Design of dams and earth retaining structures

Design of dams and earth retaining structures

1. THEORY & DESIGN OF STRUCTURES C3 -05 (5S3 NVQ 2045) Eng. Y.A.P.M Yahampath B Sc. Eng (Hons) , Dip Highway & Traffic Eng , AMIESL, AMECSL

2. Lecture Hours Allocated Lecture / Tutorial Practical Demonstrations/ Industrial visits Self Study Total 72 Hrs 00 Hrs 38 Hrs 110 Hrs Unit Title Time ( Hrs ) Combined Direct & Bending stresses 10 Principle stresses & strain 06 Structural design principles 06 Design ing of Reinforced concrete elements 26 Design of structural steel elements 18 Structural detailing 06 Total 72

3. AIM OF THE MODULE To develop fundamental understanding of the behavior of structures with particular reference to statically determinate civil engineering structures. To develop & understanding of the factors and constraints in determining suitable structural components. To develop awareness of the economical, Engineering & esthetic aspect in designing in selecting a particular structural component for given condition. To introduce students to use codes of practice and design charts/ Tables etc. for designing of structural elements.

4. Retaining Structures – Dams / Retaining walls Analysis of dams applying theory of combined stress Forces Applying on Dam Resultant force acting on base of a dam Stress variation diagram at the base of the dam

5. A P = F /A F = m.a g F = m.g P = mg /A ρ = m/V m = ρ V P = ρ V g /A h V = Ah P = ρ . A .h g / A P =h ρ g P = ϒ .h ϒ = ρ g m

6. m m m m m P = ϒ .h 1 P = ϒ .h 1 P = ϒ .h 1 P = ϒ .h 1

7. When we consider water at given depth the pressure acting on a water particle in any direction is equal .

8. P = ϒ .h P = P = ϒ .h h x P = ϒ .h x (0,0 ) P x = ϒ .x h

9. Under water, h Water surface σ σ σ σ σ = ϒ w .h At given depth the pressure acting on a water particle in any direction is equal . Water element

10. a = Top width of the dam d = Bottom width of the dam H = Height of the dam h = Height of the water retain by the dam ϒ w = Specific weight / unit weight of the water ( kN / m 3 ) ( ϒ w = ρ w g ) Γ = Unit weight of the dam masonry ( kN / m 3 ) Consider a unit length (b=1 m) of the dam, Volume (m 3 ) of unit length of the dam, V = [( a+d )/2] x H X 1 Weight (KN) of unit length of the dam, W = [( a+d )/2] x H X 1 X Γ P = F/A ( kN / m 2 ) ϒ = ρ g (g = 9.81 m/s -2 ) ρ = M/V ϒ = M/V X g F = Ma = Mg ϒ = F/V = kN / m 3 b 1m

11. a d H h ϒ w Γ A = [( a+d )/2] xH V = [( a+d )/2]x H x 1 Consider a unit length (b=1 m) of the dam, Volume (m 3 ) of unit length of the dam, W =mg m = Γ.V m = Γ. [( a+d )/2]x H x 1 W= Γ. [( a+d )/2]x H x 1 x g x

12. W2 X X/3 2X/3 X/2 X/2

13. a w1 w2 + w1 w2 = W x a d a/2 a+(d-a)/3 W.X = w 1 .a/2 + w 2 .[a+(d-a)/3] X = {w 1 .a/2 + w 2 .[a+(d-a)/3]}/W A B M A H (d-a)/3 a/2 (d-a) (d-a) Heel Toe

14. W A B

15. Water Pressure Equation P = hρ w g Hence the equation  P = hρg  represents the pressure due to the weight of any fluid having an average density ρ at any depth h below its surface. Pressure varies with depth. P = F/A F = PA dF = P . dA dA = 1 .dy dF = y . ρ w g . 1.dy F = y. ρ w g .1. dy = 1/2 ρ w g y 2 = 1/2 ρ w gh 2 y h b=1 m F h

