Further Mechanical Principles and
forces and couples The following sign convention is that which is most often used:
(i) Upward forces are positive and downward forces are negative
(ii) Horizontal forces acting to the right are positive and horizontal forces acting to the left are negative
(iii) Clockwise acting moments and couples are positive and anticlockwise acting moments and couples are negative
Forces acting at an angle create both horizontal and vertical components These components can be separated using trigonometry, making it essential to measure angles relative to the horizontal.
In Figure 1.3(a) the horizontal and vertical components are both acting in the positive directions and will be:
In Figure 1.3(b) the horizontal and vertical components are both acting in the negative directions and will be:
Forces acting upwards to the left or downwards to the right consist of one positive and one negative component By resolving all forces in a coplanar system into their horizontal and vertical components, we can algebraically sum each set to find the resultant horizontal force, F H, and the resultant vertical force.
CHAPTER 1 pull, F V The Greek letter (sigma) means ‘ the sum or total ’ of the components Pythagoras ’ theorem can then be used to fi nd the single resultant force R of the system i.e R 2 (F H ) 2 (F V ) 2
The angle which the resultant makes with the horizontal can also be found using: tan V
In non-concurrent force systems, the resultant couple or turning moment is determined by the algebraic sum of the moments of the vertical and horizontal force components around a selected point The sign of this resultant moment indicates its direction, whether clockwise or anticlockwise This information is essential for calculating the perpendicular distance from the chosen point to the line of action of the resultant force.
Find the magnitude and direction of the resultant and equilibrant of the concurrent coplanar force system shown in Figure 1.4
Space diagram (Not to scale)
When resolving forces into their horizontal and vertical components, it is crucial to apply the sign convention correctly A practical approach is to create a table that organizes the forces along with their respective horizontal and vertical components in clearly defined rows and columns.
Force Horizontal component Vertical component
Advanced Mechanical Principles and
The fi ve forces have now been reduced to two forces, F H and F V They are both negative and can be drawn as vectors ( Figure 1.5 )
The resultant R is found by Pythagoras as follows:
The angle is found from: tan tan
The equilibrant, which is required to hold the system in a state of static equilibrium, is equal to the resultant but opposite in sense
To solve Example 1.1 graphically, a force vector diagram can be created using an appropriate scale This method, known as vector addition, involves arranging the force vectors in sequence, ideally in a clockwise direction around the system, and connecting them nose to tail to form a polygon of forces.
In a system of vectors, if the final vector coincides with the start of the first vector, the system is in equilibrium with no resultant Conversely, a gap between the two indicates the magnitude, direction, and sense of the resultant vector, while the equilibrant will be equal in magnitude but opposite in direction It's crucial to note that the accuracy of graphical problem-solving relies heavily on precise measurements and drawings.
Bow's notation is an effective technique for identifying forces in a vector diagram and their direction of action In the space diagram, forces at the point of concurrence are represented with capital letters, arranged in a clockwise sequence whenever possible For example, in the provided solution, force F1 is positioned between spaces A and B, and is denoted as force ab in the vector diagram.
Use capital letters and work in a clockwise direction when lettering space diagram using Bow ’ s notation
Chapter 1 explains that the clockwise arrangement of letters on the space diagram, from A to B, indicates the direction of force in the vector diagram The letter 'a' marks the beginning of the vector, while 'b' signifies its endpoint Although arrows have been used to illustrate vector directions, they will be excluded in future graphical representations.
Alternative graphical solution ( Figures 1.6 and 1.7 )
(The angles should be measured as accurately as possible)
When applying the analytical method for solving problems, it's beneficial to measure angles from the horizontal This approach allows for the horizontal components of forces to be calculated using the formula F_H = F cos(θ), while the vertical components can be determined with F_V = F sin(θ).
3 What are the conditions necessary for a body to be in static equilibrium under the action of a coplanar force system?
4 What are the resultant and equilibrant of a coplanar force system?
To check your understanding of the preceding section, you can solve Review questions 1–3 at the end of this chapter
Force vector diagram (Drawn to a suitable scale) c b d
Applications of Mechanical Systems and
Framed structures are commonly found in everyday life, with examples including bicycles, roof trusses, electricity pylons, and tower cranes These structures consist of members that are connected at their ends While some framed structures are three-dimensional and complex to analyze, this discussion will focus on two-dimensional or coplanar structures.
There are three kinds of member in these structures ( Figure 1.8 ):
Tie Member carrying tensile forces
Strut Member carrying compressive forces Figure 1.8 Representation of structural members
● Ties , which are in tension and shown diagramatically with arrows pointing inwards You have to imagine yourself in the place of a tie
You would be pulling inwards to stop yourself from being stretched
The arrows describe the force which the tie exerts on its neighbours to keep the structure in position
Struts are structural components that operate under compression, illustrated with arrows indicating outward force Visualize yourself as a strut, exerting pressure outward to maintain the integrity of the structure and prevent collapse.
Redundant members are additional components in a framed structure that do not contribute to its stability An ideal frame consists of just enough members to maintain its integrity; any extra members intended to enhance stiffness or strength are classified as redundant.
