Distance Calculation Techniques

The skill of estimating relies on the estimator’s ability to accurately calculate lengths, heights, areas, volumes, and weights. In actual practice, however, accuracy may not be enough, especially in fast-phase construction projects. Speed is essential, so the need to know quick calculation techniques is inevitable. Here are some techniques for calculating distances.

Direct Measurement
The most direct way to measure distances like length and height is by the use of measuring instruments like measuring tape, laser, pedometer etc...  However, in situations when it is impossible to actually measure the distance, there are alternative methods that can be used.

Triangle Method
Using the Pythagorean Theorem, many hard to measure distances can be calculated. Here are some possible situations where the triangle method can be applied. 

Unknown height

H= √(D^2-L^2 )

Unknown length

L= √(D^2-H^2 )

Unknown diagonal distance

D= √(L^2+H^2 )

Pacing
Pacing is an approximate method of calculating surface distances. By knowing your pace factor (meter per pace), you can calculate the approximate distance by multiplying it by the number of steps. Do not use pacing in estimating items that need precision especially those involving expensive materials.

Getting your pace factor
Find a flat road where you can measure at least 20 meters straight. Mark the beginning and the end of 20 meters. Walk straight through the 20-meter line while counting your steps. Remember to walk on a normal pace the way you walk on a daily basis. Record the number of steps that you take to cover the distance. Repeat this step three times. Divide 20m by the number of steps and get the average result of the three trials. The number you will get is your pace factor. If it takes you 26 steps in average to walk the 20 m distance, then your pace factor is 20/26 = 0.769meter per pace.

Using your pace factor
Set eyes on the distance that you want to measure by pacing. Put your ankle at the starting point and start walking at a normal pace and count your steps as you go until the end of the distance you are measuring. Multiply the number of steps you counted by your pace factor. The resulting number is the approximate distance measurement. 100 steps, for example, is 100 x 0.769 = 76.9 meters.

Using GPS
A modern way to do approximate horizontal distance measurement is by the use of GPS or global positioning system. There are a number of apps available that allow users to measure distances using GPS. Examples of such apps are the apps used by runners to track their performance.

Using leveling hose
Leveling hose with water inside is an old yet efficient tool to measure vertical distances or level differences. Just find a flat vertical surface that you can mark on. Match the water level with the level of the higher line or surface that you are measuring and mark on the flat surface the water level at the other end. The height difference can be measured using a measuring tape.

Excavation and Backfill Calculation

Calculating excavation and backfill is not as easy as getting the volume of soil. Angle of repose and compaction need to be considered.

A. Angle of repose

Angle of repose is the angle loose materials form from the horizontal when it is placed freely on a flat surface. So when excavations are made, it is not only the volume of the soil right under the specified area that needs to be calculated but also the surrounding area around the perimeter that forms ramps because of angle of repose of the soil that became loose after excavation.

Angle of repose depends on the materials’ coefficient of internal friction. Here are the approximate average values of angle of repose for specific types of materials.

Table 1 –  Average Angle of Repose of Different Materials

Type of Material	Angle of repose (degrees)
Dry Sand	27.5
Moist Sand	37.5
Wet Sand	30
Ordinary Dry Earth	32.5
Ordinary Moist Earth	35
Ordinary Wet Earth	27.5
Gravel	39
Gravel, Sand and Clay	25

You can actually measure the angle of repose of a material. Get a sample of the loose material. Pour it on a flat surface and measure the angle it makes from the horizontal.

By applying the law of tangents, we can calculate the horizontal distance X that will be added to the volume calculation.

tanθ=  DEPTH/x

x=  DEPTH/(tanθ)

Example:

Calculate the total volume of soil to be excavated for 10m x 20m pit with the depth of 2m. Assume the soil to be ordinary moist.

Given:

            Depth = 2m

            θ = 35 degrees for ordinary moist earth

            Area = 10m x 20m

Solution:

            x = DEPTH / tan θ

               = 2m / tan35

            x = 2.86 m

Base dimension: 10m by 20m

Top dimension: (10m+2x) by (20m+2x)

Base area = A1 = 10 x 20 = 200 m²

Top area = A2 = (10+2*2.86)(20+2*2.86) = 404.3 m²

V = 2/3(200+404.3+sqrt(200+404.3)

V =  592.44 m³

B. Compaction

Backfills are usually compacted up to 25% of the fill volume. On the other hand, excavated materials are expected to expand up to 25% in volume when loaded for transport. By considering compaction and loosening up, we could get the idea.

Volumes of fills need to be multiplied by 1.25 to allow compaction.

V _fill = V _actual x 1.25

Example:

Find the required volume of fill for 10m x 10m x 3m pit. Assume the pit to have sheet piles on the edges to prevent soil from slipping down.

