Rigid Pavement Design Part 4
This section discusses the critical aspects of slab distance and the placement of joints in rigid pavement design. It emphasizes the importance of understanding the thermal expansion of concrete and provides formulas to calculate the required distances between joints. The content includes examples and solutions for determining joint widths based on varying temperatures. Proper planning of contraction and expansion joints is essential to accommodate the movements caused by temperature fluctuations.
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Rigid Pavement Design Part 4
1 1. . Distance The slab distance shall The using L = Distance between The distance slab length distance increases shall not The distance using the between Expansion distance between length and increases. . In not exceed distance between the following 2 Expansion Joints between expansion and if if slab In any exceed about between expansion following equation Joints joints (gap) length increases any way about 2 2. .5 5 cm expansion joints equation: : expansion joints slab length way the cm. . (gap) depends increases the the distance depends on the required between joints on required joints distance between joints can can be be calculated calculated t ) .(t 1 Where = = max = = thermal L L = = slab t t1 1= = temperature t t2 2=temperature increases Where: : max. . expansion thermal expansion slab length temperature casting =temperature after increases C C expansion of expansion coefficient length casting slab after casting of slab slab coefficient of cm cm concrete= 8 8 10 of concrete= 10- -6 6/ /deg deg cm cm slab C C casting slab temperature slab when when temperature
Example ( to calculate joints if if it it is is known of to half degree of 50 C C and concrete Example (9 9) ) calculate the It It is is required expansion thickness compressed the 50 concrete equals required to expansion joints thickness compressed to the degree and coefficient equals to the distance known that expansion half , , the of slab coefficient of to 8 8 10 distance between that the joint width of slab temperature of thermal 10- -6 6/deg between the material joint of gap rises from thermal expansion /deg. . material can cm , , 10 C C to of the the expansion the width temperature rises can be be gap is is 2 2. .5 5 cm from 10 expansion of to of
Solution Solution Take joint .) Take slab joint = = 35 5 slab length 35 m ( length = = distance m distance between between expansion expansion
2 2. . Distance Concrete about the between foundation Distance between Concrete slab about the the movement between the foundation of between Contraction slab contracts the temperature movement of the surface of the Contraction Joints contracts due temperature of of concrete surface of the road Joints lower temperature the pouring concrete slab of the road. . due to of the to lower pouring of slab resist the slab temperature of of concrete, resist by slab and of slab and friction slab concrete, and by the and the the friction the material material of of . . , L/2 b Slab contraction Frictional resistance
The using Case The distance using the Case plain distance between the following plain concrete c F 2 L = between contraction following equation concrete slab 10 contraction joints equation: : slab: : joints can can be be calculated calculated 4 f . w Case Case reinforced L reinforced concrete sF 00 2 = concrete slab s .A slab: : b.h.w. f Where h h = = slab L= b b = = slab f f = = friction w w = = unit F Fc c= = tensile F Fs s= = tensile A As s= = area Where: : slab thickness L= slab slab width friction coefficient unit weight tensile stress tensile stress area of thickness length and width coefficient weight of stress allowed stress allowed of steel cm cm slab length and the the distance distance between between contraction contraction joints m taken taken taken kg/cm slab width joints m m m taken taken 2400 taken 0 0. .8 8 kg/cm kg/cm2 2 width 1 1. .5 5 2400 kg/m kg/cm2 2 of the allowed in allowed in steel reinforcement concrete slab in concrete in steel reinforcement for the concrete slab kg/m3 3 concrete steel for slab cm cm2 2
Example ( the distance Example (10 distance between 10) ) It It is is required known Slab Slab Friction Slab Steel Max Max Factor Total The directions, 1 1. . Slab 2 2. . Slab required to known that Slab width Slab thickness Friction coefficient Slab concrete Steel density Max. . tensile Max. . tensile Factor of Total reinforcement The reinforced directions, for Slab of Slab of to calculate calculate the between contraction contraction joints joints if if that: : width thickness coefficient concrete density density tensile strength tensile strength of Safety reinforcement steel reinforced steel for the of plain of reinforced = = 3 3. .5 5 m = = 10 = = 1 1. .5 5 = = 2400 = = 7500 = = 1 1. .6 6 kg/cm = = 1200 = = 2 2 = = 3 3 kg/cm regular distributor m cm 10 cm density 2400 kg/m 7500 kg/m kg/cm2 2 1200 kg/cm kg/m3 3 kg/m3 3 strength for strength for Safety for concrete for steel concrete steel kg/cm2 2 steel used steel distributed the two plain concrete reinforced concrete used distributed in two cases concrete. . concrete. . kg/cm2 2 distributor in in regular in both both cases: :
Solution Solution 1 1. . Slab Slab of of plain plain concrete concrete: : 2 2. . Slab Slab of of reinforced 1200 = reinforced concrete concrete: : kg/cm 600 = 2 sF 2 Assume of using A As s Assume slide of steel using the slide of reinforcement A As sin the following of slab slab length length of in one equation: : : of 1 1m one direction m we direction in we can can find in slab find the slab width the area width area steel reinforcement following equation 1
A 3 s 4 = b 7500 2 10 3.5 10 4 3 = 7.0 = A cm 2 s 7500 2 The calculated The calculated using distance distance between the following between following equation contraction equation: : contraction joints joints can can be be using the 2 00 sF .A s = L b.h.w. f 600 7 2 00 = 6.66 = L m 10 2400 1.5 3.5 The The distance distance between between contraction contraction joints joints taken taken = = 6 6 m m
