--- title: "Benkelman beam overlay thickness numerical" subject: "Transportation Engineering II" chapter: 4 key: "semester_7/transportation_engineering_ii:benkelman-overlay-numerical-4085kg" source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/benkelman-overlay-numerical-4085kg site: BCT Engineering --- # Benkelman beam overlay thickness numerical Transportation Engineering II · Chapter 4: Highway Maintenance, Repair, and Rehabilitation - Asked in: 2082 Bhadra, 4+4 marks - Video: [Thickness of Bituminous macadam overlay| Benkelman Beam| Pavement Engineering| Transportation Engg](https://www.youtube.com/watch?v=x41FXVeW2pA) — Reinforce Engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/benkelman-overlay-numerical-4085kg - Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-4 ## Question Benkelman beam deflection studies were carried out on 15 selected points on a stretch of flexible pavement during summer season using a dual wheel load of 4085 kg, 5.6 kg/cm² pressure. The deflection values obtained in mm after making the necessary lag corrections are given below. If the present traffic consists of 750 CVPD, determine the thickness of bituminous overlay required, if the pavement temperature during the test was 30°C and the correction factor for subsequent increase in subgrade moisture content is 1.3. Assume annual rate of growth of traffic as 7.5%. | Location | Initial | Intermediate | Final | |---|---|---|---| | 1 | 145 | 84 | 82 | | 2 | 125 | 88 | 85 | | 3 | 115 | 75 | 73 | | 4 | 114 | 81 | 78 | | 5 | 137 | 100 | 100 | | 6 | 130 | 69 | 65 | | 7 | 103 | 64 | 62 | | 8 | 112 | 60 | 58 | | 9 | 111 | 52 | 50 | | 10 | 95 | 35 | 26 | (Dial gauge readings, least count 0.01 mm.) ## Short answer Compute rebound deflection (2×(initial−final)×0.01 mm), mean 1.016 mm, σ=0.258 mm. Characteristic deflection after temperature (+0.05) and seasonal (×1.3) corrections is 1.72 mm. Design traffic 1546 CVPD gives allowable 1.0 mm. Overlay thickness = 550·log10(1.72/1.0) = 130 mm bituminous macadam. ## Answer #### Method IRC:81 (Benkelman beam deflection method of overlay design), with the standard test load of 4085 kg on a dual wheel at 5.6 kg/cm2 tyre pressure. The question gives traffic only in CVPD with a growth rate, so the CVPD-based IRC:81 procedure (allowable deflection and $h=550\log_{10}(D_c/D_a)$) is used. The table gives 10 points (not 15), so 10 deflections are used. #### Step 1 - Rebound deflection at each point The dial readings are in units of 0.01 mm. Rebound deflection $=2\,(d_0-d_f)\times0.01$ mm (the factor 2 is the lever ratio of the beam). The question states that the lag corrections have already been made, so the intermediate readings are not used further and the final readings are taken. | Point | Initial | Final | $2(d_0-d_f)\times0.01$ (mm) | |---|---|---|---| | 1 | 145 | 82 | 1.26 | | 2 | 125 | 85 | 0.80 | | 3 | 115 | 73 | 0.84 | | 4 | 114 | 78 | 0.72 | | 5 | 137 | 100 | 0.74 | | 6 | 130 | 65 | 1.30 | | 7 | 103 | 62 | 0.82 | | 8 | 112 | 58 | 1.08 | | 9 | 111 | 50 | 1.22 | | 10 | 95 | 26 | 1.38 | #### Step 2 - Statistics $\sum D=10.16$ mm, so the mean $\bar D=1.016$ mm. With $\sum(D-\bar D)^2=0.598$, the standard deviation is $\sigma=\sqrt{0.598/9}=0.258$ mm. #### Step 3 - Characteristic deflection and corrections - Characteristic deflection (IRC:81, general roads): $D=\bar D+\sigma=1.016+0.258=1.274$ mm. - Temperature: the test was at 30 C and the standard is 35 C, so add 0.01 mm per degree: $1.274+0.05=1.324$ mm. - Seasonal (subgrade moisture) factor 1.3: $D_c=1.324\times1.3=1.72$ mm. #### Step 4 - Design traffic $$ A=P(1+r)^{n+10}=750\times(1.075)^{0+10}=1546\ \text{CVPD} $$ taking $n=0$ (overlay laid now); with $n=1$ or 2, $A=1662$ to 1786 CVPD. In every case the traffic is in the 1500-4500 CVPD class, for which the allowable deflection is $D_a=1.0$ mm (1.25 mm for 450-1500 CVPD, 1.5 mm for 150-450 CVPD). #### Step 5 - Overlay thickness Overlay thickness in bituminous macadam (BM): $$ h=550\log_{10}\frac{D_c}{D_a}=550\log_{10}\frac{1.72}{1.0}=550\times0.236=130\ \text{mm} $$ **Answer:** characteristic deflection $D_c\approx1.72$ mm; provide about **130 mm of BM overlay**. Since 1 mm of BC/DBM is equivalent to about 1.5 mm of BM, this is about 90 mm of bituminous concrete/DBM (e.g. 50 mm DBM plus 40 mm BC). Note: for a national or state highway, IRC:81-1997 uses $D_c=\bar D+2\sigma$. This gives $D_c=(1.016+0.516+0.05)\times1.3=2.06$ mm and $h=550\log_{10}(2.06/1.0)\approx172$ mm of BM. If the paper provides the IRC:81 overlay chart, read the thickness from it. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For point 1 (initial 145, final 82, least count 0.01 mm), the rebound deflection is: A. 0.63 mm B. 1.26 mm C. 0.82 mm D. 1.45 mm Answer: B. D = 2(145-82) x 0.01 = 1.26 mm, the factor 2 being the beam lever ratio. (Key point: Rebound deflection formula) 2. The test was at 30 C and the standard is 35 C. The mean deflection 1.016 mm becomes: A. 0.966 mm B. 1.066 mm C. 1.386 mm D. 1.016 mm unchanged Answer: B. Following the answer, add 0.01 mm per degree for 5 degrees: 1.016 + 0.05 = 1.066 mm. (Key point: Temperature correction) 3. With mean 1.386 mm and standard deviation 0.335 mm after the 1.3 factor, characteristic deflection for a national/state highway is: A. 1.386 + 0.335 = 1.72 mm B. 1.386 + 2 x 0.335 = 2.06 mm C. 1.386 x 2 = 2.77 mm D. 1.386 - 0.335 = 1.05 mm Answer: B. Dc = mean + 2 sigma = 1.386 + 0.67 = 2.06 mm. (Key point: Characteristic deflection) 4. With Dc = 2.06 mm and allowable Da = 1.1 mm, h = 550 log10(Dc/Da) gives about: A. 49 mm B. 149 mm C. 300 mm D. 2.06 mm Answer: B. log10(2.06/1.1) = 0.272, and 550 x 0.272 is about 149 mm of BM, about 150 mm. 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