Design of Roadside Barrier Systems Placed on MSE Retaining Walls (2010)

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Design guideline for 10 ft high MSE wall with 10 ft long strips design 1. Guidelines for the barrier 1.1. Sliding P L s 25.91 kips > 10 kips OK L s = 1×10 kips = 10 kips P = W tan r = 32.39 kips P = 0.8 × 32.39 kips = 25.91 kips W = 56.1 kips for three 10 ft long barrier and one 30 ft long moment slab and overburden soil. r = 30 assumed this is the same as retained fill tan r = 0.58 In order to obtain the weight of 30 ft long barrier-moment slab system, the detail calculation was conducted below. Rotation Point, B Overburden Soil h B 9" 48 " 6" 9" 5" 32 " 24 " 4.75" (1) (2) (3) (9) (10) (11) (7) (8) (6) (5) (4) 4.5" Rotation Point, B 4.25" Overburden Soil h o Rotatio n Point, O C.G. l O l B W L s H e Figure J.1. Section of a barrier-moment slab system for calculation of sliding A P P E N D I X J Example of Design Guideline 178

179 1.2. Overturning M L s h B 88.88 kips-ft > 35.83 kips-ft OK l O = 17.86 in. and h O = 26.63 in. from Table J-1 l B = 21.13 in. and h B = 43.0 in. L s h B = 1 × 10 kips × 43 in. = 430 kips-in. = 35.83 kips-ft M = W l B = 56.1 kips × 21.13 in. = 1185.1 kips-in. = 98.76 kips-ft M = 0.9 × 98.76 kips = 88.88 kips-ft 1.3 Rupture of the coping in bending (referred from AASHTO LRFD Section 5) f y = 60 ksi, f c ’ = 4 ksi Rotation Point, B Overburden Soil h C h C.G. L d H e 5" 10" Figure J.2. A barrier-moment slab system for calculation of rupture of the coping in bending Table J.1. Calculation of barrier-moment slab system weight Section Longitudinaldistance, x Vertical distance, y Area (in 2 ) weight (k) x from O y from O y* weight x*weight 1 12.00 32.00 384.0 12.00 6.00 40.00 480.00 72.00 2 12.00 9.00 108.0 3.38 6.00 19.50 65.81 20.25 3 4.50 9.00 20.25 0.63 13.50 18.00 11.39 8.54 4 16.50 10.00 165.0 5.16 8.25 10.00 51.56 42.54 5 4.25 5.00 21.25 0.66 2.13 2.50 1.66 1.41 Barrier and Coping 6 4.75 5.00 23.75 0.74 14.13 2.50 1.86 10.48 7 48.00 6.00 144.0 4.50 32.50 11.00 49.50 146.25 Moment Slab 8 48.00 9.00 432.0 13.50 40.50 4.50 60.75 546.75 9 4.50 9.00 20.25 0.53 15.00 21.00 11.07 7.91 10 48.00 9.00 432.0 11.25 40.50 19.50 219.38 455.63 Soil 11 48.00 6.00 144.0 3.75 48.50 13.00 48.75 181.88 Total 1894.5 56.10 1001.73 1493.64 h o and l o = 17.86 26.63

180 db = 0.75 in., Ab = 0.44 in2, d = 11.18 in. -2 in. -0.38 in. = 8.81 in. Therefore, use d = 9 in. Impact is resisted by the 10 ft length of a barrier unit at the moment slab As = 10 ft / 0.83 ft per bar × 0.44 in2 = 5.3 in2 2. Guidelines for the soil reinforcement The traffic live load has been neglected in this example. Please refer to Appendix A, Example 3 (pages A-15 to A-23) for detailed calculations of ps (static earth pressure) 2.1. Pullout of the soil reinforcement P s p s At + d pd At + LL p LL At 1) Top layer of reinforcement P = 1 × 2b × L × v × F* = 2.052 kips p s At = 0.688 kips (See Appendix A, Example 3) pd At = 313 psf × 2.92 ft2 = 0.914 kips (using pressure diagram) s p s At + d pd At = 1 × 0.688 kips + 1 × 0.914 kips = 1.602 kips P s p s At + d pd At 2.05 kips > 1.60 kips OK 2) Second layer of reinforcement P = 1 × 2b × L × v × F* = 3.413 kips p s At = 1.205 kips (See Appendix A, Example 3) pd At = 230 psf × 3.993 ft2 = 0.918 kips (using pressure diagram) s p s At + d pd At = 1 × 1.205 kips + 1 × 0.918 kips = 2.123 kips P s p s At + d pd At 3.413 kips > 2.12 kips OK Mult Mimpact 205.41 kip-ft > 171 kips-ft OK Mimpact = × Ld × hc = 1 × 54 kips × 38 in. = 2052 kip-in. = 171 kips-ft M = 0.9ult = × [5.3 in2 × 60 ksi × 9 in. (1-0.08662/2)] = 2464.9 kips-in. = 205.41 kips-ft k = = 0.08662 5.3 in 2 × 60 ksi 0.85 × 4 ksi × 10 ft × 9 in. As fy [As fyd (1 – –)] 0.85 fc' bd k 2 The thickness of the critical section on the coping = 11.18 in. Use No. 6 bars at 10 in. o.c.

