Mechanical Engineering Systems 2008 Part 10 doc

Example 5.1.14 Determine the reactions at the supports of the uniformhorizontal beam shown in Figure 5.1.31, with its point loadand a uniformly distributed load of 40 N/m and the beamhav

(3) Centroid of a circular arc wire

This is a body of constant cross-sectional area and weightper unit volume and so we use Equation (5.1.11) For the

circular arc of radius r shown in Figure 5.1.26 we take length

segments of length L This is an arc of radius r subtending

an angle  at the centre; hence L = r  The x co-ordinate

of the element is r cos  Hence, the position of the centroid along the x-axis is:

¯x = x dL

1

2r  – (r cos )r d = r sin

Composite bodies

For objects which can be divided into several parts for which the centres

of gravity and weights, or centroids and areas, are known, the centre ofgravity/centroid can be determined by the following procedure:(1) On a sketch of the body, divide it into a number of composite parts

A hole, i.e a part having no material, can be considered to be apart having a negative weight or area

(2) Establish the co-ordinate axes on the sketch and locate the centre

of gravity, or centroid, of each constituent part

(3) Take moments about some convenient axis of the weights or areas

of the constituent parts and equate the sum to the moment thecentre of gravity or centroid of the total weight or area about thesame axis

230 mm Taking moments about A gives, for the two parts,(160 × 100) × 50 + (260 × 60) × 230 and this is equal to theproduct of the total area, i.e 160 × 100 + 260 × 60, multiplied

by the distance of the centroid of the composite from X.Thus:

¯x = 160× 100 × 50 + 260 × 60 × 230

160× 100 + 260 × 60 = 138.9 m

Figure 5.1.26 Circular arc

Figure 5.1.27 Example 5.1.11

Figure 5.1.28 Example 5.1.12

Figure 5.1.29 Example 5.1.13

Figure 5.1.30 Distributed load

of 20 N/m over part of the beam

20 cm by 10 cm The centroid of each segment is located at

its midpoint Taking moments about the y-axis gives:

Taking moments about the left-hand end gives 10 × 240 + 50

× 400 = 22 400 N mm The total load on the beam is 100 N andthus the centre of gravity is located 22 400/100 = 22.4 mm fromthe left-hand end

Beams with distributed forces

The weight of a uniform beam can be considered to act at its centre ofgravity, this being located at its midpoint However, beams can also besubject to a distributed load spread over part of their length An example

of this is a floor beam in a building where the loading due to a machinemight be spread over just part of the length of the beam Distributedloading on beams is commonly represented on diagrams in the mannershown in Figure 5.1.30 where there is a uniformly distributed load of

20 N/m over part of it

Example 5.1.14

Determine the reactions at the supports of the uniformhorizontal beam shown in Figure 5.1.31, with its point loadand a uniformly distributed load of 40 N/m and the beamhaving a weight of 15 N

The weight of the beam can be considered to act at itscentre of gravity and thus, since the beam is uniform, thispoint will be located a distance of 0.6 m from the left-handend The distributed load can be considered to act at its

centre of gravity, i.e the midpoint of the length over which itacts, and so is 16 N at 1.0 m from the left-hand end Takingmoments about the left-hand end gives:

0.4R1+ 1.2R2 = 15 × 0.6 + 16 × 1.0 = 25

and R1= 62.5 – 3R2 The sum of the vertical forces must bezero and so:

R1 + R2 = 20 + 15 + 16 = 51Hence:

(2) A particle is in equilibrium under the action of the threecoplanar concurrent forces If two of the forces are asshown in Figure 5.1.33(a) and (b), what will be the size anddirection of the third force?