16. h F Water head F F = Area of the pressure triangle P =h ρ w g P =0 F =1/2.h.(h ρ w g ) F =1/2. ρ w .g. h 2

17. h F ? Water head F = 1/2. ( ρ w g ) .h 2 But, ϒ w = ρ w g F = 1/2. ϒ w .h 2 ϒ w = 1000 kg/m 3 x 9.81 m/s -2 9.81 10 m/s -2 ϒ w = 1000 kg/m 3 x 10 m/s -2 = 10,000 N/m 3 / 1000 ϒ w = 10 kN /m 3

18. 1m 1m 1m W = 10 kN /m 3 = ϒ w Unit weight of water ( ϒ w )

19. The deeper the water, the more horizontal pressure it exerts on the dam. So at the surface of the reservoir, the water is exerting no pressure and at the bottom of the reservoir, the water is exerting maximum pressure . The force acts  at the center of gravity of the triangle -- one-third of the way up from the bottom. h F h/3 Water head h/3 2h/3

20. Modes of dam Failure 1. Sliding 2. Over turning 3. Bearing capacity failure at Base

21. W A B F Sliding

22. W A B F Over turning

23. W A B F Bearing capacity failure at Base

24. F W G h/3 R J K N L M x J K N L M h/3 x F W X /F = h/3/W X = F.h /3W R R = √ W 2 +F 2 Φ W F R h/3 Φ = W/F X = ? Using force diagram, X = The horizontal distance between the center of gravity of the dam & point where the resultant force ( R ) pass through the dam base x A Heel B Toe h

25. F Water head h/3 h b =1m H a A B W R Toe XX-Axis x ZZ- Axis of symmetry G a/2 e K L YY-Axis P =h ρ w g P =0

26. F Water head h/3 h b =1m H a A B W R Toe XX-Axis X = F.h /3W d ZZ- Axis of symmetry G d/2 X K YY-Axis P =h ρ w g P =0 If there is no any eccentricity ( e ).

27. J K L h/3 x F W X /F = h/3/W X = F.h /3W R Φ

28. F Water head h/3 h b =1m H a A B W R Toe XX-Axis d ZZ- Axis of symmetry G d/2 e X K L YY-Axis P =h ρ w g P =0 X = F.h /3W x If there is an any eccentricity ( e ).

29. F Water head h/3 h b =1m H a A B W R Toe XX-Axis x d ZZ- Axis of symmetry G d/2 e X K L YY-Axis P =h ρ w g P =0 X = F.h /3W If there is an any eccentricity ( e ).

30. Trapezoidal Dam with Water Face Vertical h H F a d/2 h/3 A B Water head w1 W w2 R Heel Toe L x x d/2 e e = [L – d/2] σ min  = W/d { 1 – 6e/d } σ m ax  = W/d { 1 + 6e/d } Stress Distribution at base of the dam, Axis of symmetry L = Liver arm L = X+X G G2 G1 C+ C+ C+ T- Heel Toe

31. σ a σ b σ b = M/Z x σ a =W/A x Direct Stress σ a Bending Stress σ b Combine Stress σ a + σ b σ a + σ b σ min = W/A - M/Z σ max = W/A + M/Z σ min σ max σ a + σ b = W/A ± M/Z x

32. W G Y -Distance Stress R e F + - Compression Tension + C + C XX-Axis - NA ZZ-Axis YY-Axis d σ min   σ max   σ min  = W/d { 1 - 6e/d } σ max  = W/d { 1 + 6e/d }

33. Parallel to NA y y = c = d/2 Recap :

34. Z = I/ e Max I = bd 3 /12 b = 1 m & d = d m, I = 1.d 3 /12 e Max = d/2 Z = d 2 /6 A = b.d But, b = 1 m A = 1.d σ = W /A ± M/ Z M = We σ = W/d ± We/ (d 2 /6) σ = W/d ± 6We/d 2 σ = W/d { 1 ± 6e/d } σ min  = W/d { 1 - 6e/d } σ max  = W/d { 1 + 6e/d } σ = W /A ± M/Z Assume Bending around x-x, Derivation of σ min   & σ max   f or Dam Base

