Chapter 7 Structural design

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

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.

Chapter 7 Structural design 117 60 0 B t support B, the reaction is composed of two components: B v T sin 0.5 sin 0 1.5 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 0 7.5 N/mm 4 0 7.5 Ma 7.5 N/mm 0 7.5 Ma 4 7.5 N/mm 4 7.5 Ma Force Stress rea Force Stress Force Stress rea rea Therefore: 4. 10 rea required 57 mm 7.5 4. 10 rea required 57 mm 4. 10 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 00 000 N 56 604 mm σ 5. N/mm i.e. the column should be minimum 8 mm square.

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

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

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 1.5 16Q 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.

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 4 000 mm. The beams are seasoned cpress with the following properties: f w 8.0 N/mm², t w 0.7 Ma (N/mm²), E 8.400 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

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 0.78 10 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. 0.78 10 d 6 16 mm (ii) Beam mass 0.1 0.5 500 9.81 1.6 N/m b 100 0.1 kn/m 6Z vi) 6 Check 0.78 6Z for 106 shear 0.78 10 d d loading: 6 16 mm 16 mm b b100 100 Total w + 0.1.1 kn/m Q 6.4 10 τ 0.4 Ma (iii) Calculate reactions and draw shear-force and Q 100 5 10 6.4 τ 0.4 Ma bending-moment diagrams 100 5 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) - 6 4 w.1 4 M w 6.4 knm 6.4 10 6.1 4 / Nmm M 8 86.4 knm 6.4 10 6 / Nmm 8 8 M 6.4 10 6 MZ mm req 0.78 10 6 6.4 10 Z 6 fw 8 mm req 0.78 10 6 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 100 5 where: I 95 10 6 mm 4 1 1 E 8 400 bd Ma 100 (N/mm²) 5 I 95 10 6 mm 4 1 1 bd 100 bd 5 I 100 5 I 95 10 6 mm 95 4 10 6 mm 4 1 1 1 1 W.1 kn/m 4 m 1.48 kn 1.48 10 N 4 5 101.48 mm 10 4 10 δ 9 1 mm 84 8400 95 10 5 1.48 10 4 6 10 δ 9 1 mm 84 8400 95 10 6 5 1.48 10 5 1.48 4 10 10 9 4 10 δ 9 1 mm 1 mm 84 8400 84 95 8400 10 The allowable 6 95 10 deflection, 400/40 6 16.7 >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 6 0.78 10 d 6 16 mm b 100 1. ssume a purlin cross-section size of 50 mm 15 mm. Find an estimated self-load. W 0.05 0.15 600 9.81 7 N/m The total dead load becomes 50 + 7 87 N/m

10 10 5 10 8 W 00 N/m W 1 50 N/m w w M 8 10 10 10 f +.5 + 4.5 N/mm 4. 5 Ma Chapter 7 Structural design 10 10 5 10 1 10 10 10 f +.5 + 4.5 N/mm 4. 5 Ma 10 10 5 10 bd 50 100 Z 10 mm 8 6 6. 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 50100 100 mm. Z 10 mm 8 6 6 W 00 N/m + 87 N/m cos 0 448.5 N/m bd 50 bd 100 Z 100 50 8 10 mm Z 10 mm 4 W 87 N/m sin 0 14.5 N/m 6 6 6 4. The actual w stress in the.4timber must be no greater 14.5 than the allowable stress. M bd 50 15 Z 10 10 Nmm 8 8 10 6 6 M M f + z f w M MZ Z f 14.5.4 + z bd f Z 50 w Z 15 Z 10 10 10 Nmm 10 mm 6 8 6 w 5. Tr the assumed purlin size of 50 15 mm. bd 15 50 bd Z 50 15 10 mm 5 Z 10 10 mm 6 6 bd 506 15 Z 6 10 10 mm 6 6 bd 15 50 Z 5 10 mm 6 6 10 10 10 f +.5 + 4.5 N/mm 4. 5 Ma 10 10 5 10 bd 15 50 Z 5 10 mm bd 10 15 10 6 50 10 Z 6 f 5 10 mm +.5 + 4.5 N/mm 4. 5 Ma 10 610 56 10 bd 100 50 Z 4 10 mm 6 6 bd 100 50 Z 4 10 mm 10 10 10 f 6 + 6.9 +.5 6.4 N/mm 6. 4 Ma 8 10 4 10 10 This 10 is much 10 f + closer.9 +.5 to the 6.4allowable N/mm stress. 6. 4 Ma To save 8 10mone, 4 5010 mm 75 mm should also be tried. In this 10case 10 f > f w 10 and therefore 50 mm 100 mm is chosen. f +.9 +.5 6.4 N/mm 6. 4 Ma 8 10 4 10 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 448.5.4 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 448.5.4 M f + f n eample of an alternative method of notation is w 10 Z Z Nmm 05 10 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.4 448.5 f.4 + z f 10 10 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 0.. 4.0 0 8.7 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 40. 4.0 distributed load W 8.7 + 1.5 M 0.1 knm 0.1 10 Nmm 40. kn 6 8 8 W 40. 4.0 M 0.1 knm 0.1 10 6 Nmm 8 8 0.1 10 Z 6 0.1 10 6 mm 1 cm req 165 0.1 10 Z 6 0.1 10 6 mm 1 cm req 165 bd 50 100

14 Rural structures in the tropics: design and development Suitable sections as found in a handbook would be: Section Z - Mass IN 160 117 cm³ 17.9 kg/m IE 180 146 cm³ 18.8 kg/m HE 140 155 cm³ 4.7 kg/m HE 10 144 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: 4.0 17.9 9.81 70 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 -10 5 kn/m M 0 knm +10-10 -10 0 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.

