Unit 48: Structura Behaviour and Detaiing for Construction 4.1 Introduction Defection of Beams This topic investigates the deformation of beams as the direct effect of that bending tendency, which affects their serviceabiity and stabiity, and does so in terms of their defection. beam may be strong enough to resist safey the bending moments due to the appied oading and yet not be suitabe because its defection is too great. Excessive defection might not ony impair the strength and stabiity of the structure but aso give rise to minor troube such as cracking of paster ceiings, partitions and other finishes, as we as adversey affecting the functiona needs and aesthetic requirements or simpy being unsighty. The reevant BS specifications and codes of practice stipuate that the defection of a beam sha be restricted within imits appropriate to the type of structure. In the case of structura steework the maximum defection due to unfactored imposed oads for beams carrying paster or other britte finish must not exceed 1/360 of the span, but for a the other beams it may be span/200 (Cause 2.5.1 of BS 5950: Part 1: 2000). For timber beams, on the other hand, the figure is 0.003 of the span when the supporting member is fuy oaded (Cause 2.10.7 of BS 5268: Part 2: 1996). In reinforced concrete the defection is generay governed by the span/depth ratio (Cause 3.4.6.3 of BS 8110: Part 1: 1997). 4.2 Factors affecting defection For many beams in most type of buidings, e.g. fats, offices, warehouses, it wi usuay be found that, if the beams are made big enough to resist the bending stresses, the defections wi not exceed the permitted vaues. In beams of ong spans, however, it may be necessary to cacuate the defections to ensure that they are not excessive. The derivation of formuae for cacuating defections usuay invoves cacuus. In this chapter, therefore, ony a genera treatment wi be attempted and defection formuae for a few common cases of beam oadings wi be given without proof. Genera methods of cacuation of defections are given in standard books on theory of structures or strength of materias. Jesmond gius: Chapter 10 Page 1
Unit 48: Structura Behaviour and Detaiing for Construction 4.2.1 Load B (figure 1) represents a beam of span L metres supported simpy at its ends and carrying a point oad of kn at mid-span. Let us assume that the defection due to the oad is 5mm. It is obvious that, if the oad is increased, the defection wi increase. It can be proved that the defection is directy proportiona to the oad, i.e. a oad of 2 wi cause a defection of 10mm, 3 wi produce a defection of 15mm and so on. must therefore be a term in any formua for cacuating defection. ½ 5mm ½ B Figure 1: Defection of a beam under oading 4.2.2 Span In figure 2 (a) and (b) the oads are equa and the weights of the beams, which are assumed to be equa in cross-section, are ignored for the purposes of this discussion. The span of beam (b) is twice that of beam (a). It is obvious that the defection of beam (b) wi be greater than that of beam (a), but the interesting fact (which can be demonstrated experimentay or proved by mathematics) is that instead of the defection of (b) being twice that of (a), it is 8 times (e.g. 40mm in this exampe). If the span of beam (b) where 3L, its defection woud be 27 times that of beam (a). In other words, the defection of a beam is proportiona to the cube of the span, therefore L 3 is a term in the defection formua. 5mm (a) 40mm 2 (b) Figure 2: Effect of span upon defection: the span of (b) is twice that of (a) Jesmond gius: Chapter 10 Page 2
Unit 48: Structura Behaviour and Detaiing for Construction 4.2.3 Size and Shape of Beam Figure 3 (a) and (b) represents two beams (their weights being ignored) of equa spans and oading but the moment of inertia of beam (b) is twice that of beam (a). Obviousy, the greater the size of the beam, the smaer the defection (other conditions being equa). It can be proved that the defection is inversey proportiona to the moment of inertia, e.g. the defection of beam (b) wi be one-haf that of beam (a). Moment of inertia I is therefore a term in the denominator of the defection formua. (It may be noted that, since the moment of inertia of a rectanguar cross section beam is bd 3 /12, doubing the breadth of a rectanguar beam decreases the defection by one-haf, whereas doubing the depth of a beam decreases the defection to 1/8 of the previous vaue.) (a) (b) Figure 3: Effect of size and shape upon defection: (a) moment of inertia of beam = 1 unit; (b) moment of inertia of beam = 2 units 4.2.4 Stiffness of Materia The stiffer the materia of a beam, i.e. the greater its resistance to bending, the smaer wi be the defection, other conditions such as span, oad, etc., remaining constant. The measure of the stiffness of a materia is its moduus of easticity E and defection is inversey proportiona to the vaue of E. 4.3 Derivation of Defection Formuae formua for cacuating defection must therefore contain the oad, the cube of the span L 3, the moment of inertia I, and the moduus of easticity E. For standard cases of oading, the defection formua can be expressed in the form, where c is a numerica coefficient depending on the disposition of the oad and aso on the manner in which the beam is supported, that is, whether the ends of the beam are simpy supported or fixed, etc. For figure 4 the vaues of c are respectivey 1/48 and 5/384. and L 3 are in the numerator of the formua because an increase in their vaues means an increase of defection, whereas E and I are in the denominator because an increase in their vaues means a decrease of defection. Jesmond gius: Chapter 10 Page 3
