Preface Acknowledgements

Transcription

1 Contents Preface cknowledgements vi viii 1 Equilibrium of rigid structures Contents The fact sheet Symbols, units and sign conventions Worked examples Problems nswers to problems 29 2 Pin-jointed frame structures Contents The fact sheet Symbols, units and sign conventions Worked examples Problems nswers to problems 67 3 Shearing forces and bending moments Contents The fact sheet Symbols, units and sign conventions Worked examples Problems nswers to problems 95 4 Direct stress Contents The fact sheet Symbols, units and sign conventions Worked examples Problems nswers to problems Bending stress Contents The fact sheet 116 iii

2 CONTENTS iv 5.3 Symbols, units and sign conventions Worked examples Problems nswers to problems Combined bending and direct stress Contents The fact sheet Symbols, units and sign conventions Worked examples Problems nswers to problems Shear stress Contents The fact sheet Symbols, units and sign conventions Worked examples Problems nswers to problems Torsional stress Contents The fact sheet Symbols, units and sign conventions Worked examples Problems nswers to problems Mohr s circles of stress and strain Contents The fact sheet Symbols, units and sign conventions Worked examples Problems nswers to problems Composite sections Contents The fact sheet Symbols, units and sign conventions Worked examples Problems nswers to problems 250

3 v CONTENTS 11 Beam deflections and rotations Contents The fact sheet Symbols, units and sign conventions Worked examples Problems nswers to problems Strain energy and virtual work Contents The fact sheet Symbols, units and sign conventions Worked examples Problems nswers to problems 292 Index 293

4 Chapter 1 Equilibrium of rigid structures 1.1 Contents The principles of equilibrium l The equations of equilibrium l Support reactions for beams, plane frames, space frames, mass structures, arches and suspension cables The determination of the support reactions for a loaded structure is normally the first stage in any structural analysis and, hence, is normally the first part only of any examination question. Consequently, some of the following worked examples are only parts of actual examination questions; others have been specially designed to give practice in particular aspects. The reader will obtain further practice by working through the examples in later chapters. 1.2 The fact sheet (a) General definition loaded structure is in equilibrium if it does not move as a rigid body. Rigid body movement can be either a translation (movement in a straight line) or a rotation or a combination of both. In order for a structure to be in equilibrium, the effect of the loads, which tend to move or rotate the structure, must be balanced by reactive forces (reactions) developed at the supports upon which the structure is erected. (b) Plane structures structure which lies within a single plane (i.e. in two dimensions) is a plane structure. 1

5 EQUILIBRIUM OF RIGID STRUCTURES 2 (c) Conditions for equilibrium The requirement that a plane structure does not move in any direction may be specified by stating that it must not move in any two perpendicular directions. Normally, but not essentially, the two directions are taken to be horizontal and vertical. The structure will not move in any direction provided that there is no resultant force in that direction. Thus, for no movement in the horizontal direction the algebraic sum of all the forces acting horizontally must be zero that is, Similarly, for no movement vertically H ¼ 0 V ¼ 0 The requirement that a structure does not rotate in a plane implies that it can not rotate about any axis at right angles to that plane. Thus, there must be no resultant moment of force about any point in the plane. Hence, for no rotation in the plane, the algebraic sum of the moment of all forces about any point in the plane must be zero that is, M ¼ 0 where the point in the plane about which moments are taken can be either within or external to the structure. For complete equilibrium of the plane structure, H ¼ 0: V ¼ 0: M ¼ 0: the algebraic sum of all horizontal forces equals zero; the algebraic sum of all vertical forces equals zero; the algebraic sum of all the moments of all forces about any point equals zero. These are the three equations for the statical equilibrium of plane structures. Sufficient supports and corresponding support reactions must be provided to enable the above-listed equations to be satisfied. Three unique equations are involved; thus, the magnitude of three unknown reactions (or fixing moments) may be determined. If a structure is provided with just sufficient supports (no more than three unknown reactions), it may be completely analysed by use of the above equations and is externally statically determinate. If there are too many support reactions, a solution is not possible by the use of the above equations alone and the structure is statically indeterminate. Such a structure is not dealt with in this book. If a structure has too few support reactions, it will move as a rigid body. (d) Supports (Figure 1.1) (i) roller support provides one reaction only, of unknown magnitude, and acting at right angles to the direction of motion of the rollers. Such a support permits linear movement in one direction only and also permits rotation.

6 3 STRUCTURL MECHNICS (i) (ii) (iii) Figure 1.1 Support types (ii) (iii) pinned support provides a reaction of unknown magnitude and unknown direction which may be completely defined by determining the horizontal and the vertical components of reaction. Such a support prevents any linear movement but permits rotation. fixed support provides a reaction of unknown magnitude and unknown direction and also a fixing moment (a total of three unknowns). Such a support prevents any linear movement and also prevents any rotation. (e) Three-pinned plane structures If a structure contains an internal pin (hinge) (e.g. as in Figure 1.2), such that one part of the structure can rotate about the pin independently of the rotation of the other part, then one additional unique equation of equilibrium may be written, since the sum of the moments of all forces on either part of the structure about that pin must be zero. This enables one additional unknown component of support reaction to be determined. Figure 1.2 Three-pinned arch structure (f) Space structures three-dimensional structure is a space structure.

