BE Civil Engineering (IOE, TU) Surveying I (IOE, CE 453) Question Paper 2076 Nepal

(a) Two fundamental principles of surveying

1. Working from the whole to the part. A survey is always carried out by first establishing a system of control points (a framework of large, well-conditioned triangles or a main traverse) covering the whole area with high accuracy, and then locating the interior detail relative to this framework with comparatively lower accuracy. This prevents the accumulation and magnification of errors over the area; any error is confined (localised) to the small portion in which it occurs.

2. Fixing a point with respect to two reference points (location of a point). The relative position of any new point is fixed by making at least two independent measurements (e.g. two distances, two angles, or one distance and one angle) from points whose positions are already known. In chain surveying this is realised by the tie line / check line and by taking offsets, so that every plotted point can be checked.

In a chain survey these principles dictate that we first lay out the main framework of well-conditioned (near-equilateral) triangles with a baseline, take check lines to verify the framework, and only then run offsets to pick up detail.

(b) Tape too long — true length

When a tape is too long, each laid length covers more ground than its nominal value, so the true length is greater than the recorded length.

$$L_{true} = L_{measured}\times\frac{L'}{L} = 450.00\times\frac{30+0.05}{30} = 450.00\times\frac{30.05}{30}$$ $$L_{true} = 450.00\times 1.0016\overline{6} = \mathbf{450.75\ m}$$

Total correction $= L_{true}-L_{measured} = 450.75-450.00 = \mathbf{+0.75\ m}$ (additive, because the tape is too long).

(c) Distance across the obstacle

$PC$ is set perpendicular to the line $AB$ at $P$, so triangle $PQC$ is right-angled at $P$:

$$PQ = \sqrt{QC^{2}-PC^{2}} = \sqrt{40^{2}-24^{2}} = \sqrt{1600-576} = \sqrt{1024}$$ $$PQ = \mathbf{32.0\ m}$$

This is the standard 3-4-5 right triangle scaled by 8 ($24=8\times3,\ 32=8\times4,\ 40=8\times5$).

(a) Definitions

• True bearing: the horizontal angle a line makes with the true (geographic) meridian, measured clockwise. It is fixed and does not change with time/place.
• Magnetic bearing: the horizontal angle a line makes with the magnetic meridian (the direction shown by a freely suspended magnetic needle), measured clockwise. It varies because the magnetic meridian itself moves.
• Declination: the horizontal angle between the true meridian and the magnetic meridian at a place and time. Cause: the magnetic poles do not coincide with the geographic poles (and they drift), so the needle does not point to true north.
• Local attraction: the deflection of the compass needle from the magnetic meridian due to nearby magnetic material. Cause: iron/steel objects (rails, pipes, electric current, magnetic ore) close to the station.

(b) Detecting local attraction

If a line is free from local attraction at both its ends, then $BB-FB = 180^{\circ}$.

• $AB$: $304^{\circ}30'-124^{\circ}30' = 180^{\circ}00'$ ✔ (both $A$ and $B$ are free)
• $BC$: $246^{\circ}00'-68^{\circ}15' = 177^{\circ}45'$ ✘ (difference $\ne 180^{\circ}$)
• $CD$: FB$-$BB $= 310^{\circ}30'-135^{\circ}15' = 175^{\circ}15'$ ✘
• $DA$: $200^{\circ}15'-20^{\circ}15' = 180^{\circ}00'$ ✔ (both $D$ and $A$ are free)

Since $AB$ and $DA$ are perfect, stations $A$, $B$, $D$ are free from local attraction; the discrepancy must lie at station $C$ (the only common station of the two defective lines $BC$ and $CD$). Station $C$ is affected by local attraction.

