#### Basic principles of statics

Structural engineering is concerned with the strength, stiffness and stability of structures such as buildings, dams, bridges and retaining walls. Although a building is constructed from the foundation upwards, the designer has to start at the top with the roof and work his way downwards. There are two distinct stages in structural design. First the structural engineer with his experience, intuition and knowledge makes an imaginative choice of preliminary design in terms of layout, materials and erection methods. Estimates of the various forms of loading are made and then the chosen design is subjected to detailed analysis based on the principles of statics. Statics is one main branch of mechanics and deals with forces on bodies, which are 'at rest' (static equilibrium). The other main branch, dynamics, deals with moving bodies, such as parts of machines.

Static Equilibrium

Forces acting in one plane (i.e., coplanar) and in equilibrium must satisfy one of the following sets of conditions:

S Fx=0 S Fx=0 S Fy=0 S Ma=0

S Fy=0 or S Ma=0 or S Ma=0 or S Mb=0

S Ma=0 S Mb=0 S Mb=0 S Mc=0

where F refers to forces and M refers to moments of forces.

Static Determinacy

If a body is in equilibrium under the action of coplanar forces, the equations of statics above must apply. In general then, three independent unknowns can be determined from the three equations. Note that if applied and reaction forces are parallel (i.e., in one direction only) only two separate equations obtain and then only two unknowns can be determined. Such systems of forces are said to be statically determinate.

Force

A force is defined as any cause which tends to alter the state or rest of a body or its state of uniform motion in a straight line. A force can be defined quantitatively as the product of the mass of the body, which the force is acting on, and the acceleration of the force.

P = ma where
P = applied force
m= mass of the body ( kg)
a = acceleration caused by the force (m/s2)

The Sl units for force are therefore kg m/s2 which is designated a Newton (N). The following multiples are often used:

1kN = 1,000N, 1MN = 1,000,000N

All objects on earth tend to accelerate toward the centre of the earth due to gravitational attraction, hence the force of gravitation acting on a body with the mass (m) is the product of the mass and the acceleration due to gravity (g), which has a magnitude of 9.81 m/s2.

F = mg = vr g where:
F = force (N)
m= mass ( kg)
g = acceleration due to gravity (9.8m/s2)
v = volume (m³)
r = density ( kg/m³)

Vector

Most forces have magnitude and direction and can be shown as a vector. Its point of application must also be specified. A vector is illustrated by a line, whose length is proportional to the magnitude to some scale and an arrow which shows the direction.

The sum of two or more vectors is called the resultant. The resultant of two concurrent vectors is obtained by constructing a vector diagram of the two vectors.

The vectors to be added are arranged in tip-to-tail fashion. Where three or more vectors are to be added they can be arranged in the same manner and this is called a polygon. A line drawn to close the triangle or polygon (from start to finishing point) forms the resultant vector.

The subtraction of a vector is defined as the addition of the corresponding negative vector.

Resolution of a Force

In analysis and calculation it is often convenient to consider the effects of a force in other directions than that of the force itself, especially along the Cartesian (xx-yy) axes. The force effects along these axes are called vector components and are obtained by reversing the vector addition method.

Fy is the component of F in the 'y' direction Fy = F sin q

Fx is the component of F in the 'x' direction Fx = F cos q

Concurrent Coplanar Forces

Forces whose line of action meet at one point are said to be concurrent. Coplanar forces lie in the same plane, whereas non-coplanar forces have to be related to a three dimensional space and require two items of directional data together with the magnitude. Two coplanar nonparallel forces will always be concurrent.

Equilibrium of a Particle

When the resultant of all forces acting on a particle is zero, the particle is in equilibrium, i.e., it is not disturbed from its existing state of rest (or uniform movement).

The closed triangle or polygon is a graphical expression of the equilibrium of a particle.

The equilibrium of a particle to which a single force is applied may be maintained by the application of second force, which is equal in magnitude and direction, but opposite in sense, to the first force. This second force, since it restores equilibrium, is called the equilibriant. When a particle is acted upon by two or more forces, the equilibriant has to be equal and opposite to the resultant of the system. Thus the equilibriant is the vector drawn closing the vector diagram and connecting the finishing point to the starting point.

Free-body Diagram of a Particle

A sketch showing the physical conditions of a problem is known as a space diagram. This can be reduced to a diagram showing a particle and all the forces acting on it. Such a diagram is called a free-body diagram.

Example 1 Determine the tension in each of the ropes AB and AC

Example 2 A rigid rod is hinged to a vertical support and held at 50° to the horizontal by means of a cable when a weight of 250N is suspended as shown in the figure. Determine the tension in the cable and the compression in the rod, ignoring the weight of the rod.