(3) Determine the tension in the two ropes attached to the rigidsupports in the systems shown in Figure 5.1.34 if thesupported blocks are in equilibrium

(4) A vehicle of weight 10 kN is stationary on a hill which isinclined at 20° to the horizontal Determine the components

of the weight at right angles to the surface of the hill andparallel to it

(5) A horizontal rod of negligible weight is pin-jointed at oneend and supported at the other by a wire inclined at 30° tothe horizontal (Figure 5.1.35) If it supports a load of 100 N

at its midpoint, determine the tension in the wire

(6) The square plate shown in Figure 5.1.36 is pin-jointed at Dand is subject to the coplanar forces shown, the 40 N forcebeing applied at the midpoint of the upper face and the 30 Nforce along the line BA Determine the force that has to beapplied along the diagonal AC to give equilibrium and thevertical and horizontal components of the reaction force

at D

(7) Determine the magnitude and direction of the moments ofthe forces shown in Figure 5.1.37 about the axes throughpoint A

(8) Replace the system of parallel coplanar forces acting on thebeam in Figure 5.1.38 by an equivalent resultant force andcouple acting at end A

(9) Figure 5.1.39 shows a beam which has a pinned support at

A and rests on a roller support at B Determine the reactiveforces at the two supports due to the three concentratedloads shown if the beam is in equilibrium

(10) Two parallel oppositely directed forces of size 150 N areseparated by a distance of 2 m What is the moment of thecouple?

(11) Replace the force in Figure 5.1.40 with an equivalent forceand couple at A

(12) The system of coplanar forces shown in Figure 5.1.41 is in

equilibrium Determine the sizes of the forces P, Q and R.

(13) For the system shown in Figure 5.1.42, determine thedirection along which the 30 N force is to be applied to givethe maximum moment about the axis through A anddetermine the value of this moment

(14) Figure 5.1.43 shows a beam suspended horizontally bythree wires and supporting a central load of 500 N If thebeam is in equilibrium and of negligible weight, determinethe tensions in the wires

(15) Determine the positions of the centroids for the areasshown in Figure 5.1.44

(16) Determine the location of the centre of gravity of ahorizontal uniform beam of negligible weight when it haspoint loads of 20 N a distance of 2 m, 50 N at 5 m and 30 N

at 6 m from the left-hand end

(17) Determine, by integration, the location of the centroid of a

quarter circular area of radius r.

(18) Determine, by integration, the location of the centroid of a

hemispherical shell of radius r and negligible thickness.

(19) Determine the position of the centroid for the area shown inFigure 5.1.45

(20) Determine, by means of integration, the location of thecentroid of a wire bent into the form of a quarter circle.(21) A cylindrical tin can has a base but no lid and is made fromthin, uniform thickness, sheet If the base diameter is

60 mm and the can height is 160 mm, at what point is thecentre of gravity of the can?

(22) A thin uniform cross-section metal strip has part of it bent

into a semicircle of radius r with a length l tangential to the

semicircle, as illustrated in Figure 5.1.46 Show that thestrip can rest with the straight part on a horizontal bench if

l is greater than 2r.

(23) A length of wire of uniform cross-section and mass per unitlength is bent into the form of a semicircle What angle willthe diameter make with the vertical when the wire loop issuspended freely from one end of its diameter?

(24) A uniform beam of length 5.0 m rests on supports 1.0 mfrom the left-hand end and 0.5 m from the right-hand end A

Figure 5.1.43 Problem 14

Figure 5.1.44 Problem 15

Figure 5.1.45 Problem 19

Figure 5.1.46 Problem 22

uniformly distributed load of 10 kN/m is spread over itsentire length and the beam has a weight of 10 kN What will

be the reactions at the supports?

(25) Determine the reactions at the supports of the uniformhorizontal beams shown in Figure 5.1.47

(26) The tapered concrete beam shown in Figure 5.1.48 has arectangular cross-section of uniform thickness 0.30 m and adensity of 2400 kg/m3 If it is supported horizontally by twosupports, one at each end, determine the reactions at thesupports

5.2 Structures This section is about the analysis of structures The term structure is

used for an assembly of members such as bars, plates and walls whichform a stiff unit capable of supporting loads We can thus apply the term

to complex structures such as those of buildings and bridges or simplerstructures such as those of a support for a child’s swing or a cantileveredbracket to hold a pub sign (Figure 5.2.1)

In Section 5.1, structures were considered which where either a singlerigid body or a system of connected members and free-body diagramsused to analyse the forces acting on individual members or junctions ofmembers

