Manhole Analysis

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1.Design Method 1.1.Design Requirements The report aims to present necessary calculations to provide reinforcement in manholes where infiltration, traffic load, safety and risk issues cannot be addressed.

2.General The most crucial condition of the manhole need to be kept under consideration and that is when the empty and the soil surrounding the wall is wet considering infiltration is blocked.

2.1.Material Properties The properties of materials used for concrete manholes are as follows: f`c = specified compressive strength of concrete at 28 days fy Es Ec

= = = = =

30 MPa for concrete in box culverts specified minimum yield strength of reinforcement 420 MPa 200000 MPa modulus of elasticity of steel reinforcement LRFD (5.4.3.2) modulus of elasticity of concrete in box LRFD [C5.4.2.4]

=

for normal density concrete with γ c = 2320 kg/m Ec may be taken as

= =

3

4800

f`c

0.5

26290.68 MPa

3.Limit State Design Method 3.1.LFRD Requirements For manhole design, the component dimensions and the size and spacing of reinforcement shall be selected to satisfy the following equation for all appropriate limit states, as presented in LRFD [1.3.2.1] Q = Σηi γi Qi ≤ φRn = Rr Where : ηi

=

Load modifier

γi

=

Load factor

Qi

=

Force effect: moment, shear, stress range or deformation caused by

Q φ

= =

applied loads Total factored force effect Resistance factor

Rn

=

Nominal resistance: resistance of a component to force effects

Rr

=

Factored resistance = φRn

3.2.Limit States The Strength I Limit State is used to design reinforcement for flexure and checking shear in the slabs and walls, LRFD [12.5.3] . The Service I Limit State is used for checking reinforcement for crack control criteria, LRFD [12.5.2] .

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3.3.Load Factors The following Strength I load factors γ st and Service I load factors, γs1 shall be used for manhole design:

Dead Load - Components Dead Load - Wearing Surface Vertical Earth Pressure Horizontal Earth Pressure Live Load Surcharge Live Load + IM

DC DW EV EH LS LL + IM

Strength I Load Factor 1.25 0.90 1.50 0.65 1.35 0.90 1.50 0.50 1.50 1.50 1.75 1.75

#

Service I Load Factor 1.0 1.0 1.0 1.0 1.0 1.0

3.4.Strength Limit State Strength I Limit State shall be applied to ensure that strength and stability are provided to resist the significant load combinations that a structure is expected to experience during its design life LRFD [1.3.2.4]. 3.4.1 Factored Resistance The resistance factor, φ, is used to reduce the computed nominal resistance of a structural element This factor accounts for the variability of material properties, structural dimensions and workmanship and uncertainty in prediction of resistance. The resistance factors, φ, for reinforced manholes for the Strength Limit State per LRFD [Table 12.5.5-1] are as shown below: Structure Type Cast-in-place

Flexure

Shear

0.9

0.85

3.4.2 Moment Capacity For rectangular sections, the nominal moment resistance, Mn, per LRFD [5.7.3.2.3] (tension reinforcement only) equals:

2 The factored resistance, Mr, or moment capacity per LRFD [5.7.3.2.1], shall be taken as:

2 The location of the design moment will consider the haunch dimensions in accordance with LRFD [12.11.4.2 ]. No portion of the haunch shall be considered in adding to the effective depth of the section.

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3.4.3 Shear Capacity Per LRFD [12.11.4.1] ,shear in manholes shall be investigated in conformance with LRFD[5.14.5.3] The location of the critical section for shear for manholes with haunches shall be determined in conformance with LRFD [C5.13.3.6.1]  and shall be taken at a distance dv from the end of the haunch. In the absence of shear reinforcing, the nominal shear resistance is equal to the shear resistance of the concrete.The factored resistance, Vr, or shear capacity, per ACI[11.2.1.2] shall be taken as

