improve durability by lowering w/c ratio and hence porosity. High early strength with accelerator – ok if chloride free. c) add p and remove some water and cement. Will reduce cement content but will keep w/c the same. Durability will be marginally improved as paste/aggregate ratio is reduced. d) Use sp but need viscosity modifier to prevent segregation. Durability should be good because no compaction needed.

Chapter 25 1 What would the influence of the following be on the durability of the concrete in a structure: 1. Decreasing chloride diffusion rates 2. Decreasing the capacity factor for chlorides 3. Decreasing the hydroxyl ion content 4. Applying a positive voltage to the steel

5. Applying a negative voltage to the steel 6. Increasing the pore sizes in the concrete ………………………………………………………… .1 will improve durability .2 will make the structure less durable – chloride will not be bound and can thus move to the steel .3 will increase resistivity and thus improve durability but will reduce pH which will decrease durability slightly .4 Will increase corrosion (see chapter 31) and also attract chloride ions to the steel .5 Will decrease corrosion and also repel chloride ions from steel. .6 Will probably increase connectivity of pores and thus increase transport but will reduce capillary suction. Note that this is different from increasing overall porosity.

2 Describe the processes which will limit the life of a reinforced concrete beam in a road bridge.

Indicate how materials selection and construction practice may be used to extend the life. ......................................................................................

Reinforcement corrosion: chlorides: pitting/black rust on rebar and tendons carbonation: red rust on rebar only Materials: mix proportions (low w/c), ggbs, pfa, coated rebar, chloride free aggregate.

Practice: Good compaction, good pour rates, moist curing, controlled permeability formwork.

3 a. Describe the processes which take place during the carbonation of reinforced concrete. b. If a structure is constructed with 20mm of cover to the steel and the carbonation depth is 10mm after 5 years when would it be expected to reach the steel ? c. Describe the effect of applying a negative voltage to the steel after the carbonation has started to reach it. d. What will happen in (c) above if chlorides are present ? ............................................................................ a Carbon dioxide direct from the atmosphere or dissolved in water transported in concrete by diffusion, pressure driven flow etc. reacts with the lime and other alkalis to form carbonates. causes gain in strength/ shrinkage causes corrosion... red rust ... causes spalling. b 25 years (Square root relationship) c Cathodic protection stops corrosion process (Fe++ attracted back to steel by electrostatic forces) will also re-alkalise by attracting alkali ions (Na+ and k+) to steel (may even cause ASR) d Chlorides will be driven away from steel by electromigration.

4

a) State with reasons which one or more of the four main transport processes for ions in concrete will be most significant in the following circumstances. In each case also describe the effects which will be caused by the ions in the concrete. (i) The parapet of a road bridge subject to frequent applications of de-icing salt. (ii) A concrete slab placed directly over hardcore in moist conditions. (iii) External cladding panels on a building (iv) A tank containing sea water. (v) A bridge beam under a direct current power rail. b) Describe how chloride transport in concrete may be influenced by materials selection ………………………………………………………….. a (i) Absorption (capillary suction) and diffusion will transport chlorides into the concrete and cause corrosion of the reinforcement causing rust staining and loss of strength. Carbon dioxide from atmosphere may also enter by diffusion causing corrosion directly and reducing chloride binding. (ii) Diffusion, absorption, and (if head of water present) pressure driven flow will transport sulphates into concrete from hardcore if membrane under slab is absent or failed. Sulphates will cause expansion and subsequent total failure. Carbonation or chemical ingress into the top of the slab also possible. (iii) Carbonation by pressure driven flow of air and solution diffusion in pore water will cause corrosion in areas of low cover. Acid attack from diffusion of atmospheric pollutants also possible. (iv) Pressure driven flow of sea water will cause corrosion due to chlorides on steel and sulphate attack on concrete. (v) Electromigration of chlorides due to induced electrostatic fields in reinforcement may occur but polarisation of steel will probably cause more damage. b Cement replacements will tend to reduce permeability and increase adsorption.

Sulphate resisting cement has lower adsorption.

Reducing w/c will reduce all transport.

Chapter 26 1 You are advised that segregation has occurred in a concrete pour. a. If you inspect the concrete when the shutters are removed what would you expect to see ? b. What are the effects of segregation on the quality of the structure ? c. What changes should be made to subsequent pours to avoid it ? ……………………………………………………………………. a honeycombing voidage arount box outs etc laitence on surface settlement plastic cracking b Bleed will cause loss of bond to steel

segregation may make cover layer porous (lack of aggregate) dusting on slabs c Re-design mix: less water (water reducer) better proportioned aggrgate better cohesion (PFA ?) re-vibrate may have been under or over vibrated 2 a) Explain the consequences of the following in reinforced concrete construction: i) Too much cover to the reinforcement ii) Insufficient cover to the reinforcement b) Describe the two types of curing that are required in concrete construction. Outline the effects of not using them. c) Describe the difference between a shutter oil and a release agent. d) Describe the difference between air entrained concrete and foamed concrete. ………………………………………………………………………………………………………… a) i) Too much cover gives an inefficient structure. ii) Insufficient cover gives poor durability and fire resistance. b) See section 20.6 c) A shutter oil is an inert mineral oil which physically prevents adhesion of the concrete to the shutter.

A release agent is a surface retarder which stops the outer surface from setting before the shutter is removed. Release agents have the advantage that they dry on the shutter (they are often solvent based). d) Air entrained concrete has about 5% air for frost resistance and is used for structures etc. The air is entrained by adding the admixture when mixing. Foamed concrete has about 60% air and no coarse aggregate. It is made by injecting foam into the mixer before placing. It is non-structural, e.g. for trench backfill.

3 Explain why the following materials have been used in concrete and describe the consequences of their use. a. High Alumina Cement b. Calcium chloride accelerator c. Reactive siliceous aggregate d. Sea water for mixing. ……………………………………………………. a. HAC used for high strength and good chemical resistance. Converts after a few years and loses strength, some collapses and much repair work. b. Calcium chloride used as accelerator especially for winter concreting. Caused widespread corrosion, some structures still affected.

c. Reactive siliceous aggregate. Caused ASR in some structures and distress in some of them. d. Sea water for mixing. Used extensively in Middle East because nothing else available (except at high cost). Caused widespread corrosion of reinforcement. However not actually as bad as chloride transport into hydrated concrete because the chloride gets bound into the hydrating mix (and normally accelerates it slightly).

4 (a) Why is vibration important when concrete is placed ? (b) Describe the correct methods for vibrating concrete (c) State the two distinct functions of curing in concrete construction. (d) Outline the effects of not curing concrete. (e) Describe how curing is carried out. ………………………………………………………………………………….. (a) If the air is not removed by proper compaction the presence of these large voids will:  Reduce the strength of the concrete, more than 5% loss of strength for each 1% of air.  Increase the permeability and hence reduce the durability and protection to the reinforcement  Reduce the bond between the concrete and the reinforcement  Result in visual blemishes such as blowholes and honeycombing. (b) See bullet lists in section 26.4.7 (c) Curing has two distinct functions 1 2.

To stop the concrete from drying out during hydration.

To retain heat at the surface.

(d) Hydration will stop due to loss of water. If this occurs a significant loss of durability and strength will occur.

Low temperatures must be avoided : i To prevent frost damage (below 5oC) ii To increase early strength iii To reduce temperature gradients which would cause cracking (e) see the first bullet list in section 26.5 Chapter 27 1 Describe a programme to investigate the structural condition of an estate of houses with precast concrete frames which are 40 years old. ……………………………………………………………………………………………………..

Determination of objectives, is it for mortgaging ? what will satisfy the lenders ?

Desk top study, are there any records of construction ?

Visual survey, are there any obvious signs of distress, corrosion etc.

Plan of investigation, detailed survey of one or two and random checks through the rest ? Is there one in poor repair that could be demolished ?

Start with chemical test for HAC (one house only) and calcium chloride.

If possible far best to take some structural members out of a house. Can be cut up for mechanical tests in lab.

NDT on other houses with, if possible, minimum disruption to finishes. Rebound hammer and Ultrasonic [student should describe tests].

Cover depth should be measured throughout. ISAT or FIGG in some locations. Phenolphthalein indicator in several locations will give carbonation depth.

2 Very large variations in the results from strength and slump tests are observed from a particular site.

Describe the possible causes of this.

State which of these causes would indicate the presence of poor quality concrete in the construction and outline the methods which could be used to detect it ………………………………………………………………………………………….

Possible causes are: Variations in supply Water added on site Poor sampling Poor cube manufacture/ bad moulds Poor curing and despatch of cubes Poor slump procedure, affected by vibration etc.

Of these causes the first two affect the insitu concrete Detect with: Rebound hammer, Ultrasonics, cores.

3 Describe a programme of investigation to determine the condition of the reinforced concrete in a road bridge. You should give brief descriptions the tests that you would plan to use. ..............................................................................

Marks will be given for the relevance of the discussion to a road bridge.

PRELIMINARY DESK TOP STUDY Before any site work is started all available documentation should be studied. Of greatest interest are mix designs, cube test results and sources of materials.

INITIAL VISUAL SURVEY It is always worth spending sufficient time for a detailed record of all cracks, spalling, segregation or movement to be made.

PLANNING THE INVESTIGATION

see section 27.2.3 TEST METHODS (should be described) Strength tests: Rebound hammer Pull out tests Cores Load tests Durability tests: Cover depth ISAT Figg Potential mapping Resistivity Phenolphthalein 4 Describe four methods which are used to measure strength of concrete either in the laboratory or on an existing structure. For each method give typical uses and discuss the limitations of the method. ..........................................................................

CUBE STRENGTH (student to describe standard procedure) Used for all routine quality control on site Does not give the actual strength of the concrete under load in the structure

TENSILE STRENGTH (student to describe standard procedure) This can be measured (with difficulty) directly on "dog bone" samples but is normally measured indirectly by cylinder splitting Used occasionally in labs Limited use because tensile strength not used much in design

FLEXURAL STRENGTH (MODULUS OF RUPTURE) (student to describe standard procedure) This is normally measured on 500mm by 100mm by 100mm prisms .

The ends of the failed prism may be tested as equivalent cubes.

Used occasionally in labs

Limited use because tensile strength not used much in design

ULTRASONIC PULSE VELOCITY TESTING The general procedure for use is to establish acoustic contact between the concrete and the emitter and receiver and record the transit time for the pulses. The distance is then measured and the velocity calculated.

Used on site either when cubes fail or on old structures.

Very sensitive to distance measurement, cracks and moisture in concrete etc.

PULL OUT TESTS There are many of these including the Windsor Probe. In them a stud is cast, drilled or shot fired into the surface and the force required to pull it out is measured. This can give a reasonable indication of strength (better than rebound hammers but not nearly as good as cores) LOAD TESTS see section 27.3.6 CORE TESTS see section 27.3.4 REBOUND HAMMER TESTS see section 27.3.3 Chapter 28 1 (a) What are the requirements of a masonry mortar when used for the external elevation of a house? (b) Describe how these requirements may be met with good site practice and materials selection. …………………………………………………………………………………….. (a) see sections 28.2.4 and 28.2.5 (b) workability: control water content, lime, air entrainers-plasticisers, "soft" sand. cohesive: generally as workability workability retention: may use retarder strength gain: no too much retarder, w/c ratio Strength: control of w/c ratio flexibility: lime content water resistance: w/c ratio uniform appearance: control of mixing, gauging boxes etc. not "1/2 bag in mixer"

frost resistance: air entrain, control w/c sulphate resistance: sulphate resisting cement or possibly cement replacement 2 You are specifying materials to be used in the masonry mortar for a large housing project. The work is to be carried out in an area where there is a good supply of a natural pozzolanic material. Describe how you would decide on the specification. ...........................................................................

Most natural pozzolanics are worth using.

Check cost and availability Check references/past experience with this material [student should describe assessment of reference material, see section 16.3] Obtain sample: rate of strength gain likely to be critical (determine number of courses to be laid each day).

Make up samples with various pozzolanic/cement ratios.

Test for:strength gain ease of use water absorption soundness (expansion) sulphate resistance as necessary creep etc [student should outline tests] Chapter 29 1 A substantial concrete retaining wall is to be built at the top of a beach at a coastal resort. Outline the methods that could be used in the design and specification of the wall to give it: a) Good durability. b) An improved appearance. …………………………………………………………………………………… a) see section 29.8 and sections 26.2 and 26.3 Use a low w/c ratio Avoid water running down vertical surfaces - provide "drips" to keep it clear.

Avoid highly congested reinforcement which may cause voidage.

Consider specifying controlled permeability formwork Use PFA or GGBS but make sure that the curing will be adequate Under severe conditions consider using coated/stainless reinforcement. b) see section 29.9

2

a Describe the advantages and disadvantages of using high strength concrete in a ten storey office block. b. Describe the different methods which may be used to give an exposed concrete wall an attractive appearance. ……………………………………………………………………………….. a Ten storey is not very tall … HSC probably not needed (unless it is taken to mean above 60 Mpa) Advantages of High Strength Concrete (HSC) High durability (e.g. abrasion and carbonation resistance) High strength to weight ratio, particularly for lightweight HSC High early strength allows faster construction and formwork reuse Increased elastic modulus and lower creep Smaller columns allow for more floor space Disadvantages of HSC Increased cost per unit volume More stringent quality control of materials and construction needed CSF mixes may self-desiccate giving failure at day joints.

Workability difficult to define and often declines rapidly with time after mixing Timing of concrete delivery and addition of admixtures becomes critical.

High heat evolution may necessitate use of low heat binders (e.g. PFA) and cooling measures.

Stiffness (modulus) does not increase in proportion to strength.

More than 28 days may be required to reach specified strength Structural members may exhibit brittle failure.

Use of HSC not covered by codes b Concrete with good appearance (Architectural concrete) see section 29.9 Chapter 30 1 a. Explain why the grain size of steel has a significant effect on its mechanical properties. b. Describe two different methods for controlling grain size. c. For each of the two methods in (b) describe the effect of subsequent welding. …………………………………………………………………………………………………… a.

Plastic deformation and failure is by lattice dislocation... cannot propagate across grain boundaries. b.

Quenching.

Cold working.

Alloying. c Alloying unaffected but the others will be affected depending on the rate of cooling after welding.

2 Explain the following terms when applied to structural steels. For each one also describe their effect on the steel: a. Lattice dislocation b. Grain boundary c. Work hardening d. Quenching e. Annealing ........................................................................... a. A lattice dislocation is a point defect in a crystal which can move through the structure permitting shear without gross movement of atoms. b. A grain boundary is the edge of an individual crystal in the structure of a metal. The grain boundaries impede the movement of the dislocations. c. Work hardening involves repeated deformation beyond the elastic limit. It breaks up the grains and increases the number of grain boundaries. d. Quenching is rapid cooling. This will prevent the formation of large grains and will also prevent the crystalline changes that normally occur with slow cooling. e. Annealing is heating followed by slow cooling in an inert atmosphere. This will increase grain sizes and improve ductility.

