ACET 406 Mid-Term Exam B

Transcription

1 ACET 406 Mid-Term Exam B SUBJECT: ACET 406, INSTRUCTOR: Dr Antonis Michael, DATE: 24/11/09 INSTRUCTIONS You are required to answer all of the following questions within the specified time (90 minutes).you can use your calculator only. All of the questions MUST be answered on separate paper. Part A 1. Describe the formation and causes for both Joint Reflection Cracking from Concrete Slab and Longitudinal Cracking [4 points] 2. In flexible pavement design we use the horizontal tensile strain (ε t ) as a design criterion for cracking. What is the difference between ε t and the minor principal strain (ε 3 )? Which one has the highest value in tension? [2 points] 3. Rutting in flexible pavements is characterized by permanent deformation (depressions) that form in the wheel paths. Define the two types of rutting and give one or two main causes for each type? [4 points] True or False: Consider the diagram below. Use the letter T if you believe the statement is true, or the letter F if you believe the statement is false. [15 points] READ THE STATEMENTS CAREFULLY. E 1 h 1 E 2 h 2 E 3 1. In general, increasing h 2 will increase the compressive strain (ε c ) in the subgrade more than by increasing E In general, the surface deflection will be higher by increasing h 1 than by increasing E The compressive strain in the subgrade is highly affected by changes in E 3.

2 4. The horizontal tensile strain (ε t ) will increase as E 1 increases. 5. In general, decreasing either h 1 or h 2 will have a large influence on the horizontal tensile strain (ε t ), since this response is primarily affected by layer stiffness. Part B 1. [25 points] For an area of uniform subgrade conditions, a pavement engineer has set the following horizontal tensile strain criterion for design of a full-depth asphalt concrete pavement: Maximum allowable tensile strain = in/in (400 µε) The asphalt concrete to be used has the following properties: highest expected temperature = 300,000 psi lowest expected temperature = 600,000 psi P = lbs E 1 = see above h =? E 2 = 12,000 psi 2. [30 points] Determine the modulus of the subgrade given the plate-test result on the subgrade (directly on the subgrade layer): δ = RIGID in Note: δ Plate Radius = 8 in Load = 1700 lbs FLEXIBLE _ MAX 1.5 δ = 1.18 RIGID The design life (number of cycles to failure) for the above full-depth pavement is projected at 10,470 cycles. Use the Asphalt Institute s rutting equation and other charts provided to determine the thickness of asphalt concrete (h) required for the pavement to meet the design requirements. (Hint: You can assume that the effect of lateral stresses is negligible) Notes: N d = ( ε c ) When Y -2 = X, then Y = X -1/ lbs 8 AC E 1 =125,000 psi h =? E 2 =? q = 70 psi

3 3. [20 points]a pavement system consisting of a high quality crushed rock base with a thin bituminous seal coat surface on subgrade is proposed for use on a low volume road (see figure below). For the design wheel load shown, it was determined that eight inches of crushed rock base course were required to reduce the vertical compressive stress and strain on the subgrade to an acceptable level. Bituminous Seal Coat Crushed Rock Base E 1 = 70,000 q = 90 psi h = 8 Subgrade E 2 = 2,500 psi a. Based on the information provided above, determine the acceptable level of vertical compressive strain for this subgrade. You may assume that lateral stresses are negligible. b. A contractor proposes the use of a less expensive base material with the following properties: E 1 = 30,000 psi; Determine the thickness of this new base course required to offer the same level of protection to the subgrade as the crushed rock base. c. Given the following cost information, which one of the two base materials will be most cost-effective if the road is ft long and 48 ft wide? Base 1 (E 1 = 70,000 psi): 125 lb/ft 3 ; 7.00/ton Base 2 (E 1 = 30,000 psi): 110 lb/ft 3 ; 5.00/ton (1 ton = 2000 lbs, 1ft = 12 in)

4 Loaded Circular Area q = P A = P π a Hooke s Law ε z = 1 σ z ν σ x + σ y E 2 ( ( ) Asphalt Institute s Failure Equations N f = ( ε t ) ( E1) N ε ) d = ( c 4.477

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