Space diagram

Free body diagram

Force triangle

The forces may also be calculated using the law of sines:

(Compression in rod / sin 45°) = (Tension in cable / sin 40°) = (250N / sin 65°)

Point of Concurrency

Three coplanar forces that are in equilibrium, must all pass through the same point. This does not necessarily apply for more than three forces.

If two forces (which are not parallel) do not meet at their points of contact with a body such as a structural member, their lines of action can be extended until they meet.

Colinear Forces

Colinear forces are parallel and concurrent. The sum of the forces must be zero for the system to be in equilibrium.

Coplanar, Non-Concurrent, Parallel Forces Three or more parallel forces are required. They will be in equilibrium if the sum of the forces equals zero and the sum of the moments around a point in the plane equals zero. Equilibrium is also indicated by two sums of moments equal to zero.

Table 4 1 Actions and Reactions

Reactions

Structural components are usually held in equilibrium by being secured to rigid fixing points; these are often other parts of the same structure. The fixing points or supports will react against the tendency of the applied forces (loads) to cause the member to move. So the forces generated in the supports are called reactions.

In general, a structural member has to be held or supported at a minimum of two points (an exception to this is the cantilever). Anyone who has tried 'balancing' a long pole or something similar will realize that although only one support is theoretically necessary two are used to give satisfactory stability.

Resultant of Gravitation Forces

The whole weight of a body can be assumed to act at the centre of gravity of the body for the purpose of determining supporting reactions of a system of forces which are in equilibrium. Note that for other purposes the gravitation forces cannot always be treated this way.

Example 3

A ladder rests against a smooth wall and a man weighing 900N stands on it at the middle. The weight of the ladder is 100N. Determine the support reactions at the wall (RW) and at the ground (RG)

Force diagram

Since the wall is smooth the reaction RW must be at right angles to the surface of the wall and is therefore horizontal. A vertical component would have indicated a friction force between the ladder and the wall. At the bottom the ladder is resting on the ground which is not smooth, and therefore the reaction RG must have both a vertical and a horizontal component.

Since the two weight forces in this example have the same line of action, they can be combined into a single force reducing the problem from one having four forces to one having only three forces. The point of concurrency (A) can then be found, giving the direction of the ground reaction force. This in turn enables the force vector diagram to be drawn and hence the wall and ground reactions determined.

Example 4

A pin-jointed framework (truss) carries two loads as shown. The end A is pinned to a rigid support whilst the end B has a roller support. Determine the supporting reactions graphically:

• 1 Combine the two applied forces into one and find line of action.

1 Combine the two applied forces into one and find line of action.

• 2 Because of the roller support reaction RB will be vertical. Therefore the resultant line (RL) must be extended to intersect the vertical reaction of support B. This point is the point of concurrency for the resultant load, the reaction at B and the reaction at A.
• 3 From this point of concurrency, draw a line through the support pin at A. This gives the line of action of the reaction at A.
• 4 Use these three force directions and the magnitude of RL to draw the force diagram, from which RA and RB can be found.

Answer: RA = 12.2 kN at 21° to horizontal. RB = 12.7 kN vertical.

The link polygon (see an engineering handbook) may also be used to determine the reactions to a beam or a truss, though it is usually quicker and easier to obtain the reactions by calculation, the method shown in Example 4, or a combination of calculation and drawing.

The following conditions must however be satisfied.

• 1 All forces (apart from the two reactions) must be known completely, i.e., magnitude, line of action and direction.
• 2 The line of action of one of the reactions must be known.
• 3 At least one point on the line of action for the other reaction must be known. (2 and 3 reduce the number of unknowns related to the equations of equilibrium to an acceptable level.)

Moments of Forces

The effect of a force on a rigid body depends on its point of application as well as its magnitude and direction. It is common knowledge that a small force can have a large turning effect or leverage. In mechanics, the term moment is used instead of turning effect.

The moment of force with a magnitude (F) about a turning point (O) is defined as: M = F x d, where d is the perpendicular distance from O to the line of action of force F. The distance d is often called lever arm. A moment has dimensions of force times length (Nm). The direction of a moment about a point or axis is defined by the direction of the rotation that the force tends to give to the body. A clockwise moment is usually considered as having a positive sign and an anti-clockwise moment a negative sign.

The determination of the moment of a force in a coplanar system will be simplified if the force and its point of application are resolved into its horizontal and vertical components.

Free-body Diagram for a Rigid Body

In solving a problem it is essential to consider all forces acting on the body and to exclude any force which is not directly applied to the body. The first step in the solution of a problem should therefore be to draw a free-body diagram.

