COMMUNICATIONS Formulas and ConceptsFull description
COMMUNICATIONS Formulas and ConceptsFull description
ELECTRONICS Formulas and ConceptsFull description
COMMUNICATIONS Formulas and ConceptsFull description
Full description
Full description
Physics pocket diary of concepts and formulas
Full description
Rigid pavement Older’s theory Without dowels or tie bars Critical section is at the edge of a contraction joint, it will crack approximately 45˚ with the edges M = wx
fb
6M bd2
b = 2x, d = t fb
6 wx 2x t
2
;
t
3w fb
Where Fb - allowable tensile stress of concrete (psi or kg/m 2) W - wheel load (lb or kg) With dowels or tie bars The purpose of dowels or tie bars is to transmit the stresses due to the load from the adjacent pavement.
Thickness at the edge of pavement M
fb
wx 2
6M bd2
b = 2x, d = t
fb
w x 2 ; 2x t 2
6
t
3w 2 fb
Thickness at the center of pavement M
fb
wx 4
6M bd2
b = 2x, d = t
fb
w x 4 ; 2x t 2
6
3w 4 fb
t
Stress at the edge of a concrete pavement (Westergaards stress equation) Se
0.572P h2
L 4 log 10 b 0.359
Where P – wheel load h – thickness of pavement (cm) L – radius of relative stiffness b – radius of resisting section Stress at the corner region of a concrete pavement (Westergaards Stress equation)
Sc
Where
3P 1h2
a 2 L
1.2
P – wheel load h – thickness of pavement (cm) L – radius of relative stiffness b – radius of resisting section a – radius of contact area (cm) a – 14 cm Stress at interior region of a concrete pavement (Westergaards stress equation) St
0.316P 2
h
L 4 log 10 b 1.069
Where P – wheel load h – thickness of pavement (cm) L – radius of relative stiffness L – 88 cm b – radius of resisting section Warping Stress at interior of concrete pavement (kg/cm2) St
Eet Cx ρCy 2 1 - ρ2
Where e – thermal coefficient of concrete per ˚C e – 10 x 10-6 kg/cm2 E – modulus of elasticity of concrete (kg/cm2) E - 3 x 105 kg/cm2 t – temperature difference between top and bottom of the slab ρ – poisons ratio Cx and Cy – coefficient of warping stress Warping Stress at the edge region of concrete pavement (kg/cm 2) Se
CxEet 2
Warping Stress at corner region of concrete pavement (kg/cm 2)
Eet 3 1 - ρ
St
a L
Flexible pavement
A2
W f A1 = πr2
By ratio and proportion A1 A2
r2
t
r
t 0.564
2
W - r fb
;
Thickness of pavement using MCleods or Plate Load Test Method
T = K log10
P S
Where P – wheel load S – subgrade pressure K – constant value Radius of Relative stiffness of pavement
1
Eh3
4
2 12K(1- ρ)
L = Where L – radius of relative stiffness (cm) E – modulus of elasticity of concrete (kg/cm2) K – modulus of subgrade reaction (kg/cm 3) ρ – poisons ratio h – slab thickness (cm)
Radius of resisting section
√ ( 1.6 a +h ) 2
b=
2
– 0.675h
Where b – radius of resisting section (cm) a – radius of loaded area (cm) s - slab thickness (cm) when a > 1.724h, use b = a.
Thickness of pavement using U.S. Corps of Engineers 1
t=
1 2 1.75 W CBR Pπ
Where t – thickness of pavement (cm) W – wheel load (kg) CBR – California Bearing Ratio (kg/cm2) Number of vehicles per day for the analysis in the design of pavement P (1 r) n 10
A= Where A – number of vehicles per day to be used in the design of pavement r – annual rate growth of traffic n – number of years between last count and the year of completion of project P – present average daily traffic of commercial vehicles
Factor of Safety
residual strength edge load stress F.S. = Where
Residual strength = allowable flexural stress – warping stress Stress at the corner of a slab using Goldbecks Formula.
3P S=
h2
Where s – stress at the corner of slab (kg/cm 2) P – corner load (kg) h - slab thickness (cm)
California Bearing Ratio (CBR)
CBR =
CBR =
unit load at 0.10 inch penetratio n (100) 1000psi unit load at 0.20 inch penetratio n (100) 1500psi
Modulus of Subgrade (kg/cm3)
K=
F , 0.125
P A
F
Where P – load causing a settlement of 0.125cm (kg) A – area of standard plate (75cm diameter) s - slab thickness (cm) Extra widening of a curve section of road nL2 2R
w=
V 9.5
R
Where w – extra widening required (m) L – length of longest wheel base of vehicles (m) R – radius of the curve (m) V – velocity of vehicles (kph)
Varying ratio of centrifugal acceleration P
2V2 (3.6) 2 RT
Where P – varying ratio of centrifugal acceleration T – running time (sec) R – radius of the curve (m) V – velocity of vehicle (kph) Thickness of pavement by Tri-axial Test Method T
3Pxy - r 2 2E s
Es EB
1/ 3
Where P – wheel load x – traffic coefficient y – rainfall coefficient ∆ – design deflection r – radius of contact area Es – modulus of elasticity of subgrade soil EB – modulus of elasticity of base coarse material
Thickness of pavement by California Resistance Value Method T
Where
K(T.I)( 90 - R) 1 C5
K – numerical index T.I. – traffic index C – cohesiometer value R – stabilometer resistance value
Thickness of pavement by Pressure Method
T
expansion pressure average pavementdensity
Stiffener Factor of Pavement 1
S.F.