7 EQUILIBRIUM OF RIGID STRUCTURES 4 For a space structure, the sum of all the forces in each of three mutually perpendicular directions must be zero, and the sum of the moments of all forces about three mutually perpendicular axes (X, Y and Z) must be zero. Thus, X ¼ 0: the sum of the forces in the X direction equals 0; Y ¼ 0: the sum of the forces in the Y direction equals 0; Z ¼ 0: the sum of the forces in the Z direction equals 0; M XX ¼ 0: the sum of the moments about the X axis equals 0; M YY ¼ 0: the sum of the moments about the Y axis equals 0; M ZZ ¼ 0: the sum of the moments about the Z axis equals 0. (g) Mass structures T v T T H W V R O P Figure 1.3 Gravity dam mass structure, such as the gravity dam shown in Figure 1.3, is one which depends upon its own weight to ensure equilibrium. Thus, for equilibrium, V ¼ 0: H ¼ 0: M O ¼ 0: the weight (W) of the structure and the vertical components T V of any loads (T) must be balanced by the vertically upward reaction (V) of the ground (or foundation) below the structure; the tendency for the structure to slide horizontally under the action of any horizontal components (T H ) of load T must be prevented by a reaction (P) from the ground behind the structure and/or by a frictional force (R) between the structure and the ground beneath it; the overturning moment of the loads about a probable centre of rotation (O) must be balanced by the restoring moment about the same point due to the self-weight of the structure. mass structure is normally designed so that its weight is greater than the minimum required for equilibrium in order to provide a factor of safety against overturning, where factor of safety ¼ restoring moment overturning moment

8 5 STRUCTURL MECHNICS Similarly, a factor of safety against sliding will be provided, where factor of safety ¼ total reactive force resisting sliding total force tending to cause sliding (h) Principle of superposition If a structure is made of linear elastic material and is loaded by a combination of loads which do not strain the structure beyond the linear elastic range, then the resultant effect of the total load system on the structure is equivalent to the algebraic sum of the effect of each load acting separately. 1.3 Symbols, units and sign conventions H ¼ the horizontal reaction at a support (kn) M ¼ moments taken about a support (kn m) V ¼ the vertical reaction at a support (kn) Clockwise moments are assumed to be positive. Forces or components of forces acting to the right are assumed to be positive. Forces or components of forces acting vertically upward are assumed to be positive. The direction of action of known forces is indicated by solid arrowheads on the diagrams. The assumed direction of action of unknown forces is indicated by open arrowheads on the diagrams. 1.4 Worked examples Example 1.1 Beam BCD has a pinned support at and a roller support at D. It carries three concentrated loads as shown in Figure 1.4. Determine the reactions. 10 kn 20 kn 5 kn B C D E 1 m 2 m 1 m 1 m Figure 1.4 Load diagram

9 EQUILIBRIUM OF RIGID STRUCTURES 6 Support can provide two components of reaction (V and H ). Support D, being a roller, provides only one reaction (V D ), which will act vertically (that is, at right angles to the direction of motion of the rollers). These reactions are shown on the free body diagram (Figure 1.5). There are only three unknowns; thus the structure is externally statically determinate and the unknowns can be determined. 10 kn 20 kn 5 kn H V V D Figure 1.5 Free body diagram Solution 1.1 (1) To determine H (H ¼ 0) There are no horizontal loads; thus, H ¼ 0 (2) To determine V D Take moments about : (M ¼ 0) þð10 1Þþð20 3Þþð55Þ ðv D 4Þ ¼0 ; V D ¼þ23:75 kn (i.e kn upwards) (3) To determine V (V ¼ 0) þ V þ V D ¼ 0 þ V þð23:75þ 35 ¼ 0 ; V ¼þ11:25 kn (4) Check by taking moments about D: M D ¼þðV 4Þ ð10 3Þ ð20 1Þþð51Þ ¼ þðþ11:25 4Þ 45 ¼ ¼ 0 ; correct This last calculation provides a useful check on the mathematical accuracy of the previous two calculations. Example 1.2 Beam BCDE has a pinned support at and a roller support at C. It carries two concentrated loads of 15 kn each and a uniformly distributed (U.D.) load of 2 kn/m over the right-hand half, as shown in Figure 1.6. Determine the reactions.

10 7 STRUCTURL MECHNICS 15 kn 2 kn/m 15 kn B C D 1 m 2 m 2 m 2 m Figure 1.6 free body diagram has been drawn (Figure 1.7) to show the reactions which are to be determined. 15 kn 15 kn 2 kn/m H V V C Figure 1.7 Free body diagram Solution 1.2 (1) To determine H (H ¼ 0) There are no horizontal loads. ; H ¼ 0 (2) To determine V C Take moments about : Note that the moment of the U.D. load is the resultant total U.D. load (2 3 ¼ 6 kn) multiplied by the distance from to the line of action of that resultant (i.e. 4.5 m) (M ¼ 0) þð15 2Þ ðv C 4Þþð234:5Þþð15 6Þ ¼0 ; V C ¼þ36:75 kn (3) To determine V (V ¼ 0) þv 15 þ V C ð2 3Þ 15 ¼ 0 þv 15 þðþ36:75þ 6 15 ¼ 0 ; V ¼ 0:75 kn (i.e kn downwards)