• 1 Choose the free body to be used, isolate it from any other body and sketch its outline.
• 2 Locate all external forces on the free body and clearly mark their magnitude and direction. This should include the weight of the free body which is applied at the centre of gravity.
• 3 Locate and mark unknown external forces and reactions, in the free-body diagram.
• 4 Include all dimensions that indicate the location, and direction of forces.

Example 3 continued

Since the ladder in Example 3 is at rest, the conditions of equilibrium for a rigid body can be used to calculate the reactions. By taking moments around the point where the ladder rests on the ground, the moment of the reaction RG can be ignored since it has no lever arm (moment is zero). According to the third condition for equilibrium, the sum of moments must equal zero, therefore:

(6 x RW) - (900N x 1.5m) - (100N x 1.5m. = 0

RW = 250N

The vertical component of RG must, according to the second condition, be equal but opposite to the sum of the weight of the ladder and the weight of the person on the ladder, since those two are the only vertical forces and the sum of the vertical forces must equal zero. i.e.,

RGy =1000N

Using the first condition of equilibrium it can be seen that the horizontal component of RG must be equal but opposite in direction to RW i.e.;

RGX= 250N

Since RG is the third side of a force triangle, where the other two sides are the horizontal and vertical components, the magnitude of RG can be calculated as:

10002 + 2502 = 1030N

Resultant of Parallel Forces

If two or more parallel forces are applied to a horizontal beam, then theoretically the beam can be held in equilibrium by the application of a single force (reaction) which is equal and opposite to the resultant, R. The equilibrant of the downward forces must be equal and opposite to their resultant. This provides a method for calculating the resultant of a system of parallel forces. However, two reactions are required to ensure the necessary stability and a more likely arrangement will have two or more supports.

The reactions RA and RB must both be vertical, since there is no horizontal force component. Furthermore the sum of the reaction forces RA and RB must be equal to the sum of the downward acting forces.

Beam Reactions

The magnitude of the reactions may be found by the application of the third condition for equilibrium, i.e., the algebraic sum of the moments of the forces about any point must be zero.

Take the moments around point A, then:

(80 x 2) + (70 x 4) + (100 x 7) + (30 x 10) - (RB x 12)=0;

RB= 120kN

RA is now easily found with the application of the second condition for equilibrium.

RA = 75 - 70 - 100 -30+ RB=0; RB= 120kN gives:

RA=160kN.

Couples

Two equal, parallel and opposite but not colinear forces are said to be a couple.

A couple acting on a body produces rotation. Note that the couple cannot be balanced by a single force. To produce equilibrium another couple of equal and opposite moment is required.

Couples

Before any of the various load effects (tension, compression, bending etc.) can be considered, the applied loads must be rationalized into a number of ordered systems. Irregular loading is difficult to deal with exactly but even the most irregular loads may be reduced and approximated to a number of regular systems. These can then be dealt with in mathematical terms using the principle of superposition to estimate the overall combined effect.

Concentrated loads are those which can be assumed to act at a single point e.g., a weight hanging from a ceiling, or a man pushing against a box.

Concentrated loads are represented by a single arrow drawn in the direction and through the point of action of the force. The magnitude of the force is always indicated.

Uniformly distributed loads, written as u.d.l. are those which can be assumed to act uniformly over an area or along the length of a structural member, e.g., roof loads, wind loads, the effect of the weight of water on a horizontal surface, etc.

For the purpose of calculation, a u.d.l. is normally considered in a plane and is represented as shown.

In calculating reactions, uniformly distributed loads can in most, but not all cases be represented by a concentrated load equal to the total distributed load and passing through the centre of gravity of the distributed load.

This technique must not be used for calculation of shear force, bending moment or deflection.

Example 5

Consider a suspended floor where the loads are supported by a set of irregularly placed beams. Let the load arising from the weight of the floor itself and the weight of any material placed on top of it (e.g., stored grain) be 10kN/m². Determine the u.d.l. acting on beam A and beam C.

FLOOR SECTION

It can be seen from the figure below that beam A carries the floor loads contributed by half the area between the beams A and B i.e., the shaded area L. Beam C carries the loads contributed by the shaded area M.

Floor section part 2

Therefore beam A carries a total load of:

1 m x 4m x 10kN/ m² = 40kN, or 40kN / 4 = 10kN / m.

In the same way the loading of beam C can be calculated to 25kN / m. The loading per metre run can then be used to calculate the required size of the beams.

The loading shape is triangular and is the result of such actions as the pressure of water on retaining walls and dams.

