BE Civil Engineering (IOE, TU) Surveying I (IOE, CE 453) Question Paper 2080 Nepal

(a) Identifying affected stations

For a line free of local attraction at both ends, $\text{FB} - \text{BB} = \pm 180^\circ$ exactly. Checking each line:

Only line DA gives exactly $180^\circ$, so stations D and A are free of local attraction. Therefore the bearings observed at D and at A are correct. Stations B and C are affected.

(b) Corrected bearings

Since A is unaffected, $\text{FB}_{AB}=45^\circ30'$ is correct, so the correct $\text{BB}_{AB}=45^\circ30'+180^\circ = 225^\circ30'$. Observed $\text{BB}_{AB}=226^\circ30'$, hence the error at B $= +1^\circ00'$ (readings at B are $1^\circ00'$ too large).

Correct all bearings observed at B by $-1^\circ00'$:

• $\text{FB}_{BC}=113^\circ15' - 1^\circ00' = 112^\circ15'$ (corrected).

Now the correct $\text{BB}_{BC}=112^\circ15'+180^\circ=292^\circ15'$. Observed $\text{BB}_{BC}=291^\circ45'$, so the error at C $=291^\circ45'-292^\circ15' = -0^\circ30'$ (readings at C are $0^\circ30'$ too small).

Correct all bearings observed at C by $+0^\circ30'$:

• $\text{FB}_{CD}=210^\circ15' + 0^\circ30' = 210^\circ45'$ (corrected).

Correct $\text{BB}_{CD}=210^\circ45'+180^\circ-360^\circ = 30^\circ45'$ — matches the observed value at D (D unaffected). Consistent.

Line DA already correct. Summary of corrected bearings:

(c) Included angles (interior angle at a station $= \text{FB of forward line} - \text{BB of back line}$, adding $360^\circ$ where negative, then taking the interior value):

• At A: $\text{FB}_{AB} - \text{BB}_{DA} = 45^\circ30' - 135^\circ00' = -89^\circ30' \Rightarrow 270^\circ30'$; interior $=360^\circ-270^\circ30' = 89^\circ30'$.
• At B: $\text{FB}_{BC} - \text{BB}_{AB} = 112^\circ15' - 225^\circ30' = -113^\circ15' \Rightarrow 246^\circ45'$; interior $=360^\circ-246^\circ45'=113^\circ15'$.
• At C: $\text{FB}_{CD} - \text{BB}_{BC} = 210^\circ45' - 292^\circ15' = -81^\circ30' \Rightarrow 278^\circ30'$; interior $=360^\circ-278^\circ30'=81^\circ30'$.
• At D: $\text{FB}_{DA} - \text{BB}_{CD} = 315^\circ00' - 30^\circ45' = 284^\circ15'$; interior $=360^\circ-284^\circ15'=75^\circ15'$.

Sum of interior angles $= 89^\circ30' + 113^\circ15' + 81^\circ30' + 75^\circ15' = 359^\circ30'$.

Geometric check for a 4-sided closed polygon: $(2n-4)\times 90^\circ = (8-4)\times 90^\circ = 360^\circ$. The observed sum differs by $-0^\circ30'$ (angular misclosure $= 30'$), which is distributed equally as $+7'30''$ per angle in a refined adjustment.

Final corrected bearings as tabulated above; included angles A=89°30′, B=113°15′, C=81°30′, D=75°15′ (sum 359°30′ ≈ 360°).

(a) Latitudes and departures ($L=\ell\cos\theta$, $D=\ell\sin\theta$; N(+)/S(−) for latitude, E(+)/W(−) for departure):

Worked sample (AB): $\cos 58^\circ34' = 0.52150$, $211.20\times0.52150 = +110.142$ m; $\sin 58^\circ34' = 0.85325$, $211.20\times0.85325 = +180.206$ m.

(b) Closing error

$$\sum L = -0.088\text{ m}\quad(\text{closing error in latitude }e_L), \qquad \sum D = +0.287\text{ m}\quad(e_D).$$ $$e = \sqrt{e_L^2 + e_D^2} = \sqrt{(-0.088)^2 + (0.287)^2} = \sqrt{0.00774 + 0.08237} = \sqrt{0.09011} = 0.300\text{ m}.$$

Perimeter $\sum \ell = 751.88$ m.

$$\text{Relative precision} = \frac{e}{\sum \ell} = \frac{0.300}{751.88} = \frac{1}{2503} \approx \mathbf{1:2500}.$$

(c) Bowditch's rule distributes the error in proportion to line length:

$$C_{L,i} = -e_L\cdot\frac{\ell_i}{\sum\ell}, \qquad C_{D,i} = -e_D\cdot\frac{\ell_i}{\sum\ell}.$$

With $-e_L = +0.088$ and $-e_D = -0.287$ spread over $\sum\ell=751.88$:

Sample correction (AB): $C_L = 0.088\times\frac{211.20}{751.88} = +0.0247$ m; $C_D = -0.287\times\frac{211.20}{751.88} = -0.0806$ m.

The adjusted latitudes sum to $0.000$ and adjusted departures sum to $0.000$, confirming the traverse is geometrically closed.

Closing error = 0.300 m; relative precision = 1:2500; adjusted latitudes/departures tabulated above (ΣL = ΣD = 0).

(a) Rise-and-fall reduction

A rise occurs when the present reading is less than the previous reading; a fall when it is greater. Compare consecutive readings:

• BM (2.105) → A (1.875): $2.105-1.875=0.230$ → Rise 0.230
• A (1.875) → B (2.355): $1.875-2.355=-0.480$ → Fall 0.480
• B (2.355) → C-FS (0.985): $2.355-0.985=1.370$ → Rise 1.370
• C-BS (1.620) → D-FS (2.490): $1.620-2.490=-0.870$ → Fall 0.870

Reduced levels (RL of next $=$ RL of previous $\pm$ rise/fall):

(b) Arithmetic check

$$\sum BS - \sum FS = 3.725 - 3.475 = +0.250\text{ m}$$ $$\sum \text{Rise} - \sum \text{Fall} = 1.600 - 1.350 = +0.250\text{ m}$$ $$\text{Last RL} - \text{First RL} = 100.250 - 100.000 = +0.250\text{ m}$$

All three agree at $+0.250$ m, so the reduction is arithmetically correct.

(c) Difference of level

RL of D $= 100.250$ m, RL of BM $= 100.000$ m. The difference is $100.250-100.000 = \mathbf{0.250\text{ m}}$, and station D is higher than BM by 0.250 m.

Part (a) — Area between chain line and boundary

Number of offsets $=7$, so number of intervals $n=6$ (even), spacing $d=10$ m.

(i) Trapezoidal rule:

$$A = d\left[\frac{O_1+O_n}{2} + (O_2+O_3+\cdots+O_{n-1})\right]$$ $$A = 10\left[\frac{3.2+3.0}{2} + (4.8+5.6+6.1+5.9+4.3)\right] = 10\left[3.1 + 26.7\right] = 10\times29.8 = \mathbf{298.0\ m^2}.$$

(ii) Simpson's one-third rule (applicable since $n=6$ is even):

$$A = \frac{d}{3}\left[(O_1+O_7) + 4(O_2+O_4+O_6) + 2(O_3+O_5)\right]$$ $$= \frac{10}{3}\left[(3.2+3.0) + 4(4.8+6.1+4.3) + 2(5.6+5.9)\right]$$ $$= \frac{10}{3}\left[6.2 + 4(15.2) + 2(11.5)\right] = \frac{10}{3}\left[6.2 + 60.8 + 23.0\right] = \frac{10}{3}\times 90.0 = \mathbf{300.0\ m^2}.$$

Comment: Simpson's rule gives $300.0\ \text{m}^2$ versus $298.0\ \text{m}^2$ by the trapezoidal rule — a difference of $2.0\ \text{m}^2$. Because the boundary is convex (curving outward), the straight chords of the trapezoidal rule fall inside the true curve and under-estimate the area, whereas Simpson's rule fits parabolic arcs and is more accurate.

Part (b) — Volume of earthwork

Five sections ($A_1\ldots A_5$), spacing $d=30$ m, giving 4 intervals.