11 (4) Check by taking moments about C: M C ¼þðV 4Þ ð15 2Þþð230:5Þþð15 2Þ ¼ þð 0:75 4Þ 30 þ 3 þ 30 ¼ 0 ; correct Example 1.3 EQUILIBRIUM OF RIGID STRUCTURES 8 The truss shown in Figure 1.8 has a pinned support at and a roller support at B. Determine the reactions at the supports when loads act on the truss as shown. 2 kn 2 kn 2 kn 3 kn 3 m B 5 kn 5 kn 5 m 5 m 5 m Figure kn 2 kn 2 kn 3 kn H V 5 kn 5 kn V B Figure 1.9 Free body diagram Solution 1.3 (1) To determine H (H ¼ 0) þ H þ 3 ¼ 0 ; H ¼ 3:00 kn (2) To determine V B Take moments about : (M ¼ 0) þð22:5þþð55þþð27:5þþð510þþð212:5þþð33þ ðv B 15Þ ¼0 ; V B ¼þ8:60 kn

12 9 STRUCTURL MECHNICS (3) To determine V (V ¼ 0) þv þ V B ¼ 0 V 16 þðþ8:60þ ¼0 ; V ¼þ7:40 kn (4) Check by taking moments about B: M B ¼þðV 15Þ ð212:5þ ð510þ ð27:5þ ð55þ ð22:5þþð33þ ¼ þðþ7:4 15Þ 111 ¼ 0 ; correct n alternative solution might be to make use of the principle of superposition to determine the vertical reactions, as follows: (1) Vertical loading acting alone Since the vertical loading is symmetrically positioned on the truss, then V ¼ V B ¼ðtotal vertical loadþ=2 ¼ð2þ5þ2þ5þ2Þ=2¼þ16=2 ; V ¼ V B ¼þ8:00 kn (2) Horizontal load acting alone Taking moments about : (M ¼ 0) þð33þ ðv B 15Þ ¼0 ; V B ¼þ0:60 kn (V ¼ 0) V þ V B ¼ 0, V þð0:60þ ¼0 ; V ¼ 0:60 kn (3) Superimposing the effect of the two loadings: total V ¼þ8:00 0:60 ; V ¼þ7:40 kn and total V B ¼þ8:00 þ 0:60 ; V B ¼þ8:60 kn

13 EQUILIBRIUM OF RIGID STRUCTURES 10 Example 1.4 Find the reaction components at the supports for the truss shown in Figure (University of Sheffield) 30 kn 30 kn 30 kn V 30 kn 30 kn 2 m 2 m 2 m 2 m 2 m 2 m 2 m 2 m 2 m 2 m 2 m H B 50 kn 50 kn 2 m B 2 m 2 m 2 m 2 m V B Figure 1.10 This problem looks very complex, but the procedure for solution is as in the previous example. Note that the supports are at different levels; thus, the horizontal component at support B has a moment about the other support (). M will thus provide an equation with two unknowns. It is consequently better to take moments about support B. The drafting of the moment equations may be facilitated if the assumed directions of the support reactions are sketched on the diagram before proceeding. The open-headed arrows at and B have accordingly been added to the original diagram to show the presumed reactions. Solution 1.4 (1) To determine V Taking moments about B: (M B ¼ 0) þðv 22Þ ð30 20Þ ð30 16Þ ð30 10Þ ð30 6Þ ð30 2Þþð50 6Þþð50 2Þ ¼0 þðv 22Þþ ¼ 0 ; V ¼þ55:455 kn (V ¼ 0) þv þ V B ð530þ ¼0 (2) To determine V B þðþ55:455þþv B 150 ¼ 0 ; V B ¼þ94:545 kn (H ¼ 0) þ H B þ 50 þ 50 ¼ 0 (3) To determine H B ; H B ¼ 100:00 kn

14 11 STRUCTURL MECHNICS (4) Check by taking moments about : M ¼þð30 2Þþð30 6Þþð30 12Þþð30 16Þþð30 20Þ þð50 2Þ ð50 2Þ ðh B 4Þ ðv B 22Þ ¼þ1680 ð 100:00 4Þ ðþ94:545 22Þ ¼0 ; correct Note that in this example the values of the reactions have been quoted to three places of decimals. If the value of V B had been rounded up to two places of decimals, then the check calculation would not have proved to be correct. In general, however, it is quite sufficient to quote answers to two places of decimals. Example 1.5 Figure 1.11 shows a pin-jointed structure, which has pinned supports at both and C, supporting a vertical and a horizontal load. (a) It has four reactive components, two at each support and C; nevertheless, it is statically determinate both internally and externally. Explain why this is so. (b) Calculate in terms of W and L the vertical and horizontal reactive components for the loading shown. (University of Nottingham) 2 W B W 2.4 L 3 L 4 L 4 L C 1.2 L Figure 1.11 Solution 1.5 (a) The answer to the question of external determinacy follows from the Fact Sheet at the beginning of this chapter. The answer in respect of internal determinacy is related to the Fact Sheet for Chapter 2 and the reader is advised to refer to that Fact Sheet if in any doubt about the following answer. For a structure to be externally statically determinate, it must normally have no more than a total of three unknown components of reaction or moments, since only three unique equations for equilibrium can be written. In this case, however, there is a central pin and a fourth equation may be written, since the sum of the

