Find: Force on culvert in [lb/ft]

Παρουσίαση με θέμα: "Find: Force on culvert in [lb/ft]"— Μεταγράφημα παρουσίασης:

1 Find: Force on culvert in [lb/ft]
lateral strain=0 2,550 2,800 C) 3,150 D) 3,400 γT=120 [lb/ft3] 6 ft wc=12% φ=30˚ Force 5 ft Find the lateral force on the culvert in pounds per foot. In this problem 5 ft

2 Find: Force on culvert in [lb/ft]
lateral strain=0 2,550 2,800 C) 3,150 D) 3,400 γT=120 [lb/ft3] 6 ft wc=12% φ=30˚ Force 5 ft we have a 5 foot by 5 foot box culvert 5 ft

3 Find: Force on culvert in [lb/ft]
lateral strain=0 2,550 2,800 C) 3,150 D) 3,400 γT=120 [lb/ft3] 6 ft wc=12% φ=30˚ Force 5 ft With 6 feet of cover 5 ft

4 Find: Force on culvert in [lb/ft]
lateral strain=0 2,550 2,800 C) 3,150 D) 3,400 γT=120 [lb/ft3] 6 ft wc=12% φ=30˚ Force 5 ft We’ve also been provided certain soil conditions 5 ft

5 Find: Force on culvert in [lb/ft]
lateral strain=0 2,550 2,800 C) 3,150 D) 3,400 γT=120 [lb/ft3] 6 ft wc=12% φ=30˚ Force 5 ft and we’re asked to find the force. Notice the 5 ft

6 Find: Force on culvert in [lb/ft]
lateral strain=0 γT=120 [lb/ft3] 6 ft wc=12% φ=30˚ Force 5 ft groundwater table is below the culvert 5 ft

7 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] wc=12% φ=30˚ 6 ft Force 5 ft For a uniform layer of soil, 5 ft

8 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] wc=12% φ=30˚ 6 ft Force 5 ft the lateral stress distribution on a vertical body increases linearly with depth as shown here. 5 ft

9 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] wc=12% φ=30˚ 6 ft Force 5 ft For this problem we’re only interested in the force on the side of the culvert 5 ft

10 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] wc=12% φ=30˚ 6 ft σh [lb/ft2] 6 [ft]) 5 ft So we’ll compute the horizontal stress, 5 ft

11 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] wc=12% φ=30˚ 6 ft σh [lb/ft2] 6 [ft]) 5 ft σh [lb/ft2] 11 [ft]) at the top and bottom of the culvert, 5 ft

12 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] wc=12% φ=30˚ 6 ft σh [lb/ft2] 6 [ft]) 5 ft σh [lb/ft2] 11 [ft]) Since the stress profile is linear with depth, the total force per length will equal the average lateral stress at the top and bottom of the culvert, multiplied by the 5 feet of the culvert’s height. 5 ft Force = σh,ave [lb/ft2] * 5 [ft]

13 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] σv=γT * d wc=12% φ=30˚ 6 ft 5 ft So first we find our vertical stress 5 ft

14 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] σv=γT * d wc=12% φ=30˚ σv’ = σv - u 6 ft 5 ft Then we solve for the effective vertical stress 5 ft

15 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] σv=γT * d wc=12% φ=30˚ σv’ = σv - u 6 ft pore water pressure 5 ft Where ‘u’ equals the pore water pressure 5 ft

16 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] σv=γT * d wc=12% φ=30˚ σv’ = σv - u 6 ft pore water pressure 5 ft Since the groundwater table is beneath the culvert, 5 ft u = 0

17 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] σv=γT * d wc=12% φ=30˚ σv’ = σv - u 6 ft pore water pressure 5 ft the pore water pressure is zero 5 ft u = 0

18 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] σv=γT * d wc=12% φ=30˚ σv’ = σv - u 6 ft σv’ = σv 5 ft So we substitute in… 5 ft