Terms that are encountered when talking of structures are frameworks

and trusses The term framework is used for an assembly of members

which have sectional dimensions that are small compared with theirlength A framework composed of members joined at their ends to give

a rigid structure is called a truss and when the members all lie in the same plane a plane truss Bridges and roof supports are examples of

trusses, the structural members being typically I-beams, bars orchannels which are fastened together at their ends by welding, riveting

or bolts which behave like pin-jointed connections and permit forces inany direction Figure 5.2.2 shows some examples of trusses; a trussstructure that is used with bridges and one that is used to support theroof of a house

Statically determinate structures

The basic element of a plane truss is the triangle, this being threemembers pin-jointed at their ends to give a stable rigid framework

(Figure 5.2.3(a)) Such a pin-jointed structure is said to be statically determinate in that the equations of equilibrium for members, i.e no

resultant force in any direction and the sum of the moments is zero, aresufficient to enable all the forces to be determined

Figure 5.1.47 Problem 25 Figure 5.1.48 Problem 26

Figure 5.2.1 Example of

structures: (a) a child’s swing;

(b) a hanging pub sign

Figure 5.2.2 Examples of

trusses: (a) a bridge; (b) a roof

The arrangement of four members to form a rectangular framework(Figure 5.2.3(b)) gives an unstable structure since a small sidewaysforce can cause the structure to collapse To make the rectangle stable,

a diagonal member is required and this converts it into two triangularframeworks (Figure 5.2.3(c)) The addition of two diagonal members(Figure 5.2.3(d)) gives a rigid structure but one of the diagonal members

is ‘redundant’ in not being required for stability; such a structure is

termed statically indeterminate since it cannot be analysed by just the

equations for equilibrium Note that the analysis of members in trussesmight show that a member is unloaded An unloaded member must not

be confused with a redundant member; it may be that the memberbecomes loaded under different conditions

If m is the number of members and j the number of joints, then a

stable pin-jointed structure can be produced if:

be designed to cope with different loading conditions which can result

in a member being subject sometimes to tensile loading and sometimes

to compressive loading, it will have to be made thick enough to copewith the compressive loading One way this can be avoided is to designthe truss with a redundant member Figure 5.2.4 illustrates this with atruss which is designed to withstand wind loading from either the left orthe right The diagonal bracing members are slender cables which canonly operate in tension With the loading from the left, member BD is

in tension and member AC goes slack and takes no load With theloading from the right, member AC is in tension and member BD isslack and takes no load The truss can thus withstand both types ofloading without having a diagonal member designed for compression.Such a form of bracing was widely used in early biplanes to cope withthe loading on the wings changing from when they were on the ground

to when in the air Nowadays such bracing is often found in structuressubject to wind loading, e.g water towers

Example 5.2.1

Use the above criteria to determine whether the frameworksshown in Figure 5.2.3 can be stable, are unstable or containredundant members

For Figure 5.2.3(a) we have m = 3 and j = 3, hence when

we have m + 3 = 6 and 2j = 6 then the structure can be

stable

Figure 5.2.3 Frameworks: (a)

stable; (b) unstable; (c) stable;

(d) stable with a redundant

member

Figure 5.2.4 Designing with

redundancy

For Figure 5.2.3(b) we have m = 4 and j = 4, hence when we have m + 3 = 7 and 2j = 8 then the structure is unstable For Figure 5.2.3(c) we have m = 5 and j = 4, hence when

we have m + 3 = 8 and 2j = 8 then the structure can be

stable

For Figure 5.2.3(c) we have m = 6 and j = 4, hence when

we have m + 3 = 9 and 2j = 8 then the structure contains a

‘redundant’ member

Analysis of frameworks

In the following pages we discuss and apply methods that can be used

to analyse frameworks and determine the forces in individual members.For example, we might want to determine the forces in the individualmembers of the Warren bridge truss of Figure 5.2.2 when there is aheavy lorry on the bridge

The analysis of frameworks is based on several assumptions:

 Each member can be represented on a diagram as a straight linejoining its two end points where external forces are applied; externalforces are only applied at the ends of members The lines representthe longitudinal axes of the members and the joints betweenmembers are treated as points located at the intersection of themembers The weight of a member is assumed to be small comparedwith the forces acting on it

 All members are assumed to be two-force members; equilibriumoccurs under the action of just two forces with the forces being ofequal size and having the same line of action but in oppositedirections so that a member is subject to either just tension or justcompression (Figure 5.2.5) A member which is in tension is called

a tie; a member that is in compression is called a strut.

