BE Civil Engineering (IOE, TU) Surveying I (IOE, CE 453) Question Paper 2078 Nepal

(a) Fundamental principles of surveying

1. Working from the whole to the part. A framework of a few highly accurate control points is first established over the whole area (e.g. by triangulation or a precise traverse). The interior detail is then located relative to this framework with comparatively lower-order work. This prevents the accumulation and propagation of small errors over the entire survey; any error remains localised to a small region rather than growing across the map. Working from the part to the whole would let minor errors multiply and distort the whole survey.

2. Locating a point by at least two measurements. A new point must be fixed with reference to two already-fixed points by at least two independent measurements (two distances, two angles, or one distance and one angle, etc.). A third (check) measurement is taken wherever possible to detect mistakes.

(b) Distinctions

A survey can be precise but inaccurate (e.g. a chain that is uniformly too short gives repeatable but wrong lengths).

(c) Scale and area computation

(i) Representative Fraction:

$$RF = \frac{\text{map distance}}{\text{ground distance}} = \frac{1\,\text{cm}}{40\,\text{m}} = \frac{1\,\text{cm}}{4000\,\text{cm}} = \frac{1}{4000}$$

(ii) Ground dimensions:

$$L = 6.5\,\text{cm} \times 40\,\text{m/cm} = 260\,\text{m}$$ $$B = 4.0\,\text{cm} \times 40\,\text{m/cm} = 160\,\text{m}$$

Ground area:

$$A = 260 \times 160 = 41600\,\text{m}^2$$

Converting to hectares ($1\,\text{ha} = 10000\,\text{m}^2$):

$$A = \frac{41600}{10000} = \mathbf{4.16\ ha}$$

Step 1 - Test each line. A line is free from local attraction when $BB - FB = 180^\circ$ exactly.

• AB: $225^\circ30' - 45^\circ30' = 180^\circ00'$ → OK (free)
• BC: $301^\circ30' - 120^\circ00' = 181^\circ30'$ → affected
• CD: $200^\circ15' - 20^\circ15' = 180^\circ00'$ → OK (free)
• DA: $290^\circ30' - 110^\circ30' = 180^\circ00'$ → OK (free)

Only line BC shows a discrepancy ($1^\circ30'$). Both ends of AB, CD and DA satisfy the $180^\circ$ rule.

Step 2 - Locate the affected station.

Line AB is free → stations A and B carry correct meridians for AB. Line CD is free → C and D correct for CD. Line DA is free → D and A correct. The only inconsistency is on BC. Since A is free (from AB) and the line AB ends correctly at B, station B's needle is correct as read on AB. Examine BC at B: FB$_{BC}=120^\circ00'$.

From the free line CD, station C is correct. At C the BB of BC is $301^\circ30'$; if C were truly free the FB$_{BC}$ would be $301^\circ30'-180^\circ=121^\circ30'$. But B (proved free via AB) reads FB$_{BC}=120^\circ00'$. Hence the inconsistency lies at station C as recorded for line BC — i.e. C is affected by local attraction only on the BC observation set. Treating B as correct:

Correct FB$_{BC}=120^\circ00'$ (at B, free) → correct BB$_{BC}=120^\circ00'+180^\circ=300^\circ00'$. Observed BB$_{BC}=301^\circ30'$ → correction at C $= 300^\circ00'-301^\circ30' = -1^\circ30'$.

Since CD at C reads correctly ($BB-FB=180^\circ$ for CD), the $1^\circ30'$ is a local disturbance affecting the BC reading; the corrected, internally consistent set is as below.

(b) Corrected bearings

All corrected lines now satisfy $BB-FB=180^\circ$.

(c) Included angle at B

At B the two lines are BA and BC. Using corrected bearings:

• Bearing of BA (back bearing of AB) $= 225^\circ30'$
• Bearing of BC (fore bearing) $= 120^\circ00'$

Included angle $\angle ABC =$ bearing of BA $-$ bearing of BC $= 225^\circ30' - 120^\circ00' = \mathbf{105^\circ30'}$.

Identifying readings. The first reading is a Back Sight (BS) on the BM. After a shift (change point), the reading just before the shift is a Fore Sight (FS) and the next is a BS on the same change point (CP). Shifts occur after the 4th and 7th readings.

• R1 = 1.485 → BS (on BM)
• R2 = 2.120 → IS
• R3 = 0.965 → IS
• R4 = 2.745 → FS (CP1)
• R5 = 1.330 → BS (CP1)
• R6 = 0.875 → IS
• R7 = 3.045 → FS (CP2)
• R8 = 1.640 → BS (CP2)
• R9 = 0.980 → FS (last point)

Rise-and-fall: Rise/Fall = (previous staff reading) − (this staff reading); positive ⇒ Rise.

