Question

1. Figure 1 shows the earthworks for a proposed two-lane two way divided road to be constructed with a total width of 6 m. The height measured at each cross section is given in Table 1. It can be assumed that the slide slope is 1 in 10 from Chn 1490 to Chn 1500, and 1 in 5 from Chn 1520 to Chn 1530. [10 marks) a. Calculate the cross-sectional area at each chainage (3 marks) b. Estimate the total volume of: [7 marks) (1)Cut (2) fill Chainage Height of cross-section 1490 0.6 1500 2.0 1510 0.0 1520 0.5 1530 1.8 1490 1500 1510 1520 1530 Figure 1: Proposed roadway earthworks

please need answers for this I am stuck on this

Answer 1

Gm liven, Total Width Side slope s,= 1:10 from chn 1490 to chn 1500 side slope sa 1:5 from chn 1520 ts chn 1530 chainage cross-section Height 0.6 1490 1500 2.0 1510 0.0 1520 0.5 15 30 1.8 a, Cross sectional area at each chainage is At I chn 1490 Area A, = 7.2m2 52 m2 om2 (6+s, nih, = (6+1010-67b.6 AA chn 1500: Az = (b+5, 921 62 = (6+1012172 * At com 1510: Ag = 1675 ha) ng = (6+10109)(0) = + At Chn 1520: Au=[b+shaha= {G+5(0-5) 0-5) 4.25m2 * At chn 1530: Ag (6+5,9%)==[6+5 (189] (18)= 27m" b. Total volume of cut:- at chainages 1510, 1520 & 1536 [(A.-A2)+ (9,-^2)+ 21 A2 +Ag+--- + An) interval chainage Excavation id done L Volume = 2 where , L=lom

10 2 = 177-5m2 : volume volume (W) = [(0+27) + 214-25) of cut =177.5m3 in, Total volume of fill: Fill soil is observed at chn 1490, 1500 & 1510 volume e [1A,+An) + Q | Azt Agt + An-1)] = v= 10 [17-2+0) + 2(527] 3 = 556 m :. Fill volume - 556 m3