 All the joints are assumed to behave as pin-jointed and thus the joint

is capable of supporting a force in any direction (see Figure5.1.21(d) and associated text) Welded and riveted joints can usually

be assumed to behave in this way

Figure 5.2.5 Two-force

members: equilibrium occurring

under the action of just two

forces of the same size and

acting in the same straight line

but opposite directions

Figure 5.2.6 Bow’s notation,

(a) and (b) being two alternative

forms; in this book (a) is used

Bow’s notation

Bow’s notation is a useful method of labelling the forces in a truss This

is based on labelling all the spaces between the members and theirexternal forces and reactions using letters or numbers or a combination

of letters and numbers The advantage of using all letters for the spaces

is that the joints can be given numbers, as illustrated in Figure 5.2.6(a);

or, conversely, using numbers for the spaces means letters can be used

to identify joints (Figure 5.2.6(b)) The internal forces are then labelled

by the two letters or numbers on each side of it, generally taken in aclockwise direction Thus, in Figure 5.2.6(a), the force in the member

linking junctions 1 and 2 is FAF and the force in the member linking

junctions 3 and 6 is FGH In Figure 5.2.6(b), the force in the member

linking junctions A and B is F16 and the force in the member linking

junctions C and F is F

In the above illustration of Bow’s notation, the joints were labelledindependently of the spaces between forces However, the spacelabelling can be used to identify the joints without the need forindependent labelling for them The joints are labelled by the spaceletters or numbers surrounding them when read in a clockwise direction.Thus, in Figure 5.2.6(a), junction 1 could be identified as junction AFEand junction 3 as junction BCHG.

Method of joints

Each joint in a structure will be in equilibrium if the structure is in

equilibrium, thus the analysis of trusses by the method of joints involves

considering the equilibrium conditions at each joint in isolation from therest of the truss The procedure is:

(1) Draw a line diagram of the framework

(2) Label the diagram using Bow’s notation, or some other form ofnotation

(3) Determine any unknown external forces or reactions at supports byconsidering the truss at a single entity, ignoring all internal forces

be first selected for this treatment should be where there are nomore than two unknown forces

(5) Consider each junction in turn, selecting them in the order whichleaves no more than two unknown forces to be determined at ajunction

Example 5.2.2

Determine the forces acting on the members of the trussshown in Figure 5.2.7 The ends of the truss rest on smoothsurfaces and its span is 20 m

Figure 5.2.7 Example 5.2.2

Note that if the span had not been given, we could assume

an arbitrary length of 1 unit for a member and then relateother distances to this length Figure 5.2.8(a) shows Figure5.2.7 redrawn and labelled using Bow’s notation with thespaces being labelled by letters

Considering the truss as an entity we have the situationshown in Figure 5.2.8(b) Because the supporting surfacesfor the truss are smooth, the reactions at the supports will be

vertical Taking moments about the end at which reaction R1acts gives:

in the opposite directions then, when calculated, they willhave a negative sign

For joint 1, the sum of the vertical components must bezero and so:

For joint 2, the sum of the vertical components must bezero and so:

Thus the internal forces in the members of the truss are FAF=

–20.5 kN, FBG = –13.6 kN, FCH = –22.2 kN, FDH = +11.1 kN,

FFE= +10.25 kN, FFG= +6.6 kN, FGH= +4.9 kN

Example 5.2.3

For the plane truss shown in Figure 5.2.10, determine thereactions at the supports and the forces in each member Theupper end of the truss is a pin connection to the wall while thelower end is held in contact by a roller

Figure 5.2.11 shows the diagram labelled with Bow’snotation The reaction at the fixed end of the truss will be atsome angle  and thus can have both a horizontal and avertical component; the reaction at the lower end will bepurely a horizontal component

Taking moments about junction 1 gives:

Using the methods of joints, for joint 5 we have for thehorizontal components: