Chapter 7 Structural design

Transcription

1 115 Chapter 7 Structural design Introduction Structural design is the methodical investigation of the stabilit, strength and rigidit of structures. The basic objective in structural analsis and design is to produce a structure capable of resisting all applied loads without failure during its intended life. The primar purpose of a structure is to transmit or support loads. If the structure is improperl designed or fabricated, or if the actual applied loads eceed the design specifications, the device will probabl fail to perform its intended function, with possible serious consequences. wellengineered structure greatl minimizes the possibilit of costl failures. Structural design process structural design project ma be divided into three phases, i.e. planning, design and construction. lanning: This phase involves consideration of the various requirements and factors affecting the general laout and dimensions of the structure and results in the choice of one or perhaps several alternative tpes of structure, which offer the best general solution. The primar consideration is the function of the structure. Secondar considerations such as aesthetics, sociolog, law, economics and the environment ma also be taken into account. In addition there are structural and constructional requirements and limitations, which ma affect the tpe of structure to be designed. Design: This phase involves a detailed consideration of the alternative solutions defined in the planning phase and results in the determination of the most suitable proportions, dimensions and details of the structural elements and connections for constructing each alternative structural arrangement being considered. Construction: This phase involves mobilization of personnel; procurement of materials and equipment, including their transportation to the site, and actual on-site erection. During this phase, some redesign ma be required if unforeseen difficulties occur, such as unavailabilit of specified materials or foundation problems. hilosoph of designing The structural design of an structure first involves establishing the loading and other design conditions, which must be supported b the structure and therefore must be considered in its design. This is followed b the analsis and computation of internal gross forces, (i.e. thrust, shear, bending moments and twisting moments), as well as stress intensities, strain, deflection and reactions produced b loads, changes in temperature, shrinkage, creep and other design conditions. Finall comes the proportioning and selection of materials for the members and connections to respond adequatel to the effects produced b the design conditions. The criteria used to judge whether particular proportions will result in the desired behavior reflect accumulated knowledge based on field and model tests, and practical eperience. Intuition and judgment are also important to this process. The traditional basis of design called elastic design is based on allowable stress intensities which are chosen in accordance with the concept that stress or strain corresponds to the ield point of the material and should not be eceeded at the most highl stressed points of the structure, the selection of failure due to fatigue, buckling or brittle fracture or b consideration of the permissible deflection of the structure. The allowable stress method has the important disadvantage in that it does not provide a uniform overload capacit for all parts and all tpes of structures. The newer approach of design is called the strength design in reinforced concrete literature and plastic design in steel-design literature. The anticipated service loading is first multiplied b a suitable load factor, the magnitude of which depends upon uncertaint of the loading, the possibilit of it changing during the life of the structure and for a combination of loadings, the likelihood, frequenc, and duration of the particular combination. In this approach for reinforced-concrete design, theoretical capacit of a structural element is reduced b a capacitreduction factor to provide for small adverse variations in material strengths, workmanship and dimensions. The structure is then proportioned so that depending on the governing conditions, the increased load cause fatigue or buckling or a brittle-facture or just produce ielding at one internal section or sections or cause elastic-plastic displacement of the structure or cause the entire structure to be on the point of collapse. Design aids The design of an structure requires man detailed computations. Some of these are of a routine nature. n eample is the computation of allowable bending moments for standard sized, species and grades of dimension timber. The rapid development of the

2 116 Rural structures in the tropics: design and development if 100N T 1 T 58 T 1 T 1 60 T T h 1 computer in the last decade has resulted in rapid adoption of Computer Structural Design Software that has now replaced the manual computation. This has greatl reduced the compleit of the analsis and design process as well as reducing the amount of time required to finish a project. Standard construction and assembl methods have evolved through eperience and need for uniformit in the construction industr. These have resulted in standard details and standard components for building construction published in handbooks or guides. Design codes Man countries have their own structural design codes, codes of practice or technical documents which perform a similar function. It is necessar for a designer to become familiar with local requirements or recommendations in regard to correct practice. In this chapter some eamples are given, occasionall in a simplified form, in order to demonstrate procedures. The should not be assumed to appl to all areas or situations. Design of members in direct tension and compression if 100N then T 1 T 100N FORCE DIGRM FOR OINT T 1 T 10º FORCE DIGRM FOR OINT h Tensile sstems Tensile sstems allow imum use of the material because ever fibre of the cross-section can be etended to resist the applied loads up to an allowable stress. s with other structural sstems, tensile sstems require depth to transfer loads economicall across a span. s the sag (h) decreases, the tensions in the cable (T 1 and T ) increase. Further decreases in the sag would again increase the magnitudes of T 1 and T until the ultimate condition, an infinite force, would be required to transfer a vertical load across a cable that is horizontal (obviousl an impossibilit). distinguishing feature of tensile sstems is that vertical loads produce both vertical and horizontal reactions. s cables cannot resist bending or shear, the transfer all loads in tension along their lengths. The connection of a cable to its supports acts as a pin joint (hinge), with the result that the reaction (R) must be eactl equal and opposite to the tension in the cable (T). The R can be resolved into the vertical and horizontal directions producing the forces V and H. The horizontal reaction (H) is known as the thrust. The values of the components of the reactions can be obtained b using the conditions of static equilibrium and resolving the cable tensions into vertical and horizontal components at the support points. Eample 7.1 Two identical ropes support a load of 5 kn, as shown in the figure. Calculate the required diameter of the rope, if its ultimate strength is 0 Ma and a safet factor of 4.0 is applied. lso determine the horizontal support reaction at B.

3 Chapter 7 Structural design B t support B, the reaction is composed of two components: B v T sin 0.5 sin kn B H T cos 0.5 cos 0.17 kn T 1 5kN T Short columns column which is short (i.e. the height is small compared with the cross-section area) is likel to fail because of crushing of the material. Note, however, that slender columns, which are tall compared with the cross-section area, are more likel to fail from buckling under a load much smaller than that needed to cause failure from crushing. Buckling is dealt with later. Free bod diagram T 4. kn Short columns 5 kn T.5 kn The allowable stress in the rope is N/mm Ma 7.5 N/mm Ma N/mm Ma Force Stress rea Force Stress Force Stress rea rea Therefore: rea required 57 mm rea required 57 mm rea required 7.5 π d 57 mm π r 7.5 4π d π r 4 Thus: 4 57 d 7 mm (min) 4 π 57 d 7 mm (min) π Slender columns Eample 7. square concrete column, which is 0.5 m high, is made of a nominal concrete mi of 1::4, with a permissible direct compression stress of 5. Ma (N / mm²). What is the required cross-section area if the column is required to carr an aial load of 00 kn? F N mm σ 5. N/mm i.e. the column should be minimum 8 mm square.

4 118 Rural structures in the tropics: design and development Design of simple beams Bending stresses When a sponge is put across two supports and gentl pressed downwards between the supports, the pores at the top will close, indicating compression, and the pores at the bottom will open wider, indicating tension. Similarl, a beam of an elastic material, such as wood or steel, will produce a change in shape when eternal loads are acting on it. N C C T T h The moment caused b the eternal loads acting on the beam will be resisted b the moment of this internal couple. Therefore: M M R C (or T) h where: M the eternal moment M R the internal resisting moment C resultant of all compressive forces on the crosssection of the beam T resultant of all tensile forces on the cross-section of the beam h lever arm of the reaction couple Now consider a small element with the area (R) at a distance (a) from the neutral ais (N). Compression f c Tension f a Figure 7.1 Bending effects on beams N a The stresses will var from imum compression at the top to imum tension at the bottom. Where the stress changes from compressive to tensile, there will be one laer that remains unstressed and this is called the neutral laer or the neutral ais (N). This is wh beams with an I-section are so effective. The main part of the material is concentrated in the flanges, awa from the neutral ais. Hence, the imum stresses occur where there is imum material to resist them. If the material is assumed to be elastic, then the stress distribution can be represented b two triangular shapes with the line of action of the resultant force of each triangle of stress at its centroid. The couple produced b the compression and tension triangles of stress is the internal-reaction couple of the beam section. Note that it is common practice to use the smbol f for bending stress, rather than the more general smbol. Maimum compressive stress (f c ) is assumed to occur in this case at the top of the beam. Therefore, b similar triangles, the stress in the chosen element is: f a, a f c fa a f c s force stress area, then the force on the element f a R a (f c / ) R The resisting moment of the small element is: force distance (a) a (f c / ) R a Ra (f c / ) f t

5 Chapter 7 Structural design 119 The total resisting moment of all such small elements in the cross-section is: M R Ra (f c / ) N C T h But Ra I, the moment of inertia about the neutral ais, and therefore M R I (f c / ) s the section modulus Z c I /, therefore M R f c Z c M; Similarl M R f t Z t M The imum compressive stress (f c ) will occur in the cross-section area of the beam where the bending moment (M) is greatest. size and shape of crosssection, i.e. its section modulus (Z), must be selected so that the f c does not eceed an allowable value. llowable working stress values can be found in building codes or engineering handbooks. s the following diagrams show, the concept of a resisting couple can be seen in man structural members and sstems. N Rectangular beams C C T h Reinforced-concrete T-beams In summar the following equation is used to test for safe bending: f w f M / Z where: f w allowable bending stress f actual bending stress M imum bending moment Z section modulus Horizontal shear The horizontal shear force (Q) at a given cross-section in a beam induces a shearing stress that acts tangentiall to the horizontal cross-sectional plane. The average value of this shear stress is: Q τ where is the transverse cross-sectional area. This average value is used when designing rivets, bolts and welded joints. The eistence of such a horizontal stress can be illustrated b bending a paper pad. The papers will slide relative to each other, but in a beam this is prevented b the developed shear stress. N h T Girders and I beams ( 1 / 6 web area can be added to each flange area for moment resistance) Sliding of laers N C h T Rectangular reinforced-concrete beams (note that the steel bars are assumed to carr all the tensile forces). No sliding of laers Figure 7. Shearing effects on beams

6 10 Rural structures in the tropics: design and development Q Q Q τ 1.5 bd Q Q Q τ 1.5 However, the shear stresses are not equal across the bd Q Q Q cross-section. t the top and bottom edge of the beam For rectangular sections τ 1.5 the must be zero, because no horizontal shear stresses Q bd Q Q Q τ 1.5 can develop. a Q τ 1.5 bd Q Q If the shear stresses at a certain distance from For square sections τ 1.5 a the neutral ais are considered, their value can be Q Q determined according to the following formula: τ Q a4q For circular sections Q τ Q τ D 1.5 π Q a 16Q 4Q τ τ I b π D For I-shaped sections of steel 16Q beams, 4Q a convenient approimation is to assume τ Q τ that all π D shearing resistance where: is afforded b the 16Q web plus 4Q the d part t of the flanges that τ t shear stress forms a continuation Q π D of τthe web. Q shear force d t area for the part of the section being sheared off Thus: Q τ perpendicular distance from the centroid of to d t Q the neutral ais For I-sections τ d t I moment of inertia for the entire cross-section b width of the section at the place where shear stress where: is being calculated. d depth of beam t thickness of web If timber and steel beams with spans normall used in buildings are made large enough to resist the tensile and compressive stresses caused b bending, the are usuall strong enough to resist the horizontal shear stresses also. However, the size or strength of short, heavil loaded timber beams ma be limited b these stresses. Centroid for area G b Deflection of beams Ecessive deflections are unacceptable in building construction, as the can cause cracking of plaster in ceilings and can result in jamming of doors and windows. Most building codes limit the amount of allowable deflection as a proportion of the member s length, i.e. 1 / 180, 1 / 40 or 1 / 60 of the length. For standard cases of loading, the deflection formulae can be epressed as: δ W K c EI Maimum horizontal shear force in beams It can be shown that the imum shear stress t in a beam will occur at the neutral ais. Thus, the following relations for the imum shear stress in beams of different shapes can be deduced, assuming the imum shear force (Q) to be the end reaction at a beam support (column). Q where: δ imum deflection (mm) M Kf c constant w f depending on the tpe of loading and the Z end support conditions W total load (N) effective span (mm) E modulus of elasticit (N/mm²) I moment of inertia (mm 4 ) It can be seen that deflection is greatl influenced b the span, and that the best resistance is provided b beams which have the most depth (d), resulting in a large moment of inertia.

7 Chapter 7 Structural design 4 Q τ τ w 16Q πd 11 Note that the effective span is greater than the clear span. It is convenient to use the centre to centre distance of the supports as an approimation of the effective span. Some standard cases of loading and resulting deflection for beams can be found later in this section. Design criteria The design of beams is dependent upon the following factors: 1. Magnitude and tpe of loading. Duration of loading. Clear span 4. Material of the beam 5. Shape of the beam cross-section W Beams Kare designed c using the following formulae: EI 1. Bending stress δ f w f M Z where: f w allowable bending stress f actual bending stress M imum bending moment Z section modulus I N I N This relationship derives from simple beam theor and M M f f and M M I N I Z N Z I I N N f f The imum bending stress will be found in the section I I N of the beam where the imum bending N Z moment occurs. Z The imum moment can be obtained from the bending-moment diagram. Q Q τ w τ. Shear Q stressq τ w τ bd For rectangular bd cross-sections: Q Q Q τ Q τ w τ w τ bd bd 4 Q 16Q τ τ 4 Q τ τ For circular cross-sections: 16Q w w πd πd For I-shaped cross-sections of steel beams Q τw τ where: t w allowable shear stress t actual shear stress Q imum shear force cross-section area ike allowable bending stress, allowable shear stress varies for different materials and can be obtained from a building code. Maimum shear force is obtained from the shear-force diagram.. Deflection In addition, limitations are sometimes placed on imum deflection of the beam (δ ): δ W K c EI Eample 7. Consider a floor where beams are spaced at 1 00 mm and have a span of mm. The beams are seasoned cpress with the following properties: f w 8.0 N/mm², t w 0.7 Ma (N/mm²), E Ma (N/mm²), densit 500 kg/m³ oading on floor and including floor is.5 ka. llowable deflection is /40 1 m 1 m 4 m 4 4 Q τ τ Q τ 16Q w τ w πd πd 16Q 1 m 1 m 1 m 1 m 1 m Q Q τw τ τw τ

8 1 Rural structures in the tropics: design and development (i) Beam loading: w 1. m.5 kn/m kn/m Choose a 100 mm b 5 mm timber. The timber required is a little less than that assumed. No ssume a 100 mm b 50 mm cross-section for the recalculations 6Z are 6 required unless it is estimated that a beams. smaller d size 6 timber would be adequate 16 mm b 100 if a smaller size had been assumed 6Z 6initiall d 6 16 mm (ii) Beam mass N/m b kn/m 6Z vi) 6 Check Z for 106 shear d d loading: 6 16 mm 16 mm b b Total w kn/m Q τ 0.4 Ma (iii) Calculate reactions and draw shear-force and Q τ 0.4 Ma bending-moment diagrams Q s the 6.4safe 10load Q 6.4 for the 10timber is 0.7 N/mm² (Ma) the τ τ 0.4 Ma 0.4 Ma section 100 is adequate 5 100in 5 resistance to horizontal shear. SFD BMD 6 4 kn W 1 kn/m 4 m wl M 6 4 kn 8 iii) Calculate imum bending moment (M ) using the equation for a simple beam, uniforml loaded (see Table 7.1) iv) Find the required section modulus (Z) w.1 4 M w 6.4 knm / Nmm M knm / Nmm 8 8 M MZ mm req Z 6 fw 8 mm req f 8 w δ v) Find a suitable beam depth, assuming 100 mm breadths: vii) Check deflection 5 to ensure W that it is less than 1 / 40 of δ the span (from Table 847.1) EI 5 W δ 84 EI 5 W 5 W δ δ 84 EI 84 EI bd where: I mm E bd Ma 100 (N/mm²) 5 I mm bd 100 bd 5 I I mm mm W.1 kn/m 4 m 1.48 kn N mm δ 9 1 mm δ 9 1 mm δ 9 1 mm 1 mm The allowable deflection, 400/ >1. The beam is therefore satisfactor. Bending moments caused b askew loads If the beam is loaded so that the resulting bending moment is not about one of the main aes, the moment has to be resolved into components acting about the main aes. The stresses are then calculated separatel relative to each ais and the total stress is found b adding the stresses caused b the components of the moment. Eample 7.4 Design a timber purlin that will span rafters.4 m on centre. The angle of the roof slope is 0 and the purlin will support a vertical dead load of 50 N/m and a wind load of 00 N/m acting normal to the roof. The allowable bending stress (f w ) for the timber used is 8 Ma. The timber densit is 600 kg/m³. From Table 6., the section modulus for a rectangular shape is Z 1 / 6 bd 6Z d 6 16 mm b ssume a purlin cross-section size of 50 mm 15 mm. Find an estimated self-load. W N/m The total dead load becomes N/m

9 W 00 N/m W 1 50 N/m w w M f N/mm 4. 5 Ma Chapter 7 Structural design f N/mm 4. 5 Ma bd Z 10 mm Find the components of the loads relative to the This size is safe, but a smaller size ma be satisfactor. main aes. Tr 50 bdmm mm. Z 10 mm W 00 N/m + 87 N/m cos N/m bd 50 bd 100 Z mm Z 10 mm 4 W 87 N/m sin N/m The actual w stress in the.4timber must be no greater 14.5 than the allowable stress. M bd Z Nmm M M f + z f w M MZ Z f z bd f Z 50 w Z 15 Z Nmm 10 mm w 5. Tr the assumed purlin size of mm. bd bd Z mm 5 Z mm 6 6 bd Z mm 6 6 bd Z 5 10 mm f N/mm 4. 5 Ma bd Z 5 10 mm bd Z 6 f 5 10 mm N/mm 4. 5 Ma bd Z 4 10 mm 6 6 bd Z 4 10 mm f N/mm 6. 4 Ma This 10 is much 10 f + closer to the 6.4allowable N/mm stress Ma To save 8 10mone, mm 75 mm should also be tried. In this 10case 10 f > f w 10 and therefore 50 mm 100 mm is chosen. f N/mm 6. 4 Ma Universal steel beams Steel beams of various cross-sectional shapes are commerciall available. Even though the properties of their cross-sections can be calculated with the formulae given in the section Design of members in direct W w M tension and compression, it is easier to obtain them 8 8 from handbook tables. These tables will also take into consideration the effect of rounded edges, welds, etc.. Calculate the bending moments about each ais for Sections of steel beams are indicated with a a uniforml distributed Wload. w M The purlin is assumed combination of letters and a number, where the letters to be a simple beam. 8 8 represent the shape of the section and the number represents the dimension, usuall the height, of the w M section in millimetres, e.g. IE 100. In the case of HE W w M 10 Nmm 8 8 sections, the number is followed b a letter indicating 8 8 the thickness of the web and flanges, e.g. HE 180B. M M z w M f + f n eample of an alternative method of notation is w 10 Z Z Nmm UB 5, i.e. a 05 mm b 10 mm universal 8 8 beam weighing 5 kg/m. The following eample demonstrates another M 14.5 M f.4 + z f Nmm 8 Z 108 w method of taking into account the self-weight of the Z Nmm structural member being designed. 8 Eample 7.5 Design a steel beam, to be used as a lintel over a door opening, which is required to span 4.0 m between centres of simple supports. The beam will be carring a 0 mm thick and. m high brick wall, weighing 0 kn/m³. llowable bending stress is 165 Ma. Uniforml distributed load caused b brickwork is kn. ssumed self-weight for the beam is 1.5 kn. (Note: the triangular load distribution for bricks above the lintel would result in a slightl lower load value). Total W uniforml distributed load W M 0.1 knm Nmm 40. kn W M 0.1 knm Nmm Z mm 1 cm req Z mm 1 cm req 165 bd

10 14 Rural structures in the tropics: design and development Suitable sections as found in a handbook would be: Section Z - Mass IN cm³ 17.9 kg/m IE cm³ 18.8 kg/m HE cm³ 4.7 kg/m HE cm³ 6.7 kg/m lthough the total value of the load has increased, the imum shear force remains the same but the imum bending is reduced when the beam is cantilevered over the supports. Choose IN 160 because it is closest to the required section modulus and has the lowest weight. Then recalculate the required Z using the IN 160 weight: N, which is less than the assumed self-weight of 1.5 kn. recheck on the required Z reveals a value of 119 cm³, which is close enough. Continuous beams single continuous beam etending over a number of supports will safel carr a greater load than a series of simple beams etending from support to support. Consider the shear force and bending moment diagrams for the following two beam loadings: +0 8 m 5 kn/m 0 kn 0 kn M 40 knm m 8 m m +0 0 kn/m kn/m M 0 knm kn/m -0 Continuous beam Simple beam lthough continuous beams are staticall indeterminate and the calculations are comple, W BM approimate values can be found with simplified equations. 6 Conservative equations for two situations are as follows: W oad concentrated between supports: BM 6 W oad uniforml distributed: BM 1 It is best to treat the two end sections as simple beams. W Standard cases of beam loading BM 1 number of beam loading cases occur frequentl and it is useful to have standard epressions available for them. Several of these cases will be found in Table 7.1. Composite beams In small-scale buildings the spans are relativel small and, with normal loading, solid rectangular or square sections are generall the most economical beams. However, where members larger than the available sizes and/or length of solid timber are required, one of the following combinations ma be chosen: 1. rranging several pieces of timber or steel into a structural frame or truss.. Universal steel beams.. Built-up timber sections with the beam members nailed, glued or bolted together into a solid section, or with the beam members spaced apart and onl connected at intervals. 4. Strengthening the solid timber section b the addition of steel plates to form a flitch-beam. 5. lwood web beams with one or several webs. 6. Reinforced-concrete beams.

11 Chapter 7 Structural design 15 Table 7.1 Beam equations oading diagram Shear force at : Q Bending moment at : M Deflection at : δ W a b a + b B Wb Q Wa Q B - Wab M c When a b W M c 4 Wa b δ c EI Total W w B W Q W Q B - W M 8 at 5W δ 84EI at W Total W w B W w Q W w Q B M 0 m. 064 at δ at w 4 EI Total W w W B W w Q 4 W w Q B w M 1 at 4 w δ 10EI at W Q QB W M -W δ B W EI Total W w B Q W Q 0 W w M - - δ B 4 W w EI EI Total W w W B Q W Q 0 W w M δ B 4 w 0EI W a b a + b B Wb Q Wa Q B - M - M B - Wab Wa b δ C Wa b EI Total W w B W Q W Q B - M M B - W 1 δ C W 84EI Total W w W B W Q W Q B - W w M W w M B δ 4 w 764EI at W W W R 1 R W Q M W * 6 w δ 19EI W R 1 W W R W Q M W * 1 w δ 84EI

12 16 Rural structures in the tropics: design and development Built-up timber beams When designing large members, there are advantages in building up solid sections from smaller pieces because these are less epensive and easier to obtain. Smaller pieces also season properl without checking. The composite beams ma be built up in was that minimize warping and permit rigid connections between columns and beams. t the same time the importance of timber defects is decreased, because the load is distributed to several pieces, not all with defects in the same area. Built-up solid column Built-up solid beam Columns lthough the column is essentiall a compression member, the manner in which it tends to fail and the amount of load that causes failure depend on: 1. The material of which the column is made.. The shape of cross-section of the column.. The end conditions of the column. The first point is obvious: a steel column can carr a greater load than a timber column of similar size. Columns with a large cross-section area compared with the height are likel to fail b crushing. These short columns have been discussed earlier. Buckling of slender columns If a long, thin, fleible rod is loaded aiall in compression, it will deflect a noticeable amount. This phenomenon is called buckling and occurs when the stresses in the rod are still well below those required to cause a compression/shearing-tpe failure. Buckling is dangerous because it is sudden and, once started, is progressive. Rafter members spaced apart Tie member Figure 7. Built-up beams and trusses Built-up solid beams are normall formed b using vertical pieces nailed or bolted together: Nailing is satisfactor for beams up to about 50 mm in depth, although these ma require the use of bolts at the ends if the shear stresses are high. Simpl multipling the strength of one beam b the number of beams is satisfactor, provided that the staggered joints occur over supports. Built-up sections with the members spaced apart are used mainl where the forces are tensile, such as in the bottom chords of a truss. Where used in beams designed to resist bending, buckling of the individual members ma have to be considered if those members have a large depth-to-width ratio. However, this can be avoided b appropriate spacing of stiffeners that connect the spaced members at intervals. Where the loading is heav, the beam will require considerable depth, resulting in a large section modulus to keep the stresses within the allowable limit. If sufficient depth cannot be obtained in one member, it ma be achieved b combining several members, such as gluing the members together to form a laminate. lthough the buckling of a column can be compared with the bending of a beam, there is an important difference in that the designer can choose the ais about which a beam bends, but normall the column will take the line of least resistance and buckle in the direction where the column has the least lateral unsupported dimension. s the loads on columns are never perfectl aial and the columns are not perfectl straight, there will alwas be small bending moments induced in the column when it is compressed. There ma be parts of the cross-section area where the sum of the compressive stresses caused b the load on the column could reach values larger than the allowable or even the ultimate strength of the material.

13 Chapter 7 Structural design 17 Therefore the allowable compressive strength δ cw is reduced b a factor k λ, which depends on the slenderness ratio and the material used. bw k λ δ cw where: bw allowable load with respect to buckling k λ reduction factor, which depends on the slenderness ratio δ cw allowable compressive stress cross-section area of the column θ Rotation When the load on a column is not aial but eccentric, a bending stress is induced in the column as well as a direct compressive stress. This bending stress will need to be considered when designing the column with respect to buckling. Slenderness ratio s stated earlier, the relationship between the length of the column, its lateral dimensions and the end fiit conditions will strongl affect the column s resistance to buckling. n epression called the slenderness ratio has been developed to describe this relationship:. Fied in position but not in direction (pinned). K l λ r r where: λ slenderness ratio K effective length factor whose value depends on how the ends of the column are fied length of the column r radius of gration (r I / ) l effective length of the column (K ) There are four tpes of end condition for a column or strut:. Fied in direction but not in position. θ Rotation side movement 4. Fied in position and in direction. 1. Total freedom of rotation and side movement like the top of a flagpole. This is the weakest end condition. The consideration of the two end conditions together results in the following theoretical values for the effective length factor (K p is the factor usuall used in practice).