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 - - 6 M 0 m. 064 at 0. 577 δ 0. 0065 at 0. 519 w 4 EI Total W w W B W w Q 4 W w Q B - - 4 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 - - 6 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 - - 10 0 W w M B - - 15 0 δ 4 w 764EI at 0.475 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

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.

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).

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 0 65 1. 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 0 7 0 85 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.

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 1.0 000 000 mm Buckling about the - ais, both ends fied: l 0.65 000 1 950 mm Table 7. Reduction factor (k λ ) for stresses with respect to the slenderness ratio for wood columns Slenderness ratio l/b.9 5.8 8.7 11.5 14.4 17. 0..0 6.0 8.8 4.6 40.6 46. 5.0 l/r 10 0 0 40 50 60 70 80 90 100 10 140 160 180 k λ 1.0 1.00 0.91 0.81 0.7 0.6 0.5 0.44 0.5 0.8 0.0 0.14 0.11 0.40 b least dimension of cross-section; r radius of gration

l 000 8 6.1 10 Rural structures in the tropics: design and development λ r. Choose a trial cross-section, which should have l 000 λ l 1 950 8 its largest lateral dimension resisting the buckling λ r 6.1 90 gives k λ 0.16 about d 15 r r the ais with the largest effective length. Tr 6.1 mm 1.7 50 mm 15 mm. The section properties are: w 0.41 5. 9 75 19 988 N, sa 0 kn d b d 15 r d 15 506.1 15 mm 6 50 mm² r 6.1 mm w 0.5 5. 9 75 17 06 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 1 950 λ cross-section 1.6 Ma 90 75 mm 15 mm is therefore c 9 r d 15 17 kn. lthough 751.7 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 000 50 will be:. r Find λ the allowable 14.4 8 load mm with regard to buckling on the r column 6.1 b 50 for buckling in both directions. F 15 000 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 6.1 14.4 mm This is much less than the allowable compressive l l 1 950 000 stress, which makes no allowance for slenderness. λ λ 15 8gives k λ 0.16 (from Table 7.) l r r 000 14.46.1 iall loaded steel columns λ 8 lr 1 6.1 The allowable loads for steel columns with respect to l 950 λ w k λ σ c 15 w r l 1 950 λ 0.4114.4 000 15 buckling can be calculated in the same manner as for λ r 5. 14.4 8 6 50 mm² 1 5 N timber. However, the relation between the slenderness w r 0.16 6.1 5. 6 50 mm² 5 00 N 15 l ratio and the reduction factor (k 1 950 λ ) is slightl different, r λ 6.1 mm 15 as seen in Table 7.. 4. s l r the 14.4 1 950 allowable load with respect to buckling is λ smaller than 15 r the actual load, a bigger cross-section 15 14.4 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 1 950 λ and. 15 whose eternal dimensions are 10 mm 10 mm. The r 7514.4 walls of the column are 6 mm thick and the allowable r 15 1.7 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 75 75 15 6.1 mm 9 75 mm² r 75 position, but one end is also restrained in direction. r 1.7 1.7 mm mm 15 The effective length of the column l 0.85 0.85 r 6.1 mm I BD 4 bd 000 400 10 mm. 4 108 r 4 75 r 46.6 mm r 1.7 mm 1(BD bd) 1(10 108 ) 75 I BD r 1.7 mm bd 10 4 108 r 4 46.6 mm r 1(BD bd) 1(10 108 ) 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 (10 7 6.1 r 46.6-108 ) 95 kn. r D b 4 1 0.87D D Table 7. b 1 0.87D 4 Reduction factor (k l λ ) for stresses with respect to the slenderness ratio for steel columns 1 950 λ λ 10 0 90 0 40 50 60 70 80 90 100 110 10 10 140 r 1.7 k λ 0.97 0.95 0.9 0.90 0.86 0.81 0.74 0.67 0.59 0.51 0.45 0.9 0.4 0.0 λ 150 160 170 180 190 00 10 0 0 40 50 00 50 k λ 0.6 0. 0.1 0.19 0.17 0.15 0.14 0.1 0.1 0.11 0.10 0.07 0.05 σ c F 15 000 9 75 1.6 Ma

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 10 11 1 1 14 15 rescribed C10..1.0.9.8.7 C15.9.8.7.6.5.4 C0 4.8 4.6 4.5 4. 4. 4.1 Nominal 1::5.1.0.9.8.7.6 1::4.8.7.6.5.4. 1:1.5: 4.7 4.5 4.4 4. 4.1 4.0 Higher stress values ma be permitted, depending on the level of work supervision. l b 4000 1. 00 l b 4000 1. 00 σ cw σ c kf λ w σ cw l b M f + Z f w 4000 1. 00 I BDiall bd loaded 10 concrete 4 108 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 1 0.87D, 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. 1 400 mm 4 4 0 kn π D π 00 Z 785 400 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 4000 1. 00 Hence Table 7.4 gives cc.47 N/mm² b interpolation. w cc.47 00 400 416.4 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 6 00 16 50 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 000.1 6 00 mm. Tr a post with a diameter of 00 mm. The cross-sectional properties are:

K λ f w 4 Z cw 4 πd π 00 1 400 mm 41 4 Rural structures in the tropics: design and development πd π 00 1 400 mm 4 4 π D π 00 Z 785 400 mm πd π 00 1 400 mm 4 4 π D 00 e π Z 785 400 mm π D π 00 Z 785 D 400 00 r mm 50 mm 4 4 π D π 00 Z D 00 785 400 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 0.18 + cw 16 σcw K λ r f w 50 Z 1 σ cw l 6M + 00 σcw 16 K λ r Z50 f w 0 000 9.5 10 l + 6 00 6 8.17 N/mm 16 9 N/mm 0.180 1 000 400 109 785.5 400 10 r + 50 6 8.17 N/mm 9 N/mm 0.18 1 400 10 785 400 0 000 9.5 10 + 6 8.17 N/mm 9 N/mm 1 400 10 785 400 0 000 9 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.18 1 400 10 785 400 the diameter within 5 mm. Therefore a diameter of 175 mm should also be 6tried. 00 λ 144 4.75 6 00 λ 144 4.75 6 00 λ 144 k λ 0.1 4.75 6 00 λ 144 4.75 0 000 9.5 10 + 6 N/mm > 9 N/mm 0.10 4050 000 109 167.5 480 10 + 6 N/mm > 9 N/mm 0.1 4050 10 167 480 0 000 9.5 10 + 6 N/mm > 9 N/mm 4050 10 167 This 480 diameter is too small, so a diameter of 00 mm should be chosen. It will be appreciated that the choice 0 000 9.5 10 + 6 of effective length based N/mmon > end 9 N/mm fiit has a great effect 0.1 4050 10 167 480 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 800 800 is eccentric.b 0 mm. b 10. b 10. b 10 l 800 Slenderness ratio,.. b 10.8.0.7 N/mm cw.8. 7 5 Interpolation gives:. cw.8. cw cw.8.8 5 5. (.8.0).7 N/mm. 7 Ma (.8 5(.8.0.0) ).7 N/mm Ma.7 N/mm. 7. 7 Ma. e 0 (.8.0).7 N/mm cw.8. 7 Ma 0.167 llowable load 5 w cw 1.0 0.1 b 10.7 10 6 7.4 kn/m wall Ratio of eccentricit ee 0 e 0 0 0.167 b 10 0.167 b 10 0.167 b 10 double interpolation e 0 gives: 0.167 b 10 w cw 1.0 0.1 cw 1.06 N/mm² 1.06 Ma 800 10 llowable load 1.06 10 6 1.0 0.1 1.06 1.06 10 10 6 cw 1.0 0.1 6 17. kn/m wall w w cw 1.0 0.1 1 000 17. kn/m wall cw 1 000 17. kn/m wall 1 000 1.06 10 6 w 1.0 0.1 17. kn/m e Central cw reinforcement is not required because < wall 0 1 000 b b l b ( ) M 1.06 10 6 17. kn/m 1 000 e e e 0 < 0 b < < 0 b b e < 0

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 0.00 0.10 0.0 0.0 0.40 0.50 C0 5.4 1.7 0.9 - - - 0.. 1.4 0.8 0.4 0. 15 4.1.0.0 0.8 0.4 0. 10 4.8.7.7 0.8 0.4 0. C15 5.0 1. 0.7 - - - 0.8 1.9 1.1 0.7 0.5 0.5 15.4.4 1.7 0.7 0.5 0.5 10.9.0. 0.7 0.5 0.5 C10 0. 1.6 1.0 0.5 0. 0. 15.7.0 1.4 0.5 0. 0. 10..5 1.8 0.5 0. 0. 1:1: 5. 1.6 0.8 - - - 1:1: 0.. 1. 0.8 0.4 0. 15 4.1.9 1.9 0.8 0.4 0. 10 4.7.6.6 0.8 0.4 0. 1:: 0.0.1 1. 0.7 0.5 0.5 15.7.7 1.9 0.7 0.5 0.5 10 4..4.5 0.7 0.5 0.5 1::4 0.7 1.8 1.0 0.6 0. 0. 15.. 1.6 0.6 0. 0. 10.8.9.1 0.6 0. 0. 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

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

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 0 8 0 75m 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.

16 Rural structures in the tropics: design and development Consulting a structural engineering handbook reveals that a steel angle with a section of 65 mm 50 mm 6 mm and an effective length of 1.8 m can safel carr 9 kn in compression. Rafter: Using two angles back-to-back will be satisfactor because the distance between restraints is onl 1.8 m. (Note that angles must be battened together along the length of the rafter). Main Tie: The 65 mm 50 mm 6 mm section can carr the required tensile force. lthough its length is a little greater than 1.8 m, the compressive load brought about b wind uplift is safe as the design codes allow a greater slenderness ratio for intermittent loads such as wind. Finished Design: Note the use of a sole plate to safel distribute the load to the blockwork wall to ensure that the bearing stress of the blocks is not eceeded. See Figure 7.4. Frames part from the roof truss, there are a number of other structural frames commonl used in farm building construction. The include portal frames, pole barns and post-and-beam frames. single-ba portal frame consists of a horizontal beam or pitched rafters joined rigidl to vertical stanchions on either side to form a continuous plane frame. For design purposes, portal frames can be classified into three tpes: fied base, pinned base (two pins), pinned base and ridge (three pins). The rigid joints and fied bases have to withstand bending moments and all bases are subjected to horizontal, as well as vertical, reactions. Hence foundation design requires special attention. The eternall applied loads cause bending moments, shear forces and aial forces in the frame. ortal frames are staticall indeterminate structures and the compleit of the analsis precludes coverage here. However, the results of such calculations for a number of standard cases of loading are tabulated in handbooks. Using these and the principle of superposition, the designer can determine the structural section required for the frame. Determining the imum values of the bending moment, shear force and aial force acting anwhere in the frame allows the selection of an adequate section for use throughout the frame. Care must be eercised to ensure that all joints and connections are adequate. 80 50 50 160 1 mm holes in 6 mm thick soleplate 150 100 Rafter -65 60 6 50 1 75 1 75 800 750 10 itch 50 50 6 ongitudinal ties 050 950 Notes: ll welds to be 4mm fillet ll bolts to be M16 Gusset plates to be 8mm thick Internal bracing shown 65 50 6 to use common section (size can be reduced if others available) ll sections in grade 4 steel urlin supports: 70 70 6 with 6 ø holes Figure 7.6 Finished design of the roof truss