Shearing Force and Bending Moment

A beam is a structural member subject to lateral loading in which the developed resistance to deformation is of flexural character. The primary load effect that a beam is designed to resist is that of bending moments, but in addition, the effects of transverse or vertical shearing forces must be considered.

Consider the cantilever AB shown in (a). For equilibrium, the reaction force at A must be vertical and equal to the load W.

The cantilever must therefore transmit the effect of load W to the support at A by developing resistance (on vertical cross-section planes between the load and the support) to the load effect called shearing force. Failure to transmit the shearing force at any given section, e.g., section x-x, will cause the beam to fracture as in (b). The bending effect of the load will cause the beam to deform as in (c). To prevent rotation of the beam at the support A, there must be a reaction moment at A, shown as MA, which is equal to the product of load W and the distance from W to point A.

The shearing force and the bending moment transmitted across the section x-x may be considered as the force and moment respectively that are necessary to maintain equilibrium if a cut is made severing the beam at x-x. The free-body diagrams of the two portions of the beam are shown in (d).

Then the shearing force between A and C = Qx = W

and the bending moment between A and C = Mx = Wx

Note: Both the shearing force and the bending moment will be zero between C and B.

Shearing force part 1

Shearing force part 2

Definitions

Shear force (Q) is the algebraic sum of all the transverse forces acting to the left or to the right of the chosen section.

Bending moment (M) at any transverse cross section of a straight beam is the algebraic sum of the moments, taken about an axis passing through the centroid of the cross section, of all the forces applied to the beam on either side of the chosen cross section.

Table 4.2 Shearing and Bending Forces

Shear Force Variation

Concentrated loads will change the value of the shear force only at points where they occur, i.e., the shear force remains constant in between. When the load is uniformly distributed, however, the shear force will vary at a uniform rate. Thus it will be seen that uniform loads cause gradual and uniform change of shear, whilst concentrated loads bring a sudden change in the value of the shear force.

Bending Moment Variation

Concentrated loads will cause a uniform change of the bending moment between the points of action of the loads. In the case of uniformly distributed loads, the rate of change of the bending moment will be parabolic. Maximum values of bending moment will occur where the shear force is zero or where it changes sign.

Shear-Force and Bending-Moment Diagrams

Representative diagrams of the distribution of shearing force and bending moment are often required at several stages in the design process. These diagrams are obtained by plotting graphs with the beams as the base and the values of the particular effect as ordinates. It is usual to construct these diagrams in sets of three, representing the distribution of loads, shearing forces and bending moments respectively. These graphical representations provide useful information regarding:

• a the most likely section where a beam may fail in shear or in bending
• b where reinforcement may be required in certain types of beams, e.g., concrete beams
• c the shearforce diagram will provide useful information about the bending moment at any point
• d the bending-moment diagram gives useful information on the deflected shape of the beam.

The following example will show how the three diagrams are constructed:

Example 6.1

Draw a beam-loading diagram showing all loads and relevant dimensions. This is simply a free-body diagram of the beam.

2 Determine the reactions at the supports. First use the condition for equilibrium of moments about point

S ME = 0
ME=(P x a) + (w1 x b x b2 ) + w2 x c(b+c/2) - RG (b + c) = 0
ME = -(10 x 10) + (2 x 10 x 5) + 4 x 10 x (15) - RG (20) = 0
RG = 30kN
S Fy = 0 hence
S Fy=RE+RG-P-(w1 x b) - ( w2 x c)=0
S Fy = RE + 30 -10 - (2 x 10) - (4 x 10)=0
RE = 40kN

3 Draw the shear-force diagram (SFD) directly below the loading diagram and choose a convenient scale to represent the shear force.

Calculate the values of the shear force to the left and to the right of all critical points. Critical points are:

• at reactions
• at points where the magnitude of a distributed load changes.
• a Consider a section through the beam just to the left of D, and find the algebraic sum of all vertical forces to the left of this section. S Fy = 0 :. shear force to the left of D is zero
• b Consider a section just to the right of D, algebraic sum of forces to left of this section is 10kN down to the left. Hence shear force to right of D is 10kN (Negative)
• c The same result as in 2) above will be found for any such section between D and E. The shear-force diagram between D and E is thus a horizontal line at -10kN.
• d Consider a section just to the right of E, algebraic sum of forces to the left of this section is made up of P and RE given, shear force equals (-10 + 40)kN = + 30kN, i.e., up to the left of section. Thus at E shear-force diagram changes from -10kN to + 30kN.
• As we approach the right-hand end of the beam we find the mathematics easier to consider the right hand side of any section.
• e Section just to left of F. shear force = (4kN/m x 10m) -(30kN) using the sign convention to determine positive or negative. Shear force here equals + 40 - 30 = + 10kN.
• f Section just to right of F. shear force = + 40 - 30 = + 10kN (i.e., no sudden change at F). g Section just to left of G. shear force = -30kN h Variation of shear under a u.d.l. must be linear.