14 18 Rural structures in the tropics: design and development K1 0 Both ends pinned Columns and struts with both ends fied in position and effectivel restrained in direction would theoreticall have an effective length of half the actual length. However, in practice this tpe of end condition is almost never perfect and therefore somewhat higher values for K are used and can be found in building codes. In fact, in order to avoid unpleasant surprises, the ends are often considered to be pinned (K p 1.0) even when, in realit, the ends are restrained or partiall restrained in direction. The effective length can differ with respect to the different cross-sectional aes: l K 0 l l One end fied in direction and position, the other free K0 5 K p timber strut that is restrained at the centre has onl half the effective length when buckling about the - ais as when buckling about the - ais. Such a strut can therefore have a thickness of less than its width. Both ends fied in direction and position l d l l K p One end pinned, the other fied in direction and position B. In the structural framework, the braces will reduce the effective length to l when the column -B is buckling sidewas but, as there is no bracing restricting buckling forwards and backwards, the effective length for buckling in these directions is l. Similarl, the bracing struts have effective lengths of 1/ d and d respectivel.

15 Chapter 7 Structural design 19 load of 15 kn. llowable compressive stress (σ cw ) for the timber is 5. Ma. l 15 kn l in. The leg of a frame, which is pinned to the foundation, has the effective length l but, if the top is effectivel secured for sidewas movement, the effective length is reduced to l. b d 000 l d 4. In a sstem of post and lintel where the bottom of the post is effectivel held in position and secured in direction b being cast in concrete, the effective length l. b iall loaded timber columns Timber columns are designed with the following formulae: K λ and bw k λ δ r cw Note that in some building codes a value of slenderness ratio in the case of sawn timber is taken as the ratio between the effective length and the least lateral width of the column l/b. Eample 7.6 Design a timber column that is metres long, supported as shown in the figure and loaded with a compressive 1. In this case, the end conditions for buckling about the - ais are not the same as about the - ais. Therefore both directions must be designed for buckling (Where the end conditions are the same, it is sufficient to check for buckling in the direction that has the least radius of gration). Find the effective length for buckling about both aes. Buckling about the - ais, both ends pinned: l mm Buckling about the - ais, both ends fied: l mm Table 7. Reduction factor (k λ ) for stresses with respect to the slenderness ratio for wood columns Slenderness ratio l/b l/r k λ b least dimension of cross-section; r radius of gration

16 l Rural structures in the tropics: design and development λ r. Choose a trial cross-section, which should have l 000 λ l its largest lateral dimension resisting the buckling λ r gives k λ 0.16 about d 15 r r the ais with the largest effective length. Tr 6.1 mm mm 15 mm. The section properties are: w N, sa 0 kn d b d 15 r d mm 6 50 mm² r 6.1 mm w N, sa 17 kn b 50 The allowable load with respect to buckling on the r d 15 r 14.4 mm 6.1 mm F 15 l 000 σ column with λ cross-section 1.6 Ma mm 15 mm is therefore c 9 r d kn. lthough this is bigger than the actual load, r 6.1 mm b 50 further iterations to find the precise section to carr the b 50 r r 14.4 mm 14.4 mm 15 kn are not necessar. d 15 r 6.1 mm The compressive stress in the chosen cross-section l b will be:. r Find λ the allowable load mm with regard to buckling on the r column 6.1 b 50 for buckling in both directions. F r 14.4 mm σ c 1.6 Ma 9 75 l l 000 λ b 000 λ 8 gives k λ 0.41 (from Table 7.) r 50 8 r 6.1 r mm This is much less than the allowable compressive l l stress, which makes no allowance for slenderness. λ λ 15 8gives k λ 0.16 (from Table 7.) l r r iall loaded steel columns λ 8 lr The allowable loads for steel columns with respect to l 950 λ w k λ σ c 15 w r l λ buckling can be calculated in the same manner as for λ r mm² 1 5 N timber. However, the relation between the slenderness w r mm² 5 00 N 15 l ratio and the reduction factor (k λ ) is slightl different, r λ 6.1 mm 15 as seen in Table s l r the allowable load with respect to buckling is λ smaller than 15 r the actual load, a bigger cross-section Eample 7.7 r needs 15 r to 6.1 be chosen. mm Tr 75 mm 15 mm and repeat steps l 6.1 mm Calculate the safe load on a hollow square steel stanchion λ and. 15 whose eternal dimensions are 10 mm 10 mm. The r walls of the column are 6 mm thick and the allowable r mm Section r properties: 6.1 mm compressive stress σ cw 150 Ma. The column is 15 4 metres high and both ends are held effectivel in r mm 9 75 mm² r 75 position, but one end is also restrained in direction. r mm mm 15 The effective length of the column l r 6.1 mm I BD 4 bd mm r 4 75 r 46.6 mm r 1.7 mm 1(BD bd) 1( ) 75 I BD r 1.7 mm bd r mm r 1(BD bd) 1( ) 75 r 1.7 mm l 400 Find the allowable buckling load for the new crosssection: r 46.6 λ 7 gives k λ 0.7 b interpolation l 000 l 400 λ 8 gives k λ 0.41 w k λ σ cw 0.7 λ 150 ( r ) 95 kn. r D b D D Table 7. b D 4 Reduction factor (k l λ ) for stresses with respect to the slenderness ratio for steel columns λ λ r 1.7 k λ λ k λ σ c F Ma