35. σ a σ b σ b = M/Z x σ a =W/A x Direct Stress σ a Bending Stress σ b Combine Stress σ a + σ b σ a + σ b σ min = W/A - M/Z σ max = W/A + M/Z σ min σ max σ a + σ b = W/A ± M/Z x

36. W G Y -Distance Stress R e F + - Compression Tension + C + C XX-Axis - NA ZZ-Axis YY-Axis d σ min   σ max   σ min  = W/d { 1 - 6e/d } σ max  = W/d { 1 + 6e/d }

37. σ a σ b σ b = M/Z x σ a =W/A x Direct Stress σ a Bending Stress σ b Combine Stress σ a + σ b σ a + σ b σ min = W/A - M/Z σ max = W/A + M/Z σ min σ max σ a + σ b = W/A ± M/Z x

38. W G Y -Distance Stress R e F + - Compression Tension + C - T XX-Axis - NA ZZ-Axis YY-Axis d σ min   σ max   σ min  = W/d { 1 - 6e/d } σ max  = W/d { 1 + 6e/d }

39. Derivation of Middle third Rule for Dam Base The minimum stress  ( σ Min )  must be greater or equal to zero for no tensile stress at any point along the width of the column. Z = I/ e Max I = bd 3 /12 b = 1 m & d = d m, I = 1.d 3 /12 e Max = d/2 Z = d 2 /6 A = b.d But, b = 1 m A = 1.d σ min  = W /A - M/Z σ min > 0 W /A - M/ Z > M = We W/d – We/ (d 2 /6) > 0 W/d - 6We/d 2 > 0 W/d { 1 – 6e/d } > 0 W/d = 0, So, { 1 – 6e/d } > 0 1 > 6e/d d/6 > e σ = W /A ± M/Z Assume Bending around x-x,

40. Derivation of Middle third Rule for Dam Base d 1m d/3 d/3 d/3 d/6 d/6 d/2 A B A B K e L e = [L – d/2] x x W W Y Y z z Plan view of Base L = Liver arm L = X+X Axis of symmetry Z-Z,

41. F f Friction Effect on Dam Frictional force (f), f = µ R considering equilibrium of vertical forces, W = R f = µ .W considering equilibrium of horizontal forces, F – f = 0 - > F= f If, f > F Hence no sliding effect. If, f < F Failure due to sliding effect. Safety factor against water pressure = f/F f/F > F/F f/F > 1 f/F >1 - no sliding effect W G R A

42. Check for Failure Due to Tension in the base Check for Failure Due to overturning F W G F W G R R d/3 d

43. Upstream Downstream Upstream face Top level 1 m 4.8 m 24 kN /m 3 10 kN /m 3 1.2 m W G 8 m F = ? 7/3 m 7 m

44. h =7m F = ? a =1.2m h/3 =7/3m Water head W=? H =8m w1 w2 G2 G1 d =4.8 m 10kN/m 3 24kN/m 3 x A B Up stream

45. Considering a 1 m length of dam, F = 1/2. ϒ w .h 2 F = ½ X 10 kN /m 3 X 7 2 m 2 F = 245 kN /m W = (1.2+4.8)/2 X 8 X 1 X 24 W = 576kN/m Using force diagram, X = F.h /3W X = (245 x 7)/(3 x 576) Thrust on the dam wall is the area of pressure triangle, X = 0.99 m F = 245 kN /m 7 m P =h ϒ w = 7x10 =70 kN /m 2 ϒ w = ρ w g 7/3 m

46. W = m x g ρ c = m/V m = ρ c x V W = ρ c x V X g W = ρ c X g x V W = ϒ C X V ϒ C = 24 kN /m 3 W = 24 X 24 W = 576kN/m V = (1.2+4.8)/2 X 8 X 1 = 24 m 3