Given:

            V _actual = 10 x 10 x 3 = 300 m³

Solution

            V _fill = V _actual x 1.25 = 300 x 1.25 = 375m³

Bigger is NOT Always Better - Pipe Clogging and Duct Choking

Clogging and choking have always been a consistent problem with piping and duct systems whether it be hydraulic or pneumatic. The common layman's solution to this is to use a bigger pipe or duct. This intuitive solution, however, is not always effective. In contrary, using larger diameter might actually be the source of pipe clogging and duct choking problems.

The Math

There are two major factors that influence the pipe/duct size selection process - flowrate or discharge and pressure. Flowrate and pressure are included in two independently different formulas which may be hard for some to correlate.

Q = AV ... (Equation 1)

Where:
Q = Flow Rate
A = Cross-Sectional Area
V = Velocity

P = F / A  ... (Equation 2)

Where:
P = Pressure
F = Force
A = Area

By analyzing these two equations, the science behind the relationship between clogging and pipe size can be understood. As you can see, Cross-Sectional Area is common to both Equations 1 & 2. By substitution, we can say that:

A = Q/ V
A = F / P

But,

A = A

Then,
Q / V = F / P

or

P = V F/Q  ... (Equation 3)

In practice, how is clogging dealt with? Pressure is applied to force the obstruction out. To prevent clogging, sufficient pressure should be specified at the design stage. The conclusions about Pressure that can be drawn out of the equations above are as follows:

1. Pressure is inversely proportional to area (from Equation 2). This means that as the cross-sectional area increases, the pressure decreases.

2. Pressure is directly proportional to design velocity (from Equation 3). This means that as the velocity increases, the pressure increases.

3. Velocity is inversely proportional to cross-sectional area (from Equation 1). This means that as the cross-sectional area increases, velocity increases.

One can see right away that by considering pressure’s major role in declogging, using bigger pipes or duct is not the solution because using larger diameter increases the cross-sectional area and increasing cross-sectional area decreases pressure and velocity. At the design stage, the designer must specify a sufficient design velocity and pressure that will not allow clogging. It can be done by identifying the mass, density and mass flow of the possible clogging materials and calculating the necessary force and pressure to keep such materials moving across the system. The viscosity of the fluid and the roughness coefficient of the pipe or duct must always be considered to arrive at an efficient design.

Self-cleaning action in waste pipes

Waste pipes that use gravity to allow water to move are the most prone to clogging because these merely rely on the slope of the pipe. The ideal slope to allow self-cleaning action is 2% (2 units vertical and 100 units horizontal). If the slope is too steep, the water would flow faster than the solid waste so it will be left out and accumulate at the bottom of the pipe and would eventually clog the pipe. If the slope is too small, the velocity will not be enough to carry the solid waste which will result in sedimentation, accumulation, and clogging. For gravity drainage systems, the design slope which dictates the design velocity is critical. 

Simply supported or fixed end?

Simply supported or fixed end? This might seem a small decision to make but the engineer's preference on this matter has a major impact in design, economy and appearance of the structure. When to design a beam as simply supported and when to design as fixed end? May this discussion be helpful to my fellow structural designers.

A. Properties of a simply supported beam

1. There is no bending moment at the supports.
2. The maximum bending moment is somewhere at the middle of the beam depending on type of loading. 

B. Properties of a beam with both ends fixed

1. There is bending moment at the supports and at the middle.
2. The maximum bending moment is at the supports. 

C. Comparison between simply supported and fixed end 

1. Both has the same shear at support.
2. Assuming they are to carry the same load, the negative moment at the fixed end is smaller than the positive moment of a simply supported beam.
3. The point of inflection (zero moment) of a simply supported beam is at the end while the point of inflection of a fixed end beam is a few distance away from the support.

SHEAR AND MOMENT DIAGRAMS OF SIMPLY SUPPORTED AND FIXED END BEAMS

Based on these properties, we can point out the advantages and disadvantages.

A. Simply supported 

1. Since there is no bending moment at the support, the joint or support is not complicated. We can simply use a shear plate or a clip angle that would resist the shear. 
2. For rigid supports, beams can be simply laid over with no fixing. This especially applies to temporary structures during construction. Elevated concrete highways are designed as simply supported beams.
3. The supports of the beam must be rigid. A frame with two columns and a simply supported beam on top must have columns that are moment resisting at the base. 
4. Simply supported beams are of little help in resisting lateral forces like wind and earthquake.
5. Simply supported beams are usually thicker due to a big value of maximum bending moment at the center.

SIMPLE SHEAR CONNECTION
Simple shear connected beams are designed as simple beams

B. Fixed End

1. Fixed end connections are more rigid. If you are avoiding vibrations and movements at the beam, it is best to design the beam to be fixed at the end. 
2. Designing a beam in a steel frame as fixed end eliminates the need for moment resisting base. That explains the shape of PEB frame columns that are thicker at the top and thinner at base. That also explains why sometimes, you see only two bolts at the base of the column and see twelve bolts at the joint of the beam and the column.
3. Fixed end beam connections are moment resisting and therefore more costly than simply supported beam connections.
4. Fixed end beams are thinner than simply supported beams because, aside from beam center, the ends are also resisting moment.