17 Chapter 7 Structural design Table 7.4 ermissible compressive stress ( cc ) in concrete for columns (Ma or N/mm ) Concrete mi Slenderness ratio, l/b rescribed C C C Nominal 1:: :: :1.5: Higher stress values ma be permitted, depending on the level of work supervision. l b l b σ cw σ c kf λ w σ cw l b M f + Z f w I BDiall bd loaded 10 concrete r 4 columns 46.6 mm σ Obviousl, f b the law of superposition, the added 1(BD Most bd) building 1(10 codes permit 108 ) the use of plain concrete stresses + of the 1 i.e. two load effects must be below the onl in short columns, that is to sa, columns where the allowable cw f w stress. σ f + 1 i.e. ratio of the effective length to least lateral dimension cw f w does not eceed 15, i.e. l/r 15. If the slenderness ratio is σ f between 10 and 15, the allowable compressive strength Therefore + 1 i.e. must l be reduced. 400 λ 7 The tables of figures relating to l/b in cw f w place rof a 46.6 true slenderness ratio are onl approimate, aial compressive stress bending stress + 1 as radii of gration depend on both b and allowable d values compressive in stress allowable bending stress aial compressive stress bending stress the cross-section and must be used with caution. In the + 1 case of a circular column: allowable compressive stress allowable bending stress aial compressive stress bending stress D b D, where + 1 allowable compressive stress allowable bending stress 4 σ c f + 1 which can be transferred to: k λ σ cw f w σ c f D the diameter of the column. + 1 f w 1 + K λ k λ σ cw σcw 1 11 Eample 7.9 Determine within 5 mm the required diameter of a timber post loaded as shown in the figure. The bottom of the post is fied in both position and direction b being cast in a concrete foundation. llowable stresses π for D the timber π 00 used are σ cw 9 Ma and f w 10 Ma mm kn π D π 00 Z mm e 500 F 5 kn Eample 7.8 concrete column with an effective length of 4 metres has a cross-section of 00 mm b 400 mm. Calculate the allowable aial load if a nominal 1::4 concrete mi is to be used. 000 D 00 r 50 mm 4 4 Slenderness ratio l b Hence Table 7.4 gives cc.47 N/mm² b interpolation. w cc kn. σ f Eccentricall + 1 loaded i.e. timber and steel columns cw f Where a column w is eccentricall loaded, two load effects need to be considered: The aial compressive stress caused b the load and the bending stresses caused b the eccentricit of the load. ial compressive stress bending stress + 1 able compressive stress allowable bending stress l r The load of 5 kn on the cantilever causes a bending moment of M F e 5 kn 0.5 m.5 knm in the post below the cantilever. The effective length of the post K mm. Tr a post with a diameter of 00 mm. The cross-sectional properties are:

18 K λ f w 4 Z cw 4 πd π mm 41 4 Rural structures in the tropics: design and development πd π mm 4 4 π D π 00 Z mm πd π mm 4 4 π D 00 e π Z mm π D π 00 Z 785 D r mm 50 mm 4 4 π D π 00 Z D mm r 50 mm 4 4 D 00 r 50 mm 4 4 l 6 00 The slenderness ratio 16 1 σr 50 M + cw σcw D 00 K r λ 1 σf M 50 + mm wcw Z σcw 4 4K Interpolation l 6 00 in Table λ f Z 7. gives w 1 σ M k λ cw 16 σcw K λ r f w 50 Z 1 σ cw l 6M + 00 σcw 16 K λ r Z50 f w l N/mm 16 9 N/mm r N/mm 9 N/mm N/mm 9 N/mm If.5 the 10 post + 6 has a diameter of 00 mm, it will be able to carr the 8.17 loads, N/mm but the 9 N/mm task was to determine the diameter within 5 mm. Therefore a diameter of 175 mm should also be 6tried. 00 λ λ λ 144 k λ λ N/mm > 9 N/mm N/mm > 9 N/mm N/mm > 9 N/mm This 480 diameter is too small, so a diameter of 00 mm should be chosen. It will be appreciated that the choice of effective length based N/mmon > end 9 N/mm fiit has a great effect on the solution. lain and centrall reinforced concrete walls Basicall walls are designed in the same manner as columns, but there are a few differences. wall is distinguished from a column b having a length that is more than five times the thickness. lain concrete walls should have a minimum thickness of 100 mm. Where the load on the wall is eccentric, the wall must have centrall placed reinforcement of at least 0. percent of the cross-section area if the eccentricit ratio e/b eceeds 0.0. This reinforcement ma not be included in the load-carring capacit of the wall. Man agricultural buildings have walls built of blocks or bricks. The same design approach as that shown for plain concrete with aial loading can be used. The imum allowable compressive stresses must be ascertained, but the reduction ratios can be used as before.. Eample 7.10 Determine the imum allowable load per metre of a 10 mm thick wall, with an effective height of.8 metres and made from concrete grade C 15: (a) when the load is central; (b) when the l l 800 l load is eccentric.b 0 mm. b 10. b 10. b 10 l 800 Slenderness ratio,.. b N/mm cw Interpolation gives:. cw.8. cw cw (.8.0).7 N/mm. 7 Ma (.8 5(.8.0.0) ).7 N/mm Ma.7 N/mm Ma. e 0 (.8.0).7 N/mm cw.8. 7 Ma llowable load 5 w cw b kn/m wall Ratio of eccentricit ee 0 e b b b 10 double interpolation e 0 gives: b 10 w cw cw 1.06 N/mm² 1.06 Ma llowable load cw kn/m wall w w cw kn/m wall cw kn/m wall w kn/m e Central cw reinforcement is not required because < wall b b l b ( ) M kn/m e e e 0 < 0 b < < 0 b b e < 0