47. To find location of the center of gravity of the dam section (X) , Consider the dam section only & split in to two shapes (rectangle + triangle) with known center of gravity Considering moments around A, W.X = w 1 .x 1 + w 2 .x 2 576 x X = (8x1.2)x 24 x (1.2/2) + ½ x 8 x 3.6 x 24 x(1.2+3.6/3 ) X = 1.68 m L = X+X L = 1.68 + 0.99 = 2.67 m w1 w2 G G2 G1 W X = ? a/2 =0.6m a =1.2m d =4.8m 1.2+(3.6/3)= 2.4m A

48. Check the middle third rule, d =4.8 m d/3=1.6m d/6 =0.8m d/2 =2.4m A B A B K e=0.27m L=2.67m R d/3=1.6m d/3=1.6m N/A d/6 =0.8m d/6 =0.8m > e=0.27m Middle third rule satisfied. 1.6 < L=2.67m < 3.2

49. To find the eccentricity, e = [L – d/2] e = [2.67 – 4.8/2] = 0.27 m X = 1.68 d =4.8m 1.2+(3.6/3)= 2.4m a =1.2m e = ? N/A Stress Distribution at base of the dam , σ min  = W/d { 1 – 6e/d } σ min  = 576/4.8 x { 1 – 6 x 0.27/4.8 } σ min  = + 79.5 kN /m 2 σ m ax  = W/d { 1 + 6 e /d } σ m ax  = (576/4.8) x { 1 + 6 x 0.27 /4.8 } σ m ax  = + 160.5 kN /m 2 R F W G X = 0.99 L=2.67 d/2 =2.4m

50. W A B F R σ m ax   σ m in Toe   heel   + Compression   + Compression   + - Compression Tension

51. W A B F R σ m ax   σ m in   Toe   heel   σ m ax   σ m in + Compression   - Tension   WRONG + - Compression Tension

52. σ min  = 79.5 kN /m 2 σ m ax  = 160.5 kN /m 2 Stress ( kN /m 2 ) Distance ( m ) 4.8 m Toe Heel STRESS DISTRIBUTION DIAGRAM + - Compression Tension

53. Check the stability of the dam for Failure Due overturning Check for Failure Due to overturning Check if, 0< L < d – No failure L = 2.67 m d = 4.8 m So, 0 < 2.67m < 4.8 Resultant force pass within the section of the base. So, No failure due to over turning.

54. N/A R d L overturning No overturning N/A d L 0< L < d – No failure 0> L > d – failure B R

55. If the coefficient of friction between soil and dam base is 0.6 , Check the stability of the dam considering equilibrium of vertical forces, R = W R = 576 KN Frictional force (f), f = µ R = µ w = 0.6 x 576 f = 345.6 kN But, F = 245 kN /m So, 345.6 kN > 245 kN /m If, f > F Hence no sliding effect. Safety factor against water pressure = f/F = 345.6/245 = 1.4 F f W G R Check for Failure Due to sliding

56. Check the stability of the dam for Failure Due to Tension Check for Failure Due to Tension in the base Check if, d/3 < L < 2d/3 – No failure L = 2.67 m d/3 = 4.8/3 = 1.6m 2d/3 = 2 x 4.8/3 = 3.2m So, 1.6m < 2.67 m < 3.2m Resultant force pass through the middle third section of the base. So, No failure due to tension

57. A B N/A d d/3 d/3 d/3 d/2 σ m ax   σ m in Toe   heel   + Compression   + Compression   L

58. A B N/A d d/3 d/3 d/3 d/2 σ m ax   σ m in Toe   heel   + Compression   - Tension L

59. Q2. A concrete dam has its upstream face vertical & a top width of 3m. Its downstream face has a uniform batter. It stores water to a depth of 15m with a freeboard of 2 m. Calculate the minimum dam width(d) at the bottom for no tension in concrete. ϒ w = Specific weight / unit weight of the water (10kN/ m 3 ) ( ϒ w = ρ w g ) Γ = Unit weight of the dam masonry (25kN/ m 3 ) Neglect uplift force.