3 a Define the terms “0.2% proof stress” and “yield stress” and state the relative advantages of using each of them to specify the strength of steel. b. Describe the effect of increasing the carbon content in iron compounds. Your description should include the effect on the microstructure. c. Describe the effect of cold working steel and indicate how the microscopic structure is affected and how this causes the changes. ………………………………………………………………………………………….. a see section 30.6.2 b see section 30.2 c Cold working is a method of reducing the grain size. This reduces the mobility of lattice dislocations because they cannot cross grain boundaries. Thus cold working (strain hardening) will make a steel stronger but more brittle. Continuous cold working will cause failure. see section 30.3.4 Chapter 31 1 a) Describe the procedure for determining the condition of the reinforcement in a concrete structure using linear polarisation. b) The polarisation resistance and rest potential of a steel sample are measured at 1600 and 300mV.

(i) What is the corrosion current? (ii) If the rest potential falls to -400mV what does the corrosion current become?

(Assume values for the Tafel constants B1 = B2 = 0.13V.) ………………………………………………… a See sections 31.4 and 31.5.2 b (i) B = (.13  .13)/((.13 + .13)  ln(10)) = 28mV Icorr = B/Rp = 28mV/1600 = 17 A (ii) -.300 = Vc0 - B2 log(17/Ic0) and -.400 = Vco - B2 log(x/Ic0) where x is the new current. solving .1/.13 = log(x/17) x = 98 A 2 The Tafel constants for a corroding steel sample in concrete are B1 = B2 = 0.16V a. If the polarisation resistance is measured as 1700  what is the corrosion current. ? b. If the rest potential subsequently falls from -350 to -600 millivolts and the cathode conditions do not change what is the new corrosion current ? c. Discuss the relative advantages and disadvantages of linear polarisation and rest potential measurements on reinforcing steel in concrete. ………………………………………………………………………………………………………….. a B= .16 * .16 = 34.7mV (.16+.16)*2.303 Icorr =

B 1700

= 20.4A

b -.35 = Vc0 - .16 log (20.4/Ic0) -.6 = Vc0 - .16 log(Icorr/Ic0) thus .25/.16 = log(Icorr/20.4) thus Icorr/20.4 = 36.51, Icorr = 746A c Rest potential: Is fast Is cheap does not need cast in probe only tells if corrosion is possible, does not give corrosion rates.

Linear Polarisation: Much more complex Gives values for actual corrosion rates Requires probe (isolated piece of steel) Requires knowledge of Tafel constants 3 a) Contrast the different methods which may be used to assess the rate of deterioration of steel reinforcement in a concrete structure. b) Why is the word “linear” used in the term “linear polarisation”. c) The polarisation resistance and rest potential of a corroding steel sample are measured at 1300 and -350mV. (i) What is the corrosion current? (ii) If the rest potential falls to -500mV what does the corrosion current become?

(Assume values for the Tafel constants B1 = B2 = 0.12V.) ................................................................................ a Potential survey This is quick and cheap to use but only gives an indication whether corrosion is possible – not whather it is actually happening. The system is widely used and is very useful for identifying problem areas for further investigation.

Linear polarisation This can only be used if there are cast in probes (there are some guard ring systems on the market but these are not proven). This is the only system that will give a valid reading for the rate of corrosion.

Visual (with or without breaking open) Surface cracking and staining will reveal a great deal. Note that at least one area must be exposed to make a contact for potential survey.

Surface transport measurements.

Tests such as ISAT give good correlations with corrosion but do not always work. These should be used as back-up but an electrical method should always be used with them Chemical sampling Drilling out samples for chloride testing will give a good indication of the corrosion b.

Linear polarisation works be polarising both the anodic and cathodic process. These processes both have logarithmic impedances but when one is subtracted from the other the difference is linear over a small region near the rest potential.