Computation of rises/falls:

• 1→2: $1.485-2.120=-0.635$ (Fall 0.635)
• 2→3: $2.120-0.965=+1.155$ (Rise 1.155)
• 3→4: $0.965-2.745=-1.780$ (Fall 1.780)
• 4→5 (CP1): $1.330-0.875=+0.455$ (Rise 0.455)
• 5→6: $0.875-3.045=-2.170$ (Fall 2.170)
• 6→7 (CP2): $1.640-0.980=+0.660$ (Rise 0.660)

(b) Arithmetic check

$$\sum BS = 1.485+1.330+1.640 = 4.455$$ $$\sum FS = 2.745+3.045+0.980 = 6.770$$ $$\sum \text{Rise} = 1.155+0.455+0.660 = 2.270$$ $$\sum \text{Fall} = 0.635+1.780+2.170 = 4.585$$

Checks:

$$\sum BS - \sum FS = 4.455 - 6.770 = -2.315$$ $$\sum \text{Rise} - \sum \text{Fall} = 2.270 - 4.585 = -2.315$$ $$\text{Last RL} - \text{First RL} = 147.685 - 150.000 = -2.315$$

All three are equal $(-2.315\,\text{m})$ → arithmetic check satisfied.

(c) The last point RL $=147.685\,\text{m}$ is lower than the benchmark by

$$150.000 - 147.685 = \mathbf{2.315\ m}.$$

Latitude $= L\cos\theta$, Departure $= L\sin\theta$ (North & East positive).

Useful values: $\cos30^\circ=0.86603,\ \sin30^\circ=0.50000$; $\cos120^\circ=-0.50000,\ \sin120^\circ=0.86603$; $\cos210^\circ=-0.86603,\ \sin210^\circ=-0.50000$; $\cos300^\circ=0.50000,\ \sin300^\circ=-0.86603$.

Working:

• AB: $150\times0.86603=129.904$; $150\times0.5=75.000$
• BC: $200\times(-0.5)=-100.000$; $200\times0.86603=173.205$
• CD: $150\times(-0.86603)=-129.904$; $150\times(-0.5)=-75.000$
• DA: $200\times0.5=100.000$; $200\times(-0.86603)=-173.205$

(a) Latitudes and departures as tabulated.

(b) Closing error

$$\sum L = 129.904 - 100.000 - 129.904 + 100.000 = 0.000\,\text{m}$$ $$\sum D = 75.000 + 173.205 - 75.000 - 173.205 = 0.000\,\text{m}$$

Closing error:

$$e = \sqrt{(\textstyle\sum L)^2 + (\sum D)^2} = \sqrt{0^2+0^2} = \mathbf{0.000\ m}$$

The traverse closes perfectly; the closing error is zero and its direction is indeterminate, so no correction is required.

(c) Bowditch's rule

Bowditch's (compass) rule distributes the error in proportion to the length of each line:

$$C_{Lat} = \frac{\text{length of line}}{\text{perimeter}}\times(-\textstyle\sum L), \qquad C_{Dep} = \frac{\text{length of line}}{\text{perimeter}}\times(-\sum D)$$

Perimeter $= 150+200+150+200 = 700\,\text{m}$.

For line BC, since $\sum L = 0$ and $\sum D = 0$:

$$C_{Lat,\,BC} = \frac{200}{700}\times 0 = \mathbf{0.000\ m}, \qquad C_{Dep,\,BC} = \frac{200}{700}\times 0 = \mathbf{0.000\ m}$$

The corrections are zero (the geometrically balanced closed figure has no misclosure to distribute); adjusted latitudes/departures equal the computed ones.

(a) Area under irregular boundary, $d = 10\,\text{m}$, ordinates $O_1\ldots O_7$.

$O_1=2.50,\ O_2=3.80,\ O_3=4.60,\ O_4=5.20,\ O_5=6.10,\ O_6=5.40,\ O_7=4.00$ (7 ordinates ⇒ 6 intervals, even number, so Simpson's rule is applicable).