60. h =15m F = ? a =3m h/3 =5m Water head W=? H =17m w1 w2 G2 G1 d =? m 10kN/m 3 25kN/m 3 x A B Up stream 2m x L

61. Considering a 1 m length of dam, F = 1/2. ϒ w .h 2 F = ½ X 10 kN /m 3 X 15 2 m 2 F = 1125 kN /m W = [ (3+d)/2 X 17 X1] X 25 W = 212.5(3+d) kN /m Using force diagram, X = F.h/3W X = (1125 x 15)/[3 x 212.5(3+d)] X = 26.47/(3+d) m

62. w1 w2 G G2 G1 W X = ? X 1 = a/2 =1.5m a =3m d = ? m X 2 =3+[(d-3)/3]= [2+d/3]m A (d-3)m

63. To find location of the center of gravity of the dam section (X) , Consider the dam section only & split in to two shapes (rectangle + triangle) with known center of gravity Considering moments around A, W.X = w 1 .x 1 + w 2 .x 2 212.5(3+d)X = (17x3)x 25 x (3/2) + ½ x 17 x (d-3) x 25 x [2+d/3] 1912.5X + 637.5dX = 5737.5 + 212.5d 2 + 1275d - 637.5d - 3825 X = [d 2 +3d + 9 ]/3(3+d) L = X+X L = 26.47/(3+d) + [d 2 +3d + 9 ]/3(3+d) L = [88.41 +d 2 + 3d]/ 3(3+d)

64. To find the eccentricity, e = [L – d/2] e = [88.41 +d2 + 3d]/ 3(3+d) – d/2 e = [176.82 - d 2 – 3d] / (18 + 6d) Stress Distribution at base of the dam , σ min  = W/d { 1 – 6e/d } The minimum stress  ( σ Min )  must be greater or equal to zero for no tensile stress at any point along the width of the column. σ min   > = W/d { 1 – 6e/d } > = d/6 >= e Minimum d value at, d/6 = e d = 6e

65. d/6 = e = [176.82 - d 2 – 3d] / (18 + 6d) d.(18 + 6d) = [176.82 - d 2 – 3d] x 6 d 2 + 3d - 88.41 = 0 d = 8.21 m or d = -11. 021 m d can’t be a negative value. So , d = 8.21m

66. Q3. A trapezoidal shape concrete dam has a top width of 1m and base width of 7m. It stores water to a depth of 10m without a freeboard. List out assumptions clearly Find the total force due to upstream water and point of action Find the weight of the dam and point of action Calculate the ecstaticity of the resultant force at the base Check the stability of the dam for Failure Due to Tension & overturning Find the factor of safety against sliding at the base Find the maximum and minimum stress at the base Draw the stress distribution diagram µ = Friction coefficient between dam base and soil is 0.5 ϒ w = Specific weight / unit weight of the water (10kN/ m 3 ) ( ϒ w = ρ w g ) Γ = Unit weight of the dam masonry (20kN/ m 3 ) Neglect uplift force.

67. a =1m h/3 =10/3m Water head W=? h=H =10m w1 w2 G2 G1 d =7 m 10kN/m 3 20kN/m 3 A B Up stream w3 G3 w4 G4 F = ? x R L x e f E D C x 1 x 2 x 3 x 4 1m 5m

68. Considering a 1 m length of dam, F = 1/2. ϒ w .h 2 F = ½ X 10 kN /m 3 X 10 2 m 2 F = 500 kN /m W = Weight of dam + weight of the AED water mass W = [ (1+7)/2x 10 x 1x 20 ] + [½ x 10 x 1x1 x 10] W = 850kN/m Using force diagram, X = F.h/3W X = (500 x 10)/[3 x 850] X = 1.96 m Assumptions Consider 1 m length of dam. Neglect up thrust. Water pressure force at h depth is equal to any direction. So consider water force acts perpendicular to AE line. Weight of AED section due to water mass also added to total weight of the dam.