15 EQUILIBRIUM OF RIGID STRUCTURES 12 moments of all the forces on either the left- or right-hand part of the structure about that pin must be zero for equilibrium. Thus, four unknown reactive components may be determined and the structure is externally statically determinate. For a frame to be internally statically determinate, the number of members, M, and the number of joints, J, in that frame must be related by the expression M ¼ 2J 3. In this case each part of the structure is a frame in which M ¼ 13 and J ¼ 8. These numbers satisfy the requirement that M ¼ 2J 3. In addition, the members must be correctly positioned within the frame (the structure should be completely triangulated). Inspection shows this to be so; thus, both parts of the structure are, and, hence, the complete structure is, internally statically determinate. (b) (1) To determine V C, consider the whole frame and take moments about : (M ¼ 0) þð2w 3LÞþðW 1:2LÞ ðv C 8LÞ ¼0 ; V C ¼þ0:90 W (2) To determine V, consider the whole frame and resolve vertically: (V ¼ 0) þv þ V C 2W ¼ 0 þv þðþ0:90wþ 2W ¼ 0 ; V ¼þ1:10W (3) Check by taking moments about C: M C ¼þðV 8LÞþðW 1:2LÞ ð2w 5LÞ ¼ þðþ1:1w 8LÞþðW 1:2LÞ ð2w 5LÞ ¼0 ; correct (4) To determine H, consider the left-hand part only and take moments about B: 2 W B H B H V B V Figure 1.12 Left-hand part (M B ) þðv 4LÞ ð2w LÞ ðh 2:4LÞ ¼0 þðþ1:1w 4LÞ ð2w LÞ ðh 2:4LÞ ¼0 ; H ¼þ1:00W

16 13 STRUCTURL MECHNICS (5) To determine H C, consider the whole frame and resolve horizontally: (H ¼ 0) þh þ H C þ W ¼ 0 þðþ1:0wþþh C þ W ¼ 0 ; H C ¼ 2:00W Example 1.6 Figure 1.13 shows a three-pinned arch which is formed from two circular segments and carries two point loads. Construct free body diagrams for (i) part B (ii) part BC and (iii) part X, giving the forces at X in terms of shear and thrust. (University of Portsmouth) 10 kn B X 3 m 2 m 2 kn 4 m 4 m 45 2 m 6 m C Figure 1.13 Solution 1.6 (i), (ii) To be complete, the free body diagrams should show the magnitude and direction of the reactive components. Consequently, the first stage of the solution must be to determine the reactions as in the previous example. Note that, in this case, it is not possible to calculate the value of V C directly by taking moments about, since the two supports are at different levels. (1) To determine V C and H C Consider the right-hand part (BC) only (see Figure 1.14) and take moments about B: (M B ¼ 0) f2ð6 6sin 458Þg ðh C 6Þ ðv C 6Þ ¼0 ; H C ¼ V C 0:586

17 EQUILIBRIUM OF RIGID STRUCTURES 14 B 2 kn 6 sin 45 6 m 45 6 m H C C V C Figure 1.14 Right-hand part Consider the whole structure and take moments about : (M ¼ 0) þð10 2Þþf2ð6sin 458 2Þg ðh C 2Þ ðv C 10Þ ¼0 þ20 þ 4:485 2 ð V C 0:586Þ ðv C 10Þ ¼0 ; V C ¼þ3:207 kn Then H C ¼ V C 0:586 ¼ ðþ3:207þ 0:586 i.e. H C ¼ 3:793 kn (2) To determine V and H Consider the whole structure: (V ¼ 0) þ V þ V C 10 ¼ 0 þ V þðþ3:207þ 10 ¼ 0 ; V ¼þ6:793 kn (H ¼ 0) þ H þ H C þ 2 ¼ 0: þ H þð 3:793Þþ2 ¼ 0 ; H ¼þ1:793 kn (3) Check by taking moments about C for the whole structure: M C ¼þðH 2ÞþðV 10Þ ð10 8Þþð26sin 458Þ ¼ þðþ1:793 2Þþðþ6:793 10Þ 80 þ 8:845 ¼ 0 ; correct The free body diagrams for part B and part BC are shown in Figures 1.15 and 1.16, respectively. Note that the reactive forces acting at B on part B are V B ¼ 10:000 6:793 ¼ 3:207 kn and H B ¼ H ¼ 1:793 kn The reactive forces acting at B on part BC are equal in magnitude but opposite in direction to these.