φVc = φ0.17 (f'c)0.5 bw d

3.5.Service Limit State Service I Limit State shall be applied as restrictions on stress, deformation, and crack width under regular service conditions LRFD [1.3.2.2]. 3.5.1 Factored Resistance The resistance factor, φ, for Service Limit State, is found in LRFD [1.3.2.1 ] and its value is 1.00. 3.5.2 Crack Control Criteria Per LRFD [12.11.3] , the provisions of LRFD [5.7.3.4] shall apply to crack width control All reinforced concrete members are subject to cracking under any load condition, which produces tension in the gross section in excess of the cracking strength of the concrete. Provisions are provided for the distribution of tension reinforcement to control flexural cracking. Crack control criteria does not use a factored resistance, but calculates a maximum spacing for flexure reinforcement based on service load stress in bars, concrete cover and exposure condition. Crack control criteria shall be applied when the tension in the cross-section exceeds 80% of the modulus of rupture, fr, specified in LRFD [5.4.2.6] for Service I Limit State. The spacing, s, (in mm) of mild steel reinforcement in the layer closest to the tension face shall satisfy:

2

s in which :

1 γe

0.7 =

0.6

Exposure factor

(Implied crack width 0.25mm)

(1.0 for Class 1 exposure condition, 0.75 for Class 2 exposure condition, see LRFD [5.7.3.4] for guidance) dc

=

thickness of concrete cover measured from extreme tension fiber to center

of the flexural reinforcement located closest thereto(mm). fss

=

tensile stress in the steel reinforcement at the service limit state. (MPa)

h

=

overall thickness or depth of component.(mm)

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3.6.Minimum Reinforcement Check Per LRFD [12.11.4.3] , the area of reinforcement, As, in the cross-section should be checked for minimum reinforcement requirements per LRFD [5.7.3.3.2] . Unless otherwise specified, at any section of a flexural component the amount of tensile reinforcement shall be adequate to develop a factored flexural resistance at least equal to lesser of: A-1.2 times the cracking moment, Mcr, determined on the basis of elastic stress distribution and the modulus of rapture, fr, of concrete as specified in LRFD[5.4.2.6] where Mcr may be taken as:

Mcr=  fr x Sc fr= 0.52*(f'c)0.5 Sc= Ig/(0.5xh) Ig= 1/12xbw x h

3

Mpa

modulus of rapture LRFD[5.4.2.6]

m

3

section modulus

m

4

moment of inertia

B-1.33 times the factored moment required by the applicable strength load combinations.

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Structural Analysis of Manhole Ht W L Hs tts tbs twex Wc coverbot cover soil

6000 1200 1200 300 150 250 200 25 40 25 20

cell clear heigth cell clear width cell clear length depth of backfill top slab thickness bottom slab thickness exterior wall thickness weigth of concrete concrete cover(bslab) concrete cover(all ot) weigth of backfill

mm mm mm mm mm mm mm kN/m 3 mm mm kN/m 3

S= W + twex span length for cell, mm

1400

mm

Dead Load (DC) Include the structure self weight based on a unit weight of concrete of 25 kN/m3. When designing the bottom slab of a culvert do not forget that the weight of the concrete in the bottom slab acts in an opposite direction than the bottom soil pressure and thus reduces the design moments and shears. This load is designated as ,DC, dead load of structural components and nonstructural attachments, for application of load factors and limit state combinations. top slab dead load:

W dlts= W c x tts

W dlts=

3.75

kN/m

W dlbs=

6.25

kN/m

W dlsw=

144.00

kN

bottom slab dead load

W dlbs= W c x tbs Side walls dead load

W dlsw= 4 x W c x twex x Wx Ht

Linear soil bearing at bottom slab due to self weight of structure:

W bearing=

106.61 kN/m Wearing Surface (DW) the weight of the future wearing surface shall be taken as 1 kN/m

W ws=

1

kN/m

weight of future wearing surface

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Vertical Earth Load(EV) The weight of soil above the buried structure is taken as 20 kN/m 3. Calculate the modification of earth loads for soil-structure interaction per LRFD [12.11.2.2] . Embankment installations are assumed. Installation_Type="Embankment"

ϒs= S=

20 1.4

kN/m3 m

Hs=

0.3

m

unit weight of soil span of culvert (measured between outside faces of exterior walls)

depth of backfill above top edge of top slab Calculate the soil-structure interaction factor for embankment installations:

1

0.2

Fe=

1.04

Unitless

Fe shall not exceed 1.15 for installations with compacted fill along the sides of the box section: Fe= Unitless 1.04 Calculate the total unfactored earth load: W e= Fe x ϒs x S x Hs

W e=

8.76

kN/m2

Distribute the total unfactored earth load to be evenly distributed across the top of the culvert:

W sv=

6.26

kN/m

Horizontal Earth Load(EH) The weight of soil surrounding the buried structure is taken as 20 kN/m 3. A coefficient of lateral earth pressure is used for the lateral pressure from the soil. This coefficient of lateral earth pressure is based on wet and an effective friction angle of 15º The lateral earth pressure is calculated per LRFD [3.11.5.1] Soil horizontal earth load ( magnitude at bottom of wall ) It is obvious that vertical wall shall be act as one way slab.

ko=

0.6

ϒs=

20

Unitless coeffient of lateral earth pressure kN/m3

W sh_bot= ko x ϒs x (Ht+Hs)

unit weight of soil

W sh_bot=

75.60

kN/m

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Live Load Surcharge(LS) Surcharge loads are computed based on a coefficient of lateral earth pressure times the unit weight of soil times the height of surcharge. The uniform distributed load is applied to both exterior walls with the load toward the center The load is designated as, LS, live load surcharge, for application of load factors and limit state combinations. Refer to LRFD [3.11.6.4] for additional information regarding live load surcharge.

ko=

0.60

ϒs=

20 10

LSht=

Unitless coeffient of lateral earth pressure kN/m3 kN/m²

W sll= ko x ϒs x LSht

unit weight of soil live load surcharge height per

W sll=

120.00

kN/m

Live Loads(LL) Live load consists of the standard AASHTO LRFD trucks and tandem. Per LRFD [3.6.1.3.3] , design loads are always axle loads (single wheel loads should not be considered) and the lane load is not used. Where the depth of fill over the box is less than 600mm, wheel loads are distributed per LRFD[4.6.2.10] Where the depth of fill is 600 mm or more, the wheel loads shall be uniformly distributed over a rectangular area with sides equal to the dimension of the tire contact area LRFD [3.6.1.2.5] increased by the live load distribution factor (LLDF) in LRFD[Table 3.6.1.2.6a-1 ] using provisions of LRFD [3.6.1.2.6b-c] . Equivalent Strip Widths for Depth of Fill Less Than 600mm When the traffic travels primarily parallel to the span, follow LRFD [4.6.2.10.2]. Use a single lane and single lane multiple presence factor 1.2 Distribution length perpendicular to the span S= Clear span width 1.2 m Eperp= 2440 + 0.12 x S E= 2.608 m Distribution length parallel to the span LT = length of tire contact area, in LRFD[3.6.1.2.5] 250 mm LLDF= Unitless 1.15 times the depth of the fill in select granular 1.15 backfill, or the depth of the fill in all other cases. as specified in LRFD [3.6.1.2.6] Eparallel= LT + LLDF x Hs

Aeq= 1.55176

m

Eparallel=

0.595

m

2

Per LFRD [3.6.1.2.2] , the weights of the design truck wheel is below. (Note that one axle load is equal to two wheel loads.)

W wheel=

72.50

kN

center and rear wheel weights

The effect of single and multiple lanes shall be considered. In this case, a single lane with the single lane factor governs. Applying the single lane multiple presence factor:

W wheel= mpf x Wwheel

W wheel=

87.00

kN

mpf=

1.20

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LL=

LL= Wwheel/Aeq

56.07

kN/m

Dynamic load allowance for buried structures covered by section 12, in percent shall be taken as:

IM=33 x (1.0 - 4.1x10-4 Hs) >=% 0

Load kN/m Top Slab Bottom Slab Side Wall_Down

Load kNm/m Top slab center Bott. slab centre Side wall

Load kNm/m Top Slab Bottom Slab Side wall

DC 3.75 106.61 0

IM= IM=

28.94 16.23

% kN/m

Unfactored Loads Table Load Type DW EV EH LS 1 6.26 0 0 1 6.26 0 0 0 0 75.60 120.00