Chapter 7 Structural design 17 ortal frames ma be made of steel, reinforced concrete or timber. With wider spans the structural components become massive if timber or reinforced concrete is used. Hence, steel frames are most common for spans over 0 m. t the eaves, where imum bending moments occur, the section used will need a greater depth than at other points in the frame. Eample 7.1 Design the roof of a building using block walls, timber posts and rafters as shown in the figures below. Design section Knee braces Timber posts RC ORT FRME It is assumed that the knee braces reduce the effective span of the rafters between the central wall and the timber posts. The moments and forces involved are as shown in the diagram below. Self-weights and service load have been estimated. Continuit over post and brace has been disregarded. This provides a simple but safe member. 77 kn 77 kn 77 kn TIMBER ORT FRME 1 8 m 1 8 m 0 74 m Figure 7.7 ortal or rigid frame post 5 0 6 knee brace S.F.D. (kn) -0 51-8 ole barns are usuall built with a relativel simple foundation, deeper than usual, and backfilled with rammed earth. ole barns are braced between columns and rafters in each direction. The braces serve to reduce the effective length of compression members and the effective span of rafters and other beam members. This leads to a structure that is simple to analse and design, and can be a low-cost form of construction. shed-tpe building is a simple construction consisting of beams (horizontal or sloping), supported at their ends on walls or posts. There ma be one or more intermediate supports depending on the width of the building. urlins running longitudinall support the roof cover. s the principal members are simple or continuous beams (ver often timber of rectangular section), the stress-analsis aspect of the design is straightforward. When the beam is supported b timber posts, the design of the posts is not difficult because the load is assumed to be aial. ike the poles in the pole barn, the foundation can consist of a simple pad of concrete beneath the post, or the base of the post can be set into concrete. 1 19 B.M.D. (knm) Self-weights and service load have been estimated. Continuit over post and brace have been disregarded. This provides a simple safe member. Maimum shear force 5 kn Maimum bending moment 10 kn/ mm². Tr two rafters at 8 00 (back to back) Q Maimum shear stress bd 5 000 76 00 0.49 N/mm 0.49 Ma M 4 7 M Maimum bending stress I Z 10 10 6 76 00 6. N/mm 6. Ma

M M ending stress 18 Rural structures in the tropics: design and development I M Z bending stress M 10 10 I Z 6 6. N/mm 6. Ma 76 00 10 10 6 6. N/mm 6. Ma Tables 76 of 00allowable stresses indicate that most joining area than is possible with lapped members. This hardwoods, but not all softwoods, are adequate. is often an important factor in nailed and glued joints. rrangement of members on a single centre line is The load transferred to the outer wall b rafters is a usuall possible with gussets. little over kn. ssuming that the strength of the When full-length timber is not available for a blocks is at least.8 Ma (N/mm²), the area required is: member, a butt joint with cover plates can be used to 000 1 07 mm join two pieces together. This should be avoided, if.8 possible, for the top members (rafters) of a truss and 000 1 07 mm positioned near mid-span for the bottom member.8 (main tie). s the rafter underside is 76 mm, the minimum interface across the wall is: 1 07 14 mm 76 1 07 14 mm 76 Hence there is no problem of load transfer to the wall. ssume posts are 100 mm 100 mm and.5 m long, then l / b 5 and Table 7. gives K 0.8 With σ c 5. Ma (N/mm²) allowable for design, 0.8 5. N/mm² 100 0 kn. lwood gussets Cover plates Therefore the load is within the safet margin. Connections Timber structure The methods used to join members include lapped and butt connectors. Bolt and connector joints, nailed joints and glued joints, and sometimes a combination of two tpes, are eamples of lapped connections. Butt connections require the use of plates or gussets. In all cases the joints should be designed b calculating the shear forces that will occur in the members. If two members overlap, the joint is called a singlelap joint. If one is lapped b two other members, i.e. sandwiched between them, it is called a double-lap joint. With a single lap, the joint is under eccentric loading. While for small-span trusses carring light loads this is not significant, when the joints carr large loads eccentricit should be avoided b the use of double-lap joints. Double members are also used to obtain a satisfactor arrangement of members in the truss as a whole. Sandwich construction enables the necessar sectional area of a member to be obtained b the use of relativel thin timbers, with an double members in compression being blocked apart and fied in position to provide the necessar stiffness. Butt joints The use of gussets permits members to butt against each other in the same plane, avoids eccentric loading on the joints and, where necessar, provides a greater Figure 7.8 Butt joints Bolt and connector joints Simple bolted joints should onl be used for lightl loaded joints because the bearing area at the hole (hole diameter member thickness) and the relativel low bearing stress allowed for the timber compared with that of the steel bolt, ma cause the timber hole to elongate and fail. Timber connectors are metal rings or toothed plates used to increase the efficienc of bolted joints. The are embedded half into each of the adjacent members and transmit loads from one to the other. The tpe most commonl used for light structures is the toothed-plate connector, a mild-steel plate cut and stamped to form triangular teeth projecting on each side that embed in the surfaces of the members on tightening the bolt that passes through the joint. The double-sided toothed connector transmits the load and the bolt is assumed to take no load. Glued joints Glues made from snthetic resins produce the most efficient form of joint, as strong as or even stronger than the timber joint, and man are immune to attack b dampness and deca. With this tpe of joint,

