By the end of this chapter, you will be able to:
Mastering these skills will help you solve real-world engineering problems with confidence and precision.
Vectors are fundamental in welding engineering for describing forces, displacements, and directions in three-dimensional space. Understanding vector and scalar quantities and how to perform operations on vectors enables welding professionals to analyze stresses, weld bead directions, and equipment positioning accurately. This chapter focuses on the application of vector theorem concepts tailored to welding contexts commonly encountered in Kenya, such as calculating resultant forces on weld joints or determining the direction of welding torches.
In welding engineering, quantities describing physical phenomena fall into two categories: scalars and vectors. Scalars have magnitude only, such as temperature or welding current, while vectors have both magnitude and direction, such as force or displacement. Distinguishing these is crucial when analysing welding stresses or setting up welding equipment orientation.
Magnitude is the size of a quantity: how much of it there is. Every scalar and every vector has a magnitude, written as a number with its unit. For a scalar the magnitude is the whole description, for example a welding current of 150 A. For a vector the magnitude is only one half of the description and must be given together with a direction. The magnitude of a vector is never negative.
Key points about magnitude:
If a vector is given by its components along the x and y axes, its magnitude is found with Pythagoras' theorem:
|A| = √(Ax² + Ay²)
Example: A displacement vector D has components Dx = 5 m and Dy = 12 m. Find its magnitude.
Formula: |D| = √(Dx² + Dy²)
Substitution: |D| = √(5² + 12²)
Working: |D| = √(25 + 144) = √169
Result: |D| = 13 m
Example: A welding torch on a gantry at a fabrication yard moves 3 m east and 4 m north. Find the magnitude of its displacement.
Formula: |D| = √(x² + y²)
Substitution: |D| = √(3² + 4²)
Working: |D| = √(9 + 16) = √25
Result: |D| = 5 m
Scalar quantities are physical quantities that have only magnitude without any direction. They are described by a single numerical value and appropriate units. Scalars add by ordinary arithmetic: 2 kg plus 3 kg is always 5 kg.
Scalar quantity = magnitude only
Example: A welding machine operates at a current of 150 A.
Current I = 150 A is a scalar. No calculation is needed.
Answer: 150 A
Example: The temperature of a weld bead is 1200 °C.
Temperature T = 1200 °C is a scalar. No calculation is needed.
Answer: 1200 °C
Example: A welding electrode pack has a mass of 2.5 kg.
Mass m = 2.5 kg is a scalar. No calculation is needed.
Answer: 2.5 kg
Direction tells us along which line and in which sense a vector acts. Two forces of 200 N can have completely different effects on a weld joint if one pulls the plates apart and the other pushes them together. Direction is always stated relative to a reference such as an axis, a compass point or a surface. Scalars have no direction, which is the main difference between them and vectors.
Common ways of stating direction:
For a vector with components Ax and Ay, the angle from the x-axis follows from the tangent ratio:
tan θ = Ay ÷ Ax
Example: Find the direction of the displacement vector D with Dx = 5 m and Dy = 12 m.
Formula: θ = tan⁻¹(Dy ÷ Dx)
Substitution: θ = tan⁻¹(12 ÷ 5)
Working: θ = tan⁻¹(2.4)
Result: θ = 67.38° from the x-axis
Example: The torch displacement of 3 m east and 4 m north has direction θ = tan⁻¹(4 ÷ 3) = 53.13°, that is 53.13° north of east.
Vector quantities have both magnitude and direction. They are drawn as arrows, where the length shows the magnitude and the arrowhead shows the direction. Common vectors in workshop practice include force, displacement, velocity, acceleration and weight.
Vector quantity = magnitude + direction
Example: A force of 200 N acts at 30° to the horizontal on a weld joint.
Given: F = 200 N, θ = 30°
The force is a vector with magnitude 200 N and direction 30°.
Answer: F = 200 N at 30°
Example: The velocity of molten metal flow is 2 m/s vertically upward.
Given: v = 2 m/s, direction upward
The velocity is a vector with magnitude 2 m/s and a vertical direction.
Answer: v = 2 m/s upward
Scalars are represented by single numerical values. In two dimensions a vector is represented by a pair of components, one along each axis:
A = Ax i + Ay j
Here Ax and Ay are the components, and i and j are unit vectors along the x and y axes.
Example: Represent a force of 100 N at 45° in component form.
Formula: Fx = F cos θ and Fy = F sin θ
Substitution: Fx = 100 × cos 45° and Fy = 100 × sin 45°
Working: Fx = 100 × 0.7071 and Fy = 100 × 0.7071
Result: F = 70.71i + 70.71j N
A direction is positive when it points the same way as the positive end of the chosen reference axis. By convention, to the right (east) along x and upward (north) along y are positive, and angles measured anticlockwise from the positive x-axis are positive. A vector with a positive component acts in the positive sense of that axis. Choosing the positive direction is a free decision, but it must be kept the same throughout one problem.
Example: Molten metal rises through a vertical riser at 2 m/s. Taking upward as positive, the velocity is v = +2 m/s. The sign confirms that the flow is in the chosen positive direction.
Example: A force of 200 N acts on a weld joint at 30° to the horizontal, measured anticlockwise from the positive x-axis.
Formula: Fx = F cos θ and Fy = F sin θ
Substitution: Fx = 200 × cos 30° and Fy = 200 × sin 30°
Working: Fx = 200 × 0.866 and Fy = 200 × 0.5
Result: Fx = +173.2 N and Fy = +100 N
Both components are positive because the vector lies in the first quadrant, pointing right and upward.
A direction is negative when it points opposite to the chosen positive direction of the axis. A negative sign on a component or a vector means "opposite to positive", not "smaller". The magnitude of a vector stays positive, so a velocity of -4 m/s has a magnitude of 4 m/s directed along the negative axis. Reversing a vector multiplies it by -1: it keeps its magnitude and turns through 180°.
A negative sign on a scalar is different. A temperature of -5 °C in a low-temperature chamber used for toughness testing of weld samples simply lies below the zero of the scale and has no direction.
Example: A welding carriage on a straight rail reverses at 0.5 m/s. With forward as positive x, its velocity is v = -0.5 m/s, and its magnitude is |v| = 0.5 m/s.
Example: Represent a velocity of 8 m/s at 120° from the positive x-axis in components. The angle is in the second quadrant, so the x-component is negative.
Formula: vx = v cos θ and vy = v sin θ
Substitution: vx = 8 × cos 120° and vy = 8 × sin 120°
Working: vx = 8 × (-0.5) and vy = 8 × 0.866
Result: vx = -4 m/s and vy = +6.93 m/s
In component form: v = -4i + 6.93j m/s, where i and j are unit vectors along the x and y axes.
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Create a free accountThis chapter introduced the fundamental distinction between vector and scalar quantities, emphasizing that vectors possess both magnitude and direction, while scalars have only magnitude. The direction of vectors was further clarified by explaining positive and negative orientations. Various operations on vectors were covered, including addition and subtraction, which combine or compare vectors, as well as the dot product and cross product, which produce scalar and vector results respectively. The chapter also explored the resolution of vectors into components through analysis, enabling simpler calculations. Graphical methods for vector resolution were detailed, focusing on the triangle theorem, parallel theorem, and polygon theorem, each providing a visual approach to combine or break down vectors. Together, these concepts equip students with the tools to manipulate vectors effectively in engineering mathematics.
A welding torch moves with a velocity vector of magnitude 5 m/s at an angle of 30° to the horizontal. Calculate the horizontal and vertical components of this velocity. (2 marks)
Two welding rods are aligned such that one has a force vector of 8 N east and the other 6 N north. Determine the resultant force vector’s magnitude and direction. (3 marks)
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