LL 56.07 56.07 0

IM 16.23 16.23 0

DC 0.68 19.19 0

Table of Unfactored Moments Load Type DW EV EH 0.18 1.13 0 0.18 1.13 0 0 0 9.07

LS 0 0 14.40

LL 10.09 10.09 0

IM 2.92 2.92 0

DC 2.25 63.96 0

Table of Unfactored Shears Load Type DW EV EH 0.60 3.75 0 0.60 3.75 0 0 0 45.36

LS 0 0 72.00

LL 33.64 33.64 0

IM 9.74 9.74 0

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Design Reinforcement Bars Top slab Top slab shall be casted as a individual pre-cast unit and assumed to be simply support. Mstrength= 25.41 kNm/m design strength moment Mservice= 14.99 kNm/m design service moment BarD=

12

mm

diameter of reinforcement used

Sbar= Abar= Aused=

150 mm spacing of reinforcement 2 area of 1 rebar 113 mm 3 area of used reinforcement 754 mm Calculate the estimated distance from extreme compression fiber to the centroid of the nonprestressed tensile reinforcement. LRFD[5.7.3.2.2]. dc = 31 mm ds_i= h-cover-BarD/2

ds_i=

119.00

mm

The factor B1 shall be taken as 0.85 for concrete strengths not exceeding 30 Mpa B1=

0.85

Per [LRFD 5.7.2.1] , if c/ds =

13.74

kNm

OK

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Per LRFD [5.7.3.4] ,the spacing of reinforcement in the layer closest to the tension face shall satify:

2

s in which

1

0.7

Calculate the ratio of flexural strain at the extreme tension face to the strain at the centroid of the reinforcement layer nearest the tension face:

1

0.7

Bs=

1.37

ρ=

0.006

n=

7.61

Calculate the reinforcement ratio:

ρ= As/(b x ds) Calculate modular ratio:

N= Es\Ec

Calculate fss, the tensile stress in the reinforcement at the Service I Limit State. The moment arm used in the equation below to calculate fss is (j)(h-dc)

k=((ρN)2 + (2ρN))0.5 - ρN j=1-k/3 Mservice= 14.99 kN/m

k= j=

0.27 0.91

service moment

fss=Mservice/(As x j x (h-dc) =

107

kN

OK

Side Wall Horizontal Reinforcement Mstrength= Mservice=

35.21 23.47

kNm/m kNm/m

12

mm

BarD=

design strength moment design service moment diameter of reinforcement used

Sbar= Abar= Aused=

150 mm spacing of reinforcement 2 area of 1 rebar 113 mm area of used reinforcement 754 mm2 Calculate the estimated distance from extreme compression fiber to the centroid of the nonprestressed tensile reinforcement. LRFD[5.7.3.2.2]. dc = 31 mm ds_i= h-cover-BarD/2

ds_i=

169.00

mm

The factor B1 shall be taken as 0.85 for concrete strengths not exceeding 30 Mpa B1=

0.85

Per [LRFD 5.7.2.1] , if c/ds =

kNm

OK

Per LRFD [5.7.3.4] ,the spacing of reinforcement in the layer closest to the tension face shall satify:

2

s in which

1

0.7

Calculate the ratio of flexural strain at the extreme tension face to the strain at the centroid of the reinforcement layer nearest the tension face:

1

0.7

Bs=

1.26

ρ=

0.004

n=

7.61

Calculate the reinforcement ratio:

ρ= As/(b x ds) Calculate modular ratio:

N= Es\Ec

Calculate fss, the tensile stress in the reinforcement at the Service I Limit State. The moment arm used in the equation below to calculate fss is (j)(h-dc)

k=((ρN)2 + (2ρN))0.5 - ρN j=1-k/3 Mservice= 23.47 kN/m

k= j=

0.23 0.92

service moment

fss=Mservice/(As x j x (h-dc)