Vicarious Liability - Assignment Example

CIVIL ASSIGNMENT Name: Course Professor’s name University name City, State Q1: Gantt Bar Chart representing the activity Q1 (b) Vicarious liability refers to a legal doctrine where an individual or a company is held responsible for the actions or omissions done by a different person. For instance, in the context of the workplace, an employer can be liable to the actions or omissions of their employees, provides such actions or omissions are proved to have been conducted during the working time or in the course of the employment. In the case of the graduate surveyor, we understand that he set out to survey using the employers’ equipment, and vehicle and received cash payment for the services rendered. What we cannot out rightly establish is whether the work had been sanctioned by the employer, since the only evidence is the use of its equipment to perform the tasks. The key question therefore is whether the employee was acting on his personal capacity or in the course of employment. This seems to be difficult to confidently determine. Nor does the liability of the employer stops when there is no proof of employer directive. As the vicarious liability law stands, action against the employer can still be instituted even though the person the proof of the employer contribution can only be sited based on presence of his equipment and vehicle. Some of the practical steps the employer should have carried out is ensure that the graduate surveyor was effectively supervised by a seasoned surveyor or engineer Q2 (a). (i) Calculation of the RL’s of the chainage along the tangents A=| (-3)-4|=7%, from the chainage table, the pegging chainage, Pegging interval is 20m Increment at 4.0% increment at -3.0% The calculated RL’s at the tangent For 4.0% grade Chainage RL's Increment Calculated RL's 600 37.0 0 37.0 620 36.5 0.8 37.3 640 37.0 0.8 37.8 660 37.0 0.8 37.8 680 38.0 0.8 38.8 For -0.3% upgrade Chainage RL's Increment Calculated RL's 700 38.4 0 38.4 720 38.4 0.6 39.0 740 37.7 0.6 38.3 760 37.6 0.6 38.2 780 37.2 0.6 37.8 800 38.0 0.6 38.6 (ii) Reduced levels of the design surface Calculating the offsets from the design straight metrage; From the start =0.0175*0=0.00 Chainage RL formation RL's Design Surface 600 0 37.0 37.00 620 0.07 36.5 36.43 640 0.28 37.0 36.72 660 0.63 37.0 36.37 680 1.12 38.0 36.88 700 1.75 38.4 36.65 720 1.12 38.4 37.28 740 0.63 37.7 37.07 760 0.28 37.6 37.32 780 0.07 37.2 37.13 800 0 38.0 38.00 (iii) The cut to be filled on each peg along the centerline. Chainage RL's at Tangent Vc Offsets RL formation RL. Nat Surface Cut Fill 600 37.00 0 37.0 37 0.0 620 37.30 0.07 36.5 36.43 0.1 640 37.80 0.28 37.0 36.72 0.3 660 37.80 0.63 37.0 36.37 0.6 680 38.80 1.12 38.0 36.88 1.1 700 38.40 1.75 38.4 36.65 1.8 720 39.00 1.12 38.4 37.28 1.1 740 38.30 0.63 37.7 37.07 0.6 760 38.20 0.28 37.6 37.32 0.3 780 37.80 0.07 37.2 37.13 0.1 800 38.60 0 38.0 38 0.0 Q2 (b) B1=0°00’ 00” B1=20°00’ 00” R=400 Peg interval =50m Interval angle, B1=20°00’ 00” Meterage TP1=383m Calculation of Curves Short arc= = =139.626m = 3.125x10-10 H = = =0.2604*0.986 =0.2567544 Short sec= = =394.183 Short tangent= = Transitional length =400 Shift (H)= (C) Height of collimation=86.9m Design Centre-line level=86.0m Distance from the Centre-line to edge of shoulder=5.0m Pavement cross fall=-3% Side batter=1.1 (horizontal to vertical) Natural surface staff reading for batter point= 3.95m Calculate: Height difference between centre-line level to shoulder = =0.0167m R.L of shoulder = 86-0.0167= 85.9833 m R.L. of batter point = 86.9- 3.95 = 82.95m Height difference between shoulder and batter point = 85.9833 -82.95=3.0333m So, Distance from shoulder to batter point = =3.0333 m Offset from the design centre-line for the batter point = 5.0+3.0333= 8.0333m Question 3 AY2=sqrt(252+22)=25.0798m To calculate the low water I need to calculate the length of GE, and AG. Assuming that the bearing from A to B is 90 Bearing of B to Y is 1800 By plotting the diagram, we measure bearing Y to a as 355025’34” ß AB 2m ß BE 5m ß AG (1.6m) ß GA ’ (5.016m) High water N=0.026 S=0.01 A=0.5*(a+b)*h=0.5*(6+6+4)*25=200m2 P= (25.0798*2)+5=55.15974m Lower Water P= (25.0798-5.016)*2)+6= 46.1276m Q3 (b) Q4. Road Construction drawings no: 343289 and 343295 Excavation and embankment Type A roadway Table 1: Cross falls and widths for Type A for the left- hand side and right hand side lanes and shoulders Position Distance from centerline (m) Cross fall (%) Height Difference (m) Total difference from centerline (m) Traffic lane (A) Traffic lane (B) Traffic lane (C) Traffic lane (D) Traffic lane (E) Traffic lane (F) (E)+(F) Left hand side 3.5 1 3% -3% 0.105 0.075 0.075 From the plan FSL at chainage 26940 centre line = 98.082 m RL at top of stake = 98.150 m Height difference between traffic lane and centre = Total subgrade layers = 0.575 m String line adjustment Subgrade level = 98.117+ 0.105 – 0.575 = 97.612 m Distance down stake = 98.300 -97.762= 0.538m The stake should be at mark (0.538– 0.015) = 0.523 m (C. An error was detected on the line extending from CT Using close program the bearing of line CT was found to be ~285 as opposed to 220d 21’ 30 given on the plan. From the Plan, E N Coordinates of IP 50561.212 150119.664 Coordinates of CT 50324.289 150162.958 Dep Lat 50324.289-50561.212= -236.923 m 150162.958- 150119.664 = 43.294 m β = Tan (A) β of IP-CT= β of TP-IP= Intersection angle (I) = Point Line Bearing D E N TC 50755.582 149977.441 IP TC-IP 240.846 -194.369 142.22334 50561.2131 150119.6643 CT IP-CT 240.846 -236.9227 43.294694 50324.29 150162.959 Since the coordinates are the same, it can be assumed that they are correct (d). Subtract 90d from the bearing as the offset is perpendicular to centerline= Point Line Bearing D E N CT 50755.582 149977.441 26940 26940-CT 71.717 -70.549 12.892 50561.2131 150119.6643 20 to left 26940-20Left 20 -3.595 -19.674 50324.29 150162.959 Read More