Correcting the bearings. Stations $A,B,D$ being correct, all bearings observed from $A,B,D$ are correct:

• $AB$: FB $124^{\circ}30'$, BB $304^{\circ}30'$ (correct)
• $BC$: FB $68^{\circ}15'$ is observed from $B$ (correct). Correct BB of $BC = 68^{\circ}15'+180^{\circ} = \mathbf{248^{\circ}15'}$ (observed $246^{\circ}00'$, so $C$ reads $2^{\circ}15'$ low).
• $DA$: FB $200^{\circ}15'$, BB $20^{\circ}15'$ (correct).
• $CD$: BB $135^{\circ}15'$ is observed from $D$ (correct). Correct FB of $CD = 135^{\circ}15'+180^{\circ} = \mathbf{315^{\circ}15'}$ (observed $310^{\circ}30'$).

The error at $C$ from line $BC$ is $+2^{\circ}15'$ (needle reads low by $2^{\circ}15'$). Check with $CD$: corrected FB $315^{\circ}15'$ vs observed $310^{\circ}30'$ gives $+4^{\circ}45'$. The differing magnitudes simply confirm that $C$ alone is disturbed; the corrected bearings derived from the unaffected stations are the trustworthy values.

Corrected bearings table

Corrected included angle at $C$ (interior angle $\angle BCD$): use the corrected bearings of the two lines meeting at $C$, i.e. $CB$ (back bearing of $BC$) and $CD$ (fore bearing of $CD$).

$$\angle C = \text{(bearing of } CD) - \text{(bearing of } CB) = 315^{\circ}15' - 248^{\circ}15' = \mathbf{67^{\circ}00'}$$

Rise-and-fall reduction. For consecutive readings, Rise = previous reading − present reading (if positive) and Fall = present − previous (if positive). At the change point $D$ the FS ($1.875$) belongs to the first set-up; the BS ($2.340$) starts the second set-up, so no rise/fall is computed between the FS and BS at $D$.

Step computations:

• $A\to B$: $1.245-2.130 = -0.885 \Rightarrow$ Fall $0.885$
• $B\to C$: $2.130-0.985 = +1.145 \Rightarrow$ Rise $1.145$
• $C\to D$: $0.985-1.875 = -0.890 \Rightarrow$ Fall $0.890$
• $D\to E$: $2.340-1.105 = +1.235 \Rightarrow$ Rise $1.235$
• $E\to F$: $1.105-0.760 = +0.345 \Rightarrow$ Rise $0.345$

Reduced-level table

Arithmetic checks

1. $\sum BS - \sum FS = (1.245+2.340)-(1.875+0.760) = 3.585-2.635 = +0.950$
2. $\sum Rise - \sum Fall = (1.145+1.235+0.345)-(0.885+0.890) = 2.725-1.775 = +0.950$
3. $\text{Last RL} - \text{First RL} = 100.950-100.000 = +0.950$

All three agree at $+0.950\,\text{m}$, so the reduction is arithmetically correct.

RLs: $A=100.000$, $B=99.115$, $C=100.260$, $D=99.370$, $E=100.605$, $\mathbf{F=100.950\ m}$.

(a) Latitudes and departures. Latitude $L = l\cos\theta$ (N $+$, S $-$); Departure $D = l\sin\theta$ (E $+$, W $-$), $\theta=$ WCB.

Sample working for $BC$: $\cos 120^{\circ}15' = -0.50380\Rightarrow 200.8\times(-0.50380)=-101.158$; $\sin120^{\circ}15'=+0.86382\Rightarrow 200.8\times0.86382=+173.458$.

Closing error:

$$e=\sqrt{(\Sigma L)^2+(\Sigma D)^2}=\sqrt{0.096^{2}+0.674^{2}}=\sqrt{0.00917+0.45427}=\sqrt{0.46344}=\mathbf{0.681\ m}$$

Direction of closing error (measured from N): the error vector is $(\Sigma L,\Sigma D)=(+0.096,+0.674)$, so the correction points opposite (SW). Bearing of the error of closure:

$$\tan\theta_e=\frac{\Sigma D}{\Sigma L}=\frac{0.674}{0.096}=7.02\Rightarrow\theta_e=81^{\circ}53' \text{ (i.e. roughly due East, N81}^{\circ}53'\text{E).}$$

Relative precision $=\dfrac{e}{\text{perimeter}}=\dfrac{0.681}{700.1}=\dfrac{1}{1028}\approx\mathbf{1\ \text{in}\ 1029}$ — acceptable for ordinary traversing.