Chapter 7 Structural design 19 all contact surfaces must be planed smooth and the necessar pressure applied during setting of the glue. Bolts or nails that act as clamps are often used and left in place. The members ma be glued directl to each other using lapped joints, or single-thickness construction ma be used b the adoption of gussets. s with nailed joints, lapped members ma not provide sufficient gluing area and gussets must then be used to provide the etra area. Glued joints are more often used when trusses are prefabricated because control over temperature, joint fit and clamping pressure is essential. For home use, glue is often used together with nail joints. lwood gusset Figure 7.10 Truss gussets Figure 7.9 Double-sided toothed plate connector Nailed joints lthough joining b nails is the least efficient of the three methods mentioned, it is an inepensive and simple method, and can be improved upon b using glue in combination with the nails. When trusses are prefabricated in factories, nailing plates are often used to connect the member. These fasteners come in two tpes: 1. thin-gauge plate called a pierced-plate fastener, which has holes punched regularl over its surface to receive nails. The pierced plate can also be used for on-site fabrication.. heavier plate with teeth punched from the plate and bent up 90 degrees, called a toothed-plate fastener, or connector. This tpe, in which the teeth are an integral part of the plate, must be driven in b a hdraulic press or roller. ierced plate fastener Toothed plate fastener Figure 7.11 Nailing plates for truss construction

140 Rural structures in the tropics: design and development Table 7.6 Minimum nailing distances r o FORCE d 1 d 1 r o e b d 11 d 11 e b oaded edge of member r o r o Nailing area r o oaded end member 1 r b Nailing area r o d 1 d 11 r b e 0 e b 0 5d 5d 10d 5d - 15d 10 5d 5d 10d 5.5d 8d 15d 0 5d 5d 10d 6d 8d 15d 0 5d 5d 10d 6.5d 8d 15d 40 5d 5d 10d 7d 8d 15d 50 5d 5d 10d 7.5d 8d 15d 60 5d 5d 10d 8d 8d 15d Nailing area r o r o FORCE r o r b oaded edge oaded end member 1 d : Diameter of the nail (mm). r 0 : Distance from the etreme row of nails to the unloaded edge of member. d 1 : Distance between two nails in the nailing area, measured perpendicular to the ais of the member. d 11 : Distance between two nails measured parallel to the ais of the member. r b : Distance from the etreme row of nails to the loaded edge of the member. e 0 : Distance from the nearest row of nails to the unloaded end of member. e b : Distance from the nearest row of nails to the loaded end of the member. In order to permit the development of the full load at each nail and to avoid splitting the wood, minimum spacing between nails and distances from the edges and ends of the member are necessar. Nailing patterns for use on timber structures are usuall available locall. The depend on the qualit and tpe of nails and timber used, and are based on the safe lateral nail load. The Housing Research and Development Unit of the Universit of Nairobi investigated timber nailed joints made with spacings in accordance with the continental standard for timber joints, which proved to be satisfactor. The main principles are given in tables 7.6 and 7.7. Connections in steel structures Connections ma be bolted, riveted or welded. The principal design considerations are shear, tension and compression, and the calculations are relativel straightforward for the tpes of design covered. Stabilit Stabilit problems in a building are chiefl the result of horizontal loads, such as those resulting from wind pressure, storage of granular products against walls, soil pressure against foundations and sometimes earthquakes. Overturning of foundation walls and foundation piers and pads is counteracted b the width of the footing and the weight of the structure. Onl in special cases will it be necessar to give etra support in the form of buttresses. Overturning of eternal walls is counteracted b the support of perpendicular walls and partitions. Note, however, that not all tpes of wall, for eample framed walls, are adequatel rigid along their length without diagonal bracing. If supporting walls are widel spaced and/or the horizontal loads are large, etra support can be supplied b the construction of piers, columns or buttresses. See Chapter 8. Diagonal bracing is used to make framed walls and structures stiff. ong braces should preferabl transfer the load with a tensile stress to avoid buckling. Braces

Chapter 7 Structural design 141 Table 7.7 Basic lateral loads per nail Continental nail diameter (mm).1.4.8.1.4.8 4. 4.6 5.1 5.6 6.1 Kena nail diameter (mm) 1.8.0.65.5.75 4.5 5.6 6.0 Basic lateral nail load (N) 90 10 140 190 0 50 10 50 70 40 450 540 600 60 750 880 960 1 000 (If the case of pre-bored nail holes 0.8 times nail diameter, the lateral nail load can be increased b 5%) Connections in single shear at bottom boom of truss Bolts in shear and tension at ridge of portal frame Figure 7.1 Connections for steel frames are usuall supplied in pairs, i.e. on both diagonals, so that one will alwas be in tension independentl of the wind direction. If the framed wall is covered with a sheet material, such as plwood, chipboard or metal sheets, the lateral forces on the frame can be counteracted b shear in the sheets. This design requires the sheets to be securel fied to the frame, both horizontall and verticall. The sheets must be strong enough to resist buckling or failure through shear. the frames ma need etra support from longitudinal bracing. Tension rods are frequentl used. Wind force fternate direction of wind force Braces Sheet material thoroughl fied to the frame Figure 7.1 Bracing for portal frame Masonr and concrete walls that are stiff and capable of resisting lateral wind loading are called shear walls. ortal or rigid frame buildings are normall stable laterall, when the wind pressure acts on the long sides. However, when the wind loads occur at the gable ends, ost-and-beam or shed-frame buildings will, in most cases, require wind bracing, both along and across the building because there are no rigid connections at the top of the wall to transfer loads across and along the building. The same applies to buildings emploing roof trusses. End bracing should be installed.