69. To find location of the center of gravity of the dam section (X) , Consider the dam section only & split in to four shapes (water trinangle+trinangle+rectangle + triangle) with known center of gravity. Considering moments around A, W.X = w 1 .x 1 + w 2 .x 2 + w 3 .x 3 + w 4 .x 4 850X = (½ x1x10x1)x 10 x (1/3) + (½ x 1x10x1) x20x( 2/3 ) + (1x10x1)x20x(1+0.5) + (½ x1x10x1) x20x( 2+5/3) 850X = 16.667 +66.667 + 300 +1833.333 850X = 2216.667 X = 2.607 m L = X+X L = 1.96 + 2.607 L = 4.567 m

70. To find the eccentricity, e = [L – d/2] e = [ 4.567 – 7/2 ] e = 1.067m To satisfy middle third rule to avoid tension at the dam base, d/6 >= e But, d/6 = 7/6 = 1.1667m 1.1667m > 1.067m Dam base satisfy middle third rule, so no tension at the dam base.

71. Check for Over turning If, 0< L < d – No failure( Resultant force pass within the section of the base) 0 < 4.567 m < 7m So resultant force pass within the section of base, no failure due to overturning

72. Check for Sliding, considering equilibrium of vertical forces, R = W R = 850 KN Frictional force (f), f = µ R = µ w = 0.6 x 850 f = 510 kN But, F = 500 kN /m So, 510 kN > 500 kN /m If, f > F Hence no sliding effect. Safety factor against water pressure = f/F = 510/500 = 1.4

73. Stress Distribution at base of the dam , σ min  = W/d { 1 – 6e/d } σ min  = 850/7 x { 1 – 6 x 1.067/7 } σ min  = 10.37 kN /m 2 σ m ax  = W/d { 1 + 6e/d } σ m ax  = 850/7 x { 1 + 6 x 1.067/7 } σ m ax  = 232.48 kN /m 2

74. σ min  = 10.37 kN /m 2 σ m ax  = 232.48 kN /m 2 Stress ( kN /m 2 ) Distance ( m ) 7 m Toe Heel STRESS DISTRIBUTION DIAGRAM

75. Q4. Calculate the stresses of dam base for water level at D-C a =0.8m Water head W=? 0.5m w1 w2 G2 G1 d =1.4 m 10kN/m 3 20kN/m 3 Up stream F = ? R L f x 1 x 2 B D C A 0.6m 0.5m H=h=1m x w3 G3 w4 G4 x 3 x 4 0.2m 0.2m x e E I J K M

76. Considering a 1 m length of dam, F = 1/2. ϒ w .h 2 F = ½ X 10 kN /m 3 X 1 2 m 2 F = 5kN/m W = Weight of dam + weight of the AEDIJM water mass W = [ 0.8x 1 x 1x 20 ]+[0.6x0.5x1x20] + [ 0.2x 1x1 x 10] + [0.6x0.5x1x10] W = (16+6+2+3) = 27kN/m Using force diagram, X = F.h/3W X = (5x 1)/[3 x 27] X = 0.0617 m Assumptions Consider 1 m length of dam. Neglect up thrust. Water pressure force at h depth is equal to any direction. So consider water force acts perpendicular to dam. Weight of AEDIJM section due to water mass also added to total weight of the dam.

77. To find location of the center of gravity of the dam section (X) , Consider the dam section only & split in to four shapes (water trinangle+trinangle+rectangle + triangle) with known center of gravity. Considering moments around A, W.X = w 1 .x 1 + w 2 .x 2 + w 3 .x 3 + w 4 .x 4 27X = (0.2x1x1)x 10 x (0.2/2) + (0.6x0.5x1) x10x[ 0.2+(0.6/2)] + (0.6x0.5x1)x20x[0.2+(0.6/2)] + (1x0.8x1) x20x[0.2+0.6+(0.8/2)] 27X = 0.2 + 4.5 + 3 + 19.2 27X = 26.9 X = 26.9/27 X = 0.996 m L = X+X L = 0.0617 + 0.996 L = 1.057 m

78. To find the eccentricity, e = [L – d/2] e = [ 1.057 – 1.4/2 ] e = 0.3579 m To satisfy middle third rule to avoid tension at the dam base, d/6 >= e But, d/6 = 7/6 = 0.233 m 0.233 m < 0.3579 m Dam base satisfy doesn't middle third rule, so tension will built up at the dam base.