(b) Bowditch's rule. Correction to latitude (or departure) of a line $\propto$ its length:

$$C_{L}=-\Sigma L\cdot\frac{l}{\Sigma l},\qquad C_{D}=-\Sigma D\cdot\frac{l}{\Sigma l}$$

with $\Sigma L=+0.096$, $\Sigma D=+0.674$, $\Sigma l=700.1$ (so corrections are negative).

The adjusted latitudes and departures each sum to zero (to mm), so the traverse is now geometrically closed.

There are 9 offsets $\Rightarrow 8$ equal intervals of $d=10\,\text{m}$. Let the offsets be $y_0,\dots,y_8$.

(a) Trapezoidal rule

$$A=d\left[\frac{y_0+y_8}{2}+\big(y_1+y_2+\dots+y_7\big)\right]$$

Interior sum $=5.2+7.8+9.1+8.4+6.3+4.2+2.1=43.1$

$$A=10\left[\frac{0+0}{2}+43.1\right]=10\times43.1=\mathbf{431.0\ m^{2}}$$

(b) Simpson's one-third rule (applicable: number of intervals $=8$ is even):

$$A=\frac{d}{3}\Big[(y_0+y_8)+4\,(y_1+y_3+y_5+y_7)+2\,(y_2+y_4+y_6)\Big]$$

• Odd-ordinate sum $=y_1+y_3+y_5+y_7=5.2+9.1+6.3+2.1=22.7$
• Even-ordinate sum $=y_2+y_4+y_6=7.8+8.4+4.2=20.4$

$$A=\frac{10}{3}\big[(0)+4(22.7)+2(20.4)\big]=\frac{10}{3}\big[90.8+40.8\big]=\frac{10}{3}\times131.6$$ $$A=\mathbf{438.67\ m^{2}}$$

(c) Comment. Simpson's one-third rule is the more reliable. The trapezoidal rule joins the ends of successive offsets by straight chords, so for a convex curved boundary like this it systematically under-estimates the area. Simpson's rule fits a second-degree parabola through every set of three consecutive offsets, following the curvature far more closely; hence $438.67\,\text{m}^2$ is preferred (the trapezoidal value is short by about $7.7\,\text{m}^2$).

(d) Area on the drawing sheet at $1:500$. Areas scale as the square of the linear scale, so the map area $=\dfrac{\text{ground area}}{500^{2}}$.

$$A_{map}=\frac{438.67\,\text{m}^2}{250000}=0.0017547\,\text{m}^2=0.0017547\times10000\,\text{cm}^2=\mathbf{17.55\ cm^{2}}$$

Four methods of plane table surveying

1. Radiation — details are located by drawing rays from a single set-up and plotting measured distances along them.
2. Intersection (graphic triangulation) — points are fixed by intersecting rays drawn from two (or more) plotted stations; no distance to the object is measured.
3. Traversing — successive stations are occupied and a traverse is plotted directly on the sheet.
4. Resection — the plotted position of the occupied station is found from rays to already-plotted points (e.g. two-point and three-point problems).

Intersection method. Two stations $A$ and $B$ (a base line) are chosen so that the object $P$ is clearly visible from both. The base $AB$ is measured once and plotted to scale as $ab$. The table is set and oriented at $A$; with the alidade pivoted at $a$, a ray is drawn towards $P$. The table is then shifted to $B$, re-oriented (by back-sighting along $ba$), and a second ray drawn towards $P$ from $b$. The intersection of the two rays fixes $p$ on the sheet.