18 4 m 15 STRUCTURL MECHNICS 10 kn B 1.79 kn 1.79 kn B 3.21 kn 3.21 kn 2 kn 1.79 kn 6.79 kn Figure 1.15 C 3.79 kn 3.21 kn Figure 1.16 (iii) (1) To determine thrust and shear at X Refer to Figure The part X of the structure is in equilibrium under the action of V and H acting at support together with a thrust T (a direct force) acting along the longitudinal axis of the arch (i.e. tangential to the arch) at X, and a shearing force S acting at right angles to the axis of the arch (i.e. radially to the arch) at X. The direction of action of the forces S and T, as indicated in Figure 1.17, is taken as positive in the following calculations. S T 0 X H C V 3 m 0 Figure 1.17 Resolving in a direction tangential to the arch at X. ( forces ¼ 0) T þ H sin þ V cos ¼ 0 where, from the geometry of the diagram, ¼ cos 1 ð3=4þ ¼41:418. Thus, T þ 1:793 sin 41:418 þ 6:793 cos 41:418 ¼ 0 ; T ¼ 6:281 kn

19 EQUILIBRIUM OF RIGID STRUCTURES 16 Resolving in a direction radial to the arch at X: ( forces ¼ 0) S þ H cos V sin ¼ 0 S þ 1:793 cos 41:418 6:793 sin 41:418 ¼ 0 ; S ¼ 3:148 kn The free body diagram for part X is shown in Figure kn 6.28 kn X 1.79 kn 6.79 kn Figure 1.18 Example 1.7 concrete dam has a trapezoidal cross-section as shown in Figure Water is impounded to a depth of 10 m. (a) If the density of water is 1000 kg/m 3, determine: (i) the horizontal thrust on the dam expressed in newtons per metre length of dam; 1 m H = 10 m 11 m O 1 m 1 m 3 m Figure 1.19

20 17 STRUCTURL MECHNICS (b) (ii) the total thrust on the dam per metre length; (iii) the position of the centre of pressure on the wetted face of the dam; (iv) the overturning moment about the toe (O). If the density of the concrete is 2400 kg/m 3, determine: (i) (ii) the position of the centre of gravity of the dam; the factor of safety against overturning. (Coventry University) This is a mass structure problem in which the structure may fail by overturning about the toe O, or by sliding horizontally. In this case the question does not consider the possibility of failure by sliding. The wording of the question guides the candidate through the sequential steps of the solution. Solution 1.7 (a) (i) The horizontal thrust per metre length is given by ðt H Þ¼gH 2 =2 ¼ð1000 9: Þ=2 ¼ 490:50 kn The hydrostatic pressure increases linearly from zero at the surface of the water to a maximum value of gh at the base. The average pressure is gh=2 and the total horizontal thrust (average pressure the ara of the face of the dam projected onto the vertical plane) is ðgh=2þðh 1Þ ¼gH 2 =2 per metre length of dam. (ii) The total thrust (T) per metre length is given by T ¼ average pressure wetted area of face of dam The wetted area per metre length of the dam equals the length of the sloping face of the dam which is in contact with the water (i.e. length DE) multiplied by one metre (see Figure 1.20). Thus DE ¼ð10 2 þ 0:909 2 Þ 1=2 ¼ 10:0412 T ¼ðgH=2Þ ð10:0412 1Þ ¼ð1000 9:81 10=2Þ10: ¼ 492:52 kn n alternative method of determining the total thrust is to calculate the vertical component of thrust, T V, and compound it with the horizontal thrust, T H,as follows:

21 EQUILIBRIUM OF RIGID STRUCTURES 18 Vertical component of thrust ¼ weight of the wedge DEF T V ¼ volume of wedge unit weight of water ¼ 1 2 ð10 0:909 1Þð1000 9:81Þ10 3 ¼ 44:59 kn Total thrust ¼ðT 2 H þ T 2 V Þ 1=2 ¼ð490:50 2 þ 44:59 2 Þ 1=2 ¼ 492:52 kn (iii) The centre of presssure is at 2=3 total depth of water ¼ 2=3 10 ¼ 6.67 m vertically below the water surface The hydrostatic pressure acting on face DE increases linearly from zero at the surface of the water (E) to gh at the base (D). Thus, the load distribution diagram for the face DE is triangular and the resultant thrust which acts through the centroid of the load distrubution diagram will act at two-thirds of the depth. (iv) Taking moments of all overturning forces about the toe O: overturning moment ¼þðT H 3:333Þ ðt V 4:70Þ (see Figure 1.20) ¼þð490:50 3:33Þ ð44:59 4:70Þ ¼ 1423:79 knm (b) (i) To determine the position of the centre of gravity of the dam with respect to the toe (O), take moments of the area of parts, B and C about O (see Figure 1.20) x 1 = F E T V T T H 3.33 m D B C 5 1/3 (10/11) = 4.70 O Figure 1.20

22 19 STRUCTURL MECHNICS Part rea () Distance (x) from O to centroid x 1 2 ð1 11Þ ¼ 5: B ð1 11Þ ¼11: C 2 ð3 11Þ ¼16: ¼ 33:00 x ¼ 95:33 (ii) The centre of gravity is to the left of O by distance x, where x ¼ x ¼ 95:33 33:00 ¼ 2:889 m The restoring moment per metre length tending to prevent the dam rotating about O ¼ weight of dam horizontal distance from (O) ¼ volume unit weight x ¼ð33:00 1Þð2400 9:81Þ2: ¼ 2244:61 knm The factor of safety restoring moment ¼ overturning moment ¼ 2244: :79 ¼ 1:58 Example 1.8 The arrangement shown in Figure 1.21 is used to withdraw the block B, which is between the fixed surface C and the block. The self-weight of each part is 200 mm 4 kn B 500 N F F D 150 mm C 2000 mm 60 P 700 mm Small gap O Figure 1.21 Centre of gravity of D, self-weight 1 kn