79. Check for Over turning If, 0< L < d – No failure( Resultant force pass within the section of the base) 0 < 1.057 m < 1.4m So resultant force pass within the section of base, no failure due to overturning

80. Check for Sliding, considering equilibrium of vertical forces, R = W R = 27KN Frictional force (f), f = µ R = µ w = 0.6 x 27 f = 16.2 kN But, F = 5 kN /m So, 16.2 kN > 5 kN /m If, f > F Hence no sliding effect. Safety factor against water pressure = f/F =16.2/5 = 3.24

81. Stress Distribution at base of the dam , σ min  = W/d { 1 – 6e/d } σ min  = 27/71.4 x { 1 – 6 x 0.3579/1.4 } σ min  = -10.29 kN /m 2 σ m ax  = W/d { 1 + 6e/d } σ m ax  = 27/1.4 x { 1 + 6 x0.3579 /1.4 } σ m ax  = 48.86 kN /m 2

82. σ min  = -10.29 kN /m 2 σ m ax  = 48.86 kN /m 2 Stress ( kN /m 2 ) Distance ( m ) 1.4 m Toe Heel STRESS DISTRIBUTION DIAGRAM

83. h =1.8m F = ? a =0.6m 1.8/3 =0.6m Water head W=? H =2m w1 w2 G2 G1 d =2 m 10kN/m 3 22kN/m 3 x A B Up stream 0.2m x L

84. Considering a 1 m length of dam, F = 1/2. ϒ w .h 2 F = ½ X 10 kN /m 3 X 1.8 2 m 2 F = 16 kN /m W = [ (2+0.6)/2 X 2 X1] X 22 W = 57.2kN/m Using force diagram, X = F.h/3W X = (16 x 1.8)/[3 x 57.2] X = 0.168 m

85. w1 w2 G G2 G1 W X = ? X 1 = a/2 =0.3m a =0.6m d = 2 m X 2 =0.6+[1.4/3]=1.0667 m A 1.4m B

86. To find location of the center of gravity of the dam section (X) , Consider the dam section only & split in to two shapes (rectangle + triangle) with known center of gravity Considering moments around A, W.X = w 1 .x 1 + w 2 .x 2 57.2X = (2x0.6)x 22 x (0.6/2) + ½ x 2 x 1.4 x 22 x 1.0667 X = 3.96 + 32.5472 X = 0.6396 L = X+X L = 0.168 + 0.6396 L = 0.8076 m

87. To find the eccentricity, e = [L – d/2] e = [ 0.8076 – 2 /2] e = -0.1924 Stress Distribution at base of the dam , σ min  = W/d { 1 – 6e/d } The minimum stress  ( σ Min )  must be greater or equal to zero for no tensile stress at any point along the width of the column. σ min   > = W/d { 1 – 6e/d } > = d/6 >= e Minimum d value at, d/6 = e d = 6e

88. Check the middle third rule, d =2 m d/3=0.667m d/6 =0.333m d/2 =1m A B A B K e = 0.1924m L=0.8076m R N/A d/6 =0.333m d/6 =0.333m > e=0.1924m Middle third rule satisfied. 1.6 < L=2.67m < 3.2 d/3=0.667m d/3=0.667m

89. Stress Distribution at base of the dam , σ min  = W/d { 1 – 6e/d } σ min  = 27/71.4 x { 1 – 6 x 0.3579/1.4 } σ min  = 45.13632 kN /m 2 σ m ax  = W/d { 1 + 6e/d } σ m ax  = 57.2/2 x { 1 + 6 x(- 0.1924)/2 } σ m ax  = -4.41344 kN /m 2

90. σ min  = 45.13 kN /m 2 σ m ax  = -4.4 kN /m 2 Stress ( kN /m 2 ) Distance ( m ) 2 m Toe Heel STRESS DISTRIBUTION DIAGRAM

91. QUESTIONS ?