Unit 48: Structura Behaviour and Detaiing for Construction Referring to figure 4 it shoud be obvious (negecting the weights of the beams) that athough the beams are equay oaded, the defection of beam (b) wi be ess than that of beam (a). In fact, the maximum defection of beam (a) is and the maximum defection of beam (b) is 100kN 3m = 100kN UDL = 6m (a) = 6m (b) Figure 4: Defection Formuae Tabe 1 gives the vaues of c for some common types of oading, etc. Tabe 1: Vaues of coefficient c for defection formua Condition of oading Vaue of c ( max at ) ½ ½ B /2 /2 B Jesmond gius: Chapter 10 Page 4
Unit 48: Structura Behaviour and Detaiing for Construction /2 /2 UDL = Fixed Beam UDL = Fixed Beam Cantiever UDL = Cantiever hen the oad system is compicated, e.g. severa point oads of different magnitudes, or various combinations of point oads and uniformy distributed oads, the defections must be cacuated from first principes. Jesmond gius: Chapter 10 Page 5
Unit 48: Structura Behaviour and Detaiing for Construction In certain simpe cases it is possibe to derive defection formuae mathematicay without using the cacuus, and the foowing exampe is given for the more mathematicay minded student. Practica Exampe 1 ork out the vaue of the coefficient of the defection formua for the foowing scenario and verify your answers with tabe 1. ½ ½ B nswer See ecturer expanation during the esson. Try to work out the coefficient of the defection formua of the other diagrams found in tabe 1. Practica Exampe 2 406 178 UB54 simpy supported at the ends of a span of 5m carries a uniformy distributed oad of 60 kn/m. Cacuate the maximum defection. (E = 205 000 N/mm 2 ) nswer The formua for the maximum defection is (from tabe 1) here = 60 5 = 300 kn = 0.3 10 6 N = 5000 mm E = 205 000 N/mm 2 I = 18720 cm 4 = 187 200 000 mm 4 = 187.2 10 6 mm 4 (Taken from tabe of universa beams) Jesmond gius: Chapter 10 Page 6
Unit 48: Structura Behaviour and Detaiing for Construction Practica Exampe 3 Cacuate the safe incusive uniformy distributed oad for a 457 152 UB52 simpy supported at its ends if the span is 6m and if the span is 12m. The maximum permissibe bending stress is 165N/mm 2 and the maximum permissibe defection is 1/360 of the span. E is 205 000 N/mm 2. nswer Part 1 Z = 950 cm 3 = 950 000 mm 3 (Taken from tabe of universa beams) From ast year notes we have deducted that the eastic moduus can be expressed as So so, Maximum bending moment can be cacuated by using the equation. This has to be equa to. Impies So Maximum Defection for simpy supported uniformy distributed oads is equa to here = 209 000 N = 6000 mm E = 205 000 N/mm 2 I = 21370 cm 4 = 213 700 000 mm 4 = 213.7 10 6 mm 4 (Taken from tabe of universa beams) Maximum permissibe defection = safe oad = 209.0 kn Jesmond gius: Chapter 10 Page 7
Unit 48: Structura Behaviour and Detaiing for Construction nswer Part 2 For a 12m span Z = 950 cm 3 = 950 000 mm 3 (Taken from tabe of universa beams) From ast year notes we have deducted that the eastic moduus can be expressed as So so, Maximum bending moment can be cacuated by using the equation. This has to be equa to. Impies So Maximum Defection for simpy supported uniformy distributed oads is equa to here = 104 500 N = 12000 mm E = 205 000 N/mm 2 I = 21370 cm 4 = 213 700 000 mm 4 = 213.7 10 6 mm 4 (Taken from tabe of universa beams) Maximum permissibe defection = This means that, athough the beam is quite satisfactory from the strength point of view, the defection is too great, therefore the oad must be reduced. Now Jesmond gius: Chapter 10 Page 8
Unit 48: Structura Behaviour and Detaiing for Construction Giving = 64.2 kn and this is the maximum permitted oad for the beam. (Instead of being obtained from the defection formua, can aso be obtained from ) Practica Exampe 4 Cacuate the safe incusive uniformy distributed oad for a 200 mm 75 mm timber joist, simpy supported at its ends, if the span is 4m and if the span is 8m. The maximum permissibe bending stress is 6N/mm 2 and the maximum permissibe defection is 0.003 of the span. E is 9500 N/mm 2. nswer Part 1 The cross section of a timber joist is usuay a rectange. Thus (See esson 9) From esson 9 we have deducted that the eastic moduus can be expressed as So so, Maximum bending moment can be cacuated by using the equation (this is the equation for maximum bending moment of uniformy distributed oads. This has to be equa to. Impies So Maximum Defection for simpy supported uniformy distributed oads is equa to here = 6 000 N = 4000 mm E = 9500 N/mm 2 Jesmond gius: Chapter 10 Page 9
Unit 48: Structura Behaviour and Detaiing for Construction rectanguar cross section shapes) (See esson 9 for second moment of area of Maximum permissibe defection = safe UDL for the 4 m span = 6.0 kn nswer Part 2 The cross section of the second timber joist is same. Thus and so, Maximum bending moment can be cacuated by using the equation (this is the equation for maximum bending moment of uniformy distributed oads. This has to be equa to. Impies So Maximum Defection for simpy supported uniformy distributed oads is equa to here = 3 000 N = 8000 mm E = 9500 N/mm 2 (as before) Jesmond gius: Chapter 10 Page 10
Unit 48: Structura Behaviour and Detaiing for Construction Maximum permissibe defection = This means that, athough the timber joist is quite satisfactory from the strength point of view, the defection is too great, therefore the oad must be reduced. Now Impies Giving = 1.7 kn and this is the maximum permitted oad for the timber joist. (Instead of being obtained from the defection formua, can aso be obtained from ) Jesmond gius: Chapter 10 Page 11