Preferred when the points to be located are inaccessible or distant (e.g. across a river, towers, hill peaks) so that direct distance measurement is impossible or inconvenient; only the single base line need be measured.

Two essential accessories: (i) Alidade (sight rule), (ii) Spirit level (and/or trough compass, plumbing fork). The function of the alidade is to provide a line of sight to the object and, with its straight (fiducial/bevelled) edge, to draw the corresponding ray (direction) on the sheet; it thus transfers the observed direction directly onto the plan.

Reciprocal levelling eliminates collimation, curvature and refraction errors by taking the mean of the two set-ups.

Let $a=$ reading on near staff, $b=$ reading on far staff.

Instrument near $A$: difference of level $\Delta H_1 = (\text{staff }A)-(\text{staff }B)=1.585-2.230=-0.645\,\text{m}$ (i.e. $B$ higher than $A$ by $0.645$).

Instrument near $B$: $\Delta H_2 = (\text{staff }A)-(\text{staff }B)=0.985-1.560=-0.575\,\text{m}$.

(a) True difference of level:

$$\Delta H=\frac{\Delta H_1+\Delta H_2}{2}=\frac{-0.645+(-0.575)}{2}=\frac{-1.220}{2}=\mathbf{-0.610\ m}$$

So $B$ is higher than $A$ by $0.610\,\text{m}$.

(b) RL of $B$:

$$RL_B = RL_A+|\Delta H| = 150.000+0.610=\mathbf{150.610\ m}$$

(c) Combined collimation + (curvature–refraction) error. When the instrument is near $A$, the reading on the far staff $B$ carries the full error $e$; when near $B$, the far reading on $A$ carries it in the opposite sense. The error per set-up is half the difference of the two apparent values:

$$e=\frac{\Delta H_1-\Delta H_2}{2}=\frac{-0.645-(-0.575)}{2}=\frac{-0.070}{2}=\mathbf{0.035\ m}$$

The combined collimation/curvature-refraction error is $\mathbf{0.035\,\text{m}}$ (35 mm) in the far reading, automatically cancelled by the averaging.

(a) Definitions

• Contour line: an imaginary line on the ground joining points of equal reduced level (elevation); its plan projection is a contour on the map.
• Contour interval (CI): the constant vertical distance between two successive contour lines (here $2\,\text{m}$).
• Horizontal equivalent (HE): the horizontal distance between two successive contours, measured on plan/ground; unlike CI it varies with the steepness of the slope.

Two characteristics of contours: (i) Contours of different elevations never cross or merge except at a vertical/overhanging cliff. (ii) Closely-spaced contours indicate steep ground; widely-spaced contours indicate gentle ground; equally-spaced contours indicate a uniform slope.

(b) Distance for the ruling gradient

A gradient of $1\ \text{in}\ 25$ means a rise of $1\,\text{m}$ for every $25\,\text{m}$ horizontal. To rise by one contour interval ($2\,\text{m}$):

$$\text{Horizontal ground distance} = CI\times 25 = 2\times 25 = \mathbf{50\ m}$$

Distance on the map at scale $1:1000$:

$$\text{Map distance}=\frac{50\,\text{m}}{1000}=0.050\,\text{m}=\mathbf{50\ mm\ (5.0\ cm)}$$

Thus, while drawing the road, set the dividers to $50\,\text{mm}$ and step from contour to contour to lay out the uniform $1\ \text{in}\ 25$ alignment.

Five sections $\Rightarrow 4$ intervals of $d=30\,\text{m}$. Areas $A_1\dots A_5$.