19 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] σv=γT * d wc=12% φ=30˚ σv’ = σv - u 6 ft σv’ = γT * d 5 ft The horizontal effective stress 5 ft

20 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] σv=γT * d wc=12% φ=30˚ σv’ = σv - u 6 ft σv’ = γT * d σh’ = KO * σv’ 5 ft is equal to the vertical effective stress times the appropriate lateral earth pressure coefficient 5 ft

21 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] σv=γT * d wc=12% φ=30˚ σv’ = σv - u 6 ft σv’ = γT * d σh’ = KO * σv’ 5 ft For this problem we know to use “at rest” soil conditions since 5 ft At rest condition

22 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] σv=γT * d wc=12% φ=30˚ σv’ = σv - u 6 ft σv’ = γT * d σh’ = KO * σv’ 5 ft It was given the lateral strain is zero. 5 ft At rest condition lateral strain=0

23 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] σv=γT * d wc=12% φ=30˚ σv’ = σv - u 6 ft σv’ = γT * d σh’ = KO * σv’ 5 ft We substitute again, 5 ft

24 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] σv=γT * d wc=12% φ=30˚ σv’ = σv - u 6 ft σv’ = γT * d σh’ = KO * σv’ 5 ft σh’ = KO* γT * d And continue solving 5 ft

25 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] σv=γT * d wc=12% φ=30˚ σv’ = σv - u σv’ = γT * d σh’ = KO * σv’ σh’ = KO* γT * d Let’s move this over

26 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] wc=12% φ=30˚ σv=γT * d σv’ = γT * d σv’ = σv - u σh’ = KO * σv’ σh’ = KO* γT * d Now the “at-rest”

27 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] wc=12% φ=30˚ σh’ = KO* γT * d σv=γT * d σv’ = γT * d σv’ = σv - u σh’ = KO * σv’ σh’ = KO* γT * d lateral earth pressure coefficient

28 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] 1-sin φ wc=12% φ=30˚ σh’ = KO* γT * d σv=γT * d σv’ = γT * d σv’ = σv - u σh’ = KO * σv’ σh’ = KO* γT * d is equal to 1 minus the sine of the friction angle.

29 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] 1-sin φ wc=12% φ=30˚ σh’ = KO* γT * d σv=γT * d σv’ = γT * d σv’ = σv - u σh’ = KO * σv’ σh’ = KO* γT * d σh’= (1-sin φ) * γT * d The horizontal stress

30 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] 1-sin φ wc=12% φ=30˚ σh’ = KO* γT * d σv=γT * d σv’ = γT * d σv’ = σv - u σh’ = KO * σv’ σh’ = KO* γT * d σh’= (1-sin φ) * γT * d σh = σh’ + u Is equal to the effective horizontal stress plus the

31 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] 1-sin φ wc=12% φ=30˚ σh’ = KO* γT * d σv=γT * d σv’ = γT * d σv’ = σv - u σh’ = KO * σv’ σh’ = KO* γT * d σh’= (1-sin φ) * γT * d σh = σh’ + u pore water pressure. Which we’ve already determined was zero

32 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] 1-sin φ wc=12% φ=30˚ σh’ = KO* γT * d σv=γT * d σv’ = γT * d σv’ = σv - u σh’ = KO * σv’ σh’ = KO* γT * d σh’= (1-sin φ) * γT * d σh = σh’ + u σh = σh’

33 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] 1-sin φ wc=12% φ=30˚ σh’ = KO* γT * d σv=γT * d σv’ = γT * d σv’ = σv - u σh’ = KO * σv’ σh’ = KO* γT * d σh’= (1-sin φ) * γT * d σh = σh’ + u σh = σh’ So we substitute again

34 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] 1-sin φ wc=12% φ=30˚ σh’ = KO* γT * d σv=γT * d σv’ = γT * d σv’ = σv - u σh’ = KO * σv’ σh’ = KO* γT * d σh’= (1-sin φ) * γT * d σh = σh’ + u σh = σh’ And start plugging in our known data σh= (1-sin φ) * γT * d