14 Rural structures in the tropics: design and development Walls with long spans between the supporting crosswalls, partitions or buttresses tend to bend inwards under the wind load, or outwards if bulk grain or other produce is stored against the wall. t the bottom of the wall this tendenc is counteracted b the rigidit of the foundation (designed not to slide) and the support of a floor structure. The top of the wall is given stabilit b the support of the ceiling or roof structure, or a speciall designed wall beam that is securel anchored to the wall. The designer must consider the abilit of the building to withstand horizontal loading from an and all directions, without unacceptable deformation. Retaining walls Wall failure Walls are commonl used to retain soil on sloping sites, water in a pond or bulk products within a storage area. There are several limiting conditions which, if eceeded, can lead to the failure of a retaining wall. Each must be addressed in designing a wall. 1. Overturning: This occurs when the turning moment resulting from lateral forces eceeds that eerted b the self-weight of the wall. The factor of safet against overturning should be at least two.. Bearing on ground: The normal pressure between the base of the wall and the soil beneath can cause a bearing failure of the soil, if the ultimate bearing capacit is eceeded. Usuall the allowable bearing pressure will be one-third of the ultimate value. Note that the pressure distribution across the base is not constant. Bearing pressure 4. Rotational Slip: The wall and a large amount of the retained material rotate about point O if the shear resistance developed along a circular arc is eceeded. The analsis is too comple to include here. Rotation Overturning. Sliding: The wall will slide if the lateral thrust eceeds the frictional resistance developed between the base of the wall and the soil. The factor of safet against sliding should be about two. 5. Wall material failure: The structure itself must be capable of withstanding the internal stresses set up, that is to sa, the stresses must not eceed allowable values. Factors of safet used here depend on the material and the level of the designer s knowledge of the loads actuall applied. Naturall, both shear and bending must be considered, but the most critical condition is likel to be tension failure of the front facet. Sliding Joint failure in blockwork

Chapter 7 Structural design 14 Gravit walls and dams are dependent on the effect of gravit, largel from the self-weight of the wall itself, for stabilit. Other tpes of wall rel on a rigid base, combined with a wall designed against bending, to provide an adequate structure. 7H / 90 H Tension bending failure Eample 7.1 Design of a gravit wall retaining water 0 6 m ressure eerted b retained material iquid pressure The pressure in a liquid is directl proportional to both the depth and the specific weight of the liquid (w) which is the weight per unit volume, w ρg (N/m³), where: ρ densit of liquid (kg/m³) g gravitational acceleration (9.81 m/s ) The pressure at a given depth acts equall in all directions, and the resultant force on a dam or wall face is normal to the face. The pressure from the liquid can be treated as a distributed load with linear variation in a triangular load form, with a centroid two-thirds of the wa down the wet face. wh p ρgh wh (N/m ) and: wh H / acting at a depth of Free surface H It should be noted that a wall retaining a material that is saturated H (waterlogged) must resist this liquid pressure in addition to the lateral pressure from the retained material. H iquid m 1 8 m Consider a mass concrete dam with the crosssection shown, which retains water to m depth. ssume: Ground safe bearing capacit: 00 ka. Coefficient of sliding friction at base: 0.7. Specific weight of concrete: kn/m³. wh 9.8 10 44.1 kn 1. Find water force : ll calculations per metre length of wall: wh 9.8 10 44.1 kn wh 9.8 10 44.1 kn (acting one metre up face) W specific weight. Find mass ( 0.6 of one + 1.8 metre ) length 8.8of knwall: W specific weight W ( 0.6 specific + 1.8weight ) 8.8 kn ( 0.6 + 1.8) 8.8 kn. Find line of action of w: Taking moments of area about vertical face: 1 X 1 + X X 1 + ( 0.6 0.) + (0.6 1.0) 0.65 m 1 X 1 + X 1.8 + 1.8 X 1 X X X 1 + ( 0.6 1 + 0.) + (0.6 1.0) 0.65 m ( 0.6 1.8 0.) + + 1.8 (0.6 1.0) 0.65 m 1.8 + 1.8 E D

144 MI σ Rural structures in the tropics: design and development b b Y W 8.8 c 46 ka 1 1.8 Hence the self-weight of the wall acts 0.5 m to left of the base Wcentre 8.8 line. c 46 ka 1 1.8 4. Find the vertical compressive stress on the base: 10. Check overturning σ b W bd8.8 1 1.8 c I 46 ka 0.486 m 4 1 1 1.8 W 8.8 MI 1 c σ b b 46 ka 1 1.8 Y 5. Find the moment about the centre line of the base M (1 44.1) -(0.5 MI σ b b 8.8); (clockwise)-(anticlockwise) M.4 knm Y 1.8 MI 6. Find σ b the Y b bending ± stresses/pressures ± 0.9 m 1 Y Y bd 1 MI 1.8 I σ b b 0.486 where: m 4 1 bd 1 1.8 I W 0.486 M m 4 1 σ 1 + bd 1 1.8 I I 0.486 m 4 1 1 1.8 Y ± ± 0.9 m bd 1 1.8 I 0.486 m 4 1 1.4 1.8 σ Y ± 0.486 b b ± ± 0.9 m 1.6 ka 1.8 Y ± ± 0.9 0.9m σ b 7. Find the Wactual Mstresses/pressures σ + I 1.8 W M Y σ ± + ± 0.9 m I W WM σ M σ + σ E E 46 + 1.6 I I 58.6 ka (compression) σ D D 46-1.6.4 ka (compression).4 0.486 (Note: ± Compression 1.6 onl ka indicates the resultant, and b W would 0.9 intersect the base line within its middle third). W M σ + 8. Compare.4 σ ± 0.486 imum I pressure with allowable bearing b b.4 1.6 ka 0.9 0.486 σ b capacit: b ± 1.6 ka 0.9 58.6 ka This is less than the allowable safe bearing capacit of the soil. Hence the wall-soil interface is safe against.4 0.486 b ± bearing failure. 1.6 ka 0.9 9. Compare actual stresses in the wall with allowable values: 44 1 kn 1 m Overturning moment about D 44.1 1 44.1 knm Stabilising moment about D 8.8 1.15 95. knm Factor of safet against overturning 94. / 44.1.16 The wall is safe against overturning. 11. Check sliding 0 65 1 15 W Frictional resistance mw mw 0.7 8.8 58 kn Horizontal thrust 44.1 kn W 8 8 Fμ W s the required factor against sliding is, there is a deficienc of ( 44.1) - 58 0. kn. dditional anchorage against sliding should be provided. Eample 7.14 Design a circular water tank with the following dimensions/properties: Diameter 5 m, depth of water m Water weighs 9.8 10 N/m³ ressure () at a depth of m Maimum stress 58.6 ka (compression) and no tensile stress at an point across wall. Hence the wall material is safe.

Chapter 7 Structural design 145 wh 9.8 10 9.4 ka This acts verticall over the entire base; therefore the base should be designed for a uniforml distributed load (UD) of 9.4 ka. For a dr material, the angle of repose is usuall equal to the angle of shearing resistance of the material. This angle of shearing resistance is called the angle of internal friction (θ). The angle of friction is the essential propert of a granular material on which Rankine s theor is based. This theor enables the lateral pressure to be epressed as a proportion of the vertical pressure, which was shown (before) H 9.4 to depend on specific weight and depth onl. 44.1 kn/m In this case, at a depth h the active lateral pressure is given b: H 9.4 44.1 kn/m k w h where: k a constant dependent on the materials involved. lthough there 1 sinis θ k some friction between the retained material and 1 + the sinwall θ face, usuall this is disregarded, giving a relativel simple relationship for k: 1 sinθ k 1 + sinθ where: θ the angle of friction 1 sinθ p wh(n/m a ) 1 + sinθ ressure also acts laterall on the side wall at its where: bottom edge. This pressure decreases linearl to zero at 1 p sin a θtotal force per metre of wall-face (N) the water surface. p wh(n/m a ) 1 + sinθ1 sinθ wh ( 5 ) Total force on base B 9.4 π a (N/m length of wall) 577. kn 1 + sinθ 4 (acting at the centre of the base) a total force per metre of wall face (N) Total force on the side per metre of perimeter wall: H 9.4 44.1 kn/m run (acting one metre above base) ressure eerted b granular materials Granular materials such as sand soils, gravell soils and grain possess the propert of internal friction (friction between adjacent grains), but are assumed not to possess the propert of cohesion. If a quantit of such material 1 sinin θ a dr condition is tipped on to a flat surface, k 1 + it sinwill θ form a conical heap: the shape maintained b this internal friction between grains. The angle of the sloping side is known as the angle of repose. This gives the approimate horizontal resultant force on a vertical wall face when it is retaining material that is level with the top of the wall. If the surface of the retained material is sloping up from the wall at an angle equal to its angle of repose, a modification is required. Eample 7.15 1 sinθ wh Wall retaining soil 1 + sinθ SOI 1 sin5 18.6 1 + sin5 kn/m Timber beams ngle of reponse 1 sinθ p wh(n/m a ) 1 + sinθ Steel posts set in concrete

