Setting Out a Workshop Shed on Sloping Ground
Your crew must set out a rectangular workshop shed, 15.0 m × 9.0 m, on a plot that falls about 1.2 m from one corner to the opposite corner. The equipment is a 30 m steel tape (standardised at 27 °C), a dumpy level with a 4 m staff, and wooden pegs. A bench mark of known reduced level 250.000 m is available 20 m away. After the first attempt the tape measurements read AB = 15.00 m, BC = 9.00 m, CD = 15.00 m and DA = 9.00 m, but the diagonals read AC = 17.550 m and BD = 17.470 m. The air temperature at the time of measurement is 34 °C and the tape's coefficient of thermal expansion is 11.2 × 10⁻⁶ per °C. The mason is asking for a finished floor level of 250.450 m and wants the pegs marked today.
1. Check whether the corners are square. State the theoretical diagonal, interpret the readings, and describe how you would correct the setting out.
Read the worked answer
Theory. For a true rectangle of 15.0 m × 9.0 m, each diagonal should be √(15² + 9²) = √306 = 17.493 m, and both diagonals must be equal.
Interpretation. The four sides look correct (15.00, 9.00, 15.00, 9.00), but that is exactly what a rhombus or a parallelogram also gives. The diagonals give it away: 17.550 and 17.470 differ by 80 mm. If the corners were square, that difference would be zero. The shape is a parallelogram (pushed over), not a rectangle. Note also that the theoretical value 17.493 lies between the two readings, so neither diagonal is right either; the whole frame is rotated and skewed.
Correction method (do not just nudge one corner):
- Fix line AB as the baseline and re-check it; keep two pegs as the reference.
- Establish the right angle at A by the 3-4-5 method (or a theodolite), and check it by measuring 12 m and 16 m from A along the two directions, which must give 20 m between the marks.
- Set corner D from A (9.00 m) and corner C from B (9.00 m) by intersecting two tape arcs, 15.00 m and 17.493 m, and repeat from the other side.
- Accept only when both diagonals agree within about ±5 mm for a shed of this size. 80 mm is roughly 1 in 190 and would show as a visibly out-of-square building, misaligned roof sheets and a bad junction at the door frames.
- Transfer the pegs to profile boards clear of the excavation, with a nail marking each corner, and string the lines; the pegs themselves will be dug out.
2. Compute the reduced level at a peg from a backsight of 1.525 m on the bench mark and a foresight of 1.735 m on the peg, then state the staff reading needed to mark the finished floor level of 250.450 m.
Read the worked answer
Step 1 - height of instrument (collimation). HI = RL of BM + backsight = 250.000 + 1.525 = 251.525 m.
Step 2 - RL of the peg. RL = HI − foresight = 251.525 − 1.735 = 249.790 m. So the peg sits 0.210 m below the bench mark, which is consistent with the plot falling away from the BM.
Step 3 - reading for the finished floor level. Required staff reading = HI − required RL = 251.525 − 250.450 = 1.075 m. That means the floor is 1.075 m above the instrument line of sight, so the staff will be read inverted or you must move the instrument to a lower position, because a normal staff cannot show a reading above the horizontal line. In practice, shift the level to a spot where the FFL reading falls between about 0.3 m and 3.5 m on the staff: for example, if the new HI is 252.800 m, the reading becomes 252.800 − 250.450 = 2.350 m. Mark the FFL on a stout peg or on the column starter bars with paint, and re-check it from a second instrument position before the mason uses it.
Sanity checks: always close the level run on the bench mark at the end of the day; the closing error should be within about ±5√n mm for n setups, otherwise re-observe. Remember that the FFL of 250.450 is 0.660 m above the peg at 249.790, so excavation and filling depths on this sloping plot vary across the grid - set out levels at every corner, not just one.
3. The tape was used at 34 °C but is standard at 27 °C, and the ground slopes 1.2 m across the plot. Which corrections matter here and which can be ignored? Justify your answer with numbers.
Read the worked answer
Temperature correction. C = L × α × ΔT.
- For a full 30 m: 30 × 11.2 × 10⁻⁶ × (34 − 27) = 30 × 11.2 × 10⁻⁶ × 7 = 0.00235 m ≈ 2.4 mm.
- For a 15 m length: about 1.2 mm. For a shed set out to a tolerance of ±5 mm on the diagonals, this is at the edge of significance but not the cause of an 80 mm diagonal error. The real error is angular, from the corners not being set at 90°. Conclusion: note the temperature, apply the correction on long baselines or precision work, but do not blame the tape for a blunder that is 30 times larger.
Slope correction. The plot falls 1.2 m over the diagonal of about 17.5 m. The slope is small: the correction to a horizontal distance is d − s ≈ s³/(...) and for a 1.2 m rise over 17.5 m the difference between slope distance and horizontal distance is only about 40 mm over 17.5 m - larger than the temperature effect and worth correcting on the long diagonal. Practically, keep the tape horizontal and plumb the ends down with a plumb bob when measuring up or down a slope, rather than measuring along the ground and correcting afterwards; this removes the slope error entirely.
What actually matters most:
- The tape must be pulled with consistent tension and read at the same end each time.
- The tape must be straight and free of kinks; a sagging tape reads long.
- Both diagonals must be measured with the same method and the same tape.
- Pegs must be stable, on profile boards, and re-checked after rain or after a machine passes.
So: correct the slope, note the temperature, and spend your effort on the right angle - that is where the 80 mm came from.