Example 6.1

Note the following from the shear force diagram:

• Maximum shear force occurs at E and G where the values are + 30kN and 30kN respectively. These two transverse sections are the two most likely points for failure in shear.
• The maximum bending moment will occur where the shear force is zero or where the shear force changes sign. However, note that cantilivered beams always will have maximum bending at the fixed end.

The SFD in the above example has two points where the shear force is zero. One is at E and the other is H between and G. The position of H can be calculated from the fact that at F the shear force is 10kN and under the action of the u.d. 1. to the right of F it reduces at the rate of 4kN/m. It will read a value of zero after 2.5m, i.e., the point H is 2.5m to the right of F.

4 Draw the bending moment diagram (BMD) directly under the SFD and choose a convenient scale to represent the bending moment. Calculate values of the bending moment at all critical points. Critical points for bending moment are:

• ends of the beam
• where the shear force is zero or changes sign
• other points which by experience are known to be critical.

Values of bending moment are calculated using the definition and sign convention and considering each load (to one side of the point) separately. It is the effect that one load would have on the bent shape at the chosen point that determines the sign.

a For B.M. at D consider left side of this point Mp = 0

b For B. M. at E consider left side of this point ME = P x a and beam would assume a hogging shape;

MF = -(10 x 10) = -100kNm

c For B. M. at F consider loads to right of point, a sagging beam results and:

MF = -(4 x 10 x 10/2) + (30 x 10) = 100kNm

d B.M. at G is obviously zero

e At point H we have maximum bending moment: considering forces to right of this point gives

MH = -(4 x 7 512 x 7 5) + (30 x 7 5) = 112.5 (sagging)

f The variation of bending moment under a u.d.l. is parabolic

g If the inclusion of other points would be helpful in drawing the curve, they should also be plotted.

Note the following from the bending-moment diagram:

• Maximum negative bending-moment hogging (100kNm) occurs at E and maximum positive bending moment sagging ( 112.5kNm) occurs at a point between F and G. When designing beams in materials such as concrete, the steel reinforcement would have to be placed according to these moments.

Figure

• The BMD will also give an indication as to how the loaded beam will deflect. Positive bending moments (sagging) cause compression in the top fibres of the beam, hence they tend to bend the beam with the concave side downwards.
• At the supported ends of a simple beam and at the free end of a cantilevered beam, where there can be no resistance to bending, the bending moment is always zero.

Forces in Pin-jointed Frames

Designing of a framework necessitates finding the forces in the members. For the calculation of primary stresses each member is considered to be pin-jointed at each end so that it can transmit an axial force only in the direction of the line connecting the pin joints at each end. The force can be a pure tension (conventionally designated positive) in which case the member is called a tie or a pure compression (conventionally designated negative) when the member is called a strut.

These are internal forces which must be in equilibrium with the external applied forces.

To determine the forces in the members one can use a number of different techniques.

Joint analysis: This is based on considering the equilibrium of each joint in turn and using the free-body diagram for each joint.

Method of sections: The free-body diagram considered is for a portion of the framework to one side or the other of a cut section. The forces in the members cut by the section are included in the free-body diagram. Application of the equations of equilibrium will solve the unknown forces in the cut section. This provides an analytical solution and is most useful when requiring the answers for one or two members only.

Example 7

Find the forces and their direction in the members BH and HG by using the method of sections.

FHG is found by taking a moment about point C, considering the right hand section (RHS) of the cut 1-1 is in equilibrium. The forces FHC and FBC have no moment about point CBL since they intersect at and pass through the point.

S Mc=0 (FHG x CG)+(9 x CD) - (RE x 20)=0

CG = FX = 10 tan 30° = 5,774

CD = DE =FE / cos 30°

FE = EX / cos 30° =11.547m

CD = 11,547 / cos 30° = 13,333m

RE = ( 9 + 12 + 12 )/ 2 = 15kN

Hence (FHG x 5,774) + (9 x 13,333) - (15 x 20) = 0

FHG = 31,17

Take section 2-2.

Since HC = FE = 11,547 (FBH x 11,547) + (9 x 13,333) - (15 x 20) = 0 FBH = 15,59kN

It can be seen that FC;H and FBH and FBH must be clockwise to have equilibrium about point C. The members GH and HB are therefore in tension.