35 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] 1-sin φ wc=12% φ=30˚ σh’ = KO* γT * d σv=γT * d σv’ = γT * d σv’ = σv - u σh’ = KO * σv’ σh’ = KO* γT * d σh’= (1-sin φ) * γT * d σh = σh’ + u σh = σh’ The friction angle 30˚ σh= (1-sin φ) * γT * d

36 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] 1-sin φ wc=12% φ=30˚ σh’ = KO* γT * d σv=γT * d σv’ = γT * d σv’ = σv - u σh’ = KO * σv’ σh’ = KO* γT * d σh’= (1-sin φ) * γT * d σh = σh’ + u σh = σh’ The total unit weight 30˚ σh= (1-sin φ) * γT * d 120 [lb/ft3]

37 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] 1-sin φ wc=12% φ=30˚ σh’ = KO* γT * d σv=γT * d σv’ = γT * d σv’ = σv - u σh’ = KO * σv’ σh’ = KO* γT * d σh’= (1-sin φ) * γT * d σh = σh’ + u σh = σh’ 6 [ft], 11 [ft] And the depths. The horizontal stress 30˚ σh= (1-sin φ) * γT * d 120 [lb/ft3]

38 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] 1-sin φ wc=12% φ=30˚ σh’ = KO* γT * d σv=γT * d σv’ = γT * d σv’ = σv - u σh’ = KO * σv’ σh’ = KO* γT * d σh’= (1-sin φ) * γT * d σh = σh’ + u σh = σh’ 6 [ft], 11 [ft] at the top and bottom of the of the culvert is 360 and 660 pounds per square feet, respectively. 30˚ σh= (1-sin φ) * γT * d 120 [lb/ft3] σh= 360 [lb/ft2], 660 [lb/ft2]

39 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] wc=12% φ=30˚ 6 ft σh [lb/ft2] 6 [ft]) 5 ft σh [lb/ft2] 11 [ft]) So we return to our profile view 5 ft

40 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] wc=12% φ=30˚ 6 ft σh = 360 [lb/ft2] 6 [ft]) 5 ft σh = 660 [lb/ft2] 11 [ft]) and add our known values. 5 ft

41 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] wc=12% φ=30˚ 6 ft σh = 360 [lb/ft2] 6 [ft]) 5 ft σh = 660 [lb/ft2] 11 [ft]) The average stress is 510 pounds per square foot 5 ft σh,ave = 510 [lb/ft2]

42 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] wc=12% φ=30˚ 6 ft σh = 360 [lb/ft2] 6 [ft]) Force 5 ft σh = 660 [lb/ft2] 11 [ft]) Substitute this value in 5 ft 510 [lb/ft2] Force = σh,ave [lb/ft2] * 5 [ft]

43 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] wc=12% φ=30˚ 6 ft σh = 360 [lb/ft2] 6 [ft]) Force 5 ft σh = 660 [lb/ft2] 11 [ft]) And we have a force 5 ft 510 [lb/ft2] Force = 510 [lb/ft2] * 5 [ft]

44 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] wc=12% φ=30˚ 6 ft σh = 360 [lb/ft2] 6 [ft]) Force 5 ft σh = 660 [lb/ft2] 11 [ft]) Of 2,550 pounds per foot, along the culvert 5 ft 510 [lb/ft2] Force = 510 [lb/ft2] * 5 [ft] = 2,550 [lb/ft]

45 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] wc=12% φ=30˚ 2,550 2,800 C) 3,150 D) 3,400 6 ft σh = 360 [lb/ft2] 6 [ft]) Force 5 ft σh = 660 [lb/ft2] 11 [ft]) From the possible solutions, 5 ft 510 [lb/ft2] Force = 510 [lb/ft2] * 5 [ft] = 2,550 [lb/ft]