23 shown on the diagram and the coefficients of friction are BC ¼ 0:3, DC ¼ 0:5 and B ¼ 0:3. (i) Determine whether or not the base of part D will slip (and then rest against the vertical surface of C) before block B is moved. (ii) ssuming that the geometry remains essentially the same, whether or not slipping occurs between D and C, draw a free body diagram for part D at the instant that part B begins to move. (University of Portsmouth) This arrangement is a mechanism, but the problem is concerned with a situation when the parts are only on the point of moving and not actually moving. In such a situation, when the parts are static, the arrangement is effectively a structure and the analysis of forces and reactions is precisely the same as for a structure. The solution involves a clear appreciation of basic principles relating to forces in equilibrium. Solution 1.8 (i) EQUILIBRIUM OF RIGID STRUCTURES 20 If F is the tensile force in the horizontal link between B and D, then, when B is just about to move, F equals the limiting frictional force between and B plus the limiting frictional force between B and C (see Figure 1.22). 0.3 x 4000 N 4000 N 500 N F 0.3 x 4500 N 4500 N Figure 1.22 That is, F ¼ B the normal reaction between and B þ BC the normal reaction between B and C ¼ 0: þ 0: ¼ 2550 N If the part D is just about to rotate about the contact point (O) indicated on the diagram, then the anticlockwise moment of F about O will be equal to the clockwise moments about O due to force P and to the self-weight of part D. Taking moments about O: (M O ¼ 0) Fð2000 sin 608 þ 200 sin 308ÞþPð2000 cos 608 þ 150 cos 308Þ þ1000fð Þ cos 608g ¼0 ; 2550ð1832ÞþPð1130Þþ1000ð650Þ ¼0 ; P ¼ 3559 N

24 21 STRUCTURL MECHNICS For vertical equilibrium of part D the vertical component of reaction at O will equal the weight of D plus force P; thus, vertical reaction at O ¼ 1000 þ 3559 ¼ 4559 N The maximum frictional force which can be developed at O (i.e. the maximum possible horizontal component of the reaction at O) ¼ DC vertical reaction at O ¼ 0: ¼ 2279:5N The horizontal component of reaction required at O to maintain equilibrium ¼ F ¼ 2550 N. This is greater than the maximum frictional force which can be developed; thus, block D will slip to the left. (ii) fter block D has slipped and made contact with the vertical face of C, a horizontal reaction will be developed at the new point of contact to make up the deficiency in horizontal restraint. The value of this reaction is given by reaction at vertical face of C ¼ :5 ¼ 270:5N The free body diagram for part D at the instant that part B begins to move is drawn in Figure N 3559 N N 1000 N N 4559 N Figure 1.23 Example 1.9 Figure 1.24 shows a non-uniform cable hanging under gravity between two supports at the same level. The right-hand part, of projected length L, is a heavy flexible chain of weight w per unit length, while the left-hand part, also of projected length L, is a light wire of negligible weight. The central dip (d) is small in comparison with the span. The vertical and horizontal scales of the sketch are not equal. (a) Use the conditions of statical equilibrium to determine the vertical components of the end reactions; and hence, by considering the equilibrium

25 EQUILIBRIUM OF RIGID STRUCTURES 22 d L L Figure 1.24 (b) of a suitable piece, determine the horizontal component of the cable tension and thus an appropriate expression for the tension (T) throughout the cable. By considering the equilibrium of another suitable piece find the location of the lowest point and the dip of the cable at that point. (University of Cambridge) The wording of the question makes it clear that the horizontal projected lengths of the cable may be taken to be the same as the actual length of the cable. Total load wl H B H C C V 1.5 L VC 2.0 L Figure 1.25 Solution 1.9 (a) (1) Considering the whole cable Weight of cable ¼ w per unit length over right-hand half (BC) Then, by taking moments about, (M ¼ 0) þðwl 1:5LÞ ðv C 2LÞ ¼0 ; V C ¼þ0:75wL Resolving vertically: (V ¼ 0) þv þ V C wl ¼ 0 þv þ 0:75wL wl ¼ 0 ; V ¼þ0:25wL

26 23 STRUCTURL MECHNICS (2) Considering the right-hand half (see Figure 1.26) d T B V B wl H C C V C H B B L L Figure 1.26 The right-hand part is held in equilibrium by reactive components V C and H C at the support C together with a tension T B in the cable at B. T B may be determined by calculating its components H B and V B, as indicated on Figure Note that since the left-hand half is weightless, it will adopt the configuration of a straight line of slope d=l, as indicated by the dotted line in Figure Since it is known that the cable will be in tension, the direction of T B is known; hence, H B has been drawn acting to the left. Resolving vertically: (V ¼ 0) þv B þ V C wl ¼ 0 V B þ 0:75wL wl ¼ 0 ; V B ¼þ0:25wL Taking moments about C: (M C ¼ 0) þðv B LÞþðH B dþ ðwl 0:5LÞ ¼0 þ0:25wl 2 þ H B d 0:5wL 2 ¼ 0 ; H B ¼ þ0:25wl2 d But the horizontal component of the tension is constant throughout, since there is no horizontal component of loading acting anywhere along the cable. Hence, horizontal component of cable tension H ¼þ 0:25wL2 d This result could also be obtained by considering the geometry at B. The direction of T B is known to be at a slope of d=l; thus, V B =H B must equal d=l. V B has been determined; thus, H B can be calculated.

27 EQUILIBRIUM OF RIGID STRUCTURES 24 (3) The cable tension in the left-hand half (B) will be constant and of value equal to T B, where T B ¼ðH 2 B þ V 2 B Þ 1=2 ¼fð0:25wL 2 =dþ 2 þð0:25wlþ 2 g 1=2 ¼ 0:25wLfðL=dÞ 2 þ 1g 1=2 The cable tension in the right-hand half (BC) will vary along the length and an expression can be derived to give the tension at any point between B and C. Considering Figure 1.27, the tension at a point X, distance x from C, will be T X ¼ðH 2 X þ V 2 X Þ 1=2 ¼ fð0:25wl 2 =dþ 2 þðwz 0:75wLÞ 2 g 1=2 x w per unit length C H C = 0.25 wl2 d H X = 0.25 wl2 d X B V C (= 0.75 wl) V X (= wx 0.75 wl) Figure 1.27 (b) (1) To locate the lowest point Consider Figure 1.28, which shows the part of the cable to the right of the lowest point (P). The dimension D shown in the figure is the maximum dip. x D P C (0.75 wl) 0.25 wl 2 d H P = 0.25 wl2 d V P (= 0) Figure 1.28 Note that at the lowest point the cable will be horizontal, and the tension (T P ) will be horizontal and will have zero vertical component. Consequently, the component V P is zero.

28 25 STRUCTURL MECHNICS Resolving vertically: (V ¼ 0) þv P þ V C weight of cable ¼ 0 0 þ 0:75wL wx ¼ 0 ; x ¼ 0:75L and the weight of the cable CP ¼ð0:75LÞw ¼ 0:75wL. Taking moments about C: (M C ¼ 0) þðh P DÞ ð0:75wl 0:375LÞ ¼0 Example 1.10 ; 0:25wL2 D 0:281wL 2 ¼ 0 d ; D ¼ 0:281d 0:25 ¼ 1:125d ; Maximum dip is given by D ¼ 1:125d The space frame shown in Figure 1.29 has a ball-jointed support at. The support at B is free to move in any direction on the horizontal plane and the support at C is free to move in direction C only. Calculate the value of the reactions at, B and C. [Hint Take as the origin of the three axes: the X axis in direction C, the Z axis at right angles to the X axis in the horizontal plane and the Y axis vertical.] B 10 kn D 1.5 m 10 kn 10 kn D 3 m S 2 m 2 m Plan 1.5 m C S (B) Elevation S-S C Figure 1.29 Solution 1.10 The first step in the solution is to identify the reactions whose values are to be determined. Figure 1.30 shows the six components of reaction which are to be determined.

29 EQUILIBRIUM OF RIGID STRUCTURES 26 Z B Y 10 kn 10 kn D 10 kn R X R Z X R X (B) X C B C R ZC R Y R YB R YC Figure 1.30 The ball joint at provides reactive components in three directions (X, Y and Z). Support B is restrained only in the Y direction and, hence, provides only one reactive component (in the Y direction). Support C is restrained in two directions (Y and Z) and, hence, provides two reactive components (in the Y and Z directions). Taking moments about the X axis: (M XX ¼ 0) ð10 1:5ÞþðR YB 3:0Þ ¼0 ; R YB ¼þ5:00 kn Taking moments about the Z axis: (M ZZ ¼ 0) þð10 2Þ ð10 3Þ ðr YB 2Þ ðr YC 4Þ ¼0 þ20 30 ðþ52þ ðr YC 4Þ ¼0 ; R YC ¼ 5:00 kn Taking moments about the Y axis: (M YY ¼ 0) þð10 1:5ÞþðR ZC 4Þ ¼0 ; R ZC ¼ 3:75 kn Resolving in the X direction: (X ¼ 0) þ R X 10 ¼ 0 ; R X ¼þ10:00 kn Resolving in the Z direction: (Z ¼ 0) þ R Z þ R ZC ¼ 0 þ R Z þð 3:75Þ ¼0 ; R Z ¼ 3:75 kn

30 27 STRUCTURL MECHNICS Resolving in the Y direction (vertical): (Y ¼ 0) þr Y þ R YB þ R YC 10 ¼ 0 þr Y þðþ5:00þþð 5:00Þ 10 ¼ 0 ; R Y ¼þ10:00 kn Moments have been taken as positive if they act in a clockwise direction about an axis when viewed from in the positive direction of the axis. 1.5 Problems 1.1 The plane pin-jointed truss shown in Figure P1.1 carries a horizontal load at Dof2W and a vertical load at G of W. Evaluate all the support reactions. (University of Portsmouth) 2 m 2 m 2 m B C D 2W 4 m E F G 2 m W H Figure P Figure P1.2 shows a beam C with a pinned support at and on roller supports at B and C. pin at P connects the two parts of the beam together. Determine the values of the reactions at, B and C. 10 kn 2 kn/m 2 m B P C 4 m 1 m 3 m Figure P Determine the magnitudes and directions of the horizontal and vertical components of reaction at and F in the pin-jointed frame shown in Figure P1.3. (University of Nottingham)

31 EQUILIBRIUM OF RIGID STRUCTURES 28 C 4 kn B 30º 30º 30º 30º 5 kn D 8.66 m 30º 30º E 60º 60º 10 m Figure P1.3 F 1.4 Figure P1.4 shows a concrete retaining wall supporting soil to a depth of 2 m. The soil may be considered as exerting a total horizontal thrust on the face of the wall of 5.1 kn per metre length of wall. If the density of concrete is 2400 kg/m 3 and if the coefficient of friction between the base of the wall and the soil is 0.35, determine the factors of safety against overturning and sliding. 0.3 m 0.5 m 5.1 kn 0.9 m 1.7 m 0.3 m 2.0 m Figure P Figure P1.5 shows a cantilevered pin-jointed frame supporting a vertical load at D. Support is pinned (not capable of linear movement in any direction), B is free to move vertically only and C is free to move in any direction in the vertical plane. Determine the values and directions of all the support reactions. [Hint Take B as the origin of three axes; the X axis in direction BC, the Z axis at right angles to the X axis and in the horizontal plane, and the Y axis vertical.]

32 29 STRUCTURL MECHNICS 3 m D D 12 kn 1 m 2 m C B (C) Side elevation SS B End elevation C D 1 m S S B Plan Figure P nswers to problems 1.1 H ¼ 3W (3W to the left); H H ¼þW (W to the right); V H ¼þW (W vertically upwards) 1.2 V ¼þ4:25 kn; V B ¼þ8:75 kn; V C ¼þ3:00 kn 1.3 H ¼ 3:44 kn; V ¼ 5:96 kn; H F ¼ 0:56 kn; V F ¼þ10:96 kn 1.4 Factor of safety against overturning is 2.19; Factor of safety against sliding is R Z ¼ 18:0 kn; R X ¼ 0:0 kn; R Y ¼þ12:0kN R ZB ¼þ9:0kN; R XB ¼ 0:0kN R ZC ¼þ9:0kN (see Figure P1.6) 18 kn 12 kn Y 0 kn 9 kn X C D 9 kn 12 kn B 0 kn Z Figure P1.6

33 Index rches 13 rea centroid of 18 polar second moment of 189 second moment of , xial stress see Stress, direct Bending biaxial 140, 144, combined with direct stress equations of 116 single axis 139, 142 Bending moment definition of 70 standard formulae 71 Bending moment diagrams of beams with external couples of beams with internal pins 87 of cranked beams and bents 85, 89 maximum values of 71, of non-uniformly distributed loads 76 position of contraflexure point in 72 Bending stresses Complementary shear stress 166 Composite sections axially loaded 231, deflection of 257 flexural members 232, reinforced concrete 244 thermal stresses in 246 Contraflexure 71 Deflection of beams Deflection of pin-jointed frames using strain energy 276, 280 using virtual work 277, Eccentrically loaded sections 140, , 160 Elastic section modulus 117 Equilibrium 1 29 conditions for 2 Factor of safety 4 5, 16 Force diagrams 32, 60 Foundations, stresses beneath 147, 155 Hoop stress 99, 100, 112 Hydrostatic thrust 16, 105, 112, 155 Linear deformation Macaulay s method for deflection and rotation of beams 253, for treatment of couples use of dummy loads in 254, Mass structures 4, 16 Method of resolution at joints 31, 33 44, 45 Method of sections 31, Modular ratio Modulus of rigidity 165, 189 Mohr s circle of strain Mohr s circle of stress Moment bending see Bending moment relationship with curvature 252 Moment of resistance 117, 123 Neutral axis 116, Parallel axis theorem 117, Pin-jointed frames 30 69, deflections of graphical methods for 32,

34 INDEX 294 Pin-jointed frames (cont.) method of resolution at joints 31, 33 44, 55 method of sections 31, Plane structures 1 Poisson s ratio 99, Polar second moment of area 189 Power, transmitted in torsion 189, Prestressed concrete beams 160 Principal plane 209, Principal stress 209, Reactions 1 calculation of 5 29 Reinforced concrete sections 244 Rotation, of beams Second moment of area , Shear modulus 165, 188 Shear strain 165 Shear stress average values of 165, 167, 170 in beams 166, complementary 166 distribution diagrams of 172, 175, 179 formula for 166 in non-rectangular sections relationship with shear strain 165 Shearing force Shearing force diagrams 70, beams with internal pins 87 cranked beams 85, 89 Space structures 3, 25, 58 Springs, strain energy within 277 Statical determinacy external 2, 11, 42 internal 12, 30, 42 Strain direct 98 lateral 99 principal shear 165 Strain energy for calculation of deflections in springs 277, 278 Strain gauges 222, 225 Strengthening, of beams, using flange plates 123, 125, 130 Stress bending combined bending and direct complementary shear 166 in composite sections direct principal 209, relationship with strain 99 shear thermal 233, 246 in thin-walled cylinders 99, 100, 112 torsional Superposition, principle of 5, 9, 78, 196, 257 Supports 1 3 fixed 3 pinned 3 roller 2 Suspension cables 21 Tension coefficients 50, 58 Theory of bending 116 Thermal stress 233, 246 Three-pinned arches 13 Thrust 85, 89 Torsion equations of 188 of non-prismatic sections 189, 199 power transmitted 189, stress Transformed sections 232, , 257 Virtual work for calculation of deflections principle of 277 Volumetric strain 110 Young s modulus of elasticity 99