If you have completed the BTEC First Diploma unit Mathematics and Science for Technicians and the BTEC National Certifi cate/Diploma unit Mechanical Principles and Applications, you will
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 which act at an angle exert a pull which is part horizontal and part vertical They can be split into their horizontal and vertical parts or components , by the use of trigonometry When you are doing this, it is a useful rule to always measure angles to the horizontal ( Figure 1.3 )
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 which act upwards to the left or downwards to the right will have one component which is positive and one which is negative Having resolved all of the forces in a coplanar system into their horizontal and vertical components, each set can then be added algebraically to determine the resultant horizontal pull, F H , and the resultant vertical
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
With non-concurrent force systems, the algebraic sum of the moments of the vertical and horizontal components of the forces, taken about some convenient point, gives the resultant couple or turning moment Its sign, positive or negative, indicates whether its direction is clockwise or anticlockwise This in turn can be used to fi nd the perpendicular distance of the line of action of the resultant from the chosen point.
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 you resolve the forces into their horizontal and vertical components it is essential to use the sign convention A logical way is to draw up a table as follows with the forces and their horizontal and vertical components, set out in 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
Example 1.1 can be solved graphically by means of a force vector diagram drawn to a suitable scale The process is known as vector addition The force vectors are taken in order, preferably working clockwise around the system, and added nose to tail to produce a polygon of forces
Should the fi nal vector be found to end at the start of the fi rst, there will be no resultant and the system will be in equilibrium If however there is a gap between the two, this when measured from the start of the fi rst vector to the end of the last represents the magnitude, direction and sense of the resultant The equilibrant will of course be equal and opposite It must be remembered that when you solve problems graphically, the accuracy of the answers will depend on the accuracy of your measurement and drawing
Bow ’ s notation is a useful method of identifying the forces on a vector diagram and also the sense in which they act In the space diagram, which shows the forces acting at the point of concurrence, the spaces between the forces are each given a capital letter Wherever possible, the letters should follow a clockwise sequence around the diagram In the solution shown below, the force F 1 is between the spaces A and B and when drawn on the vector diagram it is identifi ed by the lower case letters as force ab
Use capital letters and work in a clockwise direction when lettering space diagram using Bow ’ s notation
CHAPTER 1 The clockwise sequence of letters on the space diagram, i.e A to B, gives the direction of the force on the vector diagram The letter a is at the start of the vector and the letter b is at its end Although arrows have been drawn to show the directions of the vectors they are not really necessary and will be omitted on future graphical solutions.
Alternative graphical solution ( Figures 1.6 and 1.7 )
Space diagram (The angles should be measured as accurately as possible)
When using the analytical method of solution it is a good idea to measure angles from the horizontal All of the horizontal components of the forces will then be given by F H F cos and all of the vertical components will be given by F V F sin
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
Examples of framed structures which you see in everyday life are bicycles, roof trusses, electricity pylons and tower cranes They are made up of members which are joined at their ends Some of these are three-dimensional structures whose analysis is complex and it is only the two-dimensional or coplanar structures, which we will consider
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 , which are in compression and shown diagramatically with arrows pointing outwards Once again, you have to imagine the way you would be pushing if you were in the place of a strut You would be pushing outwards to keep the structure in position and to stop yourself from being squashed
● The third type is redundant members A perfect framed structure is one which has just suffi cient members to prevent it from becoming unstable Any additional members, which may have been added to create a stiffer or stronger frame, are known as redundant members
Redundant members may be struts or ties or they may carry no load in normal circumstances We shall avoid framed structures with redundant members as very often they cannot be solved by the ordinary methods of statics
In reality the members are bolted, riveted or welded together at their ends but in our analysis we assume that they are pin-jointed or hinged at their ends, with frictionless pins We further assume that because of this, the only forces present in the members are tensile and compressive forces These are called primary forces In practice there might also be bending and twisting forces present but we will leave these for study at a higher level
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
We can also safely assume that if a structure is in a state of static equilibrium, each of its members will also be in equilibrium It follows that the above three conditions can also be applied to individual members, groups of members and indeed to any internal part or section of a structure
A full analysis of the external and internal forces acting on and within a structure can be carried out using mathematics or graphically by drawing a force vector diagram We will use the mathematical method 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
Next, we will apply the conditions for equilibrium to each joint in the structure, starting with one which has only two unknown forces acting on it This will enable us to fi nd the magnitude and direction of the force in each member.
Determine the magnitude and direction of the support reactions at X and Y for the pin-jointed cantilever shown in Figure 1.9 , together with the magnitude and nature of the force acting in each member
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 wall support reaction R X must be horizontal because it is equal and opposite to the force carried by the top member CA Also, the three external forces must be concurrent with lines of action meeting at the point P
Begin by locating the point P and drawing in the support reactions at X and Y in the sense which you think they are acting If you guess wrongly, you will obtain a negative answer from your calculations, telling you that your arrow should be pointing in the opposite direction Letter the diagram using Bow ’ s notation, with capital letters in the spaces between the external forces and inside the structure
Now you can begin 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:
You can now turn your attention to the forces in the individual members You already know the force F AC acting in member AC because it is equal and opposite to the support reaction R X Choose a joint where you know the magnitude and direction of one of the forces and the directions of the other two, i.e where there are only two unknown forces Joint ABD will be ideal It is good practice to assume that the unknown forces are tensile, with arrows pulling away from the joint A negative answer will tell you that the force is compressive ( Figure 1.10 )
Take vertical components of the forces to fi nd the force F DA in member DA
The positive answer denotes that the force F DA in member DA is tensile and that the member is a tie Now take horizontal components of the forces to fi nd F BD in member BD
The negative sign denotes that the force F BD in member BD is compressive and that the member is a strut Now go to joint ADC and take vertical components of the forces to fi nd F DC , the force in member DC ( Figure 1.11 ).
Engineering Design
The negative sign denotes that the force F DC in member DC is compressive and that the member is a strut You have now found the values of all the external and internal forces which can be tabulated and their directions indicated on the structure diagram as follows ( Figure 1.12 ):
Reaction/member Force (kN) Nature
You may also solve framed structure problems graphically Sometimes this is quicker but the results may not be so accurate As before, it depends on the accuracy of your drawing and measurement An alternative solution to the above framed structure is as follows Once again, use is made of Bow ’ s notation to identify the members.
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°
Now each of the joints is in equilibrium Choose one where you know the magnitude and direction of one force and the directions of the other two The joint ABD is in fact the only one where you know these conditions
Draw the force vector diagram for this joint to a suitable scale, beginning with the
5 kN load It will be a triangle of forces, which you can quite easily construct because you know the length of one side and the directions of the other two The force vectors are added together nose to tail but do not put any arrows on 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 ABD gives the directions in which the forces act on the joint in the vector diagram, i.e the force ab acts from a to b , the force bd acts from b to d and the force da acts from d to a
Now go to another joint where you know the magnitude and direction of one of the forces and the directions of the other two Such a joint is ADC where the same 10 kN force da , which acts on joint ADB, also acts in the opposite direction on joint ADC We now call it force ad , and the triangle of forces is as shown in Figure 1.15 a 30 ° 30 ° c
CHAPTER 1 All of the internal forces have now been found and also the reaction R X , which is equal and opposite to ca , and can thus be written as ac A fi nal triangle of forces ABC , can now be drawn, representing the three external forces (see 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
Figure 1.18 Accurate measurement gives, Reaction kN Angle
You will note that the above three vector diagram triangles have sides in common and to save time it is usual to draw the second and third diagrams as additions to the fi rst The combined vector diagram appears as in Figure 1.17
The directions of the forces, acting towards or away from the joints on which they act, can be drawn on the space diagram This will immediately tell you which members are struts and which are ties The measured values of the support reactions and the forces in the members can then be tabulated ( Figure 1.18 )
Reaction/member Force (kN) Nature
The jib-crane shown in Figure 1.19 carries a load of 10 kN Making use of Bow ’ s notation fi nd the reactions of the supports at X and Y and the magnitude and nature of the force in each member graphically by means of a force vector diagram Apply the conditions for static equilibrium to the structure and check your results by calculation
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 Bow ’ s notation?
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 in such a way that it is allowed to expand and bend freely In practice the supports are often rollers The loads on the beam may be concentrated at different points or uniformly distributed along the beam Figure 1.20 shows concentrated loads only
The downward forces on a beam are said to be active loads, due to the force of gravity, whilst the loads carried by the supports are said to be reactive When investigating the effects of loading, we often have to begin by calculating the 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, F V 0
Uniformly distributed loads, or UDLs, are evenly spread out along a beam They might be due to the beam ’ s own weight, paving slabs or an asphalt surface UDLs are generally expressed in kN per metre length, i.e kN m 1 This is also known as the ‘ loading rate ’ UDLs are shown diagrammatically as in 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
Begin by calculating the total UDL Then, replace it by an equal concentrated load acting at its centroid, which is the midpoint of the span This is shown with dotted line in the fi gure
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
As can be seen in the cantilever in Figure 1.24(a) , the load F has a shearing effect and thus sets up shear stress at section Y–Y The load also has a bending effect and at any section Y–Y , this produces tensile stress in the upper layers of the beam and compressive stress in the lower layers
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