HNC Construction
Analytical Methods
Unit 3 Analytical Methods
Task 4 – Analytical methods to analyse building services
P4a)
a) Define the following terms, specifying units and equations. Include illustrative diagrams,
- Stress;
(Taken from http://www.en.wikipedia.org)
Stress, defined as force per unit area, is a measure of the intensity of the total internal forces acting within a body across imaginary internal surfaces, as a reaction to external applied forces and body forces.
The SI unit for stress is the Pascal (symbol Pa), the same as that of pressure. Since the Pascal is so small (1 N/m2), engineering quantities are usually measured in megapascals (MPa) or gigapascals (GPa). In Imperial units, stress is expressed in pounds-force per square inch (psi) or kilopounds-force per square inch (ksi).
The formula to work out the force being applied to an area is;
σ = F - Stress = Force
A Area
Tensile strength (stress) σUTS, or SU measures the stress required to pull something such as rope, wire, or a structural beam to the point where it breaks.
The tensile strength of a material is the maximum amount of tensile stress that it can be subjected to before failure. The definition of failure can vary according to material type and design methodology. This is an important concept in engineering, especially in the fields of material science, mechanical engineering and structural engineering.
- Compressive Stress is when a structure is compressed or squashed and the actual length of the material is shortened. Examples of areas that suffer from compressive stress are columns and walls. If a material is weakened because of compression then it can become brittle and very unstable
Pictures from
Shear strength (stress) in engineering is a term used to describe the strength of a material or component against the type of yield or structural failure where the material or component fails in shear.
In structural and mechanical engineering the shear strength of a component is important for designing the dimensions and materials to be used for the manufacture/construction of the component. For example, beams, plates, bolt etc. In a reinforced concrete beam, the main purpose of stirrups is to increase the shear strength.
This large beam is under stress from the weight of the roof.
- Torque;
When there are two equal forces being applied to an object at either end it can cause the object to turn in a rotational effect. This type of force is called a couple and this rotational moment translates as torque.
The formula to work out torque is;
Torque = F x r Nm
= Force x perpendicular distance
b) Use Trigonometric problems to solve problems such as static forces. Relative motion, framework, Friction torque
For the beam shown below,
i/ determine the force acting on the beam support at B
ii/ determine the force acting on the beam support at A
Drawing 1
To calculate the force acting on B;
(2 X 0.2) + (7 X 0.5) + (3 X 0.8) = Rr x 1.0
0.4 + 3.5 = Rr x 1.0
6.3 = Rr x 1.0
Rr = 6.3
1.0
= 6.3 KN
The force acting on beam B is 6.3KN
To calculate the force acting on A;
Rl+ Rr = Total Force
So, Total force = 2 + 7 + 3 = 12
Rl = Total force – force at Support B
Rl = 12 – 6.3
= 5.7KN
The force acting on beam A is 5.7 KN
c) Below is a diagram of part of a roof truss. You as the structural Engineer have been asked to determine the resultant force
Drawing 2
To determine the resultant force I must convert the diagram shown in drawing 2 and change it to a parallelogram, shown in drawing 3
Drawing 3
To continue with the calculation I need to find the angles I have shown this below in drawing 4.
Drawing 4
So, to calculate the force on the resultant I must use the cosine law, as I now have the angles relevant to the Resultant.
The cosine law is:
a2 = b2 + c2 - 2bc cos A
So,
a2 = 82 + 52 – 2 x 8 x 5 cos 100°
= 64 + 25 – ( -13.89)
= 89 + 13.89
= 102.89
= √ 102.89
= 10.14KN
The resultant force is 10.14 KN
P4b) Use the principals of calculus to solve problems appropriate to construction, civil engineering and building services engineering.
a) Integrate each of the following with respect to x:
.i) 6x2 – 4x + 3 ii) 1 / √x iii) √x3
- When using the principals of calculus the equation looks like this;
dy = 6 x 2 - 4 x 2 + 3
dx
When integrated it becomes
Y= 6x 3 – 4 x 2 + 3c
3 2
- The equation;
dy = 1 (This can be simplified to x -1/2 )
dx √x
When integrated;
Y = x ½
½
- The equation;
√ x 3 This can be written to a simpler form by - x3 + ½ = x3/2