Minimum Thickness of Concrete Beams and Slabs

One of the common mistakes being committed by some builders, who rely on experience and not on structural calculations, is not observing the minimum thickness or depth of concrete beams and slabs specified in the code.

The code specified minimum thicknesses for different types of support in order to prevent excessive deflection or sagging on site.

NSCP 2010 states the following...

...where L is the length of beam or one way slab and the denominator is the dividing factor. Length in beams is obvious. It is simply the center to center distance of the columns that support the beam. In one way slabs, however, the design length is not the longer side of the slab. It is wiser to design one way slabs considering the shorter side. For example, if you have a 2m x 4m slab, we take the 2m as the length of the slab and design it as a 1m-wide strip beam.

Example: Calculate the minimum thickness of 6m long beam with both ends continuous.

Given:

     Type of Structure:    Beam
     Type of Support:      Both ends continuous
     Length of beam:      6m

Solution:

     Based on the table, the minimum thickness for a beam with both ends continuous is L/21.

Therefore:

     T = L/21
        = 6m / 21
        = 0.286 m or 286 mm

Here is the program that automatically calculates minimum thickness of concrete beams and slabs:
Click here and download the file -> Concrete Beam and Slab Thickness Calculator

Calculating Maximum Moments of Cantilever Structures

Cantilever is the most daring of all the structures being built. The beauty of a legless front entrance canopy or shed is really stunning. However, wrong calculation of the maximum design moment of the structure may lead to structural failure when ultimate loading condition is reached.

Cantilever shed failed during 4.6 magnitude earthquake. Age might be one of the factors.

For basic cantilevers, earthquake is the main lateral (horizontal) load to be considered in addition to the vertical dead and live loads. Wind might prevail if the structure is too high and the columns are covered with wind resisting panels.

Considering dead load, live load and earthquake loads, here are the load combinations that we need to consider based on NSCP 2010.

1. U = 1.2D +1.6L
2. U = 1.2D + E + 0.5L

I eliminated other combinations which will obviously yield lower result.

From the load diagram below, we can calculate the maximum base moment based on the given equations.

By taking summation of moments at point A...

Mu = X^2/2(1.2D + 1.6L)                -> Equation 1
Mu = X^2/2(1.2D + 0.5L) + EH      -> Equation 2

We just need to substitute actual values here to see which equation would yield a greater result. Whichever resulting value is greater, that would be the design moment that should be used.

Example:

Given:
            D = 15kN/m
            L = 6 kN/m
            E = 3kN
            H = 4m
            x = 2m


Solution:
Equation 1     Mu = X^2/2(1.2D + 1.6L)
                             = 2^2/2(1.2*15 + 1.6*6)
                             = 64.2 kNm

Equation 2     Mu = X^2/2(1.2D + 0.5L) + EH
                             = 2^2/2(1.2*15 + 0.5*6) + 3*4
                             = 63 kN

Since Equation 1 yielded greater result, therefore Mu = 64.2 kNm

Assuming we will use a 300 mm x 300 mm concrete column, by applying reinforced concrete design formulas, use:

300 x 300 concrete column with 12-Ø20 RSB.

1.4DL + 1.7LL or 1.2DL + 1.6LL ?

For designers like me who were trapped in the past (used to using the old codes), the introduction of new and smaller load factors might be quite hard to accept. In school, we were taught to use 1.4DL + 1.7LL in calculating the ultimate load involving basic dead and live loads. However in the latest NSCP which was released a few years ago, the new load factors are smaller - 1.2DL + 1.6LL.

Below is the table showing the comparison of the new and old codes:

Where:

D = dead load

E = earthquake load 

F = load due to fluids

H = load due to lateral pressure of soil and water in soil

L = live load, except roof live load, including any permitted live load reduction

Lr = roof live load , including any permitted live load reduction

R = rain load on the undef1ected roof

T = self-straining force and effects arising from contraction or expansion resulting from temperature change, shrinkage, moisture change, creep in component materials, movement due to differential settlement, or combinations thereof .

W = load due to wind  pressure

f1 = 1.0 for floors in places of public assembly, for live loads in excess of 4.8 kPa, and for garage live load.

= 0.5 for other live loads

Em = the maximum effect of horizontal and vertical forces as set forth in Section 208.6.1

Which one should we use? In practice, we could use any of these based on engineers discretion because presumably, both of them are safe to apply and there has been lots of studies, researches and experimentation done before structural scientists came up with the numbers. In schools, however, I am not sure if they are using the new code. If the professor says use the new code then use the new code otherwise you'll get the wrong answer because of using the load factors in the old code.