Thus the relationship between voltage V and external current Ix is equivalent to a resistance Rp and a voltage step Eo in series. c (i) B = (.12  .12)/((.12 + .12)  ln(10)) = 26mV Icorr = B/Rp = 26mV/1300 = 20 A (ii)

-.350 = Vc0 - B2 log(20/Ic0) and -.500 = Vco - B2 log(x/Ic0) where x is the new current. solving .15/.12 = log(x/20) x = 356 A 4 A steel sample is cast into a concrete structure and is kept electrically insulated from the reinforcing cage. By applying a voltage relative to the reinforcing cage the following observations are made: Voltage relative to reference electrode at Current through sample and reinforcing cage. concrete surface mV A -380 35 -390 0 -400 -35 -410 -70 -420 -105 a. Assuming B = 26 mV, calculate the corrosion current in the bar. b. Assuming the mass and charge of an Fe ++ ion are 9.3  10-26 kg and 3.2  10-19 Coulomb calculate the rate of loss of mass from the bar. c. Give the name of the test described above and discuss its relative merits when compared to other electrical methods for measuring corrosion of steel in concrete. ............................................................................

From the table: when Ix = 35 A V-E0 = 10 mV Thus Rp = (v-E0)/ Ix = 285 Ohms Thus Icorr = B/Rp = 91 A b. Rate = 91  10-6 Coulomb/s = 91  10-6  9.3  10-26 / 3.2  10-19 = 2.64  10-11 kg/s c.

Test is linear polarisation.

Gives reading for actual corrosion rate but needs cast in probe from time of construction (as in Channel Tunnel segments) or complex and unreliable guard ring system.

Potential mapping:  Potential mapping is quick, non-destructive, and cheap. Large areas may be covered.  The results must be treated with caution. For example carbonation increases concrete resistivity and therefore increases apparent rest potential.  The results only indicate that corrosion is possible and give no information about corrosion rates.

Resistivity.

Good test gives fair prediction of corrosion and is easy to do but only measures the surface to give indirect reading for corrosion rates.

Chapter 32 1.

Describe the processes that take place when each of the following are cooled from a liquid state to a solid state: a. A mixture of two metals which are completely insoluble in one another. b. An alloy of two metals which are partially soluble in one another. c. An alloy of two metals which are completely soluble in one another. d. Discuss the use of two different non-ferrous metals (or alloys) in construction and explain why they are used in preference to ferrous metals. ……………………………………………………………………….. a (i). The two metals will separate with one below the other and solidify separately. (ii). Result will depend on the relative proportions of each metal. Structure will resemble that of metal in highest proportion, this will start to solidify first. If there is more than a minor amount of other metal present eutectic will form at lower temperatures and solid will have some of this in it. (iii). When the alloy starts to cool it will form a solid with a high proportion of the metal with the highest melting point in it leaving the remaining liquid rich in the metal with the lower melting point. With further cooling the proportions in the solid will move towards those in the original liquid. b. Any of the following could be discussed: Aluminium,e.g. windows, low corrosion, light weight, high thermal cond, lower strength than steel.

Copper, e.g. pipes, low corrosion Brass, e.g. pipe fittings Lead, e.g. roofing (restoration).

2.

Some properties of some non-ferrous metals are as follows: Property Lead Zinc Copper Standard electrode -0.12 -0.76 +0.34 potential (V) Relative density 11.3 7.1 8.7 o Melting Point C 327 419 1083 Elastic Modulus GPa 16.2 90 130 Tensile Strength MPa 18 37 210 Thermal Conductivity 35 113 300 o W/m C

Aluminium -1.7 2.7 659 70.5 45 200

Describe a major application of each metal which takes advantage of one of the properties listed and describe how the property makes the metal suitable for the application. Also describe the influence of the other properties affects the performance of the metal in the application. ........................................................................

Typical examples Lead Roof flashings

High potential gives low corrosion (unless copper nails used)/ High density causes larger loadings if large areas but also prevents uplift./ low melting point gives easy welding /low modulus gives easy working/ low strength means easy cutting.