19 Chapter 7 Structural design 1 Table 7.5 llowable compressive stress, cw for concrete used in walls (N/mm²) Concrete grade Slenderness Ratio of eccentricit of the load e/b or mi ratio l/b lain concrete walls Centrall reinforced concrete walls C C C :1: :1: :: :: Higher values of stress ma be permitted, depending on the level of work supervision. Trusses It can be seen from the stress distribution of a loaded beam that the greatest stress occurs at the top and bottom etremities of the beam. For these situations where bending is high but shear is low, for eample in roof design, material can be saved b raising a framework design. truss is a pinpointed framework. f c C C h N N T T f t This led to the improvement on a rectangular section b introducing the I-section in which the large flanges were situated at a distance from the neutral ais. In effect, the flanges carried the bending in the form of tension stress in one flange and compression stress in the other, while the shear was carried b the web. truss concentrates the imum amount of materials as far awa as possible from the neutral ais. With the resulting greater moment arm (h), much larger moments can be resisted. Resistance of a truss at a section is provided b: M C h T h

20 14 Rural structures in the tropics: design and development where: C T in parallel chords and: C compression in the top chord of the truss. T tension in bottom chord of a simpl supported truss. h vertical height of truss section. If either C or T or h can be increased, then the truss will be capable of resisting heavier loads. The value of h can be increased b making a deeper truss. llowable C- or T-stresses can be increased b choosing a larger cross-section for the chords of the truss, or b changing to a stronger material. framework or truss can be considered as a beam with the major part of the web removed. This is possible where bending stresses are more significant than shear stresses. The simple beam has a constant section along its length, et the bending and shear stresses var. The truss, comprising a number of simple members, can be fabricated to take into account this change in stress along its length. The pitched-roof truss is the best eample of this, although the original shape was probabl designed to shed rainwater. Roof trusses consist of sloping rafters that meet at the ridge, a main tie connecting the feet of the rafters and internal bracing members. The are used to support a roof covering in conjunction with purlins, which are laid longitudinall across the rafters, with the roof cover attached to the purlin. The arrangement of the internal bracing depends on the span. Rafters are normall divided into equal lengths and, ideall, the purlins are supported at the joints so that the rafters are onl subjected to aial forces. This is not alwas practicable because purlin spacing is dependent on the tpe of roof covering. When the purlins are not supported at the panel joints, the rafter members must be designed for bending as well as aial force. See Figure 7.. The internal bracing members of a truss should be triangulated and, as far as possible, arranged so that long members are in tension and compression members are short to avoid buckling problems. The outlines in Figure 7. give tpical forms for various spans. The thick lines indicate struts. The lattice girder, also called a truss, is a plane frame of open web construction, usuall with parallel chords or booms at top and bottom. There are two main tpes, the N- (or ratt) girder and the Warren girder. The are ver useful in long-span construction, in which their small depth-to-span ratio, generall about 1 / 10 to 1 / 14, gives them a distinct advantage over roof trusses. Steel and timber trusses are usuall designed assuming pin-jointed members. In practice, timber trusses are assembled with bolts, nails or special connectors, and steel trusses are bolted, riveted or welded. lthough these rigid joints impose secondar stresses, it is seldom necessar to consider them in the design procedure. The following steps should be considered when designing a truss: 1. Select general laout of truss members and truss spacing.. Estimate eternal loads to be applied including self-weight of truss, purlins and roof covering, together with wind loads.. Determine critical (worst combinations) loading. It is usual to consider dead loads alone, and then dead and imposed loads combined. 4. nalse the framework to find forces in all members. 5. Select the material and section to produce in each member a stress value that does not eceed the permissible value. articular care must be taken with compression members (struts), or members normall in tension but subject to stress reversal caused b wind uplift. Unless there are particular constructional requirements, roof trusses should, as far as possible, be spaced to achieve minimum weight and econom of materials used in the total roof structure. s the distance between trusses is increased, the weight of the purlins tends to increase more rapidl than that of the trusses. For spans up to around 0 m, the spacing of steel trusses is likel to be about 4 metres and, in the case of timber, metres. The pitch, or slope, of a roof depends on localit, imposed loading and tpe of covering. Heav rainfall ma require steep slopes for rapid drainage; a slope of is common for corrugated steel and asbestos roofing sheets. Manufacturers of roofing material usuall make recommendations regarding suitable slopes and fiings. urlin Ridge Internal bracing Rafter urlin occurs between joints Main tie Eaves Bending Figure 7.4 Truss components

21 Chapter 7 Structural design 15 DIN TRUSS RTT TRUSS W or BEGIN TRUSS U TO 8 m span FN TRUSS RTT TRUSS U TO 1 m span N - GIRDER WRREN GIRDER ONG SN CONSTRUCTION Figure 7.5 Tpes of trusses To enable the designer to determine the imum design load for each member, the member forces can be evaluated either b calculation or graphical means, and the results tabulated as shown: 1 8m 1 8m m Member Dead Imposed Dead + imposed Wind Design 8 0m oad oad oad oad oad D I D + I W simplified approach can be taken if the intention is to use a common section throughout. Once the laout has been chosen, the member that will carr the imum load can be established. n understanding of the problems of instabilit of compression members will lead the designer to concentrate on the top chord or rafter members. force diagram or method of sections can then be used to determine the load on these members and the necessar size. Eample 7.11 farm building comprising block walls carries steel roof trusses over a span of 8 metres. Roofing sheets determine the purlin spacings. Design the roof trusses. 0m YOUT CHOSEN (nodes at purlin points) ssume a force analsis shows imum rafter forces of approimatel 50 kn in compression (D + I) and 0 kn in tension (D + W), outer main tie member 50 kn tension (D + I) and 0 kn compression (D + W). reversal of forces caused b the uplift action of wind will cause the outer main tie member to have 50 kn of tension and 0 kn of compression.