46 Find: Force on culvert in [lb/ft]
γT=120 [lb/ft3] wc=12% φ=30˚ 2,550 2,800 C) 3,150 D) 3,400 6 ft σh = 360 [lb/ft2] 6 [ft]) Force 5 ft σh = 660 [lb/ft2] 11 [ft]) The answer is A…. Another question which 5 ft 510 [lb/ft2] Force = 510 [lb/ft2] * 5 [ft] = 2,550 [lb/ft] Answer: A

47 σh = 360 [lb/ft2] (@ 6 [ft]) σh = 660 [lb/ft2] (@ 11 [ft]) Force
At what depth, d, would the equivalent point force act? 6 ft σh = 360 [lb/ft2] 6 [ft]) Force 5 ft σh = 660 [lb/ft2] 11 [ft]) may be asked in this sort of problem is, 5 ft Force = 510 [lb/ft2] * 5 [ft] = 2,550 [lb/ft]

48 σh = 360 [lb/ft2] (@ 6 [ft]) σh = 660 [lb/ft2] (@ 11 [ft]) Force
At what depth, d, would the equivalent point force act? 6 ft σh = 360 [lb/ft2] 6 [ft]) Force 5 ft σh = 660 [lb/ft2] 11 [ft]) At what depth, d, would the equivalent point force act 5 ft Force = 510 [lb/ft2] * 5 [ft] = 2,550 [lb/ft]

49 At what depth, d, would the equivalent point force act?
6 ft on the side of the culvert? 5 ft 5 ft

50 σh = 360 [lb/ft2] σh = 660 [lb/ft2] At what depth, d, would the
equivalent point force act? 6 ft σh = 360 [lb/ft2] From our two horizontal stress values, we can 5 ft 5 ft σh = 660 [lb/ft2]

51 σh = 360 [lb/ft2] σh = 660 [lb/ft2] At what depth, d, would the
equivalent point force act? 6 ft σh = 360 [lb/ft2] Visualize this stress as a rectangle 5 ft 5 ft σh = 660 [lb/ft2]

52 σh = 360 [lb/ft2] σh = 660 [lb/ft2] At what depth, d, would the
equivalent point force act? 6 ft σh = 360 [lb/ft2] And a triangle, and solve for 5 ft 5 ft σh = 660 [lb/ft2]

53 σh = 360 [lb/ft2] σh = 660 [lb/ft2] At what depth, d, would the
equivalent point force act? d=area*c+area*c 6 ft area+area σh = 360 [lb/ft2] the depth of the equivalent point force based on their magnitudes and centroids 5 ft 5 ft σh = 660 [lb/ft2]

54 σh = 360 [lb/ft2] σh = 660 [lb/ft2] At what depth, d, would the
equivalent point force act? d=area*c+area*c 6 ft area+area area σh = 360 [lb/ft2] Where the magnitudes are represented by the areas 5 ft 5 ft area σh = 660 [lb/ft2]

55 c c σh = 360 [lb/ft2] σh = 660 [lb/ft2] At what depth, d, would the
equivalent point force act? d=area*c+area*c 6 ft area+area c c area σh = 360 [lb/ft2] And the centroids of each area represent is the depth to be averaged. For this example, 5 ft 5 ft area σh = 660 [lb/ft2]

56 d=8.75 [ft] c c σh = 360 [lb/ft2] σh = 660 [lb/ft2]
At what depth, d, would the equivalent point force act? d=8.75 [ft] d=area*c+area*c 6 ft area+area c c area σh = 360 [lb/ft2] The depth, d, would be 8.75 feet 5 ft 5 ft area σh = 660 [lb/ft2]

57 ? Index σ’v = Σ γ d γT=100 [lb/ft3] +γclay dclay 1
Find: σ’v at d = 30 feet (1+wc)*γw wc+(1/SG) σ’v = Σ γ d d Sand 10 ft γT=100 [lb/ft3] 100 [lb/ft3] 10 [ft] 20 ft Clay = γsand dsand +γclay dclay A W S V [ft3] W [lb] 40 ft text wc = 37% ? 20 [ft]