Zinc Galvanising low potential protects steel/ low melting poins reduces energy demand in process Copper pipes high potential prevents corrosion/ high strength means good spacing between supports/ high thermal conductivity useful in calorifiers but means insulation often needed.

Aluminium window frames low potential promotes stable oxide layer/ high thermal conductinity often gives condensation.

Chapter 33 1 A client is considering the use of tropical hardwood for the front facade of a public building. a. Briefly discuss this choice of materials in the context of current public concern about the destruction of tropical forests. b. Describe the processes which will affect the long term performance of the wood. c. Indicate how materials selection and detailing may be used to improve this long term performance. ........................................................................................................................................ a Some timber comes from sustainable forestry - use this and keep records of source if possible.

50% of tropical hardwood is burnt at source so increasing use may not increase cutting rate. provides income for 3rd world. see section 33.2.1. b see section 33.7. c Select durable species (and not exotic).

Use heartwood Season to moisture content found insitu.

Pressure impregnate with preservative Try to ensure that water does not penetrate joints.

Do not leave end grain exposed Minimise contact with ground Seal joints with masonry and concrete Ventilate enclosed voids see section 33.8 Chapter 34 1 a. Describe what is meant by the term "efflorescence" on brickwork and discuss how the problem may be avoided.

b. Describe the methods that should be used to achieve good appearance and durability of brickwork. ……………………………………………………………………. a Efflorescence is leaching of sulphate and other salts from bricks or lime from mortar. Causes unsightly white deposits on surface. avoid by: keeping dry if possible exhausting by washing using low efflorescence bricks avoiding calcium chloride in mortar using low w/c mortar b See summary in section 34.11.

Chapter 35 1 a. What is meant by the following (i) Organic materials (ii) Polymeric materials (iii) Thermoplastics (iv) Thermosetting plastics b. What are the processes which limit the long term performance of polymeric materials in construction? c. Identify two thermosetting resins and describe how good site practice may maximise their long term performance. ………………………………………………………… a.  Organic materials: These are materials originating from living organisms. Strictly all materials containing carbon are defined as organic.  Polymeric materials: A material formed of large molecules which are built up (polymerised) from a large number of small molecules (monomers). The usual (but not only) example is the organic polymers.  Thermoplastics always soften when heated.  Thermosetting plastics which set, i.e. polymerise when mixed with a "hardener" and will not soften when heated. b. see section 35.5 c Thermosetting resins.

These are mixed on sites, e.g. for setting in holding down bolts into existing slabs.  Epoxies. Two components must be mixed in exact proportions ("hooks and eyes"). Will cure down to 5oC.  Polyesters. Sets with a catalyst. More catalyst gives faster setting. Can be recognised by the characteristic "fibreglass" smell.

For all Thermosetting resins.  For low permeability the resins must be cured correctly in VERY DRY conditions. A white "bloom" on the surface indicates moisture during curing.  The setting reactions are exothermic (especially polyesters). If use is delayed after mixing the set may be delayed by placing the material in a shallow metal container. see section 35.3 Chapter 36 1 Describe the properties of the following materials and explain how these properties are used in construction. a. Toughened glass b. Laminated glass c. Low-emissivity glass d. Tinted glass. e. Clear acrylic sheet. f. Clear polycarbonate sheet. ……………………………………………………………….. a.

Toughened glass has higher strength and, if it fails, it produces small fragments. Is used for safety and security in low level windows and doors. b.

Laminated glass also has high strength and is safe when broken. Heavier than toughened and more expensive but can be cut so is used where special cutting needed. c.

Low emissivity has high reflection of i.r. frequencies emitted by warm objects. Thus used in glazing where heat retention needed. d.

Tinted glass absorbs light at most frequencies. Can be used to reduce solar gain but gets hot itself so allowance must be made for expansion. e.

Acrylic sheet (Perspex) cheap, light and easily cut. Cracks easily. Used for applications where fracture unlikely. f.