Redundant members in structural frames, whether they are struts or ties, often do not carry any load under normal conditions To ensure solvability using standard static methods, it is advisable to avoid framed structures that include these redundant members.
In structural analysis, we simplify the behavior of members by assuming they are pin-jointed at their ends with frictionless pins, despite being actually bolted, riveted, or welded This assumption allows us to focus solely on the primary forces, which are tensile and compressive, while neglecting the complexities of bending and twisting forces that may occur in real-world applications.
When a structure is in a state of static equilibrium, the external active loads which it carries will be balanced by the reactions of its supports
The conditions for external equilibrium are:
1 The vector sum of the horizontal forces or horizontal components of the forces is zero
2 The vector sum of the vertical forces or vertical components of the forces is zero
When a body or structure is in a static equilibrium under the action of three external forces, the forces must be concurrent
CHAPTER 1 3 The vector sum of the turning moments of the forces taken about any point in the plane of the structure is zero i.e F H 0 F V 0 M 0
In a structure at static equilibrium, all members are also in equilibrium, allowing the three fundamental conditions of equilibrium to apply not only to the entire structure but also to individual members, groups of members, and any internal sections.
A comprehensive analysis of the external and internal forces affecting a structure can be conducted through mathematical methods or by creating a force vector diagram In this discussion, we will focus on employing the mathematical approach for our analysis.
fi rst and begin by applying the conditions for equilibrium to the external forces This will enable us to fi nd the magnitude and direction of the support reactions
We will begin by applying the equilibrium conditions to each joint in the structure, focusing first on those with only two unknown forces This approach will allow us to determine both the magnitude and direction of the forces acting in each member.
To analyze the pin-jointed cantilever depicted in Figure 1.9, it is essential to determine the magnitude and direction of the support reactions at points X and Y Additionally, it is important to calculate the magnitude and nature of the forces acting in each member of the structure This evaluation will provide insight into the structural integrity and load distribution within the cantilever system.
Whichever method of solution you adopt, it is good practice to fi nd all of the external forces acting on a structure before investigating the internal forces in the members
When you are fi nding the internal forces start with a joint which has only two unknown forces acting on it whose directions you know
The horizontal wall support reaction R X counteracts the force exerted by the top member CA Additionally, the three external forces must converge at point P, ensuring that their lines of action intersect at this critical juncture.
To start, identify point P and illustrate the support reactions at points X and Y based on your assumptions of their direction If your initial guess is incorrect, the calculations will yield a negative result, indicating the need to reverse the arrow's direction Label the diagram using Bow's notation, employing capital letters in the areas between external forces and within the structure You can then proceed with the calculations.
Finding distances XY , YZ and the angle :
XY 1 tan 30 0.577 m YZ 2 0 m tan XY
Properties and Applications of
Take moments about Y to fi nd R X For equilibrium, M Y 0:
The force in member AC will also be 17.3 kN because it is equal and opposite to R X
It will be a tensile force, and this member will be a tie
Take vertical components of external forces to fi nd R Y For equilibrium, F V 0:
To analyze the forces in individual members, focus on the force F AC in member AC, which is equal and opposite to the support reaction R X Select a joint with a known force magnitude and direction, such as joint ABD, where only two unknown forces exist It is advisable to initially assume that the unknown forces are tensile, indicated by arrows pulling away from the joint; a negative result will indicate that the force is compressive.
Take vertical components of the forces to fi nd the force F DA in member DA
The positive response indicates that the force F DA in member DA is tensile, classifying the member as a tie Next, we will analyze the horizontal components of the forces to determine F BD in member BD.
The negative sign indicates that the force F BD in member BD is compressive, classifying it as a strut Next, analyze joint ADC by examining the vertical components of the forces to determine F DC, the force in member DC (see Figure 1.11).
Engineering Design
The negative sign indicates that the force F DC in member DC is compressive, classifying the member as a strut You have successfully determined the values of all external and internal forces, which can be organized into a table and illustrated with their directions on the structure diagram (Figure 1.12).
Reaction/member Force (kN) Nature
Framed structure problems can also be solved graphically, which may offer quicker results, though accuracy can vary based on the precision of your drawing and measurements An alternative approach involves utilizing Bow's notation to clearly identify the members of the structure.
Begin by drawing the space diagram shown in Figure 1.13 to some suitable scale Take care to measure the angles as accurately as possible The angle R y can be measured, giving 16.1°
In the analysis of joint equilibrium, joint ABD is unique as it is the only joint where the magnitude and direction of one force are known, along with the directions of the other two forces.
Draw the force vector diagram for this joint to a suitable scale, beginning with the
To analyze a 5 kN load, you can easily construct a triangle of forces using the known length of one side and the directions of the other two forces By adding the force vectors together in a nose-to-tail arrangement, you can effectively visualize the system without the need for arrows in the diagram.
Bow ’ s notation is suffi cient to indicate the directions of the forces ( Figure 1.14 )
Remember, the clockwise sequence of letters around the joint on the space diagram
The ABD vector diagram illustrates the directions of forces acting on the joint, with force ab exerting influence from point a to point b, force bd acting from point b to point d, and force da operating from point d back to point a.