(a) Trapezoidal (end-area) rule

$$V=d\left[\frac{A_1+A_5}{2}+(A_2+A_3+A_4)\right]$$

Interior sum $=18.2+25.6+21.3=65.1$

$$V=30\left[\frac{12.5+15.8}{2}+65.1\right]=30\left[14.15+65.1\right]=30\times79.25=\mathbf{2377.5\ m^{3}}$$

(b) Prismoidal (Simpson's) rule (number of intervals $=4$, even — applicable):

$$V=\frac{d}{3}\Big[A_1+A_5+4(A_2+A_4)+2A_3\Big]$$

• $4(A_2+A_4)=4(18.2+21.3)=4\times39.5=158.0$
• $2A_3=2\times25.6=51.2$

$$V=\frac{30}{3}\big[(12.5+15.8)+158.0+51.2\big]=10\big[28.3+158.0+51.2\big]=10\times237.5=\mathbf{2375.0\ m^{3}}$$

(c) Prismoidal correction. The prismoidal volume is the more accurate; the correction to be applied to the (over-estimated) trapezoidal value is

$$C_p = V_{trap}-V_{prism}=2377.5-2375.0=\mathbf{2.5\ m^{3}}\ (\text{deduct}).$$

Hence the adopted earthwork volume is $\mathbf{2375.0\ m^{3}}$.

(a) Systematic vs accidental errors

• Systematic (cumulative) error: follows a definite law, has the same sign/magnitude under the same conditions, and accumulates; it can be eliminated by calibration or correction. Example: a tape that is too long (or sag, temperature, slope corrections in chaining).
• Accidental (random) error: small, of variable sign and magnitude, beyond the surveyor's control, obeying the laws of probability; it tends to partly cancel and is treated by the theory of least squares. Example: small errors in bisecting the target or reading the last fraction on the staff/vernier.

(b) Most probable value and errors. $n=6$.

Most probable value (arithmetic mean):

$$\bar{x}=\frac{100.25+100.18+100.30+100.22+100.27+100.19}{6}=\frac{601.41}{6}=\mathbf{100.235\ m}$$

Residuals $v=x-\bar x$ (in m) and $v^2$:

Standard deviation of a single observation:

$$\sigma=\sqrt{\frac{\Sigma v^2}{n-1}}=\sqrt{\frac{0.010950}{5}}=\sqrt{0.002190}=\mathbf{0.0468\ m}$$

Probable error of a single observation: $E_s=0.6745\,\sigma=0.6745\times0.0468=0.0316\,\text{m}$.

Probable error of the mean:

$$E_m=\frac{E_s}{\sqrt{n}}=\frac{0.0316}{\sqrt{6}}=\frac{0.0316}{2.449}=\mathbf{\pm 0.0129\ m}$$

Result: length $=100.235\pm0.013\,\text{m}$.

(a) Well-conditioned triangle. A triangle in which no angle is too small or too large, so that the position of its apex is not sensitive to small errors in the measured sides; the intersection of the locating arcs is sharp and well-defined. It is preferred because plotting/fixing accuracy is highest when the sides meet at angles close to $60^{\circ}$. Practical limits: no angle smaller than about $30^{\circ}$ and none greater than about $120^{\circ}$; the ideal is the equilateral triangle ($60^{\circ}$ each). (An ill-conditioned triangle has an angle near $0^{\circ}$ or $180^{\circ}$.)

(b) Direct vs indirect ranging.

• Direct ranging: intermediate points are fixed directly in line, by eye or with a line ranger, when both ends of the line are intervisible and the ground is fairly level.
• Indirect / reciprocal ranging: used when the two ends are not intervisible (e.g. a hill or rise between them). Two assistants, one near each end, alternately direct each other into line by successive approximation until all four points lie on the straight line.

(c) Offset. An offset is the lateral (perpendicular or oblique) distance measured from the survey/chain line to locate a detail point relative to the line.

• Perpendicular (right-angled) offset: measured at $90^{\circ}$ to the chain line; it fixes the point with a single distance (the chainage of the foot plus the offset length) and is quicker to plot.
• Oblique offset: measured at any convenient angle other than $90^{\circ}$ (typically two tie distances from two known chainages); used when a right angle cannot be set or to check the position of important points.