March 2017

Structural Analysis

Equilibrium of Structures
Plane Trusses
Space Trusses
Shear and Moment in Beams
Shear and Moment in Frames
Deflection of Beams
Deflection of Trusses
Deflection of Frames
Influence Lines

Geotechnical Engineering

Symbols and Notations
$e$   = void ratio
$n$   = porosity
$w$   = moisture content, water content
$s$   = specific gravity of any substance
$G$   = specific gravity of solids
$S$   = degree of saturation
$V$   = volume of soil mass
$V_a$   = volume of air
$V_w$   = volume of water
$V_s$   = volume of solids
$V_v$   = volume of voids
$W$   = total weight of soil
$W_w$   = weight of water
$W_s$   = weight of solids
$D_r$   = relative density

Physical Properties of Soil

Phase Diagram of Soil
Soil is composed of solids, liquids, and gases. Liquids and gases are mostly water and air, respectively. These two (water and air) are called voids which occupy between soil particles. The figure shown below is an idealized soil drawn into phases of solids, water, and air.
 

001-phase-diagram-of-soil.gif

 

Weight-Volume Relationship from the Phase Diagram of Soil
total volume = volume of soilds + volume of voids
$V = V_s + V_v$

volume of voids = volume of water + volume of air
$V_v = V_w + V_a$

total weight = weight of solids + weight of water
$W = W_s + W_w$
 

Consistency of Soil (Atterberg Limits)

Consistency is the term used to describe the ability of the soil to resist rupture and deformation. It is commonly describe as soft, stiff or firm, and hard.
 

Water content greatly affects the engineering behavior of fine-grained soils. In the order of increasing moisture content (see Figure 2 below), a dry soil will exist into four distinct states: from solid state, to semisolid state, to plastic state, and to liquid state. The water contents at the boundary of these states are known as Atterberg limits. Between the solid and semisolid states is shrinkage limit, between semisolid and plastic states is plastic limit, and between plastic and liquid states is liquid limit.
 

002-atterberg-limits.gif

 

Atterberg limits, then, are water contents at critical stages of soil behavior. They, together with natural water content, are essential descriptions of fine-grained soils.
 

Unit Weights and Densities of Soil

Symbols and Notations
γ, γm = Unit weight, bulk unit weight, moist unit weight
γd = Dry unit weight
γsat = Saturated unit weight
γb, γ' = Buoyant unit weight or effective unit weight
γs = Unit weight of solids
γw = Unit weight of water (equal to 9810 N/m3)
W = Total weight of soil
Ws = Weight of solid particles
Ww = Weight of water
V = Volume of soil
Vs = Volume of solid particles
Vv = Volume of voids
Vw = Volume of water
S = Degree of saturation
w = Water content or moisture content
G = Specific gravity of solid particles
 

Reinforced Concrete Design

Analysis and design of concrete members subject to flexure.
Analysis and design of concrete members subject to shear.
Analysis and design of concrete members subject to axial load.
Analysis and design of concrete members subject to combined loadings.
Deflection of concrete members
Working stress design
Design specifications of concrete structures
Ultimate strength design
Load factors
Pre-stressed concrete
Method of construction
Concrete technology
Seismic design
Serviceability

Working Stress Design of Reinforced Concrete

Working Stress Design is called Alternate Design Method by NSCP (National Structural Code of the Philippines) and ACI (American Concrete Institute, ACI).
 

Code Reference
NSCP 2010 - Section 424: Alternate Design Method
ACI 318 - Appendix A: Alternate Design Method
 

Notation

fc = allowable compressive stress of concrete
fs = allowable tesnile stress of steel reinforcement
f'c = specified compressive strength of concrete
fy = specified yield strength of steel reinforcement
Ec = modulus of elasticity of concrete
Es = modulus of elasticity of steel
n = modular ratio
M = design moment
d = distance from extreme concrete fiber to centroid of steel reinforcement
kd = distance from the neutral axis to the extreme fiber of concrete
jd = distance between compressive force C and tensile force T
ρ = ratio of the area of steel to the effective area of concrete
As = area of steel reinforcement

 

wsd-assumptions.jpg

 

Working Stress Analysis for Concrete Beams

Consider a relatively long simply supported beam shown below. Assume the load wo to be increasing progressively until the beam fails. The beam will go into the following three stages:

  1. Uncrack Concrete Stage – at this stage, the gross section of the concrete will resist the bending which means that the beam will behave like a solid beam made entirely of concrete.
  2. Crack Concrete Stage – Elastic Stress range
  3. Ultimate Stress Stage – Beam Failure
wsd-beam-analysis-crack-uncrack.jpg

 

Design of Steel Reinforcement of Concrete Beams by WSD Method

Steps is for finding the required steel reinforcements of beam with known Mmax and other beam properties using Working Stress Design method.

Given the following, direct or indirect:

Width or breadth = b
Effective depth = d
Allowable stress for concrete = fc
Allowable stress for steel = fs
Modular ratio = n
Maximum moment carried by the beam = Mmax

 

wsd-doubly-reinforced-beam.jpg

Example 01: Required Steel Area of Reinforced Concrete Beam

Problem
A rectangular concrete beam is reinforced in tension only. The width is 300 mm and the effective depth is 600 mm. The beam carries a moment of 80 kN·m which causes a stress of 5 MPa in the extreme compression fiber of concrete. Use n = 9.
1.   What is the distance of the neutral axis from the top of the beam?
2.   Calculate the required area for steel reinforcement.
3.   Find the stress developed in the steel.
 

wsd-example-01-unknown-steel-area.jpg

 

Example 01: Total Compression Force at the Section of Concrete Beam

Problem
A rectangular reinforced concrete beam with width of 250 mm and effective depth of 500 mm is subjected to 150 kN·m bending moment. The beam is reinforced with 4 – 25 mm ø bars. Use alternate design method and modular ratio n = 9.