At joint ADC, where one force is known in magnitude and direction, the 10 kN force (denoted as force ad) acts in the opposite direction to the force at joint ADB This setup allows for the analysis of the triangle of forces, as illustrated in Figure 1.15, with angles of 30° and 30°.
In Chapter 1, all internal forces have been identified, along with the reaction force R X, which is equal and opposite to ca, allowing it to be expressed as ac Consequently, a final triangle of forces ABC can be illustrated, depicting the three external forces involved (refer to Figure 1.16).
Two of these, th e 5 kN load and R X , are now known in both magnitude and direction together with the angle which R Y makes with the horizontal a c θ b
Accurate measurement gives, Reaction kN Angle
The three vector diagram triangles share common sides, allowing for an efficient representation by drawing the second and third diagrams as additions to the first This results in a combined vector diagram, as illustrated in Figure 1.17.
The forces acting on the joints can be illustrated on a space diagram, indicating the direction of each force This visualization helps identify which members function as struts and which as ties Subsequently, the measured values of the support reactions and the forces within the members can be organized in a table format (Figure 1.18).
Reaction/member Force (kN) Nature
The jib-crane depicted in Figure 1.19 supports a load of 10 kN Using Bow's notation, determine the reactions at supports X and Y, as well as the magnitude and nature of the forces in each member through a graphical force vector diagram Apply static equilibrium conditions to the structure and verify the results through calculations.
Figure 1.20 Simply supported beam with concentrated loads
1 What are the three different types of member found in framed structures?
2 What are the primary forces which act in structural members?
3 How do you indicate using arrows, which members of a structure are ties and which are struts?
4 How do you letter the space diagram of a coplanar force system using
To check your understanding of the preceding section, you can solve
Review question 4 at the end of this chapter
A simply supported beam is supported at two points, allowing for free expansion and bending, typically using roller supports The loads applied to the beam can either be concentrated at specific points or uniformly distributed along its length, as illustrated in Figure 1.20, which depicts only concentrated loads.
Active loads on a beam are the downward forces caused by gravity, whereas the loads that the supports carry are referred to as reactive loads To analyze the impact of these loads, it is essential to start with the calculation of support reactions.
CHAPTER 1 The beam is in static equilibrium under the action of these external forces, and so we proceed as follows
1 Equate the sum of the turning moments, taken about the right-hand support D, to zero i.e.M D 0
R l A W l 1 1 W l 2 2 0 You can fi nd R A from this condition
2 Equate vector sum of the vertical forces to zero i.e F V 0
R A R D W 1 W 2 0 You can fi nd R D from this condition.
Calculate the support reactions of the simply supported beam shown in
Take moments about the point D, remembering to use the sign convention that clockwise moments are positive and anticlockwise moments are negative For equilibrium, M D 0
Equate the vector sum of the vertical forces to zero, remembering the sign convention that upward forces are positive and downward forces are negative For equilibrium,
Uniformly distributed loads (UDLs) refer to loads that are evenly spread across a beam, often resulting from the beam's own weight, paving slabs, or asphalt surfaces These loads are typically measured in kilonewtons per meter (kN/m), also referred to as the loading rate UDLs can be visually represented in diagrams, such as Figure 1.22.
The total UDL over a particular length l of a beam is given by the product of the loading rate and the length i.e Total UDL wl
When you are equating moments to fi nd the beam reactions, the total UDL is assumed to act at its centroid, i.e at the centre of the length l
You can then treat it as just another concentrated load and calculate the support reactions in the same way as before.
Example 1.6 Calculate the support reactions of the simply supported beam shown in Figure 1.23
Uniformly distributed load, w kN m − 1 Concentrated load, W kN l/2
Figure 1.22 Simply supported beam with concentrated and distributed load
A uniformly distributed load may be considered to act at its centroid
2.5 m Distributed load, w = 2 kN m − 1 5 kN 7 kN
To calculate the total uniformly distributed load (UDL), first determine its magnitude Next, replace the UDL with an equivalent concentrated load acting at its centroid, located at the midpoint of the span, as illustrated by the dotted line in the figure.
You can now apply the conditions for static equilibrium Begin by taking moments about the point D For equilibrium, M D 0
Now equate the vector sum of the vertical forces to zero For equilibrium, F V 0
When structural engineers are designing beams to carry given loads, they have to make sure that the maximum allowable stresses will not
CHAPTER 1 be exceeded On any transverse section of a loaded beam or cantilever, shear stress, tensile stress and compressive stress are all usually present
The cantilever in Figure 1.24(a) demonstrates that the applied load F induces shear stress at section Y–Y, resulting in a shearing effect Additionally, this load generates bending effects, creating tensile stress in the upper layers of the beam and compressive stress in the lower layers at the same section Y–Y.
Tensile stress in upper layers
Shear stress Compressive stress in lower layers
Figure 1.24 Stress in beams: (a) shearing effect of load, (b) bending effect of load