Polycarbonate much tougher and more expensive than acrylic. Bullet-proof when laminated. Used for security and vandal proof applications.

Chapter 37 1 Discuss the factors which contribute to the relative durability of concrete and bituminous road pavement construction materials ........................................................................................................................

Bitumens (asphalt) intrinsically durable - occur as a natural material.

Will polymerise, oxidise and loose light oil components when exposed to air and heat. This will make them harder and possibly brittle.

Virtually impermeable Intrinsically resistant to plant growth Softened by high temperatures - may cause rutting softened by solvents (e.g. fuel spillage) - will take a long time to harden so may require removal.

Concrete Corrosion due to carbonation and chloride ingress not very common in road slabs.

Cracking and spalling common Joints (if present) often not durable Continuous pavement can be damaged in hot weather.

Overall.

Design life for concrete is higher. Concrete is rigid and thus theoretically does not fail in fatigue due to heavy traffic. In practice concrete now rarely used.

2 a) Define the following terms: (i) Binder (ii) Bitumen (iii) Tar (iv) Asphalt (v) Mastic (vi) Macadam (b) Describe the components of a typical flexible road construction. (c) Describe the common failure mechanisms for the components. ………………………………………………………………………….. (a) see definitions in 37.1.2 (vi) Macadam: A road construction material with binder and coarse aggregate. (b) see section 37.8 (c) Failure mechanisms.

Wearing course: Rutting (shear failure) Potholes (bond failure) loss of skid resistance (viscous flow) Lower courses Cracking (Fatigue failure) Breakdown due to failure of lower courses (e.g. reflected cracking from dry-lean concrete) Chapter 38 1 Give 3 examples of composites that are used in construction and discuss why they are better than equivalent components made from a single material.

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Typical examples could be: ...Reinforced concrete Steel in tension, concrete in compression. Gives lower cost solution and higher durability than plain steel. ...Composite bridge deck – concrete slab on steel beams with shear studs.

Easy to build because steel carries formwork (with propping). ... Timber composites (e.g. plywood) – can maintain good appearance but have a far more useful shape. Can have different materials on surface Chapter 39 1 a. Describe the common failure mechanisms for adhesive joints b. Explain why detailing of joints with sealant is important and give an example of correct practice.

Illustrate your answer with a sketch. …………………………………………………………………………… a see sections 39.2.5 and 39.2.6 b The detail must make sure that when the joint moves the strain of the sealant remains below about 50%. In the sketch shown the backing strip below the sealant (e.g. paper) which prevents adhesion is vital to achieve this. (figure 39.4).

Chapter 40 1 Discuss the factors which should be taken into consideration when choosing one of the following materials for the frame of a small industrial building: a. Timber b. Steel c. Precast concrete d. Brick structural walls with steel roof structure …………………………………………………………………………..

Timber.

Limited span (without glue-lam at high cost), will not support cranes. Reasonable fire resistance.

Durability ok if treated. Light weight for site handling. Good chemical resistance.

Steel Good for long spans and cranes. Poor in fire. Probably cheapest. Needs cranes for erection. Delivery times may be longest.

Precast concrete. ok for long spans. Good durability and fire resistance. May require heavy cranes for erection.

Delivery probably quick.

Brick/steel Fastest start (quick delivery on bricks). Most attractive finish. Will support cranes. Fire resistance good for bricks but expanding steel may push them over.

2 Describe the processes which will cause loss of long term performance of each of the following materials if used to construct a pedestrian access bridge to a residential building. a. Reinforced concrete. b. Steel. c. Timber. d. Masonry. .............................................................................................................................. a.

Rebar corrosion from carbonation and chlorides sulphate attack (unlikely) frost ASR b. corrosion due to breakdown of paint dissimilar metals c Wet rot/ water penetration at joints etc. d. frost, espescially on mortar sulphate on mortar