  1. What is the maximum stress of concrete?
  2. What is the maximum stress of steel?
  3. What is the total compressive force in concrete?

 

wsd-example-01-flexural-stresses-concrete-steel.jpg

 

Example 02: Moment Capacity of a Concrete Beam

Problem
A reinforced concrete beam 300 mm wide has an effective depth of 600 mm. It is reinforced with 4-32 mm diameter bars for tension. f’c = 21 MPa and fy = 275 MPa. Find the moment capacity of the beam.
 

wsd-example-02-moment-capacity-beam.jpg

 

Example 03: Compressive Force at the Section of Concrete T-Beam

Problem
The following are the dimensions of a concrete T-beam section

Width of flange, bf = 600 mm
Thickness of flange, tf = 80 mm
Width of web, bw = 300 mm
Effective depth, d = 500 mm

The beam is reinforced with 3-32 mm diameter bars in tension and is carrying a moment of 100 kN·m. Find the total compressive force in the concrete. Use n = 9.
 

wsd-example-03-strength-of-t-beam.jpg

 

Example 04: Stress of Tension Steel, Stress of Compression Steel, and Stress of Concrete in Doubly Reinforced Beam

Problem
A 300 mm × 600 mm reinforced concrete beam section is reinforced with 4 - 28-mm-diameter tension steel at d = 536 mm and 2 - 28-mm-diameter compression steel at d' = 64 mm. The section is subjected to a bending moment of 150 kN·m. Use n = 9.

1. Find the maximum stress in concrete.
2. Determine the stress in the compression steel.
3. Calculate the stress in the tension steel.
 

wsd-example-04-doubly-reinforced-beam-analysis.jpg

Example 02: Finding the Number of 28-mm Steel Bars of Singly-Reinforced Concrete Cantilever Beam

Problem
A reinforced concrete cantilever beam 4 m long has a cross-sectional dimensions of 400 mm by 750 mm. The steel reinforcement has an effective depth of 685 mm. The beam is to support a superimposed load of 29.05 kN/m including its own weight. Use f’c = 21 MPa, fs = 165 MPa, and n = 9. Determine the required number of 28 mm ø reinforcing bars using Working Stress Design method.
 

wsd-example-02-cantilever-beam.jpg

 

Example 03: Finding the Number of 32-mm Steel Bars for Doubly-Reinforced Concrete Propped Beam

Problem
A propped beam 8 m long is to support a total load of 28.8 kN/m. It is desired to find the steel reinforcements at the most critical section in bending. The cross section of the concrete beam is 400 mm by 600 mm with an effective cover of 60 mm for the reinforcements. f’c = 21 MPa, fs = 140 MPa, n = 9. Determine the required number of 32 mm ø tension bars and the required number of 32 mm ø compression bars.
 

wsd-example-03-propped-beam.jpg

 

Surveying and Transportation Engineering

Fundamentals of Surveying
Tape Measurements
Leveling
Angles and Directions
Transits and Theodolites
Traverse
Highway Curves
Topographic Surveys
Control Surveys
Construction Surveys
Hydrographic Surveys
Land Surveys
Photogrammetry
 

Simple Curves

Formulas for Circular Curves
The formulas we are about to present need not be memorized. All we need is geometry plus names of all elements in simple curve. Note that we are only dealing with circular arc, it is in our great advantage if we deal it at geometry level rather than memorize these formulas.

001-circular-simple-curve.gif

 

Compound and Reversed Curves

Compound Curves
A compound curve consists of two (or more) circular curves between two main tangents joined at point of compound curve (PCC). Curve at PC is designated as 1 (R1, L1, T1, etc) and curve at PT is designated as 2 (R2, L2, T2, etc).
 

002-compound-simple-curves.gif

 

Spiral Curve

Spirals are used to overcome the abrupt change in curvature and superelevation that occurs between tangent and circular curve. The spiral curve is used to gradually change the curvature and superelevation of the road, thus called transition curve.
 

003-spiral-curve-transition-curve.gif

Parabolic Curve

Vertical Parabolic Curve
Vertical curves are used to provide gradual change between two adjacent vertical grade lines. The curve used to connect the two adjacent grades is parabola. Parabola offers smooth transition because its second derivative is constant. For a downward parabola with vertex at the origin, the standard equation is
 

$x^2 = -4ay$   or   $y = -\dfrac{x^2}{4a}$.

 

004-verical-symmetrical-parabolic-curve.gif

 

Problem 01 - Simple Curve

Problem
The angle of intersection of a circular curve is 45° 30' and its radius is 198.17 m. PC is at Sta. 0 + 700. Compute the right angle offset from Sta. 0 + 736.58 on the curve to tangent through PC.
 

01-001-offset-distance-simple-curve.gif

 

Problem 02 - Simple Curve

Problem
The angle of intersection of a circular curve is 36° 30'. Compute the radius if the external distance is 12.02 m.
 

01-002-radius-of-circular-curve.gif

 

Problem 03 - Simple Curve

Problem
Given the following elements of a circular curve: middle ordinate = 2 m; length of long chord = 70 m. Find its degree of curve, use arc basis.
 

01-003-degree-of-circular-curve.gif

 

Pages