E 3 B EP
Where EB – modulus of elasticity of subgrade EP – modulus of elasticity of pavement Bulk specific gravity of a core of compacted asphalt concrete pavement d
A
D - A F
D - E - Where
d – bulk specific gravity of core A – weight of dry specimen in air D – weight of specimen + paraffin coating in air E – weight of specimen + paraffin coating in water d – bulk specific gravity of core
F – bulk specific gravity of paraffin
Asphalt Absorption
G - Gsb (G ) Pba 100 se b Gse Gsb Where Pba – asphalt absorption Gse – effective specific gravity of aggregates Gsb – bulk specific gravity of aggregates Gb – specific gravity of asphalt
Effective Asphalt Content Pbe Pb -
Pba Ps 100
Where Pbe – effective asphalt content Pb – percentage weight of fine aggregates Ps – sum of percentage weight of fine and coarse aggregates Pba – asphalt absorption Problem A flexible pavement carries a static wheel load of 60 kN. The circular contact area of the tire is 85806 mm2 and the transmitted load is distributed across a wide area of the subgrade at an angle of 45˚. The subgrade bearing value is 0.14MPa, while the that of the base is 0.41MPa. Design the thickness of the pavement.
A pavement carries a static wheel load 53.5 kN. The circular contact area of the tire is 85806 mm2 and the transmitted load is distributed across a wide area of the sugrade at an angle of 45˚. Design the thickness of pavement using a. Asphalt (or flexible pavement) when the subgrade bearing value is 0.14MPa, while the that of the base is 0.41MPa. b. Plain cement concrete pavement without dowels. Allowable tensile stress of concrete is 1.40MPa. c. If sufficient dowels are used across the joint.
DESIGN OF TIE BARS Problem 1. A cement concrete pavement has a thickness of 18 cm and has two lanes of 7 meters with a longitudinal joint. Design the spacing of the tie bar the length of bars. Allowable tensile stress of steel = 1600 kg/m2. Unit weight of concrete = 2400 kg/m 3 Coefficient of friction between the pavement and sub grade = 1.5 Allowable bond stress in concrete = 24 kg/cm2 Use 16 mm ф bars. Solution: Consider 1 meter length of slab W
= (0.18)(3.5)(2400) = 1512 kg
ΣFv
= 0; W = N F = ʯN =
ΣFh
(1.5)(1512) = 2268 kg
= 0; P= F P = Asfs F = Asfs, As(1600) = 2268 As = 1.42 sqm/meter
Asb = 2.011 cm2
Spacing of ties =
2.01 1.415 m 1.42
Allowable spacing = 0.0051(1000)(105) = 535.5 mm Adopt center to center spacing s = 500 mm Length of bars Note : Length of bars shall be at least twice the computed value Asfs = (πDL)(Bond stress) (1.42)(1600) = (π)(1.6)(24)L L = 18.83 cm say 20 cm Use L = 2(20) = 40 cm
Problem No. 2 Determine the spacing between contraction joints for a 3.00 meter slab having a thickness of 20 cm. Allowable tensile stress of steel = 800 kg/m2. Unit weight of concrete = 2400 kg/m 3 Unit weight of steel = 7500 kg/m3 Total reinforcement = 4 kg/m2 and is equally distributed in both directions Coefficient of friction between the pavement and sub grade = 1.5 Allowable tensile stress of concrete = 0.80 Bond stress in concrete = 24 kg/cm 2 Use 16 mm ф bars.
Solution: Consider only one half of the section
(Using principles of mechanics) W
L 20 3 2400 720 L 2 100
ΣFv = 0; W = N F=ʯN = (1.5)(720(L) F = 1080L T = (300)(20)(0.80) = 4800 kg T=F 4800 = 1080 L L = 4.44m Using equation L
2 f x 10 4 μD
Problem No. 3. A concrete pavement 8 meter wide and 150 mm thick is to be provided with a center longitudinal joint using 12mm ф bars. Using the following design criteria, compute the spacing of the longitudinal bars. Allowable tensile stress of steel = 138 MPa. Unit weight of concrete = 23.54 kN/m3 Coefficient of friction between the pavement and sub grade = 2.0 Allowable bond stress in concrete = 24 kg/cm2
Solution: W = bts = (4.00)(0.15)S (23.54) = 14.124S W = N = 14.124S
F=ʯN F = (2.0)(14.124S) = 28.248S T = Asfs (113.1)(138) = 15607 T=F 15607 = 28.248S S = 552.53 mm
Problem No. 4. Design the thickness of a highway pavement using the plate load test method if the base coarse has a constant value of 90. The wheel load is 4 100 kg. The total subgrade pressure for the same contact area, deflection and number of repetitions of the load is equal to 2100 kg. Provide 5 cm of bituminous surfacing on the top of the computed thickness. Use MCleod method.