(i) End-area (trapezoidal) formula:

$$V = d\left[\frac{A_1+A_5}{2} + (A_2+A_3+A_4)\right] = 30\left[\frac{120+150}{2} + (168+196+180)\right]$$ $$= 30\left[135 + 544\right] = 30\times 679 = \mathbf{20\,370\ m^3}.$$

(ii) Prismoidal formula (number of sections odd $=5$, so intervals even $=4$, applicable):

$$V = \frac{d}{3}\left[A_1 + A_5 + 4(A_2+A_4) + 2(A_3)\right] = \frac{30}{3}\left[120 + 150 + 4(168+180) + 2(196)\right]$$ $$= 10\left[270 + 4(348) + 392\right] = 10\left[270 + 1392 + 392\right] = 10\times 2054 = \mathbf{20\,540\ m^3}.$$

Prismoidal correction $= V_{\text{end-area}} - V_{\text{prismoidal}} = 20\,370 - 20\,540 = -170\ \text{m}^3$. The end-area formula here under-estimates the volume by $170\ \text{m}^3$; the prismoidal value $\mathbf{20\,540\ m^3}$ is the more accurate estimate.

Part (a) — Most probable value and errors

Work with the seconds part (the $42^\circ21'$ are common). Observed seconds: $28, 31, 25, 30, 27$.

Most probable value (arithmetic mean) of the seconds:

$$\bar{s} = \frac{28+31+25+30+27}{5} = \frac{141}{5} = 28.2''.$$

Most probable angle $= \mathbf{42^\circ\,21'\,28.2''}.$

Residuals $v_i = s_i - \bar{s}$ and their squares:

Standard deviation of a single observation ($n=5$):

$$\sigma = \sqrt{\frac{\sum v^2}{n-1}} = \sqrt{\frac{22.80}{4}} = \sqrt{5.70} = 2.387'' \approx \mathbf{2.39''}.$$

Standard error of the mean:

$$\sigma_{\bar{s}} = \frac{\sigma}{\sqrt{n}} = \frac{2.387}{\sqrt{5}} = \frac{2.387}{2.236} = 1.068''.$$

Probable error of the mean (factor $0.6745$):

$$E_{\bar{s}} = 0.6745\,\sigma_{\bar{s}} = 0.6745\times 1.068 = 0.720'' \approx \mathbf{\pm 0.72''}.$$

Hence the angle may be stated as $42^\circ21'28.2'' \pm 0.72''$ (probable error of the mean).

Part (b) — Area and its standard error

$$A = L\times B = 150.00\times 80.00 = 12\,000\ \text{m}^2.$$

For $A = LB$, $\dfrac{\partial A}{\partial L}=B$ and $\dfrac{\partial A}{\partial B}=L$. By the propagation (general) law:

$$\sigma_A = \sqrt{\left(B\,\sigma_L\right)^2 + \left(L\,\sigma_B\right)^2} = \sqrt{(80.00\times 0.03)^2 + (150.00\times 0.02)^2}$$ $$= \sqrt{(2.40)^2 + (3.00)^2} = \sqrt{5.76 + 9.00} = \sqrt{14.76} = 3.842\ \text{m}^2.$$

Area $= 12\,000 \pm 3.84\ \text{m}^2$ (standard error).

Part (c) — Accuracy vs. precision

• Accuracy is the closeness of a measured (or computed) value to the true value. It reflects the absence of systematic errors and mistakes.
• Precision is the closeness of repeated measurements to one another (their degree of mutual agreement / refinement of instrument and method). It reflects small random scatter.

Measurements can be precise but inaccurate (tightly grouped but biased by a systematic error such as a wrongly standardised tape), or accurate on average but imprecise (widely scattered about the true value). Good surveying aims for both.

(a) Two fundamental principles of surveying

1. Working from the whole to the part. A survey must first establish a system of well-distributed control points of high precision (a framework), and then locate the minor details with respect to this framework using lesser precision. This prevents the accumulation and magnification of errors and localises any error within a small area. The reverse (part to whole) would let small errors expand over the whole survey.

2. Locating (fixing) a point by at least two independent measurements/two control points. Any new point should be fixed relative to two already-fixed reference points so that it is uniquely determined and can be checked. Common methods: two distances (linear), one distance + one angle (polar), or two angles (intersection). The redundant/independent measurement provides a check on accuracy.