(i) Trapezoidal rule

$$A = d\left[\frac{O_1+O_7}{2} + (O_2+O_3+O_4+O_5+O_6)\right]$$

Sum of intermediate ordinates $= 3.80+4.60+5.20+6.10+5.40 = 25.10$.

$$A = 10\left[\frac{2.50+4.00}{2} + 25.10\right] = 10\left[3.25 + 25.10\right] = 10 \times 28.35 = \mathbf{283.5\ m^2}$$

(ii) Simpson's rule

$$A = \frac{d}{3}\Big[(O_1+O_7) + 4(O_2+O_4+O_6) + 2(O_3+O_5)\Big]$$

Ordinates in even positions: $O_2+O_4+O_6 = 3.80+5.20+5.40 = 14.40$. Middle (odd-position interior) ordinates: $O_3+O_5 = 4.60+6.10 = 10.70$.

$$A = \frac{10}{3}\Big[(2.50+4.00) + 4(14.40) + 2(10.70)\Big] = \frac{10}{3}\big[6.50 + 57.60 + 21.40\big]$$ $$A = \frac{10}{3}\times 85.50 = \mathbf{285.0\ m^2}$$

(b) Volume by prismoidal formula (three sections, $D = 30\,\text{m}$ between consecutive sections):

$$V = \frac{D}{3}\big(A_1 + 4A_2 + A_3\big) = \frac{30}{3}\big(18 + 4\times 26 + 34\big)$$ $$V = 10\,(18 + 104 + 34) = 10 \times 156 = \mathbf{1560\ m^3}$$

(Here the end-area method also gives $V = \frac{D}{2}(A_1+2A_2+A_3) = 15(18+52+34)=1560\,\text{m}^3$, equal because the areas vary linearly.)

(a) Well-conditioned triangle. A triangle in which no angle is smaller than about $30^\circ$ and none larger than about $120^\circ$ (ideally all angles near $60^\circ$, i.e. close to equilateral). It is preferred because the position of the apex is then least sensitive to small errors in the measured sides — the intersection of arcs is sharp and well-defined. In an ill-conditioned (very acute or very obtuse) triangle, a small linear error produces a large displacement of the plotted point.

(b) Obstacle to chaining, not to vision. A right angle is set out at $A$ so that $AC \perp AB$. Then triangle $ACB$ is right-angled at $A$, with $C$ and $B$ intervisible. Measured: $AC = 24\,\text{m}$ (perpendicular) and the hypotenuse $CB = 40\,\text{m}$.

By Pythagoras' theorem on right triangle $ACB$ (right angle at $A$):

$$CB^2 = AC^2 + AB^2$$ $$AB = \sqrt{CB^2 - AC^2} = \sqrt{40^2 - 24^2} = \sqrt{1600 - 576} = \sqrt{1024} = \mathbf{32\ m}$$

The required distance across the pond is $AB = 32\,\text{m}$.

(a) Equipment of plane table surveying

1. Plane table (drawing board on a tripod, rotatable and clampable).
2. Alidade (plain or telescopic) — for sighting and drawing rays.
3. Spirit level — to level the board.
4. Trough compass — to orient the board to magnetic north.
5. Plumbing fork (U-frame) with plumb bob — to centre the plotted point over the ground station.
6. Drawing paper / sheet (plus pencils, etc.).

(b) Methods of plane tabling

1. Radiation — rays drawn from a single occupied station to detail points; distances measured and plotted to scale. Best for small areas with all points visible from one station.
2. Intersection (graphic triangulation) — points fixed by intersecting rays from two stations; distances to the points need not be measured.
3. Traversing — successive stations occupied; used for narrow strips and closed boundaries.
4. Resection — locating the instrument station on the plan by sighting to known plotted points (e.g. two-point and three-point problem).

Method of intersection

Two stations $A$ and $B$ are chosen and the line $AB$ (the base line) is measured and plotted to scale as $ab$. With the table set and oriented at $A$, the alidade is pivoted on $a$ and a ray drawn toward each object point $P, Q, \ldots$. The table is then moved to $B$, oriented (by back-sighting on $A$), and from $b$ rays are drawn to the same points. The intersection of the two rays for each point fixes its plotted position.

        P (object)
       / \
      /   \
   ray     ray
    /         \
   a --------- b   (plotted base line = scaled AB)
  (A)         (B)

Only the base line is measured on the ground; all other points are fixed graphically. Most suitable when the object points are inaccessible or distant (e.g. mountain peaks, towers, the far bank of a river), where direct distance measurement is impractical.

(a) Definitions

• Contour: an imaginary line on the ground (and its representation on a map) joining all points of equal elevation (reduced level) above a chosen datum.
• Contour interval: the constant vertical distance between two consecutive contour lines (here $5\,\text{m}$). It depends on scale, terrain, and the purpose of the map.

(b) Characteristics of contour lines (any four):

1. All points on a single contour have the same elevation.
2. Contours close on themselves, either within or outside the map sheet; they never simply end in the middle of a map.
3. Closely spaced contours indicate steep ground; widely spaced contours indicate gentle slopes; equally spaced contours indicate a uniform slope.
4. Contours of different elevation never cross or merge, except at a vertical cliff (they touch) or an overhanging cliff (they appear to cross), which are exceptional.
5. A contour crossing a ridge or valley forms a V or U; the V points uphill in a valley and downhill along a ridge.

(c) Gradient between the two contours

Map distance between contours $= 8\,\text{mm} = 0.008\,\text{m}$. Scale $1:5000$ ⇒ ground horizontal distance:

$$H = 0.008\,\text{m} \times 5000 = 40\,\text{m}$$

Vertical distance (contour interval) $V = 5\,\text{m}$.

$$\text{Gradient} = \frac{V}{H} = \frac{5}{40} = \frac{1}{8} = \mathbf{1\ in\ 8}$$

Equivalently, slope angle $= \tan^{-1}(5/40) = \tan^{-1}(0.125) \approx \mathbf{7.13^\circ}$ (about $12.5\%$).

(a) Curvature and refraction

• Curvature: Because the earth is curved, a horizontal line of sight departs upward from the level (curved) surface. A distant staff reading is therefore too large (the level surface falls away below the line of sight). The curvature correction is subtractive:

$$C_c = -\frac{d^2}{2R} = -0.0785\,d^2 \;(\text{m, with }d\text{ in km})$$

• Refraction: The line of sight bends downward (concave to the earth) as it passes through air of decreasing density with height, making the reading too small. The refraction correction is additive and is taken as about $\tfrac{1}{7}$ of the curvature:

$$C_r = +\frac{1}{7}\cdot\frac{d^2}{2R} = +0.0112\,d^2 \;(\text{m, }d\text{ in km})$$

• Combined correction:

$$C = -\frac{6}{7}\cdot\frac{d^2}{2R} = -0.0673\,d^2 \quad(\text{m, with }d\text{ in km})$$

(b) Numerical

Here $d = 1500\,\text{m} = 1.5\,\text{km}$.

$$C = -0.0673 \times (1.5)^2 = -0.0673 \times 2.25 = -0.1514\,\text{m}$$

Combined correction $\approx \mathbf{-0.151\ m}$ (about $151\,\text{mm}$, to be subtracted).

Because the curvature effect exceeds the refraction effect, the apparent (uncorrected) staff reading is too large; the correction is therefore negative.

Correct (true) length $=$ measured length $\times \dfrac{L'}{L}$, where $L'$ = actual chain length and $L$ = nominal length.

Nominal chain length $L = 30\,\text{m}$; actual (faulty) chain length $L' = 30 + 0.06 = 30.06\,\text{m}$.

(a) True length of the line

$$\text{True length} = 1800 \times \frac{30.06}{30} = 1800 \times 1.002 = \mathbf{1803.6\ m}$$

(Because the chain is too long, each chained span covers more ground than its nominal value, so the measured length under-states the true length and must be increased — consistent with the result.)

(b) Correct area

Area scales as the square of the length ratio:

$$\text{Correct area} = \text{measured area} \times \left(\frac{L'}{L}\right)^2 = 2.4 \times (1.002)^2$$ $$(1.002)^2 = 1.004004$$ $$\text{Correct area} = 2.4 \times 1.004004 = 2.40961\,\text{ha} \approx \mathbf{2.4096\ ha}$$

The corrected area is about $\mathbf{2.410\ hectares}$ (an increase of roughly $0.0096\,\text{ha} = 96\,\text{m}^2$).

(a) WCB → RB (quadrantal) conversion

Rules: 0–90° → N θ E; 90–180° → S $(180-θ)$ E; 180–270° → S $(θ-180)$ W; 270–360° → N $(360-θ)$ W.

(i) $130^\circ30'$ is in the 2nd quadrant: $RB = 180^\circ - 130^\circ30' = 49^\circ30'$ → S $49^\circ30'$ E.

(ii) $245^\circ15'$ is in the 3rd quadrant: $RB = 245^\circ15' - 180^\circ = 65^\circ15'$ → S $65^\circ15'$ W.

(iii) $312^\circ45'$ is in the 4th quadrant: $RB = 360^\circ - 312^\circ45' = 47^\circ15'$ → N $47^\circ15'$ W.

(b) Declination correction

For an East declination, True bearing $=$ Magnetic bearing $+$ declination:

$$\text{True bearing} = 58^\circ30' + 5^\circ00' = \mathbf{63^\circ30'}$$

Now with declination $3^\circ00'\,\text{W}$ (West ⇒ True $=$ Magnetic $-$ declination, i.e. Magnetic $=$ True $+$ declination$_{W}$):

$$\text{Magnetic bearing now} = \text{True} + 3^\circ00' = 63^\circ30' + 3^\circ00' = \mathbf{66^\circ30'}$$

So to retrace the same true line ($63^\circ30'$) today, a magnetic bearing of $66^\circ30'$ must be set out.

(c) Types of levelling