181 2) Second layer of reinforcement R = t A s = t × b × Ec = 60 ksi × 50 mm × 1.984 mm = 9.226 kips for 100 year corrosion p s A t = 1.205 kips (See Appendix A, Example 3) p d A t = 230 psf × 3.993 ft 2 = 0.918 kips (using pressure diagram) s p s A t + d p d A t = 1 × 1.205 kips + 1 × 0.918 kips = 2.123 kips R s p s A t + d p d A t 9.226 kips > 2.123 kips OK 3. Guidelines for the wall panels 3.1. Check Moment Stability M u M i 3.1.1 Find M u b = 12 in. (unit length) f y = 60000 psi h = 5.5 in. E y = 29000000 psi f ' c = 4000 psi d = 2.75 in. A s = 0.22 in 2 1) Cracking h b h /2 cr cr f r T C = 28697.67 lbs-in/ft = 2.39 kips-ft/ft I g (2 nd moment of area) = bh 3 /12 = 166.38 in 4 c b = h/2 = 2.75 in. f r = 7.5 = 474.34 psi E cr = 57000 = 3605 ksi cr = f r /E cr = 0.000132 strain cr = cr /c b = 0.000574 strain/ft 2.2. Rupture of the soil reinforcement R s p s A t + d p d A t + LL p LL A t 1) Top layer of reinforcement R = t A s = t × b × Ec = 60 ksi × 50 mm × 1.984 mm = 9.226 kips for 100 year corrosion p s A t = 0.688 kips (See Appendix A, Example 3) p d A t = 1200 psf × 3.993 ft 2 = 4.792 kips (using pressure diagram) s p s A t + d p d A t = 1 × 0.688 kips + 1 × 4.792 kips = 6.566 kips R s p s A t + d p d A t 9.226 kips > 6.566 kips OK Mcr = = 166.38 in4 × 474.34 psi 2.75 in. Ig fr cb √fc' √fc'

182 = 0.22 in2 × 60 ksi × 2.75 in × (1-0.21/3) = 33803.62 lbs-in/ft = 2.82 kips-ft/ft = As/Ac = 0.333% n = Es/Ec = 8.04 n = 0.027 = cr = fr/Es = 0.00207 strain cr = cr/(d- d) = 0.0114 strain/ft 3) Ultimate h b h/2 Strain Stress cr s f y More cracking As d = 0.003cr T C Force kd kd 0.85fc' kd 2 y Yielding = 0.22 in2 × 60 ksi × 2.75 in × (1-0.1176/2) = 34,164.7 lbs-in/ft = 2.85 kips-ft/ft u = 0.003 strain 1 = 0.85 u = cr/( d/ 1) = 0.0946 strain/ft 2) Yield h b Strain Stress y s T C Force More cracking fc'c x=kd sy cr f fs y d As √(ρn)2 + 2ρn – ρn = 0.21 ′ = × × A f f bd in ksi ksi s y c0 85 0 22 60 0 85 4 2 . . . . . . . × × = 12 2 75 01176 in in κ = M A f dn s y k = −( )1 3 M A f dn s y k = −( )1 3

183 3.1.2 Find Mi P2=230 psf l2=2.5 ft l3=1.2 ft P1=1200 psf l1=0.54 ft -648 1200.6 -299.98 -586.88 276 F1=1848.62 lb F2=862.88 lb -174.96 425.66388.31 -165.6 BA Mi has been selected maximum positive moment. Shear Force (kips) Bending Moment (kips-ft) 0 0.5 1 1.5 2 2.5 3 0 0.02 0.04 0.06 0.08 0.1 Curvature (strain/ft) M o m en t (k ip - ft /ft ) Cracking Yielding Ultimate Figure J.3. Moment and curvature relationship for a wall panel Mu = 0.9 × 2.85 kip-ft/ft = 2.56 kip-ft/ft

184 M i = 1 × 0.43 kips-ft/ft = 0.43 kip-ft/ft M u M i 2.56 kip-ft/ft > 0.43 kip-ft/ft OK 3.2. Check Shear Stability V ul t V im pact V ult = b w d = 0.9 × 2 × 63.25 × 12 × 2.75 = 3756.79 lbs = 3.76 kips 1/2 V ult = 1.88 kips > V im pact = 1.2 kip OK ′fc