(b) Plane vs. geodetic surveying

The classification is based on whether the curvature of the Earth is taken into account:

• In plane surveying, the Earth's surface is treated as a flat horizontal plane; curvature is neglected and all triangles are plane triangles.
• In geodetic surveying, the curvature of the Earth is considered; lines are arcs and computations use spherical/spheroidal trigonometry, with very high precision.

Plane surveying is generally regarded as valid for areas up to about $250\ \text{km}^2$ (the error from neglecting curvature over such an area is within permissible limits — e.g. the arc–chord difference over $18.5$ km is only about $1/100\,000$).

(c) Plan vs. map

(a) Correct length of the line

When a chain is too long, each chain laid down covers more ground than its nominal length, so the recorded length is too small; the correction is positive. The rule:

$$\text{Correct length} = \text{Measured length} \times \frac{\text{Actual (true) chain length}}{\text{Nominal chain length}}.$$ $$L_{\text{correct}} = 728.45 \times \frac{20.06}{20.00} = 728.45 \times 1.0030 = \mathbf{730.635\ m}.$$

(The line is actually about $2.19$ m longer than recorded.)

(b) Correct area

Area scales as the square of the length ratio:

$$A_{\text{correct}} = A_{\text{measured}} \times \left(\frac{\text{Actual chain length}}{\text{Nominal chain length}}\right)^2 = 4.50 \times \left(\frac{20.06}{20.00}\right)^2.$$ $$\left(\frac{20.06}{20.00}\right)^2 = (1.0030)^2 = 1.006009.$$ $$A_{\text{correct}} = 4.50 \times 1.006009 = \mathbf{4.527\ hectares}\ (\approx 45\,270\ \text{m}^2).$$

(c) Effect of a long chain

A chain that is too long makes the measured (recorded) length too small, because fewer chain-lengths are needed to span the line. Since the true ground covered per chain exceeds the nominal $20$ m, the count of chains under-represents the true distance. Hence a positive correction must be applied, exactly as done above.

(a) Combined curvature and refraction correction

Curvature makes a horizontal line of sight rise above the level (curved) surface, so the staff reading is too large; the curvature correction is negative:

$$C_c = -\frac{D^2}{2R} = -0.0785\,D^2 \ \text{(m, with $D$ in km, $R \approx 6370$ km)}.$$

Refraction bends the line of sight downward toward the Earth, reducing the reading; it acts opposite to curvature and is taken as about $1/7$ of curvature, correction positive:

$$C_r = +\frac{1}{7}\cdot\frac{D^2}{2R} = +0.0112\,D^2.$$

Combined correction:

$$C = C_c + C_r = -\left(0.0785 - 0.0112\right)D^2 = -0.0673\,D^2 \ \text{m}\quad(D\text{ in km}).$$

For $D = 2.5$ km:

$$C = -0.0673\times(2.5)^2 = -0.0673\times 6.25 = -0.4206\ \text{m} \approx \mathbf{-0.421\ m}.$$

That is, the observed staff reading at $2.5$ km must be reduced by about $0.421$ m.

(b) Reciprocal leveling

Reciprocal leveling is used to find the true difference in level between two points (such as the banks of a wide river or a deep valley) where the instrument cannot be set up midway between them. The level is set up close to one point and a pair of staff readings is taken; the instrument is then moved close to the other point and the pair is repeated. The true difference of level is the mean of the two computed differences.

The two principal errors eliminated by reciprocal leveling are:

1. Collimation (instrumental) error — due to the line of collimation not being truly horizontal; and
2. The combined error of Earth's curvature and atmospheric refraction.

(Because in the two set-ups the back-sight and fore-sight lengths are interchanged, these errors are equal and opposite in the two computed differences and cancel on averaging.)

(a) Accessories of plane table surveying and their functions

Essential equipment: the plane table with tripod, the alidade, a spirit level, a trough compass, a plumbing fork (U-fork) with plumb bob, drawing paper, and pins/pencils.

• Alidade: a straight-edged ruler carrying sighting vanes (plain alidade) or a small telescope (telescopic alidade). It is used to sight the object and to draw the ray (direction line) on the paper along its fiducial edge; the point lies somewhere along this ray.
• Trough compass: a narrow box compass placed on the table to mark the magnetic north (meridian) direction on the sheet and to orient the table so that successive set-ups give a consistent orientation.