1 sinθ wh 146 a 1 + sinθ Rural structures in the tropics: design and development 1 sinθ wh Consider the wall shown retaining loose sand soil to 1 + sinθ a depth of metres. Tables provide angle of friction equal to 5 and specific weight equal to 18.6 kn/m³. ssuming a smooth vertical surface and horizontal soil surface, Rankine s theor gives: m 1 sinθ wh 1 + sinθ S 4 m 1 sin5 18.6 kn/m length of wall 1 + sin5 10.1 kn/m length of wall. If steel posts are placed at.5 m centres, each post can S tan 45 + θ be approimated to a vertical cantilever beam.5 m 1 sin5 18.6 long, carring a total kn/mdistributed load of 10.1.5 1 + sin5 5.5 kn of linear variation from zero at the top to a imum at the base. The steel post and foundation S 4 concrete must be capable of resisting the applied load, tan 45 + θ tan 45 + 1.5 principall in bending but also in shear. Critical height is: Eample 7.16 Grain storage bin (The theor given does not appl to deep bins). shallow bin can be defined as one with a sidewall height of less than S θ tan 45 + for a square bin of side length S. Consider a square bin of side length 4 metres retaining shelled maize/corn to a depth of metres. ssume θ 7 ; specific weight is 7.7 kn/m³. 4 tan ( 45 + 1.5 ) 1.6.6 m RESSURE DIGRM ( ) 1.6.6 m The timber crossbeams can be analsed as beams simpl supported over a span of.5 m, Seach carring 4 S 4 tan tan 45 + θ 45 + θ tan( 45 1 + 1.5 sinθ tan ) 1.6.6 m a uniforml distributed load. This load is equal to the ( 45 + 1.5 ) 1.6.6 m wh 1 + sinθ product of the face area of the beam and the pressure in the soil at a depth indicated b the centroid of the area Design in the same wa as the shallow bin because the of the beam face. depth of grain 1 is sin onl θ wh metres. 1 + sinθ Maimum pressure at the base of the wall: 1 sinθ p a wh 1 + sinθ 1 sin7 7.7 5.78 kn/m 1 sinθ 1 sinθ 1 + sin7 wh wh 1 + sinθ if beam face is 0. m high, 1 + sinθ 1 sin7 h.0 0.15 1.85 m 7.7 5.78 kn/m 1 + sin7 0.7 18.6 1.85 9.9 kn/m 5.78 S θ or resultant force 11.56 kn/m Total tan uniforml 45 + distributed for a square load bin on of the side beam length S. 1 sin7 1 sin7 7.7 5.78 kn/m 9.9 0..5 6.97 kn 7.7 1 5.78 + sin7 kn/m 1 + sin7 (acting / m above the base of the wall). The 1imum sinθ bending moment at the centre of the Note 5.78 p wh span a that the design 11.56 of the kn/m wall is comple if it consists 1can + sinbe θ determined and the beam section checked. of a plate of uniform thickness, but if the wall is thought of as comprising a number of vertical members cantilevered from the floor, an approach similar to that 5.78 for the soil-retaining 5.78 wall can be used. 11.56 kn/m 11.56 kn/m Designing for earthqukes In areas where earthquakes occur frequentl, buildings must be designed to resist the stresses caused b tremors. While the intensit of tremors can be much greater in loosel compacted soil than in firm soil or solid bedrock, one- and two-store buildings are at greater risk on firm ground or bedrock because of the shorter resonance periods.

Chapter 7 Structural design 147 Casualties are most likel to be caused b the collapse of walls causing the roof to fall, and the failure of projecting elements such as parapets, water tanks, non-monolithic chimnes and loose roof coverings. Outbreaks of fire caused b fractures in chimnes or breaks in mains suppl lines present an additional hazard. While small buildings with timber frame walls, or a wooden ring beam supported b the posts of a mudand-pole wall, can resist quite violent earthquakes, the following measures will increase the resistance of a large building to collapse from earth tremors: Use a round or rectangular shape for the building. Other shapes such as T or U should be divided into separate units. To be effective, this separation must be carried down through to the foundation. void large spans, greatl elongated walls, vaultand-dome construction and wall openings in ecess of one-third of the total wall area. Construct a continuousl reinforced footing that rests on uniform soil at a uniform depth even on sloping ground. Fi the roof securel, either to a continuousl reinforced ring beam on top of the walls, or to independent supports, which will not fail even if the walls collapse. void projecting elements, brittle materials and heav materials on weak supports. void combustible materials near chimnes and power lines. Ductile structures have man joints that can move slightl without failing, e.g. bolted trusses. Such structures have a greater capacit to absorb the energ of earthquake waves. smmetrical, uniforml distributed ductile framework with the walls securel fied to the frame is suitable for large buildings. Masonr walls are sensitive to earthquake loads and tend to crack through the joints. It is therefore important to use a good mortar and occasionall reinforcing will be required. Review questions 1. Define structural design.. Briefl describe the structural design process.. Wh is it important to take into account deflection of structural elements during design phase? 4. Outline factors that influence design of beams. 5. Which measures improve the resistance of buildings to earthquake? 6. Calculate section moduli for a T section, flange 150 mm b 5 mm, web thickness 5 mm and overall depth, 150 mm. 7. 10 m long T section beam is simpl supported, with the flange uppermost, from the right-hand end and at a point.5 m from the left-hand end. The beam is to carr a uniforml distributed load of 8 kn/m over the entire length. The allowable flange and web thickness is 5 mm. If the allowable imum tensile strength and compressive stress are 15 Ma and 70 Ma respectivel. Determine the size of the flange. 8. short hollow clindrical column with an internal diameter of 00 mm and eternal diameter of 50 mm carries a compressive load of 600 kn. Find the imum permissible eccentricit of the load if (a) the tensile stress in the column must not eceed 15 Ma; (b) the compressive stress must not eceed 76 Ma. 9. Design a section of a trapezoidal masonr retaining wall 10 metres high, to retain earth weighing 16 000 N/m. The angle of repose for the earth is 5 when the earth surface is horizontal and level with the top of the wall. The weight of the masonr wall is 5 000 N/m. 10. reinforced concrete beam is 00 mm wide, has an effective depth of 450 mm and four 0 mm diameter reinforcing bars. If the section has to resist a bending moment of 50 10 6 N mm, calculate the stresses in steel and concrete. The modular ratio of steel to concrete is equal to 18. Further reading l Nageim, H., Durka, F., Morgan, W. & Williams, D.T. 010. Structural mechanics: loads, analsis, materials and design of structural elements. 7 th edition. ondon, earson Education. Nelson, G.., Manbeck, H.B. & Meador, N.F. 1988. ight agricultural and industrial structures: analsis and design. VI Book Co. rasad, I.B. 000. tet book of strength of materials. 0 th edition. B, Nath Market, Nai Sarak, Delhi, Khanna ublishers. Ro, S.K. & Chakrabart, S. 009. Fundamentals of structural analsis with computer analsis and applications. Ram Nagar, New Delhi, S. Chand and Co. td. Salvadori, M. & Heller, R. 1986. Structure in architecture: the building of buildings. rd edition. Englewood Cliffs, New Jerse, rentice-hall. Whitlow, R. 197. Materials and structures. New York, ongman.