(b) Radiation and intersection

Radiation: The table is set up at one station $O$ within or near the area. The alidade is centred on the plotted position $o$ of $O$, rays are drawn to each detail point, the distance to each point is measured, and the point is plotted by scaling that distance along its ray.

• Preferred when the area is small, all points are visible from a single station, and distances can be measured easily (e.g. by tape or by tacheometry). It is essentially a polar-coordinate method.

Intersection (graphical triangulation): The table is set up successively at two stations forming a measured base line $AB$ (plotted to scale as $ab$). From $a$, a ray is drawn to a detail point $P$; from $b$, another ray is drawn to the same point. The intersection of the two rays fixes $p$ on the sheet; no linear measurement to the point is needed.

• Preferred when the points are distant or inaccessible (across a river, on a hilltop) so that direct distance measurement is impractical. Used for locating detail and broken-boundary points and for small triangulation.

(a) Definitions

• A contour is an imaginary line on the ground joining points of equal elevation (reduced level) above a chosen datum; its plan representation is a contour line.
• The contour interval is the constant vertical distance between two consecutive contours.

Two factors governing the choice of contour interval:

1. the scale of the map (smaller scale → larger interval);
2. the nature/ruggedness of the terrain (steep/hilly ground → larger interval, flat ground → smaller interval). (Other valid factors: purpose of the survey and time/funds available.)

(b) Four characteristics of contour lines

1. All points on a single contour line have the same elevation.
2. Contour lines close on themselves, either within the map or beyond its borders; they never simply end in the middle of a map.
3. Closely spaced contours indicate a steep slope; widely spaced contours indicate a gentle slope; equally spaced contours indicate a uniform slope.
4. Contour lines never cross or merge except at a vertical cliff (where they coincide) or an overhanging cliff (where they appear to cross). A set of closed contours with higher values inside represents a hill; with lower values inside, a depression. Contours crossing a valley form a V pointing upstream (uphill), and crossing a ridge form a V or U pointing downhill.

(c) Horizontal equivalent

A gradient of $1$ in $20$ means a rise of $1$ m vertically for every $20$ m horizontally. For a contour interval (vertical rise) of $1$ m:

$$\text{Horizontal equivalent} = \text{Contour interval} \times \frac{\text{horizontal}}{\text{vertical}} = 1 \times 20 = \mathbf{20\ m}.$$

Since the slope is uniform, the contours are equally spaced at a horizontal distance of 20 m on the plan.

(a) Classification of errors

1. Mistakes (gross errors/blunders): caused by carelessness or inexperience of the observer — e.g. reading $6$ for $9$ on the staff, omitting a whole chain length, or transposing digits. These are not errors of measurement; they can be detected and eliminated by careful checking and redundant measurements.

2. Systematic (cumulative) errors: follow a definite physical law and have the same sign and magnitude under the same conditions, so they accumulate — e.g. using a tape that is too long, or neglecting temperature/sag/pull corrections. These can be eliminated/computed out by applying corrections or by proper procedure, since their cause is known.

3. Random (accidental/compensating) errors: remain after mistakes and systematic errors are removed; they obey the laws of probability, are equally likely to be positive or negative, and tend to partly cancel — e.g. small errors in bisecting a target or in the final fraction of a staff reading. They cannot be eliminated, only minimised and treated by the theory of least squares/probable error.

(b) Temperature correction of the tape

The temperature correction per measured length $L$ is:

$$C_t = \alpha\,(T_m - T_0)\,L,$$

where $T_m = 35^\circ\text{C}$ (field), $T_0 = 20^\circ\text{C}$ (standard), $L = 645.30$ m.

$$C_t = (1.15\times10^{-5})\times(35-20)\times 645.30 = (1.15\times10^{-5})\times 15 \times 645.30.$$ $$C_t = 1.15\times10^{-5}\times 9679.5 = 0.11131\ \text{m} \approx +0.1113\ \text{m}.$$

Because the field temperature is higher than the standard, the steel tape has expanded, so it is actually longer than its nominal $30$ m; each laid length covers more ground than recorded and the correction is positive (additive).

Corrected length:

$$L_{\text{correct}} = L + C_t = 645.30 + 0.1113 = \mathbf{645.411\ m}.$$